Odd Dunkl Operators and nilHecke Algebras

Ritesh RagavenderMay 30 2014

Abstract

Symmetric functions appear in many areas of mathematics and physics including enumerative combi-natorics the representation theory of symmetric groups statistical mechanics and the quantum statis-tics of ideal gases In the commutative (or ldquoevenrdquo) case of these symmetric functions Kostant andKumar introduced a nilHecke algebra that categorifies the quantum group Uq(sl2) This categorificationhelps to better understand Khovanov homology which has important applications in studying knotpolynomials and gauge theory Recently Ellis and Khovanov initiated the program of ldquooddificationrdquo asan effort to create a representation theoretic understanding of a new ldquooddrdquo Khovanov homology whichoften yields more powerful results than regular Khovanov homology In this paper we contribute to-wards the project of oddification by studying the odd Dunkl operators of Khongsap and Wang in thesetting of the odd nilHecke algebra Specifically we show that odd divided difference operators canbe used to construct odd Dunkl operators which we use to give a representation of sl2 on the algebraof skew polynomials and evaluate the odd Dunkl Laplacian We then investigate q-analogs of divideddifference operators to introduce new algebras that are similar to the even and odd nilHecke algebrasand act on q-symmetric polynomials We describe such algebras for all previously unstudied values ofq We conclude by generalizing a diagrammatic method and developing the novel method of insertionin order to study q-symmetric polynomials from the perspective of bialgebras

Contents

1 Introduction 211 The Commutative (ldquoEvenrdquo) Case 212 The ldquoOddrdquo Case 213 Outline of the Present Paper 3

2 Odd Dunkl operators and the Odd nilHecke algebra 521 Preliminaries Even Dunkl Operators 522 Introduction to the Odd nilHecke Algebra 623 Some Operations on Skew Polynomials 7

3 Classical Yang-Baxter Equation and the Dunkl Laplacian 9

4 A Variant of the Khongsap-Wang Odd Dunkl Operator 12

5 q-nilHecke Algebras 15

6 A Diagrammatic Approach to q-Symmetric Polynomials 1761 Introduction to a q-Bialgebra 1762 The Elementary q-Symmetric Functions 1963 Relations Between Elementary q-Symmetric Polynomials 2164 Insertion 23

7 Conclusion and Further Research 25

8 Acknowledgements 26

1

1 Introduction

11 The Commutative (ldquoEvenrdquo) Case

Symmetric polynomials are polynomials in n independent commutative variables x1 x2 xn that areinvariant under the action of any permutation acting on the indices They arise in enumerative com-binatorics algebraic combinatorics Galois theory quantum statistics and the quantum mechanics ofidentical particles [25 27] The even nilHecke algebra NHn introduced by Kostant and Kumar in [15]is important in studying these symmetric polynomials NHn is graded Morita equivalent to the sym-metric polynomials in n variables and is generated by n commuting variables x1 xn and n divideddifference operators parti = (xi minus xi+1)

minus1(1minus si) for 1 le i le n Here si is the simple transposition in thesymmetric group that swaps xi and xi+1

Combining these divided differences with partial derivatives one obtains a commuting family ofoperators originally introduced by Dunkl [7] These Dunkl operators denoted ηi have a major role inmathematical physics and conformal field theory In particular they relate to the study of quantummany-body problems in the Calogero-Moser-Sutherland model which describes integrable systems ofone dimension [11 24] Dunkl operators can also be used to define three operators which arise inphysics and harmonic analysis that satisfy the defining relations of the Lie algebra sl2 [13] These threeoperators found by Heckman and called an sl2-triple play a crucial role in studying Fischer decomposi-tion which has importance not only in representation theory but also in the algebraic Dirichlet problem[1 6 23]

The Cherednik operators denoted by Yi are defined in terms of Dunkl operators and have importantapplications in representation theory [2 20] They have non-degenerate simultaneous eigenfunctionsknown as Jack polynomials These polynomials are a specific case of the well-known Macdonald poly-nomials and contribute to representation theory statistical mechanics and the study of the quantumfractional Hall problem important in condensed matter physics [3 20]

The diagram below depicts the relationship between the three operators discussed so far

Divided Differences parti

nilHecke AlgebraSchubert polynomials

Cohomology

Dunkl Operators ηi

sl2 triple

Cherednik Operators Yi

Jack PolynomialsCherednik Algebras

Affine Hecke Algebras(harmonic analysis)

quantum CMS Model

ldquocan be used tordquoldquoused to definerdquo

Figure 1 Operators in the study of symmetric polynomials

12 The ldquoOddrdquo Case

The divided difference operators Dunkl operators and Cherednik operators all study the commutativesymmetric functions Ellis and Khovanov however sought to study different kinds of symmetric func-tions They recently introduced the quantum case of symmetric functions where xjxi = qxixj for j gt i[8] In the ldquooddrdquo case q = minus1 they describe the ldquoodd symmetric polynomialsrdquo which are polynomialsin the n variables x1 xn where xixj +xjxi = 0 for i 6= j This type of noncommutativity arises in thestudy of exterior algebras and parastatistics

The motivation for considering these odd symmetric polynomials and their corresponding odd nil-

2

Hecke algebra involves the categorification of quantum groups Categorification introduced by Craneand Frenkel is generally the process of replacing algebras and representations by categories and highercategories in order to make quantum 3-manifold invariants into 4-manifold invariants [4] In physicscategorification corresponds to increasing dimensions which allows one to understand symmetries inlower dimensions and then use categorification to better understand higher dimensions In mathemat-ics categorified quantum groups give a higher representation theoretic construction of link homologieswhich in turn categorify quantum link polynomials

The original example of link homology is Khovanov homology a bigraded abelian group which cat-egorifies the well-known Jones polynomial It has major applications in studying knot polynomialsquantum field theory and classical gauge theory [19 28] Since the quantum group Uq(sl2) plays a rolein understanding the Jones polynomial a categorification of Uq(sl2) would be useful in better under-standing Khovanov homology This precise categorification is achieved through the ldquoevenrdquo nilHeckealgebra NHn described in subsection 11

Recently Ozsvath Rasmussen and Szabo found an odd analog of Khovanov homology [22] Theirodd Khovanov homology also categorifies the Jones polynomial and agrees modulo 2 with Khovanovhomology However both theories can detect knots that the other theory cannot [26] The subject ofodd Khovanov homology has yet to be fully understood despite its crucial connections with Khovanovhomology

Knowing this Ellis Lauda and Khovanov developed the odd nilHecke algebra to provide an oddcategorification of Uq(sl2) and give a construction of odd Khovanov homology from a representationtheoretic standpoint [9 10] In addition to being useful in the categorification of quantum groups theodd nilHecke algebra is also related to Hecke-Clifford superalgebras [16 17] and has been used to con-struct odd analogs of the cohomology groups of Springer varieties [21]

The below diagram summarizes the categorifications that motivate the present work NH standsfor nilHecke Cat stands for categorification and KH stands for Khovanov homology

Jones Polynomial Uq(sl2)Uq(sl2)

KH Odd KHCat of Uq(sl2) Odd Cat of Uq(sl2)

odd NH algebraNH algebra

ldquocategorifiesrdquoldquohelps to explainrdquo

ldquoequivalent modulo 2rdquo

Figure 2 Odd Khovanov Homology and Categorification

13 Outline of the Present Paper

The main goal of the present paper is to make progress towards giving a representation theoretic con-struction of odd Khovanov homology Despite being relatively new odd Khovanov homology seemsto have great importance in knot theory It has connections to Heegaard-Floer homology and yieldsstronger results than Khovanov homology in bounding the Thurston-Bennequin number and detectingquasi-alternating knots [26] Odd Khovanov homology is also related to signed hyperplane arrange-ments which have many implications in graph theory and topology [5]

We study odd Khovanov homology by looking for new representation theoretic structures that arise

3

from identifying ldquooddrdquo analogs of structures that play important algebraic roles in the even case EllisKhovanov and Lauda started this program of ldquooddificationrdquo by finding an odd analog of NHn andusing it to categorify Uq(sl2) Searching for the geometry underlying these odd constructions would alsoprovide a very new approach to noncommutative geometry For example we study potential generatorsof certain Cherednik algebras Since spherical rational Cherednik algebras fit nicely into a family ofalgebras from the geometry of symplectic resolutions including Websterrsquos tensor product algebras andcyclotomic KLR algebras this project may be used in the context of ldquooddrdquo noncommutative geometry

In Subsection 11 we discussed the (even) divided difference operators Dunkl operators and Chered-nik operators Although analogs of Dunkl operators have been found in the odd case they have not beenwell-studied Odd Cherednik operators have not even been defined

As a result the first goal of the present paper is to unify certain results in the odd case and to furtherstudy the odd Dunkl operators of Khongsap and Wang In Section 2 we introduce an operator rikrelated to the generalized odd divided difference operator partodd

ik and study its properties One of ourmain results (Equation 26) is that the odd Dunkl operator ηi may be expressed in terms of the odddivided difference operators of Ellis Khovanov and Lauda

ηoddi = tδi + u

sumk 6=i

partoddik sik (11)

This result connects odd Dunkl operators and the odd nilHecke algebra both of which play importantroles in the project of oddification

In the even case one can introduce an operator known as the Dunkl Laplacian given bysumn

i=1 η2i This

operator has important applications in spherical harmonics and heat semigroups [24] Our next goal inthe present paper is to express the Dunkl Laplacian in the odd case In Section 3 we show that therik satisfy the classical Yang-Baxter equation and use this result to evaluate the odd Dunkl LaplacianSpecifically we show that

nsumi=1

η2i = t2sum

1leilenxminus2i (1minus τi)

In Section 4 we find an odd analog of Heckmanrsquos important sl2-triple in [13] by showing that avariant Di of the odd Dunkl operator can be used to construct three operators r2 E and ∆ that satisfythe defining relations of the Lie algebra sl2

r2 = (2t)minus1nsumi=1

x2i

E =nsumi=1

xipi +n

2+u

t

sumk 6=i

sik

∆ = minus(2t)minus1nsumi=1

D2i

Since even Dunkl operators play an important role in the representation theory of symmetric groupsour study of odd Dunkl operators should result in a better understanding of the representation theoryof odd symmetric functions which correspondingly results in a better understanding of odd Khovanovhomology

The second goal of this paper is to study a generalization of the odd symmetric functions knownas q-symmetric functions for which xjxi = qxixj when j gt i Previous authors have described a nil-Hecke algebra structure only for the odd case q = minus1 and the even case q = 1 [9 15] In Section 5we find a q-divided difference operator for all previously unstudied values of q and explore its propertiesWe show for example that twisted elementary symmetric polynomials are in the kernel of q-divided

4

difference operators just as odd elementary symmetric functions are in the kernel of odd divided differ-ence operators We then use q-divided difference operators to construct algebras acting on q-symmetricpolynomials that have many similarities to the even and odd nilHecke algebras We call these q-nilHeckealgebras These algebras are nontrivial generalizations of the even and odd nilHecke algebras becausethe q-twist map introduced in Section 5 is not its own inverse when q2 6= 1 In Section 6 we present theelementary q-symmetric polynomials using a generalization of a clever diagrammatic method arisingin the context of bialgebras We use these diagrams to study relations between elementary q-symmetricpolynomials when q is a root of unity These methods can be combined with the algebras of Section 5 inorder to continue studying q-symmetric polynomials including q-Schur and q-monomial functions

In the conclusion we also define the odd Cherednik operators and outline a procedure for findingand studying odd analogs of Jack polynomials This makes progress towards answering a question ofEllis about the existence of Macdonald-like polynomials in the odd case Since the Jack polynomialshave importance in representation theory their study would enhance our knowledge about the oddalgebraic theory

2 Odd Dunkl operators and the Odd nilHecke algebra

21 Preliminaries Even Dunkl Operators

In the even case we work with the ring C[x1 xn] and a root system of type An where xixj = xjxifor all 1 le i j le n and α isin C We first introduce some notation involving the symmetric group

1 Let sik be the simple transposition in Sn swapping xi and xk We let si = sii+1

2 Let sij(k `) be the result of applying sij to the pair (k `) Similarly define sij(k)

In [7] Dunkl introduced the remarkable operator

ηeveni =

part

partxi+ α

sumk 6=i

partevenik

where partpartxi

is the partial derivative with respect to xi and partevenik is the even divided difference operator

partevenik = (xi minus xk)minus1(1minus sik)

Since xi minus xk always divides f minus sik(f) for f isin C[x1 xn] partik sends polynomials to polynomialsThese Dunkl operators have various important properties one of which is that they commute (ηiηj =

ηjηi) In [18] Khongsap and Wang introduced anti-commuting odd Dunkl operators on skew polynomi-als In Section 2 we will develop the connection between these operators and the odd nilHecke algebraintroduced in [9]

Returning to the even case introduce operators r2 E (the Euler operator) and ∆k

r2 =1

2

nsumi=1

x2i

E =

nsumi=1

xipart

partxi+micro

2

∆ =1

2

nsumi=1

η2i

where micro is the Dunkl dimension which is defined by the relation ∆|x|2 = 2micro as in [6]

5

Let [p q] = pq minus qp be the commutator Heckman showed that r2 E and ∆k satisfy the definingrelations of the Lie algebra sl2 [13]

[E r2] = 2r2

[E∆k] = minus2∆k

[r2∆k] = E

Remark 21 If one were to replace ∆k with the classical Laplacian on flat Rn (replacing the Dunkloperator with the partial derivative) these three operators still satisfy the sl2 relations

Remark 22 From now on we will use ηi to denote the odd Dunkl operator of Khongsap and Wangdefined in equation 25

In Section 4 we will focus on finding analogous results in the odd case

22 Introduction to the Odd nilHecke Algebra

We will now discuss operators with the algebra Pminus = C〈x1 xn〉〈xjxi + xixj = 0 for i 6= j〉 We callPminus the skew polynomial ring We can define linear operators called the odd divided difference operatorsas below

Definition 23 For i = 1 nminus1 the i-th odd divided difference operator parti is the linear operator Pminus rarr Pminus

defined by parti(xi) = 1 parti(xi+1) = 1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + (minus1)|f |si(f)parti(g)

for all functions f g isin Pminus We call this last relation the Leibniz rule

It is shown in [9] that the odd divided difference operators can be used to construct an odd nilHeckealgebra generated by xi and parti for 1 le i le n subject to the following relations

1 part2i = 0

2 partipartj + partjparti = 0 for |iminus j| ge 2

3 partiparti+1parti = parti+1partiparti+1

4 xixj + xjxi = 0 for i 6= j

5 xiparti + partixi+1 = 1 partixi + xi+1parti = 1

6 xipartj + partjxi = 0 for i 6= j j + 1

Due to [14] we have the following explicit definition of the odd divided difference operator

parti(f) = (x2i+1 minus x2i )minus1[(xi+1 minus xi)f minus (minus1)|f |si(f)(xi+1 minus xi)] (21)

Although this formula a priori involves denominators it does take skew polynomials to skew polyno-mials We extend this definition to non-consecutive indices by replacing i + 1 with any index k 6= i for1 le k le n and by replacing si with sik Equation 21 then becomes

partik(f) = (x2k minus x2i )minus1[(xk minus xi)f minus (minus1)|f |sik(f)(xk minus xi)] (22)

This extended odd divided difference operator satisfies the Leibniz rule partik(fg) = partik(f)g+(minus1)|f |sikpartik(g)[9]

6

23 Some Operations on Skew Polynomials

First we introduce a common operator in the study of Dunkl operators

Definition 24 Let the (minus1)-shift operator τi be the automorphism of Pminus which sends xi to minusxi and xj toxj for j 6= i

Suppose 1 le i 6= j le n and 1 le k 6= ` le n where f is an element in C〈x1 xn〉〈xixj + xjxi =0 for i 6= j〉 Then one has

sijτk` = τsij(k`)sij

fxi = (minus1)|f |xiτi(f)

Remark 25 Since skew polynomials are not super-commutative we cannot say that fg = (minus1)|f ||g|gf But the operator τi allows us to track the discrepancy from super-commutativity since xif = (minus1)|f |τi(f)ximaking it useful in this context

We now introduce the operator rik = partiksik for k 6= i which will serve as another odd divideddifference operator that we will use to study odd Dunkl operators For simplicity let ri = rii+1 In thefollowing lemma we study the action of the transposition and (minus1)-shift operator on rik

Lemma 26 The operators sik and τi act on rik as follows

rijsk` = sk`rsk`(ij) (23)

We also have that

1 sirik = ri+1ksi if k 6= i+ 1

2 siri = risi

3 siri+1 = rii+2si

4 si+1ri = rii+2si+1

5 sirj = rjsi for |iminus j| ge 2

6 τirj = rjτi for |iminus j| ge 2

Proof Recall the following relationship between partij and sk` for i 6= j and k 6= ` from Lemma 219 (1) of[9]

partijsk` = sk`partsk`(ij) (24)

Multiplying both sides by sij we obtain that

sijpartijsk` = sijsk`partsk`(ij)

= sk`sk`(i j)partsk`(ij)

which implies the desired result since rij = partijsij = sijpartij Properties 1-5 are special cases of equation23 Property 6 follows from τisj = sjτi and the fact that τi(xj) = xj for i 6= j

Remark 27 Differences between our formulas and those of [9] are due to a difference of sign conventionin the action of sij on Pminus

We now show that the properties of the rik are similar to those of the odd divided difference operatorpartik

Lemma 28 The following relations hold

7

1 r2i = 0

2 rirj + rjri = 0 for |iminus j| ge 2

3 riri+1ri = ri+1riri+1

4 rik(fg) = rik(f)sik(g) + (minus1)|f |frik(g)

5 rixi+1 + xi+1ri = rixi + xiri = si

6 rjxi + xjri = 0 for i 6= j j + 1

Proof Since siri = risi and ri = partisi it follows that siparti = partisi Then since part2i = 0 r2i = 0 as well Dueto Equation 5 from Lemma 26 we have that sirj = rjsi for |i minus j| ge 2 so sipartj = partjsi Thus ri andrj anti-commute since partipartj + partjparti = 0 The operators ri also satisfy braid relations which we show byinductively reducing to i = 1 and then using 24 and siparti = partisi repeatedly

r1r2r1 = s1part12part2s1part1 = s1s2part13s1part13part1 = s1s2s1part23part13part12

r2r1r2 = s2part2s1part1s2part2 = s2s1part13s2part13part2 = s2s1s2part12part13part23

Since s1s2s1 = s2s1s2 and part23part13part12 = part12part13part23 by symmetry we conclude that r1r2r1 = r2r1r2 TheLeibniz rule for rik (equation 4 of this lemma) follows immediately from the Leibniz rule for partik Sinceri(xi) = ri(xi+1) = 1 and ri(xj) = 0 for j 6= i i + 1 equations 5 and 6 follow from the Leibniz rule forrik

We also desire an explicit definition of the rik analogous to that of the odd divided difference oper-ator of [9] To find such an expression we use a preparatory lemma

Lemma 29 For all f isinPminus and 1 le i 6= k le n we have

sikxiτi(f)minus sikxkτk(f) = (minus1)|f |sik(f)(xi minus xk)

Proof It suffices to prove the result for a monomial xλ = xλ11 xλii xλkk xλnn where i lt k Wecalculate that

sikxiτi(xλ) = (minus1)λ1++λixλ11 xλi+1

k xλki xλnn

sikxkτk(xλ) = (minus1)λ1++λkminus1xλ11 xλik xλk+1

i xλnn

sik(xλ)xi = (minus1)λk+1++λnxλ11 xλik x

λk+1

i xλnn

sik(xλ)xk = (minus1)λi+1++λnxλ11 x

λi+1

k xλki xλnn

Since |f | = λ1 + + λn the desired result follows

Lemma 210 The operator rik has explicit form rik = (x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Proof Follows from Lemma 29 and Equation 22

We will now connect the above results to the odd Dunkl operator introduced by Khongsap and Wangin [18]

Definition 211 Define an operator δi by δi = (2xi)minus1(1minus τi)

The above super-derivative can also be defined inductively by imposing that δi(xj) = 1 if i = j and0 otherwise We then extend the action to monomials as follows

δi(xa1xa2 xa`) =sumk=1

(minus1)kminus1xa1 δi(xak)xak+1 xa`

8

The operator δi is a priori from Laurent skew polynomials to Laurent skew polynomials but it is easy tocheck that it preserves the subalgebra of skew polynomials Khongsap and Wang found an odd analogof the Dunkl operator given by

ηi = tδi + usumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk] (25)

where t u isin Ctimes Their operators anti-commute ηiηj + ηjηi = 0 for i 6= jBy Lemma 210 this odd Dunkl operator may be expressed as

ηi = tδi + usumk 6=i

partiksik (26)

By analogy with the commutative case discussed in Section 21 the operator rik plays the same role inthe odd theory that the even divided difference operator plays in the even theory

3 Classical Yang-Baxter Equation and the Dunkl Laplacian

Theorem 31 LetH123 = [r12 r13]+ + [r13 r23]+ + [r12 r23]+ (31)

where [p q]+ = pq + qp is the anti-commutator Then the operators rik satisfy the classical Yang-Baxterequation

H123 = 0 (32)

Proof To avoid a cumbersome direct calculation we instead use an inductive approach Namely sup-pose that H123(f) = 0 for some function f isin Pminus Then we show that H123(xif) = 0 for all integersi ge 1 Note that for i = 1

r12r13x1 = r12(s13 minus x1r13)= r12s13 minus r12(x1r13)= r12s13 minus (s12r13 minus x1r12r13)= r12s13 minus s12r13 + x1r12r13

(33)

where we have used Equation 4 twice Similarly we find that

r13r12x1 = r13s12 minus s13r12 + x1r13r12 (34)r23r12x1 = r23s12 + x1r23r12 (35)r12r23x1 = minuss12r23 + x1r12r23 (36)r13r23x1 = minuss13r23 + x1r13r23 (37)r23r13x1 = r23s13 + x1r23r13 (38)

By our inductive hypothesis

x1(r12r13 + r13r12 + r23r12 + r12r23 + r13r23 + r23r13) = 0

Keeping this in mind add Equations 33-38 to show that

H123x1 = r12s13 minus s12r13 + r13s12 minus s13r12 + r23s12 minus s12r23 minus s13r23 + r23s13

= r12s13 minus r23s12 + r13s12 minus r23s13 + r23s12 minus r13s12 minus r12s13 + r23s13

= 0

where we have repeatedly used Lemma 26 to slide rij past sk`We can similarly show that H123x2 = H123x3 = 0 Since rjkxi = xirjk for i gt 3 and j k isin (1 2 3)

it also follows that H123xi = 0 for i gt 3 proving the desired result

9

Corollary 32 The double summationsumn

i=1

(sumk 6=i rik

)2= 0

Proof The expansion of this double summation has n(nminus1)2 total terms Since r2ij = 0 n(nminus1) of theseterms are immediately zero leaving n(n minus 1)(n minus 2) terms of the form rijrk` where i = k and j = `are not both true By Theorem 31 the sum of all six terms of the form rijrk` where i j k ` isin a b c fordistinct integers 1 le a b c le n i 6= j and k 6= ` is zero There are

(n3

)ways to choose integers a b c

and for each choice of a b c six terms of the form rijrk` vanish This fact eliminates all the remaining6(n3

)= n(nminus 1)(nminus 2) terms of the double summation

As an application of the results in this section we will compute the odd Dunkl Laplaciansumn

i=1 η2i

We will first require a lemma involving the commutator of τi and rik

Lemma 33 The equation xminus1i [rik τi] = (x2i minus x2k)minus1(sik(τi + τk)minus xminus1i xksik(τi minus τk)minus 2) holds

Proof By Lemma 210

rikτi = (x2i minus x2k)minus1((xi minus xk)sikτi minus xiτ2i + xkτkτi)

τirik = (x2i minus x2k)minus1(minus(xi + xk)sikτk + xiτ2i + xkτiτk)

Since τiτk = τkτi and τ2i = 1 the result follows by subtraction

Now define Ai = (2xi)minus1(1minus τi) and Bi =

sumk 6=i rik so that the odd Dunkl operator ηi of Khongsap

and Wang may be expressed as ηi = tAi + uBi Note that

A2i =

1

4xminus1i (1minus τi)xminus1i (1minus τi)

=1

2xminus2i (1minus τi)(1minus τi)

= xminus2i (1minus τi)

(39)

since τi(xminus1i ) = minusxminus1i and τ2i = 1

Lemma 34 The relationsumn

i=1(AiBi +BiAi) = 0 holds

Proof Due to the Leibniz Rule for rik (equation 4) we can find that 0 = rik(xixminus1i ) = xminus1k minus xirik(x

minus1i )

so rik(xminus1i ) = xminus1i xminus1k It follows that rik(xminus1i f) = xminus1i xminus1k sik(f)minus xminus1i rik(f) Using this factnsumi=1

BiAi =1

2

nsumi=1

sumk 6=i

rikxminus1i (1minus τi)

=1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi)minus xminus1i rik(1minus τi))

By definitionnsumi=1

AiBi =1

2

nsumi=1

sumk 6=i

xminus1i (1minus τi)rik

Adding the above two equations we find thatnsumi=1

[Ai Bi]+ =1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi) + xminus1i (rikτi minus τirik))

=1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi) + (x2i minus x2k)minus1(sik(τi + τk)minus xminus1i xksik(τi minus τk)minus 2))

(310)

10

where we have used Lemma 33Each double summation repeats the pair of indices (p q) twice when 1 le p q le n one time when

i = p and k = q and once more when i = q and k = p Note that

(x2i minus x2k)minus1sik(τi + τk) = minus(x2k minus x2i )ski(τk + τi)

As a result the sum 12

sumni=1

sumk 6=i(x

2i minus x2k)minus1sik(τi + τk) = 0 Similarly

1

2

nsumi=1

sumk 6=i

xminus1i xminus1k sik = 0 and1

2

nsumi=1

sumk 6=i

(x2i minus x2k)minus1(minus2) = 0

since xminus1i xminus1k + xminus1k xminus1i = 0 Equation 310 then becomes

nsumi=1

(AiBi +BiAi) =1

2

nsumi=1

sumk 6=i

(minusxminus1i xminus1k sikτi minus (x2i minus x2k)minus1(xminus1i xksik(τi minus τk)) (311)

However note thatxminus1i xk + xminus1k xi = minus(x2i minus x2k)xminus1i xminus1k

which implies

(x2i minus x2k)minus1xminus1i xksik(τi minus τk) + (x2k minus x2i )minus1xminus1k xiski(τk minus τi) = minusxminus1i xminus1k sik(τi minus τk) (312)

Similarly we find that

xminus1i xminus1k sikτi + xminus1k xminus1i sikτk = xminus1i xminus1k sik(τi minus τk) (313)

As a result equation 311 becomes

nsumi=1

(AiBi +BiAi) = minus1

2

sum1leiltklen

(xminus1i xminus1k sik(τi minus τk)minus xminus1i sminus1k sik(τi minus τk)) = 0

We are now equipped to compute the Dunkl Laplacian in the odd case

Theorem 35 The equationsumn

i=1 η2i = t2

sum1leilen x

minus2i (1minus τi) holds

Proof Since ηi = tAi + uBi we have that

nsumi=1

η2i = t2nsumi=1

A2i + tu

nsumi=1

(AiBi +BiAi) + u2nsumi=1

B2i

By Lemma 34sumn

i=1(AiBi +BiAi) = 0 By Corollary 32sumn

i=1B2i = 0 Therefore

nsumi=1

η2i = t2nsumi=1

A2i = t2

sum1leilen

xminus2i (1minus τi)

by equation 39

11

4 A Variant of the Khongsap-Wang Odd Dunkl Operator

In this section we will show that a close variant of the odd Dunkl operator introduced by Khongsapand Wang can be used in the construction of three operators that satisfy the defining relations of the Liealgebra sl2 First we will consider an operator pi which is different from δi but plays a similar role

Definition 41 The operator pi is a C-linear map Pminus rarr Pminus which acts on monomials as follows

pi(xλ11 xλii xλnn ) = λi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn

Remark 42 One may also introduce pi by using a nice Leibniz-like expression involving τi by defining

pi(xj) = δij

pi(fg) = pi(f)g + (minus1)|fτi(f)pi(g)

where f g isin Pminus and δij is the Kronecker delta Now note the analogous relationship between thedegree-preserving operators si and τi in their respective Leibniz rules for the (minus1)-degree operators partiand pi This provides motivation for the definition of pi and suggests its natural role in our theory

Now consider a modified version of ηi

Definition 43 LetDi = tpi + u

sumk 6=i

rik (41)

Definition 44 Introduce the odd r2 Euler and ∆ operators as below

r2 = (2t)minus1nsumi=1

x2i (42)

E =nsumi=1

xipi +n

2+u

t

sumk 6=i

sik (43)

∆ = minus(2t)minus1nsumi=1

D2i (44)

Remark 45 Heckman who used the even Dunkl operators to find a sl2-triple useful in harmonic anal-ysis uses the convention t = 1 [13] For now we will consider t to be a fixed constant in Ctimes

Remark 46 The commutator in the setting of superalgebras is usually defined as [a b] = abminus(minus1)|a||b|bawhere |a| and |b| are the degrees of a and b respectively However since all of the operators we will beconsidering in this section have even degree there is no need to distinguish between commutators andsuper-commutators

To construct an sl2 action from these operators we will require a series of lemmas regarding theaction of portions of the odd Euler operator E In the next lemma we investigate the action of the firstterm of the odd Euler operator on skew polynomials

Lemma 47 The operatorsumn

i=1 xipi acts by multiplication by |f | on the space of homogenous functionsf isinPminus

Proof It suffices to show the result for a monomial xλ = xλ11 xλii xλnn Note that

xipi(xλ) = λixi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn = λix

λ

12

By summing over all indices i we obtain that

nsumi=1

xipi(xλ) = (λ1 + λ2 + + λn)xλ

which implies the desired result

The above lemma holds true in the even case as well where pi is replaced by the partial derivativewith respect to xi We now prove some properties about the action of the third term of the odd Euleroperator on r2 and ∆

Lemma 48 The commutation relation[sum

k 6=i sik∆]

= 0 holds

Proof Note that sjkpi = pjsjk if i = k sjkpi = pksjk if i = j and sjkpi = pisjk otherwise Indeedthese relations can be verified by checking if they are true for xai x

bjxck a b c isin Z+ and then extending

by linearity We prove that sjkpi = pjsjk if i = k and the other two cases are similar Without loss ofgenerality let j lt k and observe that

sjkpk(xajx

bk) = b(minus1)asjk(x

ajx

bminus1k ) = b(minus1)axakx

bminus1j = b(minus1)abxbminus1j xak

pjsjk(xajx

bk) = (minus1)abpj(x

bjxak) = b(minus1)abxbminus1j xak

By our work in Lemma 26 one can deduce that sjkr`m = rsjk(`m)sij As a consequence we find thatsjkDi = Dsjk(i)sjk By an easy induction we now have that sjk∆ = ∆sjk Using the above equationmultiple times proves the desired result

Lemma 49 The commutation relation[sum

k 6=i sik r2]

= 0 holds

Proof Follows since sjkxj = xksjk sjkxk = xjsjk and sjkxi = xisjk if i 6= j k

We are now ready to obtain two commutativity relations involving the odd Euler operator E

Theorem 410 The odd Euler operator and r2 satisfy the following commutation relations

[E r2] = 2r2 (45)[E∆] = minus2∆ (46)

Proof Since r2 has degree 2 and ∆ has degreeminus2 the theorem follows from Lemmas 47 48 and 49

We also need to investigate what the third commutativity relation [r2∆] turns out to be We willprove one lemma before doing so

Lemma 411 For i = 1 to n the equation xiDi +Dixi = 2txipi + t+ usum

k 6=i sik holds

Proof Recall thatDi = tpi + u

sumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Therefore since pixi = xipi + 1

Dixi = txipi + t+ usumk 6=i

(x2i minus x2k)minus1(xixk minus x2k)sik +sumk 6=i

(x2i minus x2k)[x2i τi minus xixkτk]

xiDi = txipi + usumk 6=i

(x2i minus x2k)minus1(x2i minus xixk)sik +sumk 6=i

(x2i minus x2k)minus1[minusx2i τi + xixkτk]

Adding we obtain the desired result

13

We now have the tools to find the third relation between r2 E and ∆

Theorem 412 The commutation relation [r2∆] = E holds

Proof We will first find [r2 Di] The derivative pi much like the partial derivative in the even casesatisfies the properties pixj = minusxjpi for i 6= j and pixi = xipi + 1 Now suppose that i 6= j Then

Dix2j = tpix

2j + ux2j

sumk 6=i 6=j

rik + x2k(x2i minus x2k)minus1[minusxiτi + xkτk] + x2i (x

2i minus x2k)minus1[(xi minus xk)sik]

= tx2jpi + ux2jsumk 6=i

rik + (xi minus xj)sij

Now we will find Dix2i

Dix2i = tpix

2i +

sumk 6=i

x2k(x2i minus x2k)minus1(xi minus xk)sik + x2i

sumk 6=i

(x2i minus x2k)minus1[minusxiτi + xkτk]

= tx2i pi + 2txi + x2isumk 6=i

rik minussumk 6=i

(xi minus xk)sik

Therefore[sumn

i=1 x2i Di

]= minus2txi This implies that

r2Di minusDir2 = minusxi (47)

As a result we find that

[r2∆] = minus(2t)minus1nsumi=1

[r2 D2j ] = minus(2t)minus1

nsumi=1

(r2D2j minusD2

j r2)

= minus(2t)minus1nsumi=1

[(Djr2Dj minus xjDj)minus (Djr

2Dj +Djxj)]

= (2t)minus1nsumi=1

(xiDi +Dixi)

where we have used 47 Now by Lemma 411

[r2∆] =

nsumi=1

xipi +n

2+u

t

sumk 6=i

sik = E

as desired

To summarize we have found operators Er2 and ∆ similar to their even counterparts whichsatisfy the defining relations of the Lie algebra sl2

[E r2] = 2r2

[E∆] = minus2∆

[r2∆] = E

Remark 413 If one uses the odd Dunkl operator ηi as found in [18] instead of the Di introduced herethe r2 E and ∆ operators do not generate sl2

Remark 414 Although our results hold true for all t and u in C one typically sets t = 1 and u = αminus1

for some α isin Ctimes since without loss of generality one of t and u may equal 1

14

Remark 415 In the even case let X be a Euclidean vector space with dimension n and let C[X] be thealgebra of C-valued functions on X Then this result about sl2 plays a major role in the study of higherdifferential operators on C[X] This is because the representation theory of sl2 allows for the reductionof degree to the second order [13] As a result our results in this section should correspondingly have arole in further studying differential operators in the odd case

5 q-nilHecke Algebras

Until now we have been concerned with the odd symmetric polynomials in variables x1 x2 xnwhere xixj = (minus1)xjxi for 1 le i 6= j le n This immediately suggests the question what if one replacesthe minus1 by any constant q isin Ctimes Specifically we ask the following questions

Question 51 Is it possible to study q-symmetric polynomials for which xixj = qxjxi when i gt j

Question 52 Are there q-analogs of evenodd divided difference operators and nilHecke algebras Sofar such structures are known only for the even case (q = 1) and the odd case (q = minus1)

In this section we answer both questions in the affirmativeWe work in the Z-graded q-braided setting throughout LetC be a commutative ring and let q isin Ctimes

be a unit If VW are graded C-modules and v isin V w isin W are homogeneous the braiding is theldquoq-twistrdquo

τq V otimesW rarrW otimes Vv otimes w 7rarr q|v||w|w otimes v

(51)

where | middot | is the degree function By q-algebra we mean an algebra object in the category of gradedC-modules equipped with this braided monoidal structure likewise for q-bialgebras q-Hopf algebrasand so forth

Remark 53 Note that the q-twist described above is its own inverse only when q2 = 1 which correlatesto the even and odd cases When q2 6= 1 the corresponding theory becomes more complex There-fore the q-nilHecke algebras that we introduce later in this section are nontrivial generalizations of thepreviously studied even and odd nilHecke algebras

Definition 54 The q-algebra P qn is defined to be

P qn = C〈x1 xn〉(xjxi minus qxixj = 0 if i lt j) (52)

where |xi| = 1 for i = 1 n

Note that P qn sim= otimesni=1Pq1 There are two interesting subalgebras of P qn that can be thought of as q-

analogs of the symmetric polynomials Define the k-th elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middotxin

and define the k-th twisted elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middot xin

where xj = qjminus1xj

Definition 55 The q-algebra of q-symmetric polynomials in n variables denoted Λqn is the subalgebra ofP qn generated by e1 en Likewise for the twisted q-symmetric polynomials Λqn and e1 en

The type A braid group on n strands acts on P qn by setting

15

1 σi(xj) = qxi+1 if j = i

2 σi(xj) = qminus1xi if j = i+ 1

3 σi(xj) = qxj if j gt i+ 1

4 σi(xj) = qminus1xj if j lt i

and extending multiplicatively

Definition 56 For i = 1 nminus 1 the i-th q-divided difference operator parti is the linear operator P qn rarr P qndefined by parti(xi) = q parti(xi+1) = minus1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + σi(f)parti(g) (53)

for all functions f g isin P qn We call Equation 53 the q-Leibniz rule

Lemma 57 For every i and every j lt k parti(xkxj minus qxjxk) = 0

Proof Since parti(xj) = 0 for j gt i+1 one may reduce the lemma to having to prove that part1(x2x1minusqx1x2) =0 part1(x3x1 minus qx1x3) = 0 and part1(x3x2 minus qx2x3) = 0 These statements follow from the q-Leibniz rule

Therefore parti is a well-defined operator on P qn

Lemma 58 The following relations hold

parti(xki ) =

kminus1sumj=0

qjkminus2jminusj2+kxjix

kminus1minusji+1

parti(xki+1) = minus

kminus1sumj=0

qminusjxjixkminus1minusji+1

Proof We induct on k The base case (k = 1) follows from the definition of the parti and the powers of qarise mostly from xni+1x

mi = qmnxmi x

ni+1 for all mn isin Z+

Our q-divided difference operators also annihilate the twisted elementary q-symmetric polynomialsjust as the even divided difference operators annihilate the elementary symmetric functions

Lemma 59 For every i = 1 nminus 1 and every k parti(ek) = 0 Hence Λqn sube⋂nminus1i=1 ker(parti)

Proof We can express ek as

ek =sum|J|=kii+1isinJ

xJ +sum|J|=kminus1ii+1isinJ

qf(Jik)xJ(xi + qxi+1) +sum|J|=kminus2ii+1isinJ

qg(Jik)xJxixi+1

for certain Z-valued functions f g The result then follows from parti(xi + qxi+1) = parti(xixi+1) = 0 and theq-Leibniz rule

Having discussed q-divided difference operators we can now construct algebras for every q 6=0 1minus1 that have many similarities to the even and odd nilHecke algebras For every such q we de-fine a q-nilHecke algebra generated by xi and parti for 1 le i le n subject to the relations found in thefollowing two lemmas (510 and 511)

Lemma 510 The following relations hold among the operators parti and xi (left multiplication by xi)

1 part2i = 0

2 partjparti minus qpartipartj = 0 for j gt i+ 1

3 xjxi = qxixj for i lt j

4 partixj minus qxjparti = 0 for j gt i+ 1

5 qpartixj minus xjparti = 0 for j lt i

6 partixi minus qxi+1parti = q

7 xiparti minus qpartixi+1 = q

16

Proof To show that part2i = 0 note that we can reduce to i = 1 and proceed by induction Since parti(1) = 0the base case follows Suppose that part2i (f) = 0 Then note that

part21(x1f) = part1(qf + qx2part1(f)) = qpart1(f)minus qpart1(f) + x1part21(f) = 0

part21(x2f) = part1(minusf + qminus1x1part1(f)) = minuspart1(f) + part1(f) + x2part21(f) = 0

part21(x3f) = part1(qx3part1(f)) = q2x3part21(f) = 0

which completes the proof of the first statement in the lemmaStatement 3 follows by definition Statements 4 5 6 and 7 follow from a suitable application of the

q-Leibniz rule Statement 2 follows from an inductive argument We can reduce to i = 1 and j = 3Suppose that partjparti = qpartipartj if j gt i+ 1 Then

part3part1(x1f)minus qpart1part3(x1f) = (qpart3(f) + x2part3part1(f))minus q(part3(f) + x2part1part3(f)) = 0

part3part1(x2f)minus qpart1part3(x2f) = (minuspart3(f) + qminus2x1part3part1(f))minus q(minusqminus1part3(f) + qminus2x1part1part3(f)) = 0

part3part1(x3f)minus qpart1part3(x3f) = (q2part1(f) + q2x4part3part1(f))minus q(qpart1(f) + q2x4part1part3(f)) = 0

part3part1(x4f)minus qpart1part3(x4f) = (minusqpart1(f) + x3part3part1(f))minus q(minuspart1(f) + x3part1part3(f)) = 0

part3part1(x5f)minus qpart1part3(x5f) = q2x5part3part1(f)minus q(q2x5part1part3(f)) = 0

thereby completing the induction

Lemma 511 partiparti+1partiparti+1partiparti+1 + parti+1partiparti+1partiparti+1parti = 0

Proof This result follows from an inductive argument we reduce to i = 1 and assume that the braidrelation holds true for some function f Then we check that the braid relation is true for x1f x2f x3f and x4f (since the behavior of xjf for j ge 4 is the same as that of x4f ) For brevity we will show theargument for x2f only

part1part2(x2f) = qpart1(f) + q2x3part1part2(f) part2part1(x2f) = minuspart2(f) + qminus2x1part2part1(f)

part212(x2f) = qpart2part1(f)minus q2part1part2(f) + qx2part212(f) qpart121(x2f) = minusqpart1part2(f) + part2part1(f) + part121(f)

part1212(x2f) = qpart121(f)minus qpart212(f) + x1part1212(f) part2121(x2f) = minuspart212(f) + part121(f) + x3part2121(f)

part21212(x2f) = qpart2121(f) + qminus1x2part21212(f) part12121(x2f) = minuspart1212(f) + qx3part12121(f)

We continue the above calculations to find that

part121212(x2f) = qpart12121(f) + part21212(f) + x2part121212(f)

part212121(x2f) = minusqpart12121(f)minus part21212(f) + x2part121212(f)

and the braid relation for x2f follows from the inductive hypothesis

6 A Diagrammatic Approach to q-Symmetric Polynomials

61 Introduction to a q-Bialgebra

In the previous section we answered Question 51 in an algebraic way by defining q-analogs of theclassical elementary and complete symmetric functions In this section we generalize the diagrammaticmethod used in [8] in order to study this question from the perspective of bialgebras

Let NΛq be a free associative Z-graded C-algebra with generators hm for m ge 0 We define h0 = 1and hm = 0 for m lt 0 and let q isin Ctimes The homogenous part of NΛq of degree ` has a basis hααkwhere

hα = hα1 middot middot middothαz for a composition α = (α1 αz) of `

17

Define a multiplication for homogenous x and y on NΛqotimes2 as follows where deg(x) denotes the degreeof x

(w otimes x)(y otimes z) = qdeg(x)deg(y)(wy otimes xz)

We can make NΛq into a q-bialgebra by letting the comultiplication on generators be

∆(hn) =

nsumk=0

hk otimes hnminusk

and by letting the counit be ε(x) = 0 if x is homogenous and deg(x)gt 0We can impose through the braiding structure that

∆(hahb) =asumj=0

bsumk=0

(hj otimes haminusj)(hk otimes hbminusk) =asumj=0

bsumk=0

qk(aminusj)(hjhk otimes haminusjhbminusk)

For any partitions λ and micro of n consider the set of double cosets of subroups Sλ and Smicro of Sn SλSnSmicroFor every C in this set let wC be the minimal length representative of C and let `(wC) be the length ofthis minimal length representative We will now attribute a bilinear form to NΛq

(hλ hmicro) =sum

CisinSλSnSmicro

q`(wC)

This bilinear form admits a diagrammatic description Let hn be an orange platform with n non-intersecting strands coming out of it When computing (hλ hmicro) with `(λ) = z and `(micro) = y drawz orange platforms at the top of the diagram representing λ1 λ2middot middot middot λz Draw y orange platforms atthe bottom of the diagram representative of micro1 micro2middot middot middot microy We require that |λ| = |micro| so that the topplatforms and bottom platforms have the same number of strands

Consider the example (h121 h22) In the following diagram snippets of the strands from each plat-form are shown

Every strand must start at one platform at the top and end on another platform at the bottom Nostrands that have originated from one platform may intersect The strands themselves have no criticalpoints with respect to the height function no two strands ever intersect more than once and there are notriple-intersections where three strands are concurrent Diagrams are considered up to isotopy Withoutany restrictions there would be n such diagrams if |λ| = n since there would be no limitations onthe ordering of the strands However due to the above rules there are only 4 possible diagrams in thecomputation of (h121 h22) shown below

Define(hλ hmicro) =

sumall diagrams D representing (hλhmicro)

q number of of crossings in D (61)

In the above example (h121 h22) = 1 + 2q2 + q3

18

We can extend the bilinear form toNΛqotimes2 by stating that any diagram in which strands from distincttensor factors intersect contributes 0 to the bilinear form

(w otimes x y otimes z) = (w y)(x z)

Let I be the radical of the bilinear form inNΛq In [8] the authors prove for any q that multiplicationand comultiplication are adjoint In other words for all xy1 y2 in NΛq

(y1 otimes y2∆(x)) = (y1y2 x) (62)

62 The Elementary q-Symmetric Functions

We now use the bilinear form of q-symmetric functions to study one of their important bases the ele-mentary q-symmetric functions

Define elements ek isin NΛq by ek = 0 for k lt 0 e0 = 1 and

ksumi=0

(minus1)iq(i2)eihkminusi = 0 for k ge 1 (63)

Equivalently leten = qminus(n2)

sumαn

(minus1)`(α)minusnhα (64)

Lemma 61

1 The coproduct of an elementary function is given by ∆(en) =

nsumk=0

ek otimes enminusk

2 If λ n then (hλ en) =

1 if λ = (1 1)

0 otherwise

Proof We begin by demonstrating (2) from which (1) will follow To show (2) it suffices to show that

(hmx en) =

(x enminus1) if m = 10 otherwise

We will utilize strong induction on n in order to find (hmx ekhnminusk) The base cases n = 0 1 are easyto show There are two cases to consider by the inductive hypothesis applied to k lt n Either there isa strand connecting hm and ek or there is not Just as we used an orange platform to denote hn wewill use a blue platform to denote ek The rules of the diagrammatic notation are the same for the blueplatforms as they are for the orange platforms

k

m

nminus k minusm

k nminus k

mlowast lowast lowast lowast lowast

x

If there is not a strand connecting hm and ek the configuration contributes qkm(x ekhnminuskminusm)

19

k minus 1

mminus 1

nminus k minusm + 1

k nminus k

mlowast lowast lowast lowast lowast

x

If a stand connects hm and ek this configuration contributes q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) We have thusshown that (hmx ekhnminusk) = qkm(x ekhnminuskminusm) + q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) Now we are equippedto consider (hmx ek)

(minus1)n+1q(n2)(hmx en) =

nminus1sumk=0

(minus1)kq(k2)(hmx ekhnminusk)

=nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus1sumk=0

(minus1)kq(k2)+(mminus1)(kminus1)(x ekminus1hnminuskminusm+1)

=

nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus2sumk=0

(minus1)k+1q(k+12 )+(mminus1)(k)(x ekhnminuskminusm)

= (minus1)nminus1q(nminus12 )+nm(x enminus1h1minusm)

Corresponding terms from the two sums cancel in pairs since q(k2)+km = q(

k+12 )+k(mminus1) leaving only the

k = nminus 1 term in the first sum The second statement of the lemma thus followsWe will now use (2) to prove (1) This follows from equation 62

(∆(ek) hλ otimes hmicro) = (ek hλhmicro) =

1 λ = (1`) micro = (1p) `+ p = k

0 otherwise

We now calculate the sign incurred when strands connect two blue (ek) platforms

(minus1)n+1q(n2)(en en) =

nminus1sumk=0

(minus1)kq(k2)(en ekhnminusk)

= (minus1)nminus1q(nminus12 )(en enminus1h1)

= (minus1)nminus1q(nminus12 )(∆(en) enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )

nsumk=0

(ek otimes enminusk enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )(enminus1 enminus1)

One may solve this recursion to find that (en en) = qminus(n2)Here the second equality follows from not-ing that at most one strand can connect hnminusk and en (so that k = n minus 1) the third equality followsfrom adjointness and the fourth and fifth equalities follow from the diagrammatic considerations of theprevious lemma

To summarize the diagrammatics of the bilinear form thus developed

20

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

1 Introduction

11 The Commutative (ldquoEvenrdquo) Case

Symmetric polynomials are polynomials in n independent commutative variables x1 x2 xn that areinvariant under the action of any permutation acting on the indices They arise in enumerative com-binatorics algebraic combinatorics Galois theory quantum statistics and the quantum mechanics ofidentical particles [25 27] The even nilHecke algebra NHn introduced by Kostant and Kumar in [15]is important in studying these symmetric polynomials NHn is graded Morita equivalent to the sym-metric polynomials in n variables and is generated by n commuting variables x1 xn and n divideddifference operators parti = (xi minus xi+1)

minus1(1minus si) for 1 le i le n Here si is the simple transposition in thesymmetric group that swaps xi and xi+1

Combining these divided differences with partial derivatives one obtains a commuting family ofoperators originally introduced by Dunkl [7] These Dunkl operators denoted ηi have a major role inmathematical physics and conformal field theory In particular they relate to the study of quantummany-body problems in the Calogero-Moser-Sutherland model which describes integrable systems ofone dimension [11 24] Dunkl operators can also be used to define three operators which arise inphysics and harmonic analysis that satisfy the defining relations of the Lie algebra sl2 [13] These threeoperators found by Heckman and called an sl2-triple play a crucial role in studying Fischer decomposi-tion which has importance not only in representation theory but also in the algebraic Dirichlet problem[1 6 23]

The Cherednik operators denoted by Yi are defined in terms of Dunkl operators and have importantapplications in representation theory [2 20] They have non-degenerate simultaneous eigenfunctionsknown as Jack polynomials These polynomials are a specific case of the well-known Macdonald poly-nomials and contribute to representation theory statistical mechanics and the study of the quantumfractional Hall problem important in condensed matter physics [3 20]

The diagram below depicts the relationship between the three operators discussed so far

Divided Differences parti

nilHecke AlgebraSchubert polynomials

Cohomology

Dunkl Operators ηi

sl2 triple

Cherednik Operators Yi

Jack PolynomialsCherednik Algebras

Affine Hecke Algebras(harmonic analysis)

quantum CMS Model

ldquocan be used tordquoldquoused to definerdquo

Figure 1 Operators in the study of symmetric polynomials

12 The ldquoOddrdquo Case

The divided difference operators Dunkl operators and Cherednik operators all study the commutativesymmetric functions Ellis and Khovanov however sought to study different kinds of symmetric func-tions They recently introduced the quantum case of symmetric functions where xjxi = qxixj for j gt i[8] In the ldquooddrdquo case q = minus1 they describe the ldquoodd symmetric polynomialsrdquo which are polynomialsin the n variables x1 xn where xixj +xjxi = 0 for i 6= j This type of noncommutativity arises in thestudy of exterior algebras and parastatistics

The motivation for considering these odd symmetric polynomials and their corresponding odd nil-

2

Hecke algebra involves the categorification of quantum groups Categorification introduced by Craneand Frenkel is generally the process of replacing algebras and representations by categories and highercategories in order to make quantum 3-manifold invariants into 4-manifold invariants [4] In physicscategorification corresponds to increasing dimensions which allows one to understand symmetries inlower dimensions and then use categorification to better understand higher dimensions In mathemat-ics categorified quantum groups give a higher representation theoretic construction of link homologieswhich in turn categorify quantum link polynomials

The original example of link homology is Khovanov homology a bigraded abelian group which cat-egorifies the well-known Jones polynomial It has major applications in studying knot polynomialsquantum field theory and classical gauge theory [19 28] Since the quantum group Uq(sl2) plays a rolein understanding the Jones polynomial a categorification of Uq(sl2) would be useful in better under-standing Khovanov homology This precise categorification is achieved through the ldquoevenrdquo nilHeckealgebra NHn described in subsection 11

Recently Ozsvath Rasmussen and Szabo found an odd analog of Khovanov homology [22] Theirodd Khovanov homology also categorifies the Jones polynomial and agrees modulo 2 with Khovanovhomology However both theories can detect knots that the other theory cannot [26] The subject ofodd Khovanov homology has yet to be fully understood despite its crucial connections with Khovanovhomology

Knowing this Ellis Lauda and Khovanov developed the odd nilHecke algebra to provide an oddcategorification of Uq(sl2) and give a construction of odd Khovanov homology from a representationtheoretic standpoint [9 10] In addition to being useful in the categorification of quantum groups theodd nilHecke algebra is also related to Hecke-Clifford superalgebras [16 17] and has been used to con-struct odd analogs of the cohomology groups of Springer varieties [21]

The below diagram summarizes the categorifications that motivate the present work NH standsfor nilHecke Cat stands for categorification and KH stands for Khovanov homology

Jones Polynomial Uq(sl2)Uq(sl2)

KH Odd KHCat of Uq(sl2) Odd Cat of Uq(sl2)

odd NH algebraNH algebra

ldquocategorifiesrdquoldquohelps to explainrdquo

ldquoequivalent modulo 2rdquo

Figure 2 Odd Khovanov Homology and Categorification

13 Outline of the Present Paper

The main goal of the present paper is to make progress towards giving a representation theoretic con-struction of odd Khovanov homology Despite being relatively new odd Khovanov homology seemsto have great importance in knot theory It has connections to Heegaard-Floer homology and yieldsstronger results than Khovanov homology in bounding the Thurston-Bennequin number and detectingquasi-alternating knots [26] Odd Khovanov homology is also related to signed hyperplane arrange-ments which have many implications in graph theory and topology [5]

We study odd Khovanov homology by looking for new representation theoretic structures that arise

3

from identifying ldquooddrdquo analogs of structures that play important algebraic roles in the even case EllisKhovanov and Lauda started this program of ldquooddificationrdquo by finding an odd analog of NHn andusing it to categorify Uq(sl2) Searching for the geometry underlying these odd constructions would alsoprovide a very new approach to noncommutative geometry For example we study potential generatorsof certain Cherednik algebras Since spherical rational Cherednik algebras fit nicely into a family ofalgebras from the geometry of symplectic resolutions including Websterrsquos tensor product algebras andcyclotomic KLR algebras this project may be used in the context of ldquooddrdquo noncommutative geometry

In Subsection 11 we discussed the (even) divided difference operators Dunkl operators and Chered-nik operators Although analogs of Dunkl operators have been found in the odd case they have not beenwell-studied Odd Cherednik operators have not even been defined

As a result the first goal of the present paper is to unify certain results in the odd case and to furtherstudy the odd Dunkl operators of Khongsap and Wang In Section 2 we introduce an operator rikrelated to the generalized odd divided difference operator partodd

ik and study its properties One of ourmain results (Equation 26) is that the odd Dunkl operator ηi may be expressed in terms of the odddivided difference operators of Ellis Khovanov and Lauda

ηoddi = tδi + u

sumk 6=i

partoddik sik (11)

This result connects odd Dunkl operators and the odd nilHecke algebra both of which play importantroles in the project of oddification

In the even case one can introduce an operator known as the Dunkl Laplacian given bysumn

i=1 η2i This

operator has important applications in spherical harmonics and heat semigroups [24] Our next goal inthe present paper is to express the Dunkl Laplacian in the odd case In Section 3 we show that therik satisfy the classical Yang-Baxter equation and use this result to evaluate the odd Dunkl LaplacianSpecifically we show that

nsumi=1

η2i = t2sum

1leilenxminus2i (1minus τi)

In Section 4 we find an odd analog of Heckmanrsquos important sl2-triple in [13] by showing that avariant Di of the odd Dunkl operator can be used to construct three operators r2 E and ∆ that satisfythe defining relations of the Lie algebra sl2

r2 = (2t)minus1nsumi=1

x2i

E =nsumi=1

xipi +n

2+u

t

sumk 6=i

sik

∆ = minus(2t)minus1nsumi=1

D2i

Since even Dunkl operators play an important role in the representation theory of symmetric groupsour study of odd Dunkl operators should result in a better understanding of the representation theoryof odd symmetric functions which correspondingly results in a better understanding of odd Khovanovhomology

The second goal of this paper is to study a generalization of the odd symmetric functions knownas q-symmetric functions for which xjxi = qxixj when j gt i Previous authors have described a nil-Hecke algebra structure only for the odd case q = minus1 and the even case q = 1 [9 15] In Section 5we find a q-divided difference operator for all previously unstudied values of q and explore its propertiesWe show for example that twisted elementary symmetric polynomials are in the kernel of q-divided

4

difference operators just as odd elementary symmetric functions are in the kernel of odd divided differ-ence operators We then use q-divided difference operators to construct algebras acting on q-symmetricpolynomials that have many similarities to the even and odd nilHecke algebras We call these q-nilHeckealgebras These algebras are nontrivial generalizations of the even and odd nilHecke algebras becausethe q-twist map introduced in Section 5 is not its own inverse when q2 6= 1 In Section 6 we present theelementary q-symmetric polynomials using a generalization of a clever diagrammatic method arisingin the context of bialgebras We use these diagrams to study relations between elementary q-symmetricpolynomials when q is a root of unity These methods can be combined with the algebras of Section 5 inorder to continue studying q-symmetric polynomials including q-Schur and q-monomial functions

In the conclusion we also define the odd Cherednik operators and outline a procedure for findingand studying odd analogs of Jack polynomials This makes progress towards answering a question ofEllis about the existence of Macdonald-like polynomials in the odd case Since the Jack polynomialshave importance in representation theory their study would enhance our knowledge about the oddalgebraic theory

2 Odd Dunkl operators and the Odd nilHecke algebra

21 Preliminaries Even Dunkl Operators

In the even case we work with the ring C[x1 xn] and a root system of type An where xixj = xjxifor all 1 le i j le n and α isin C We first introduce some notation involving the symmetric group

1 Let sik be the simple transposition in Sn swapping xi and xk We let si = sii+1

2 Let sij(k `) be the result of applying sij to the pair (k `) Similarly define sij(k)

In [7] Dunkl introduced the remarkable operator

ηeveni =

part

partxi+ α

sumk 6=i

partevenik

where partpartxi

is the partial derivative with respect to xi and partevenik is the even divided difference operator

partevenik = (xi minus xk)minus1(1minus sik)

Since xi minus xk always divides f minus sik(f) for f isin C[x1 xn] partik sends polynomials to polynomialsThese Dunkl operators have various important properties one of which is that they commute (ηiηj =

ηjηi) In [18] Khongsap and Wang introduced anti-commuting odd Dunkl operators on skew polynomi-als In Section 2 we will develop the connection between these operators and the odd nilHecke algebraintroduced in [9]

Returning to the even case introduce operators r2 E (the Euler operator) and ∆k

r2 =1

2

nsumi=1

x2i

E =

nsumi=1

xipart

partxi+micro

2

∆ =1

2

nsumi=1

η2i

where micro is the Dunkl dimension which is defined by the relation ∆|x|2 = 2micro as in [6]

5

Let [p q] = pq minus qp be the commutator Heckman showed that r2 E and ∆k satisfy the definingrelations of the Lie algebra sl2 [13]

[E r2] = 2r2

[E∆k] = minus2∆k

[r2∆k] = E

Remark 21 If one were to replace ∆k with the classical Laplacian on flat Rn (replacing the Dunkloperator with the partial derivative) these three operators still satisfy the sl2 relations

Remark 22 From now on we will use ηi to denote the odd Dunkl operator of Khongsap and Wangdefined in equation 25

In Section 4 we will focus on finding analogous results in the odd case

22 Introduction to the Odd nilHecke Algebra

We will now discuss operators with the algebra Pminus = C〈x1 xn〉〈xjxi + xixj = 0 for i 6= j〉 We callPminus the skew polynomial ring We can define linear operators called the odd divided difference operatorsas below

Definition 23 For i = 1 nminus1 the i-th odd divided difference operator parti is the linear operator Pminus rarr Pminus

defined by parti(xi) = 1 parti(xi+1) = 1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + (minus1)|f |si(f)parti(g)

for all functions f g isin Pminus We call this last relation the Leibniz rule

It is shown in [9] that the odd divided difference operators can be used to construct an odd nilHeckealgebra generated by xi and parti for 1 le i le n subject to the following relations

1 part2i = 0

2 partipartj + partjparti = 0 for |iminus j| ge 2

3 partiparti+1parti = parti+1partiparti+1

4 xixj + xjxi = 0 for i 6= j

5 xiparti + partixi+1 = 1 partixi + xi+1parti = 1

6 xipartj + partjxi = 0 for i 6= j j + 1

Due to [14] we have the following explicit definition of the odd divided difference operator

parti(f) = (x2i+1 minus x2i )minus1[(xi+1 minus xi)f minus (minus1)|f |si(f)(xi+1 minus xi)] (21)

Although this formula a priori involves denominators it does take skew polynomials to skew polyno-mials We extend this definition to non-consecutive indices by replacing i + 1 with any index k 6= i for1 le k le n and by replacing si with sik Equation 21 then becomes

partik(f) = (x2k minus x2i )minus1[(xk minus xi)f minus (minus1)|f |sik(f)(xk minus xi)] (22)

This extended odd divided difference operator satisfies the Leibniz rule partik(fg) = partik(f)g+(minus1)|f |sikpartik(g)[9]

6

23 Some Operations on Skew Polynomials

First we introduce a common operator in the study of Dunkl operators

Definition 24 Let the (minus1)-shift operator τi be the automorphism of Pminus which sends xi to minusxi and xj toxj for j 6= i

Suppose 1 le i 6= j le n and 1 le k 6= ` le n where f is an element in C〈x1 xn〉〈xixj + xjxi =0 for i 6= j〉 Then one has

sijτk` = τsij(k`)sij

fxi = (minus1)|f |xiτi(f)

Remark 25 Since skew polynomials are not super-commutative we cannot say that fg = (minus1)|f ||g|gf But the operator τi allows us to track the discrepancy from super-commutativity since xif = (minus1)|f |τi(f)ximaking it useful in this context

We now introduce the operator rik = partiksik for k 6= i which will serve as another odd divideddifference operator that we will use to study odd Dunkl operators For simplicity let ri = rii+1 In thefollowing lemma we study the action of the transposition and (minus1)-shift operator on rik

Lemma 26 The operators sik and τi act on rik as follows

rijsk` = sk`rsk`(ij) (23)

We also have that

1 sirik = ri+1ksi if k 6= i+ 1

2 siri = risi

3 siri+1 = rii+2si

4 si+1ri = rii+2si+1

5 sirj = rjsi for |iminus j| ge 2

6 τirj = rjτi for |iminus j| ge 2

Proof Recall the following relationship between partij and sk` for i 6= j and k 6= ` from Lemma 219 (1) of[9]

partijsk` = sk`partsk`(ij) (24)

Multiplying both sides by sij we obtain that

sijpartijsk` = sijsk`partsk`(ij)

= sk`sk`(i j)partsk`(ij)

which implies the desired result since rij = partijsij = sijpartij Properties 1-5 are special cases of equation23 Property 6 follows from τisj = sjτi and the fact that τi(xj) = xj for i 6= j

Remark 27 Differences between our formulas and those of [9] are due to a difference of sign conventionin the action of sij on Pminus

We now show that the properties of the rik are similar to those of the odd divided difference operatorpartik

Lemma 28 The following relations hold

7

1 r2i = 0

2 rirj + rjri = 0 for |iminus j| ge 2

3 riri+1ri = ri+1riri+1

4 rik(fg) = rik(f)sik(g) + (minus1)|f |frik(g)

5 rixi+1 + xi+1ri = rixi + xiri = si

6 rjxi + xjri = 0 for i 6= j j + 1

Proof Since siri = risi and ri = partisi it follows that siparti = partisi Then since part2i = 0 r2i = 0 as well Dueto Equation 5 from Lemma 26 we have that sirj = rjsi for |i minus j| ge 2 so sipartj = partjsi Thus ri andrj anti-commute since partipartj + partjparti = 0 The operators ri also satisfy braid relations which we show byinductively reducing to i = 1 and then using 24 and siparti = partisi repeatedly

r1r2r1 = s1part12part2s1part1 = s1s2part13s1part13part1 = s1s2s1part23part13part12

r2r1r2 = s2part2s1part1s2part2 = s2s1part13s2part13part2 = s2s1s2part12part13part23

Since s1s2s1 = s2s1s2 and part23part13part12 = part12part13part23 by symmetry we conclude that r1r2r1 = r2r1r2 TheLeibniz rule for rik (equation 4 of this lemma) follows immediately from the Leibniz rule for partik Sinceri(xi) = ri(xi+1) = 1 and ri(xj) = 0 for j 6= i i + 1 equations 5 and 6 follow from the Leibniz rule forrik

We also desire an explicit definition of the rik analogous to that of the odd divided difference oper-ator of [9] To find such an expression we use a preparatory lemma

Lemma 29 For all f isinPminus and 1 le i 6= k le n we have

sikxiτi(f)minus sikxkτk(f) = (minus1)|f |sik(f)(xi minus xk)

Proof It suffices to prove the result for a monomial xλ = xλ11 xλii xλkk xλnn where i lt k Wecalculate that

sikxiτi(xλ) = (minus1)λ1++λixλ11 xλi+1

k xλki xλnn

sikxkτk(xλ) = (minus1)λ1++λkminus1xλ11 xλik xλk+1

i xλnn

sik(xλ)xi = (minus1)λk+1++λnxλ11 xλik x

λk+1

i xλnn

sik(xλ)xk = (minus1)λi+1++λnxλ11 x

λi+1

k xλki xλnn

Since |f | = λ1 + + λn the desired result follows

Lemma 210 The operator rik has explicit form rik = (x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Proof Follows from Lemma 29 and Equation 22

We will now connect the above results to the odd Dunkl operator introduced by Khongsap and Wangin [18]

Definition 211 Define an operator δi by δi = (2xi)minus1(1minus τi)

The above super-derivative can also be defined inductively by imposing that δi(xj) = 1 if i = j and0 otherwise We then extend the action to monomials as follows

δi(xa1xa2 xa`) =sumk=1

(minus1)kminus1xa1 δi(xak)xak+1 xa`

8

The operator δi is a priori from Laurent skew polynomials to Laurent skew polynomials but it is easy tocheck that it preserves the subalgebra of skew polynomials Khongsap and Wang found an odd analogof the Dunkl operator given by

ηi = tδi + usumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk] (25)

where t u isin Ctimes Their operators anti-commute ηiηj + ηjηi = 0 for i 6= jBy Lemma 210 this odd Dunkl operator may be expressed as

ηi = tδi + usumk 6=i

partiksik (26)

By analogy with the commutative case discussed in Section 21 the operator rik plays the same role inthe odd theory that the even divided difference operator plays in the even theory

3 Classical Yang-Baxter Equation and the Dunkl Laplacian

Theorem 31 LetH123 = [r12 r13]+ + [r13 r23]+ + [r12 r23]+ (31)

where [p q]+ = pq + qp is the anti-commutator Then the operators rik satisfy the classical Yang-Baxterequation

H123 = 0 (32)

Proof To avoid a cumbersome direct calculation we instead use an inductive approach Namely sup-pose that H123(f) = 0 for some function f isin Pminus Then we show that H123(xif) = 0 for all integersi ge 1 Note that for i = 1

r12r13x1 = r12(s13 minus x1r13)= r12s13 minus r12(x1r13)= r12s13 minus (s12r13 minus x1r12r13)= r12s13 minus s12r13 + x1r12r13

(33)

where we have used Equation 4 twice Similarly we find that

r13r12x1 = r13s12 minus s13r12 + x1r13r12 (34)r23r12x1 = r23s12 + x1r23r12 (35)r12r23x1 = minuss12r23 + x1r12r23 (36)r13r23x1 = minuss13r23 + x1r13r23 (37)r23r13x1 = r23s13 + x1r23r13 (38)

By our inductive hypothesis

x1(r12r13 + r13r12 + r23r12 + r12r23 + r13r23 + r23r13) = 0

Keeping this in mind add Equations 33-38 to show that

H123x1 = r12s13 minus s12r13 + r13s12 minus s13r12 + r23s12 minus s12r23 minus s13r23 + r23s13

= r12s13 minus r23s12 + r13s12 minus r23s13 + r23s12 minus r13s12 minus r12s13 + r23s13

= 0

where we have repeatedly used Lemma 26 to slide rij past sk`We can similarly show that H123x2 = H123x3 = 0 Since rjkxi = xirjk for i gt 3 and j k isin (1 2 3)

it also follows that H123xi = 0 for i gt 3 proving the desired result

9

Corollary 32 The double summationsumn

i=1

(sumk 6=i rik

)2= 0

Proof The expansion of this double summation has n(nminus1)2 total terms Since r2ij = 0 n(nminus1) of theseterms are immediately zero leaving n(n minus 1)(n minus 2) terms of the form rijrk` where i = k and j = `are not both true By Theorem 31 the sum of all six terms of the form rijrk` where i j k ` isin a b c fordistinct integers 1 le a b c le n i 6= j and k 6= ` is zero There are

(n3

)ways to choose integers a b c

and for each choice of a b c six terms of the form rijrk` vanish This fact eliminates all the remaining6(n3

)= n(nminus 1)(nminus 2) terms of the double summation

As an application of the results in this section we will compute the odd Dunkl Laplaciansumn

i=1 η2i

We will first require a lemma involving the commutator of τi and rik

Lemma 33 The equation xminus1i [rik τi] = (x2i minus x2k)minus1(sik(τi + τk)minus xminus1i xksik(τi minus τk)minus 2) holds

Proof By Lemma 210

rikτi = (x2i minus x2k)minus1((xi minus xk)sikτi minus xiτ2i + xkτkτi)

τirik = (x2i minus x2k)minus1(minus(xi + xk)sikτk + xiτ2i + xkτiτk)

Since τiτk = τkτi and τ2i = 1 the result follows by subtraction

Now define Ai = (2xi)minus1(1minus τi) and Bi =

sumk 6=i rik so that the odd Dunkl operator ηi of Khongsap

and Wang may be expressed as ηi = tAi + uBi Note that

A2i =

1

4xminus1i (1minus τi)xminus1i (1minus τi)

=1

2xminus2i (1minus τi)(1minus τi)

= xminus2i (1minus τi)

(39)

since τi(xminus1i ) = minusxminus1i and τ2i = 1

Lemma 34 The relationsumn

i=1(AiBi +BiAi) = 0 holds

Proof Due to the Leibniz Rule for rik (equation 4) we can find that 0 = rik(xixminus1i ) = xminus1k minus xirik(x

minus1i )

so rik(xminus1i ) = xminus1i xminus1k It follows that rik(xminus1i f) = xminus1i xminus1k sik(f)minus xminus1i rik(f) Using this factnsumi=1

BiAi =1

2

nsumi=1

sumk 6=i

rikxminus1i (1minus τi)

=1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi)minus xminus1i rik(1minus τi))

By definitionnsumi=1

AiBi =1

2

nsumi=1

sumk 6=i

xminus1i (1minus τi)rik

Adding the above two equations we find thatnsumi=1

[Ai Bi]+ =1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi) + xminus1i (rikτi minus τirik))

=1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi) + (x2i minus x2k)minus1(sik(τi + τk)minus xminus1i xksik(τi minus τk)minus 2))

(310)

10

where we have used Lemma 33Each double summation repeats the pair of indices (p q) twice when 1 le p q le n one time when

i = p and k = q and once more when i = q and k = p Note that

(x2i minus x2k)minus1sik(τi + τk) = minus(x2k minus x2i )ski(τk + τi)

As a result the sum 12

sumni=1

sumk 6=i(x

2i minus x2k)minus1sik(τi + τk) = 0 Similarly

1

2

nsumi=1

sumk 6=i

xminus1i xminus1k sik = 0 and1

2

nsumi=1

sumk 6=i

(x2i minus x2k)minus1(minus2) = 0

since xminus1i xminus1k + xminus1k xminus1i = 0 Equation 310 then becomes

nsumi=1

(AiBi +BiAi) =1

2

nsumi=1

sumk 6=i

(minusxminus1i xminus1k sikτi minus (x2i minus x2k)minus1(xminus1i xksik(τi minus τk)) (311)

However note thatxminus1i xk + xminus1k xi = minus(x2i minus x2k)xminus1i xminus1k

which implies

(x2i minus x2k)minus1xminus1i xksik(τi minus τk) + (x2k minus x2i )minus1xminus1k xiski(τk minus τi) = minusxminus1i xminus1k sik(τi minus τk) (312)

Similarly we find that

xminus1i xminus1k sikτi + xminus1k xminus1i sikτk = xminus1i xminus1k sik(τi minus τk) (313)

As a result equation 311 becomes

nsumi=1

(AiBi +BiAi) = minus1

2

sum1leiltklen

(xminus1i xminus1k sik(τi minus τk)minus xminus1i sminus1k sik(τi minus τk)) = 0

We are now equipped to compute the Dunkl Laplacian in the odd case

Theorem 35 The equationsumn

i=1 η2i = t2

sum1leilen x

minus2i (1minus τi) holds

Proof Since ηi = tAi + uBi we have that

nsumi=1

η2i = t2nsumi=1

A2i + tu

nsumi=1

(AiBi +BiAi) + u2nsumi=1

B2i

By Lemma 34sumn

i=1(AiBi +BiAi) = 0 By Corollary 32sumn

i=1B2i = 0 Therefore

nsumi=1

η2i = t2nsumi=1

A2i = t2

sum1leilen

xminus2i (1minus τi)

by equation 39

11

4 A Variant of the Khongsap-Wang Odd Dunkl Operator

In this section we will show that a close variant of the odd Dunkl operator introduced by Khongsapand Wang can be used in the construction of three operators that satisfy the defining relations of the Liealgebra sl2 First we will consider an operator pi which is different from δi but plays a similar role

Definition 41 The operator pi is a C-linear map Pminus rarr Pminus which acts on monomials as follows

pi(xλ11 xλii xλnn ) = λi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn

Remark 42 One may also introduce pi by using a nice Leibniz-like expression involving τi by defining

pi(xj) = δij

pi(fg) = pi(f)g + (minus1)|fτi(f)pi(g)

where f g isin Pminus and δij is the Kronecker delta Now note the analogous relationship between thedegree-preserving operators si and τi in their respective Leibniz rules for the (minus1)-degree operators partiand pi This provides motivation for the definition of pi and suggests its natural role in our theory

Now consider a modified version of ηi

Definition 43 LetDi = tpi + u

sumk 6=i

rik (41)

Definition 44 Introduce the odd r2 Euler and ∆ operators as below

r2 = (2t)minus1nsumi=1

x2i (42)

E =nsumi=1

xipi +n

2+u

t

sumk 6=i

sik (43)

∆ = minus(2t)minus1nsumi=1

D2i (44)

Remark 45 Heckman who used the even Dunkl operators to find a sl2-triple useful in harmonic anal-ysis uses the convention t = 1 [13] For now we will consider t to be a fixed constant in Ctimes

Remark 46 The commutator in the setting of superalgebras is usually defined as [a b] = abminus(minus1)|a||b|bawhere |a| and |b| are the degrees of a and b respectively However since all of the operators we will beconsidering in this section have even degree there is no need to distinguish between commutators andsuper-commutators

To construct an sl2 action from these operators we will require a series of lemmas regarding theaction of portions of the odd Euler operator E In the next lemma we investigate the action of the firstterm of the odd Euler operator on skew polynomials

Lemma 47 The operatorsumn

i=1 xipi acts by multiplication by |f | on the space of homogenous functionsf isinPminus

Proof It suffices to show the result for a monomial xλ = xλ11 xλii xλnn Note that

xipi(xλ) = λixi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn = λix

λ

12

By summing over all indices i we obtain that

nsumi=1

xipi(xλ) = (λ1 + λ2 + + λn)xλ

which implies the desired result

The above lemma holds true in the even case as well where pi is replaced by the partial derivativewith respect to xi We now prove some properties about the action of the third term of the odd Euleroperator on r2 and ∆

Lemma 48 The commutation relation[sum

k 6=i sik∆]

= 0 holds

Proof Note that sjkpi = pjsjk if i = k sjkpi = pksjk if i = j and sjkpi = pisjk otherwise Indeedthese relations can be verified by checking if they are true for xai x

bjxck a b c isin Z+ and then extending

by linearity We prove that sjkpi = pjsjk if i = k and the other two cases are similar Without loss ofgenerality let j lt k and observe that

sjkpk(xajx

bk) = b(minus1)asjk(x

ajx

bminus1k ) = b(minus1)axakx

bminus1j = b(minus1)abxbminus1j xak

pjsjk(xajx

bk) = (minus1)abpj(x

bjxak) = b(minus1)abxbminus1j xak

By our work in Lemma 26 one can deduce that sjkr`m = rsjk(`m)sij As a consequence we find thatsjkDi = Dsjk(i)sjk By an easy induction we now have that sjk∆ = ∆sjk Using the above equationmultiple times proves the desired result

Lemma 49 The commutation relation[sum

k 6=i sik r2]

= 0 holds

Proof Follows since sjkxj = xksjk sjkxk = xjsjk and sjkxi = xisjk if i 6= j k

We are now ready to obtain two commutativity relations involving the odd Euler operator E

Theorem 410 The odd Euler operator and r2 satisfy the following commutation relations

[E r2] = 2r2 (45)[E∆] = minus2∆ (46)

Proof Since r2 has degree 2 and ∆ has degreeminus2 the theorem follows from Lemmas 47 48 and 49

We also need to investigate what the third commutativity relation [r2∆] turns out to be We willprove one lemma before doing so

Lemma 411 For i = 1 to n the equation xiDi +Dixi = 2txipi + t+ usum

k 6=i sik holds

Proof Recall thatDi = tpi + u

sumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Therefore since pixi = xipi + 1

Dixi = txipi + t+ usumk 6=i

(x2i minus x2k)minus1(xixk minus x2k)sik +sumk 6=i

(x2i minus x2k)[x2i τi minus xixkτk]

xiDi = txipi + usumk 6=i

(x2i minus x2k)minus1(x2i minus xixk)sik +sumk 6=i

(x2i minus x2k)minus1[minusx2i τi + xixkτk]

Adding we obtain the desired result

13

We now have the tools to find the third relation between r2 E and ∆

Theorem 412 The commutation relation [r2∆] = E holds

Proof We will first find [r2 Di] The derivative pi much like the partial derivative in the even casesatisfies the properties pixj = minusxjpi for i 6= j and pixi = xipi + 1 Now suppose that i 6= j Then

Dix2j = tpix

2j + ux2j

sumk 6=i 6=j

rik + x2k(x2i minus x2k)minus1[minusxiτi + xkτk] + x2i (x

2i minus x2k)minus1[(xi minus xk)sik]

= tx2jpi + ux2jsumk 6=i

rik + (xi minus xj)sij

Now we will find Dix2i

Dix2i = tpix

2i +

sumk 6=i

x2k(x2i minus x2k)minus1(xi minus xk)sik + x2i

sumk 6=i

(x2i minus x2k)minus1[minusxiτi + xkτk]

= tx2i pi + 2txi + x2isumk 6=i

rik minussumk 6=i

(xi minus xk)sik

Therefore[sumn

i=1 x2i Di

]= minus2txi This implies that

r2Di minusDir2 = minusxi (47)

As a result we find that

[r2∆] = minus(2t)minus1nsumi=1

[r2 D2j ] = minus(2t)minus1

nsumi=1

(r2D2j minusD2

j r2)

= minus(2t)minus1nsumi=1

[(Djr2Dj minus xjDj)minus (Djr

2Dj +Djxj)]

= (2t)minus1nsumi=1

(xiDi +Dixi)

where we have used 47 Now by Lemma 411

[r2∆] =

nsumi=1

xipi +n

2+u

t

sumk 6=i

sik = E

as desired

To summarize we have found operators Er2 and ∆ similar to their even counterparts whichsatisfy the defining relations of the Lie algebra sl2

[E r2] = 2r2

[E∆] = minus2∆

[r2∆] = E

Remark 413 If one uses the odd Dunkl operator ηi as found in [18] instead of the Di introduced herethe r2 E and ∆ operators do not generate sl2

Remark 414 Although our results hold true for all t and u in C one typically sets t = 1 and u = αminus1

for some α isin Ctimes since without loss of generality one of t and u may equal 1

14

Remark 415 In the even case let X be a Euclidean vector space with dimension n and let C[X] be thealgebra of C-valued functions on X Then this result about sl2 plays a major role in the study of higherdifferential operators on C[X] This is because the representation theory of sl2 allows for the reductionof degree to the second order [13] As a result our results in this section should correspondingly have arole in further studying differential operators in the odd case

5 q-nilHecke Algebras

Until now we have been concerned with the odd symmetric polynomials in variables x1 x2 xnwhere xixj = (minus1)xjxi for 1 le i 6= j le n This immediately suggests the question what if one replacesthe minus1 by any constant q isin Ctimes Specifically we ask the following questions

Question 51 Is it possible to study q-symmetric polynomials for which xixj = qxjxi when i gt j

Question 52 Are there q-analogs of evenodd divided difference operators and nilHecke algebras Sofar such structures are known only for the even case (q = 1) and the odd case (q = minus1)

In this section we answer both questions in the affirmativeWe work in the Z-graded q-braided setting throughout LetC be a commutative ring and let q isin Ctimes

be a unit If VW are graded C-modules and v isin V w isin W are homogeneous the braiding is theldquoq-twistrdquo

τq V otimesW rarrW otimes Vv otimes w 7rarr q|v||w|w otimes v

(51)

where | middot | is the degree function By q-algebra we mean an algebra object in the category of gradedC-modules equipped with this braided monoidal structure likewise for q-bialgebras q-Hopf algebrasand so forth

Remark 53 Note that the q-twist described above is its own inverse only when q2 = 1 which correlatesto the even and odd cases When q2 6= 1 the corresponding theory becomes more complex There-fore the q-nilHecke algebras that we introduce later in this section are nontrivial generalizations of thepreviously studied even and odd nilHecke algebras

Definition 54 The q-algebra P qn is defined to be

P qn = C〈x1 xn〉(xjxi minus qxixj = 0 if i lt j) (52)

where |xi| = 1 for i = 1 n

Note that P qn sim= otimesni=1Pq1 There are two interesting subalgebras of P qn that can be thought of as q-

analogs of the symmetric polynomials Define the k-th elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middotxin

and define the k-th twisted elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middot xin

where xj = qjminus1xj

Definition 55 The q-algebra of q-symmetric polynomials in n variables denoted Λqn is the subalgebra ofP qn generated by e1 en Likewise for the twisted q-symmetric polynomials Λqn and e1 en

The type A braid group on n strands acts on P qn by setting

15

1 σi(xj) = qxi+1 if j = i

2 σi(xj) = qminus1xi if j = i+ 1

3 σi(xj) = qxj if j gt i+ 1

4 σi(xj) = qminus1xj if j lt i

and extending multiplicatively

Definition 56 For i = 1 nminus 1 the i-th q-divided difference operator parti is the linear operator P qn rarr P qndefined by parti(xi) = q parti(xi+1) = minus1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + σi(f)parti(g) (53)

for all functions f g isin P qn We call Equation 53 the q-Leibniz rule

Lemma 57 For every i and every j lt k parti(xkxj minus qxjxk) = 0

Proof Since parti(xj) = 0 for j gt i+1 one may reduce the lemma to having to prove that part1(x2x1minusqx1x2) =0 part1(x3x1 minus qx1x3) = 0 and part1(x3x2 minus qx2x3) = 0 These statements follow from the q-Leibniz rule

Therefore parti is a well-defined operator on P qn

Lemma 58 The following relations hold

parti(xki ) =

kminus1sumj=0

qjkminus2jminusj2+kxjix

kminus1minusji+1

parti(xki+1) = minus

kminus1sumj=0

qminusjxjixkminus1minusji+1

Proof We induct on k The base case (k = 1) follows from the definition of the parti and the powers of qarise mostly from xni+1x

mi = qmnxmi x

ni+1 for all mn isin Z+

Our q-divided difference operators also annihilate the twisted elementary q-symmetric polynomialsjust as the even divided difference operators annihilate the elementary symmetric functions

Lemma 59 For every i = 1 nminus 1 and every k parti(ek) = 0 Hence Λqn sube⋂nminus1i=1 ker(parti)

Proof We can express ek as

ek =sum|J|=kii+1isinJ

xJ +sum|J|=kminus1ii+1isinJ

qf(Jik)xJ(xi + qxi+1) +sum|J|=kminus2ii+1isinJ

qg(Jik)xJxixi+1

for certain Z-valued functions f g The result then follows from parti(xi + qxi+1) = parti(xixi+1) = 0 and theq-Leibniz rule

Having discussed q-divided difference operators we can now construct algebras for every q 6=0 1minus1 that have many similarities to the even and odd nilHecke algebras For every such q we de-fine a q-nilHecke algebra generated by xi and parti for 1 le i le n subject to the relations found in thefollowing two lemmas (510 and 511)

Lemma 510 The following relations hold among the operators parti and xi (left multiplication by xi)

1 part2i = 0

2 partjparti minus qpartipartj = 0 for j gt i+ 1

3 xjxi = qxixj for i lt j

4 partixj minus qxjparti = 0 for j gt i+ 1

5 qpartixj minus xjparti = 0 for j lt i

6 partixi minus qxi+1parti = q

7 xiparti minus qpartixi+1 = q

16

Proof To show that part2i = 0 note that we can reduce to i = 1 and proceed by induction Since parti(1) = 0the base case follows Suppose that part2i (f) = 0 Then note that

part21(x1f) = part1(qf + qx2part1(f)) = qpart1(f)minus qpart1(f) + x1part21(f) = 0

part21(x2f) = part1(minusf + qminus1x1part1(f)) = minuspart1(f) + part1(f) + x2part21(f) = 0

part21(x3f) = part1(qx3part1(f)) = q2x3part21(f) = 0

which completes the proof of the first statement in the lemmaStatement 3 follows by definition Statements 4 5 6 and 7 follow from a suitable application of the

q-Leibniz rule Statement 2 follows from an inductive argument We can reduce to i = 1 and j = 3Suppose that partjparti = qpartipartj if j gt i+ 1 Then

part3part1(x1f)minus qpart1part3(x1f) = (qpart3(f) + x2part3part1(f))minus q(part3(f) + x2part1part3(f)) = 0

part3part1(x2f)minus qpart1part3(x2f) = (minuspart3(f) + qminus2x1part3part1(f))minus q(minusqminus1part3(f) + qminus2x1part1part3(f)) = 0

part3part1(x3f)minus qpart1part3(x3f) = (q2part1(f) + q2x4part3part1(f))minus q(qpart1(f) + q2x4part1part3(f)) = 0

part3part1(x4f)minus qpart1part3(x4f) = (minusqpart1(f) + x3part3part1(f))minus q(minuspart1(f) + x3part1part3(f)) = 0

part3part1(x5f)minus qpart1part3(x5f) = q2x5part3part1(f)minus q(q2x5part1part3(f)) = 0

thereby completing the induction

Lemma 511 partiparti+1partiparti+1partiparti+1 + parti+1partiparti+1partiparti+1parti = 0

Proof This result follows from an inductive argument we reduce to i = 1 and assume that the braidrelation holds true for some function f Then we check that the braid relation is true for x1f x2f x3f and x4f (since the behavior of xjf for j ge 4 is the same as that of x4f ) For brevity we will show theargument for x2f only

part1part2(x2f) = qpart1(f) + q2x3part1part2(f) part2part1(x2f) = minuspart2(f) + qminus2x1part2part1(f)

part212(x2f) = qpart2part1(f)minus q2part1part2(f) + qx2part212(f) qpart121(x2f) = minusqpart1part2(f) + part2part1(f) + part121(f)

part1212(x2f) = qpart121(f)minus qpart212(f) + x1part1212(f) part2121(x2f) = minuspart212(f) + part121(f) + x3part2121(f)

part21212(x2f) = qpart2121(f) + qminus1x2part21212(f) part12121(x2f) = minuspart1212(f) + qx3part12121(f)

We continue the above calculations to find that

part121212(x2f) = qpart12121(f) + part21212(f) + x2part121212(f)

part212121(x2f) = minusqpart12121(f)minus part21212(f) + x2part121212(f)

and the braid relation for x2f follows from the inductive hypothesis

6 A Diagrammatic Approach to q-Symmetric Polynomials

61 Introduction to a q-Bialgebra

In the previous section we answered Question 51 in an algebraic way by defining q-analogs of theclassical elementary and complete symmetric functions In this section we generalize the diagrammaticmethod used in [8] in order to study this question from the perspective of bialgebras

Let NΛq be a free associative Z-graded C-algebra with generators hm for m ge 0 We define h0 = 1and hm = 0 for m lt 0 and let q isin Ctimes The homogenous part of NΛq of degree ` has a basis hααkwhere

hα = hα1 middot middot middothαz for a composition α = (α1 αz) of `

17

Define a multiplication for homogenous x and y on NΛqotimes2 as follows where deg(x) denotes the degreeof x

(w otimes x)(y otimes z) = qdeg(x)deg(y)(wy otimes xz)

We can make NΛq into a q-bialgebra by letting the comultiplication on generators be

∆(hn) =

nsumk=0

hk otimes hnminusk

and by letting the counit be ε(x) = 0 if x is homogenous and deg(x)gt 0We can impose through the braiding structure that

∆(hahb) =asumj=0

bsumk=0

(hj otimes haminusj)(hk otimes hbminusk) =asumj=0

bsumk=0

qk(aminusj)(hjhk otimes haminusjhbminusk)

For any partitions λ and micro of n consider the set of double cosets of subroups Sλ and Smicro of Sn SλSnSmicroFor every C in this set let wC be the minimal length representative of C and let `(wC) be the length ofthis minimal length representative We will now attribute a bilinear form to NΛq

(hλ hmicro) =sum

CisinSλSnSmicro

q`(wC)

This bilinear form admits a diagrammatic description Let hn be an orange platform with n non-intersecting strands coming out of it When computing (hλ hmicro) with `(λ) = z and `(micro) = y drawz orange platforms at the top of the diagram representing λ1 λ2middot middot middot λz Draw y orange platforms atthe bottom of the diagram representative of micro1 micro2middot middot middot microy We require that |λ| = |micro| so that the topplatforms and bottom platforms have the same number of strands

Consider the example (h121 h22) In the following diagram snippets of the strands from each plat-form are shown

Every strand must start at one platform at the top and end on another platform at the bottom Nostrands that have originated from one platform may intersect The strands themselves have no criticalpoints with respect to the height function no two strands ever intersect more than once and there are notriple-intersections where three strands are concurrent Diagrams are considered up to isotopy Withoutany restrictions there would be n such diagrams if |λ| = n since there would be no limitations onthe ordering of the strands However due to the above rules there are only 4 possible diagrams in thecomputation of (h121 h22) shown below

Define(hλ hmicro) =

sumall diagrams D representing (hλhmicro)

q number of of crossings in D (61)

In the above example (h121 h22) = 1 + 2q2 + q3

18

We can extend the bilinear form toNΛqotimes2 by stating that any diagram in which strands from distincttensor factors intersect contributes 0 to the bilinear form

(w otimes x y otimes z) = (w y)(x z)

Let I be the radical of the bilinear form inNΛq In [8] the authors prove for any q that multiplicationand comultiplication are adjoint In other words for all xy1 y2 in NΛq

(y1 otimes y2∆(x)) = (y1y2 x) (62)

62 The Elementary q-Symmetric Functions

We now use the bilinear form of q-symmetric functions to study one of their important bases the ele-mentary q-symmetric functions

Define elements ek isin NΛq by ek = 0 for k lt 0 e0 = 1 and

ksumi=0

(minus1)iq(i2)eihkminusi = 0 for k ge 1 (63)

Equivalently leten = qminus(n2)

sumαn

(minus1)`(α)minusnhα (64)

Lemma 61

1 The coproduct of an elementary function is given by ∆(en) =

nsumk=0

ek otimes enminusk

2 If λ n then (hλ en) =

1 if λ = (1 1)

0 otherwise

Proof We begin by demonstrating (2) from which (1) will follow To show (2) it suffices to show that

(hmx en) =

(x enminus1) if m = 10 otherwise

We will utilize strong induction on n in order to find (hmx ekhnminusk) The base cases n = 0 1 are easyto show There are two cases to consider by the inductive hypothesis applied to k lt n Either there isa strand connecting hm and ek or there is not Just as we used an orange platform to denote hn wewill use a blue platform to denote ek The rules of the diagrammatic notation are the same for the blueplatforms as they are for the orange platforms

k

m

nminus k minusm

k nminus k

mlowast lowast lowast lowast lowast

x

If there is not a strand connecting hm and ek the configuration contributes qkm(x ekhnminuskminusm)

19

k minus 1

mminus 1

nminus k minusm + 1

k nminus k

mlowast lowast lowast lowast lowast

x

If a stand connects hm and ek this configuration contributes q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) We have thusshown that (hmx ekhnminusk) = qkm(x ekhnminuskminusm) + q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) Now we are equippedto consider (hmx ek)

(minus1)n+1q(n2)(hmx en) =

nminus1sumk=0

(minus1)kq(k2)(hmx ekhnminusk)

=nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus1sumk=0

(minus1)kq(k2)+(mminus1)(kminus1)(x ekminus1hnminuskminusm+1)

=

nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus2sumk=0

(minus1)k+1q(k+12 )+(mminus1)(k)(x ekhnminuskminusm)

= (minus1)nminus1q(nminus12 )+nm(x enminus1h1minusm)

Corresponding terms from the two sums cancel in pairs since q(k2)+km = q(

k+12 )+k(mminus1) leaving only the

k = nminus 1 term in the first sum The second statement of the lemma thus followsWe will now use (2) to prove (1) This follows from equation 62

(∆(ek) hλ otimes hmicro) = (ek hλhmicro) =

1 λ = (1`) micro = (1p) `+ p = k

0 otherwise

We now calculate the sign incurred when strands connect two blue (ek) platforms

(minus1)n+1q(n2)(en en) =

nminus1sumk=0

(minus1)kq(k2)(en ekhnminusk)

= (minus1)nminus1q(nminus12 )(en enminus1h1)

= (minus1)nminus1q(nminus12 )(∆(en) enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )

nsumk=0

(ek otimes enminusk enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )(enminus1 enminus1)

One may solve this recursion to find that (en en) = qminus(n2)Here the second equality follows from not-ing that at most one strand can connect hnminusk and en (so that k = n minus 1) the third equality followsfrom adjointness and the fourth and fifth equalities follow from the diagrammatic considerations of theprevious lemma

To summarize the diagrammatics of the bilinear form thus developed

20

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

Hecke algebra involves the categorification of quantum groups Categorification introduced by Craneand Frenkel is generally the process of replacing algebras and representations by categories and highercategories in order to make quantum 3-manifold invariants into 4-manifold invariants [4] In physicscategorification corresponds to increasing dimensions which allows one to understand symmetries inlower dimensions and then use categorification to better understand higher dimensions In mathemat-ics categorified quantum groups give a higher representation theoretic construction of link homologieswhich in turn categorify quantum link polynomials

The original example of link homology is Khovanov homology a bigraded abelian group which cat-egorifies the well-known Jones polynomial It has major applications in studying knot polynomialsquantum field theory and classical gauge theory [19 28] Since the quantum group Uq(sl2) plays a rolein understanding the Jones polynomial a categorification of Uq(sl2) would be useful in better under-standing Khovanov homology This precise categorification is achieved through the ldquoevenrdquo nilHeckealgebra NHn described in subsection 11

Recently Ozsvath Rasmussen and Szabo found an odd analog of Khovanov homology [22] Theirodd Khovanov homology also categorifies the Jones polynomial and agrees modulo 2 with Khovanovhomology However both theories can detect knots that the other theory cannot [26] The subject ofodd Khovanov homology has yet to be fully understood despite its crucial connections with Khovanovhomology

Knowing this Ellis Lauda and Khovanov developed the odd nilHecke algebra to provide an oddcategorification of Uq(sl2) and give a construction of odd Khovanov homology from a representationtheoretic standpoint [9 10] In addition to being useful in the categorification of quantum groups theodd nilHecke algebra is also related to Hecke-Clifford superalgebras [16 17] and has been used to con-struct odd analogs of the cohomology groups of Springer varieties [21]

The below diagram summarizes the categorifications that motivate the present work NH standsfor nilHecke Cat stands for categorification and KH stands for Khovanov homology

Jones Polynomial Uq(sl2)Uq(sl2)

KH Odd KHCat of Uq(sl2) Odd Cat of Uq(sl2)

odd NH algebraNH algebra

ldquocategorifiesrdquoldquohelps to explainrdquo

ldquoequivalent modulo 2rdquo

Figure 2 Odd Khovanov Homology and Categorification

13 Outline of the Present Paper

The main goal of the present paper is to make progress towards giving a representation theoretic con-struction of odd Khovanov homology Despite being relatively new odd Khovanov homology seemsto have great importance in knot theory It has connections to Heegaard-Floer homology and yieldsstronger results than Khovanov homology in bounding the Thurston-Bennequin number and detectingquasi-alternating knots [26] Odd Khovanov homology is also related to signed hyperplane arrange-ments which have many implications in graph theory and topology [5]

We study odd Khovanov homology by looking for new representation theoretic structures that arise

3

from identifying ldquooddrdquo analogs of structures that play important algebraic roles in the even case EllisKhovanov and Lauda started this program of ldquooddificationrdquo by finding an odd analog of NHn andusing it to categorify Uq(sl2) Searching for the geometry underlying these odd constructions would alsoprovide a very new approach to noncommutative geometry For example we study potential generatorsof certain Cherednik algebras Since spherical rational Cherednik algebras fit nicely into a family ofalgebras from the geometry of symplectic resolutions including Websterrsquos tensor product algebras andcyclotomic KLR algebras this project may be used in the context of ldquooddrdquo noncommutative geometry

In Subsection 11 we discussed the (even) divided difference operators Dunkl operators and Chered-nik operators Although analogs of Dunkl operators have been found in the odd case they have not beenwell-studied Odd Cherednik operators have not even been defined

As a result the first goal of the present paper is to unify certain results in the odd case and to furtherstudy the odd Dunkl operators of Khongsap and Wang In Section 2 we introduce an operator rikrelated to the generalized odd divided difference operator partodd

ik and study its properties One of ourmain results (Equation 26) is that the odd Dunkl operator ηi may be expressed in terms of the odddivided difference operators of Ellis Khovanov and Lauda

ηoddi = tδi + u

sumk 6=i

partoddik sik (11)

This result connects odd Dunkl operators and the odd nilHecke algebra both of which play importantroles in the project of oddification

In the even case one can introduce an operator known as the Dunkl Laplacian given bysumn

i=1 η2i This

operator has important applications in spherical harmonics and heat semigroups [24] Our next goal inthe present paper is to express the Dunkl Laplacian in the odd case In Section 3 we show that therik satisfy the classical Yang-Baxter equation and use this result to evaluate the odd Dunkl LaplacianSpecifically we show that

nsumi=1

η2i = t2sum

1leilenxminus2i (1minus τi)

In Section 4 we find an odd analog of Heckmanrsquos important sl2-triple in [13] by showing that avariant Di of the odd Dunkl operator can be used to construct three operators r2 E and ∆ that satisfythe defining relations of the Lie algebra sl2

r2 = (2t)minus1nsumi=1

x2i

E =nsumi=1

xipi +n

2+u

t

sumk 6=i

sik

∆ = minus(2t)minus1nsumi=1

D2i

Since even Dunkl operators play an important role in the representation theory of symmetric groupsour study of odd Dunkl operators should result in a better understanding of the representation theoryof odd symmetric functions which correspondingly results in a better understanding of odd Khovanovhomology

The second goal of this paper is to study a generalization of the odd symmetric functions knownas q-symmetric functions for which xjxi = qxixj when j gt i Previous authors have described a nil-Hecke algebra structure only for the odd case q = minus1 and the even case q = 1 [9 15] In Section 5we find a q-divided difference operator for all previously unstudied values of q and explore its propertiesWe show for example that twisted elementary symmetric polynomials are in the kernel of q-divided

4

difference operators just as odd elementary symmetric functions are in the kernel of odd divided differ-ence operators We then use q-divided difference operators to construct algebras acting on q-symmetricpolynomials that have many similarities to the even and odd nilHecke algebras We call these q-nilHeckealgebras These algebras are nontrivial generalizations of the even and odd nilHecke algebras becausethe q-twist map introduced in Section 5 is not its own inverse when q2 6= 1 In Section 6 we present theelementary q-symmetric polynomials using a generalization of a clever diagrammatic method arisingin the context of bialgebras We use these diagrams to study relations between elementary q-symmetricpolynomials when q is a root of unity These methods can be combined with the algebras of Section 5 inorder to continue studying q-symmetric polynomials including q-Schur and q-monomial functions

In the conclusion we also define the odd Cherednik operators and outline a procedure for findingand studying odd analogs of Jack polynomials This makes progress towards answering a question ofEllis about the existence of Macdonald-like polynomials in the odd case Since the Jack polynomialshave importance in representation theory their study would enhance our knowledge about the oddalgebraic theory

2 Odd Dunkl operators and the Odd nilHecke algebra

21 Preliminaries Even Dunkl Operators

In the even case we work with the ring C[x1 xn] and a root system of type An where xixj = xjxifor all 1 le i j le n and α isin C We first introduce some notation involving the symmetric group

1 Let sik be the simple transposition in Sn swapping xi and xk We let si = sii+1

2 Let sij(k `) be the result of applying sij to the pair (k `) Similarly define sij(k)

In [7] Dunkl introduced the remarkable operator

ηeveni =

part

partxi+ α

sumk 6=i

partevenik

where partpartxi

is the partial derivative with respect to xi and partevenik is the even divided difference operator

partevenik = (xi minus xk)minus1(1minus sik)

Since xi minus xk always divides f minus sik(f) for f isin C[x1 xn] partik sends polynomials to polynomialsThese Dunkl operators have various important properties one of which is that they commute (ηiηj =

ηjηi) In [18] Khongsap and Wang introduced anti-commuting odd Dunkl operators on skew polynomi-als In Section 2 we will develop the connection between these operators and the odd nilHecke algebraintroduced in [9]

Returning to the even case introduce operators r2 E (the Euler operator) and ∆k

r2 =1

2

nsumi=1

x2i

E =

nsumi=1

xipart

partxi+micro

2

∆ =1

2

nsumi=1

η2i

where micro is the Dunkl dimension which is defined by the relation ∆|x|2 = 2micro as in [6]

5

Let [p q] = pq minus qp be the commutator Heckman showed that r2 E and ∆k satisfy the definingrelations of the Lie algebra sl2 [13]

[E r2] = 2r2

[E∆k] = minus2∆k

[r2∆k] = E

Remark 21 If one were to replace ∆k with the classical Laplacian on flat Rn (replacing the Dunkloperator with the partial derivative) these three operators still satisfy the sl2 relations

Remark 22 From now on we will use ηi to denote the odd Dunkl operator of Khongsap and Wangdefined in equation 25

In Section 4 we will focus on finding analogous results in the odd case

22 Introduction to the Odd nilHecke Algebra

We will now discuss operators with the algebra Pminus = C〈x1 xn〉〈xjxi + xixj = 0 for i 6= j〉 We callPminus the skew polynomial ring We can define linear operators called the odd divided difference operatorsas below

Definition 23 For i = 1 nminus1 the i-th odd divided difference operator parti is the linear operator Pminus rarr Pminus

defined by parti(xi) = 1 parti(xi+1) = 1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + (minus1)|f |si(f)parti(g)

for all functions f g isin Pminus We call this last relation the Leibniz rule

It is shown in [9] that the odd divided difference operators can be used to construct an odd nilHeckealgebra generated by xi and parti for 1 le i le n subject to the following relations

1 part2i = 0

2 partipartj + partjparti = 0 for |iminus j| ge 2

3 partiparti+1parti = parti+1partiparti+1

4 xixj + xjxi = 0 for i 6= j

5 xiparti + partixi+1 = 1 partixi + xi+1parti = 1

6 xipartj + partjxi = 0 for i 6= j j + 1

Due to [14] we have the following explicit definition of the odd divided difference operator

parti(f) = (x2i+1 minus x2i )minus1[(xi+1 minus xi)f minus (minus1)|f |si(f)(xi+1 minus xi)] (21)

Although this formula a priori involves denominators it does take skew polynomials to skew polyno-mials We extend this definition to non-consecutive indices by replacing i + 1 with any index k 6= i for1 le k le n and by replacing si with sik Equation 21 then becomes

partik(f) = (x2k minus x2i )minus1[(xk minus xi)f minus (minus1)|f |sik(f)(xk minus xi)] (22)

This extended odd divided difference operator satisfies the Leibniz rule partik(fg) = partik(f)g+(minus1)|f |sikpartik(g)[9]

6

23 Some Operations on Skew Polynomials

First we introduce a common operator in the study of Dunkl operators

Definition 24 Let the (minus1)-shift operator τi be the automorphism of Pminus which sends xi to minusxi and xj toxj for j 6= i

Suppose 1 le i 6= j le n and 1 le k 6= ` le n where f is an element in C〈x1 xn〉〈xixj + xjxi =0 for i 6= j〉 Then one has

sijτk` = τsij(k`)sij

fxi = (minus1)|f |xiτi(f)

Remark 25 Since skew polynomials are not super-commutative we cannot say that fg = (minus1)|f ||g|gf But the operator τi allows us to track the discrepancy from super-commutativity since xif = (minus1)|f |τi(f)ximaking it useful in this context

We now introduce the operator rik = partiksik for k 6= i which will serve as another odd divideddifference operator that we will use to study odd Dunkl operators For simplicity let ri = rii+1 In thefollowing lemma we study the action of the transposition and (minus1)-shift operator on rik

Lemma 26 The operators sik and τi act on rik as follows

rijsk` = sk`rsk`(ij) (23)

We also have that

1 sirik = ri+1ksi if k 6= i+ 1

2 siri = risi

3 siri+1 = rii+2si

4 si+1ri = rii+2si+1

5 sirj = rjsi for |iminus j| ge 2

6 τirj = rjτi for |iminus j| ge 2

Proof Recall the following relationship between partij and sk` for i 6= j and k 6= ` from Lemma 219 (1) of[9]

partijsk` = sk`partsk`(ij) (24)

Multiplying both sides by sij we obtain that

sijpartijsk` = sijsk`partsk`(ij)

= sk`sk`(i j)partsk`(ij)

which implies the desired result since rij = partijsij = sijpartij Properties 1-5 are special cases of equation23 Property 6 follows from τisj = sjτi and the fact that τi(xj) = xj for i 6= j

Remark 27 Differences between our formulas and those of [9] are due to a difference of sign conventionin the action of sij on Pminus

We now show that the properties of the rik are similar to those of the odd divided difference operatorpartik

Lemma 28 The following relations hold

7

1 r2i = 0

2 rirj + rjri = 0 for |iminus j| ge 2

3 riri+1ri = ri+1riri+1

4 rik(fg) = rik(f)sik(g) + (minus1)|f |frik(g)

5 rixi+1 + xi+1ri = rixi + xiri = si

6 rjxi + xjri = 0 for i 6= j j + 1

Proof Since siri = risi and ri = partisi it follows that siparti = partisi Then since part2i = 0 r2i = 0 as well Dueto Equation 5 from Lemma 26 we have that sirj = rjsi for |i minus j| ge 2 so sipartj = partjsi Thus ri andrj anti-commute since partipartj + partjparti = 0 The operators ri also satisfy braid relations which we show byinductively reducing to i = 1 and then using 24 and siparti = partisi repeatedly

r1r2r1 = s1part12part2s1part1 = s1s2part13s1part13part1 = s1s2s1part23part13part12

r2r1r2 = s2part2s1part1s2part2 = s2s1part13s2part13part2 = s2s1s2part12part13part23

Since s1s2s1 = s2s1s2 and part23part13part12 = part12part13part23 by symmetry we conclude that r1r2r1 = r2r1r2 TheLeibniz rule for rik (equation 4 of this lemma) follows immediately from the Leibniz rule for partik Sinceri(xi) = ri(xi+1) = 1 and ri(xj) = 0 for j 6= i i + 1 equations 5 and 6 follow from the Leibniz rule forrik

We also desire an explicit definition of the rik analogous to that of the odd divided difference oper-ator of [9] To find such an expression we use a preparatory lemma

Lemma 29 For all f isinPminus and 1 le i 6= k le n we have

sikxiτi(f)minus sikxkτk(f) = (minus1)|f |sik(f)(xi minus xk)

Proof It suffices to prove the result for a monomial xλ = xλ11 xλii xλkk xλnn where i lt k Wecalculate that

sikxiτi(xλ) = (minus1)λ1++λixλ11 xλi+1

k xλki xλnn

sikxkτk(xλ) = (minus1)λ1++λkminus1xλ11 xλik xλk+1

i xλnn

sik(xλ)xi = (minus1)λk+1++λnxλ11 xλik x

λk+1

i xλnn

sik(xλ)xk = (minus1)λi+1++λnxλ11 x

λi+1

k xλki xλnn

Since |f | = λ1 + + λn the desired result follows

Lemma 210 The operator rik has explicit form rik = (x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Proof Follows from Lemma 29 and Equation 22

We will now connect the above results to the odd Dunkl operator introduced by Khongsap and Wangin [18]

Definition 211 Define an operator δi by δi = (2xi)minus1(1minus τi)

The above super-derivative can also be defined inductively by imposing that δi(xj) = 1 if i = j and0 otherwise We then extend the action to monomials as follows

δi(xa1xa2 xa`) =sumk=1

(minus1)kminus1xa1 δi(xak)xak+1 xa`

8

The operator δi is a priori from Laurent skew polynomials to Laurent skew polynomials but it is easy tocheck that it preserves the subalgebra of skew polynomials Khongsap and Wang found an odd analogof the Dunkl operator given by

ηi = tδi + usumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk] (25)

where t u isin Ctimes Their operators anti-commute ηiηj + ηjηi = 0 for i 6= jBy Lemma 210 this odd Dunkl operator may be expressed as

ηi = tδi + usumk 6=i

partiksik (26)

By analogy with the commutative case discussed in Section 21 the operator rik plays the same role inthe odd theory that the even divided difference operator plays in the even theory

3 Classical Yang-Baxter Equation and the Dunkl Laplacian

Theorem 31 LetH123 = [r12 r13]+ + [r13 r23]+ + [r12 r23]+ (31)

where [p q]+ = pq + qp is the anti-commutator Then the operators rik satisfy the classical Yang-Baxterequation

H123 = 0 (32)

Proof To avoid a cumbersome direct calculation we instead use an inductive approach Namely sup-pose that H123(f) = 0 for some function f isin Pminus Then we show that H123(xif) = 0 for all integersi ge 1 Note that for i = 1

r12r13x1 = r12(s13 minus x1r13)= r12s13 minus r12(x1r13)= r12s13 minus (s12r13 minus x1r12r13)= r12s13 minus s12r13 + x1r12r13

(33)

where we have used Equation 4 twice Similarly we find that

r13r12x1 = r13s12 minus s13r12 + x1r13r12 (34)r23r12x1 = r23s12 + x1r23r12 (35)r12r23x1 = minuss12r23 + x1r12r23 (36)r13r23x1 = minuss13r23 + x1r13r23 (37)r23r13x1 = r23s13 + x1r23r13 (38)

By our inductive hypothesis

x1(r12r13 + r13r12 + r23r12 + r12r23 + r13r23 + r23r13) = 0

Keeping this in mind add Equations 33-38 to show that

H123x1 = r12s13 minus s12r13 + r13s12 minus s13r12 + r23s12 minus s12r23 minus s13r23 + r23s13

= r12s13 minus r23s12 + r13s12 minus r23s13 + r23s12 minus r13s12 minus r12s13 + r23s13

= 0

where we have repeatedly used Lemma 26 to slide rij past sk`We can similarly show that H123x2 = H123x3 = 0 Since rjkxi = xirjk for i gt 3 and j k isin (1 2 3)

it also follows that H123xi = 0 for i gt 3 proving the desired result

9

Corollary 32 The double summationsumn

i=1

(sumk 6=i rik

)2= 0

Proof The expansion of this double summation has n(nminus1)2 total terms Since r2ij = 0 n(nminus1) of theseterms are immediately zero leaving n(n minus 1)(n minus 2) terms of the form rijrk` where i = k and j = `are not both true By Theorem 31 the sum of all six terms of the form rijrk` where i j k ` isin a b c fordistinct integers 1 le a b c le n i 6= j and k 6= ` is zero There are

(n3

)ways to choose integers a b c

and for each choice of a b c six terms of the form rijrk` vanish This fact eliminates all the remaining6(n3

)= n(nminus 1)(nminus 2) terms of the double summation

As an application of the results in this section we will compute the odd Dunkl Laplaciansumn

i=1 η2i

We will first require a lemma involving the commutator of τi and rik

Lemma 33 The equation xminus1i [rik τi] = (x2i minus x2k)minus1(sik(τi + τk)minus xminus1i xksik(τi minus τk)minus 2) holds

Proof By Lemma 210

rikτi = (x2i minus x2k)minus1((xi minus xk)sikτi minus xiτ2i + xkτkτi)

τirik = (x2i minus x2k)minus1(minus(xi + xk)sikτk + xiτ2i + xkτiτk)

Since τiτk = τkτi and τ2i = 1 the result follows by subtraction

Now define Ai = (2xi)minus1(1minus τi) and Bi =

sumk 6=i rik so that the odd Dunkl operator ηi of Khongsap

and Wang may be expressed as ηi = tAi + uBi Note that

A2i =

1

4xminus1i (1minus τi)xminus1i (1minus τi)

=1

2xminus2i (1minus τi)(1minus τi)

= xminus2i (1minus τi)

(39)

since τi(xminus1i ) = minusxminus1i and τ2i = 1

Lemma 34 The relationsumn

i=1(AiBi +BiAi) = 0 holds

Proof Due to the Leibniz Rule for rik (equation 4) we can find that 0 = rik(xixminus1i ) = xminus1k minus xirik(x

minus1i )

so rik(xminus1i ) = xminus1i xminus1k It follows that rik(xminus1i f) = xminus1i xminus1k sik(f)minus xminus1i rik(f) Using this factnsumi=1

BiAi =1

2

nsumi=1

sumk 6=i

rikxminus1i (1minus τi)

=1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi)minus xminus1i rik(1minus τi))

By definitionnsumi=1

AiBi =1

2

nsumi=1

sumk 6=i

xminus1i (1minus τi)rik

Adding the above two equations we find thatnsumi=1

[Ai Bi]+ =1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi) + xminus1i (rikτi minus τirik))

=1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi) + (x2i minus x2k)minus1(sik(τi + τk)minus xminus1i xksik(τi minus τk)minus 2))

(310)

10

where we have used Lemma 33Each double summation repeats the pair of indices (p q) twice when 1 le p q le n one time when

i = p and k = q and once more when i = q and k = p Note that

(x2i minus x2k)minus1sik(τi + τk) = minus(x2k minus x2i )ski(τk + τi)

As a result the sum 12

sumni=1

sumk 6=i(x

2i minus x2k)minus1sik(τi + τk) = 0 Similarly

1

2

nsumi=1

sumk 6=i

xminus1i xminus1k sik = 0 and1

2

nsumi=1

sumk 6=i

(x2i minus x2k)minus1(minus2) = 0

since xminus1i xminus1k + xminus1k xminus1i = 0 Equation 310 then becomes

nsumi=1

(AiBi +BiAi) =1

2

nsumi=1

sumk 6=i

(minusxminus1i xminus1k sikτi minus (x2i minus x2k)minus1(xminus1i xksik(τi minus τk)) (311)

However note thatxminus1i xk + xminus1k xi = minus(x2i minus x2k)xminus1i xminus1k

which implies

(x2i minus x2k)minus1xminus1i xksik(τi minus τk) + (x2k minus x2i )minus1xminus1k xiski(τk minus τi) = minusxminus1i xminus1k sik(τi minus τk) (312)

Similarly we find that

xminus1i xminus1k sikτi + xminus1k xminus1i sikτk = xminus1i xminus1k sik(τi minus τk) (313)

As a result equation 311 becomes

nsumi=1

(AiBi +BiAi) = minus1

2

sum1leiltklen

(xminus1i xminus1k sik(τi minus τk)minus xminus1i sminus1k sik(τi minus τk)) = 0

We are now equipped to compute the Dunkl Laplacian in the odd case

Theorem 35 The equationsumn

i=1 η2i = t2

sum1leilen x

minus2i (1minus τi) holds

Proof Since ηi = tAi + uBi we have that

nsumi=1

η2i = t2nsumi=1

A2i + tu

nsumi=1

(AiBi +BiAi) + u2nsumi=1

B2i

By Lemma 34sumn

i=1(AiBi +BiAi) = 0 By Corollary 32sumn

i=1B2i = 0 Therefore

nsumi=1

η2i = t2nsumi=1

A2i = t2

sum1leilen

xminus2i (1minus τi)

by equation 39

11

4 A Variant of the Khongsap-Wang Odd Dunkl Operator

In this section we will show that a close variant of the odd Dunkl operator introduced by Khongsapand Wang can be used in the construction of three operators that satisfy the defining relations of the Liealgebra sl2 First we will consider an operator pi which is different from δi but plays a similar role

Definition 41 The operator pi is a C-linear map Pminus rarr Pminus which acts on monomials as follows

pi(xλ11 xλii xλnn ) = λi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn

Remark 42 One may also introduce pi by using a nice Leibniz-like expression involving τi by defining

pi(xj) = δij

pi(fg) = pi(f)g + (minus1)|fτi(f)pi(g)

where f g isin Pminus and δij is the Kronecker delta Now note the analogous relationship between thedegree-preserving operators si and τi in their respective Leibniz rules for the (minus1)-degree operators partiand pi This provides motivation for the definition of pi and suggests its natural role in our theory

Now consider a modified version of ηi

Definition 43 LetDi = tpi + u

sumk 6=i

rik (41)

Definition 44 Introduce the odd r2 Euler and ∆ operators as below

r2 = (2t)minus1nsumi=1

x2i (42)

E =nsumi=1

xipi +n

2+u

t

sumk 6=i

sik (43)

∆ = minus(2t)minus1nsumi=1

D2i (44)

Remark 45 Heckman who used the even Dunkl operators to find a sl2-triple useful in harmonic anal-ysis uses the convention t = 1 [13] For now we will consider t to be a fixed constant in Ctimes

Remark 46 The commutator in the setting of superalgebras is usually defined as [a b] = abminus(minus1)|a||b|bawhere |a| and |b| are the degrees of a and b respectively However since all of the operators we will beconsidering in this section have even degree there is no need to distinguish between commutators andsuper-commutators

To construct an sl2 action from these operators we will require a series of lemmas regarding theaction of portions of the odd Euler operator E In the next lemma we investigate the action of the firstterm of the odd Euler operator on skew polynomials

Lemma 47 The operatorsumn

i=1 xipi acts by multiplication by |f | on the space of homogenous functionsf isinPminus

Proof It suffices to show the result for a monomial xλ = xλ11 xλii xλnn Note that

xipi(xλ) = λixi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn = λix

λ

12

By summing over all indices i we obtain that

nsumi=1

xipi(xλ) = (λ1 + λ2 + + λn)xλ

which implies the desired result

The above lemma holds true in the even case as well where pi is replaced by the partial derivativewith respect to xi We now prove some properties about the action of the third term of the odd Euleroperator on r2 and ∆

Lemma 48 The commutation relation[sum

k 6=i sik∆]

= 0 holds

Proof Note that sjkpi = pjsjk if i = k sjkpi = pksjk if i = j and sjkpi = pisjk otherwise Indeedthese relations can be verified by checking if they are true for xai x

bjxck a b c isin Z+ and then extending

by linearity We prove that sjkpi = pjsjk if i = k and the other two cases are similar Without loss ofgenerality let j lt k and observe that

sjkpk(xajx

bk) = b(minus1)asjk(x

ajx

bminus1k ) = b(minus1)axakx

bminus1j = b(minus1)abxbminus1j xak

pjsjk(xajx

bk) = (minus1)abpj(x

bjxak) = b(minus1)abxbminus1j xak

By our work in Lemma 26 one can deduce that sjkr`m = rsjk(`m)sij As a consequence we find thatsjkDi = Dsjk(i)sjk By an easy induction we now have that sjk∆ = ∆sjk Using the above equationmultiple times proves the desired result

Lemma 49 The commutation relation[sum

k 6=i sik r2]

= 0 holds

Proof Follows since sjkxj = xksjk sjkxk = xjsjk and sjkxi = xisjk if i 6= j k

We are now ready to obtain two commutativity relations involving the odd Euler operator E

Theorem 410 The odd Euler operator and r2 satisfy the following commutation relations

[E r2] = 2r2 (45)[E∆] = minus2∆ (46)

Proof Since r2 has degree 2 and ∆ has degreeminus2 the theorem follows from Lemmas 47 48 and 49

We also need to investigate what the third commutativity relation [r2∆] turns out to be We willprove one lemma before doing so

Lemma 411 For i = 1 to n the equation xiDi +Dixi = 2txipi + t+ usum

k 6=i sik holds

Proof Recall thatDi = tpi + u

sumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Therefore since pixi = xipi + 1

Dixi = txipi + t+ usumk 6=i

(x2i minus x2k)minus1(xixk minus x2k)sik +sumk 6=i

(x2i minus x2k)[x2i τi minus xixkτk]

xiDi = txipi + usumk 6=i

(x2i minus x2k)minus1(x2i minus xixk)sik +sumk 6=i

(x2i minus x2k)minus1[minusx2i τi + xixkτk]

Adding we obtain the desired result

13

We now have the tools to find the third relation between r2 E and ∆

Theorem 412 The commutation relation [r2∆] = E holds

Proof We will first find [r2 Di] The derivative pi much like the partial derivative in the even casesatisfies the properties pixj = minusxjpi for i 6= j and pixi = xipi + 1 Now suppose that i 6= j Then

Dix2j = tpix

2j + ux2j

sumk 6=i 6=j

rik + x2k(x2i minus x2k)minus1[minusxiτi + xkτk] + x2i (x

2i minus x2k)minus1[(xi minus xk)sik]

= tx2jpi + ux2jsumk 6=i

rik + (xi minus xj)sij

Now we will find Dix2i

Dix2i = tpix

2i +

sumk 6=i

x2k(x2i minus x2k)minus1(xi minus xk)sik + x2i

sumk 6=i

(x2i minus x2k)minus1[minusxiτi + xkτk]

= tx2i pi + 2txi + x2isumk 6=i

rik minussumk 6=i

(xi minus xk)sik

Therefore[sumn

i=1 x2i Di

]= minus2txi This implies that

r2Di minusDir2 = minusxi (47)

As a result we find that

[r2∆] = minus(2t)minus1nsumi=1

[r2 D2j ] = minus(2t)minus1

nsumi=1

(r2D2j minusD2

j r2)

= minus(2t)minus1nsumi=1

[(Djr2Dj minus xjDj)minus (Djr

2Dj +Djxj)]

= (2t)minus1nsumi=1

(xiDi +Dixi)

where we have used 47 Now by Lemma 411

[r2∆] =

nsumi=1

xipi +n

2+u

t

sumk 6=i

sik = E

as desired

To summarize we have found operators Er2 and ∆ similar to their even counterparts whichsatisfy the defining relations of the Lie algebra sl2

[E r2] = 2r2

[E∆] = minus2∆

[r2∆] = E

Remark 413 If one uses the odd Dunkl operator ηi as found in [18] instead of the Di introduced herethe r2 E and ∆ operators do not generate sl2

Remark 414 Although our results hold true for all t and u in C one typically sets t = 1 and u = αminus1

for some α isin Ctimes since without loss of generality one of t and u may equal 1

14

Remark 415 In the even case let X be a Euclidean vector space with dimension n and let C[X] be thealgebra of C-valued functions on X Then this result about sl2 plays a major role in the study of higherdifferential operators on C[X] This is because the representation theory of sl2 allows for the reductionof degree to the second order [13] As a result our results in this section should correspondingly have arole in further studying differential operators in the odd case

5 q-nilHecke Algebras

Until now we have been concerned with the odd symmetric polynomials in variables x1 x2 xnwhere xixj = (minus1)xjxi for 1 le i 6= j le n This immediately suggests the question what if one replacesthe minus1 by any constant q isin Ctimes Specifically we ask the following questions

Question 51 Is it possible to study q-symmetric polynomials for which xixj = qxjxi when i gt j

Question 52 Are there q-analogs of evenodd divided difference operators and nilHecke algebras Sofar such structures are known only for the even case (q = 1) and the odd case (q = minus1)

In this section we answer both questions in the affirmativeWe work in the Z-graded q-braided setting throughout LetC be a commutative ring and let q isin Ctimes

be a unit If VW are graded C-modules and v isin V w isin W are homogeneous the braiding is theldquoq-twistrdquo

τq V otimesW rarrW otimes Vv otimes w 7rarr q|v||w|w otimes v

(51)

where | middot | is the degree function By q-algebra we mean an algebra object in the category of gradedC-modules equipped with this braided monoidal structure likewise for q-bialgebras q-Hopf algebrasand so forth

Remark 53 Note that the q-twist described above is its own inverse only when q2 = 1 which correlatesto the even and odd cases When q2 6= 1 the corresponding theory becomes more complex There-fore the q-nilHecke algebras that we introduce later in this section are nontrivial generalizations of thepreviously studied even and odd nilHecke algebras

Definition 54 The q-algebra P qn is defined to be

P qn = C〈x1 xn〉(xjxi minus qxixj = 0 if i lt j) (52)

where |xi| = 1 for i = 1 n

Note that P qn sim= otimesni=1Pq1 There are two interesting subalgebras of P qn that can be thought of as q-

analogs of the symmetric polynomials Define the k-th elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middotxin

and define the k-th twisted elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middot xin

where xj = qjminus1xj

Definition 55 The q-algebra of q-symmetric polynomials in n variables denoted Λqn is the subalgebra ofP qn generated by e1 en Likewise for the twisted q-symmetric polynomials Λqn and e1 en

The type A braid group on n strands acts on P qn by setting

15

1 σi(xj) = qxi+1 if j = i

2 σi(xj) = qminus1xi if j = i+ 1

3 σi(xj) = qxj if j gt i+ 1

4 σi(xj) = qminus1xj if j lt i

and extending multiplicatively

Definition 56 For i = 1 nminus 1 the i-th q-divided difference operator parti is the linear operator P qn rarr P qndefined by parti(xi) = q parti(xi+1) = minus1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + σi(f)parti(g) (53)

for all functions f g isin P qn We call Equation 53 the q-Leibniz rule

Lemma 57 For every i and every j lt k parti(xkxj minus qxjxk) = 0

Proof Since parti(xj) = 0 for j gt i+1 one may reduce the lemma to having to prove that part1(x2x1minusqx1x2) =0 part1(x3x1 minus qx1x3) = 0 and part1(x3x2 minus qx2x3) = 0 These statements follow from the q-Leibniz rule

Therefore parti is a well-defined operator on P qn

Lemma 58 The following relations hold

parti(xki ) =

kminus1sumj=0

qjkminus2jminusj2+kxjix

kminus1minusji+1

parti(xki+1) = minus

kminus1sumj=0

qminusjxjixkminus1minusji+1

Proof We induct on k The base case (k = 1) follows from the definition of the parti and the powers of qarise mostly from xni+1x

mi = qmnxmi x

ni+1 for all mn isin Z+

Our q-divided difference operators also annihilate the twisted elementary q-symmetric polynomialsjust as the even divided difference operators annihilate the elementary symmetric functions

Lemma 59 For every i = 1 nminus 1 and every k parti(ek) = 0 Hence Λqn sube⋂nminus1i=1 ker(parti)

Proof We can express ek as

ek =sum|J|=kii+1isinJ

xJ +sum|J|=kminus1ii+1isinJ

qf(Jik)xJ(xi + qxi+1) +sum|J|=kminus2ii+1isinJ

qg(Jik)xJxixi+1

for certain Z-valued functions f g The result then follows from parti(xi + qxi+1) = parti(xixi+1) = 0 and theq-Leibniz rule

Having discussed q-divided difference operators we can now construct algebras for every q 6=0 1minus1 that have many similarities to the even and odd nilHecke algebras For every such q we de-fine a q-nilHecke algebra generated by xi and parti for 1 le i le n subject to the relations found in thefollowing two lemmas (510 and 511)

Lemma 510 The following relations hold among the operators parti and xi (left multiplication by xi)

1 part2i = 0

2 partjparti minus qpartipartj = 0 for j gt i+ 1

3 xjxi = qxixj for i lt j

4 partixj minus qxjparti = 0 for j gt i+ 1

5 qpartixj minus xjparti = 0 for j lt i

6 partixi minus qxi+1parti = q

7 xiparti minus qpartixi+1 = q

16

Proof To show that part2i = 0 note that we can reduce to i = 1 and proceed by induction Since parti(1) = 0the base case follows Suppose that part2i (f) = 0 Then note that

part21(x1f) = part1(qf + qx2part1(f)) = qpart1(f)minus qpart1(f) + x1part21(f) = 0

part21(x2f) = part1(minusf + qminus1x1part1(f)) = minuspart1(f) + part1(f) + x2part21(f) = 0

part21(x3f) = part1(qx3part1(f)) = q2x3part21(f) = 0

which completes the proof of the first statement in the lemmaStatement 3 follows by definition Statements 4 5 6 and 7 follow from a suitable application of the

q-Leibniz rule Statement 2 follows from an inductive argument We can reduce to i = 1 and j = 3Suppose that partjparti = qpartipartj if j gt i+ 1 Then

part3part1(x1f)minus qpart1part3(x1f) = (qpart3(f) + x2part3part1(f))minus q(part3(f) + x2part1part3(f)) = 0

part3part1(x2f)minus qpart1part3(x2f) = (minuspart3(f) + qminus2x1part3part1(f))minus q(minusqminus1part3(f) + qminus2x1part1part3(f)) = 0

part3part1(x3f)minus qpart1part3(x3f) = (q2part1(f) + q2x4part3part1(f))minus q(qpart1(f) + q2x4part1part3(f)) = 0

part3part1(x4f)minus qpart1part3(x4f) = (minusqpart1(f) + x3part3part1(f))minus q(minuspart1(f) + x3part1part3(f)) = 0

part3part1(x5f)minus qpart1part3(x5f) = q2x5part3part1(f)minus q(q2x5part1part3(f)) = 0

thereby completing the induction

Lemma 511 partiparti+1partiparti+1partiparti+1 + parti+1partiparti+1partiparti+1parti = 0

Proof This result follows from an inductive argument we reduce to i = 1 and assume that the braidrelation holds true for some function f Then we check that the braid relation is true for x1f x2f x3f and x4f (since the behavior of xjf for j ge 4 is the same as that of x4f ) For brevity we will show theargument for x2f only

part1part2(x2f) = qpart1(f) + q2x3part1part2(f) part2part1(x2f) = minuspart2(f) + qminus2x1part2part1(f)

part212(x2f) = qpart2part1(f)minus q2part1part2(f) + qx2part212(f) qpart121(x2f) = minusqpart1part2(f) + part2part1(f) + part121(f)

part1212(x2f) = qpart121(f)minus qpart212(f) + x1part1212(f) part2121(x2f) = minuspart212(f) + part121(f) + x3part2121(f)

part21212(x2f) = qpart2121(f) + qminus1x2part21212(f) part12121(x2f) = minuspart1212(f) + qx3part12121(f)

We continue the above calculations to find that

part121212(x2f) = qpart12121(f) + part21212(f) + x2part121212(f)

part212121(x2f) = minusqpart12121(f)minus part21212(f) + x2part121212(f)

and the braid relation for x2f follows from the inductive hypothesis

6 A Diagrammatic Approach to q-Symmetric Polynomials

61 Introduction to a q-Bialgebra

In the previous section we answered Question 51 in an algebraic way by defining q-analogs of theclassical elementary and complete symmetric functions In this section we generalize the diagrammaticmethod used in [8] in order to study this question from the perspective of bialgebras

Let NΛq be a free associative Z-graded C-algebra with generators hm for m ge 0 We define h0 = 1and hm = 0 for m lt 0 and let q isin Ctimes The homogenous part of NΛq of degree ` has a basis hααkwhere

hα = hα1 middot middot middothαz for a composition α = (α1 αz) of `

17

Define a multiplication for homogenous x and y on NΛqotimes2 as follows where deg(x) denotes the degreeof x

(w otimes x)(y otimes z) = qdeg(x)deg(y)(wy otimes xz)

We can make NΛq into a q-bialgebra by letting the comultiplication on generators be

∆(hn) =

nsumk=0

hk otimes hnminusk

and by letting the counit be ε(x) = 0 if x is homogenous and deg(x)gt 0We can impose through the braiding structure that

∆(hahb) =asumj=0

bsumk=0

(hj otimes haminusj)(hk otimes hbminusk) =asumj=0

bsumk=0

qk(aminusj)(hjhk otimes haminusjhbminusk)

For any partitions λ and micro of n consider the set of double cosets of subroups Sλ and Smicro of Sn SλSnSmicroFor every C in this set let wC be the minimal length representative of C and let `(wC) be the length ofthis minimal length representative We will now attribute a bilinear form to NΛq

(hλ hmicro) =sum

CisinSλSnSmicro

q`(wC)

This bilinear form admits a diagrammatic description Let hn be an orange platform with n non-intersecting strands coming out of it When computing (hλ hmicro) with `(λ) = z and `(micro) = y drawz orange platforms at the top of the diagram representing λ1 λ2middot middot middot λz Draw y orange platforms atthe bottom of the diagram representative of micro1 micro2middot middot middot microy We require that |λ| = |micro| so that the topplatforms and bottom platforms have the same number of strands

Consider the example (h121 h22) In the following diagram snippets of the strands from each plat-form are shown

Every strand must start at one platform at the top and end on another platform at the bottom Nostrands that have originated from one platform may intersect The strands themselves have no criticalpoints with respect to the height function no two strands ever intersect more than once and there are notriple-intersections where three strands are concurrent Diagrams are considered up to isotopy Withoutany restrictions there would be n such diagrams if |λ| = n since there would be no limitations onthe ordering of the strands However due to the above rules there are only 4 possible diagrams in thecomputation of (h121 h22) shown below

Define(hλ hmicro) =

sumall diagrams D representing (hλhmicro)

q number of of crossings in D (61)

In the above example (h121 h22) = 1 + 2q2 + q3

18

We can extend the bilinear form toNΛqotimes2 by stating that any diagram in which strands from distincttensor factors intersect contributes 0 to the bilinear form

(w otimes x y otimes z) = (w y)(x z)

Let I be the radical of the bilinear form inNΛq In [8] the authors prove for any q that multiplicationand comultiplication are adjoint In other words for all xy1 y2 in NΛq

(y1 otimes y2∆(x)) = (y1y2 x) (62)

62 The Elementary q-Symmetric Functions

We now use the bilinear form of q-symmetric functions to study one of their important bases the ele-mentary q-symmetric functions

Define elements ek isin NΛq by ek = 0 for k lt 0 e0 = 1 and

ksumi=0

(minus1)iq(i2)eihkminusi = 0 for k ge 1 (63)

Equivalently leten = qminus(n2)

sumαn

(minus1)`(α)minusnhα (64)

Lemma 61

1 The coproduct of an elementary function is given by ∆(en) =

nsumk=0

ek otimes enminusk

2 If λ n then (hλ en) =

1 if λ = (1 1)

0 otherwise

Proof We begin by demonstrating (2) from which (1) will follow To show (2) it suffices to show that

(hmx en) =

(x enminus1) if m = 10 otherwise

We will utilize strong induction on n in order to find (hmx ekhnminusk) The base cases n = 0 1 are easyto show There are two cases to consider by the inductive hypothesis applied to k lt n Either there isa strand connecting hm and ek or there is not Just as we used an orange platform to denote hn wewill use a blue platform to denote ek The rules of the diagrammatic notation are the same for the blueplatforms as they are for the orange platforms

k

m

nminus k minusm

k nminus k

mlowast lowast lowast lowast lowast

x

If there is not a strand connecting hm and ek the configuration contributes qkm(x ekhnminuskminusm)

19

k minus 1

mminus 1

nminus k minusm + 1

k nminus k

mlowast lowast lowast lowast lowast

x

If a stand connects hm and ek this configuration contributes q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) We have thusshown that (hmx ekhnminusk) = qkm(x ekhnminuskminusm) + q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) Now we are equippedto consider (hmx ek)

(minus1)n+1q(n2)(hmx en) =

nminus1sumk=0

(minus1)kq(k2)(hmx ekhnminusk)

=nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus1sumk=0

(minus1)kq(k2)+(mminus1)(kminus1)(x ekminus1hnminuskminusm+1)

=

nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus2sumk=0

(minus1)k+1q(k+12 )+(mminus1)(k)(x ekhnminuskminusm)

= (minus1)nminus1q(nminus12 )+nm(x enminus1h1minusm)

Corresponding terms from the two sums cancel in pairs since q(k2)+km = q(

k+12 )+k(mminus1) leaving only the

k = nminus 1 term in the first sum The second statement of the lemma thus followsWe will now use (2) to prove (1) This follows from equation 62

(∆(ek) hλ otimes hmicro) = (ek hλhmicro) =

1 λ = (1`) micro = (1p) `+ p = k

0 otherwise

We now calculate the sign incurred when strands connect two blue (ek) platforms

(minus1)n+1q(n2)(en en) =

nminus1sumk=0

(minus1)kq(k2)(en ekhnminusk)

= (minus1)nminus1q(nminus12 )(en enminus1h1)

= (minus1)nminus1q(nminus12 )(∆(en) enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )

nsumk=0

(ek otimes enminusk enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )(enminus1 enminus1)

One may solve this recursion to find that (en en) = qminus(n2)Here the second equality follows from not-ing that at most one strand can connect hnminusk and en (so that k = n minus 1) the third equality followsfrom adjointness and the fourth and fifth equalities follow from the diagrammatic considerations of theprevious lemma

To summarize the diagrammatics of the bilinear form thus developed

20

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

from identifying ldquooddrdquo analogs of structures that play important algebraic roles in the even case EllisKhovanov and Lauda started this program of ldquooddificationrdquo by finding an odd analog of NHn andusing it to categorify Uq(sl2) Searching for the geometry underlying these odd constructions would alsoprovide a very new approach to noncommutative geometry For example we study potential generatorsof certain Cherednik algebras Since spherical rational Cherednik algebras fit nicely into a family ofalgebras from the geometry of symplectic resolutions including Websterrsquos tensor product algebras andcyclotomic KLR algebras this project may be used in the context of ldquooddrdquo noncommutative geometry

In Subsection 11 we discussed the (even) divided difference operators Dunkl operators and Chered-nik operators Although analogs of Dunkl operators have been found in the odd case they have not beenwell-studied Odd Cherednik operators have not even been defined

As a result the first goal of the present paper is to unify certain results in the odd case and to furtherstudy the odd Dunkl operators of Khongsap and Wang In Section 2 we introduce an operator rikrelated to the generalized odd divided difference operator partodd

ik and study its properties One of ourmain results (Equation 26) is that the odd Dunkl operator ηi may be expressed in terms of the odddivided difference operators of Ellis Khovanov and Lauda

ηoddi = tδi + u

sumk 6=i

partoddik sik (11)

This result connects odd Dunkl operators and the odd nilHecke algebra both of which play importantroles in the project of oddification

In the even case one can introduce an operator known as the Dunkl Laplacian given bysumn

i=1 η2i This

operator has important applications in spherical harmonics and heat semigroups [24] Our next goal inthe present paper is to express the Dunkl Laplacian in the odd case In Section 3 we show that therik satisfy the classical Yang-Baxter equation and use this result to evaluate the odd Dunkl LaplacianSpecifically we show that

nsumi=1

η2i = t2sum

1leilenxminus2i (1minus τi)

In Section 4 we find an odd analog of Heckmanrsquos important sl2-triple in [13] by showing that avariant Di of the odd Dunkl operator can be used to construct three operators r2 E and ∆ that satisfythe defining relations of the Lie algebra sl2

r2 = (2t)minus1nsumi=1

x2i

E =nsumi=1

xipi +n

2+u

t

sumk 6=i

sik

∆ = minus(2t)minus1nsumi=1

D2i

Since even Dunkl operators play an important role in the representation theory of symmetric groupsour study of odd Dunkl operators should result in a better understanding of the representation theoryof odd symmetric functions which correspondingly results in a better understanding of odd Khovanovhomology

The second goal of this paper is to study a generalization of the odd symmetric functions knownas q-symmetric functions for which xjxi = qxixj when j gt i Previous authors have described a nil-Hecke algebra structure only for the odd case q = minus1 and the even case q = 1 [9 15] In Section 5we find a q-divided difference operator for all previously unstudied values of q and explore its propertiesWe show for example that twisted elementary symmetric polynomials are in the kernel of q-divided

4

difference operators just as odd elementary symmetric functions are in the kernel of odd divided differ-ence operators We then use q-divided difference operators to construct algebras acting on q-symmetricpolynomials that have many similarities to the even and odd nilHecke algebras We call these q-nilHeckealgebras These algebras are nontrivial generalizations of the even and odd nilHecke algebras becausethe q-twist map introduced in Section 5 is not its own inverse when q2 6= 1 In Section 6 we present theelementary q-symmetric polynomials using a generalization of a clever diagrammatic method arisingin the context of bialgebras We use these diagrams to study relations between elementary q-symmetricpolynomials when q is a root of unity These methods can be combined with the algebras of Section 5 inorder to continue studying q-symmetric polynomials including q-Schur and q-monomial functions

In the conclusion we also define the odd Cherednik operators and outline a procedure for findingand studying odd analogs of Jack polynomials This makes progress towards answering a question ofEllis about the existence of Macdonald-like polynomials in the odd case Since the Jack polynomialshave importance in representation theory their study would enhance our knowledge about the oddalgebraic theory

2 Odd Dunkl operators and the Odd nilHecke algebra

21 Preliminaries Even Dunkl Operators

In the even case we work with the ring C[x1 xn] and a root system of type An where xixj = xjxifor all 1 le i j le n and α isin C We first introduce some notation involving the symmetric group

1 Let sik be the simple transposition in Sn swapping xi and xk We let si = sii+1

2 Let sij(k `) be the result of applying sij to the pair (k `) Similarly define sij(k)

In [7] Dunkl introduced the remarkable operator

ηeveni =

part

partxi+ α

sumk 6=i

partevenik

where partpartxi

is the partial derivative with respect to xi and partevenik is the even divided difference operator

partevenik = (xi minus xk)minus1(1minus sik)

Since xi minus xk always divides f minus sik(f) for f isin C[x1 xn] partik sends polynomials to polynomialsThese Dunkl operators have various important properties one of which is that they commute (ηiηj =

ηjηi) In [18] Khongsap and Wang introduced anti-commuting odd Dunkl operators on skew polynomi-als In Section 2 we will develop the connection between these operators and the odd nilHecke algebraintroduced in [9]

Returning to the even case introduce operators r2 E (the Euler operator) and ∆k

r2 =1

2

nsumi=1

x2i

E =

nsumi=1

xipart

partxi+micro

2

∆ =1

2

nsumi=1

η2i

where micro is the Dunkl dimension which is defined by the relation ∆|x|2 = 2micro as in [6]

5

Let [p q] = pq minus qp be the commutator Heckman showed that r2 E and ∆k satisfy the definingrelations of the Lie algebra sl2 [13]

[E r2] = 2r2

[E∆k] = minus2∆k

[r2∆k] = E

Remark 21 If one were to replace ∆k with the classical Laplacian on flat Rn (replacing the Dunkloperator with the partial derivative) these three operators still satisfy the sl2 relations

Remark 22 From now on we will use ηi to denote the odd Dunkl operator of Khongsap and Wangdefined in equation 25

In Section 4 we will focus on finding analogous results in the odd case

22 Introduction to the Odd nilHecke Algebra

We will now discuss operators with the algebra Pminus = C〈x1 xn〉〈xjxi + xixj = 0 for i 6= j〉 We callPminus the skew polynomial ring We can define linear operators called the odd divided difference operatorsas below

Definition 23 For i = 1 nminus1 the i-th odd divided difference operator parti is the linear operator Pminus rarr Pminus

defined by parti(xi) = 1 parti(xi+1) = 1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + (minus1)|f |si(f)parti(g)

for all functions f g isin Pminus We call this last relation the Leibniz rule

It is shown in [9] that the odd divided difference operators can be used to construct an odd nilHeckealgebra generated by xi and parti for 1 le i le n subject to the following relations

1 part2i = 0

2 partipartj + partjparti = 0 for |iminus j| ge 2

3 partiparti+1parti = parti+1partiparti+1

4 xixj + xjxi = 0 for i 6= j

5 xiparti + partixi+1 = 1 partixi + xi+1parti = 1

6 xipartj + partjxi = 0 for i 6= j j + 1

Due to [14] we have the following explicit definition of the odd divided difference operator

parti(f) = (x2i+1 minus x2i )minus1[(xi+1 minus xi)f minus (minus1)|f |si(f)(xi+1 minus xi)] (21)

Although this formula a priori involves denominators it does take skew polynomials to skew polyno-mials We extend this definition to non-consecutive indices by replacing i + 1 with any index k 6= i for1 le k le n and by replacing si with sik Equation 21 then becomes

partik(f) = (x2k minus x2i )minus1[(xk minus xi)f minus (minus1)|f |sik(f)(xk minus xi)] (22)

This extended odd divided difference operator satisfies the Leibniz rule partik(fg) = partik(f)g+(minus1)|f |sikpartik(g)[9]

6

23 Some Operations on Skew Polynomials

First we introduce a common operator in the study of Dunkl operators

Definition 24 Let the (minus1)-shift operator τi be the automorphism of Pminus which sends xi to minusxi and xj toxj for j 6= i

Suppose 1 le i 6= j le n and 1 le k 6= ` le n where f is an element in C〈x1 xn〉〈xixj + xjxi =0 for i 6= j〉 Then one has

sijτk` = τsij(k`)sij

fxi = (minus1)|f |xiτi(f)

Remark 25 Since skew polynomials are not super-commutative we cannot say that fg = (minus1)|f ||g|gf But the operator τi allows us to track the discrepancy from super-commutativity since xif = (minus1)|f |τi(f)ximaking it useful in this context

We now introduce the operator rik = partiksik for k 6= i which will serve as another odd divideddifference operator that we will use to study odd Dunkl operators For simplicity let ri = rii+1 In thefollowing lemma we study the action of the transposition and (minus1)-shift operator on rik

Lemma 26 The operators sik and τi act on rik as follows

rijsk` = sk`rsk`(ij) (23)

We also have that

1 sirik = ri+1ksi if k 6= i+ 1

2 siri = risi

3 siri+1 = rii+2si

4 si+1ri = rii+2si+1

5 sirj = rjsi for |iminus j| ge 2

6 τirj = rjτi for |iminus j| ge 2

Proof Recall the following relationship between partij and sk` for i 6= j and k 6= ` from Lemma 219 (1) of[9]

partijsk` = sk`partsk`(ij) (24)

Multiplying both sides by sij we obtain that

sijpartijsk` = sijsk`partsk`(ij)

= sk`sk`(i j)partsk`(ij)

which implies the desired result since rij = partijsij = sijpartij Properties 1-5 are special cases of equation23 Property 6 follows from τisj = sjτi and the fact that τi(xj) = xj for i 6= j

Remark 27 Differences between our formulas and those of [9] are due to a difference of sign conventionin the action of sij on Pminus

We now show that the properties of the rik are similar to those of the odd divided difference operatorpartik

Lemma 28 The following relations hold

7

1 r2i = 0

2 rirj + rjri = 0 for |iminus j| ge 2

3 riri+1ri = ri+1riri+1

4 rik(fg) = rik(f)sik(g) + (minus1)|f |frik(g)

5 rixi+1 + xi+1ri = rixi + xiri = si

6 rjxi + xjri = 0 for i 6= j j + 1

Proof Since siri = risi and ri = partisi it follows that siparti = partisi Then since part2i = 0 r2i = 0 as well Dueto Equation 5 from Lemma 26 we have that sirj = rjsi for |i minus j| ge 2 so sipartj = partjsi Thus ri andrj anti-commute since partipartj + partjparti = 0 The operators ri also satisfy braid relations which we show byinductively reducing to i = 1 and then using 24 and siparti = partisi repeatedly

r1r2r1 = s1part12part2s1part1 = s1s2part13s1part13part1 = s1s2s1part23part13part12

r2r1r2 = s2part2s1part1s2part2 = s2s1part13s2part13part2 = s2s1s2part12part13part23

Since s1s2s1 = s2s1s2 and part23part13part12 = part12part13part23 by symmetry we conclude that r1r2r1 = r2r1r2 TheLeibniz rule for rik (equation 4 of this lemma) follows immediately from the Leibniz rule for partik Sinceri(xi) = ri(xi+1) = 1 and ri(xj) = 0 for j 6= i i + 1 equations 5 and 6 follow from the Leibniz rule forrik

We also desire an explicit definition of the rik analogous to that of the odd divided difference oper-ator of [9] To find such an expression we use a preparatory lemma

Lemma 29 For all f isinPminus and 1 le i 6= k le n we have

sikxiτi(f)minus sikxkτk(f) = (minus1)|f |sik(f)(xi minus xk)

Proof It suffices to prove the result for a monomial xλ = xλ11 xλii xλkk xλnn where i lt k Wecalculate that

sikxiτi(xλ) = (minus1)λ1++λixλ11 xλi+1

k xλki xλnn

sikxkτk(xλ) = (minus1)λ1++λkminus1xλ11 xλik xλk+1

i xλnn

sik(xλ)xi = (minus1)λk+1++λnxλ11 xλik x

λk+1

i xλnn

sik(xλ)xk = (minus1)λi+1++λnxλ11 x

λi+1

k xλki xλnn

Since |f | = λ1 + + λn the desired result follows

Lemma 210 The operator rik has explicit form rik = (x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Proof Follows from Lemma 29 and Equation 22

We will now connect the above results to the odd Dunkl operator introduced by Khongsap and Wangin [18]

Definition 211 Define an operator δi by δi = (2xi)minus1(1minus τi)

The above super-derivative can also be defined inductively by imposing that δi(xj) = 1 if i = j and0 otherwise We then extend the action to monomials as follows

δi(xa1xa2 xa`) =sumk=1

(minus1)kminus1xa1 δi(xak)xak+1 xa`

8

The operator δi is a priori from Laurent skew polynomials to Laurent skew polynomials but it is easy tocheck that it preserves the subalgebra of skew polynomials Khongsap and Wang found an odd analogof the Dunkl operator given by

ηi = tδi + usumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk] (25)

where t u isin Ctimes Their operators anti-commute ηiηj + ηjηi = 0 for i 6= jBy Lemma 210 this odd Dunkl operator may be expressed as

ηi = tδi + usumk 6=i

partiksik (26)

By analogy with the commutative case discussed in Section 21 the operator rik plays the same role inthe odd theory that the even divided difference operator plays in the even theory

3 Classical Yang-Baxter Equation and the Dunkl Laplacian

Theorem 31 LetH123 = [r12 r13]+ + [r13 r23]+ + [r12 r23]+ (31)

where [p q]+ = pq + qp is the anti-commutator Then the operators rik satisfy the classical Yang-Baxterequation

H123 = 0 (32)

Proof To avoid a cumbersome direct calculation we instead use an inductive approach Namely sup-pose that H123(f) = 0 for some function f isin Pminus Then we show that H123(xif) = 0 for all integersi ge 1 Note that for i = 1

r12r13x1 = r12(s13 minus x1r13)= r12s13 minus r12(x1r13)= r12s13 minus (s12r13 minus x1r12r13)= r12s13 minus s12r13 + x1r12r13

(33)

where we have used Equation 4 twice Similarly we find that

r13r12x1 = r13s12 minus s13r12 + x1r13r12 (34)r23r12x1 = r23s12 + x1r23r12 (35)r12r23x1 = minuss12r23 + x1r12r23 (36)r13r23x1 = minuss13r23 + x1r13r23 (37)r23r13x1 = r23s13 + x1r23r13 (38)

By our inductive hypothesis

x1(r12r13 + r13r12 + r23r12 + r12r23 + r13r23 + r23r13) = 0

Keeping this in mind add Equations 33-38 to show that

H123x1 = r12s13 minus s12r13 + r13s12 minus s13r12 + r23s12 minus s12r23 minus s13r23 + r23s13

= r12s13 minus r23s12 + r13s12 minus r23s13 + r23s12 minus r13s12 minus r12s13 + r23s13

= 0

where we have repeatedly used Lemma 26 to slide rij past sk`We can similarly show that H123x2 = H123x3 = 0 Since rjkxi = xirjk for i gt 3 and j k isin (1 2 3)

it also follows that H123xi = 0 for i gt 3 proving the desired result

9

Corollary 32 The double summationsumn

i=1

(sumk 6=i rik

)2= 0

Proof The expansion of this double summation has n(nminus1)2 total terms Since r2ij = 0 n(nminus1) of theseterms are immediately zero leaving n(n minus 1)(n minus 2) terms of the form rijrk` where i = k and j = `are not both true By Theorem 31 the sum of all six terms of the form rijrk` where i j k ` isin a b c fordistinct integers 1 le a b c le n i 6= j and k 6= ` is zero There are

(n3

)ways to choose integers a b c

and for each choice of a b c six terms of the form rijrk` vanish This fact eliminates all the remaining6(n3

)= n(nminus 1)(nminus 2) terms of the double summation

As an application of the results in this section we will compute the odd Dunkl Laplaciansumn

i=1 η2i

We will first require a lemma involving the commutator of τi and rik

Lemma 33 The equation xminus1i [rik τi] = (x2i minus x2k)minus1(sik(τi + τk)minus xminus1i xksik(τi minus τk)minus 2) holds

Proof By Lemma 210

rikτi = (x2i minus x2k)minus1((xi minus xk)sikτi minus xiτ2i + xkτkτi)

τirik = (x2i minus x2k)minus1(minus(xi + xk)sikτk + xiτ2i + xkτiτk)

Since τiτk = τkτi and τ2i = 1 the result follows by subtraction

Now define Ai = (2xi)minus1(1minus τi) and Bi =

sumk 6=i rik so that the odd Dunkl operator ηi of Khongsap

and Wang may be expressed as ηi = tAi + uBi Note that

A2i =

1

4xminus1i (1minus τi)xminus1i (1minus τi)

=1

2xminus2i (1minus τi)(1minus τi)

= xminus2i (1minus τi)

(39)

since τi(xminus1i ) = minusxminus1i and τ2i = 1

Lemma 34 The relationsumn

i=1(AiBi +BiAi) = 0 holds

Proof Due to the Leibniz Rule for rik (equation 4) we can find that 0 = rik(xixminus1i ) = xminus1k minus xirik(x

minus1i )

so rik(xminus1i ) = xminus1i xminus1k It follows that rik(xminus1i f) = xminus1i xminus1k sik(f)minus xminus1i rik(f) Using this factnsumi=1

BiAi =1

2

nsumi=1

sumk 6=i

rikxminus1i (1minus τi)

=1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi)minus xminus1i rik(1minus τi))

By definitionnsumi=1

AiBi =1

2

nsumi=1

sumk 6=i

xminus1i (1minus τi)rik

Adding the above two equations we find thatnsumi=1

[Ai Bi]+ =1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi) + xminus1i (rikτi minus τirik))

=1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi) + (x2i minus x2k)minus1(sik(τi + τk)minus xminus1i xksik(τi minus τk)minus 2))

(310)

10

where we have used Lemma 33Each double summation repeats the pair of indices (p q) twice when 1 le p q le n one time when

i = p and k = q and once more when i = q and k = p Note that

(x2i minus x2k)minus1sik(τi + τk) = minus(x2k minus x2i )ski(τk + τi)

As a result the sum 12

sumni=1

sumk 6=i(x

2i minus x2k)minus1sik(τi + τk) = 0 Similarly

1

2

nsumi=1

sumk 6=i

xminus1i xminus1k sik = 0 and1

2

nsumi=1

sumk 6=i

(x2i minus x2k)minus1(minus2) = 0

since xminus1i xminus1k + xminus1k xminus1i = 0 Equation 310 then becomes

nsumi=1

(AiBi +BiAi) =1

2

nsumi=1

sumk 6=i

(minusxminus1i xminus1k sikτi minus (x2i minus x2k)minus1(xminus1i xksik(τi minus τk)) (311)

However note thatxminus1i xk + xminus1k xi = minus(x2i minus x2k)xminus1i xminus1k

which implies

(x2i minus x2k)minus1xminus1i xksik(τi minus τk) + (x2k minus x2i )minus1xminus1k xiski(τk minus τi) = minusxminus1i xminus1k sik(τi minus τk) (312)

Similarly we find that

xminus1i xminus1k sikτi + xminus1k xminus1i sikτk = xminus1i xminus1k sik(τi minus τk) (313)

As a result equation 311 becomes

nsumi=1

(AiBi +BiAi) = minus1

2

sum1leiltklen

(xminus1i xminus1k sik(τi minus τk)minus xminus1i sminus1k sik(τi minus τk)) = 0

We are now equipped to compute the Dunkl Laplacian in the odd case

Theorem 35 The equationsumn

i=1 η2i = t2

sum1leilen x

minus2i (1minus τi) holds

Proof Since ηi = tAi + uBi we have that

nsumi=1

η2i = t2nsumi=1

A2i + tu

nsumi=1

(AiBi +BiAi) + u2nsumi=1

B2i

By Lemma 34sumn

i=1(AiBi +BiAi) = 0 By Corollary 32sumn

i=1B2i = 0 Therefore

nsumi=1

η2i = t2nsumi=1

A2i = t2

sum1leilen

xminus2i (1minus τi)

by equation 39

11

4 A Variant of the Khongsap-Wang Odd Dunkl Operator

In this section we will show that a close variant of the odd Dunkl operator introduced by Khongsapand Wang can be used in the construction of three operators that satisfy the defining relations of the Liealgebra sl2 First we will consider an operator pi which is different from δi but plays a similar role

Definition 41 The operator pi is a C-linear map Pminus rarr Pminus which acts on monomials as follows

pi(xλ11 xλii xλnn ) = λi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn

Remark 42 One may also introduce pi by using a nice Leibniz-like expression involving τi by defining

pi(xj) = δij

pi(fg) = pi(f)g + (minus1)|fτi(f)pi(g)

where f g isin Pminus and δij is the Kronecker delta Now note the analogous relationship between thedegree-preserving operators si and τi in their respective Leibniz rules for the (minus1)-degree operators partiand pi This provides motivation for the definition of pi and suggests its natural role in our theory

Now consider a modified version of ηi

Definition 43 LetDi = tpi + u

sumk 6=i

rik (41)

Definition 44 Introduce the odd r2 Euler and ∆ operators as below

r2 = (2t)minus1nsumi=1

x2i (42)

E =nsumi=1

xipi +n

2+u

t

sumk 6=i

sik (43)

∆ = minus(2t)minus1nsumi=1

D2i (44)

Remark 45 Heckman who used the even Dunkl operators to find a sl2-triple useful in harmonic anal-ysis uses the convention t = 1 [13] For now we will consider t to be a fixed constant in Ctimes

Remark 46 The commutator in the setting of superalgebras is usually defined as [a b] = abminus(minus1)|a||b|bawhere |a| and |b| are the degrees of a and b respectively However since all of the operators we will beconsidering in this section have even degree there is no need to distinguish between commutators andsuper-commutators

To construct an sl2 action from these operators we will require a series of lemmas regarding theaction of portions of the odd Euler operator E In the next lemma we investigate the action of the firstterm of the odd Euler operator on skew polynomials

Lemma 47 The operatorsumn

i=1 xipi acts by multiplication by |f | on the space of homogenous functionsf isinPminus

Proof It suffices to show the result for a monomial xλ = xλ11 xλii xλnn Note that

xipi(xλ) = λixi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn = λix

λ

12

By summing over all indices i we obtain that

nsumi=1

xipi(xλ) = (λ1 + λ2 + + λn)xλ

which implies the desired result

The above lemma holds true in the even case as well where pi is replaced by the partial derivativewith respect to xi We now prove some properties about the action of the third term of the odd Euleroperator on r2 and ∆

Lemma 48 The commutation relation[sum

k 6=i sik∆]

= 0 holds

Proof Note that sjkpi = pjsjk if i = k sjkpi = pksjk if i = j and sjkpi = pisjk otherwise Indeedthese relations can be verified by checking if they are true for xai x

bjxck a b c isin Z+ and then extending

by linearity We prove that sjkpi = pjsjk if i = k and the other two cases are similar Without loss ofgenerality let j lt k and observe that

sjkpk(xajx

bk) = b(minus1)asjk(x

ajx

bminus1k ) = b(minus1)axakx

bminus1j = b(minus1)abxbminus1j xak

pjsjk(xajx

bk) = (minus1)abpj(x

bjxak) = b(minus1)abxbminus1j xak

By our work in Lemma 26 one can deduce that sjkr`m = rsjk(`m)sij As a consequence we find thatsjkDi = Dsjk(i)sjk By an easy induction we now have that sjk∆ = ∆sjk Using the above equationmultiple times proves the desired result

Lemma 49 The commutation relation[sum

k 6=i sik r2]

= 0 holds

Proof Follows since sjkxj = xksjk sjkxk = xjsjk and sjkxi = xisjk if i 6= j k

We are now ready to obtain two commutativity relations involving the odd Euler operator E

Theorem 410 The odd Euler operator and r2 satisfy the following commutation relations

[E r2] = 2r2 (45)[E∆] = minus2∆ (46)

Proof Since r2 has degree 2 and ∆ has degreeminus2 the theorem follows from Lemmas 47 48 and 49

We also need to investigate what the third commutativity relation [r2∆] turns out to be We willprove one lemma before doing so

Lemma 411 For i = 1 to n the equation xiDi +Dixi = 2txipi + t+ usum

k 6=i sik holds

Proof Recall thatDi = tpi + u

sumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Therefore since pixi = xipi + 1

Dixi = txipi + t+ usumk 6=i

(x2i minus x2k)minus1(xixk minus x2k)sik +sumk 6=i

(x2i minus x2k)[x2i τi minus xixkτk]

xiDi = txipi + usumk 6=i

(x2i minus x2k)minus1(x2i minus xixk)sik +sumk 6=i

(x2i minus x2k)minus1[minusx2i τi + xixkτk]

Adding we obtain the desired result

13

We now have the tools to find the third relation between r2 E and ∆

Theorem 412 The commutation relation [r2∆] = E holds

Proof We will first find [r2 Di] The derivative pi much like the partial derivative in the even casesatisfies the properties pixj = minusxjpi for i 6= j and pixi = xipi + 1 Now suppose that i 6= j Then

Dix2j = tpix

2j + ux2j

sumk 6=i 6=j

rik + x2k(x2i minus x2k)minus1[minusxiτi + xkτk] + x2i (x

2i minus x2k)minus1[(xi minus xk)sik]

= tx2jpi + ux2jsumk 6=i

rik + (xi minus xj)sij

Now we will find Dix2i

Dix2i = tpix

2i +

sumk 6=i

x2k(x2i minus x2k)minus1(xi minus xk)sik + x2i

sumk 6=i

(x2i minus x2k)minus1[minusxiτi + xkτk]

= tx2i pi + 2txi + x2isumk 6=i

rik minussumk 6=i

(xi minus xk)sik

Therefore[sumn

i=1 x2i Di

]= minus2txi This implies that

r2Di minusDir2 = minusxi (47)

As a result we find that

[r2∆] = minus(2t)minus1nsumi=1

[r2 D2j ] = minus(2t)minus1

nsumi=1

(r2D2j minusD2

j r2)

= minus(2t)minus1nsumi=1

[(Djr2Dj minus xjDj)minus (Djr

2Dj +Djxj)]

= (2t)minus1nsumi=1

(xiDi +Dixi)

where we have used 47 Now by Lemma 411

[r2∆] =

nsumi=1

xipi +n

2+u

t

sumk 6=i

sik = E

as desired

To summarize we have found operators Er2 and ∆ similar to their even counterparts whichsatisfy the defining relations of the Lie algebra sl2

[E r2] = 2r2

[E∆] = minus2∆

[r2∆] = E

Remark 413 If one uses the odd Dunkl operator ηi as found in [18] instead of the Di introduced herethe r2 E and ∆ operators do not generate sl2

Remark 414 Although our results hold true for all t and u in C one typically sets t = 1 and u = αminus1

for some α isin Ctimes since without loss of generality one of t and u may equal 1

14

Remark 415 In the even case let X be a Euclidean vector space with dimension n and let C[X] be thealgebra of C-valued functions on X Then this result about sl2 plays a major role in the study of higherdifferential operators on C[X] This is because the representation theory of sl2 allows for the reductionof degree to the second order [13] As a result our results in this section should correspondingly have arole in further studying differential operators in the odd case

5 q-nilHecke Algebras

Until now we have been concerned with the odd symmetric polynomials in variables x1 x2 xnwhere xixj = (minus1)xjxi for 1 le i 6= j le n This immediately suggests the question what if one replacesthe minus1 by any constant q isin Ctimes Specifically we ask the following questions

Question 51 Is it possible to study q-symmetric polynomials for which xixj = qxjxi when i gt j

Question 52 Are there q-analogs of evenodd divided difference operators and nilHecke algebras Sofar such structures are known only for the even case (q = 1) and the odd case (q = minus1)

In this section we answer both questions in the affirmativeWe work in the Z-graded q-braided setting throughout LetC be a commutative ring and let q isin Ctimes

be a unit If VW are graded C-modules and v isin V w isin W are homogeneous the braiding is theldquoq-twistrdquo

τq V otimesW rarrW otimes Vv otimes w 7rarr q|v||w|w otimes v

(51)

where | middot | is the degree function By q-algebra we mean an algebra object in the category of gradedC-modules equipped with this braided monoidal structure likewise for q-bialgebras q-Hopf algebrasand so forth

Remark 53 Note that the q-twist described above is its own inverse only when q2 = 1 which correlatesto the even and odd cases When q2 6= 1 the corresponding theory becomes more complex There-fore the q-nilHecke algebras that we introduce later in this section are nontrivial generalizations of thepreviously studied even and odd nilHecke algebras

Definition 54 The q-algebra P qn is defined to be

P qn = C〈x1 xn〉(xjxi minus qxixj = 0 if i lt j) (52)

where |xi| = 1 for i = 1 n

Note that P qn sim= otimesni=1Pq1 There are two interesting subalgebras of P qn that can be thought of as q-

analogs of the symmetric polynomials Define the k-th elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middotxin

and define the k-th twisted elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middot xin

where xj = qjminus1xj

Definition 55 The q-algebra of q-symmetric polynomials in n variables denoted Λqn is the subalgebra ofP qn generated by e1 en Likewise for the twisted q-symmetric polynomials Λqn and e1 en

The type A braid group on n strands acts on P qn by setting

15

1 σi(xj) = qxi+1 if j = i

2 σi(xj) = qminus1xi if j = i+ 1

3 σi(xj) = qxj if j gt i+ 1

4 σi(xj) = qminus1xj if j lt i

and extending multiplicatively

Definition 56 For i = 1 nminus 1 the i-th q-divided difference operator parti is the linear operator P qn rarr P qndefined by parti(xi) = q parti(xi+1) = minus1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + σi(f)parti(g) (53)

for all functions f g isin P qn We call Equation 53 the q-Leibniz rule

Lemma 57 For every i and every j lt k parti(xkxj minus qxjxk) = 0

Proof Since parti(xj) = 0 for j gt i+1 one may reduce the lemma to having to prove that part1(x2x1minusqx1x2) =0 part1(x3x1 minus qx1x3) = 0 and part1(x3x2 minus qx2x3) = 0 These statements follow from the q-Leibniz rule

Therefore parti is a well-defined operator on P qn

Lemma 58 The following relations hold

parti(xki ) =

kminus1sumj=0

qjkminus2jminusj2+kxjix

kminus1minusji+1

parti(xki+1) = minus

kminus1sumj=0

qminusjxjixkminus1minusji+1

Proof We induct on k The base case (k = 1) follows from the definition of the parti and the powers of qarise mostly from xni+1x

mi = qmnxmi x

ni+1 for all mn isin Z+

Our q-divided difference operators also annihilate the twisted elementary q-symmetric polynomialsjust as the even divided difference operators annihilate the elementary symmetric functions

Lemma 59 For every i = 1 nminus 1 and every k parti(ek) = 0 Hence Λqn sube⋂nminus1i=1 ker(parti)

Proof We can express ek as

ek =sum|J|=kii+1isinJ

xJ +sum|J|=kminus1ii+1isinJ

qf(Jik)xJ(xi + qxi+1) +sum|J|=kminus2ii+1isinJ

qg(Jik)xJxixi+1

for certain Z-valued functions f g The result then follows from parti(xi + qxi+1) = parti(xixi+1) = 0 and theq-Leibniz rule

Having discussed q-divided difference operators we can now construct algebras for every q 6=0 1minus1 that have many similarities to the even and odd nilHecke algebras For every such q we de-fine a q-nilHecke algebra generated by xi and parti for 1 le i le n subject to the relations found in thefollowing two lemmas (510 and 511)

Lemma 510 The following relations hold among the operators parti and xi (left multiplication by xi)

1 part2i = 0

2 partjparti minus qpartipartj = 0 for j gt i+ 1

3 xjxi = qxixj for i lt j

4 partixj minus qxjparti = 0 for j gt i+ 1

5 qpartixj minus xjparti = 0 for j lt i

6 partixi minus qxi+1parti = q

7 xiparti minus qpartixi+1 = q

16

Proof To show that part2i = 0 note that we can reduce to i = 1 and proceed by induction Since parti(1) = 0the base case follows Suppose that part2i (f) = 0 Then note that

part21(x1f) = part1(qf + qx2part1(f)) = qpart1(f)minus qpart1(f) + x1part21(f) = 0

part21(x2f) = part1(minusf + qminus1x1part1(f)) = minuspart1(f) + part1(f) + x2part21(f) = 0

part21(x3f) = part1(qx3part1(f)) = q2x3part21(f) = 0

which completes the proof of the first statement in the lemmaStatement 3 follows by definition Statements 4 5 6 and 7 follow from a suitable application of the

q-Leibniz rule Statement 2 follows from an inductive argument We can reduce to i = 1 and j = 3Suppose that partjparti = qpartipartj if j gt i+ 1 Then

part3part1(x1f)minus qpart1part3(x1f) = (qpart3(f) + x2part3part1(f))minus q(part3(f) + x2part1part3(f)) = 0

part3part1(x2f)minus qpart1part3(x2f) = (minuspart3(f) + qminus2x1part3part1(f))minus q(minusqminus1part3(f) + qminus2x1part1part3(f)) = 0

part3part1(x3f)minus qpart1part3(x3f) = (q2part1(f) + q2x4part3part1(f))minus q(qpart1(f) + q2x4part1part3(f)) = 0

part3part1(x4f)minus qpart1part3(x4f) = (minusqpart1(f) + x3part3part1(f))minus q(minuspart1(f) + x3part1part3(f)) = 0

part3part1(x5f)minus qpart1part3(x5f) = q2x5part3part1(f)minus q(q2x5part1part3(f)) = 0

thereby completing the induction

Lemma 511 partiparti+1partiparti+1partiparti+1 + parti+1partiparti+1partiparti+1parti = 0

Proof This result follows from an inductive argument we reduce to i = 1 and assume that the braidrelation holds true for some function f Then we check that the braid relation is true for x1f x2f x3f and x4f (since the behavior of xjf for j ge 4 is the same as that of x4f ) For brevity we will show theargument for x2f only

part1part2(x2f) = qpart1(f) + q2x3part1part2(f) part2part1(x2f) = minuspart2(f) + qminus2x1part2part1(f)

part212(x2f) = qpart2part1(f)minus q2part1part2(f) + qx2part212(f) qpart121(x2f) = minusqpart1part2(f) + part2part1(f) + part121(f)

part1212(x2f) = qpart121(f)minus qpart212(f) + x1part1212(f) part2121(x2f) = minuspart212(f) + part121(f) + x3part2121(f)

part21212(x2f) = qpart2121(f) + qminus1x2part21212(f) part12121(x2f) = minuspart1212(f) + qx3part12121(f)

We continue the above calculations to find that

part121212(x2f) = qpart12121(f) + part21212(f) + x2part121212(f)

part212121(x2f) = minusqpart12121(f)minus part21212(f) + x2part121212(f)

and the braid relation for x2f follows from the inductive hypothesis

6 A Diagrammatic Approach to q-Symmetric Polynomials

61 Introduction to a q-Bialgebra

In the previous section we answered Question 51 in an algebraic way by defining q-analogs of theclassical elementary and complete symmetric functions In this section we generalize the diagrammaticmethod used in [8] in order to study this question from the perspective of bialgebras

Let NΛq be a free associative Z-graded C-algebra with generators hm for m ge 0 We define h0 = 1and hm = 0 for m lt 0 and let q isin Ctimes The homogenous part of NΛq of degree ` has a basis hααkwhere

hα = hα1 middot middot middothαz for a composition α = (α1 αz) of `

17

Define a multiplication for homogenous x and y on NΛqotimes2 as follows where deg(x) denotes the degreeof x

(w otimes x)(y otimes z) = qdeg(x)deg(y)(wy otimes xz)

We can make NΛq into a q-bialgebra by letting the comultiplication on generators be

∆(hn) =

nsumk=0

hk otimes hnminusk

and by letting the counit be ε(x) = 0 if x is homogenous and deg(x)gt 0We can impose through the braiding structure that

∆(hahb) =asumj=0

bsumk=0

(hj otimes haminusj)(hk otimes hbminusk) =asumj=0

bsumk=0

qk(aminusj)(hjhk otimes haminusjhbminusk)

For any partitions λ and micro of n consider the set of double cosets of subroups Sλ and Smicro of Sn SλSnSmicroFor every C in this set let wC be the minimal length representative of C and let `(wC) be the length ofthis minimal length representative We will now attribute a bilinear form to NΛq

(hλ hmicro) =sum

CisinSλSnSmicro

q`(wC)

This bilinear form admits a diagrammatic description Let hn be an orange platform with n non-intersecting strands coming out of it When computing (hλ hmicro) with `(λ) = z and `(micro) = y drawz orange platforms at the top of the diagram representing λ1 λ2middot middot middot λz Draw y orange platforms atthe bottom of the diagram representative of micro1 micro2middot middot middot microy We require that |λ| = |micro| so that the topplatforms and bottom platforms have the same number of strands

Consider the example (h121 h22) In the following diagram snippets of the strands from each plat-form are shown

Every strand must start at one platform at the top and end on another platform at the bottom Nostrands that have originated from one platform may intersect The strands themselves have no criticalpoints with respect to the height function no two strands ever intersect more than once and there are notriple-intersections where three strands are concurrent Diagrams are considered up to isotopy Withoutany restrictions there would be n such diagrams if |λ| = n since there would be no limitations onthe ordering of the strands However due to the above rules there are only 4 possible diagrams in thecomputation of (h121 h22) shown below

Define(hλ hmicro) =

sumall diagrams D representing (hλhmicro)

q number of of crossings in D (61)

In the above example (h121 h22) = 1 + 2q2 + q3

18

We can extend the bilinear form toNΛqotimes2 by stating that any diagram in which strands from distincttensor factors intersect contributes 0 to the bilinear form

(w otimes x y otimes z) = (w y)(x z)

Let I be the radical of the bilinear form inNΛq In [8] the authors prove for any q that multiplicationand comultiplication are adjoint In other words for all xy1 y2 in NΛq

(y1 otimes y2∆(x)) = (y1y2 x) (62)

62 The Elementary q-Symmetric Functions

We now use the bilinear form of q-symmetric functions to study one of their important bases the ele-mentary q-symmetric functions

Define elements ek isin NΛq by ek = 0 for k lt 0 e0 = 1 and

ksumi=0

(minus1)iq(i2)eihkminusi = 0 for k ge 1 (63)

Equivalently leten = qminus(n2)

sumαn

(minus1)`(α)minusnhα (64)

Lemma 61

1 The coproduct of an elementary function is given by ∆(en) =

nsumk=0

ek otimes enminusk

2 If λ n then (hλ en) =

1 if λ = (1 1)

0 otherwise

Proof We begin by demonstrating (2) from which (1) will follow To show (2) it suffices to show that

(hmx en) =

(x enminus1) if m = 10 otherwise

We will utilize strong induction on n in order to find (hmx ekhnminusk) The base cases n = 0 1 are easyto show There are two cases to consider by the inductive hypothesis applied to k lt n Either there isa strand connecting hm and ek or there is not Just as we used an orange platform to denote hn wewill use a blue platform to denote ek The rules of the diagrammatic notation are the same for the blueplatforms as they are for the orange platforms

k

m

nminus k minusm

k nminus k

mlowast lowast lowast lowast lowast

x

If there is not a strand connecting hm and ek the configuration contributes qkm(x ekhnminuskminusm)

19

k minus 1

mminus 1

nminus k minusm + 1

k nminus k

mlowast lowast lowast lowast lowast

x

If a stand connects hm and ek this configuration contributes q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) We have thusshown that (hmx ekhnminusk) = qkm(x ekhnminuskminusm) + q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) Now we are equippedto consider (hmx ek)

(minus1)n+1q(n2)(hmx en) =

nminus1sumk=0

(minus1)kq(k2)(hmx ekhnminusk)

=nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus1sumk=0

(minus1)kq(k2)+(mminus1)(kminus1)(x ekminus1hnminuskminusm+1)

=

nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus2sumk=0

(minus1)k+1q(k+12 )+(mminus1)(k)(x ekhnminuskminusm)

= (minus1)nminus1q(nminus12 )+nm(x enminus1h1minusm)

Corresponding terms from the two sums cancel in pairs since q(k2)+km = q(

k+12 )+k(mminus1) leaving only the

k = nminus 1 term in the first sum The second statement of the lemma thus followsWe will now use (2) to prove (1) This follows from equation 62

(∆(ek) hλ otimes hmicro) = (ek hλhmicro) =

1 λ = (1`) micro = (1p) `+ p = k

0 otherwise

We now calculate the sign incurred when strands connect two blue (ek) platforms

(minus1)n+1q(n2)(en en) =

nminus1sumk=0

(minus1)kq(k2)(en ekhnminusk)

= (minus1)nminus1q(nminus12 )(en enminus1h1)

= (minus1)nminus1q(nminus12 )(∆(en) enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )

nsumk=0

(ek otimes enminusk enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )(enminus1 enminus1)

One may solve this recursion to find that (en en) = qminus(n2)Here the second equality follows from not-ing that at most one strand can connect hnminusk and en (so that k = n minus 1) the third equality followsfrom adjointness and the fourth and fifth equalities follow from the diagrammatic considerations of theprevious lemma

To summarize the diagrammatics of the bilinear form thus developed

20

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

difference operators just as odd elementary symmetric functions are in the kernel of odd divided differ-ence operators We then use q-divided difference operators to construct algebras acting on q-symmetricpolynomials that have many similarities to the even and odd nilHecke algebras We call these q-nilHeckealgebras These algebras are nontrivial generalizations of the even and odd nilHecke algebras becausethe q-twist map introduced in Section 5 is not its own inverse when q2 6= 1 In Section 6 we present theelementary q-symmetric polynomials using a generalization of a clever diagrammatic method arisingin the context of bialgebras We use these diagrams to study relations between elementary q-symmetricpolynomials when q is a root of unity These methods can be combined with the algebras of Section 5 inorder to continue studying q-symmetric polynomials including q-Schur and q-monomial functions

In the conclusion we also define the odd Cherednik operators and outline a procedure for findingand studying odd analogs of Jack polynomials This makes progress towards answering a question ofEllis about the existence of Macdonald-like polynomials in the odd case Since the Jack polynomialshave importance in representation theory their study would enhance our knowledge about the oddalgebraic theory

2 Odd Dunkl operators and the Odd nilHecke algebra

21 Preliminaries Even Dunkl Operators

In the even case we work with the ring C[x1 xn] and a root system of type An where xixj = xjxifor all 1 le i j le n and α isin C We first introduce some notation involving the symmetric group

1 Let sik be the simple transposition in Sn swapping xi and xk We let si = sii+1

2 Let sij(k `) be the result of applying sij to the pair (k `) Similarly define sij(k)

In [7] Dunkl introduced the remarkable operator

ηeveni =

part

partxi+ α

sumk 6=i

partevenik

where partpartxi

is the partial derivative with respect to xi and partevenik is the even divided difference operator

partevenik = (xi minus xk)minus1(1minus sik)

Since xi minus xk always divides f minus sik(f) for f isin C[x1 xn] partik sends polynomials to polynomialsThese Dunkl operators have various important properties one of which is that they commute (ηiηj =

ηjηi) In [18] Khongsap and Wang introduced anti-commuting odd Dunkl operators on skew polynomi-als In Section 2 we will develop the connection between these operators and the odd nilHecke algebraintroduced in [9]

Returning to the even case introduce operators r2 E (the Euler operator) and ∆k

r2 =1

2

nsumi=1

x2i

E =

nsumi=1

xipart

partxi+micro

2

∆ =1

2

nsumi=1

η2i

where micro is the Dunkl dimension which is defined by the relation ∆|x|2 = 2micro as in [6]

5

Let [p q] = pq minus qp be the commutator Heckman showed that r2 E and ∆k satisfy the definingrelations of the Lie algebra sl2 [13]

[E r2] = 2r2

[E∆k] = minus2∆k

[r2∆k] = E

Remark 21 If one were to replace ∆k with the classical Laplacian on flat Rn (replacing the Dunkloperator with the partial derivative) these three operators still satisfy the sl2 relations

Remark 22 From now on we will use ηi to denote the odd Dunkl operator of Khongsap and Wangdefined in equation 25

In Section 4 we will focus on finding analogous results in the odd case

22 Introduction to the Odd nilHecke Algebra

We will now discuss operators with the algebra Pminus = C〈x1 xn〉〈xjxi + xixj = 0 for i 6= j〉 We callPminus the skew polynomial ring We can define linear operators called the odd divided difference operatorsas below

Definition 23 For i = 1 nminus1 the i-th odd divided difference operator parti is the linear operator Pminus rarr Pminus

defined by parti(xi) = 1 parti(xi+1) = 1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + (minus1)|f |si(f)parti(g)

for all functions f g isin Pminus We call this last relation the Leibniz rule

It is shown in [9] that the odd divided difference operators can be used to construct an odd nilHeckealgebra generated by xi and parti for 1 le i le n subject to the following relations

1 part2i = 0

2 partipartj + partjparti = 0 for |iminus j| ge 2

3 partiparti+1parti = parti+1partiparti+1

4 xixj + xjxi = 0 for i 6= j

5 xiparti + partixi+1 = 1 partixi + xi+1parti = 1

6 xipartj + partjxi = 0 for i 6= j j + 1

Due to [14] we have the following explicit definition of the odd divided difference operator

parti(f) = (x2i+1 minus x2i )minus1[(xi+1 minus xi)f minus (minus1)|f |si(f)(xi+1 minus xi)] (21)

Although this formula a priori involves denominators it does take skew polynomials to skew polyno-mials We extend this definition to non-consecutive indices by replacing i + 1 with any index k 6= i for1 le k le n and by replacing si with sik Equation 21 then becomes

partik(f) = (x2k minus x2i )minus1[(xk minus xi)f minus (minus1)|f |sik(f)(xk minus xi)] (22)

This extended odd divided difference operator satisfies the Leibniz rule partik(fg) = partik(f)g+(minus1)|f |sikpartik(g)[9]

6

23 Some Operations on Skew Polynomials

First we introduce a common operator in the study of Dunkl operators

Definition 24 Let the (minus1)-shift operator τi be the automorphism of Pminus which sends xi to minusxi and xj toxj for j 6= i

Suppose 1 le i 6= j le n and 1 le k 6= ` le n where f is an element in C〈x1 xn〉〈xixj + xjxi =0 for i 6= j〉 Then one has

sijτk` = τsij(k`)sij

fxi = (minus1)|f |xiτi(f)

Remark 25 Since skew polynomials are not super-commutative we cannot say that fg = (minus1)|f ||g|gf But the operator τi allows us to track the discrepancy from super-commutativity since xif = (minus1)|f |τi(f)ximaking it useful in this context

We now introduce the operator rik = partiksik for k 6= i which will serve as another odd divideddifference operator that we will use to study odd Dunkl operators For simplicity let ri = rii+1 In thefollowing lemma we study the action of the transposition and (minus1)-shift operator on rik

Lemma 26 The operators sik and τi act on rik as follows

rijsk` = sk`rsk`(ij) (23)

We also have that

1 sirik = ri+1ksi if k 6= i+ 1

2 siri = risi

3 siri+1 = rii+2si

4 si+1ri = rii+2si+1

5 sirj = rjsi for |iminus j| ge 2

6 τirj = rjτi for |iminus j| ge 2

Proof Recall the following relationship between partij and sk` for i 6= j and k 6= ` from Lemma 219 (1) of[9]

partijsk` = sk`partsk`(ij) (24)

Multiplying both sides by sij we obtain that

sijpartijsk` = sijsk`partsk`(ij)

= sk`sk`(i j)partsk`(ij)

which implies the desired result since rij = partijsij = sijpartij Properties 1-5 are special cases of equation23 Property 6 follows from τisj = sjτi and the fact that τi(xj) = xj for i 6= j

Remark 27 Differences between our formulas and those of [9] are due to a difference of sign conventionin the action of sij on Pminus

We now show that the properties of the rik are similar to those of the odd divided difference operatorpartik

Lemma 28 The following relations hold

7

1 r2i = 0

2 rirj + rjri = 0 for |iminus j| ge 2

3 riri+1ri = ri+1riri+1

4 rik(fg) = rik(f)sik(g) + (minus1)|f |frik(g)

5 rixi+1 + xi+1ri = rixi + xiri = si

6 rjxi + xjri = 0 for i 6= j j + 1

Proof Since siri = risi and ri = partisi it follows that siparti = partisi Then since part2i = 0 r2i = 0 as well Dueto Equation 5 from Lemma 26 we have that sirj = rjsi for |i minus j| ge 2 so sipartj = partjsi Thus ri andrj anti-commute since partipartj + partjparti = 0 The operators ri also satisfy braid relations which we show byinductively reducing to i = 1 and then using 24 and siparti = partisi repeatedly

r1r2r1 = s1part12part2s1part1 = s1s2part13s1part13part1 = s1s2s1part23part13part12

r2r1r2 = s2part2s1part1s2part2 = s2s1part13s2part13part2 = s2s1s2part12part13part23

Since s1s2s1 = s2s1s2 and part23part13part12 = part12part13part23 by symmetry we conclude that r1r2r1 = r2r1r2 TheLeibniz rule for rik (equation 4 of this lemma) follows immediately from the Leibniz rule for partik Sinceri(xi) = ri(xi+1) = 1 and ri(xj) = 0 for j 6= i i + 1 equations 5 and 6 follow from the Leibniz rule forrik

We also desire an explicit definition of the rik analogous to that of the odd divided difference oper-ator of [9] To find such an expression we use a preparatory lemma

Lemma 29 For all f isinPminus and 1 le i 6= k le n we have

sikxiτi(f)minus sikxkτk(f) = (minus1)|f |sik(f)(xi minus xk)

Proof It suffices to prove the result for a monomial xλ = xλ11 xλii xλkk xλnn where i lt k Wecalculate that

sikxiτi(xλ) = (minus1)λ1++λixλ11 xλi+1

k xλki xλnn

sikxkτk(xλ) = (minus1)λ1++λkminus1xλ11 xλik xλk+1

i xλnn

sik(xλ)xi = (minus1)λk+1++λnxλ11 xλik x

λk+1

i xλnn

sik(xλ)xk = (minus1)λi+1++λnxλ11 x

λi+1

k xλki xλnn

Since |f | = λ1 + + λn the desired result follows

Lemma 210 The operator rik has explicit form rik = (x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Proof Follows from Lemma 29 and Equation 22

We will now connect the above results to the odd Dunkl operator introduced by Khongsap and Wangin [18]

Definition 211 Define an operator δi by δi = (2xi)minus1(1minus τi)

The above super-derivative can also be defined inductively by imposing that δi(xj) = 1 if i = j and0 otherwise We then extend the action to monomials as follows

δi(xa1xa2 xa`) =sumk=1

(minus1)kminus1xa1 δi(xak)xak+1 xa`

8

The operator δi is a priori from Laurent skew polynomials to Laurent skew polynomials but it is easy tocheck that it preserves the subalgebra of skew polynomials Khongsap and Wang found an odd analogof the Dunkl operator given by

ηi = tδi + usumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk] (25)

where t u isin Ctimes Their operators anti-commute ηiηj + ηjηi = 0 for i 6= jBy Lemma 210 this odd Dunkl operator may be expressed as

ηi = tδi + usumk 6=i

partiksik (26)

By analogy with the commutative case discussed in Section 21 the operator rik plays the same role inthe odd theory that the even divided difference operator plays in the even theory

3 Classical Yang-Baxter Equation and the Dunkl Laplacian

Theorem 31 LetH123 = [r12 r13]+ + [r13 r23]+ + [r12 r23]+ (31)

where [p q]+ = pq + qp is the anti-commutator Then the operators rik satisfy the classical Yang-Baxterequation

H123 = 0 (32)

Proof To avoid a cumbersome direct calculation we instead use an inductive approach Namely sup-pose that H123(f) = 0 for some function f isin Pminus Then we show that H123(xif) = 0 for all integersi ge 1 Note that for i = 1

r12r13x1 = r12(s13 minus x1r13)= r12s13 minus r12(x1r13)= r12s13 minus (s12r13 minus x1r12r13)= r12s13 minus s12r13 + x1r12r13

(33)

where we have used Equation 4 twice Similarly we find that

r13r12x1 = r13s12 minus s13r12 + x1r13r12 (34)r23r12x1 = r23s12 + x1r23r12 (35)r12r23x1 = minuss12r23 + x1r12r23 (36)r13r23x1 = minuss13r23 + x1r13r23 (37)r23r13x1 = r23s13 + x1r23r13 (38)

By our inductive hypothesis

x1(r12r13 + r13r12 + r23r12 + r12r23 + r13r23 + r23r13) = 0

Keeping this in mind add Equations 33-38 to show that

H123x1 = r12s13 minus s12r13 + r13s12 minus s13r12 + r23s12 minus s12r23 minus s13r23 + r23s13

= r12s13 minus r23s12 + r13s12 minus r23s13 + r23s12 minus r13s12 minus r12s13 + r23s13

= 0

where we have repeatedly used Lemma 26 to slide rij past sk`We can similarly show that H123x2 = H123x3 = 0 Since rjkxi = xirjk for i gt 3 and j k isin (1 2 3)

it also follows that H123xi = 0 for i gt 3 proving the desired result

9

Corollary 32 The double summationsumn

i=1

(sumk 6=i rik

)2= 0

Proof The expansion of this double summation has n(nminus1)2 total terms Since r2ij = 0 n(nminus1) of theseterms are immediately zero leaving n(n minus 1)(n minus 2) terms of the form rijrk` where i = k and j = `are not both true By Theorem 31 the sum of all six terms of the form rijrk` where i j k ` isin a b c fordistinct integers 1 le a b c le n i 6= j and k 6= ` is zero There are

(n3

)ways to choose integers a b c

and for each choice of a b c six terms of the form rijrk` vanish This fact eliminates all the remaining6(n3

)= n(nminus 1)(nminus 2) terms of the double summation

As an application of the results in this section we will compute the odd Dunkl Laplaciansumn

i=1 η2i

We will first require a lemma involving the commutator of τi and rik

Lemma 33 The equation xminus1i [rik τi] = (x2i minus x2k)minus1(sik(τi + τk)minus xminus1i xksik(τi minus τk)minus 2) holds

Proof By Lemma 210

rikτi = (x2i minus x2k)minus1((xi minus xk)sikτi minus xiτ2i + xkτkτi)

τirik = (x2i minus x2k)minus1(minus(xi + xk)sikτk + xiτ2i + xkτiτk)

Since τiτk = τkτi and τ2i = 1 the result follows by subtraction

Now define Ai = (2xi)minus1(1minus τi) and Bi =

sumk 6=i rik so that the odd Dunkl operator ηi of Khongsap

and Wang may be expressed as ηi = tAi + uBi Note that

A2i =

1

4xminus1i (1minus τi)xminus1i (1minus τi)

=1

2xminus2i (1minus τi)(1minus τi)

= xminus2i (1minus τi)

(39)

since τi(xminus1i ) = minusxminus1i and τ2i = 1

Lemma 34 The relationsumn

i=1(AiBi +BiAi) = 0 holds

Proof Due to the Leibniz Rule for rik (equation 4) we can find that 0 = rik(xixminus1i ) = xminus1k minus xirik(x

minus1i )

so rik(xminus1i ) = xminus1i xminus1k It follows that rik(xminus1i f) = xminus1i xminus1k sik(f)minus xminus1i rik(f) Using this factnsumi=1

BiAi =1

2

nsumi=1

sumk 6=i

rikxminus1i (1minus τi)

=1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi)minus xminus1i rik(1minus τi))

By definitionnsumi=1

AiBi =1

2

nsumi=1

sumk 6=i

xminus1i (1minus τi)rik

Adding the above two equations we find thatnsumi=1

[Ai Bi]+ =1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi) + xminus1i (rikτi minus τirik))

=1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi) + (x2i minus x2k)minus1(sik(τi + τk)minus xminus1i xksik(τi minus τk)minus 2))

(310)

10

where we have used Lemma 33Each double summation repeats the pair of indices (p q) twice when 1 le p q le n one time when

i = p and k = q and once more when i = q and k = p Note that

(x2i minus x2k)minus1sik(τi + τk) = minus(x2k minus x2i )ski(τk + τi)

As a result the sum 12

sumni=1

sumk 6=i(x

2i minus x2k)minus1sik(τi + τk) = 0 Similarly

1

2

nsumi=1

sumk 6=i

xminus1i xminus1k sik = 0 and1

2

nsumi=1

sumk 6=i

(x2i minus x2k)minus1(minus2) = 0

since xminus1i xminus1k + xminus1k xminus1i = 0 Equation 310 then becomes

nsumi=1

(AiBi +BiAi) =1

2

nsumi=1

sumk 6=i

(minusxminus1i xminus1k sikτi minus (x2i minus x2k)minus1(xminus1i xksik(τi minus τk)) (311)

However note thatxminus1i xk + xminus1k xi = minus(x2i minus x2k)xminus1i xminus1k

which implies

(x2i minus x2k)minus1xminus1i xksik(τi minus τk) + (x2k minus x2i )minus1xminus1k xiski(τk minus τi) = minusxminus1i xminus1k sik(τi minus τk) (312)

Similarly we find that

xminus1i xminus1k sikτi + xminus1k xminus1i sikτk = xminus1i xminus1k sik(τi minus τk) (313)

As a result equation 311 becomes

nsumi=1

(AiBi +BiAi) = minus1

2

sum1leiltklen

(xminus1i xminus1k sik(τi minus τk)minus xminus1i sminus1k sik(τi minus τk)) = 0

We are now equipped to compute the Dunkl Laplacian in the odd case

Theorem 35 The equationsumn

i=1 η2i = t2

sum1leilen x

minus2i (1minus τi) holds

Proof Since ηi = tAi + uBi we have that

nsumi=1

η2i = t2nsumi=1

A2i + tu

nsumi=1

(AiBi +BiAi) + u2nsumi=1

B2i

By Lemma 34sumn

i=1(AiBi +BiAi) = 0 By Corollary 32sumn

i=1B2i = 0 Therefore

nsumi=1

η2i = t2nsumi=1

A2i = t2

sum1leilen

xminus2i (1minus τi)

by equation 39

11

4 A Variant of the Khongsap-Wang Odd Dunkl Operator

In this section we will show that a close variant of the odd Dunkl operator introduced by Khongsapand Wang can be used in the construction of three operators that satisfy the defining relations of the Liealgebra sl2 First we will consider an operator pi which is different from δi but plays a similar role

Definition 41 The operator pi is a C-linear map Pminus rarr Pminus which acts on monomials as follows

pi(xλ11 xλii xλnn ) = λi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn

Remark 42 One may also introduce pi by using a nice Leibniz-like expression involving τi by defining

pi(xj) = δij

pi(fg) = pi(f)g + (minus1)|fτi(f)pi(g)

where f g isin Pminus and δij is the Kronecker delta Now note the analogous relationship between thedegree-preserving operators si and τi in their respective Leibniz rules for the (minus1)-degree operators partiand pi This provides motivation for the definition of pi and suggests its natural role in our theory

Now consider a modified version of ηi

Definition 43 LetDi = tpi + u

sumk 6=i

rik (41)

Definition 44 Introduce the odd r2 Euler and ∆ operators as below

r2 = (2t)minus1nsumi=1

x2i (42)

E =nsumi=1

xipi +n

2+u

t

sumk 6=i

sik (43)

∆ = minus(2t)minus1nsumi=1

D2i (44)

Remark 45 Heckman who used the even Dunkl operators to find a sl2-triple useful in harmonic anal-ysis uses the convention t = 1 [13] For now we will consider t to be a fixed constant in Ctimes

Remark 46 The commutator in the setting of superalgebras is usually defined as [a b] = abminus(minus1)|a||b|bawhere |a| and |b| are the degrees of a and b respectively However since all of the operators we will beconsidering in this section have even degree there is no need to distinguish between commutators andsuper-commutators

To construct an sl2 action from these operators we will require a series of lemmas regarding theaction of portions of the odd Euler operator E In the next lemma we investigate the action of the firstterm of the odd Euler operator on skew polynomials

Lemma 47 The operatorsumn

i=1 xipi acts by multiplication by |f | on the space of homogenous functionsf isinPminus

Proof It suffices to show the result for a monomial xλ = xλ11 xλii xλnn Note that

xipi(xλ) = λixi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn = λix

λ

12

By summing over all indices i we obtain that

nsumi=1

xipi(xλ) = (λ1 + λ2 + + λn)xλ

which implies the desired result

The above lemma holds true in the even case as well where pi is replaced by the partial derivativewith respect to xi We now prove some properties about the action of the third term of the odd Euleroperator on r2 and ∆

Lemma 48 The commutation relation[sum

k 6=i sik∆]

= 0 holds

Proof Note that sjkpi = pjsjk if i = k sjkpi = pksjk if i = j and sjkpi = pisjk otherwise Indeedthese relations can be verified by checking if they are true for xai x

bjxck a b c isin Z+ and then extending

by linearity We prove that sjkpi = pjsjk if i = k and the other two cases are similar Without loss ofgenerality let j lt k and observe that

sjkpk(xajx

bk) = b(minus1)asjk(x

ajx

bminus1k ) = b(minus1)axakx

bminus1j = b(minus1)abxbminus1j xak

pjsjk(xajx

bk) = (minus1)abpj(x

bjxak) = b(minus1)abxbminus1j xak

By our work in Lemma 26 one can deduce that sjkr`m = rsjk(`m)sij As a consequence we find thatsjkDi = Dsjk(i)sjk By an easy induction we now have that sjk∆ = ∆sjk Using the above equationmultiple times proves the desired result

Lemma 49 The commutation relation[sum

k 6=i sik r2]

= 0 holds

Proof Follows since sjkxj = xksjk sjkxk = xjsjk and sjkxi = xisjk if i 6= j k

We are now ready to obtain two commutativity relations involving the odd Euler operator E

Theorem 410 The odd Euler operator and r2 satisfy the following commutation relations

[E r2] = 2r2 (45)[E∆] = minus2∆ (46)

Proof Since r2 has degree 2 and ∆ has degreeminus2 the theorem follows from Lemmas 47 48 and 49

We also need to investigate what the third commutativity relation [r2∆] turns out to be We willprove one lemma before doing so

Lemma 411 For i = 1 to n the equation xiDi +Dixi = 2txipi + t+ usum

k 6=i sik holds

Proof Recall thatDi = tpi + u

sumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Therefore since pixi = xipi + 1

Dixi = txipi + t+ usumk 6=i

(x2i minus x2k)minus1(xixk minus x2k)sik +sumk 6=i

(x2i minus x2k)[x2i τi minus xixkτk]

xiDi = txipi + usumk 6=i

(x2i minus x2k)minus1(x2i minus xixk)sik +sumk 6=i

(x2i minus x2k)minus1[minusx2i τi + xixkτk]

Adding we obtain the desired result

13

We now have the tools to find the third relation between r2 E and ∆

Theorem 412 The commutation relation [r2∆] = E holds

Proof We will first find [r2 Di] The derivative pi much like the partial derivative in the even casesatisfies the properties pixj = minusxjpi for i 6= j and pixi = xipi + 1 Now suppose that i 6= j Then

Dix2j = tpix

2j + ux2j

sumk 6=i 6=j

rik + x2k(x2i minus x2k)minus1[minusxiτi + xkτk] + x2i (x

2i minus x2k)minus1[(xi minus xk)sik]

= tx2jpi + ux2jsumk 6=i

rik + (xi minus xj)sij

Now we will find Dix2i

Dix2i = tpix

2i +

sumk 6=i

x2k(x2i minus x2k)minus1(xi minus xk)sik + x2i

sumk 6=i

(x2i minus x2k)minus1[minusxiτi + xkτk]

= tx2i pi + 2txi + x2isumk 6=i

rik minussumk 6=i

(xi minus xk)sik

Therefore[sumn

i=1 x2i Di

]= minus2txi This implies that

r2Di minusDir2 = minusxi (47)

As a result we find that

[r2∆] = minus(2t)minus1nsumi=1

[r2 D2j ] = minus(2t)minus1

nsumi=1

(r2D2j minusD2

j r2)

= minus(2t)minus1nsumi=1

[(Djr2Dj minus xjDj)minus (Djr

2Dj +Djxj)]

= (2t)minus1nsumi=1

(xiDi +Dixi)

where we have used 47 Now by Lemma 411

[r2∆] =

nsumi=1

xipi +n

2+u

t

sumk 6=i

sik = E

as desired

To summarize we have found operators Er2 and ∆ similar to their even counterparts whichsatisfy the defining relations of the Lie algebra sl2

[E r2] = 2r2

[E∆] = minus2∆

[r2∆] = E

Remark 413 If one uses the odd Dunkl operator ηi as found in [18] instead of the Di introduced herethe r2 E and ∆ operators do not generate sl2

Remark 414 Although our results hold true for all t and u in C one typically sets t = 1 and u = αminus1

for some α isin Ctimes since without loss of generality one of t and u may equal 1

14

Remark 415 In the even case let X be a Euclidean vector space with dimension n and let C[X] be thealgebra of C-valued functions on X Then this result about sl2 plays a major role in the study of higherdifferential operators on C[X] This is because the representation theory of sl2 allows for the reductionof degree to the second order [13] As a result our results in this section should correspondingly have arole in further studying differential operators in the odd case

5 q-nilHecke Algebras

Until now we have been concerned with the odd symmetric polynomials in variables x1 x2 xnwhere xixj = (minus1)xjxi for 1 le i 6= j le n This immediately suggests the question what if one replacesthe minus1 by any constant q isin Ctimes Specifically we ask the following questions

Question 51 Is it possible to study q-symmetric polynomials for which xixj = qxjxi when i gt j

Question 52 Are there q-analogs of evenodd divided difference operators and nilHecke algebras Sofar such structures are known only for the even case (q = 1) and the odd case (q = minus1)

In this section we answer both questions in the affirmativeWe work in the Z-graded q-braided setting throughout LetC be a commutative ring and let q isin Ctimes

be a unit If VW are graded C-modules and v isin V w isin W are homogeneous the braiding is theldquoq-twistrdquo

τq V otimesW rarrW otimes Vv otimes w 7rarr q|v||w|w otimes v

(51)

where | middot | is the degree function By q-algebra we mean an algebra object in the category of gradedC-modules equipped with this braided monoidal structure likewise for q-bialgebras q-Hopf algebrasand so forth

Remark 53 Note that the q-twist described above is its own inverse only when q2 = 1 which correlatesto the even and odd cases When q2 6= 1 the corresponding theory becomes more complex There-fore the q-nilHecke algebras that we introduce later in this section are nontrivial generalizations of thepreviously studied even and odd nilHecke algebras

Definition 54 The q-algebra P qn is defined to be

P qn = C〈x1 xn〉(xjxi minus qxixj = 0 if i lt j) (52)

where |xi| = 1 for i = 1 n

Note that P qn sim= otimesni=1Pq1 There are two interesting subalgebras of P qn that can be thought of as q-

analogs of the symmetric polynomials Define the k-th elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middotxin

and define the k-th twisted elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middot xin

where xj = qjminus1xj

Definition 55 The q-algebra of q-symmetric polynomials in n variables denoted Λqn is the subalgebra ofP qn generated by e1 en Likewise for the twisted q-symmetric polynomials Λqn and e1 en

The type A braid group on n strands acts on P qn by setting

15

1 σi(xj) = qxi+1 if j = i

2 σi(xj) = qminus1xi if j = i+ 1

3 σi(xj) = qxj if j gt i+ 1

4 σi(xj) = qminus1xj if j lt i

and extending multiplicatively

Definition 56 For i = 1 nminus 1 the i-th q-divided difference operator parti is the linear operator P qn rarr P qndefined by parti(xi) = q parti(xi+1) = minus1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + σi(f)parti(g) (53)

for all functions f g isin P qn We call Equation 53 the q-Leibniz rule

Lemma 57 For every i and every j lt k parti(xkxj minus qxjxk) = 0

Proof Since parti(xj) = 0 for j gt i+1 one may reduce the lemma to having to prove that part1(x2x1minusqx1x2) =0 part1(x3x1 minus qx1x3) = 0 and part1(x3x2 minus qx2x3) = 0 These statements follow from the q-Leibniz rule

Therefore parti is a well-defined operator on P qn

Lemma 58 The following relations hold

parti(xki ) =

kminus1sumj=0

qjkminus2jminusj2+kxjix

kminus1minusji+1

parti(xki+1) = minus

kminus1sumj=0

qminusjxjixkminus1minusji+1

Proof We induct on k The base case (k = 1) follows from the definition of the parti and the powers of qarise mostly from xni+1x

mi = qmnxmi x

ni+1 for all mn isin Z+

Our q-divided difference operators also annihilate the twisted elementary q-symmetric polynomialsjust as the even divided difference operators annihilate the elementary symmetric functions

Lemma 59 For every i = 1 nminus 1 and every k parti(ek) = 0 Hence Λqn sube⋂nminus1i=1 ker(parti)

Proof We can express ek as

ek =sum|J|=kii+1isinJ

xJ +sum|J|=kminus1ii+1isinJ

qf(Jik)xJ(xi + qxi+1) +sum|J|=kminus2ii+1isinJ

qg(Jik)xJxixi+1

for certain Z-valued functions f g The result then follows from parti(xi + qxi+1) = parti(xixi+1) = 0 and theq-Leibniz rule

Having discussed q-divided difference operators we can now construct algebras for every q 6=0 1minus1 that have many similarities to the even and odd nilHecke algebras For every such q we de-fine a q-nilHecke algebra generated by xi and parti for 1 le i le n subject to the relations found in thefollowing two lemmas (510 and 511)

Lemma 510 The following relations hold among the operators parti and xi (left multiplication by xi)

1 part2i = 0

2 partjparti minus qpartipartj = 0 for j gt i+ 1

3 xjxi = qxixj for i lt j

4 partixj minus qxjparti = 0 for j gt i+ 1

5 qpartixj minus xjparti = 0 for j lt i

6 partixi minus qxi+1parti = q

7 xiparti minus qpartixi+1 = q

16

Proof To show that part2i = 0 note that we can reduce to i = 1 and proceed by induction Since parti(1) = 0the base case follows Suppose that part2i (f) = 0 Then note that

part21(x1f) = part1(qf + qx2part1(f)) = qpart1(f)minus qpart1(f) + x1part21(f) = 0

part21(x2f) = part1(minusf + qminus1x1part1(f)) = minuspart1(f) + part1(f) + x2part21(f) = 0

part21(x3f) = part1(qx3part1(f)) = q2x3part21(f) = 0

which completes the proof of the first statement in the lemmaStatement 3 follows by definition Statements 4 5 6 and 7 follow from a suitable application of the

q-Leibniz rule Statement 2 follows from an inductive argument We can reduce to i = 1 and j = 3Suppose that partjparti = qpartipartj if j gt i+ 1 Then

part3part1(x1f)minus qpart1part3(x1f) = (qpart3(f) + x2part3part1(f))minus q(part3(f) + x2part1part3(f)) = 0

part3part1(x2f)minus qpart1part3(x2f) = (minuspart3(f) + qminus2x1part3part1(f))minus q(minusqminus1part3(f) + qminus2x1part1part3(f)) = 0

part3part1(x3f)minus qpart1part3(x3f) = (q2part1(f) + q2x4part3part1(f))minus q(qpart1(f) + q2x4part1part3(f)) = 0

part3part1(x4f)minus qpart1part3(x4f) = (minusqpart1(f) + x3part3part1(f))minus q(minuspart1(f) + x3part1part3(f)) = 0

part3part1(x5f)minus qpart1part3(x5f) = q2x5part3part1(f)minus q(q2x5part1part3(f)) = 0

thereby completing the induction

Lemma 511 partiparti+1partiparti+1partiparti+1 + parti+1partiparti+1partiparti+1parti = 0

Proof This result follows from an inductive argument we reduce to i = 1 and assume that the braidrelation holds true for some function f Then we check that the braid relation is true for x1f x2f x3f and x4f (since the behavior of xjf for j ge 4 is the same as that of x4f ) For brevity we will show theargument for x2f only

part1part2(x2f) = qpart1(f) + q2x3part1part2(f) part2part1(x2f) = minuspart2(f) + qminus2x1part2part1(f)

part212(x2f) = qpart2part1(f)minus q2part1part2(f) + qx2part212(f) qpart121(x2f) = minusqpart1part2(f) + part2part1(f) + part121(f)

part1212(x2f) = qpart121(f)minus qpart212(f) + x1part1212(f) part2121(x2f) = minuspart212(f) + part121(f) + x3part2121(f)

part21212(x2f) = qpart2121(f) + qminus1x2part21212(f) part12121(x2f) = minuspart1212(f) + qx3part12121(f)

We continue the above calculations to find that

part121212(x2f) = qpart12121(f) + part21212(f) + x2part121212(f)

part212121(x2f) = minusqpart12121(f)minus part21212(f) + x2part121212(f)

and the braid relation for x2f follows from the inductive hypothesis

6 A Diagrammatic Approach to q-Symmetric Polynomials

61 Introduction to a q-Bialgebra

In the previous section we answered Question 51 in an algebraic way by defining q-analogs of theclassical elementary and complete symmetric functions In this section we generalize the diagrammaticmethod used in [8] in order to study this question from the perspective of bialgebras

Let NΛq be a free associative Z-graded C-algebra with generators hm for m ge 0 We define h0 = 1and hm = 0 for m lt 0 and let q isin Ctimes The homogenous part of NΛq of degree ` has a basis hααkwhere

hα = hα1 middot middot middothαz for a composition α = (α1 αz) of `

17

Define a multiplication for homogenous x and y on NΛqotimes2 as follows where deg(x) denotes the degreeof x

(w otimes x)(y otimes z) = qdeg(x)deg(y)(wy otimes xz)

We can make NΛq into a q-bialgebra by letting the comultiplication on generators be

∆(hn) =

nsumk=0

hk otimes hnminusk

and by letting the counit be ε(x) = 0 if x is homogenous and deg(x)gt 0We can impose through the braiding structure that

∆(hahb) =asumj=0

bsumk=0

(hj otimes haminusj)(hk otimes hbminusk) =asumj=0

bsumk=0

qk(aminusj)(hjhk otimes haminusjhbminusk)

For any partitions λ and micro of n consider the set of double cosets of subroups Sλ and Smicro of Sn SλSnSmicroFor every C in this set let wC be the minimal length representative of C and let `(wC) be the length ofthis minimal length representative We will now attribute a bilinear form to NΛq

(hλ hmicro) =sum

CisinSλSnSmicro

q`(wC)

This bilinear form admits a diagrammatic description Let hn be an orange platform with n non-intersecting strands coming out of it When computing (hλ hmicro) with `(λ) = z and `(micro) = y drawz orange platforms at the top of the diagram representing λ1 λ2middot middot middot λz Draw y orange platforms atthe bottom of the diagram representative of micro1 micro2middot middot middot microy We require that |λ| = |micro| so that the topplatforms and bottom platforms have the same number of strands

Consider the example (h121 h22) In the following diagram snippets of the strands from each plat-form are shown

Every strand must start at one platform at the top and end on another platform at the bottom Nostrands that have originated from one platform may intersect The strands themselves have no criticalpoints with respect to the height function no two strands ever intersect more than once and there are notriple-intersections where three strands are concurrent Diagrams are considered up to isotopy Withoutany restrictions there would be n such diagrams if |λ| = n since there would be no limitations onthe ordering of the strands However due to the above rules there are only 4 possible diagrams in thecomputation of (h121 h22) shown below

Define(hλ hmicro) =

sumall diagrams D representing (hλhmicro)

q number of of crossings in D (61)

In the above example (h121 h22) = 1 + 2q2 + q3

18

We can extend the bilinear form toNΛqotimes2 by stating that any diagram in which strands from distincttensor factors intersect contributes 0 to the bilinear form

(w otimes x y otimes z) = (w y)(x z)

Let I be the radical of the bilinear form inNΛq In [8] the authors prove for any q that multiplicationand comultiplication are adjoint In other words for all xy1 y2 in NΛq

(y1 otimes y2∆(x)) = (y1y2 x) (62)

62 The Elementary q-Symmetric Functions

We now use the bilinear form of q-symmetric functions to study one of their important bases the ele-mentary q-symmetric functions

Define elements ek isin NΛq by ek = 0 for k lt 0 e0 = 1 and

ksumi=0

(minus1)iq(i2)eihkminusi = 0 for k ge 1 (63)

Equivalently leten = qminus(n2)

sumαn

(minus1)`(α)minusnhα (64)

Lemma 61

1 The coproduct of an elementary function is given by ∆(en) =

nsumk=0

ek otimes enminusk

2 If λ n then (hλ en) =

1 if λ = (1 1)

0 otherwise

Proof We begin by demonstrating (2) from which (1) will follow To show (2) it suffices to show that

(hmx en) =

(x enminus1) if m = 10 otherwise

We will utilize strong induction on n in order to find (hmx ekhnminusk) The base cases n = 0 1 are easyto show There are two cases to consider by the inductive hypothesis applied to k lt n Either there isa strand connecting hm and ek or there is not Just as we used an orange platform to denote hn wewill use a blue platform to denote ek The rules of the diagrammatic notation are the same for the blueplatforms as they are for the orange platforms

k

m

nminus k minusm

k nminus k

mlowast lowast lowast lowast lowast

x

If there is not a strand connecting hm and ek the configuration contributes qkm(x ekhnminuskminusm)

19

k minus 1

mminus 1

nminus k minusm + 1

k nminus k

mlowast lowast lowast lowast lowast

x

If a stand connects hm and ek this configuration contributes q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) We have thusshown that (hmx ekhnminusk) = qkm(x ekhnminuskminusm) + q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) Now we are equippedto consider (hmx ek)

(minus1)n+1q(n2)(hmx en) =

nminus1sumk=0

(minus1)kq(k2)(hmx ekhnminusk)

=nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus1sumk=0

(minus1)kq(k2)+(mminus1)(kminus1)(x ekminus1hnminuskminusm+1)

=

nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus2sumk=0

(minus1)k+1q(k+12 )+(mminus1)(k)(x ekhnminuskminusm)

= (minus1)nminus1q(nminus12 )+nm(x enminus1h1minusm)

Corresponding terms from the two sums cancel in pairs since q(k2)+km = q(

k+12 )+k(mminus1) leaving only the

k = nminus 1 term in the first sum The second statement of the lemma thus followsWe will now use (2) to prove (1) This follows from equation 62

(∆(ek) hλ otimes hmicro) = (ek hλhmicro) =

1 λ = (1`) micro = (1p) `+ p = k

0 otherwise

We now calculate the sign incurred when strands connect two blue (ek) platforms

(minus1)n+1q(n2)(en en) =

nminus1sumk=0

(minus1)kq(k2)(en ekhnminusk)

= (minus1)nminus1q(nminus12 )(en enminus1h1)

= (minus1)nminus1q(nminus12 )(∆(en) enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )

nsumk=0

(ek otimes enminusk enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )(enminus1 enminus1)

One may solve this recursion to find that (en en) = qminus(n2)Here the second equality follows from not-ing that at most one strand can connect hnminusk and en (so that k = n minus 1) the third equality followsfrom adjointness and the fourth and fifth equalities follow from the diagrammatic considerations of theprevious lemma

To summarize the diagrammatics of the bilinear form thus developed

20

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

Let [p q] = pq minus qp be the commutator Heckman showed that r2 E and ∆k satisfy the definingrelations of the Lie algebra sl2 [13]

[E r2] = 2r2

[E∆k] = minus2∆k

[r2∆k] = E

Remark 21 If one were to replace ∆k with the classical Laplacian on flat Rn (replacing the Dunkloperator with the partial derivative) these three operators still satisfy the sl2 relations

Remark 22 From now on we will use ηi to denote the odd Dunkl operator of Khongsap and Wangdefined in equation 25

In Section 4 we will focus on finding analogous results in the odd case

22 Introduction to the Odd nilHecke Algebra

We will now discuss operators with the algebra Pminus = C〈x1 xn〉〈xjxi + xixj = 0 for i 6= j〉 We callPminus the skew polynomial ring We can define linear operators called the odd divided difference operatorsas below

Definition 23 For i = 1 nminus1 the i-th odd divided difference operator parti is the linear operator Pminus rarr Pminus

defined by parti(xi) = 1 parti(xi+1) = 1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + (minus1)|f |si(f)parti(g)

for all functions f g isin Pminus We call this last relation the Leibniz rule

It is shown in [9] that the odd divided difference operators can be used to construct an odd nilHeckealgebra generated by xi and parti for 1 le i le n subject to the following relations

1 part2i = 0

2 partipartj + partjparti = 0 for |iminus j| ge 2

3 partiparti+1parti = parti+1partiparti+1

4 xixj + xjxi = 0 for i 6= j

5 xiparti + partixi+1 = 1 partixi + xi+1parti = 1

6 xipartj + partjxi = 0 for i 6= j j + 1

Due to [14] we have the following explicit definition of the odd divided difference operator

parti(f) = (x2i+1 minus x2i )minus1[(xi+1 minus xi)f minus (minus1)|f |si(f)(xi+1 minus xi)] (21)

Although this formula a priori involves denominators it does take skew polynomials to skew polyno-mials We extend this definition to non-consecutive indices by replacing i + 1 with any index k 6= i for1 le k le n and by replacing si with sik Equation 21 then becomes

partik(f) = (x2k minus x2i )minus1[(xk minus xi)f minus (minus1)|f |sik(f)(xk minus xi)] (22)

This extended odd divided difference operator satisfies the Leibniz rule partik(fg) = partik(f)g+(minus1)|f |sikpartik(g)[9]

6

23 Some Operations on Skew Polynomials

First we introduce a common operator in the study of Dunkl operators

Definition 24 Let the (minus1)-shift operator τi be the automorphism of Pminus which sends xi to minusxi and xj toxj for j 6= i

Suppose 1 le i 6= j le n and 1 le k 6= ` le n where f is an element in C〈x1 xn〉〈xixj + xjxi =0 for i 6= j〉 Then one has

sijτk` = τsij(k`)sij

fxi = (minus1)|f |xiτi(f)

Remark 25 Since skew polynomials are not super-commutative we cannot say that fg = (minus1)|f ||g|gf But the operator τi allows us to track the discrepancy from super-commutativity since xif = (minus1)|f |τi(f)ximaking it useful in this context

We now introduce the operator rik = partiksik for k 6= i which will serve as another odd divideddifference operator that we will use to study odd Dunkl operators For simplicity let ri = rii+1 In thefollowing lemma we study the action of the transposition and (minus1)-shift operator on rik

Lemma 26 The operators sik and τi act on rik as follows

rijsk` = sk`rsk`(ij) (23)

We also have that

1 sirik = ri+1ksi if k 6= i+ 1

2 siri = risi

3 siri+1 = rii+2si

4 si+1ri = rii+2si+1

5 sirj = rjsi for |iminus j| ge 2

6 τirj = rjτi for |iminus j| ge 2

Proof Recall the following relationship between partij and sk` for i 6= j and k 6= ` from Lemma 219 (1) of[9]

partijsk` = sk`partsk`(ij) (24)

Multiplying both sides by sij we obtain that

sijpartijsk` = sijsk`partsk`(ij)

= sk`sk`(i j)partsk`(ij)

which implies the desired result since rij = partijsij = sijpartij Properties 1-5 are special cases of equation23 Property 6 follows from τisj = sjτi and the fact that τi(xj) = xj for i 6= j

Remark 27 Differences between our formulas and those of [9] are due to a difference of sign conventionin the action of sij on Pminus

We now show that the properties of the rik are similar to those of the odd divided difference operatorpartik

Lemma 28 The following relations hold

7

1 r2i = 0

2 rirj + rjri = 0 for |iminus j| ge 2

3 riri+1ri = ri+1riri+1

4 rik(fg) = rik(f)sik(g) + (minus1)|f |frik(g)

5 rixi+1 + xi+1ri = rixi + xiri = si

6 rjxi + xjri = 0 for i 6= j j + 1

Proof Since siri = risi and ri = partisi it follows that siparti = partisi Then since part2i = 0 r2i = 0 as well Dueto Equation 5 from Lemma 26 we have that sirj = rjsi for |i minus j| ge 2 so sipartj = partjsi Thus ri andrj anti-commute since partipartj + partjparti = 0 The operators ri also satisfy braid relations which we show byinductively reducing to i = 1 and then using 24 and siparti = partisi repeatedly

r1r2r1 = s1part12part2s1part1 = s1s2part13s1part13part1 = s1s2s1part23part13part12

r2r1r2 = s2part2s1part1s2part2 = s2s1part13s2part13part2 = s2s1s2part12part13part23

Since s1s2s1 = s2s1s2 and part23part13part12 = part12part13part23 by symmetry we conclude that r1r2r1 = r2r1r2 TheLeibniz rule for rik (equation 4 of this lemma) follows immediately from the Leibniz rule for partik Sinceri(xi) = ri(xi+1) = 1 and ri(xj) = 0 for j 6= i i + 1 equations 5 and 6 follow from the Leibniz rule forrik

We also desire an explicit definition of the rik analogous to that of the odd divided difference oper-ator of [9] To find such an expression we use a preparatory lemma

Lemma 29 For all f isinPminus and 1 le i 6= k le n we have

sikxiτi(f)minus sikxkτk(f) = (minus1)|f |sik(f)(xi minus xk)

Proof It suffices to prove the result for a monomial xλ = xλ11 xλii xλkk xλnn where i lt k Wecalculate that

sikxiτi(xλ) = (minus1)λ1++λixλ11 xλi+1

k xλki xλnn

sikxkτk(xλ) = (minus1)λ1++λkminus1xλ11 xλik xλk+1

i xλnn

sik(xλ)xi = (minus1)λk+1++λnxλ11 xλik x

λk+1

i xλnn

sik(xλ)xk = (minus1)λi+1++λnxλ11 x

λi+1

k xλki xλnn

Since |f | = λ1 + + λn the desired result follows

Lemma 210 The operator rik has explicit form rik = (x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Proof Follows from Lemma 29 and Equation 22

We will now connect the above results to the odd Dunkl operator introduced by Khongsap and Wangin [18]

Definition 211 Define an operator δi by δi = (2xi)minus1(1minus τi)

The above super-derivative can also be defined inductively by imposing that δi(xj) = 1 if i = j and0 otherwise We then extend the action to monomials as follows

δi(xa1xa2 xa`) =sumk=1

(minus1)kminus1xa1 δi(xak)xak+1 xa`

8

The operator δi is a priori from Laurent skew polynomials to Laurent skew polynomials but it is easy tocheck that it preserves the subalgebra of skew polynomials Khongsap and Wang found an odd analogof the Dunkl operator given by

ηi = tδi + usumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk] (25)

where t u isin Ctimes Their operators anti-commute ηiηj + ηjηi = 0 for i 6= jBy Lemma 210 this odd Dunkl operator may be expressed as

ηi = tδi + usumk 6=i

partiksik (26)

By analogy with the commutative case discussed in Section 21 the operator rik plays the same role inthe odd theory that the even divided difference operator plays in the even theory

3 Classical Yang-Baxter Equation and the Dunkl Laplacian

Theorem 31 LetH123 = [r12 r13]+ + [r13 r23]+ + [r12 r23]+ (31)

where [p q]+ = pq + qp is the anti-commutator Then the operators rik satisfy the classical Yang-Baxterequation

H123 = 0 (32)

Proof To avoid a cumbersome direct calculation we instead use an inductive approach Namely sup-pose that H123(f) = 0 for some function f isin Pminus Then we show that H123(xif) = 0 for all integersi ge 1 Note that for i = 1

r12r13x1 = r12(s13 minus x1r13)= r12s13 minus r12(x1r13)= r12s13 minus (s12r13 minus x1r12r13)= r12s13 minus s12r13 + x1r12r13

(33)

where we have used Equation 4 twice Similarly we find that

r13r12x1 = r13s12 minus s13r12 + x1r13r12 (34)r23r12x1 = r23s12 + x1r23r12 (35)r12r23x1 = minuss12r23 + x1r12r23 (36)r13r23x1 = minuss13r23 + x1r13r23 (37)r23r13x1 = r23s13 + x1r23r13 (38)

By our inductive hypothesis

x1(r12r13 + r13r12 + r23r12 + r12r23 + r13r23 + r23r13) = 0

Keeping this in mind add Equations 33-38 to show that

H123x1 = r12s13 minus s12r13 + r13s12 minus s13r12 + r23s12 minus s12r23 minus s13r23 + r23s13

= r12s13 minus r23s12 + r13s12 minus r23s13 + r23s12 minus r13s12 minus r12s13 + r23s13

= 0

where we have repeatedly used Lemma 26 to slide rij past sk`We can similarly show that H123x2 = H123x3 = 0 Since rjkxi = xirjk for i gt 3 and j k isin (1 2 3)

it also follows that H123xi = 0 for i gt 3 proving the desired result

9

Corollary 32 The double summationsumn

i=1

(sumk 6=i rik

)2= 0

Proof The expansion of this double summation has n(nminus1)2 total terms Since r2ij = 0 n(nminus1) of theseterms are immediately zero leaving n(n minus 1)(n minus 2) terms of the form rijrk` where i = k and j = `are not both true By Theorem 31 the sum of all six terms of the form rijrk` where i j k ` isin a b c fordistinct integers 1 le a b c le n i 6= j and k 6= ` is zero There are

(n3

)ways to choose integers a b c

and for each choice of a b c six terms of the form rijrk` vanish This fact eliminates all the remaining6(n3

)= n(nminus 1)(nminus 2) terms of the double summation

As an application of the results in this section we will compute the odd Dunkl Laplaciansumn

i=1 η2i

We will first require a lemma involving the commutator of τi and rik

Lemma 33 The equation xminus1i [rik τi] = (x2i minus x2k)minus1(sik(τi + τk)minus xminus1i xksik(τi minus τk)minus 2) holds

Proof By Lemma 210

rikτi = (x2i minus x2k)minus1((xi minus xk)sikτi minus xiτ2i + xkτkτi)

τirik = (x2i minus x2k)minus1(minus(xi + xk)sikτk + xiτ2i + xkτiτk)

Since τiτk = τkτi and τ2i = 1 the result follows by subtraction

Now define Ai = (2xi)minus1(1minus τi) and Bi =

sumk 6=i rik so that the odd Dunkl operator ηi of Khongsap

and Wang may be expressed as ηi = tAi + uBi Note that

A2i =

1

4xminus1i (1minus τi)xminus1i (1minus τi)

=1

2xminus2i (1minus τi)(1minus τi)

= xminus2i (1minus τi)

(39)

since τi(xminus1i ) = minusxminus1i and τ2i = 1

Lemma 34 The relationsumn

i=1(AiBi +BiAi) = 0 holds

Proof Due to the Leibniz Rule for rik (equation 4) we can find that 0 = rik(xixminus1i ) = xminus1k minus xirik(x

minus1i )

so rik(xminus1i ) = xminus1i xminus1k It follows that rik(xminus1i f) = xminus1i xminus1k sik(f)minus xminus1i rik(f) Using this factnsumi=1

BiAi =1

2

nsumi=1

sumk 6=i

rikxminus1i (1minus τi)

=1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi)minus xminus1i rik(1minus τi))

By definitionnsumi=1

AiBi =1

2

nsumi=1

sumk 6=i

xminus1i (1minus τi)rik

Adding the above two equations we find thatnsumi=1

[Ai Bi]+ =1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi) + xminus1i (rikτi minus τirik))

=1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi) + (x2i minus x2k)minus1(sik(τi + τk)minus xminus1i xksik(τi minus τk)minus 2))

(310)

10

where we have used Lemma 33Each double summation repeats the pair of indices (p q) twice when 1 le p q le n one time when

i = p and k = q and once more when i = q and k = p Note that

(x2i minus x2k)minus1sik(τi + τk) = minus(x2k minus x2i )ski(τk + τi)

As a result the sum 12

sumni=1

sumk 6=i(x

2i minus x2k)minus1sik(τi + τk) = 0 Similarly

1

2

nsumi=1

sumk 6=i

xminus1i xminus1k sik = 0 and1

2

nsumi=1

sumk 6=i

(x2i minus x2k)minus1(minus2) = 0

since xminus1i xminus1k + xminus1k xminus1i = 0 Equation 310 then becomes

nsumi=1

(AiBi +BiAi) =1

2

nsumi=1

sumk 6=i

(minusxminus1i xminus1k sikτi minus (x2i minus x2k)minus1(xminus1i xksik(τi minus τk)) (311)

However note thatxminus1i xk + xminus1k xi = minus(x2i minus x2k)xminus1i xminus1k

which implies

(x2i minus x2k)minus1xminus1i xksik(τi minus τk) + (x2k minus x2i )minus1xminus1k xiski(τk minus τi) = minusxminus1i xminus1k sik(τi minus τk) (312)

Similarly we find that

xminus1i xminus1k sikτi + xminus1k xminus1i sikτk = xminus1i xminus1k sik(τi minus τk) (313)

As a result equation 311 becomes

nsumi=1

(AiBi +BiAi) = minus1

2

sum1leiltklen

(xminus1i xminus1k sik(τi minus τk)minus xminus1i sminus1k sik(τi minus τk)) = 0

We are now equipped to compute the Dunkl Laplacian in the odd case

Theorem 35 The equationsumn

i=1 η2i = t2

sum1leilen x

minus2i (1minus τi) holds

Proof Since ηi = tAi + uBi we have that

nsumi=1

η2i = t2nsumi=1

A2i + tu

nsumi=1

(AiBi +BiAi) + u2nsumi=1

B2i

By Lemma 34sumn

i=1(AiBi +BiAi) = 0 By Corollary 32sumn

i=1B2i = 0 Therefore

nsumi=1

η2i = t2nsumi=1

A2i = t2

sum1leilen

xminus2i (1minus τi)

by equation 39

11

4 A Variant of the Khongsap-Wang Odd Dunkl Operator

In this section we will show that a close variant of the odd Dunkl operator introduced by Khongsapand Wang can be used in the construction of three operators that satisfy the defining relations of the Liealgebra sl2 First we will consider an operator pi which is different from δi but plays a similar role

Definition 41 The operator pi is a C-linear map Pminus rarr Pminus which acts on monomials as follows

pi(xλ11 xλii xλnn ) = λi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn

Remark 42 One may also introduce pi by using a nice Leibniz-like expression involving τi by defining

pi(xj) = δij

pi(fg) = pi(f)g + (minus1)|fτi(f)pi(g)

where f g isin Pminus and δij is the Kronecker delta Now note the analogous relationship between thedegree-preserving operators si and τi in their respective Leibniz rules for the (minus1)-degree operators partiand pi This provides motivation for the definition of pi and suggests its natural role in our theory

Now consider a modified version of ηi

Definition 43 LetDi = tpi + u

sumk 6=i

rik (41)

Definition 44 Introduce the odd r2 Euler and ∆ operators as below

r2 = (2t)minus1nsumi=1

x2i (42)

E =nsumi=1

xipi +n

2+u

t

sumk 6=i

sik (43)

∆ = minus(2t)minus1nsumi=1

D2i (44)

Remark 45 Heckman who used the even Dunkl operators to find a sl2-triple useful in harmonic anal-ysis uses the convention t = 1 [13] For now we will consider t to be a fixed constant in Ctimes

Remark 46 The commutator in the setting of superalgebras is usually defined as [a b] = abminus(minus1)|a||b|bawhere |a| and |b| are the degrees of a and b respectively However since all of the operators we will beconsidering in this section have even degree there is no need to distinguish between commutators andsuper-commutators

To construct an sl2 action from these operators we will require a series of lemmas regarding theaction of portions of the odd Euler operator E In the next lemma we investigate the action of the firstterm of the odd Euler operator on skew polynomials

Lemma 47 The operatorsumn

i=1 xipi acts by multiplication by |f | on the space of homogenous functionsf isinPminus

Proof It suffices to show the result for a monomial xλ = xλ11 xλii xλnn Note that

xipi(xλ) = λixi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn = λix

λ

12

By summing over all indices i we obtain that

nsumi=1

xipi(xλ) = (λ1 + λ2 + + λn)xλ

which implies the desired result

The above lemma holds true in the even case as well where pi is replaced by the partial derivativewith respect to xi We now prove some properties about the action of the third term of the odd Euleroperator on r2 and ∆

Lemma 48 The commutation relation[sum

k 6=i sik∆]

= 0 holds

Proof Note that sjkpi = pjsjk if i = k sjkpi = pksjk if i = j and sjkpi = pisjk otherwise Indeedthese relations can be verified by checking if they are true for xai x

bjxck a b c isin Z+ and then extending

by linearity We prove that sjkpi = pjsjk if i = k and the other two cases are similar Without loss ofgenerality let j lt k and observe that

sjkpk(xajx

bk) = b(minus1)asjk(x

ajx

bminus1k ) = b(minus1)axakx

bminus1j = b(minus1)abxbminus1j xak

pjsjk(xajx

bk) = (minus1)abpj(x

bjxak) = b(minus1)abxbminus1j xak

By our work in Lemma 26 one can deduce that sjkr`m = rsjk(`m)sij As a consequence we find thatsjkDi = Dsjk(i)sjk By an easy induction we now have that sjk∆ = ∆sjk Using the above equationmultiple times proves the desired result

Lemma 49 The commutation relation[sum

k 6=i sik r2]

= 0 holds

Proof Follows since sjkxj = xksjk sjkxk = xjsjk and sjkxi = xisjk if i 6= j k

We are now ready to obtain two commutativity relations involving the odd Euler operator E

Theorem 410 The odd Euler operator and r2 satisfy the following commutation relations

[E r2] = 2r2 (45)[E∆] = minus2∆ (46)

Proof Since r2 has degree 2 and ∆ has degreeminus2 the theorem follows from Lemmas 47 48 and 49

We also need to investigate what the third commutativity relation [r2∆] turns out to be We willprove one lemma before doing so

Lemma 411 For i = 1 to n the equation xiDi +Dixi = 2txipi + t+ usum

k 6=i sik holds

Proof Recall thatDi = tpi + u

sumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Therefore since pixi = xipi + 1

Dixi = txipi + t+ usumk 6=i

(x2i minus x2k)minus1(xixk minus x2k)sik +sumk 6=i

(x2i minus x2k)[x2i τi minus xixkτk]

xiDi = txipi + usumk 6=i

(x2i minus x2k)minus1(x2i minus xixk)sik +sumk 6=i

(x2i minus x2k)minus1[minusx2i τi + xixkτk]

Adding we obtain the desired result

13

We now have the tools to find the third relation between r2 E and ∆

Theorem 412 The commutation relation [r2∆] = E holds

Proof We will first find [r2 Di] The derivative pi much like the partial derivative in the even casesatisfies the properties pixj = minusxjpi for i 6= j and pixi = xipi + 1 Now suppose that i 6= j Then

Dix2j = tpix

2j + ux2j

sumk 6=i 6=j

rik + x2k(x2i minus x2k)minus1[minusxiτi + xkτk] + x2i (x

2i minus x2k)minus1[(xi minus xk)sik]

= tx2jpi + ux2jsumk 6=i

rik + (xi minus xj)sij

Now we will find Dix2i

Dix2i = tpix

2i +

sumk 6=i

x2k(x2i minus x2k)minus1(xi minus xk)sik + x2i

sumk 6=i

(x2i minus x2k)minus1[minusxiτi + xkτk]

= tx2i pi + 2txi + x2isumk 6=i

rik minussumk 6=i

(xi minus xk)sik

Therefore[sumn

i=1 x2i Di

]= minus2txi This implies that

r2Di minusDir2 = minusxi (47)

As a result we find that

[r2∆] = minus(2t)minus1nsumi=1

[r2 D2j ] = minus(2t)minus1

nsumi=1

(r2D2j minusD2

j r2)

= minus(2t)minus1nsumi=1

[(Djr2Dj minus xjDj)minus (Djr

2Dj +Djxj)]

= (2t)minus1nsumi=1

(xiDi +Dixi)

where we have used 47 Now by Lemma 411

[r2∆] =

nsumi=1

xipi +n

2+u

t

sumk 6=i

sik = E

as desired

To summarize we have found operators Er2 and ∆ similar to their even counterparts whichsatisfy the defining relations of the Lie algebra sl2

[E r2] = 2r2

[E∆] = minus2∆

[r2∆] = E

Remark 413 If one uses the odd Dunkl operator ηi as found in [18] instead of the Di introduced herethe r2 E and ∆ operators do not generate sl2

Remark 414 Although our results hold true for all t and u in C one typically sets t = 1 and u = αminus1

for some α isin Ctimes since without loss of generality one of t and u may equal 1

14

Remark 415 In the even case let X be a Euclidean vector space with dimension n and let C[X] be thealgebra of C-valued functions on X Then this result about sl2 plays a major role in the study of higherdifferential operators on C[X] This is because the representation theory of sl2 allows for the reductionof degree to the second order [13] As a result our results in this section should correspondingly have arole in further studying differential operators in the odd case

5 q-nilHecke Algebras

Until now we have been concerned with the odd symmetric polynomials in variables x1 x2 xnwhere xixj = (minus1)xjxi for 1 le i 6= j le n This immediately suggests the question what if one replacesthe minus1 by any constant q isin Ctimes Specifically we ask the following questions

Question 51 Is it possible to study q-symmetric polynomials for which xixj = qxjxi when i gt j

Question 52 Are there q-analogs of evenodd divided difference operators and nilHecke algebras Sofar such structures are known only for the even case (q = 1) and the odd case (q = minus1)

In this section we answer both questions in the affirmativeWe work in the Z-graded q-braided setting throughout LetC be a commutative ring and let q isin Ctimes

be a unit If VW are graded C-modules and v isin V w isin W are homogeneous the braiding is theldquoq-twistrdquo

τq V otimesW rarrW otimes Vv otimes w 7rarr q|v||w|w otimes v

(51)

where | middot | is the degree function By q-algebra we mean an algebra object in the category of gradedC-modules equipped with this braided monoidal structure likewise for q-bialgebras q-Hopf algebrasand so forth

Remark 53 Note that the q-twist described above is its own inverse only when q2 = 1 which correlatesto the even and odd cases When q2 6= 1 the corresponding theory becomes more complex There-fore the q-nilHecke algebras that we introduce later in this section are nontrivial generalizations of thepreviously studied even and odd nilHecke algebras

Definition 54 The q-algebra P qn is defined to be

P qn = C〈x1 xn〉(xjxi minus qxixj = 0 if i lt j) (52)

where |xi| = 1 for i = 1 n

Note that P qn sim= otimesni=1Pq1 There are two interesting subalgebras of P qn that can be thought of as q-

analogs of the symmetric polynomials Define the k-th elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middotxin

and define the k-th twisted elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middot xin

where xj = qjminus1xj

Definition 55 The q-algebra of q-symmetric polynomials in n variables denoted Λqn is the subalgebra ofP qn generated by e1 en Likewise for the twisted q-symmetric polynomials Λqn and e1 en

The type A braid group on n strands acts on P qn by setting

15

1 σi(xj) = qxi+1 if j = i

2 σi(xj) = qminus1xi if j = i+ 1

3 σi(xj) = qxj if j gt i+ 1

4 σi(xj) = qminus1xj if j lt i

and extending multiplicatively

Definition 56 For i = 1 nminus 1 the i-th q-divided difference operator parti is the linear operator P qn rarr P qndefined by parti(xi) = q parti(xi+1) = minus1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + σi(f)parti(g) (53)

for all functions f g isin P qn We call Equation 53 the q-Leibniz rule

Lemma 57 For every i and every j lt k parti(xkxj minus qxjxk) = 0

Proof Since parti(xj) = 0 for j gt i+1 one may reduce the lemma to having to prove that part1(x2x1minusqx1x2) =0 part1(x3x1 minus qx1x3) = 0 and part1(x3x2 minus qx2x3) = 0 These statements follow from the q-Leibniz rule

Therefore parti is a well-defined operator on P qn

Lemma 58 The following relations hold

parti(xki ) =

kminus1sumj=0

qjkminus2jminusj2+kxjix

kminus1minusji+1

parti(xki+1) = minus

kminus1sumj=0

qminusjxjixkminus1minusji+1

Proof We induct on k The base case (k = 1) follows from the definition of the parti and the powers of qarise mostly from xni+1x

mi = qmnxmi x

ni+1 for all mn isin Z+

Our q-divided difference operators also annihilate the twisted elementary q-symmetric polynomialsjust as the even divided difference operators annihilate the elementary symmetric functions

Lemma 59 For every i = 1 nminus 1 and every k parti(ek) = 0 Hence Λqn sube⋂nminus1i=1 ker(parti)

Proof We can express ek as

ek =sum|J|=kii+1isinJ

xJ +sum|J|=kminus1ii+1isinJ

qf(Jik)xJ(xi + qxi+1) +sum|J|=kminus2ii+1isinJ

qg(Jik)xJxixi+1

for certain Z-valued functions f g The result then follows from parti(xi + qxi+1) = parti(xixi+1) = 0 and theq-Leibniz rule

Having discussed q-divided difference operators we can now construct algebras for every q 6=0 1minus1 that have many similarities to the even and odd nilHecke algebras For every such q we de-fine a q-nilHecke algebra generated by xi and parti for 1 le i le n subject to the relations found in thefollowing two lemmas (510 and 511)

Lemma 510 The following relations hold among the operators parti and xi (left multiplication by xi)

1 part2i = 0

2 partjparti minus qpartipartj = 0 for j gt i+ 1

3 xjxi = qxixj for i lt j

4 partixj minus qxjparti = 0 for j gt i+ 1

5 qpartixj minus xjparti = 0 for j lt i

6 partixi minus qxi+1parti = q

7 xiparti minus qpartixi+1 = q

16

Proof To show that part2i = 0 note that we can reduce to i = 1 and proceed by induction Since parti(1) = 0the base case follows Suppose that part2i (f) = 0 Then note that

part21(x1f) = part1(qf + qx2part1(f)) = qpart1(f)minus qpart1(f) + x1part21(f) = 0

part21(x2f) = part1(minusf + qminus1x1part1(f)) = minuspart1(f) + part1(f) + x2part21(f) = 0

part21(x3f) = part1(qx3part1(f)) = q2x3part21(f) = 0

which completes the proof of the first statement in the lemmaStatement 3 follows by definition Statements 4 5 6 and 7 follow from a suitable application of the

q-Leibniz rule Statement 2 follows from an inductive argument We can reduce to i = 1 and j = 3Suppose that partjparti = qpartipartj if j gt i+ 1 Then

part3part1(x1f)minus qpart1part3(x1f) = (qpart3(f) + x2part3part1(f))minus q(part3(f) + x2part1part3(f)) = 0

part3part1(x2f)minus qpart1part3(x2f) = (minuspart3(f) + qminus2x1part3part1(f))minus q(minusqminus1part3(f) + qminus2x1part1part3(f)) = 0

part3part1(x3f)minus qpart1part3(x3f) = (q2part1(f) + q2x4part3part1(f))minus q(qpart1(f) + q2x4part1part3(f)) = 0

part3part1(x4f)minus qpart1part3(x4f) = (minusqpart1(f) + x3part3part1(f))minus q(minuspart1(f) + x3part1part3(f)) = 0

part3part1(x5f)minus qpart1part3(x5f) = q2x5part3part1(f)minus q(q2x5part1part3(f)) = 0

thereby completing the induction

Lemma 511 partiparti+1partiparti+1partiparti+1 + parti+1partiparti+1partiparti+1parti = 0

Proof This result follows from an inductive argument we reduce to i = 1 and assume that the braidrelation holds true for some function f Then we check that the braid relation is true for x1f x2f x3f and x4f (since the behavior of xjf for j ge 4 is the same as that of x4f ) For brevity we will show theargument for x2f only

part1part2(x2f) = qpart1(f) + q2x3part1part2(f) part2part1(x2f) = minuspart2(f) + qminus2x1part2part1(f)

part212(x2f) = qpart2part1(f)minus q2part1part2(f) + qx2part212(f) qpart121(x2f) = minusqpart1part2(f) + part2part1(f) + part121(f)

part1212(x2f) = qpart121(f)minus qpart212(f) + x1part1212(f) part2121(x2f) = minuspart212(f) + part121(f) + x3part2121(f)

part21212(x2f) = qpart2121(f) + qminus1x2part21212(f) part12121(x2f) = minuspart1212(f) + qx3part12121(f)

We continue the above calculations to find that

part121212(x2f) = qpart12121(f) + part21212(f) + x2part121212(f)

part212121(x2f) = minusqpart12121(f)minus part21212(f) + x2part121212(f)

and the braid relation for x2f follows from the inductive hypothesis

6 A Diagrammatic Approach to q-Symmetric Polynomials

61 Introduction to a q-Bialgebra

In the previous section we answered Question 51 in an algebraic way by defining q-analogs of theclassical elementary and complete symmetric functions In this section we generalize the diagrammaticmethod used in [8] in order to study this question from the perspective of bialgebras

Let NΛq be a free associative Z-graded C-algebra with generators hm for m ge 0 We define h0 = 1and hm = 0 for m lt 0 and let q isin Ctimes The homogenous part of NΛq of degree ` has a basis hααkwhere

hα = hα1 middot middot middothαz for a composition α = (α1 αz) of `

17

Define a multiplication for homogenous x and y on NΛqotimes2 as follows where deg(x) denotes the degreeof x

(w otimes x)(y otimes z) = qdeg(x)deg(y)(wy otimes xz)

We can make NΛq into a q-bialgebra by letting the comultiplication on generators be

∆(hn) =

nsumk=0

hk otimes hnminusk

and by letting the counit be ε(x) = 0 if x is homogenous and deg(x)gt 0We can impose through the braiding structure that

∆(hahb) =asumj=0

bsumk=0

(hj otimes haminusj)(hk otimes hbminusk) =asumj=0

bsumk=0

qk(aminusj)(hjhk otimes haminusjhbminusk)

For any partitions λ and micro of n consider the set of double cosets of subroups Sλ and Smicro of Sn SλSnSmicroFor every C in this set let wC be the minimal length representative of C and let `(wC) be the length ofthis minimal length representative We will now attribute a bilinear form to NΛq

(hλ hmicro) =sum

CisinSλSnSmicro

q`(wC)

This bilinear form admits a diagrammatic description Let hn be an orange platform with n non-intersecting strands coming out of it When computing (hλ hmicro) with `(λ) = z and `(micro) = y drawz orange platforms at the top of the diagram representing λ1 λ2middot middot middot λz Draw y orange platforms atthe bottom of the diagram representative of micro1 micro2middot middot middot microy We require that |λ| = |micro| so that the topplatforms and bottom platforms have the same number of strands

Consider the example (h121 h22) In the following diagram snippets of the strands from each plat-form are shown

Every strand must start at one platform at the top and end on another platform at the bottom Nostrands that have originated from one platform may intersect The strands themselves have no criticalpoints with respect to the height function no two strands ever intersect more than once and there are notriple-intersections where three strands are concurrent Diagrams are considered up to isotopy Withoutany restrictions there would be n such diagrams if |λ| = n since there would be no limitations onthe ordering of the strands However due to the above rules there are only 4 possible diagrams in thecomputation of (h121 h22) shown below

Define(hλ hmicro) =

sumall diagrams D representing (hλhmicro)

q number of of crossings in D (61)

In the above example (h121 h22) = 1 + 2q2 + q3

18

We can extend the bilinear form toNΛqotimes2 by stating that any diagram in which strands from distincttensor factors intersect contributes 0 to the bilinear form

(w otimes x y otimes z) = (w y)(x z)

Let I be the radical of the bilinear form inNΛq In [8] the authors prove for any q that multiplicationand comultiplication are adjoint In other words for all xy1 y2 in NΛq

(y1 otimes y2∆(x)) = (y1y2 x) (62)

62 The Elementary q-Symmetric Functions

We now use the bilinear form of q-symmetric functions to study one of their important bases the ele-mentary q-symmetric functions

Define elements ek isin NΛq by ek = 0 for k lt 0 e0 = 1 and

ksumi=0

(minus1)iq(i2)eihkminusi = 0 for k ge 1 (63)

Equivalently leten = qminus(n2)

sumαn

(minus1)`(α)minusnhα (64)

Lemma 61

1 The coproduct of an elementary function is given by ∆(en) =

nsumk=0

ek otimes enminusk

2 If λ n then (hλ en) =

1 if λ = (1 1)

0 otherwise

Proof We begin by demonstrating (2) from which (1) will follow To show (2) it suffices to show that

(hmx en) =

(x enminus1) if m = 10 otherwise

We will utilize strong induction on n in order to find (hmx ekhnminusk) The base cases n = 0 1 are easyto show There are two cases to consider by the inductive hypothesis applied to k lt n Either there isa strand connecting hm and ek or there is not Just as we used an orange platform to denote hn wewill use a blue platform to denote ek The rules of the diagrammatic notation are the same for the blueplatforms as they are for the orange platforms

k

m

nminus k minusm

k nminus k

mlowast lowast lowast lowast lowast

x

If there is not a strand connecting hm and ek the configuration contributes qkm(x ekhnminuskminusm)

19

k minus 1

mminus 1

nminus k minusm + 1

k nminus k

mlowast lowast lowast lowast lowast

x

If a stand connects hm and ek this configuration contributes q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) We have thusshown that (hmx ekhnminusk) = qkm(x ekhnminuskminusm) + q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) Now we are equippedto consider (hmx ek)

(minus1)n+1q(n2)(hmx en) =

nminus1sumk=0

(minus1)kq(k2)(hmx ekhnminusk)

=nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus1sumk=0

(minus1)kq(k2)+(mminus1)(kminus1)(x ekminus1hnminuskminusm+1)

=

nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus2sumk=0

(minus1)k+1q(k+12 )+(mminus1)(k)(x ekhnminuskminusm)

= (minus1)nminus1q(nminus12 )+nm(x enminus1h1minusm)

Corresponding terms from the two sums cancel in pairs since q(k2)+km = q(

k+12 )+k(mminus1) leaving only the

k = nminus 1 term in the first sum The second statement of the lemma thus followsWe will now use (2) to prove (1) This follows from equation 62

(∆(ek) hλ otimes hmicro) = (ek hλhmicro) =

1 λ = (1`) micro = (1p) `+ p = k

0 otherwise

We now calculate the sign incurred when strands connect two blue (ek) platforms

(minus1)n+1q(n2)(en en) =

nminus1sumk=0

(minus1)kq(k2)(en ekhnminusk)

= (minus1)nminus1q(nminus12 )(en enminus1h1)

= (minus1)nminus1q(nminus12 )(∆(en) enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )

nsumk=0

(ek otimes enminusk enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )(enminus1 enminus1)

One may solve this recursion to find that (en en) = qminus(n2)Here the second equality follows from not-ing that at most one strand can connect hnminusk and en (so that k = n minus 1) the third equality followsfrom adjointness and the fourth and fifth equalities follow from the diagrammatic considerations of theprevious lemma

To summarize the diagrammatics of the bilinear form thus developed

20

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

23 Some Operations on Skew Polynomials

First we introduce a common operator in the study of Dunkl operators

Definition 24 Let the (minus1)-shift operator τi be the automorphism of Pminus which sends xi to minusxi and xj toxj for j 6= i

Suppose 1 le i 6= j le n and 1 le k 6= ` le n where f is an element in C〈x1 xn〉〈xixj + xjxi =0 for i 6= j〉 Then one has

sijτk` = τsij(k`)sij

fxi = (minus1)|f |xiτi(f)

Remark 25 Since skew polynomials are not super-commutative we cannot say that fg = (minus1)|f ||g|gf But the operator τi allows us to track the discrepancy from super-commutativity since xif = (minus1)|f |τi(f)ximaking it useful in this context

We now introduce the operator rik = partiksik for k 6= i which will serve as another odd divideddifference operator that we will use to study odd Dunkl operators For simplicity let ri = rii+1 In thefollowing lemma we study the action of the transposition and (minus1)-shift operator on rik

Lemma 26 The operators sik and τi act on rik as follows

rijsk` = sk`rsk`(ij) (23)

We also have that

1 sirik = ri+1ksi if k 6= i+ 1

2 siri = risi

3 siri+1 = rii+2si

4 si+1ri = rii+2si+1

5 sirj = rjsi for |iminus j| ge 2

6 τirj = rjτi for |iminus j| ge 2

Proof Recall the following relationship between partij and sk` for i 6= j and k 6= ` from Lemma 219 (1) of[9]

partijsk` = sk`partsk`(ij) (24)

Multiplying both sides by sij we obtain that

sijpartijsk` = sijsk`partsk`(ij)

= sk`sk`(i j)partsk`(ij)

which implies the desired result since rij = partijsij = sijpartij Properties 1-5 are special cases of equation23 Property 6 follows from τisj = sjτi and the fact that τi(xj) = xj for i 6= j

Remark 27 Differences between our formulas and those of [9] are due to a difference of sign conventionin the action of sij on Pminus

We now show that the properties of the rik are similar to those of the odd divided difference operatorpartik

Lemma 28 The following relations hold

7

1 r2i = 0

2 rirj + rjri = 0 for |iminus j| ge 2

3 riri+1ri = ri+1riri+1

4 rik(fg) = rik(f)sik(g) + (minus1)|f |frik(g)

5 rixi+1 + xi+1ri = rixi + xiri = si

6 rjxi + xjri = 0 for i 6= j j + 1

Proof Since siri = risi and ri = partisi it follows that siparti = partisi Then since part2i = 0 r2i = 0 as well Dueto Equation 5 from Lemma 26 we have that sirj = rjsi for |i minus j| ge 2 so sipartj = partjsi Thus ri andrj anti-commute since partipartj + partjparti = 0 The operators ri also satisfy braid relations which we show byinductively reducing to i = 1 and then using 24 and siparti = partisi repeatedly

r1r2r1 = s1part12part2s1part1 = s1s2part13s1part13part1 = s1s2s1part23part13part12

r2r1r2 = s2part2s1part1s2part2 = s2s1part13s2part13part2 = s2s1s2part12part13part23

Since s1s2s1 = s2s1s2 and part23part13part12 = part12part13part23 by symmetry we conclude that r1r2r1 = r2r1r2 TheLeibniz rule for rik (equation 4 of this lemma) follows immediately from the Leibniz rule for partik Sinceri(xi) = ri(xi+1) = 1 and ri(xj) = 0 for j 6= i i + 1 equations 5 and 6 follow from the Leibniz rule forrik

We also desire an explicit definition of the rik analogous to that of the odd divided difference oper-ator of [9] To find such an expression we use a preparatory lemma

Lemma 29 For all f isinPminus and 1 le i 6= k le n we have

sikxiτi(f)minus sikxkτk(f) = (minus1)|f |sik(f)(xi minus xk)

Proof It suffices to prove the result for a monomial xλ = xλ11 xλii xλkk xλnn where i lt k Wecalculate that

sikxiτi(xλ) = (minus1)λ1++λixλ11 xλi+1

k xλki xλnn

sikxkτk(xλ) = (minus1)λ1++λkminus1xλ11 xλik xλk+1

i xλnn

sik(xλ)xi = (minus1)λk+1++λnxλ11 xλik x

λk+1

i xλnn

sik(xλ)xk = (minus1)λi+1++λnxλ11 x

λi+1

k xλki xλnn

Since |f | = λ1 + + λn the desired result follows

Lemma 210 The operator rik has explicit form rik = (x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Proof Follows from Lemma 29 and Equation 22

We will now connect the above results to the odd Dunkl operator introduced by Khongsap and Wangin [18]

Definition 211 Define an operator δi by δi = (2xi)minus1(1minus τi)

The above super-derivative can also be defined inductively by imposing that δi(xj) = 1 if i = j and0 otherwise We then extend the action to monomials as follows

δi(xa1xa2 xa`) =sumk=1

(minus1)kminus1xa1 δi(xak)xak+1 xa`

8

The operator δi is a priori from Laurent skew polynomials to Laurent skew polynomials but it is easy tocheck that it preserves the subalgebra of skew polynomials Khongsap and Wang found an odd analogof the Dunkl operator given by

ηi = tδi + usumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk] (25)

where t u isin Ctimes Their operators anti-commute ηiηj + ηjηi = 0 for i 6= jBy Lemma 210 this odd Dunkl operator may be expressed as

ηi = tδi + usumk 6=i

partiksik (26)

By analogy with the commutative case discussed in Section 21 the operator rik plays the same role inthe odd theory that the even divided difference operator plays in the even theory

3 Classical Yang-Baxter Equation and the Dunkl Laplacian

Theorem 31 LetH123 = [r12 r13]+ + [r13 r23]+ + [r12 r23]+ (31)

where [p q]+ = pq + qp is the anti-commutator Then the operators rik satisfy the classical Yang-Baxterequation

H123 = 0 (32)

Proof To avoid a cumbersome direct calculation we instead use an inductive approach Namely sup-pose that H123(f) = 0 for some function f isin Pminus Then we show that H123(xif) = 0 for all integersi ge 1 Note that for i = 1

r12r13x1 = r12(s13 minus x1r13)= r12s13 minus r12(x1r13)= r12s13 minus (s12r13 minus x1r12r13)= r12s13 minus s12r13 + x1r12r13

(33)

where we have used Equation 4 twice Similarly we find that

r13r12x1 = r13s12 minus s13r12 + x1r13r12 (34)r23r12x1 = r23s12 + x1r23r12 (35)r12r23x1 = minuss12r23 + x1r12r23 (36)r13r23x1 = minuss13r23 + x1r13r23 (37)r23r13x1 = r23s13 + x1r23r13 (38)

By our inductive hypothesis

x1(r12r13 + r13r12 + r23r12 + r12r23 + r13r23 + r23r13) = 0

Keeping this in mind add Equations 33-38 to show that

H123x1 = r12s13 minus s12r13 + r13s12 minus s13r12 + r23s12 minus s12r23 minus s13r23 + r23s13

= r12s13 minus r23s12 + r13s12 minus r23s13 + r23s12 minus r13s12 minus r12s13 + r23s13

= 0

where we have repeatedly used Lemma 26 to slide rij past sk`We can similarly show that H123x2 = H123x3 = 0 Since rjkxi = xirjk for i gt 3 and j k isin (1 2 3)

it also follows that H123xi = 0 for i gt 3 proving the desired result

9

Corollary 32 The double summationsumn

i=1

(sumk 6=i rik

)2= 0

Proof The expansion of this double summation has n(nminus1)2 total terms Since r2ij = 0 n(nminus1) of theseterms are immediately zero leaving n(n minus 1)(n minus 2) terms of the form rijrk` where i = k and j = `are not both true By Theorem 31 the sum of all six terms of the form rijrk` where i j k ` isin a b c fordistinct integers 1 le a b c le n i 6= j and k 6= ` is zero There are

(n3

)ways to choose integers a b c

and for each choice of a b c six terms of the form rijrk` vanish This fact eliminates all the remaining6(n3

)= n(nminus 1)(nminus 2) terms of the double summation

As an application of the results in this section we will compute the odd Dunkl Laplaciansumn

i=1 η2i

We will first require a lemma involving the commutator of τi and rik

Lemma 33 The equation xminus1i [rik τi] = (x2i minus x2k)minus1(sik(τi + τk)minus xminus1i xksik(τi minus τk)minus 2) holds

Proof By Lemma 210

rikτi = (x2i minus x2k)minus1((xi minus xk)sikτi minus xiτ2i + xkτkτi)

τirik = (x2i minus x2k)minus1(minus(xi + xk)sikτk + xiτ2i + xkτiτk)

Since τiτk = τkτi and τ2i = 1 the result follows by subtraction

Now define Ai = (2xi)minus1(1minus τi) and Bi =

sumk 6=i rik so that the odd Dunkl operator ηi of Khongsap

and Wang may be expressed as ηi = tAi + uBi Note that

A2i =

1

4xminus1i (1minus τi)xminus1i (1minus τi)

=1

2xminus2i (1minus τi)(1minus τi)

= xminus2i (1minus τi)

(39)

since τi(xminus1i ) = minusxminus1i and τ2i = 1

Lemma 34 The relationsumn

i=1(AiBi +BiAi) = 0 holds

Proof Due to the Leibniz Rule for rik (equation 4) we can find that 0 = rik(xixminus1i ) = xminus1k minus xirik(x

minus1i )

so rik(xminus1i ) = xminus1i xminus1k It follows that rik(xminus1i f) = xminus1i xminus1k sik(f)minus xminus1i rik(f) Using this factnsumi=1

BiAi =1

2

nsumi=1

sumk 6=i

rikxminus1i (1minus τi)

=1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi)minus xminus1i rik(1minus τi))

By definitionnsumi=1

AiBi =1

2

nsumi=1

sumk 6=i

xminus1i (1minus τi)rik

Adding the above two equations we find thatnsumi=1

[Ai Bi]+ =1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi) + xminus1i (rikτi minus τirik))

=1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi) + (x2i minus x2k)minus1(sik(τi + τk)minus xminus1i xksik(τi minus τk)minus 2))

(310)

10

where we have used Lemma 33Each double summation repeats the pair of indices (p q) twice when 1 le p q le n one time when

i = p and k = q and once more when i = q and k = p Note that

(x2i minus x2k)minus1sik(τi + τk) = minus(x2k minus x2i )ski(τk + τi)

As a result the sum 12

sumni=1

sumk 6=i(x

2i minus x2k)minus1sik(τi + τk) = 0 Similarly

1

2

nsumi=1

sumk 6=i

xminus1i xminus1k sik = 0 and1

2

nsumi=1

sumk 6=i

(x2i minus x2k)minus1(minus2) = 0

since xminus1i xminus1k + xminus1k xminus1i = 0 Equation 310 then becomes

nsumi=1

(AiBi +BiAi) =1

2

nsumi=1

sumk 6=i

(minusxminus1i xminus1k sikτi minus (x2i minus x2k)minus1(xminus1i xksik(τi minus τk)) (311)

However note thatxminus1i xk + xminus1k xi = minus(x2i minus x2k)xminus1i xminus1k

which implies

(x2i minus x2k)minus1xminus1i xksik(τi minus τk) + (x2k minus x2i )minus1xminus1k xiski(τk minus τi) = minusxminus1i xminus1k sik(τi minus τk) (312)

Similarly we find that

xminus1i xminus1k sikτi + xminus1k xminus1i sikτk = xminus1i xminus1k sik(τi minus τk) (313)

As a result equation 311 becomes

nsumi=1

(AiBi +BiAi) = minus1

2

sum1leiltklen

(xminus1i xminus1k sik(τi minus τk)minus xminus1i sminus1k sik(τi minus τk)) = 0

We are now equipped to compute the Dunkl Laplacian in the odd case

Theorem 35 The equationsumn

i=1 η2i = t2

sum1leilen x

minus2i (1minus τi) holds

Proof Since ηi = tAi + uBi we have that

nsumi=1

η2i = t2nsumi=1

A2i + tu

nsumi=1

(AiBi +BiAi) + u2nsumi=1

B2i

By Lemma 34sumn

i=1(AiBi +BiAi) = 0 By Corollary 32sumn

i=1B2i = 0 Therefore

nsumi=1

η2i = t2nsumi=1

A2i = t2

sum1leilen

xminus2i (1minus τi)

by equation 39

11

4 A Variant of the Khongsap-Wang Odd Dunkl Operator

In this section we will show that a close variant of the odd Dunkl operator introduced by Khongsapand Wang can be used in the construction of three operators that satisfy the defining relations of the Liealgebra sl2 First we will consider an operator pi which is different from δi but plays a similar role

Definition 41 The operator pi is a C-linear map Pminus rarr Pminus which acts on monomials as follows

pi(xλ11 xλii xλnn ) = λi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn

Remark 42 One may also introduce pi by using a nice Leibniz-like expression involving τi by defining

pi(xj) = δij

pi(fg) = pi(f)g + (minus1)|fτi(f)pi(g)

where f g isin Pminus and δij is the Kronecker delta Now note the analogous relationship between thedegree-preserving operators si and τi in their respective Leibniz rules for the (minus1)-degree operators partiand pi This provides motivation for the definition of pi and suggests its natural role in our theory

Now consider a modified version of ηi

Definition 43 LetDi = tpi + u

sumk 6=i

rik (41)

Definition 44 Introduce the odd r2 Euler and ∆ operators as below

r2 = (2t)minus1nsumi=1

x2i (42)

E =nsumi=1

xipi +n

2+u

t

sumk 6=i

sik (43)

∆ = minus(2t)minus1nsumi=1

D2i (44)

Remark 45 Heckman who used the even Dunkl operators to find a sl2-triple useful in harmonic anal-ysis uses the convention t = 1 [13] For now we will consider t to be a fixed constant in Ctimes

Remark 46 The commutator in the setting of superalgebras is usually defined as [a b] = abminus(minus1)|a||b|bawhere |a| and |b| are the degrees of a and b respectively However since all of the operators we will beconsidering in this section have even degree there is no need to distinguish between commutators andsuper-commutators

To construct an sl2 action from these operators we will require a series of lemmas regarding theaction of portions of the odd Euler operator E In the next lemma we investigate the action of the firstterm of the odd Euler operator on skew polynomials

Lemma 47 The operatorsumn

i=1 xipi acts by multiplication by |f | on the space of homogenous functionsf isinPminus

Proof It suffices to show the result for a monomial xλ = xλ11 xλii xλnn Note that

xipi(xλ) = λixi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn = λix

λ

12

By summing over all indices i we obtain that

nsumi=1

xipi(xλ) = (λ1 + λ2 + + λn)xλ

which implies the desired result

The above lemma holds true in the even case as well where pi is replaced by the partial derivativewith respect to xi We now prove some properties about the action of the third term of the odd Euleroperator on r2 and ∆

Lemma 48 The commutation relation[sum

k 6=i sik∆]

= 0 holds

Proof Note that sjkpi = pjsjk if i = k sjkpi = pksjk if i = j and sjkpi = pisjk otherwise Indeedthese relations can be verified by checking if they are true for xai x

bjxck a b c isin Z+ and then extending

by linearity We prove that sjkpi = pjsjk if i = k and the other two cases are similar Without loss ofgenerality let j lt k and observe that

sjkpk(xajx

bk) = b(minus1)asjk(x

ajx

bminus1k ) = b(minus1)axakx

bminus1j = b(minus1)abxbminus1j xak

pjsjk(xajx

bk) = (minus1)abpj(x

bjxak) = b(minus1)abxbminus1j xak

By our work in Lemma 26 one can deduce that sjkr`m = rsjk(`m)sij As a consequence we find thatsjkDi = Dsjk(i)sjk By an easy induction we now have that sjk∆ = ∆sjk Using the above equationmultiple times proves the desired result

Lemma 49 The commutation relation[sum

k 6=i sik r2]

= 0 holds

Proof Follows since sjkxj = xksjk sjkxk = xjsjk and sjkxi = xisjk if i 6= j k

We are now ready to obtain two commutativity relations involving the odd Euler operator E

Theorem 410 The odd Euler operator and r2 satisfy the following commutation relations

[E r2] = 2r2 (45)[E∆] = minus2∆ (46)

Proof Since r2 has degree 2 and ∆ has degreeminus2 the theorem follows from Lemmas 47 48 and 49

We also need to investigate what the third commutativity relation [r2∆] turns out to be We willprove one lemma before doing so

Lemma 411 For i = 1 to n the equation xiDi +Dixi = 2txipi + t+ usum

k 6=i sik holds

Proof Recall thatDi = tpi + u

sumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Therefore since pixi = xipi + 1

Dixi = txipi + t+ usumk 6=i

(x2i minus x2k)minus1(xixk minus x2k)sik +sumk 6=i

(x2i minus x2k)[x2i τi minus xixkτk]

xiDi = txipi + usumk 6=i

(x2i minus x2k)minus1(x2i minus xixk)sik +sumk 6=i

(x2i minus x2k)minus1[minusx2i τi + xixkτk]

Adding we obtain the desired result

13

We now have the tools to find the third relation between r2 E and ∆

Theorem 412 The commutation relation [r2∆] = E holds

Proof We will first find [r2 Di] The derivative pi much like the partial derivative in the even casesatisfies the properties pixj = minusxjpi for i 6= j and pixi = xipi + 1 Now suppose that i 6= j Then

Dix2j = tpix

2j + ux2j

sumk 6=i 6=j

rik + x2k(x2i minus x2k)minus1[minusxiτi + xkτk] + x2i (x

2i minus x2k)minus1[(xi minus xk)sik]

= tx2jpi + ux2jsumk 6=i

rik + (xi minus xj)sij

Now we will find Dix2i

Dix2i = tpix

2i +

sumk 6=i

x2k(x2i minus x2k)minus1(xi minus xk)sik + x2i

sumk 6=i

(x2i minus x2k)minus1[minusxiτi + xkτk]

= tx2i pi + 2txi + x2isumk 6=i

rik minussumk 6=i

(xi minus xk)sik

Therefore[sumn

i=1 x2i Di

]= minus2txi This implies that

r2Di minusDir2 = minusxi (47)

As a result we find that

[r2∆] = minus(2t)minus1nsumi=1

[r2 D2j ] = minus(2t)minus1

nsumi=1

(r2D2j minusD2

j r2)

= minus(2t)minus1nsumi=1

[(Djr2Dj minus xjDj)minus (Djr

2Dj +Djxj)]

= (2t)minus1nsumi=1

(xiDi +Dixi)

where we have used 47 Now by Lemma 411

[r2∆] =

nsumi=1

xipi +n

2+u

t

sumk 6=i

sik = E

as desired

To summarize we have found operators Er2 and ∆ similar to their even counterparts whichsatisfy the defining relations of the Lie algebra sl2

[E r2] = 2r2

[E∆] = minus2∆

[r2∆] = E

Remark 413 If one uses the odd Dunkl operator ηi as found in [18] instead of the Di introduced herethe r2 E and ∆ operators do not generate sl2

Remark 414 Although our results hold true for all t and u in C one typically sets t = 1 and u = αminus1

for some α isin Ctimes since without loss of generality one of t and u may equal 1

14

Remark 415 In the even case let X be a Euclidean vector space with dimension n and let C[X] be thealgebra of C-valued functions on X Then this result about sl2 plays a major role in the study of higherdifferential operators on C[X] This is because the representation theory of sl2 allows for the reductionof degree to the second order [13] As a result our results in this section should correspondingly have arole in further studying differential operators in the odd case

5 q-nilHecke Algebras

Until now we have been concerned with the odd symmetric polynomials in variables x1 x2 xnwhere xixj = (minus1)xjxi for 1 le i 6= j le n This immediately suggests the question what if one replacesthe minus1 by any constant q isin Ctimes Specifically we ask the following questions

Question 51 Is it possible to study q-symmetric polynomials for which xixj = qxjxi when i gt j

Question 52 Are there q-analogs of evenodd divided difference operators and nilHecke algebras Sofar such structures are known only for the even case (q = 1) and the odd case (q = minus1)

In this section we answer both questions in the affirmativeWe work in the Z-graded q-braided setting throughout LetC be a commutative ring and let q isin Ctimes

be a unit If VW are graded C-modules and v isin V w isin W are homogeneous the braiding is theldquoq-twistrdquo

τq V otimesW rarrW otimes Vv otimes w 7rarr q|v||w|w otimes v

(51)

where | middot | is the degree function By q-algebra we mean an algebra object in the category of gradedC-modules equipped with this braided monoidal structure likewise for q-bialgebras q-Hopf algebrasand so forth

Remark 53 Note that the q-twist described above is its own inverse only when q2 = 1 which correlatesto the even and odd cases When q2 6= 1 the corresponding theory becomes more complex There-fore the q-nilHecke algebras that we introduce later in this section are nontrivial generalizations of thepreviously studied even and odd nilHecke algebras

Definition 54 The q-algebra P qn is defined to be

P qn = C〈x1 xn〉(xjxi minus qxixj = 0 if i lt j) (52)

where |xi| = 1 for i = 1 n

Note that P qn sim= otimesni=1Pq1 There are two interesting subalgebras of P qn that can be thought of as q-

analogs of the symmetric polynomials Define the k-th elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middotxin

and define the k-th twisted elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middot xin

where xj = qjminus1xj

Definition 55 The q-algebra of q-symmetric polynomials in n variables denoted Λqn is the subalgebra ofP qn generated by e1 en Likewise for the twisted q-symmetric polynomials Λqn and e1 en

The type A braid group on n strands acts on P qn by setting

15

1 σi(xj) = qxi+1 if j = i

2 σi(xj) = qminus1xi if j = i+ 1

3 σi(xj) = qxj if j gt i+ 1

4 σi(xj) = qminus1xj if j lt i

and extending multiplicatively

Definition 56 For i = 1 nminus 1 the i-th q-divided difference operator parti is the linear operator P qn rarr P qndefined by parti(xi) = q parti(xi+1) = minus1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + σi(f)parti(g) (53)

for all functions f g isin P qn We call Equation 53 the q-Leibniz rule

Lemma 57 For every i and every j lt k parti(xkxj minus qxjxk) = 0

Proof Since parti(xj) = 0 for j gt i+1 one may reduce the lemma to having to prove that part1(x2x1minusqx1x2) =0 part1(x3x1 minus qx1x3) = 0 and part1(x3x2 minus qx2x3) = 0 These statements follow from the q-Leibniz rule

Therefore parti is a well-defined operator on P qn

Lemma 58 The following relations hold

parti(xki ) =

kminus1sumj=0

qjkminus2jminusj2+kxjix

kminus1minusji+1

parti(xki+1) = minus

kminus1sumj=0

qminusjxjixkminus1minusji+1

Proof We induct on k The base case (k = 1) follows from the definition of the parti and the powers of qarise mostly from xni+1x

mi = qmnxmi x

ni+1 for all mn isin Z+

Our q-divided difference operators also annihilate the twisted elementary q-symmetric polynomialsjust as the even divided difference operators annihilate the elementary symmetric functions

Lemma 59 For every i = 1 nminus 1 and every k parti(ek) = 0 Hence Λqn sube⋂nminus1i=1 ker(parti)

Proof We can express ek as

ek =sum|J|=kii+1isinJ

xJ +sum|J|=kminus1ii+1isinJ

qf(Jik)xJ(xi + qxi+1) +sum|J|=kminus2ii+1isinJ

qg(Jik)xJxixi+1

for certain Z-valued functions f g The result then follows from parti(xi + qxi+1) = parti(xixi+1) = 0 and theq-Leibniz rule

Having discussed q-divided difference operators we can now construct algebras for every q 6=0 1minus1 that have many similarities to the even and odd nilHecke algebras For every such q we de-fine a q-nilHecke algebra generated by xi and parti for 1 le i le n subject to the relations found in thefollowing two lemmas (510 and 511)

Lemma 510 The following relations hold among the operators parti and xi (left multiplication by xi)

1 part2i = 0

2 partjparti minus qpartipartj = 0 for j gt i+ 1

3 xjxi = qxixj for i lt j

4 partixj minus qxjparti = 0 for j gt i+ 1

5 qpartixj minus xjparti = 0 for j lt i

6 partixi minus qxi+1parti = q

7 xiparti minus qpartixi+1 = q

16

Proof To show that part2i = 0 note that we can reduce to i = 1 and proceed by induction Since parti(1) = 0the base case follows Suppose that part2i (f) = 0 Then note that

part21(x1f) = part1(qf + qx2part1(f)) = qpart1(f)minus qpart1(f) + x1part21(f) = 0

part21(x2f) = part1(minusf + qminus1x1part1(f)) = minuspart1(f) + part1(f) + x2part21(f) = 0

part21(x3f) = part1(qx3part1(f)) = q2x3part21(f) = 0

which completes the proof of the first statement in the lemmaStatement 3 follows by definition Statements 4 5 6 and 7 follow from a suitable application of the

q-Leibniz rule Statement 2 follows from an inductive argument We can reduce to i = 1 and j = 3Suppose that partjparti = qpartipartj if j gt i+ 1 Then

part3part1(x1f)minus qpart1part3(x1f) = (qpart3(f) + x2part3part1(f))minus q(part3(f) + x2part1part3(f)) = 0

part3part1(x2f)minus qpart1part3(x2f) = (minuspart3(f) + qminus2x1part3part1(f))minus q(minusqminus1part3(f) + qminus2x1part1part3(f)) = 0

part3part1(x3f)minus qpart1part3(x3f) = (q2part1(f) + q2x4part3part1(f))minus q(qpart1(f) + q2x4part1part3(f)) = 0

part3part1(x4f)minus qpart1part3(x4f) = (minusqpart1(f) + x3part3part1(f))minus q(minuspart1(f) + x3part1part3(f)) = 0

part3part1(x5f)minus qpart1part3(x5f) = q2x5part3part1(f)minus q(q2x5part1part3(f)) = 0

thereby completing the induction

Lemma 511 partiparti+1partiparti+1partiparti+1 + parti+1partiparti+1partiparti+1parti = 0

Proof This result follows from an inductive argument we reduce to i = 1 and assume that the braidrelation holds true for some function f Then we check that the braid relation is true for x1f x2f x3f and x4f (since the behavior of xjf for j ge 4 is the same as that of x4f ) For brevity we will show theargument for x2f only

part1part2(x2f) = qpart1(f) + q2x3part1part2(f) part2part1(x2f) = minuspart2(f) + qminus2x1part2part1(f)

part212(x2f) = qpart2part1(f)minus q2part1part2(f) + qx2part212(f) qpart121(x2f) = minusqpart1part2(f) + part2part1(f) + part121(f)

part1212(x2f) = qpart121(f)minus qpart212(f) + x1part1212(f) part2121(x2f) = minuspart212(f) + part121(f) + x3part2121(f)

part21212(x2f) = qpart2121(f) + qminus1x2part21212(f) part12121(x2f) = minuspart1212(f) + qx3part12121(f)

We continue the above calculations to find that

part121212(x2f) = qpart12121(f) + part21212(f) + x2part121212(f)

part212121(x2f) = minusqpart12121(f)minus part21212(f) + x2part121212(f)

and the braid relation for x2f follows from the inductive hypothesis

6 A Diagrammatic Approach to q-Symmetric Polynomials

61 Introduction to a q-Bialgebra

In the previous section we answered Question 51 in an algebraic way by defining q-analogs of theclassical elementary and complete symmetric functions In this section we generalize the diagrammaticmethod used in [8] in order to study this question from the perspective of bialgebras

Let NΛq be a free associative Z-graded C-algebra with generators hm for m ge 0 We define h0 = 1and hm = 0 for m lt 0 and let q isin Ctimes The homogenous part of NΛq of degree ` has a basis hααkwhere

hα = hα1 middot middot middothαz for a composition α = (α1 αz) of `

17

Define a multiplication for homogenous x and y on NΛqotimes2 as follows where deg(x) denotes the degreeof x

(w otimes x)(y otimes z) = qdeg(x)deg(y)(wy otimes xz)

We can make NΛq into a q-bialgebra by letting the comultiplication on generators be

∆(hn) =

nsumk=0

hk otimes hnminusk

and by letting the counit be ε(x) = 0 if x is homogenous and deg(x)gt 0We can impose through the braiding structure that

∆(hahb) =asumj=0

bsumk=0

(hj otimes haminusj)(hk otimes hbminusk) =asumj=0

bsumk=0

qk(aminusj)(hjhk otimes haminusjhbminusk)

For any partitions λ and micro of n consider the set of double cosets of subroups Sλ and Smicro of Sn SλSnSmicroFor every C in this set let wC be the minimal length representative of C and let `(wC) be the length ofthis minimal length representative We will now attribute a bilinear form to NΛq

(hλ hmicro) =sum

CisinSλSnSmicro

q`(wC)

This bilinear form admits a diagrammatic description Let hn be an orange platform with n non-intersecting strands coming out of it When computing (hλ hmicro) with `(λ) = z and `(micro) = y drawz orange platforms at the top of the diagram representing λ1 λ2middot middot middot λz Draw y orange platforms atthe bottom of the diagram representative of micro1 micro2middot middot middot microy We require that |λ| = |micro| so that the topplatforms and bottom platforms have the same number of strands

Consider the example (h121 h22) In the following diagram snippets of the strands from each plat-form are shown

Every strand must start at one platform at the top and end on another platform at the bottom Nostrands that have originated from one platform may intersect The strands themselves have no criticalpoints with respect to the height function no two strands ever intersect more than once and there are notriple-intersections where three strands are concurrent Diagrams are considered up to isotopy Withoutany restrictions there would be n such diagrams if |λ| = n since there would be no limitations onthe ordering of the strands However due to the above rules there are only 4 possible diagrams in thecomputation of (h121 h22) shown below

Define(hλ hmicro) =

sumall diagrams D representing (hλhmicro)

q number of of crossings in D (61)

In the above example (h121 h22) = 1 + 2q2 + q3

18

We can extend the bilinear form toNΛqotimes2 by stating that any diagram in which strands from distincttensor factors intersect contributes 0 to the bilinear form

(w otimes x y otimes z) = (w y)(x z)

Let I be the radical of the bilinear form inNΛq In [8] the authors prove for any q that multiplicationand comultiplication are adjoint In other words for all xy1 y2 in NΛq

(y1 otimes y2∆(x)) = (y1y2 x) (62)

62 The Elementary q-Symmetric Functions

We now use the bilinear form of q-symmetric functions to study one of their important bases the ele-mentary q-symmetric functions

Define elements ek isin NΛq by ek = 0 for k lt 0 e0 = 1 and

ksumi=0

(minus1)iq(i2)eihkminusi = 0 for k ge 1 (63)

Equivalently leten = qminus(n2)

sumαn

(minus1)`(α)minusnhα (64)

Lemma 61

1 The coproduct of an elementary function is given by ∆(en) =

nsumk=0

ek otimes enminusk

2 If λ n then (hλ en) =

1 if λ = (1 1)

0 otherwise

Proof We begin by demonstrating (2) from which (1) will follow To show (2) it suffices to show that

(hmx en) =

(x enminus1) if m = 10 otherwise

We will utilize strong induction on n in order to find (hmx ekhnminusk) The base cases n = 0 1 are easyto show There are two cases to consider by the inductive hypothesis applied to k lt n Either there isa strand connecting hm and ek or there is not Just as we used an orange platform to denote hn wewill use a blue platform to denote ek The rules of the diagrammatic notation are the same for the blueplatforms as they are for the orange platforms

k

m

nminus k minusm

k nminus k

mlowast lowast lowast lowast lowast

x

If there is not a strand connecting hm and ek the configuration contributes qkm(x ekhnminuskminusm)

19

k minus 1

mminus 1

nminus k minusm + 1

k nminus k

mlowast lowast lowast lowast lowast

x

If a stand connects hm and ek this configuration contributes q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) We have thusshown that (hmx ekhnminusk) = qkm(x ekhnminuskminusm) + q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) Now we are equippedto consider (hmx ek)

(minus1)n+1q(n2)(hmx en) =

nminus1sumk=0

(minus1)kq(k2)(hmx ekhnminusk)

=nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus1sumk=0

(minus1)kq(k2)+(mminus1)(kminus1)(x ekminus1hnminuskminusm+1)

=

nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus2sumk=0

(minus1)k+1q(k+12 )+(mminus1)(k)(x ekhnminuskminusm)

= (minus1)nminus1q(nminus12 )+nm(x enminus1h1minusm)

Corresponding terms from the two sums cancel in pairs since q(k2)+km = q(

k+12 )+k(mminus1) leaving only the

k = nminus 1 term in the first sum The second statement of the lemma thus followsWe will now use (2) to prove (1) This follows from equation 62

(∆(ek) hλ otimes hmicro) = (ek hλhmicro) =

1 λ = (1`) micro = (1p) `+ p = k

0 otherwise

We now calculate the sign incurred when strands connect two blue (ek) platforms

(minus1)n+1q(n2)(en en) =

nminus1sumk=0

(minus1)kq(k2)(en ekhnminusk)

= (minus1)nminus1q(nminus12 )(en enminus1h1)

= (minus1)nminus1q(nminus12 )(∆(en) enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )

nsumk=0

(ek otimes enminusk enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )(enminus1 enminus1)

One may solve this recursion to find that (en en) = qminus(n2)Here the second equality follows from not-ing that at most one strand can connect hnminusk and en (so that k = n minus 1) the third equality followsfrom adjointness and the fourth and fifth equalities follow from the diagrammatic considerations of theprevious lemma

To summarize the diagrammatics of the bilinear form thus developed

20

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

1 r2i = 0

2 rirj + rjri = 0 for |iminus j| ge 2

3 riri+1ri = ri+1riri+1

4 rik(fg) = rik(f)sik(g) + (minus1)|f |frik(g)

5 rixi+1 + xi+1ri = rixi + xiri = si

6 rjxi + xjri = 0 for i 6= j j + 1

Proof Since siri = risi and ri = partisi it follows that siparti = partisi Then since part2i = 0 r2i = 0 as well Dueto Equation 5 from Lemma 26 we have that sirj = rjsi for |i minus j| ge 2 so sipartj = partjsi Thus ri andrj anti-commute since partipartj + partjparti = 0 The operators ri also satisfy braid relations which we show byinductively reducing to i = 1 and then using 24 and siparti = partisi repeatedly

r1r2r1 = s1part12part2s1part1 = s1s2part13s1part13part1 = s1s2s1part23part13part12

r2r1r2 = s2part2s1part1s2part2 = s2s1part13s2part13part2 = s2s1s2part12part13part23

Since s1s2s1 = s2s1s2 and part23part13part12 = part12part13part23 by symmetry we conclude that r1r2r1 = r2r1r2 TheLeibniz rule for rik (equation 4 of this lemma) follows immediately from the Leibniz rule for partik Sinceri(xi) = ri(xi+1) = 1 and ri(xj) = 0 for j 6= i i + 1 equations 5 and 6 follow from the Leibniz rule forrik

We also desire an explicit definition of the rik analogous to that of the odd divided difference oper-ator of [9] To find such an expression we use a preparatory lemma

Lemma 29 For all f isinPminus and 1 le i 6= k le n we have

sikxiτi(f)minus sikxkτk(f) = (minus1)|f |sik(f)(xi minus xk)

Proof It suffices to prove the result for a monomial xλ = xλ11 xλii xλkk xλnn where i lt k Wecalculate that

sikxiτi(xλ) = (minus1)λ1++λixλ11 xλi+1

k xλki xλnn

sikxkτk(xλ) = (minus1)λ1++λkminus1xλ11 xλik xλk+1

i xλnn

sik(xλ)xi = (minus1)λk+1++λnxλ11 xλik x

λk+1

i xλnn

sik(xλ)xk = (minus1)λi+1++λnxλ11 x

λi+1

k xλki xλnn

Since |f | = λ1 + + λn the desired result follows

Lemma 210 The operator rik has explicit form rik = (x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Proof Follows from Lemma 29 and Equation 22

We will now connect the above results to the odd Dunkl operator introduced by Khongsap and Wangin [18]

Definition 211 Define an operator δi by δi = (2xi)minus1(1minus τi)

The above super-derivative can also be defined inductively by imposing that δi(xj) = 1 if i = j and0 otherwise We then extend the action to monomials as follows

δi(xa1xa2 xa`) =sumk=1

(minus1)kminus1xa1 δi(xak)xak+1 xa`

8

The operator δi is a priori from Laurent skew polynomials to Laurent skew polynomials but it is easy tocheck that it preserves the subalgebra of skew polynomials Khongsap and Wang found an odd analogof the Dunkl operator given by

ηi = tδi + usumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk] (25)

where t u isin Ctimes Their operators anti-commute ηiηj + ηjηi = 0 for i 6= jBy Lemma 210 this odd Dunkl operator may be expressed as

ηi = tδi + usumk 6=i

partiksik (26)

By analogy with the commutative case discussed in Section 21 the operator rik plays the same role inthe odd theory that the even divided difference operator plays in the even theory

3 Classical Yang-Baxter Equation and the Dunkl Laplacian

Theorem 31 LetH123 = [r12 r13]+ + [r13 r23]+ + [r12 r23]+ (31)

where [p q]+ = pq + qp is the anti-commutator Then the operators rik satisfy the classical Yang-Baxterequation

H123 = 0 (32)

Proof To avoid a cumbersome direct calculation we instead use an inductive approach Namely sup-pose that H123(f) = 0 for some function f isin Pminus Then we show that H123(xif) = 0 for all integersi ge 1 Note that for i = 1

r12r13x1 = r12(s13 minus x1r13)= r12s13 minus r12(x1r13)= r12s13 minus (s12r13 minus x1r12r13)= r12s13 minus s12r13 + x1r12r13

(33)

where we have used Equation 4 twice Similarly we find that

r13r12x1 = r13s12 minus s13r12 + x1r13r12 (34)r23r12x1 = r23s12 + x1r23r12 (35)r12r23x1 = minuss12r23 + x1r12r23 (36)r13r23x1 = minuss13r23 + x1r13r23 (37)r23r13x1 = r23s13 + x1r23r13 (38)

By our inductive hypothesis

x1(r12r13 + r13r12 + r23r12 + r12r23 + r13r23 + r23r13) = 0

Keeping this in mind add Equations 33-38 to show that

H123x1 = r12s13 minus s12r13 + r13s12 minus s13r12 + r23s12 minus s12r23 minus s13r23 + r23s13

= r12s13 minus r23s12 + r13s12 minus r23s13 + r23s12 minus r13s12 minus r12s13 + r23s13

= 0

where we have repeatedly used Lemma 26 to slide rij past sk`We can similarly show that H123x2 = H123x3 = 0 Since rjkxi = xirjk for i gt 3 and j k isin (1 2 3)

it also follows that H123xi = 0 for i gt 3 proving the desired result

9

Corollary 32 The double summationsumn

i=1

(sumk 6=i rik

)2= 0

Proof The expansion of this double summation has n(nminus1)2 total terms Since r2ij = 0 n(nminus1) of theseterms are immediately zero leaving n(n minus 1)(n minus 2) terms of the form rijrk` where i = k and j = `are not both true By Theorem 31 the sum of all six terms of the form rijrk` where i j k ` isin a b c fordistinct integers 1 le a b c le n i 6= j and k 6= ` is zero There are

(n3

)ways to choose integers a b c

and for each choice of a b c six terms of the form rijrk` vanish This fact eliminates all the remaining6(n3

)= n(nminus 1)(nminus 2) terms of the double summation

As an application of the results in this section we will compute the odd Dunkl Laplaciansumn

i=1 η2i

We will first require a lemma involving the commutator of τi and rik

Lemma 33 The equation xminus1i [rik τi] = (x2i minus x2k)minus1(sik(τi + τk)minus xminus1i xksik(τi minus τk)minus 2) holds

Proof By Lemma 210

rikτi = (x2i minus x2k)minus1((xi minus xk)sikτi minus xiτ2i + xkτkτi)

τirik = (x2i minus x2k)minus1(minus(xi + xk)sikτk + xiτ2i + xkτiτk)

Since τiτk = τkτi and τ2i = 1 the result follows by subtraction

Now define Ai = (2xi)minus1(1minus τi) and Bi =

sumk 6=i rik so that the odd Dunkl operator ηi of Khongsap

and Wang may be expressed as ηi = tAi + uBi Note that

A2i =

1

4xminus1i (1minus τi)xminus1i (1minus τi)

=1

2xminus2i (1minus τi)(1minus τi)

= xminus2i (1minus τi)

(39)

since τi(xminus1i ) = minusxminus1i and τ2i = 1

Lemma 34 The relationsumn

i=1(AiBi +BiAi) = 0 holds

Proof Due to the Leibniz Rule for rik (equation 4) we can find that 0 = rik(xixminus1i ) = xminus1k minus xirik(x

minus1i )

so rik(xminus1i ) = xminus1i xminus1k It follows that rik(xminus1i f) = xminus1i xminus1k sik(f)minus xminus1i rik(f) Using this factnsumi=1

BiAi =1

2

nsumi=1

sumk 6=i

rikxminus1i (1minus τi)

=1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi)minus xminus1i rik(1minus τi))

By definitionnsumi=1

AiBi =1

2

nsumi=1

sumk 6=i

xminus1i (1minus τi)rik

Adding the above two equations we find thatnsumi=1

[Ai Bi]+ =1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi) + xminus1i (rikτi minus τirik))

=1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi) + (x2i minus x2k)minus1(sik(τi + τk)minus xminus1i xksik(τi minus τk)minus 2))

(310)

10

where we have used Lemma 33Each double summation repeats the pair of indices (p q) twice when 1 le p q le n one time when

i = p and k = q and once more when i = q and k = p Note that

(x2i minus x2k)minus1sik(τi + τk) = minus(x2k minus x2i )ski(τk + τi)

As a result the sum 12

sumni=1

sumk 6=i(x

2i minus x2k)minus1sik(τi + τk) = 0 Similarly

1

2

nsumi=1

sumk 6=i

xminus1i xminus1k sik = 0 and1

2

nsumi=1

sumk 6=i

(x2i minus x2k)minus1(minus2) = 0

since xminus1i xminus1k + xminus1k xminus1i = 0 Equation 310 then becomes

nsumi=1

(AiBi +BiAi) =1

2

nsumi=1

sumk 6=i

(minusxminus1i xminus1k sikτi minus (x2i minus x2k)minus1(xminus1i xksik(τi minus τk)) (311)

However note thatxminus1i xk + xminus1k xi = minus(x2i minus x2k)xminus1i xminus1k

which implies

(x2i minus x2k)minus1xminus1i xksik(τi minus τk) + (x2k minus x2i )minus1xminus1k xiski(τk minus τi) = minusxminus1i xminus1k sik(τi minus τk) (312)

Similarly we find that

xminus1i xminus1k sikτi + xminus1k xminus1i sikτk = xminus1i xminus1k sik(τi minus τk) (313)

As a result equation 311 becomes

nsumi=1

(AiBi +BiAi) = minus1

2

sum1leiltklen

(xminus1i xminus1k sik(τi minus τk)minus xminus1i sminus1k sik(τi minus τk)) = 0

We are now equipped to compute the Dunkl Laplacian in the odd case

Theorem 35 The equationsumn

i=1 η2i = t2

sum1leilen x

minus2i (1minus τi) holds

Proof Since ηi = tAi + uBi we have that

nsumi=1

η2i = t2nsumi=1

A2i + tu

nsumi=1

(AiBi +BiAi) + u2nsumi=1

B2i

By Lemma 34sumn

i=1(AiBi +BiAi) = 0 By Corollary 32sumn

i=1B2i = 0 Therefore

nsumi=1

η2i = t2nsumi=1

A2i = t2

sum1leilen

xminus2i (1minus τi)

by equation 39

11

4 A Variant of the Khongsap-Wang Odd Dunkl Operator

In this section we will show that a close variant of the odd Dunkl operator introduced by Khongsapand Wang can be used in the construction of three operators that satisfy the defining relations of the Liealgebra sl2 First we will consider an operator pi which is different from δi but plays a similar role

Definition 41 The operator pi is a C-linear map Pminus rarr Pminus which acts on monomials as follows

pi(xλ11 xλii xλnn ) = λi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn

Remark 42 One may also introduce pi by using a nice Leibniz-like expression involving τi by defining

pi(xj) = δij

pi(fg) = pi(f)g + (minus1)|fτi(f)pi(g)

where f g isin Pminus and δij is the Kronecker delta Now note the analogous relationship between thedegree-preserving operators si and τi in their respective Leibniz rules for the (minus1)-degree operators partiand pi This provides motivation for the definition of pi and suggests its natural role in our theory

Now consider a modified version of ηi

Definition 43 LetDi = tpi + u

sumk 6=i

rik (41)

Definition 44 Introduce the odd r2 Euler and ∆ operators as below

r2 = (2t)minus1nsumi=1

x2i (42)

E =nsumi=1

xipi +n

2+u

t

sumk 6=i

sik (43)

∆ = minus(2t)minus1nsumi=1

D2i (44)

Remark 45 Heckman who used the even Dunkl operators to find a sl2-triple useful in harmonic anal-ysis uses the convention t = 1 [13] For now we will consider t to be a fixed constant in Ctimes

Remark 46 The commutator in the setting of superalgebras is usually defined as [a b] = abminus(minus1)|a||b|bawhere |a| and |b| are the degrees of a and b respectively However since all of the operators we will beconsidering in this section have even degree there is no need to distinguish between commutators andsuper-commutators

To construct an sl2 action from these operators we will require a series of lemmas regarding theaction of portions of the odd Euler operator E In the next lemma we investigate the action of the firstterm of the odd Euler operator on skew polynomials

Lemma 47 The operatorsumn

i=1 xipi acts by multiplication by |f | on the space of homogenous functionsf isinPminus

Proof It suffices to show the result for a monomial xλ = xλ11 xλii xλnn Note that

xipi(xλ) = λixi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn = λix

λ

12

By summing over all indices i we obtain that

nsumi=1

xipi(xλ) = (λ1 + λ2 + + λn)xλ

which implies the desired result

The above lemma holds true in the even case as well where pi is replaced by the partial derivativewith respect to xi We now prove some properties about the action of the third term of the odd Euleroperator on r2 and ∆

Lemma 48 The commutation relation[sum

k 6=i sik∆]

= 0 holds

Proof Note that sjkpi = pjsjk if i = k sjkpi = pksjk if i = j and sjkpi = pisjk otherwise Indeedthese relations can be verified by checking if they are true for xai x

bjxck a b c isin Z+ and then extending

by linearity We prove that sjkpi = pjsjk if i = k and the other two cases are similar Without loss ofgenerality let j lt k and observe that

sjkpk(xajx

bk) = b(minus1)asjk(x

ajx

bminus1k ) = b(minus1)axakx

bminus1j = b(minus1)abxbminus1j xak

pjsjk(xajx

bk) = (minus1)abpj(x

bjxak) = b(minus1)abxbminus1j xak

By our work in Lemma 26 one can deduce that sjkr`m = rsjk(`m)sij As a consequence we find thatsjkDi = Dsjk(i)sjk By an easy induction we now have that sjk∆ = ∆sjk Using the above equationmultiple times proves the desired result

Lemma 49 The commutation relation[sum

k 6=i sik r2]

= 0 holds

Proof Follows since sjkxj = xksjk sjkxk = xjsjk and sjkxi = xisjk if i 6= j k

We are now ready to obtain two commutativity relations involving the odd Euler operator E

Theorem 410 The odd Euler operator and r2 satisfy the following commutation relations

[E r2] = 2r2 (45)[E∆] = minus2∆ (46)

Proof Since r2 has degree 2 and ∆ has degreeminus2 the theorem follows from Lemmas 47 48 and 49

We also need to investigate what the third commutativity relation [r2∆] turns out to be We willprove one lemma before doing so

Lemma 411 For i = 1 to n the equation xiDi +Dixi = 2txipi + t+ usum

k 6=i sik holds

Proof Recall thatDi = tpi + u

sumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Therefore since pixi = xipi + 1

Dixi = txipi + t+ usumk 6=i

(x2i minus x2k)minus1(xixk minus x2k)sik +sumk 6=i

(x2i minus x2k)[x2i τi minus xixkτk]

xiDi = txipi + usumk 6=i

(x2i minus x2k)minus1(x2i minus xixk)sik +sumk 6=i

(x2i minus x2k)minus1[minusx2i τi + xixkτk]

Adding we obtain the desired result

13

We now have the tools to find the third relation between r2 E and ∆

Theorem 412 The commutation relation [r2∆] = E holds

Proof We will first find [r2 Di] The derivative pi much like the partial derivative in the even casesatisfies the properties pixj = minusxjpi for i 6= j and pixi = xipi + 1 Now suppose that i 6= j Then

Dix2j = tpix

2j + ux2j

sumk 6=i 6=j

rik + x2k(x2i minus x2k)minus1[minusxiτi + xkτk] + x2i (x

2i minus x2k)minus1[(xi minus xk)sik]

= tx2jpi + ux2jsumk 6=i

rik + (xi minus xj)sij

Now we will find Dix2i

Dix2i = tpix

2i +

sumk 6=i

x2k(x2i minus x2k)minus1(xi minus xk)sik + x2i

sumk 6=i

(x2i minus x2k)minus1[minusxiτi + xkτk]

= tx2i pi + 2txi + x2isumk 6=i

rik minussumk 6=i

(xi minus xk)sik

Therefore[sumn

i=1 x2i Di

]= minus2txi This implies that

r2Di minusDir2 = minusxi (47)

As a result we find that

[r2∆] = minus(2t)minus1nsumi=1

[r2 D2j ] = minus(2t)minus1

nsumi=1

(r2D2j minusD2

j r2)

= minus(2t)minus1nsumi=1

[(Djr2Dj minus xjDj)minus (Djr

2Dj +Djxj)]

= (2t)minus1nsumi=1

(xiDi +Dixi)

where we have used 47 Now by Lemma 411

[r2∆] =

nsumi=1

xipi +n

2+u

t

sumk 6=i

sik = E

as desired

To summarize we have found operators Er2 and ∆ similar to their even counterparts whichsatisfy the defining relations of the Lie algebra sl2

[E r2] = 2r2

[E∆] = minus2∆

[r2∆] = E

Remark 413 If one uses the odd Dunkl operator ηi as found in [18] instead of the Di introduced herethe r2 E and ∆ operators do not generate sl2

Remark 414 Although our results hold true for all t and u in C one typically sets t = 1 and u = αminus1

for some α isin Ctimes since without loss of generality one of t and u may equal 1

14

Remark 415 In the even case let X be a Euclidean vector space with dimension n and let C[X] be thealgebra of C-valued functions on X Then this result about sl2 plays a major role in the study of higherdifferential operators on C[X] This is because the representation theory of sl2 allows for the reductionof degree to the second order [13] As a result our results in this section should correspondingly have arole in further studying differential operators in the odd case

5 q-nilHecke Algebras

Until now we have been concerned with the odd symmetric polynomials in variables x1 x2 xnwhere xixj = (minus1)xjxi for 1 le i 6= j le n This immediately suggests the question what if one replacesthe minus1 by any constant q isin Ctimes Specifically we ask the following questions

Question 51 Is it possible to study q-symmetric polynomials for which xixj = qxjxi when i gt j

Question 52 Are there q-analogs of evenodd divided difference operators and nilHecke algebras Sofar such structures are known only for the even case (q = 1) and the odd case (q = minus1)

In this section we answer both questions in the affirmativeWe work in the Z-graded q-braided setting throughout LetC be a commutative ring and let q isin Ctimes

be a unit If VW are graded C-modules and v isin V w isin W are homogeneous the braiding is theldquoq-twistrdquo

τq V otimesW rarrW otimes Vv otimes w 7rarr q|v||w|w otimes v

(51)

where | middot | is the degree function By q-algebra we mean an algebra object in the category of gradedC-modules equipped with this braided monoidal structure likewise for q-bialgebras q-Hopf algebrasand so forth

Remark 53 Note that the q-twist described above is its own inverse only when q2 = 1 which correlatesto the even and odd cases When q2 6= 1 the corresponding theory becomes more complex There-fore the q-nilHecke algebras that we introduce later in this section are nontrivial generalizations of thepreviously studied even and odd nilHecke algebras

Definition 54 The q-algebra P qn is defined to be

P qn = C〈x1 xn〉(xjxi minus qxixj = 0 if i lt j) (52)

where |xi| = 1 for i = 1 n

Note that P qn sim= otimesni=1Pq1 There are two interesting subalgebras of P qn that can be thought of as q-

analogs of the symmetric polynomials Define the k-th elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middotxin

and define the k-th twisted elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middot xin

where xj = qjminus1xj

Definition 55 The q-algebra of q-symmetric polynomials in n variables denoted Λqn is the subalgebra ofP qn generated by e1 en Likewise for the twisted q-symmetric polynomials Λqn and e1 en

The type A braid group on n strands acts on P qn by setting

15

1 σi(xj) = qxi+1 if j = i

2 σi(xj) = qminus1xi if j = i+ 1

3 σi(xj) = qxj if j gt i+ 1

4 σi(xj) = qminus1xj if j lt i

and extending multiplicatively

Definition 56 For i = 1 nminus 1 the i-th q-divided difference operator parti is the linear operator P qn rarr P qndefined by parti(xi) = q parti(xi+1) = minus1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + σi(f)parti(g) (53)

for all functions f g isin P qn We call Equation 53 the q-Leibniz rule

Lemma 57 For every i and every j lt k parti(xkxj minus qxjxk) = 0

Proof Since parti(xj) = 0 for j gt i+1 one may reduce the lemma to having to prove that part1(x2x1minusqx1x2) =0 part1(x3x1 minus qx1x3) = 0 and part1(x3x2 minus qx2x3) = 0 These statements follow from the q-Leibniz rule

Therefore parti is a well-defined operator on P qn

Lemma 58 The following relations hold

parti(xki ) =

kminus1sumj=0

qjkminus2jminusj2+kxjix

kminus1minusji+1

parti(xki+1) = minus

kminus1sumj=0

qminusjxjixkminus1minusji+1

Proof We induct on k The base case (k = 1) follows from the definition of the parti and the powers of qarise mostly from xni+1x

mi = qmnxmi x

ni+1 for all mn isin Z+

Our q-divided difference operators also annihilate the twisted elementary q-symmetric polynomialsjust as the even divided difference operators annihilate the elementary symmetric functions

Lemma 59 For every i = 1 nminus 1 and every k parti(ek) = 0 Hence Λqn sube⋂nminus1i=1 ker(parti)

Proof We can express ek as

ek =sum|J|=kii+1isinJ

xJ +sum|J|=kminus1ii+1isinJ

qf(Jik)xJ(xi + qxi+1) +sum|J|=kminus2ii+1isinJ

qg(Jik)xJxixi+1

for certain Z-valued functions f g The result then follows from parti(xi + qxi+1) = parti(xixi+1) = 0 and theq-Leibniz rule

Having discussed q-divided difference operators we can now construct algebras for every q 6=0 1minus1 that have many similarities to the even and odd nilHecke algebras For every such q we de-fine a q-nilHecke algebra generated by xi and parti for 1 le i le n subject to the relations found in thefollowing two lemmas (510 and 511)

Lemma 510 The following relations hold among the operators parti and xi (left multiplication by xi)

1 part2i = 0

2 partjparti minus qpartipartj = 0 for j gt i+ 1

3 xjxi = qxixj for i lt j

4 partixj minus qxjparti = 0 for j gt i+ 1

5 qpartixj minus xjparti = 0 for j lt i

6 partixi minus qxi+1parti = q

7 xiparti minus qpartixi+1 = q

16

Proof To show that part2i = 0 note that we can reduce to i = 1 and proceed by induction Since parti(1) = 0the base case follows Suppose that part2i (f) = 0 Then note that

part21(x1f) = part1(qf + qx2part1(f)) = qpart1(f)minus qpart1(f) + x1part21(f) = 0

part21(x2f) = part1(minusf + qminus1x1part1(f)) = minuspart1(f) + part1(f) + x2part21(f) = 0

part21(x3f) = part1(qx3part1(f)) = q2x3part21(f) = 0

which completes the proof of the first statement in the lemmaStatement 3 follows by definition Statements 4 5 6 and 7 follow from a suitable application of the

q-Leibniz rule Statement 2 follows from an inductive argument We can reduce to i = 1 and j = 3Suppose that partjparti = qpartipartj if j gt i+ 1 Then

part3part1(x1f)minus qpart1part3(x1f) = (qpart3(f) + x2part3part1(f))minus q(part3(f) + x2part1part3(f)) = 0

part3part1(x2f)minus qpart1part3(x2f) = (minuspart3(f) + qminus2x1part3part1(f))minus q(minusqminus1part3(f) + qminus2x1part1part3(f)) = 0

part3part1(x3f)minus qpart1part3(x3f) = (q2part1(f) + q2x4part3part1(f))minus q(qpart1(f) + q2x4part1part3(f)) = 0

part3part1(x4f)minus qpart1part3(x4f) = (minusqpart1(f) + x3part3part1(f))minus q(minuspart1(f) + x3part1part3(f)) = 0

part3part1(x5f)minus qpart1part3(x5f) = q2x5part3part1(f)minus q(q2x5part1part3(f)) = 0

thereby completing the induction

Lemma 511 partiparti+1partiparti+1partiparti+1 + parti+1partiparti+1partiparti+1parti = 0

Proof This result follows from an inductive argument we reduce to i = 1 and assume that the braidrelation holds true for some function f Then we check that the braid relation is true for x1f x2f x3f and x4f (since the behavior of xjf for j ge 4 is the same as that of x4f ) For brevity we will show theargument for x2f only

part1part2(x2f) = qpart1(f) + q2x3part1part2(f) part2part1(x2f) = minuspart2(f) + qminus2x1part2part1(f)

part212(x2f) = qpart2part1(f)minus q2part1part2(f) + qx2part212(f) qpart121(x2f) = minusqpart1part2(f) + part2part1(f) + part121(f)

part1212(x2f) = qpart121(f)minus qpart212(f) + x1part1212(f) part2121(x2f) = minuspart212(f) + part121(f) + x3part2121(f)

part21212(x2f) = qpart2121(f) + qminus1x2part21212(f) part12121(x2f) = minuspart1212(f) + qx3part12121(f)

We continue the above calculations to find that

part121212(x2f) = qpart12121(f) + part21212(f) + x2part121212(f)

part212121(x2f) = minusqpart12121(f)minus part21212(f) + x2part121212(f)

and the braid relation for x2f follows from the inductive hypothesis

6 A Diagrammatic Approach to q-Symmetric Polynomials

61 Introduction to a q-Bialgebra

In the previous section we answered Question 51 in an algebraic way by defining q-analogs of theclassical elementary and complete symmetric functions In this section we generalize the diagrammaticmethod used in [8] in order to study this question from the perspective of bialgebras

Let NΛq be a free associative Z-graded C-algebra with generators hm for m ge 0 We define h0 = 1and hm = 0 for m lt 0 and let q isin Ctimes The homogenous part of NΛq of degree ` has a basis hααkwhere

hα = hα1 middot middot middothαz for a composition α = (α1 αz) of `

17

Define a multiplication for homogenous x and y on NΛqotimes2 as follows where deg(x) denotes the degreeof x

(w otimes x)(y otimes z) = qdeg(x)deg(y)(wy otimes xz)

We can make NΛq into a q-bialgebra by letting the comultiplication on generators be

∆(hn) =

nsumk=0

hk otimes hnminusk

and by letting the counit be ε(x) = 0 if x is homogenous and deg(x)gt 0We can impose through the braiding structure that

∆(hahb) =asumj=0

bsumk=0

(hj otimes haminusj)(hk otimes hbminusk) =asumj=0

bsumk=0

qk(aminusj)(hjhk otimes haminusjhbminusk)

For any partitions λ and micro of n consider the set of double cosets of subroups Sλ and Smicro of Sn SλSnSmicroFor every C in this set let wC be the minimal length representative of C and let `(wC) be the length ofthis minimal length representative We will now attribute a bilinear form to NΛq

(hλ hmicro) =sum

CisinSλSnSmicro

q`(wC)

This bilinear form admits a diagrammatic description Let hn be an orange platform with n non-intersecting strands coming out of it When computing (hλ hmicro) with `(λ) = z and `(micro) = y drawz orange platforms at the top of the diagram representing λ1 λ2middot middot middot λz Draw y orange platforms atthe bottom of the diagram representative of micro1 micro2middot middot middot microy We require that |λ| = |micro| so that the topplatforms and bottom platforms have the same number of strands

Consider the example (h121 h22) In the following diagram snippets of the strands from each plat-form are shown

Every strand must start at one platform at the top and end on another platform at the bottom Nostrands that have originated from one platform may intersect The strands themselves have no criticalpoints with respect to the height function no two strands ever intersect more than once and there are notriple-intersections where three strands are concurrent Diagrams are considered up to isotopy Withoutany restrictions there would be n such diagrams if |λ| = n since there would be no limitations onthe ordering of the strands However due to the above rules there are only 4 possible diagrams in thecomputation of (h121 h22) shown below

Define(hλ hmicro) =

sumall diagrams D representing (hλhmicro)

q number of of crossings in D (61)

In the above example (h121 h22) = 1 + 2q2 + q3

18

We can extend the bilinear form toNΛqotimes2 by stating that any diagram in which strands from distincttensor factors intersect contributes 0 to the bilinear form

(w otimes x y otimes z) = (w y)(x z)

Let I be the radical of the bilinear form inNΛq In [8] the authors prove for any q that multiplicationand comultiplication are adjoint In other words for all xy1 y2 in NΛq

(y1 otimes y2∆(x)) = (y1y2 x) (62)

62 The Elementary q-Symmetric Functions

We now use the bilinear form of q-symmetric functions to study one of their important bases the ele-mentary q-symmetric functions

Define elements ek isin NΛq by ek = 0 for k lt 0 e0 = 1 and

ksumi=0

(minus1)iq(i2)eihkminusi = 0 for k ge 1 (63)

Equivalently leten = qminus(n2)

sumαn

(minus1)`(α)minusnhα (64)

Lemma 61

1 The coproduct of an elementary function is given by ∆(en) =

nsumk=0

ek otimes enminusk

2 If λ n then (hλ en) =

1 if λ = (1 1)

0 otherwise

Proof We begin by demonstrating (2) from which (1) will follow To show (2) it suffices to show that

(hmx en) =

(x enminus1) if m = 10 otherwise

We will utilize strong induction on n in order to find (hmx ekhnminusk) The base cases n = 0 1 are easyto show There are two cases to consider by the inductive hypothesis applied to k lt n Either there isa strand connecting hm and ek or there is not Just as we used an orange platform to denote hn wewill use a blue platform to denote ek The rules of the diagrammatic notation are the same for the blueplatforms as they are for the orange platforms

k

m

nminus k minusm

k nminus k

mlowast lowast lowast lowast lowast

x

If there is not a strand connecting hm and ek the configuration contributes qkm(x ekhnminuskminusm)

19

k minus 1

mminus 1

nminus k minusm + 1

k nminus k

mlowast lowast lowast lowast lowast

x

If a stand connects hm and ek this configuration contributes q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) We have thusshown that (hmx ekhnminusk) = qkm(x ekhnminuskminusm) + q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) Now we are equippedto consider (hmx ek)

(minus1)n+1q(n2)(hmx en) =

nminus1sumk=0

(minus1)kq(k2)(hmx ekhnminusk)

=nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus1sumk=0

(minus1)kq(k2)+(mminus1)(kminus1)(x ekminus1hnminuskminusm+1)

=

nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus2sumk=0

(minus1)k+1q(k+12 )+(mminus1)(k)(x ekhnminuskminusm)

= (minus1)nminus1q(nminus12 )+nm(x enminus1h1minusm)

Corresponding terms from the two sums cancel in pairs since q(k2)+km = q(

k+12 )+k(mminus1) leaving only the

k = nminus 1 term in the first sum The second statement of the lemma thus followsWe will now use (2) to prove (1) This follows from equation 62

(∆(ek) hλ otimes hmicro) = (ek hλhmicro) =

1 λ = (1`) micro = (1p) `+ p = k

0 otherwise

We now calculate the sign incurred when strands connect two blue (ek) platforms

(minus1)n+1q(n2)(en en) =

nminus1sumk=0

(minus1)kq(k2)(en ekhnminusk)

= (minus1)nminus1q(nminus12 )(en enminus1h1)

= (minus1)nminus1q(nminus12 )(∆(en) enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )

nsumk=0

(ek otimes enminusk enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )(enminus1 enminus1)

One may solve this recursion to find that (en en) = qminus(n2)Here the second equality follows from not-ing that at most one strand can connect hnminusk and en (so that k = n minus 1) the third equality followsfrom adjointness and the fourth and fifth equalities follow from the diagrammatic considerations of theprevious lemma

To summarize the diagrammatics of the bilinear form thus developed

20

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

The operator δi is a priori from Laurent skew polynomials to Laurent skew polynomials but it is easy tocheck that it preserves the subalgebra of skew polynomials Khongsap and Wang found an odd analogof the Dunkl operator given by

ηi = tδi + usumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk] (25)

where t u isin Ctimes Their operators anti-commute ηiηj + ηjηi = 0 for i 6= jBy Lemma 210 this odd Dunkl operator may be expressed as

ηi = tδi + usumk 6=i

partiksik (26)

By analogy with the commutative case discussed in Section 21 the operator rik plays the same role inthe odd theory that the even divided difference operator plays in the even theory

3 Classical Yang-Baxter Equation and the Dunkl Laplacian

Theorem 31 LetH123 = [r12 r13]+ + [r13 r23]+ + [r12 r23]+ (31)

where [p q]+ = pq + qp is the anti-commutator Then the operators rik satisfy the classical Yang-Baxterequation

H123 = 0 (32)

Proof To avoid a cumbersome direct calculation we instead use an inductive approach Namely sup-pose that H123(f) = 0 for some function f isin Pminus Then we show that H123(xif) = 0 for all integersi ge 1 Note that for i = 1

r12r13x1 = r12(s13 minus x1r13)= r12s13 minus r12(x1r13)= r12s13 minus (s12r13 minus x1r12r13)= r12s13 minus s12r13 + x1r12r13

(33)

where we have used Equation 4 twice Similarly we find that

r13r12x1 = r13s12 minus s13r12 + x1r13r12 (34)r23r12x1 = r23s12 + x1r23r12 (35)r12r23x1 = minuss12r23 + x1r12r23 (36)r13r23x1 = minuss13r23 + x1r13r23 (37)r23r13x1 = r23s13 + x1r23r13 (38)

By our inductive hypothesis

x1(r12r13 + r13r12 + r23r12 + r12r23 + r13r23 + r23r13) = 0

Keeping this in mind add Equations 33-38 to show that

H123x1 = r12s13 minus s12r13 + r13s12 minus s13r12 + r23s12 minus s12r23 minus s13r23 + r23s13

= r12s13 minus r23s12 + r13s12 minus r23s13 + r23s12 minus r13s12 minus r12s13 + r23s13

= 0

where we have repeatedly used Lemma 26 to slide rij past sk`We can similarly show that H123x2 = H123x3 = 0 Since rjkxi = xirjk for i gt 3 and j k isin (1 2 3)

it also follows that H123xi = 0 for i gt 3 proving the desired result

9

Corollary 32 The double summationsumn

i=1

(sumk 6=i rik

)2= 0

Proof The expansion of this double summation has n(nminus1)2 total terms Since r2ij = 0 n(nminus1) of theseterms are immediately zero leaving n(n minus 1)(n minus 2) terms of the form rijrk` where i = k and j = `are not both true By Theorem 31 the sum of all six terms of the form rijrk` where i j k ` isin a b c fordistinct integers 1 le a b c le n i 6= j and k 6= ` is zero There are

(n3

)ways to choose integers a b c

and for each choice of a b c six terms of the form rijrk` vanish This fact eliminates all the remaining6(n3

)= n(nminus 1)(nminus 2) terms of the double summation

As an application of the results in this section we will compute the odd Dunkl Laplaciansumn

i=1 η2i

We will first require a lemma involving the commutator of τi and rik

Lemma 33 The equation xminus1i [rik τi] = (x2i minus x2k)minus1(sik(τi + τk)minus xminus1i xksik(τi minus τk)minus 2) holds

Proof By Lemma 210

rikτi = (x2i minus x2k)minus1((xi minus xk)sikτi minus xiτ2i + xkτkτi)

τirik = (x2i minus x2k)minus1(minus(xi + xk)sikτk + xiτ2i + xkτiτk)

Since τiτk = τkτi and τ2i = 1 the result follows by subtraction

Now define Ai = (2xi)minus1(1minus τi) and Bi =

sumk 6=i rik so that the odd Dunkl operator ηi of Khongsap

and Wang may be expressed as ηi = tAi + uBi Note that

A2i =

1

4xminus1i (1minus τi)xminus1i (1minus τi)

=1

2xminus2i (1minus τi)(1minus τi)

= xminus2i (1minus τi)

(39)

since τi(xminus1i ) = minusxminus1i and τ2i = 1

Lemma 34 The relationsumn

i=1(AiBi +BiAi) = 0 holds

Proof Due to the Leibniz Rule for rik (equation 4) we can find that 0 = rik(xixminus1i ) = xminus1k minus xirik(x

minus1i )

so rik(xminus1i ) = xminus1i xminus1k It follows that rik(xminus1i f) = xminus1i xminus1k sik(f)minus xminus1i rik(f) Using this factnsumi=1

BiAi =1

2

nsumi=1

sumk 6=i

rikxminus1i (1minus τi)

=1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi)minus xminus1i rik(1minus τi))

By definitionnsumi=1

AiBi =1

2

nsumi=1

sumk 6=i

xminus1i (1minus τi)rik

Adding the above two equations we find thatnsumi=1

[Ai Bi]+ =1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi) + xminus1i (rikτi minus τirik))

=1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi) + (x2i minus x2k)minus1(sik(τi + τk)minus xminus1i xksik(τi minus τk)minus 2))

(310)

10

where we have used Lemma 33Each double summation repeats the pair of indices (p q) twice when 1 le p q le n one time when

i = p and k = q and once more when i = q and k = p Note that

(x2i minus x2k)minus1sik(τi + τk) = minus(x2k minus x2i )ski(τk + τi)

As a result the sum 12

sumni=1

sumk 6=i(x

2i minus x2k)minus1sik(τi + τk) = 0 Similarly

1

2

nsumi=1

sumk 6=i

xminus1i xminus1k sik = 0 and1

2

nsumi=1

sumk 6=i

(x2i minus x2k)minus1(minus2) = 0

since xminus1i xminus1k + xminus1k xminus1i = 0 Equation 310 then becomes

nsumi=1

(AiBi +BiAi) =1

2

nsumi=1

sumk 6=i

(minusxminus1i xminus1k sikτi minus (x2i minus x2k)minus1(xminus1i xksik(τi minus τk)) (311)

However note thatxminus1i xk + xminus1k xi = minus(x2i minus x2k)xminus1i xminus1k

which implies

(x2i minus x2k)minus1xminus1i xksik(τi minus τk) + (x2k minus x2i )minus1xminus1k xiski(τk minus τi) = minusxminus1i xminus1k sik(τi minus τk) (312)

Similarly we find that

xminus1i xminus1k sikτi + xminus1k xminus1i sikτk = xminus1i xminus1k sik(τi minus τk) (313)

As a result equation 311 becomes

nsumi=1

(AiBi +BiAi) = minus1

2

sum1leiltklen

(xminus1i xminus1k sik(τi minus τk)minus xminus1i sminus1k sik(τi minus τk)) = 0

We are now equipped to compute the Dunkl Laplacian in the odd case

Theorem 35 The equationsumn

i=1 η2i = t2

sum1leilen x

minus2i (1minus τi) holds

Proof Since ηi = tAi + uBi we have that

nsumi=1

η2i = t2nsumi=1

A2i + tu

nsumi=1

(AiBi +BiAi) + u2nsumi=1

B2i

By Lemma 34sumn

i=1(AiBi +BiAi) = 0 By Corollary 32sumn

i=1B2i = 0 Therefore

nsumi=1

η2i = t2nsumi=1

A2i = t2

sum1leilen

xminus2i (1minus τi)

by equation 39

11

4 A Variant of the Khongsap-Wang Odd Dunkl Operator

In this section we will show that a close variant of the odd Dunkl operator introduced by Khongsapand Wang can be used in the construction of three operators that satisfy the defining relations of the Liealgebra sl2 First we will consider an operator pi which is different from δi but plays a similar role

Definition 41 The operator pi is a C-linear map Pminus rarr Pminus which acts on monomials as follows

pi(xλ11 xλii xλnn ) = λi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn

Remark 42 One may also introduce pi by using a nice Leibniz-like expression involving τi by defining

pi(xj) = δij

pi(fg) = pi(f)g + (minus1)|fτi(f)pi(g)

where f g isin Pminus and δij is the Kronecker delta Now note the analogous relationship between thedegree-preserving operators si and τi in their respective Leibniz rules for the (minus1)-degree operators partiand pi This provides motivation for the definition of pi and suggests its natural role in our theory

Now consider a modified version of ηi

Definition 43 LetDi = tpi + u

sumk 6=i

rik (41)

Definition 44 Introduce the odd r2 Euler and ∆ operators as below

r2 = (2t)minus1nsumi=1

x2i (42)

E =nsumi=1

xipi +n

2+u

t

sumk 6=i

sik (43)

∆ = minus(2t)minus1nsumi=1

D2i (44)

Remark 45 Heckman who used the even Dunkl operators to find a sl2-triple useful in harmonic anal-ysis uses the convention t = 1 [13] For now we will consider t to be a fixed constant in Ctimes

Remark 46 The commutator in the setting of superalgebras is usually defined as [a b] = abminus(minus1)|a||b|bawhere |a| and |b| are the degrees of a and b respectively However since all of the operators we will beconsidering in this section have even degree there is no need to distinguish between commutators andsuper-commutators

To construct an sl2 action from these operators we will require a series of lemmas regarding theaction of portions of the odd Euler operator E In the next lemma we investigate the action of the firstterm of the odd Euler operator on skew polynomials

Lemma 47 The operatorsumn

i=1 xipi acts by multiplication by |f | on the space of homogenous functionsf isinPminus

Proof It suffices to show the result for a monomial xλ = xλ11 xλii xλnn Note that

xipi(xλ) = λixi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn = λix

λ

12

By summing over all indices i we obtain that

nsumi=1

xipi(xλ) = (λ1 + λ2 + + λn)xλ

which implies the desired result

The above lemma holds true in the even case as well where pi is replaced by the partial derivativewith respect to xi We now prove some properties about the action of the third term of the odd Euleroperator on r2 and ∆

Lemma 48 The commutation relation[sum

k 6=i sik∆]

= 0 holds

Proof Note that sjkpi = pjsjk if i = k sjkpi = pksjk if i = j and sjkpi = pisjk otherwise Indeedthese relations can be verified by checking if they are true for xai x

bjxck a b c isin Z+ and then extending

by linearity We prove that sjkpi = pjsjk if i = k and the other two cases are similar Without loss ofgenerality let j lt k and observe that

sjkpk(xajx

bk) = b(minus1)asjk(x

ajx

bminus1k ) = b(minus1)axakx

bminus1j = b(minus1)abxbminus1j xak

pjsjk(xajx

bk) = (minus1)abpj(x

bjxak) = b(minus1)abxbminus1j xak

By our work in Lemma 26 one can deduce that sjkr`m = rsjk(`m)sij As a consequence we find thatsjkDi = Dsjk(i)sjk By an easy induction we now have that sjk∆ = ∆sjk Using the above equationmultiple times proves the desired result

Lemma 49 The commutation relation[sum

k 6=i sik r2]

= 0 holds

Proof Follows since sjkxj = xksjk sjkxk = xjsjk and sjkxi = xisjk if i 6= j k

We are now ready to obtain two commutativity relations involving the odd Euler operator E

Theorem 410 The odd Euler operator and r2 satisfy the following commutation relations

[E r2] = 2r2 (45)[E∆] = minus2∆ (46)

Proof Since r2 has degree 2 and ∆ has degreeminus2 the theorem follows from Lemmas 47 48 and 49

We also need to investigate what the third commutativity relation [r2∆] turns out to be We willprove one lemma before doing so

Lemma 411 For i = 1 to n the equation xiDi +Dixi = 2txipi + t+ usum

k 6=i sik holds

Proof Recall thatDi = tpi + u

sumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Therefore since pixi = xipi + 1

Dixi = txipi + t+ usumk 6=i

(x2i minus x2k)minus1(xixk minus x2k)sik +sumk 6=i

(x2i minus x2k)[x2i τi minus xixkτk]

xiDi = txipi + usumk 6=i

(x2i minus x2k)minus1(x2i minus xixk)sik +sumk 6=i

(x2i minus x2k)minus1[minusx2i τi + xixkτk]

Adding we obtain the desired result

13

We now have the tools to find the third relation between r2 E and ∆

Theorem 412 The commutation relation [r2∆] = E holds

Proof We will first find [r2 Di] The derivative pi much like the partial derivative in the even casesatisfies the properties pixj = minusxjpi for i 6= j and pixi = xipi + 1 Now suppose that i 6= j Then

Dix2j = tpix

2j + ux2j

sumk 6=i 6=j

rik + x2k(x2i minus x2k)minus1[minusxiτi + xkτk] + x2i (x

2i minus x2k)minus1[(xi minus xk)sik]

= tx2jpi + ux2jsumk 6=i

rik + (xi minus xj)sij

Now we will find Dix2i

Dix2i = tpix

2i +

sumk 6=i

x2k(x2i minus x2k)minus1(xi minus xk)sik + x2i

sumk 6=i

(x2i minus x2k)minus1[minusxiτi + xkτk]

= tx2i pi + 2txi + x2isumk 6=i

rik minussumk 6=i

(xi minus xk)sik

Therefore[sumn

i=1 x2i Di

]= minus2txi This implies that

r2Di minusDir2 = minusxi (47)

As a result we find that

[r2∆] = minus(2t)minus1nsumi=1

[r2 D2j ] = minus(2t)minus1

nsumi=1

(r2D2j minusD2

j r2)

= minus(2t)minus1nsumi=1

[(Djr2Dj minus xjDj)minus (Djr

2Dj +Djxj)]

= (2t)minus1nsumi=1

(xiDi +Dixi)

where we have used 47 Now by Lemma 411

[r2∆] =

nsumi=1

xipi +n

2+u

t

sumk 6=i

sik = E

as desired

To summarize we have found operators Er2 and ∆ similar to their even counterparts whichsatisfy the defining relations of the Lie algebra sl2

[E r2] = 2r2

[E∆] = minus2∆

[r2∆] = E

Remark 413 If one uses the odd Dunkl operator ηi as found in [18] instead of the Di introduced herethe r2 E and ∆ operators do not generate sl2

Remark 414 Although our results hold true for all t and u in C one typically sets t = 1 and u = αminus1

for some α isin Ctimes since without loss of generality one of t and u may equal 1

14

Remark 415 In the even case let X be a Euclidean vector space with dimension n and let C[X] be thealgebra of C-valued functions on X Then this result about sl2 plays a major role in the study of higherdifferential operators on C[X] This is because the representation theory of sl2 allows for the reductionof degree to the second order [13] As a result our results in this section should correspondingly have arole in further studying differential operators in the odd case

5 q-nilHecke Algebras

Until now we have been concerned with the odd symmetric polynomials in variables x1 x2 xnwhere xixj = (minus1)xjxi for 1 le i 6= j le n This immediately suggests the question what if one replacesthe minus1 by any constant q isin Ctimes Specifically we ask the following questions

Question 51 Is it possible to study q-symmetric polynomials for which xixj = qxjxi when i gt j

Question 52 Are there q-analogs of evenodd divided difference operators and nilHecke algebras Sofar such structures are known only for the even case (q = 1) and the odd case (q = minus1)

In this section we answer both questions in the affirmativeWe work in the Z-graded q-braided setting throughout LetC be a commutative ring and let q isin Ctimes

be a unit If VW are graded C-modules and v isin V w isin W are homogeneous the braiding is theldquoq-twistrdquo

τq V otimesW rarrW otimes Vv otimes w 7rarr q|v||w|w otimes v

(51)

where | middot | is the degree function By q-algebra we mean an algebra object in the category of gradedC-modules equipped with this braided monoidal structure likewise for q-bialgebras q-Hopf algebrasand so forth

Remark 53 Note that the q-twist described above is its own inverse only when q2 = 1 which correlatesto the even and odd cases When q2 6= 1 the corresponding theory becomes more complex There-fore the q-nilHecke algebras that we introduce later in this section are nontrivial generalizations of thepreviously studied even and odd nilHecke algebras

Definition 54 The q-algebra P qn is defined to be

P qn = C〈x1 xn〉(xjxi minus qxixj = 0 if i lt j) (52)

where |xi| = 1 for i = 1 n

Note that P qn sim= otimesni=1Pq1 There are two interesting subalgebras of P qn that can be thought of as q-

analogs of the symmetric polynomials Define the k-th elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middotxin

and define the k-th twisted elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middot xin

where xj = qjminus1xj

Definition 55 The q-algebra of q-symmetric polynomials in n variables denoted Λqn is the subalgebra ofP qn generated by e1 en Likewise for the twisted q-symmetric polynomials Λqn and e1 en

The type A braid group on n strands acts on P qn by setting

15

1 σi(xj) = qxi+1 if j = i

2 σi(xj) = qminus1xi if j = i+ 1

3 σi(xj) = qxj if j gt i+ 1

4 σi(xj) = qminus1xj if j lt i

and extending multiplicatively

Definition 56 For i = 1 nminus 1 the i-th q-divided difference operator parti is the linear operator P qn rarr P qndefined by parti(xi) = q parti(xi+1) = minus1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + σi(f)parti(g) (53)

for all functions f g isin P qn We call Equation 53 the q-Leibniz rule

Lemma 57 For every i and every j lt k parti(xkxj minus qxjxk) = 0

Proof Since parti(xj) = 0 for j gt i+1 one may reduce the lemma to having to prove that part1(x2x1minusqx1x2) =0 part1(x3x1 minus qx1x3) = 0 and part1(x3x2 minus qx2x3) = 0 These statements follow from the q-Leibniz rule

Therefore parti is a well-defined operator on P qn

Lemma 58 The following relations hold

parti(xki ) =

kminus1sumj=0

qjkminus2jminusj2+kxjix

kminus1minusji+1

parti(xki+1) = minus

kminus1sumj=0

qminusjxjixkminus1minusji+1

Proof We induct on k The base case (k = 1) follows from the definition of the parti and the powers of qarise mostly from xni+1x

mi = qmnxmi x

ni+1 for all mn isin Z+

Our q-divided difference operators also annihilate the twisted elementary q-symmetric polynomialsjust as the even divided difference operators annihilate the elementary symmetric functions

Lemma 59 For every i = 1 nminus 1 and every k parti(ek) = 0 Hence Λqn sube⋂nminus1i=1 ker(parti)

Proof We can express ek as

ek =sum|J|=kii+1isinJ

xJ +sum|J|=kminus1ii+1isinJ

qf(Jik)xJ(xi + qxi+1) +sum|J|=kminus2ii+1isinJ

qg(Jik)xJxixi+1

for certain Z-valued functions f g The result then follows from parti(xi + qxi+1) = parti(xixi+1) = 0 and theq-Leibniz rule

Having discussed q-divided difference operators we can now construct algebras for every q 6=0 1minus1 that have many similarities to the even and odd nilHecke algebras For every such q we de-fine a q-nilHecke algebra generated by xi and parti for 1 le i le n subject to the relations found in thefollowing two lemmas (510 and 511)

Lemma 510 The following relations hold among the operators parti and xi (left multiplication by xi)

1 part2i = 0

2 partjparti minus qpartipartj = 0 for j gt i+ 1

3 xjxi = qxixj for i lt j

4 partixj minus qxjparti = 0 for j gt i+ 1

5 qpartixj minus xjparti = 0 for j lt i

6 partixi minus qxi+1parti = q

7 xiparti minus qpartixi+1 = q

16

Proof To show that part2i = 0 note that we can reduce to i = 1 and proceed by induction Since parti(1) = 0the base case follows Suppose that part2i (f) = 0 Then note that

part21(x1f) = part1(qf + qx2part1(f)) = qpart1(f)minus qpart1(f) + x1part21(f) = 0

part21(x2f) = part1(minusf + qminus1x1part1(f)) = minuspart1(f) + part1(f) + x2part21(f) = 0

part21(x3f) = part1(qx3part1(f)) = q2x3part21(f) = 0

which completes the proof of the first statement in the lemmaStatement 3 follows by definition Statements 4 5 6 and 7 follow from a suitable application of the

q-Leibniz rule Statement 2 follows from an inductive argument We can reduce to i = 1 and j = 3Suppose that partjparti = qpartipartj if j gt i+ 1 Then

part3part1(x1f)minus qpart1part3(x1f) = (qpart3(f) + x2part3part1(f))minus q(part3(f) + x2part1part3(f)) = 0

part3part1(x2f)minus qpart1part3(x2f) = (minuspart3(f) + qminus2x1part3part1(f))minus q(minusqminus1part3(f) + qminus2x1part1part3(f)) = 0

part3part1(x3f)minus qpart1part3(x3f) = (q2part1(f) + q2x4part3part1(f))minus q(qpart1(f) + q2x4part1part3(f)) = 0

part3part1(x4f)minus qpart1part3(x4f) = (minusqpart1(f) + x3part3part1(f))minus q(minuspart1(f) + x3part1part3(f)) = 0

part3part1(x5f)minus qpart1part3(x5f) = q2x5part3part1(f)minus q(q2x5part1part3(f)) = 0

thereby completing the induction

Lemma 511 partiparti+1partiparti+1partiparti+1 + parti+1partiparti+1partiparti+1parti = 0

Proof This result follows from an inductive argument we reduce to i = 1 and assume that the braidrelation holds true for some function f Then we check that the braid relation is true for x1f x2f x3f and x4f (since the behavior of xjf for j ge 4 is the same as that of x4f ) For brevity we will show theargument for x2f only

part1part2(x2f) = qpart1(f) + q2x3part1part2(f) part2part1(x2f) = minuspart2(f) + qminus2x1part2part1(f)

part212(x2f) = qpart2part1(f)minus q2part1part2(f) + qx2part212(f) qpart121(x2f) = minusqpart1part2(f) + part2part1(f) + part121(f)

part1212(x2f) = qpart121(f)minus qpart212(f) + x1part1212(f) part2121(x2f) = minuspart212(f) + part121(f) + x3part2121(f)

part21212(x2f) = qpart2121(f) + qminus1x2part21212(f) part12121(x2f) = minuspart1212(f) + qx3part12121(f)

We continue the above calculations to find that

part121212(x2f) = qpart12121(f) + part21212(f) + x2part121212(f)

part212121(x2f) = minusqpart12121(f)minus part21212(f) + x2part121212(f)

and the braid relation for x2f follows from the inductive hypothesis

6 A Diagrammatic Approach to q-Symmetric Polynomials

61 Introduction to a q-Bialgebra

In the previous section we answered Question 51 in an algebraic way by defining q-analogs of theclassical elementary and complete symmetric functions In this section we generalize the diagrammaticmethod used in [8] in order to study this question from the perspective of bialgebras

Let NΛq be a free associative Z-graded C-algebra with generators hm for m ge 0 We define h0 = 1and hm = 0 for m lt 0 and let q isin Ctimes The homogenous part of NΛq of degree ` has a basis hααkwhere

hα = hα1 middot middot middothαz for a composition α = (α1 αz) of `

17

Define a multiplication for homogenous x and y on NΛqotimes2 as follows where deg(x) denotes the degreeof x

(w otimes x)(y otimes z) = qdeg(x)deg(y)(wy otimes xz)

We can make NΛq into a q-bialgebra by letting the comultiplication on generators be

∆(hn) =

nsumk=0

hk otimes hnminusk

and by letting the counit be ε(x) = 0 if x is homogenous and deg(x)gt 0We can impose through the braiding structure that

∆(hahb) =asumj=0

bsumk=0

(hj otimes haminusj)(hk otimes hbminusk) =asumj=0

bsumk=0

qk(aminusj)(hjhk otimes haminusjhbminusk)

For any partitions λ and micro of n consider the set of double cosets of subroups Sλ and Smicro of Sn SλSnSmicroFor every C in this set let wC be the minimal length representative of C and let `(wC) be the length ofthis minimal length representative We will now attribute a bilinear form to NΛq

(hλ hmicro) =sum

CisinSλSnSmicro

q`(wC)

This bilinear form admits a diagrammatic description Let hn be an orange platform with n non-intersecting strands coming out of it When computing (hλ hmicro) with `(λ) = z and `(micro) = y drawz orange platforms at the top of the diagram representing λ1 λ2middot middot middot λz Draw y orange platforms atthe bottom of the diagram representative of micro1 micro2middot middot middot microy We require that |λ| = |micro| so that the topplatforms and bottom platforms have the same number of strands

Consider the example (h121 h22) In the following diagram snippets of the strands from each plat-form are shown

Every strand must start at one platform at the top and end on another platform at the bottom Nostrands that have originated from one platform may intersect The strands themselves have no criticalpoints with respect to the height function no two strands ever intersect more than once and there are notriple-intersections where three strands are concurrent Diagrams are considered up to isotopy Withoutany restrictions there would be n such diagrams if |λ| = n since there would be no limitations onthe ordering of the strands However due to the above rules there are only 4 possible diagrams in thecomputation of (h121 h22) shown below

Define(hλ hmicro) =

sumall diagrams D representing (hλhmicro)

q number of of crossings in D (61)

In the above example (h121 h22) = 1 + 2q2 + q3

18

We can extend the bilinear form toNΛqotimes2 by stating that any diagram in which strands from distincttensor factors intersect contributes 0 to the bilinear form

(w otimes x y otimes z) = (w y)(x z)

Let I be the radical of the bilinear form inNΛq In [8] the authors prove for any q that multiplicationand comultiplication are adjoint In other words for all xy1 y2 in NΛq

(y1 otimes y2∆(x)) = (y1y2 x) (62)

62 The Elementary q-Symmetric Functions

We now use the bilinear form of q-symmetric functions to study one of their important bases the ele-mentary q-symmetric functions

Define elements ek isin NΛq by ek = 0 for k lt 0 e0 = 1 and

ksumi=0

(minus1)iq(i2)eihkminusi = 0 for k ge 1 (63)

Equivalently leten = qminus(n2)

sumαn

(minus1)`(α)minusnhα (64)

Lemma 61

1 The coproduct of an elementary function is given by ∆(en) =

nsumk=0

ek otimes enminusk

2 If λ n then (hλ en) =

1 if λ = (1 1)

0 otherwise

Proof We begin by demonstrating (2) from which (1) will follow To show (2) it suffices to show that

(hmx en) =

(x enminus1) if m = 10 otherwise

We will utilize strong induction on n in order to find (hmx ekhnminusk) The base cases n = 0 1 are easyto show There are two cases to consider by the inductive hypothesis applied to k lt n Either there isa strand connecting hm and ek or there is not Just as we used an orange platform to denote hn wewill use a blue platform to denote ek The rules of the diagrammatic notation are the same for the blueplatforms as they are for the orange platforms

k

m

nminus k minusm

k nminus k

mlowast lowast lowast lowast lowast

x

If there is not a strand connecting hm and ek the configuration contributes qkm(x ekhnminuskminusm)

19

k minus 1

mminus 1

nminus k minusm + 1

k nminus k

mlowast lowast lowast lowast lowast

x

If a stand connects hm and ek this configuration contributes q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) We have thusshown that (hmx ekhnminusk) = qkm(x ekhnminuskminusm) + q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) Now we are equippedto consider (hmx ek)

(minus1)n+1q(n2)(hmx en) =

nminus1sumk=0

(minus1)kq(k2)(hmx ekhnminusk)

=nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus1sumk=0

(minus1)kq(k2)+(mminus1)(kminus1)(x ekminus1hnminuskminusm+1)

=

nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus2sumk=0

(minus1)k+1q(k+12 )+(mminus1)(k)(x ekhnminuskminusm)

= (minus1)nminus1q(nminus12 )+nm(x enminus1h1minusm)

Corresponding terms from the two sums cancel in pairs since q(k2)+km = q(

k+12 )+k(mminus1) leaving only the

k = nminus 1 term in the first sum The second statement of the lemma thus followsWe will now use (2) to prove (1) This follows from equation 62

(∆(ek) hλ otimes hmicro) = (ek hλhmicro) =

1 λ = (1`) micro = (1p) `+ p = k

0 otherwise

We now calculate the sign incurred when strands connect two blue (ek) platforms

(minus1)n+1q(n2)(en en) =

nminus1sumk=0

(minus1)kq(k2)(en ekhnminusk)

= (minus1)nminus1q(nminus12 )(en enminus1h1)

= (minus1)nminus1q(nminus12 )(∆(en) enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )

nsumk=0

(ek otimes enminusk enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )(enminus1 enminus1)

One may solve this recursion to find that (en en) = qminus(n2)Here the second equality follows from not-ing that at most one strand can connect hnminusk and en (so that k = n minus 1) the third equality followsfrom adjointness and the fourth and fifth equalities follow from the diagrammatic considerations of theprevious lemma

To summarize the diagrammatics of the bilinear form thus developed

20

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

Corollary 32 The double summationsumn

i=1

(sumk 6=i rik

)2= 0

Proof The expansion of this double summation has n(nminus1)2 total terms Since r2ij = 0 n(nminus1) of theseterms are immediately zero leaving n(n minus 1)(n minus 2) terms of the form rijrk` where i = k and j = `are not both true By Theorem 31 the sum of all six terms of the form rijrk` where i j k ` isin a b c fordistinct integers 1 le a b c le n i 6= j and k 6= ` is zero There are

(n3

)ways to choose integers a b c

and for each choice of a b c six terms of the form rijrk` vanish This fact eliminates all the remaining6(n3

)= n(nminus 1)(nminus 2) terms of the double summation

As an application of the results in this section we will compute the odd Dunkl Laplaciansumn

i=1 η2i

We will first require a lemma involving the commutator of τi and rik

Lemma 33 The equation xminus1i [rik τi] = (x2i minus x2k)minus1(sik(τi + τk)minus xminus1i xksik(τi minus τk)minus 2) holds

Proof By Lemma 210

rikτi = (x2i minus x2k)minus1((xi minus xk)sikτi minus xiτ2i + xkτkτi)

τirik = (x2i minus x2k)minus1(minus(xi + xk)sikτk + xiτ2i + xkτiτk)

Since τiτk = τkτi and τ2i = 1 the result follows by subtraction

Now define Ai = (2xi)minus1(1minus τi) and Bi =

sumk 6=i rik so that the odd Dunkl operator ηi of Khongsap

and Wang may be expressed as ηi = tAi + uBi Note that

A2i =

1

4xminus1i (1minus τi)xminus1i (1minus τi)

=1

2xminus2i (1minus τi)(1minus τi)

= xminus2i (1minus τi)

(39)

since τi(xminus1i ) = minusxminus1i and τ2i = 1

Lemma 34 The relationsumn

i=1(AiBi +BiAi) = 0 holds

Proof Due to the Leibniz Rule for rik (equation 4) we can find that 0 = rik(xixminus1i ) = xminus1k minus xirik(x

minus1i )

so rik(xminus1i ) = xminus1i xminus1k It follows that rik(xminus1i f) = xminus1i xminus1k sik(f)minus xminus1i rik(f) Using this factnsumi=1

BiAi =1

2

nsumi=1

sumk 6=i

rikxminus1i (1minus τi)

=1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi)minus xminus1i rik(1minus τi))

By definitionnsumi=1

AiBi =1

2

nsumi=1

sumk 6=i

xminus1i (1minus τi)rik

Adding the above two equations we find thatnsumi=1

[Ai Bi]+ =1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi) + xminus1i (rikτi minus τirik))

=1

2

nsumi=1

sumk 6=i

(xminus1i xminus1k sik(1minus τi) + (x2i minus x2k)minus1(sik(τi + τk)minus xminus1i xksik(τi minus τk)minus 2))

(310)

10

where we have used Lemma 33Each double summation repeats the pair of indices (p q) twice when 1 le p q le n one time when

i = p and k = q and once more when i = q and k = p Note that

(x2i minus x2k)minus1sik(τi + τk) = minus(x2k minus x2i )ski(τk + τi)

As a result the sum 12

sumni=1

sumk 6=i(x

2i minus x2k)minus1sik(τi + τk) = 0 Similarly

1

2

nsumi=1

sumk 6=i

xminus1i xminus1k sik = 0 and1

2

nsumi=1

sumk 6=i

(x2i minus x2k)minus1(minus2) = 0

since xminus1i xminus1k + xminus1k xminus1i = 0 Equation 310 then becomes

nsumi=1

(AiBi +BiAi) =1

2

nsumi=1

sumk 6=i

(minusxminus1i xminus1k sikτi minus (x2i minus x2k)minus1(xminus1i xksik(τi minus τk)) (311)

However note thatxminus1i xk + xminus1k xi = minus(x2i minus x2k)xminus1i xminus1k

which implies

(x2i minus x2k)minus1xminus1i xksik(τi minus τk) + (x2k minus x2i )minus1xminus1k xiski(τk minus τi) = minusxminus1i xminus1k sik(τi minus τk) (312)

Similarly we find that

xminus1i xminus1k sikτi + xminus1k xminus1i sikτk = xminus1i xminus1k sik(τi minus τk) (313)

As a result equation 311 becomes

nsumi=1

(AiBi +BiAi) = minus1

2

sum1leiltklen

(xminus1i xminus1k sik(τi minus τk)minus xminus1i sminus1k sik(τi minus τk)) = 0

We are now equipped to compute the Dunkl Laplacian in the odd case

Theorem 35 The equationsumn

i=1 η2i = t2

sum1leilen x

minus2i (1minus τi) holds

Proof Since ηi = tAi + uBi we have that

nsumi=1

η2i = t2nsumi=1

A2i + tu

nsumi=1

(AiBi +BiAi) + u2nsumi=1

B2i

By Lemma 34sumn

i=1(AiBi +BiAi) = 0 By Corollary 32sumn

i=1B2i = 0 Therefore

nsumi=1

η2i = t2nsumi=1

A2i = t2

sum1leilen

xminus2i (1minus τi)

by equation 39

11

4 A Variant of the Khongsap-Wang Odd Dunkl Operator

In this section we will show that a close variant of the odd Dunkl operator introduced by Khongsapand Wang can be used in the construction of three operators that satisfy the defining relations of the Liealgebra sl2 First we will consider an operator pi which is different from δi but plays a similar role

Definition 41 The operator pi is a C-linear map Pminus rarr Pminus which acts on monomials as follows

pi(xλ11 xλii xλnn ) = λi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn

Remark 42 One may also introduce pi by using a nice Leibniz-like expression involving τi by defining

pi(xj) = δij

pi(fg) = pi(f)g + (minus1)|fτi(f)pi(g)

where f g isin Pminus and δij is the Kronecker delta Now note the analogous relationship between thedegree-preserving operators si and τi in their respective Leibniz rules for the (minus1)-degree operators partiand pi This provides motivation for the definition of pi and suggests its natural role in our theory

Now consider a modified version of ηi

Definition 43 LetDi = tpi + u

sumk 6=i

rik (41)

Definition 44 Introduce the odd r2 Euler and ∆ operators as below

r2 = (2t)minus1nsumi=1

x2i (42)

E =nsumi=1

xipi +n

2+u

t

sumk 6=i

sik (43)

∆ = minus(2t)minus1nsumi=1

D2i (44)

Remark 45 Heckman who used the even Dunkl operators to find a sl2-triple useful in harmonic anal-ysis uses the convention t = 1 [13] For now we will consider t to be a fixed constant in Ctimes

Remark 46 The commutator in the setting of superalgebras is usually defined as [a b] = abminus(minus1)|a||b|bawhere |a| and |b| are the degrees of a and b respectively However since all of the operators we will beconsidering in this section have even degree there is no need to distinguish between commutators andsuper-commutators

To construct an sl2 action from these operators we will require a series of lemmas regarding theaction of portions of the odd Euler operator E In the next lemma we investigate the action of the firstterm of the odd Euler operator on skew polynomials

Lemma 47 The operatorsumn

i=1 xipi acts by multiplication by |f | on the space of homogenous functionsf isinPminus

Proof It suffices to show the result for a monomial xλ = xλ11 xλii xλnn Note that

xipi(xλ) = λixi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn = λix

λ

12

By summing over all indices i we obtain that

nsumi=1

xipi(xλ) = (λ1 + λ2 + + λn)xλ

which implies the desired result

The above lemma holds true in the even case as well where pi is replaced by the partial derivativewith respect to xi We now prove some properties about the action of the third term of the odd Euleroperator on r2 and ∆

Lemma 48 The commutation relation[sum

k 6=i sik∆]

= 0 holds

Proof Note that sjkpi = pjsjk if i = k sjkpi = pksjk if i = j and sjkpi = pisjk otherwise Indeedthese relations can be verified by checking if they are true for xai x

bjxck a b c isin Z+ and then extending

by linearity We prove that sjkpi = pjsjk if i = k and the other two cases are similar Without loss ofgenerality let j lt k and observe that

sjkpk(xajx

bk) = b(minus1)asjk(x

ajx

bminus1k ) = b(minus1)axakx

bminus1j = b(minus1)abxbminus1j xak

pjsjk(xajx

bk) = (minus1)abpj(x

bjxak) = b(minus1)abxbminus1j xak

By our work in Lemma 26 one can deduce that sjkr`m = rsjk(`m)sij As a consequence we find thatsjkDi = Dsjk(i)sjk By an easy induction we now have that sjk∆ = ∆sjk Using the above equationmultiple times proves the desired result

Lemma 49 The commutation relation[sum

k 6=i sik r2]

= 0 holds

Proof Follows since sjkxj = xksjk sjkxk = xjsjk and sjkxi = xisjk if i 6= j k

We are now ready to obtain two commutativity relations involving the odd Euler operator E

Theorem 410 The odd Euler operator and r2 satisfy the following commutation relations

[E r2] = 2r2 (45)[E∆] = minus2∆ (46)

Proof Since r2 has degree 2 and ∆ has degreeminus2 the theorem follows from Lemmas 47 48 and 49

We also need to investigate what the third commutativity relation [r2∆] turns out to be We willprove one lemma before doing so

Lemma 411 For i = 1 to n the equation xiDi +Dixi = 2txipi + t+ usum

k 6=i sik holds

Proof Recall thatDi = tpi + u

sumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Therefore since pixi = xipi + 1

Dixi = txipi + t+ usumk 6=i

(x2i minus x2k)minus1(xixk minus x2k)sik +sumk 6=i

(x2i minus x2k)[x2i τi minus xixkτk]

xiDi = txipi + usumk 6=i

(x2i minus x2k)minus1(x2i minus xixk)sik +sumk 6=i

(x2i minus x2k)minus1[minusx2i τi + xixkτk]

Adding we obtain the desired result

13

We now have the tools to find the third relation between r2 E and ∆

Theorem 412 The commutation relation [r2∆] = E holds

Proof We will first find [r2 Di] The derivative pi much like the partial derivative in the even casesatisfies the properties pixj = minusxjpi for i 6= j and pixi = xipi + 1 Now suppose that i 6= j Then

Dix2j = tpix

2j + ux2j

sumk 6=i 6=j

rik + x2k(x2i minus x2k)minus1[minusxiτi + xkτk] + x2i (x

2i minus x2k)minus1[(xi minus xk)sik]

= tx2jpi + ux2jsumk 6=i

rik + (xi minus xj)sij

Now we will find Dix2i

Dix2i = tpix

2i +

sumk 6=i

x2k(x2i minus x2k)minus1(xi minus xk)sik + x2i

sumk 6=i

(x2i minus x2k)minus1[minusxiτi + xkτk]

= tx2i pi + 2txi + x2isumk 6=i

rik minussumk 6=i

(xi minus xk)sik

Therefore[sumn

i=1 x2i Di

]= minus2txi This implies that

r2Di minusDir2 = minusxi (47)

As a result we find that

[r2∆] = minus(2t)minus1nsumi=1

[r2 D2j ] = minus(2t)minus1

nsumi=1

(r2D2j minusD2

j r2)

= minus(2t)minus1nsumi=1

[(Djr2Dj minus xjDj)minus (Djr

2Dj +Djxj)]

= (2t)minus1nsumi=1

(xiDi +Dixi)

where we have used 47 Now by Lemma 411

[r2∆] =

nsumi=1

xipi +n

2+u

t

sumk 6=i

sik = E

as desired

To summarize we have found operators Er2 and ∆ similar to their even counterparts whichsatisfy the defining relations of the Lie algebra sl2

[E r2] = 2r2

[E∆] = minus2∆

[r2∆] = E

Remark 413 If one uses the odd Dunkl operator ηi as found in [18] instead of the Di introduced herethe r2 E and ∆ operators do not generate sl2

Remark 414 Although our results hold true for all t and u in C one typically sets t = 1 and u = αminus1

for some α isin Ctimes since without loss of generality one of t and u may equal 1

14

Remark 415 In the even case let X be a Euclidean vector space with dimension n and let C[X] be thealgebra of C-valued functions on X Then this result about sl2 plays a major role in the study of higherdifferential operators on C[X] This is because the representation theory of sl2 allows for the reductionof degree to the second order [13] As a result our results in this section should correspondingly have arole in further studying differential operators in the odd case

5 q-nilHecke Algebras

Until now we have been concerned with the odd symmetric polynomials in variables x1 x2 xnwhere xixj = (minus1)xjxi for 1 le i 6= j le n This immediately suggests the question what if one replacesthe minus1 by any constant q isin Ctimes Specifically we ask the following questions

Question 51 Is it possible to study q-symmetric polynomials for which xixj = qxjxi when i gt j

Question 52 Are there q-analogs of evenodd divided difference operators and nilHecke algebras Sofar such structures are known only for the even case (q = 1) and the odd case (q = minus1)

In this section we answer both questions in the affirmativeWe work in the Z-graded q-braided setting throughout LetC be a commutative ring and let q isin Ctimes

be a unit If VW are graded C-modules and v isin V w isin W are homogeneous the braiding is theldquoq-twistrdquo

τq V otimesW rarrW otimes Vv otimes w 7rarr q|v||w|w otimes v

(51)

where | middot | is the degree function By q-algebra we mean an algebra object in the category of gradedC-modules equipped with this braided monoidal structure likewise for q-bialgebras q-Hopf algebrasand so forth

Remark 53 Note that the q-twist described above is its own inverse only when q2 = 1 which correlatesto the even and odd cases When q2 6= 1 the corresponding theory becomes more complex There-fore the q-nilHecke algebras that we introduce later in this section are nontrivial generalizations of thepreviously studied even and odd nilHecke algebras

Definition 54 The q-algebra P qn is defined to be

P qn = C〈x1 xn〉(xjxi minus qxixj = 0 if i lt j) (52)

where |xi| = 1 for i = 1 n

Note that P qn sim= otimesni=1Pq1 There are two interesting subalgebras of P qn that can be thought of as q-

analogs of the symmetric polynomials Define the k-th elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middotxin

and define the k-th twisted elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middot xin

where xj = qjminus1xj

Definition 55 The q-algebra of q-symmetric polynomials in n variables denoted Λqn is the subalgebra ofP qn generated by e1 en Likewise for the twisted q-symmetric polynomials Λqn and e1 en

The type A braid group on n strands acts on P qn by setting

15

1 σi(xj) = qxi+1 if j = i

2 σi(xj) = qminus1xi if j = i+ 1

3 σi(xj) = qxj if j gt i+ 1

4 σi(xj) = qminus1xj if j lt i

and extending multiplicatively

Definition 56 For i = 1 nminus 1 the i-th q-divided difference operator parti is the linear operator P qn rarr P qndefined by parti(xi) = q parti(xi+1) = minus1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + σi(f)parti(g) (53)

for all functions f g isin P qn We call Equation 53 the q-Leibniz rule

Lemma 57 For every i and every j lt k parti(xkxj minus qxjxk) = 0

Proof Since parti(xj) = 0 for j gt i+1 one may reduce the lemma to having to prove that part1(x2x1minusqx1x2) =0 part1(x3x1 minus qx1x3) = 0 and part1(x3x2 minus qx2x3) = 0 These statements follow from the q-Leibniz rule

Therefore parti is a well-defined operator on P qn

Lemma 58 The following relations hold

parti(xki ) =

kminus1sumj=0

qjkminus2jminusj2+kxjix

kminus1minusji+1

parti(xki+1) = minus

kminus1sumj=0

qminusjxjixkminus1minusji+1

Proof We induct on k The base case (k = 1) follows from the definition of the parti and the powers of qarise mostly from xni+1x

mi = qmnxmi x

ni+1 for all mn isin Z+

Our q-divided difference operators also annihilate the twisted elementary q-symmetric polynomialsjust as the even divided difference operators annihilate the elementary symmetric functions

Lemma 59 For every i = 1 nminus 1 and every k parti(ek) = 0 Hence Λqn sube⋂nminus1i=1 ker(parti)

Proof We can express ek as

ek =sum|J|=kii+1isinJ

xJ +sum|J|=kminus1ii+1isinJ

qf(Jik)xJ(xi + qxi+1) +sum|J|=kminus2ii+1isinJ

qg(Jik)xJxixi+1

for certain Z-valued functions f g The result then follows from parti(xi + qxi+1) = parti(xixi+1) = 0 and theq-Leibniz rule

Having discussed q-divided difference operators we can now construct algebras for every q 6=0 1minus1 that have many similarities to the even and odd nilHecke algebras For every such q we de-fine a q-nilHecke algebra generated by xi and parti for 1 le i le n subject to the relations found in thefollowing two lemmas (510 and 511)

Lemma 510 The following relations hold among the operators parti and xi (left multiplication by xi)

1 part2i = 0

2 partjparti minus qpartipartj = 0 for j gt i+ 1

3 xjxi = qxixj for i lt j

4 partixj minus qxjparti = 0 for j gt i+ 1

5 qpartixj minus xjparti = 0 for j lt i

6 partixi minus qxi+1parti = q

7 xiparti minus qpartixi+1 = q

16

Proof To show that part2i = 0 note that we can reduce to i = 1 and proceed by induction Since parti(1) = 0the base case follows Suppose that part2i (f) = 0 Then note that

part21(x1f) = part1(qf + qx2part1(f)) = qpart1(f)minus qpart1(f) + x1part21(f) = 0

part21(x2f) = part1(minusf + qminus1x1part1(f)) = minuspart1(f) + part1(f) + x2part21(f) = 0

part21(x3f) = part1(qx3part1(f)) = q2x3part21(f) = 0

which completes the proof of the first statement in the lemmaStatement 3 follows by definition Statements 4 5 6 and 7 follow from a suitable application of the

q-Leibniz rule Statement 2 follows from an inductive argument We can reduce to i = 1 and j = 3Suppose that partjparti = qpartipartj if j gt i+ 1 Then

part3part1(x1f)minus qpart1part3(x1f) = (qpart3(f) + x2part3part1(f))minus q(part3(f) + x2part1part3(f)) = 0

part3part1(x2f)minus qpart1part3(x2f) = (minuspart3(f) + qminus2x1part3part1(f))minus q(minusqminus1part3(f) + qminus2x1part1part3(f)) = 0

part3part1(x3f)minus qpart1part3(x3f) = (q2part1(f) + q2x4part3part1(f))minus q(qpart1(f) + q2x4part1part3(f)) = 0

part3part1(x4f)minus qpart1part3(x4f) = (minusqpart1(f) + x3part3part1(f))minus q(minuspart1(f) + x3part1part3(f)) = 0

part3part1(x5f)minus qpart1part3(x5f) = q2x5part3part1(f)minus q(q2x5part1part3(f)) = 0

thereby completing the induction

Lemma 511 partiparti+1partiparti+1partiparti+1 + parti+1partiparti+1partiparti+1parti = 0

Proof This result follows from an inductive argument we reduce to i = 1 and assume that the braidrelation holds true for some function f Then we check that the braid relation is true for x1f x2f x3f and x4f (since the behavior of xjf for j ge 4 is the same as that of x4f ) For brevity we will show theargument for x2f only

part1part2(x2f) = qpart1(f) + q2x3part1part2(f) part2part1(x2f) = minuspart2(f) + qminus2x1part2part1(f)

part212(x2f) = qpart2part1(f)minus q2part1part2(f) + qx2part212(f) qpart121(x2f) = minusqpart1part2(f) + part2part1(f) + part121(f)

part1212(x2f) = qpart121(f)minus qpart212(f) + x1part1212(f) part2121(x2f) = minuspart212(f) + part121(f) + x3part2121(f)

part21212(x2f) = qpart2121(f) + qminus1x2part21212(f) part12121(x2f) = minuspart1212(f) + qx3part12121(f)

We continue the above calculations to find that

part121212(x2f) = qpart12121(f) + part21212(f) + x2part121212(f)

part212121(x2f) = minusqpart12121(f)minus part21212(f) + x2part121212(f)

and the braid relation for x2f follows from the inductive hypothesis

6 A Diagrammatic Approach to q-Symmetric Polynomials

61 Introduction to a q-Bialgebra

In the previous section we answered Question 51 in an algebraic way by defining q-analogs of theclassical elementary and complete symmetric functions In this section we generalize the diagrammaticmethod used in [8] in order to study this question from the perspective of bialgebras

Let NΛq be a free associative Z-graded C-algebra with generators hm for m ge 0 We define h0 = 1and hm = 0 for m lt 0 and let q isin Ctimes The homogenous part of NΛq of degree ` has a basis hααkwhere

hα = hα1 middot middot middothαz for a composition α = (α1 αz) of `

17

Define a multiplication for homogenous x and y on NΛqotimes2 as follows where deg(x) denotes the degreeof x

(w otimes x)(y otimes z) = qdeg(x)deg(y)(wy otimes xz)

We can make NΛq into a q-bialgebra by letting the comultiplication on generators be

∆(hn) =

nsumk=0

hk otimes hnminusk

and by letting the counit be ε(x) = 0 if x is homogenous and deg(x)gt 0We can impose through the braiding structure that

∆(hahb) =asumj=0

bsumk=0

(hj otimes haminusj)(hk otimes hbminusk) =asumj=0

bsumk=0

qk(aminusj)(hjhk otimes haminusjhbminusk)

For any partitions λ and micro of n consider the set of double cosets of subroups Sλ and Smicro of Sn SλSnSmicroFor every C in this set let wC be the minimal length representative of C and let `(wC) be the length ofthis minimal length representative We will now attribute a bilinear form to NΛq

(hλ hmicro) =sum

CisinSλSnSmicro

q`(wC)

This bilinear form admits a diagrammatic description Let hn be an orange platform with n non-intersecting strands coming out of it When computing (hλ hmicro) with `(λ) = z and `(micro) = y drawz orange platforms at the top of the diagram representing λ1 λ2middot middot middot λz Draw y orange platforms atthe bottom of the diagram representative of micro1 micro2middot middot middot microy We require that |λ| = |micro| so that the topplatforms and bottom platforms have the same number of strands

Consider the example (h121 h22) In the following diagram snippets of the strands from each plat-form are shown

Every strand must start at one platform at the top and end on another platform at the bottom Nostrands that have originated from one platform may intersect The strands themselves have no criticalpoints with respect to the height function no two strands ever intersect more than once and there are notriple-intersections where three strands are concurrent Diagrams are considered up to isotopy Withoutany restrictions there would be n such diagrams if |λ| = n since there would be no limitations onthe ordering of the strands However due to the above rules there are only 4 possible diagrams in thecomputation of (h121 h22) shown below

Define(hλ hmicro) =

sumall diagrams D representing (hλhmicro)

q number of of crossings in D (61)

In the above example (h121 h22) = 1 + 2q2 + q3

18

We can extend the bilinear form toNΛqotimes2 by stating that any diagram in which strands from distincttensor factors intersect contributes 0 to the bilinear form

(w otimes x y otimes z) = (w y)(x z)

Let I be the radical of the bilinear form inNΛq In [8] the authors prove for any q that multiplicationand comultiplication are adjoint In other words for all xy1 y2 in NΛq

(y1 otimes y2∆(x)) = (y1y2 x) (62)

62 The Elementary q-Symmetric Functions

We now use the bilinear form of q-symmetric functions to study one of their important bases the ele-mentary q-symmetric functions

Define elements ek isin NΛq by ek = 0 for k lt 0 e0 = 1 and

ksumi=0

(minus1)iq(i2)eihkminusi = 0 for k ge 1 (63)

Equivalently leten = qminus(n2)

sumαn

(minus1)`(α)minusnhα (64)

Lemma 61

1 The coproduct of an elementary function is given by ∆(en) =

nsumk=0

ek otimes enminusk

2 If λ n then (hλ en) =

1 if λ = (1 1)

0 otherwise

Proof We begin by demonstrating (2) from which (1) will follow To show (2) it suffices to show that

(hmx en) =

(x enminus1) if m = 10 otherwise

We will utilize strong induction on n in order to find (hmx ekhnminusk) The base cases n = 0 1 are easyto show There are two cases to consider by the inductive hypothesis applied to k lt n Either there isa strand connecting hm and ek or there is not Just as we used an orange platform to denote hn wewill use a blue platform to denote ek The rules of the diagrammatic notation are the same for the blueplatforms as they are for the orange platforms

k

m

nminus k minusm

k nminus k

mlowast lowast lowast lowast lowast

x

If there is not a strand connecting hm and ek the configuration contributes qkm(x ekhnminuskminusm)

19

k minus 1

mminus 1

nminus k minusm + 1

k nminus k

mlowast lowast lowast lowast lowast

x

If a stand connects hm and ek this configuration contributes q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) We have thusshown that (hmx ekhnminusk) = qkm(x ekhnminuskminusm) + q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) Now we are equippedto consider (hmx ek)

(minus1)n+1q(n2)(hmx en) =

nminus1sumk=0

(minus1)kq(k2)(hmx ekhnminusk)

=nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus1sumk=0

(minus1)kq(k2)+(mminus1)(kminus1)(x ekminus1hnminuskminusm+1)

=

nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus2sumk=0

(minus1)k+1q(k+12 )+(mminus1)(k)(x ekhnminuskminusm)

= (minus1)nminus1q(nminus12 )+nm(x enminus1h1minusm)

Corresponding terms from the two sums cancel in pairs since q(k2)+km = q(

k+12 )+k(mminus1) leaving only the

k = nminus 1 term in the first sum The second statement of the lemma thus followsWe will now use (2) to prove (1) This follows from equation 62

(∆(ek) hλ otimes hmicro) = (ek hλhmicro) =

1 λ = (1`) micro = (1p) `+ p = k

0 otherwise

We now calculate the sign incurred when strands connect two blue (ek) platforms

(minus1)n+1q(n2)(en en) =

nminus1sumk=0

(minus1)kq(k2)(en ekhnminusk)

= (minus1)nminus1q(nminus12 )(en enminus1h1)

= (minus1)nminus1q(nminus12 )(∆(en) enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )

nsumk=0

(ek otimes enminusk enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )(enminus1 enminus1)

One may solve this recursion to find that (en en) = qminus(n2)Here the second equality follows from not-ing that at most one strand can connect hnminusk and en (so that k = n minus 1) the third equality followsfrom adjointness and the fourth and fifth equalities follow from the diagrammatic considerations of theprevious lemma

To summarize the diagrammatics of the bilinear form thus developed

20

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

where we have used Lemma 33Each double summation repeats the pair of indices (p q) twice when 1 le p q le n one time when

i = p and k = q and once more when i = q and k = p Note that

(x2i minus x2k)minus1sik(τi + τk) = minus(x2k minus x2i )ski(τk + τi)

As a result the sum 12

sumni=1

sumk 6=i(x

2i minus x2k)minus1sik(τi + τk) = 0 Similarly

1

2

nsumi=1

sumk 6=i

xminus1i xminus1k sik = 0 and1

2

nsumi=1

sumk 6=i

(x2i minus x2k)minus1(minus2) = 0

since xminus1i xminus1k + xminus1k xminus1i = 0 Equation 310 then becomes

nsumi=1

(AiBi +BiAi) =1

2

nsumi=1

sumk 6=i

(minusxminus1i xminus1k sikτi minus (x2i minus x2k)minus1(xminus1i xksik(τi minus τk)) (311)

However note thatxminus1i xk + xminus1k xi = minus(x2i minus x2k)xminus1i xminus1k

which implies

(x2i minus x2k)minus1xminus1i xksik(τi minus τk) + (x2k minus x2i )minus1xminus1k xiski(τk minus τi) = minusxminus1i xminus1k sik(τi minus τk) (312)

Similarly we find that

xminus1i xminus1k sikτi + xminus1k xminus1i sikτk = xminus1i xminus1k sik(τi minus τk) (313)

As a result equation 311 becomes

nsumi=1

(AiBi +BiAi) = minus1

2

sum1leiltklen

(xminus1i xminus1k sik(τi minus τk)minus xminus1i sminus1k sik(τi minus τk)) = 0

We are now equipped to compute the Dunkl Laplacian in the odd case

Theorem 35 The equationsumn

i=1 η2i = t2

sum1leilen x

minus2i (1minus τi) holds

Proof Since ηi = tAi + uBi we have that

nsumi=1

η2i = t2nsumi=1

A2i + tu

nsumi=1

(AiBi +BiAi) + u2nsumi=1

B2i

By Lemma 34sumn

i=1(AiBi +BiAi) = 0 By Corollary 32sumn

i=1B2i = 0 Therefore

nsumi=1

η2i = t2nsumi=1

A2i = t2

sum1leilen

xminus2i (1minus τi)

by equation 39

11

4 A Variant of the Khongsap-Wang Odd Dunkl Operator

In this section we will show that a close variant of the odd Dunkl operator introduced by Khongsapand Wang can be used in the construction of three operators that satisfy the defining relations of the Liealgebra sl2 First we will consider an operator pi which is different from δi but plays a similar role

Definition 41 The operator pi is a C-linear map Pminus rarr Pminus which acts on monomials as follows

pi(xλ11 xλii xλnn ) = λi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn

Remark 42 One may also introduce pi by using a nice Leibniz-like expression involving τi by defining

pi(xj) = δij

pi(fg) = pi(f)g + (minus1)|fτi(f)pi(g)

where f g isin Pminus and δij is the Kronecker delta Now note the analogous relationship between thedegree-preserving operators si and τi in their respective Leibniz rules for the (minus1)-degree operators partiand pi This provides motivation for the definition of pi and suggests its natural role in our theory

Now consider a modified version of ηi

Definition 43 LetDi = tpi + u

sumk 6=i

rik (41)

Definition 44 Introduce the odd r2 Euler and ∆ operators as below

r2 = (2t)minus1nsumi=1

x2i (42)

E =nsumi=1

xipi +n

2+u

t

sumk 6=i

sik (43)

∆ = minus(2t)minus1nsumi=1

D2i (44)

Remark 45 Heckman who used the even Dunkl operators to find a sl2-triple useful in harmonic anal-ysis uses the convention t = 1 [13] For now we will consider t to be a fixed constant in Ctimes

Remark 46 The commutator in the setting of superalgebras is usually defined as [a b] = abminus(minus1)|a||b|bawhere |a| and |b| are the degrees of a and b respectively However since all of the operators we will beconsidering in this section have even degree there is no need to distinguish between commutators andsuper-commutators

To construct an sl2 action from these operators we will require a series of lemmas regarding theaction of portions of the odd Euler operator E In the next lemma we investigate the action of the firstterm of the odd Euler operator on skew polynomials

Lemma 47 The operatorsumn

i=1 xipi acts by multiplication by |f | on the space of homogenous functionsf isinPminus

Proof It suffices to show the result for a monomial xλ = xλ11 xλii xλnn Note that

xipi(xλ) = λixi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn = λix

λ

12

By summing over all indices i we obtain that

nsumi=1

xipi(xλ) = (λ1 + λ2 + + λn)xλ

which implies the desired result

The above lemma holds true in the even case as well where pi is replaced by the partial derivativewith respect to xi We now prove some properties about the action of the third term of the odd Euleroperator on r2 and ∆

Lemma 48 The commutation relation[sum

k 6=i sik∆]

= 0 holds

Proof Note that sjkpi = pjsjk if i = k sjkpi = pksjk if i = j and sjkpi = pisjk otherwise Indeedthese relations can be verified by checking if they are true for xai x

bjxck a b c isin Z+ and then extending

by linearity We prove that sjkpi = pjsjk if i = k and the other two cases are similar Without loss ofgenerality let j lt k and observe that

sjkpk(xajx

bk) = b(minus1)asjk(x

ajx

bminus1k ) = b(minus1)axakx

bminus1j = b(minus1)abxbminus1j xak

pjsjk(xajx

bk) = (minus1)abpj(x

bjxak) = b(minus1)abxbminus1j xak

By our work in Lemma 26 one can deduce that sjkr`m = rsjk(`m)sij As a consequence we find thatsjkDi = Dsjk(i)sjk By an easy induction we now have that sjk∆ = ∆sjk Using the above equationmultiple times proves the desired result

Lemma 49 The commutation relation[sum

k 6=i sik r2]

= 0 holds

Proof Follows since sjkxj = xksjk sjkxk = xjsjk and sjkxi = xisjk if i 6= j k

We are now ready to obtain two commutativity relations involving the odd Euler operator E

Theorem 410 The odd Euler operator and r2 satisfy the following commutation relations

[E r2] = 2r2 (45)[E∆] = minus2∆ (46)

Proof Since r2 has degree 2 and ∆ has degreeminus2 the theorem follows from Lemmas 47 48 and 49

We also need to investigate what the third commutativity relation [r2∆] turns out to be We willprove one lemma before doing so

Lemma 411 For i = 1 to n the equation xiDi +Dixi = 2txipi + t+ usum

k 6=i sik holds

Proof Recall thatDi = tpi + u

sumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Therefore since pixi = xipi + 1

Dixi = txipi + t+ usumk 6=i

(x2i minus x2k)minus1(xixk minus x2k)sik +sumk 6=i

(x2i minus x2k)[x2i τi minus xixkτk]

xiDi = txipi + usumk 6=i

(x2i minus x2k)minus1(x2i minus xixk)sik +sumk 6=i

(x2i minus x2k)minus1[minusx2i τi + xixkτk]

Adding we obtain the desired result

13

We now have the tools to find the third relation between r2 E and ∆

Theorem 412 The commutation relation [r2∆] = E holds

Proof We will first find [r2 Di] The derivative pi much like the partial derivative in the even casesatisfies the properties pixj = minusxjpi for i 6= j and pixi = xipi + 1 Now suppose that i 6= j Then

Dix2j = tpix

2j + ux2j

sumk 6=i 6=j

rik + x2k(x2i minus x2k)minus1[minusxiτi + xkτk] + x2i (x

2i minus x2k)minus1[(xi minus xk)sik]

= tx2jpi + ux2jsumk 6=i

rik + (xi minus xj)sij

Now we will find Dix2i

Dix2i = tpix

2i +

sumk 6=i

x2k(x2i minus x2k)minus1(xi minus xk)sik + x2i

sumk 6=i

(x2i minus x2k)minus1[minusxiτi + xkτk]

= tx2i pi + 2txi + x2isumk 6=i

rik minussumk 6=i

(xi minus xk)sik

Therefore[sumn

i=1 x2i Di

]= minus2txi This implies that

r2Di minusDir2 = minusxi (47)

As a result we find that

[r2∆] = minus(2t)minus1nsumi=1

[r2 D2j ] = minus(2t)minus1

nsumi=1

(r2D2j minusD2

j r2)

= minus(2t)minus1nsumi=1

[(Djr2Dj minus xjDj)minus (Djr

2Dj +Djxj)]

= (2t)minus1nsumi=1

(xiDi +Dixi)

where we have used 47 Now by Lemma 411

[r2∆] =

nsumi=1

xipi +n

2+u

t

sumk 6=i

sik = E

as desired

To summarize we have found operators Er2 and ∆ similar to their even counterparts whichsatisfy the defining relations of the Lie algebra sl2

[E r2] = 2r2

[E∆] = minus2∆

[r2∆] = E

Remark 413 If one uses the odd Dunkl operator ηi as found in [18] instead of the Di introduced herethe r2 E and ∆ operators do not generate sl2

Remark 414 Although our results hold true for all t and u in C one typically sets t = 1 and u = αminus1

for some α isin Ctimes since without loss of generality one of t and u may equal 1

14

Remark 415 In the even case let X be a Euclidean vector space with dimension n and let C[X] be thealgebra of C-valued functions on X Then this result about sl2 plays a major role in the study of higherdifferential operators on C[X] This is because the representation theory of sl2 allows for the reductionof degree to the second order [13] As a result our results in this section should correspondingly have arole in further studying differential operators in the odd case

5 q-nilHecke Algebras

Until now we have been concerned with the odd symmetric polynomials in variables x1 x2 xnwhere xixj = (minus1)xjxi for 1 le i 6= j le n This immediately suggests the question what if one replacesthe minus1 by any constant q isin Ctimes Specifically we ask the following questions

Question 51 Is it possible to study q-symmetric polynomials for which xixj = qxjxi when i gt j

Question 52 Are there q-analogs of evenodd divided difference operators and nilHecke algebras Sofar such structures are known only for the even case (q = 1) and the odd case (q = minus1)

In this section we answer both questions in the affirmativeWe work in the Z-graded q-braided setting throughout LetC be a commutative ring and let q isin Ctimes

be a unit If VW are graded C-modules and v isin V w isin W are homogeneous the braiding is theldquoq-twistrdquo

τq V otimesW rarrW otimes Vv otimes w 7rarr q|v||w|w otimes v

(51)

where | middot | is the degree function By q-algebra we mean an algebra object in the category of gradedC-modules equipped with this braided monoidal structure likewise for q-bialgebras q-Hopf algebrasand so forth

Remark 53 Note that the q-twist described above is its own inverse only when q2 = 1 which correlatesto the even and odd cases When q2 6= 1 the corresponding theory becomes more complex There-fore the q-nilHecke algebras that we introduce later in this section are nontrivial generalizations of thepreviously studied even and odd nilHecke algebras

Definition 54 The q-algebra P qn is defined to be

P qn = C〈x1 xn〉(xjxi minus qxixj = 0 if i lt j) (52)

where |xi| = 1 for i = 1 n

Note that P qn sim= otimesni=1Pq1 There are two interesting subalgebras of P qn that can be thought of as q-

analogs of the symmetric polynomials Define the k-th elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middotxin

and define the k-th twisted elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middot xin

where xj = qjminus1xj

Definition 55 The q-algebra of q-symmetric polynomials in n variables denoted Λqn is the subalgebra ofP qn generated by e1 en Likewise for the twisted q-symmetric polynomials Λqn and e1 en

The type A braid group on n strands acts on P qn by setting

15

1 σi(xj) = qxi+1 if j = i

2 σi(xj) = qminus1xi if j = i+ 1

3 σi(xj) = qxj if j gt i+ 1

4 σi(xj) = qminus1xj if j lt i

and extending multiplicatively

Definition 56 For i = 1 nminus 1 the i-th q-divided difference operator parti is the linear operator P qn rarr P qndefined by parti(xi) = q parti(xi+1) = minus1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + σi(f)parti(g) (53)

for all functions f g isin P qn We call Equation 53 the q-Leibniz rule

Lemma 57 For every i and every j lt k parti(xkxj minus qxjxk) = 0

Proof Since parti(xj) = 0 for j gt i+1 one may reduce the lemma to having to prove that part1(x2x1minusqx1x2) =0 part1(x3x1 minus qx1x3) = 0 and part1(x3x2 minus qx2x3) = 0 These statements follow from the q-Leibniz rule

Therefore parti is a well-defined operator on P qn

Lemma 58 The following relations hold

parti(xki ) =

kminus1sumj=0

qjkminus2jminusj2+kxjix

kminus1minusji+1

parti(xki+1) = minus

kminus1sumj=0

qminusjxjixkminus1minusji+1

Proof We induct on k The base case (k = 1) follows from the definition of the parti and the powers of qarise mostly from xni+1x

mi = qmnxmi x

ni+1 for all mn isin Z+

Our q-divided difference operators also annihilate the twisted elementary q-symmetric polynomialsjust as the even divided difference operators annihilate the elementary symmetric functions

Lemma 59 For every i = 1 nminus 1 and every k parti(ek) = 0 Hence Λqn sube⋂nminus1i=1 ker(parti)

Proof We can express ek as

ek =sum|J|=kii+1isinJ

xJ +sum|J|=kminus1ii+1isinJ

qf(Jik)xJ(xi + qxi+1) +sum|J|=kminus2ii+1isinJ

qg(Jik)xJxixi+1

for certain Z-valued functions f g The result then follows from parti(xi + qxi+1) = parti(xixi+1) = 0 and theq-Leibniz rule

Having discussed q-divided difference operators we can now construct algebras for every q 6=0 1minus1 that have many similarities to the even and odd nilHecke algebras For every such q we de-fine a q-nilHecke algebra generated by xi and parti for 1 le i le n subject to the relations found in thefollowing two lemmas (510 and 511)

Lemma 510 The following relations hold among the operators parti and xi (left multiplication by xi)

1 part2i = 0

2 partjparti minus qpartipartj = 0 for j gt i+ 1

3 xjxi = qxixj for i lt j

4 partixj minus qxjparti = 0 for j gt i+ 1

5 qpartixj minus xjparti = 0 for j lt i

6 partixi minus qxi+1parti = q

7 xiparti minus qpartixi+1 = q

16

Proof To show that part2i = 0 note that we can reduce to i = 1 and proceed by induction Since parti(1) = 0the base case follows Suppose that part2i (f) = 0 Then note that

part21(x1f) = part1(qf + qx2part1(f)) = qpart1(f)minus qpart1(f) + x1part21(f) = 0

part21(x2f) = part1(minusf + qminus1x1part1(f)) = minuspart1(f) + part1(f) + x2part21(f) = 0

part21(x3f) = part1(qx3part1(f)) = q2x3part21(f) = 0

which completes the proof of the first statement in the lemmaStatement 3 follows by definition Statements 4 5 6 and 7 follow from a suitable application of the

q-Leibniz rule Statement 2 follows from an inductive argument We can reduce to i = 1 and j = 3Suppose that partjparti = qpartipartj if j gt i+ 1 Then

part3part1(x1f)minus qpart1part3(x1f) = (qpart3(f) + x2part3part1(f))minus q(part3(f) + x2part1part3(f)) = 0

part3part1(x2f)minus qpart1part3(x2f) = (minuspart3(f) + qminus2x1part3part1(f))minus q(minusqminus1part3(f) + qminus2x1part1part3(f)) = 0

part3part1(x3f)minus qpart1part3(x3f) = (q2part1(f) + q2x4part3part1(f))minus q(qpart1(f) + q2x4part1part3(f)) = 0

part3part1(x4f)minus qpart1part3(x4f) = (minusqpart1(f) + x3part3part1(f))minus q(minuspart1(f) + x3part1part3(f)) = 0

part3part1(x5f)minus qpart1part3(x5f) = q2x5part3part1(f)minus q(q2x5part1part3(f)) = 0

thereby completing the induction

Lemma 511 partiparti+1partiparti+1partiparti+1 + parti+1partiparti+1partiparti+1parti = 0

Proof This result follows from an inductive argument we reduce to i = 1 and assume that the braidrelation holds true for some function f Then we check that the braid relation is true for x1f x2f x3f and x4f (since the behavior of xjf for j ge 4 is the same as that of x4f ) For brevity we will show theargument for x2f only

part1part2(x2f) = qpart1(f) + q2x3part1part2(f) part2part1(x2f) = minuspart2(f) + qminus2x1part2part1(f)

part212(x2f) = qpart2part1(f)minus q2part1part2(f) + qx2part212(f) qpart121(x2f) = minusqpart1part2(f) + part2part1(f) + part121(f)

part1212(x2f) = qpart121(f)minus qpart212(f) + x1part1212(f) part2121(x2f) = minuspart212(f) + part121(f) + x3part2121(f)

part21212(x2f) = qpart2121(f) + qminus1x2part21212(f) part12121(x2f) = minuspart1212(f) + qx3part12121(f)

We continue the above calculations to find that

part121212(x2f) = qpart12121(f) + part21212(f) + x2part121212(f)

part212121(x2f) = minusqpart12121(f)minus part21212(f) + x2part121212(f)

and the braid relation for x2f follows from the inductive hypothesis

6 A Diagrammatic Approach to q-Symmetric Polynomials

61 Introduction to a q-Bialgebra

In the previous section we answered Question 51 in an algebraic way by defining q-analogs of theclassical elementary and complete symmetric functions In this section we generalize the diagrammaticmethod used in [8] in order to study this question from the perspective of bialgebras

Let NΛq be a free associative Z-graded C-algebra with generators hm for m ge 0 We define h0 = 1and hm = 0 for m lt 0 and let q isin Ctimes The homogenous part of NΛq of degree ` has a basis hααkwhere

hα = hα1 middot middot middothαz for a composition α = (α1 αz) of `

17

Define a multiplication for homogenous x and y on NΛqotimes2 as follows where deg(x) denotes the degreeof x

(w otimes x)(y otimes z) = qdeg(x)deg(y)(wy otimes xz)

We can make NΛq into a q-bialgebra by letting the comultiplication on generators be

∆(hn) =

nsumk=0

hk otimes hnminusk

and by letting the counit be ε(x) = 0 if x is homogenous and deg(x)gt 0We can impose through the braiding structure that

∆(hahb) =asumj=0

bsumk=0

(hj otimes haminusj)(hk otimes hbminusk) =asumj=0

bsumk=0

qk(aminusj)(hjhk otimes haminusjhbminusk)

For any partitions λ and micro of n consider the set of double cosets of subroups Sλ and Smicro of Sn SλSnSmicroFor every C in this set let wC be the minimal length representative of C and let `(wC) be the length ofthis minimal length representative We will now attribute a bilinear form to NΛq

(hλ hmicro) =sum

CisinSλSnSmicro

q`(wC)

This bilinear form admits a diagrammatic description Let hn be an orange platform with n non-intersecting strands coming out of it When computing (hλ hmicro) with `(λ) = z and `(micro) = y drawz orange platforms at the top of the diagram representing λ1 λ2middot middot middot λz Draw y orange platforms atthe bottom of the diagram representative of micro1 micro2middot middot middot microy We require that |λ| = |micro| so that the topplatforms and bottom platforms have the same number of strands

Consider the example (h121 h22) In the following diagram snippets of the strands from each plat-form are shown

Every strand must start at one platform at the top and end on another platform at the bottom Nostrands that have originated from one platform may intersect The strands themselves have no criticalpoints with respect to the height function no two strands ever intersect more than once and there are notriple-intersections where three strands are concurrent Diagrams are considered up to isotopy Withoutany restrictions there would be n such diagrams if |λ| = n since there would be no limitations onthe ordering of the strands However due to the above rules there are only 4 possible diagrams in thecomputation of (h121 h22) shown below

Define(hλ hmicro) =

sumall diagrams D representing (hλhmicro)

q number of of crossings in D (61)

In the above example (h121 h22) = 1 + 2q2 + q3

18

We can extend the bilinear form toNΛqotimes2 by stating that any diagram in which strands from distincttensor factors intersect contributes 0 to the bilinear form

(w otimes x y otimes z) = (w y)(x z)

Let I be the radical of the bilinear form inNΛq In [8] the authors prove for any q that multiplicationand comultiplication are adjoint In other words for all xy1 y2 in NΛq

(y1 otimes y2∆(x)) = (y1y2 x) (62)

62 The Elementary q-Symmetric Functions

We now use the bilinear form of q-symmetric functions to study one of their important bases the ele-mentary q-symmetric functions

Define elements ek isin NΛq by ek = 0 for k lt 0 e0 = 1 and

ksumi=0

(minus1)iq(i2)eihkminusi = 0 for k ge 1 (63)

Equivalently leten = qminus(n2)

sumαn

(minus1)`(α)minusnhα (64)

Lemma 61

1 The coproduct of an elementary function is given by ∆(en) =

nsumk=0

ek otimes enminusk

2 If λ n then (hλ en) =

1 if λ = (1 1)

0 otherwise

Proof We begin by demonstrating (2) from which (1) will follow To show (2) it suffices to show that

(hmx en) =

(x enminus1) if m = 10 otherwise

We will utilize strong induction on n in order to find (hmx ekhnminusk) The base cases n = 0 1 are easyto show There are two cases to consider by the inductive hypothesis applied to k lt n Either there isa strand connecting hm and ek or there is not Just as we used an orange platform to denote hn wewill use a blue platform to denote ek The rules of the diagrammatic notation are the same for the blueplatforms as they are for the orange platforms

k

m

nminus k minusm

k nminus k

mlowast lowast lowast lowast lowast

x

If there is not a strand connecting hm and ek the configuration contributes qkm(x ekhnminuskminusm)

19

k minus 1

mminus 1

nminus k minusm + 1

k nminus k

mlowast lowast lowast lowast lowast

x

If a stand connects hm and ek this configuration contributes q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) We have thusshown that (hmx ekhnminusk) = qkm(x ekhnminuskminusm) + q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) Now we are equippedto consider (hmx ek)

(minus1)n+1q(n2)(hmx en) =

nminus1sumk=0

(minus1)kq(k2)(hmx ekhnminusk)

=nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus1sumk=0

(minus1)kq(k2)+(mminus1)(kminus1)(x ekminus1hnminuskminusm+1)

=

nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus2sumk=0

(minus1)k+1q(k+12 )+(mminus1)(k)(x ekhnminuskminusm)

= (minus1)nminus1q(nminus12 )+nm(x enminus1h1minusm)

Corresponding terms from the two sums cancel in pairs since q(k2)+km = q(

k+12 )+k(mminus1) leaving only the

k = nminus 1 term in the first sum The second statement of the lemma thus followsWe will now use (2) to prove (1) This follows from equation 62

(∆(ek) hλ otimes hmicro) = (ek hλhmicro) =

1 λ = (1`) micro = (1p) `+ p = k

0 otherwise

We now calculate the sign incurred when strands connect two blue (ek) platforms

(minus1)n+1q(n2)(en en) =

nminus1sumk=0

(minus1)kq(k2)(en ekhnminusk)

= (minus1)nminus1q(nminus12 )(en enminus1h1)

= (minus1)nminus1q(nminus12 )(∆(en) enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )

nsumk=0

(ek otimes enminusk enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )(enminus1 enminus1)

One may solve this recursion to find that (en en) = qminus(n2)Here the second equality follows from not-ing that at most one strand can connect hnminusk and en (so that k = n minus 1) the third equality followsfrom adjointness and the fourth and fifth equalities follow from the diagrammatic considerations of theprevious lemma

To summarize the diagrammatics of the bilinear form thus developed

20

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

4 A Variant of the Khongsap-Wang Odd Dunkl Operator

In this section we will show that a close variant of the odd Dunkl operator introduced by Khongsapand Wang can be used in the construction of three operators that satisfy the defining relations of the Liealgebra sl2 First we will consider an operator pi which is different from δi but plays a similar role

Definition 41 The operator pi is a C-linear map Pminus rarr Pminus which acts on monomials as follows

pi(xλ11 xλii xλnn ) = λi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn

Remark 42 One may also introduce pi by using a nice Leibniz-like expression involving τi by defining

pi(xj) = δij

pi(fg) = pi(f)g + (minus1)|fτi(f)pi(g)

where f g isin Pminus and δij is the Kronecker delta Now note the analogous relationship between thedegree-preserving operators si and τi in their respective Leibniz rules for the (minus1)-degree operators partiand pi This provides motivation for the definition of pi and suggests its natural role in our theory

Now consider a modified version of ηi

Definition 43 LetDi = tpi + u

sumk 6=i

rik (41)

Definition 44 Introduce the odd r2 Euler and ∆ operators as below

r2 = (2t)minus1nsumi=1

x2i (42)

E =nsumi=1

xipi +n

2+u

t

sumk 6=i

sik (43)

∆ = minus(2t)minus1nsumi=1

D2i (44)

Remark 45 Heckman who used the even Dunkl operators to find a sl2-triple useful in harmonic anal-ysis uses the convention t = 1 [13] For now we will consider t to be a fixed constant in Ctimes

Remark 46 The commutator in the setting of superalgebras is usually defined as [a b] = abminus(minus1)|a||b|bawhere |a| and |b| are the degrees of a and b respectively However since all of the operators we will beconsidering in this section have even degree there is no need to distinguish between commutators andsuper-commutators

To construct an sl2 action from these operators we will require a series of lemmas regarding theaction of portions of the odd Euler operator E In the next lemma we investigate the action of the firstterm of the odd Euler operator on skew polynomials

Lemma 47 The operatorsumn

i=1 xipi acts by multiplication by |f | on the space of homogenous functionsf isinPminus

Proof It suffices to show the result for a monomial xλ = xλ11 xλii xλnn Note that

xipi(xλ) = λixi(minus1)λ1++λiminus1xλ11 xλiminus1i xλnn = λix

λ

12

By summing over all indices i we obtain that

nsumi=1

xipi(xλ) = (λ1 + λ2 + + λn)xλ

which implies the desired result

The above lemma holds true in the even case as well where pi is replaced by the partial derivativewith respect to xi We now prove some properties about the action of the third term of the odd Euleroperator on r2 and ∆

Lemma 48 The commutation relation[sum

k 6=i sik∆]

= 0 holds

Proof Note that sjkpi = pjsjk if i = k sjkpi = pksjk if i = j and sjkpi = pisjk otherwise Indeedthese relations can be verified by checking if they are true for xai x

bjxck a b c isin Z+ and then extending

by linearity We prove that sjkpi = pjsjk if i = k and the other two cases are similar Without loss ofgenerality let j lt k and observe that

sjkpk(xajx

bk) = b(minus1)asjk(x

ajx

bminus1k ) = b(minus1)axakx

bminus1j = b(minus1)abxbminus1j xak

pjsjk(xajx

bk) = (minus1)abpj(x

bjxak) = b(minus1)abxbminus1j xak

By our work in Lemma 26 one can deduce that sjkr`m = rsjk(`m)sij As a consequence we find thatsjkDi = Dsjk(i)sjk By an easy induction we now have that sjk∆ = ∆sjk Using the above equationmultiple times proves the desired result

Lemma 49 The commutation relation[sum

k 6=i sik r2]

= 0 holds

Proof Follows since sjkxj = xksjk sjkxk = xjsjk and sjkxi = xisjk if i 6= j k

We are now ready to obtain two commutativity relations involving the odd Euler operator E

Theorem 410 The odd Euler operator and r2 satisfy the following commutation relations

[E r2] = 2r2 (45)[E∆] = minus2∆ (46)

Proof Since r2 has degree 2 and ∆ has degreeminus2 the theorem follows from Lemmas 47 48 and 49

We also need to investigate what the third commutativity relation [r2∆] turns out to be We willprove one lemma before doing so

Lemma 411 For i = 1 to n the equation xiDi +Dixi = 2txipi + t+ usum

k 6=i sik holds

Proof Recall thatDi = tpi + u

sumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Therefore since pixi = xipi + 1

Dixi = txipi + t+ usumk 6=i

(x2i minus x2k)minus1(xixk minus x2k)sik +sumk 6=i

(x2i minus x2k)[x2i τi minus xixkτk]

xiDi = txipi + usumk 6=i

(x2i minus x2k)minus1(x2i minus xixk)sik +sumk 6=i

(x2i minus x2k)minus1[minusx2i τi + xixkτk]

Adding we obtain the desired result

13

We now have the tools to find the third relation between r2 E and ∆

Theorem 412 The commutation relation [r2∆] = E holds

Proof We will first find [r2 Di] The derivative pi much like the partial derivative in the even casesatisfies the properties pixj = minusxjpi for i 6= j and pixi = xipi + 1 Now suppose that i 6= j Then

Dix2j = tpix

2j + ux2j

sumk 6=i 6=j

rik + x2k(x2i minus x2k)minus1[minusxiτi + xkτk] + x2i (x

2i minus x2k)minus1[(xi minus xk)sik]

= tx2jpi + ux2jsumk 6=i

rik + (xi minus xj)sij

Now we will find Dix2i

Dix2i = tpix

2i +

sumk 6=i

x2k(x2i minus x2k)minus1(xi minus xk)sik + x2i

sumk 6=i

(x2i minus x2k)minus1[minusxiτi + xkτk]

= tx2i pi + 2txi + x2isumk 6=i

rik minussumk 6=i

(xi minus xk)sik

Therefore[sumn

i=1 x2i Di

]= minus2txi This implies that

r2Di minusDir2 = minusxi (47)

As a result we find that

[r2∆] = minus(2t)minus1nsumi=1

[r2 D2j ] = minus(2t)minus1

nsumi=1

(r2D2j minusD2

j r2)

= minus(2t)minus1nsumi=1

[(Djr2Dj minus xjDj)minus (Djr

2Dj +Djxj)]

= (2t)minus1nsumi=1

(xiDi +Dixi)

where we have used 47 Now by Lemma 411

[r2∆] =

nsumi=1

xipi +n

2+u

t

sumk 6=i

sik = E

as desired

To summarize we have found operators Er2 and ∆ similar to their even counterparts whichsatisfy the defining relations of the Lie algebra sl2

[E r2] = 2r2

[E∆] = minus2∆

[r2∆] = E

Remark 413 If one uses the odd Dunkl operator ηi as found in [18] instead of the Di introduced herethe r2 E and ∆ operators do not generate sl2

Remark 414 Although our results hold true for all t and u in C one typically sets t = 1 and u = αminus1

for some α isin Ctimes since without loss of generality one of t and u may equal 1

14

Remark 415 In the even case let X be a Euclidean vector space with dimension n and let C[X] be thealgebra of C-valued functions on X Then this result about sl2 plays a major role in the study of higherdifferential operators on C[X] This is because the representation theory of sl2 allows for the reductionof degree to the second order [13] As a result our results in this section should correspondingly have arole in further studying differential operators in the odd case

5 q-nilHecke Algebras

Until now we have been concerned with the odd symmetric polynomials in variables x1 x2 xnwhere xixj = (minus1)xjxi for 1 le i 6= j le n This immediately suggests the question what if one replacesthe minus1 by any constant q isin Ctimes Specifically we ask the following questions

Question 51 Is it possible to study q-symmetric polynomials for which xixj = qxjxi when i gt j

Question 52 Are there q-analogs of evenodd divided difference operators and nilHecke algebras Sofar such structures are known only for the even case (q = 1) and the odd case (q = minus1)

In this section we answer both questions in the affirmativeWe work in the Z-graded q-braided setting throughout LetC be a commutative ring and let q isin Ctimes

be a unit If VW are graded C-modules and v isin V w isin W are homogeneous the braiding is theldquoq-twistrdquo

τq V otimesW rarrW otimes Vv otimes w 7rarr q|v||w|w otimes v

(51)

where | middot | is the degree function By q-algebra we mean an algebra object in the category of gradedC-modules equipped with this braided monoidal structure likewise for q-bialgebras q-Hopf algebrasand so forth

Remark 53 Note that the q-twist described above is its own inverse only when q2 = 1 which correlatesto the even and odd cases When q2 6= 1 the corresponding theory becomes more complex There-fore the q-nilHecke algebras that we introduce later in this section are nontrivial generalizations of thepreviously studied even and odd nilHecke algebras

Definition 54 The q-algebra P qn is defined to be

P qn = C〈x1 xn〉(xjxi minus qxixj = 0 if i lt j) (52)

where |xi| = 1 for i = 1 n

Note that P qn sim= otimesni=1Pq1 There are two interesting subalgebras of P qn that can be thought of as q-

analogs of the symmetric polynomials Define the k-th elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middotxin

and define the k-th twisted elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middot xin

where xj = qjminus1xj

Definition 55 The q-algebra of q-symmetric polynomials in n variables denoted Λqn is the subalgebra ofP qn generated by e1 en Likewise for the twisted q-symmetric polynomials Λqn and e1 en

The type A braid group on n strands acts on P qn by setting

15

1 σi(xj) = qxi+1 if j = i

2 σi(xj) = qminus1xi if j = i+ 1

3 σi(xj) = qxj if j gt i+ 1

4 σi(xj) = qminus1xj if j lt i

and extending multiplicatively

Definition 56 For i = 1 nminus 1 the i-th q-divided difference operator parti is the linear operator P qn rarr P qndefined by parti(xi) = q parti(xi+1) = minus1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + σi(f)parti(g) (53)

for all functions f g isin P qn We call Equation 53 the q-Leibniz rule

Lemma 57 For every i and every j lt k parti(xkxj minus qxjxk) = 0

Proof Since parti(xj) = 0 for j gt i+1 one may reduce the lemma to having to prove that part1(x2x1minusqx1x2) =0 part1(x3x1 minus qx1x3) = 0 and part1(x3x2 minus qx2x3) = 0 These statements follow from the q-Leibniz rule

Therefore parti is a well-defined operator on P qn

Lemma 58 The following relations hold

parti(xki ) =

kminus1sumj=0

qjkminus2jminusj2+kxjix

kminus1minusji+1

parti(xki+1) = minus

kminus1sumj=0

qminusjxjixkminus1minusji+1

Proof We induct on k The base case (k = 1) follows from the definition of the parti and the powers of qarise mostly from xni+1x

mi = qmnxmi x

ni+1 for all mn isin Z+

Our q-divided difference operators also annihilate the twisted elementary q-symmetric polynomialsjust as the even divided difference operators annihilate the elementary symmetric functions

Lemma 59 For every i = 1 nminus 1 and every k parti(ek) = 0 Hence Λqn sube⋂nminus1i=1 ker(parti)

Proof We can express ek as

ek =sum|J|=kii+1isinJ

xJ +sum|J|=kminus1ii+1isinJ

qf(Jik)xJ(xi + qxi+1) +sum|J|=kminus2ii+1isinJ

qg(Jik)xJxixi+1

for certain Z-valued functions f g The result then follows from parti(xi + qxi+1) = parti(xixi+1) = 0 and theq-Leibniz rule

Having discussed q-divided difference operators we can now construct algebras for every q 6=0 1minus1 that have many similarities to the even and odd nilHecke algebras For every such q we de-fine a q-nilHecke algebra generated by xi and parti for 1 le i le n subject to the relations found in thefollowing two lemmas (510 and 511)

Lemma 510 The following relations hold among the operators parti and xi (left multiplication by xi)

1 part2i = 0

2 partjparti minus qpartipartj = 0 for j gt i+ 1

3 xjxi = qxixj for i lt j

4 partixj minus qxjparti = 0 for j gt i+ 1

5 qpartixj minus xjparti = 0 for j lt i

6 partixi minus qxi+1parti = q

7 xiparti minus qpartixi+1 = q

16

Proof To show that part2i = 0 note that we can reduce to i = 1 and proceed by induction Since parti(1) = 0the base case follows Suppose that part2i (f) = 0 Then note that

part21(x1f) = part1(qf + qx2part1(f)) = qpart1(f)minus qpart1(f) + x1part21(f) = 0

part21(x2f) = part1(minusf + qminus1x1part1(f)) = minuspart1(f) + part1(f) + x2part21(f) = 0

part21(x3f) = part1(qx3part1(f)) = q2x3part21(f) = 0

which completes the proof of the first statement in the lemmaStatement 3 follows by definition Statements 4 5 6 and 7 follow from a suitable application of the

q-Leibniz rule Statement 2 follows from an inductive argument We can reduce to i = 1 and j = 3Suppose that partjparti = qpartipartj if j gt i+ 1 Then

part3part1(x1f)minus qpart1part3(x1f) = (qpart3(f) + x2part3part1(f))minus q(part3(f) + x2part1part3(f)) = 0

part3part1(x2f)minus qpart1part3(x2f) = (minuspart3(f) + qminus2x1part3part1(f))minus q(minusqminus1part3(f) + qminus2x1part1part3(f)) = 0

part3part1(x3f)minus qpart1part3(x3f) = (q2part1(f) + q2x4part3part1(f))minus q(qpart1(f) + q2x4part1part3(f)) = 0

part3part1(x4f)minus qpart1part3(x4f) = (minusqpart1(f) + x3part3part1(f))minus q(minuspart1(f) + x3part1part3(f)) = 0

part3part1(x5f)minus qpart1part3(x5f) = q2x5part3part1(f)minus q(q2x5part1part3(f)) = 0

thereby completing the induction

Lemma 511 partiparti+1partiparti+1partiparti+1 + parti+1partiparti+1partiparti+1parti = 0

Proof This result follows from an inductive argument we reduce to i = 1 and assume that the braidrelation holds true for some function f Then we check that the braid relation is true for x1f x2f x3f and x4f (since the behavior of xjf for j ge 4 is the same as that of x4f ) For brevity we will show theargument for x2f only

part1part2(x2f) = qpart1(f) + q2x3part1part2(f) part2part1(x2f) = minuspart2(f) + qminus2x1part2part1(f)

part212(x2f) = qpart2part1(f)minus q2part1part2(f) + qx2part212(f) qpart121(x2f) = minusqpart1part2(f) + part2part1(f) + part121(f)

part1212(x2f) = qpart121(f)minus qpart212(f) + x1part1212(f) part2121(x2f) = minuspart212(f) + part121(f) + x3part2121(f)

part21212(x2f) = qpart2121(f) + qminus1x2part21212(f) part12121(x2f) = minuspart1212(f) + qx3part12121(f)

We continue the above calculations to find that

part121212(x2f) = qpart12121(f) + part21212(f) + x2part121212(f)

part212121(x2f) = minusqpart12121(f)minus part21212(f) + x2part121212(f)

and the braid relation for x2f follows from the inductive hypothesis

6 A Diagrammatic Approach to q-Symmetric Polynomials

61 Introduction to a q-Bialgebra

In the previous section we answered Question 51 in an algebraic way by defining q-analogs of theclassical elementary and complete symmetric functions In this section we generalize the diagrammaticmethod used in [8] in order to study this question from the perspective of bialgebras

Let NΛq be a free associative Z-graded C-algebra with generators hm for m ge 0 We define h0 = 1and hm = 0 for m lt 0 and let q isin Ctimes The homogenous part of NΛq of degree ` has a basis hααkwhere

hα = hα1 middot middot middothαz for a composition α = (α1 αz) of `

17

Define a multiplication for homogenous x and y on NΛqotimes2 as follows where deg(x) denotes the degreeof x

(w otimes x)(y otimes z) = qdeg(x)deg(y)(wy otimes xz)

We can make NΛq into a q-bialgebra by letting the comultiplication on generators be

∆(hn) =

nsumk=0

hk otimes hnminusk

and by letting the counit be ε(x) = 0 if x is homogenous and deg(x)gt 0We can impose through the braiding structure that

∆(hahb) =asumj=0

bsumk=0

(hj otimes haminusj)(hk otimes hbminusk) =asumj=0

bsumk=0

qk(aminusj)(hjhk otimes haminusjhbminusk)

For any partitions λ and micro of n consider the set of double cosets of subroups Sλ and Smicro of Sn SλSnSmicroFor every C in this set let wC be the minimal length representative of C and let `(wC) be the length ofthis minimal length representative We will now attribute a bilinear form to NΛq

(hλ hmicro) =sum

CisinSλSnSmicro

q`(wC)

This bilinear form admits a diagrammatic description Let hn be an orange platform with n non-intersecting strands coming out of it When computing (hλ hmicro) with `(λ) = z and `(micro) = y drawz orange platforms at the top of the diagram representing λ1 λ2middot middot middot λz Draw y orange platforms atthe bottom of the diagram representative of micro1 micro2middot middot middot microy We require that |λ| = |micro| so that the topplatforms and bottom platforms have the same number of strands

Consider the example (h121 h22) In the following diagram snippets of the strands from each plat-form are shown

Every strand must start at one platform at the top and end on another platform at the bottom Nostrands that have originated from one platform may intersect The strands themselves have no criticalpoints with respect to the height function no two strands ever intersect more than once and there are notriple-intersections where three strands are concurrent Diagrams are considered up to isotopy Withoutany restrictions there would be n such diagrams if |λ| = n since there would be no limitations onthe ordering of the strands However due to the above rules there are only 4 possible diagrams in thecomputation of (h121 h22) shown below

Define(hλ hmicro) =

sumall diagrams D representing (hλhmicro)

q number of of crossings in D (61)

In the above example (h121 h22) = 1 + 2q2 + q3

18

We can extend the bilinear form toNΛqotimes2 by stating that any diagram in which strands from distincttensor factors intersect contributes 0 to the bilinear form

(w otimes x y otimes z) = (w y)(x z)

Let I be the radical of the bilinear form inNΛq In [8] the authors prove for any q that multiplicationand comultiplication are adjoint In other words for all xy1 y2 in NΛq

(y1 otimes y2∆(x)) = (y1y2 x) (62)

62 The Elementary q-Symmetric Functions

We now use the bilinear form of q-symmetric functions to study one of their important bases the ele-mentary q-symmetric functions

Define elements ek isin NΛq by ek = 0 for k lt 0 e0 = 1 and

ksumi=0

(minus1)iq(i2)eihkminusi = 0 for k ge 1 (63)

Equivalently leten = qminus(n2)

sumαn

(minus1)`(α)minusnhα (64)

Lemma 61

1 The coproduct of an elementary function is given by ∆(en) =

nsumk=0

ek otimes enminusk

2 If λ n then (hλ en) =

1 if λ = (1 1)

0 otherwise

Proof We begin by demonstrating (2) from which (1) will follow To show (2) it suffices to show that

(hmx en) =

(x enminus1) if m = 10 otherwise

We will utilize strong induction on n in order to find (hmx ekhnminusk) The base cases n = 0 1 are easyto show There are two cases to consider by the inductive hypothesis applied to k lt n Either there isa strand connecting hm and ek or there is not Just as we used an orange platform to denote hn wewill use a blue platform to denote ek The rules of the diagrammatic notation are the same for the blueplatforms as they are for the orange platforms

k

m

nminus k minusm

k nminus k

mlowast lowast lowast lowast lowast

x

If there is not a strand connecting hm and ek the configuration contributes qkm(x ekhnminuskminusm)

19

k minus 1

mminus 1

nminus k minusm + 1

k nminus k

mlowast lowast lowast lowast lowast

x

If a stand connects hm and ek this configuration contributes q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) We have thusshown that (hmx ekhnminusk) = qkm(x ekhnminuskminusm) + q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) Now we are equippedto consider (hmx ek)

(minus1)n+1q(n2)(hmx en) =

nminus1sumk=0

(minus1)kq(k2)(hmx ekhnminusk)

=nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus1sumk=0

(minus1)kq(k2)+(mminus1)(kminus1)(x ekminus1hnminuskminusm+1)

=

nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus2sumk=0

(minus1)k+1q(k+12 )+(mminus1)(k)(x ekhnminuskminusm)

= (minus1)nminus1q(nminus12 )+nm(x enminus1h1minusm)

Corresponding terms from the two sums cancel in pairs since q(k2)+km = q(

k+12 )+k(mminus1) leaving only the

k = nminus 1 term in the first sum The second statement of the lemma thus followsWe will now use (2) to prove (1) This follows from equation 62

(∆(ek) hλ otimes hmicro) = (ek hλhmicro) =

1 λ = (1`) micro = (1p) `+ p = k

0 otherwise

We now calculate the sign incurred when strands connect two blue (ek) platforms

(minus1)n+1q(n2)(en en) =

nminus1sumk=0

(minus1)kq(k2)(en ekhnminusk)

= (minus1)nminus1q(nminus12 )(en enminus1h1)

= (minus1)nminus1q(nminus12 )(∆(en) enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )

nsumk=0

(ek otimes enminusk enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )(enminus1 enminus1)

One may solve this recursion to find that (en en) = qminus(n2)Here the second equality follows from not-ing that at most one strand can connect hnminusk and en (so that k = n minus 1) the third equality followsfrom adjointness and the fourth and fifth equalities follow from the diagrammatic considerations of theprevious lemma

To summarize the diagrammatics of the bilinear form thus developed

20

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

By summing over all indices i we obtain that

nsumi=1

xipi(xλ) = (λ1 + λ2 + + λn)xλ

which implies the desired result

The above lemma holds true in the even case as well where pi is replaced by the partial derivativewith respect to xi We now prove some properties about the action of the third term of the odd Euleroperator on r2 and ∆

Lemma 48 The commutation relation[sum

k 6=i sik∆]

= 0 holds

Proof Note that sjkpi = pjsjk if i = k sjkpi = pksjk if i = j and sjkpi = pisjk otherwise Indeedthese relations can be verified by checking if they are true for xai x

bjxck a b c isin Z+ and then extending

by linearity We prove that sjkpi = pjsjk if i = k and the other two cases are similar Without loss ofgenerality let j lt k and observe that

sjkpk(xajx

bk) = b(minus1)asjk(x

ajx

bminus1k ) = b(minus1)axakx

bminus1j = b(minus1)abxbminus1j xak

pjsjk(xajx

bk) = (minus1)abpj(x

bjxak) = b(minus1)abxbminus1j xak

By our work in Lemma 26 one can deduce that sjkr`m = rsjk(`m)sij As a consequence we find thatsjkDi = Dsjk(i)sjk By an easy induction we now have that sjk∆ = ∆sjk Using the above equationmultiple times proves the desired result

Lemma 49 The commutation relation[sum

k 6=i sik r2]

= 0 holds

Proof Follows since sjkxj = xksjk sjkxk = xjsjk and sjkxi = xisjk if i 6= j k

We are now ready to obtain two commutativity relations involving the odd Euler operator E

Theorem 410 The odd Euler operator and r2 satisfy the following commutation relations

[E r2] = 2r2 (45)[E∆] = minus2∆ (46)

Proof Since r2 has degree 2 and ∆ has degreeminus2 the theorem follows from Lemmas 47 48 and 49

We also need to investigate what the third commutativity relation [r2∆] turns out to be We willprove one lemma before doing so

Lemma 411 For i = 1 to n the equation xiDi +Dixi = 2txipi + t+ usum

k 6=i sik holds

Proof Recall thatDi = tpi + u

sumk 6=i

(x2i minus x2k)minus1[(xi minus xk)sik minus xiτi + xkτk]

Therefore since pixi = xipi + 1

Dixi = txipi + t+ usumk 6=i

(x2i minus x2k)minus1(xixk minus x2k)sik +sumk 6=i

(x2i minus x2k)[x2i τi minus xixkτk]

xiDi = txipi + usumk 6=i

(x2i minus x2k)minus1(x2i minus xixk)sik +sumk 6=i

(x2i minus x2k)minus1[minusx2i τi + xixkτk]

Adding we obtain the desired result

13

We now have the tools to find the third relation between r2 E and ∆

Theorem 412 The commutation relation [r2∆] = E holds

Proof We will first find [r2 Di] The derivative pi much like the partial derivative in the even casesatisfies the properties pixj = minusxjpi for i 6= j and pixi = xipi + 1 Now suppose that i 6= j Then

Dix2j = tpix

2j + ux2j

sumk 6=i 6=j

rik + x2k(x2i minus x2k)minus1[minusxiτi + xkτk] + x2i (x

2i minus x2k)minus1[(xi minus xk)sik]

= tx2jpi + ux2jsumk 6=i

rik + (xi minus xj)sij

Now we will find Dix2i

Dix2i = tpix

2i +

sumk 6=i

x2k(x2i minus x2k)minus1(xi minus xk)sik + x2i

sumk 6=i

(x2i minus x2k)minus1[minusxiτi + xkτk]

= tx2i pi + 2txi + x2isumk 6=i

rik minussumk 6=i

(xi minus xk)sik

Therefore[sumn

i=1 x2i Di

]= minus2txi This implies that

r2Di minusDir2 = minusxi (47)

As a result we find that

[r2∆] = minus(2t)minus1nsumi=1

[r2 D2j ] = minus(2t)minus1

nsumi=1

(r2D2j minusD2

j r2)

= minus(2t)minus1nsumi=1

[(Djr2Dj minus xjDj)minus (Djr

2Dj +Djxj)]

= (2t)minus1nsumi=1

(xiDi +Dixi)

where we have used 47 Now by Lemma 411

[r2∆] =

nsumi=1

xipi +n

2+u

t

sumk 6=i

sik = E

as desired

To summarize we have found operators Er2 and ∆ similar to their even counterparts whichsatisfy the defining relations of the Lie algebra sl2

[E r2] = 2r2

[E∆] = minus2∆

[r2∆] = E

Remark 413 If one uses the odd Dunkl operator ηi as found in [18] instead of the Di introduced herethe r2 E and ∆ operators do not generate sl2

Remark 414 Although our results hold true for all t and u in C one typically sets t = 1 and u = αminus1

for some α isin Ctimes since without loss of generality one of t and u may equal 1

14

Remark 415 In the even case let X be a Euclidean vector space with dimension n and let C[X] be thealgebra of C-valued functions on X Then this result about sl2 plays a major role in the study of higherdifferential operators on C[X] This is because the representation theory of sl2 allows for the reductionof degree to the second order [13] As a result our results in this section should correspondingly have arole in further studying differential operators in the odd case

5 q-nilHecke Algebras

Until now we have been concerned with the odd symmetric polynomials in variables x1 x2 xnwhere xixj = (minus1)xjxi for 1 le i 6= j le n This immediately suggests the question what if one replacesthe minus1 by any constant q isin Ctimes Specifically we ask the following questions

Question 51 Is it possible to study q-symmetric polynomials for which xixj = qxjxi when i gt j

Question 52 Are there q-analogs of evenodd divided difference operators and nilHecke algebras Sofar such structures are known only for the even case (q = 1) and the odd case (q = minus1)

In this section we answer both questions in the affirmativeWe work in the Z-graded q-braided setting throughout LetC be a commutative ring and let q isin Ctimes

be a unit If VW are graded C-modules and v isin V w isin W are homogeneous the braiding is theldquoq-twistrdquo

τq V otimesW rarrW otimes Vv otimes w 7rarr q|v||w|w otimes v

(51)

where | middot | is the degree function By q-algebra we mean an algebra object in the category of gradedC-modules equipped with this braided monoidal structure likewise for q-bialgebras q-Hopf algebrasand so forth

Remark 53 Note that the q-twist described above is its own inverse only when q2 = 1 which correlatesto the even and odd cases When q2 6= 1 the corresponding theory becomes more complex There-fore the q-nilHecke algebras that we introduce later in this section are nontrivial generalizations of thepreviously studied even and odd nilHecke algebras

Definition 54 The q-algebra P qn is defined to be

P qn = C〈x1 xn〉(xjxi minus qxixj = 0 if i lt j) (52)

where |xi| = 1 for i = 1 n

Note that P qn sim= otimesni=1Pq1 There are two interesting subalgebras of P qn that can be thought of as q-

analogs of the symmetric polynomials Define the k-th elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middotxin

and define the k-th twisted elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middot xin

where xj = qjminus1xj

Definition 55 The q-algebra of q-symmetric polynomials in n variables denoted Λqn is the subalgebra ofP qn generated by e1 en Likewise for the twisted q-symmetric polynomials Λqn and e1 en

The type A braid group on n strands acts on P qn by setting

15

1 σi(xj) = qxi+1 if j = i

2 σi(xj) = qminus1xi if j = i+ 1

3 σi(xj) = qxj if j gt i+ 1

4 σi(xj) = qminus1xj if j lt i

and extending multiplicatively

Definition 56 For i = 1 nminus 1 the i-th q-divided difference operator parti is the linear operator P qn rarr P qndefined by parti(xi) = q parti(xi+1) = minus1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + σi(f)parti(g) (53)

for all functions f g isin P qn We call Equation 53 the q-Leibniz rule

Lemma 57 For every i and every j lt k parti(xkxj minus qxjxk) = 0

Proof Since parti(xj) = 0 for j gt i+1 one may reduce the lemma to having to prove that part1(x2x1minusqx1x2) =0 part1(x3x1 minus qx1x3) = 0 and part1(x3x2 minus qx2x3) = 0 These statements follow from the q-Leibniz rule

Therefore parti is a well-defined operator on P qn

Lemma 58 The following relations hold

parti(xki ) =

kminus1sumj=0

qjkminus2jminusj2+kxjix

kminus1minusji+1

parti(xki+1) = minus

kminus1sumj=0

qminusjxjixkminus1minusji+1

Proof We induct on k The base case (k = 1) follows from the definition of the parti and the powers of qarise mostly from xni+1x

mi = qmnxmi x

ni+1 for all mn isin Z+

Our q-divided difference operators also annihilate the twisted elementary q-symmetric polynomialsjust as the even divided difference operators annihilate the elementary symmetric functions

Lemma 59 For every i = 1 nminus 1 and every k parti(ek) = 0 Hence Λqn sube⋂nminus1i=1 ker(parti)

Proof We can express ek as

ek =sum|J|=kii+1isinJ

xJ +sum|J|=kminus1ii+1isinJ

qf(Jik)xJ(xi + qxi+1) +sum|J|=kminus2ii+1isinJ

qg(Jik)xJxixi+1

for certain Z-valued functions f g The result then follows from parti(xi + qxi+1) = parti(xixi+1) = 0 and theq-Leibniz rule

Having discussed q-divided difference operators we can now construct algebras for every q 6=0 1minus1 that have many similarities to the even and odd nilHecke algebras For every such q we de-fine a q-nilHecke algebra generated by xi and parti for 1 le i le n subject to the relations found in thefollowing two lemmas (510 and 511)

Lemma 510 The following relations hold among the operators parti and xi (left multiplication by xi)

1 part2i = 0

2 partjparti minus qpartipartj = 0 for j gt i+ 1

3 xjxi = qxixj for i lt j

4 partixj minus qxjparti = 0 for j gt i+ 1

5 qpartixj minus xjparti = 0 for j lt i

6 partixi minus qxi+1parti = q

7 xiparti minus qpartixi+1 = q

16

Proof To show that part2i = 0 note that we can reduce to i = 1 and proceed by induction Since parti(1) = 0the base case follows Suppose that part2i (f) = 0 Then note that

part21(x1f) = part1(qf + qx2part1(f)) = qpart1(f)minus qpart1(f) + x1part21(f) = 0

part21(x2f) = part1(minusf + qminus1x1part1(f)) = minuspart1(f) + part1(f) + x2part21(f) = 0

part21(x3f) = part1(qx3part1(f)) = q2x3part21(f) = 0

which completes the proof of the first statement in the lemmaStatement 3 follows by definition Statements 4 5 6 and 7 follow from a suitable application of the

q-Leibniz rule Statement 2 follows from an inductive argument We can reduce to i = 1 and j = 3Suppose that partjparti = qpartipartj if j gt i+ 1 Then

part3part1(x1f)minus qpart1part3(x1f) = (qpart3(f) + x2part3part1(f))minus q(part3(f) + x2part1part3(f)) = 0

part3part1(x2f)minus qpart1part3(x2f) = (minuspart3(f) + qminus2x1part3part1(f))minus q(minusqminus1part3(f) + qminus2x1part1part3(f)) = 0

part3part1(x3f)minus qpart1part3(x3f) = (q2part1(f) + q2x4part3part1(f))minus q(qpart1(f) + q2x4part1part3(f)) = 0

part3part1(x4f)minus qpart1part3(x4f) = (minusqpart1(f) + x3part3part1(f))minus q(minuspart1(f) + x3part1part3(f)) = 0

part3part1(x5f)minus qpart1part3(x5f) = q2x5part3part1(f)minus q(q2x5part1part3(f)) = 0

thereby completing the induction

Lemma 511 partiparti+1partiparti+1partiparti+1 + parti+1partiparti+1partiparti+1parti = 0

Proof This result follows from an inductive argument we reduce to i = 1 and assume that the braidrelation holds true for some function f Then we check that the braid relation is true for x1f x2f x3f and x4f (since the behavior of xjf for j ge 4 is the same as that of x4f ) For brevity we will show theargument for x2f only

part1part2(x2f) = qpart1(f) + q2x3part1part2(f) part2part1(x2f) = minuspart2(f) + qminus2x1part2part1(f)

part212(x2f) = qpart2part1(f)minus q2part1part2(f) + qx2part212(f) qpart121(x2f) = minusqpart1part2(f) + part2part1(f) + part121(f)

part1212(x2f) = qpart121(f)minus qpart212(f) + x1part1212(f) part2121(x2f) = minuspart212(f) + part121(f) + x3part2121(f)

part21212(x2f) = qpart2121(f) + qminus1x2part21212(f) part12121(x2f) = minuspart1212(f) + qx3part12121(f)

We continue the above calculations to find that

part121212(x2f) = qpart12121(f) + part21212(f) + x2part121212(f)

part212121(x2f) = minusqpart12121(f)minus part21212(f) + x2part121212(f)

and the braid relation for x2f follows from the inductive hypothesis

6 A Diagrammatic Approach to q-Symmetric Polynomials

61 Introduction to a q-Bialgebra

In the previous section we answered Question 51 in an algebraic way by defining q-analogs of theclassical elementary and complete symmetric functions In this section we generalize the diagrammaticmethod used in [8] in order to study this question from the perspective of bialgebras

Let NΛq be a free associative Z-graded C-algebra with generators hm for m ge 0 We define h0 = 1and hm = 0 for m lt 0 and let q isin Ctimes The homogenous part of NΛq of degree ` has a basis hααkwhere

hα = hα1 middot middot middothαz for a composition α = (α1 αz) of `

17

Define a multiplication for homogenous x and y on NΛqotimes2 as follows where deg(x) denotes the degreeof x

(w otimes x)(y otimes z) = qdeg(x)deg(y)(wy otimes xz)

We can make NΛq into a q-bialgebra by letting the comultiplication on generators be

∆(hn) =

nsumk=0

hk otimes hnminusk

and by letting the counit be ε(x) = 0 if x is homogenous and deg(x)gt 0We can impose through the braiding structure that

∆(hahb) =asumj=0

bsumk=0

(hj otimes haminusj)(hk otimes hbminusk) =asumj=0

bsumk=0

qk(aminusj)(hjhk otimes haminusjhbminusk)

For any partitions λ and micro of n consider the set of double cosets of subroups Sλ and Smicro of Sn SλSnSmicroFor every C in this set let wC be the minimal length representative of C and let `(wC) be the length ofthis minimal length representative We will now attribute a bilinear form to NΛq

(hλ hmicro) =sum

CisinSλSnSmicro

q`(wC)

This bilinear form admits a diagrammatic description Let hn be an orange platform with n non-intersecting strands coming out of it When computing (hλ hmicro) with `(λ) = z and `(micro) = y drawz orange platforms at the top of the diagram representing λ1 λ2middot middot middot λz Draw y orange platforms atthe bottom of the diagram representative of micro1 micro2middot middot middot microy We require that |λ| = |micro| so that the topplatforms and bottom platforms have the same number of strands

Consider the example (h121 h22) In the following diagram snippets of the strands from each plat-form are shown

Every strand must start at one platform at the top and end on another platform at the bottom Nostrands that have originated from one platform may intersect The strands themselves have no criticalpoints with respect to the height function no two strands ever intersect more than once and there are notriple-intersections where three strands are concurrent Diagrams are considered up to isotopy Withoutany restrictions there would be n such diagrams if |λ| = n since there would be no limitations onthe ordering of the strands However due to the above rules there are only 4 possible diagrams in thecomputation of (h121 h22) shown below

Define(hλ hmicro) =

sumall diagrams D representing (hλhmicro)

q number of of crossings in D (61)

In the above example (h121 h22) = 1 + 2q2 + q3

18

We can extend the bilinear form toNΛqotimes2 by stating that any diagram in which strands from distincttensor factors intersect contributes 0 to the bilinear form

(w otimes x y otimes z) = (w y)(x z)

Let I be the radical of the bilinear form inNΛq In [8] the authors prove for any q that multiplicationand comultiplication are adjoint In other words for all xy1 y2 in NΛq

(y1 otimes y2∆(x)) = (y1y2 x) (62)

62 The Elementary q-Symmetric Functions

We now use the bilinear form of q-symmetric functions to study one of their important bases the ele-mentary q-symmetric functions

Define elements ek isin NΛq by ek = 0 for k lt 0 e0 = 1 and

ksumi=0

(minus1)iq(i2)eihkminusi = 0 for k ge 1 (63)

Equivalently leten = qminus(n2)

sumαn

(minus1)`(α)minusnhα (64)

Lemma 61

1 The coproduct of an elementary function is given by ∆(en) =

nsumk=0

ek otimes enminusk

2 If λ n then (hλ en) =

1 if λ = (1 1)

0 otherwise

Proof We begin by demonstrating (2) from which (1) will follow To show (2) it suffices to show that

(hmx en) =

(x enminus1) if m = 10 otherwise

We will utilize strong induction on n in order to find (hmx ekhnminusk) The base cases n = 0 1 are easyto show There are two cases to consider by the inductive hypothesis applied to k lt n Either there isa strand connecting hm and ek or there is not Just as we used an orange platform to denote hn wewill use a blue platform to denote ek The rules of the diagrammatic notation are the same for the blueplatforms as they are for the orange platforms

k

m

nminus k minusm

k nminus k

mlowast lowast lowast lowast lowast

x

If there is not a strand connecting hm and ek the configuration contributes qkm(x ekhnminuskminusm)

19

k minus 1

mminus 1

nminus k minusm + 1

k nminus k

mlowast lowast lowast lowast lowast

x

If a stand connects hm and ek this configuration contributes q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) We have thusshown that (hmx ekhnminusk) = qkm(x ekhnminuskminusm) + q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) Now we are equippedto consider (hmx ek)

(minus1)n+1q(n2)(hmx en) =

nminus1sumk=0

(minus1)kq(k2)(hmx ekhnminusk)

=nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus1sumk=0

(minus1)kq(k2)+(mminus1)(kminus1)(x ekminus1hnminuskminusm+1)

=

nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus2sumk=0

(minus1)k+1q(k+12 )+(mminus1)(k)(x ekhnminuskminusm)

= (minus1)nminus1q(nminus12 )+nm(x enminus1h1minusm)

Corresponding terms from the two sums cancel in pairs since q(k2)+km = q(

k+12 )+k(mminus1) leaving only the

k = nminus 1 term in the first sum The second statement of the lemma thus followsWe will now use (2) to prove (1) This follows from equation 62

(∆(ek) hλ otimes hmicro) = (ek hλhmicro) =

1 λ = (1`) micro = (1p) `+ p = k

0 otherwise

We now calculate the sign incurred when strands connect two blue (ek) platforms

(minus1)n+1q(n2)(en en) =

nminus1sumk=0

(minus1)kq(k2)(en ekhnminusk)

= (minus1)nminus1q(nminus12 )(en enminus1h1)

= (minus1)nminus1q(nminus12 )(∆(en) enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )

nsumk=0

(ek otimes enminusk enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )(enminus1 enminus1)

One may solve this recursion to find that (en en) = qminus(n2)Here the second equality follows from not-ing that at most one strand can connect hnminusk and en (so that k = n minus 1) the third equality followsfrom adjointness and the fourth and fifth equalities follow from the diagrammatic considerations of theprevious lemma

To summarize the diagrammatics of the bilinear form thus developed

20

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

We now have the tools to find the third relation between r2 E and ∆

Theorem 412 The commutation relation [r2∆] = E holds

Proof We will first find [r2 Di] The derivative pi much like the partial derivative in the even casesatisfies the properties pixj = minusxjpi for i 6= j and pixi = xipi + 1 Now suppose that i 6= j Then

Dix2j = tpix

2j + ux2j

sumk 6=i 6=j

rik + x2k(x2i minus x2k)minus1[minusxiτi + xkτk] + x2i (x

2i minus x2k)minus1[(xi minus xk)sik]

= tx2jpi + ux2jsumk 6=i

rik + (xi minus xj)sij

Now we will find Dix2i

Dix2i = tpix

2i +

sumk 6=i

x2k(x2i minus x2k)minus1(xi minus xk)sik + x2i

sumk 6=i

(x2i minus x2k)minus1[minusxiτi + xkτk]

= tx2i pi + 2txi + x2isumk 6=i

rik minussumk 6=i

(xi minus xk)sik

Therefore[sumn

i=1 x2i Di

]= minus2txi This implies that

r2Di minusDir2 = minusxi (47)

As a result we find that

[r2∆] = minus(2t)minus1nsumi=1

[r2 D2j ] = minus(2t)minus1

nsumi=1

(r2D2j minusD2

j r2)

= minus(2t)minus1nsumi=1

[(Djr2Dj minus xjDj)minus (Djr

2Dj +Djxj)]

= (2t)minus1nsumi=1

(xiDi +Dixi)

where we have used 47 Now by Lemma 411

[r2∆] =

nsumi=1

xipi +n

2+u

t

sumk 6=i

sik = E

as desired

To summarize we have found operators Er2 and ∆ similar to their even counterparts whichsatisfy the defining relations of the Lie algebra sl2

[E r2] = 2r2

[E∆] = minus2∆

[r2∆] = E

Remark 413 If one uses the odd Dunkl operator ηi as found in [18] instead of the Di introduced herethe r2 E and ∆ operators do not generate sl2

Remark 414 Although our results hold true for all t and u in C one typically sets t = 1 and u = αminus1

for some α isin Ctimes since without loss of generality one of t and u may equal 1

14

Remark 415 In the even case let X be a Euclidean vector space with dimension n and let C[X] be thealgebra of C-valued functions on X Then this result about sl2 plays a major role in the study of higherdifferential operators on C[X] This is because the representation theory of sl2 allows for the reductionof degree to the second order [13] As a result our results in this section should correspondingly have arole in further studying differential operators in the odd case

5 q-nilHecke Algebras

Until now we have been concerned with the odd symmetric polynomials in variables x1 x2 xnwhere xixj = (minus1)xjxi for 1 le i 6= j le n This immediately suggests the question what if one replacesthe minus1 by any constant q isin Ctimes Specifically we ask the following questions

Question 51 Is it possible to study q-symmetric polynomials for which xixj = qxjxi when i gt j

Question 52 Are there q-analogs of evenodd divided difference operators and nilHecke algebras Sofar such structures are known only for the even case (q = 1) and the odd case (q = minus1)

In this section we answer both questions in the affirmativeWe work in the Z-graded q-braided setting throughout LetC be a commutative ring and let q isin Ctimes

be a unit If VW are graded C-modules and v isin V w isin W are homogeneous the braiding is theldquoq-twistrdquo

τq V otimesW rarrW otimes Vv otimes w 7rarr q|v||w|w otimes v

(51)

where | middot | is the degree function By q-algebra we mean an algebra object in the category of gradedC-modules equipped with this braided monoidal structure likewise for q-bialgebras q-Hopf algebrasand so forth

Remark 53 Note that the q-twist described above is its own inverse only when q2 = 1 which correlatesto the even and odd cases When q2 6= 1 the corresponding theory becomes more complex There-fore the q-nilHecke algebras that we introduce later in this section are nontrivial generalizations of thepreviously studied even and odd nilHecke algebras

Definition 54 The q-algebra P qn is defined to be

P qn = C〈x1 xn〉(xjxi minus qxixj = 0 if i lt j) (52)

where |xi| = 1 for i = 1 n

Note that P qn sim= otimesni=1Pq1 There are two interesting subalgebras of P qn that can be thought of as q-

analogs of the symmetric polynomials Define the k-th elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middotxin

and define the k-th twisted elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middot xin

where xj = qjminus1xj

Definition 55 The q-algebra of q-symmetric polynomials in n variables denoted Λqn is the subalgebra ofP qn generated by e1 en Likewise for the twisted q-symmetric polynomials Λqn and e1 en

The type A braid group on n strands acts on P qn by setting

15

1 σi(xj) = qxi+1 if j = i

2 σi(xj) = qminus1xi if j = i+ 1

3 σi(xj) = qxj if j gt i+ 1

4 σi(xj) = qminus1xj if j lt i

and extending multiplicatively

Definition 56 For i = 1 nminus 1 the i-th q-divided difference operator parti is the linear operator P qn rarr P qndefined by parti(xi) = q parti(xi+1) = minus1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + σi(f)parti(g) (53)

for all functions f g isin P qn We call Equation 53 the q-Leibniz rule

Lemma 57 For every i and every j lt k parti(xkxj minus qxjxk) = 0

Proof Since parti(xj) = 0 for j gt i+1 one may reduce the lemma to having to prove that part1(x2x1minusqx1x2) =0 part1(x3x1 minus qx1x3) = 0 and part1(x3x2 minus qx2x3) = 0 These statements follow from the q-Leibniz rule

Therefore parti is a well-defined operator on P qn

Lemma 58 The following relations hold

parti(xki ) =

kminus1sumj=0

qjkminus2jminusj2+kxjix

kminus1minusji+1

parti(xki+1) = minus

kminus1sumj=0

qminusjxjixkminus1minusji+1

Proof We induct on k The base case (k = 1) follows from the definition of the parti and the powers of qarise mostly from xni+1x

mi = qmnxmi x

ni+1 for all mn isin Z+

Our q-divided difference operators also annihilate the twisted elementary q-symmetric polynomialsjust as the even divided difference operators annihilate the elementary symmetric functions

Lemma 59 For every i = 1 nminus 1 and every k parti(ek) = 0 Hence Λqn sube⋂nminus1i=1 ker(parti)

Proof We can express ek as

ek =sum|J|=kii+1isinJ

xJ +sum|J|=kminus1ii+1isinJ

qf(Jik)xJ(xi + qxi+1) +sum|J|=kminus2ii+1isinJ

qg(Jik)xJxixi+1

for certain Z-valued functions f g The result then follows from parti(xi + qxi+1) = parti(xixi+1) = 0 and theq-Leibniz rule

Having discussed q-divided difference operators we can now construct algebras for every q 6=0 1minus1 that have many similarities to the even and odd nilHecke algebras For every such q we de-fine a q-nilHecke algebra generated by xi and parti for 1 le i le n subject to the relations found in thefollowing two lemmas (510 and 511)

Lemma 510 The following relations hold among the operators parti and xi (left multiplication by xi)

1 part2i = 0

2 partjparti minus qpartipartj = 0 for j gt i+ 1

3 xjxi = qxixj for i lt j

4 partixj minus qxjparti = 0 for j gt i+ 1

5 qpartixj minus xjparti = 0 for j lt i

6 partixi minus qxi+1parti = q

7 xiparti minus qpartixi+1 = q

16

Proof To show that part2i = 0 note that we can reduce to i = 1 and proceed by induction Since parti(1) = 0the base case follows Suppose that part2i (f) = 0 Then note that

part21(x1f) = part1(qf + qx2part1(f)) = qpart1(f)minus qpart1(f) + x1part21(f) = 0

part21(x2f) = part1(minusf + qminus1x1part1(f)) = minuspart1(f) + part1(f) + x2part21(f) = 0

part21(x3f) = part1(qx3part1(f)) = q2x3part21(f) = 0

which completes the proof of the first statement in the lemmaStatement 3 follows by definition Statements 4 5 6 and 7 follow from a suitable application of the

q-Leibniz rule Statement 2 follows from an inductive argument We can reduce to i = 1 and j = 3Suppose that partjparti = qpartipartj if j gt i+ 1 Then

part3part1(x1f)minus qpart1part3(x1f) = (qpart3(f) + x2part3part1(f))minus q(part3(f) + x2part1part3(f)) = 0

part3part1(x2f)minus qpart1part3(x2f) = (minuspart3(f) + qminus2x1part3part1(f))minus q(minusqminus1part3(f) + qminus2x1part1part3(f)) = 0

part3part1(x3f)minus qpart1part3(x3f) = (q2part1(f) + q2x4part3part1(f))minus q(qpart1(f) + q2x4part1part3(f)) = 0

part3part1(x4f)minus qpart1part3(x4f) = (minusqpart1(f) + x3part3part1(f))minus q(minuspart1(f) + x3part1part3(f)) = 0

part3part1(x5f)minus qpart1part3(x5f) = q2x5part3part1(f)minus q(q2x5part1part3(f)) = 0

thereby completing the induction

Lemma 511 partiparti+1partiparti+1partiparti+1 + parti+1partiparti+1partiparti+1parti = 0

Proof This result follows from an inductive argument we reduce to i = 1 and assume that the braidrelation holds true for some function f Then we check that the braid relation is true for x1f x2f x3f and x4f (since the behavior of xjf for j ge 4 is the same as that of x4f ) For brevity we will show theargument for x2f only

part1part2(x2f) = qpart1(f) + q2x3part1part2(f) part2part1(x2f) = minuspart2(f) + qminus2x1part2part1(f)

part212(x2f) = qpart2part1(f)minus q2part1part2(f) + qx2part212(f) qpart121(x2f) = minusqpart1part2(f) + part2part1(f) + part121(f)

part1212(x2f) = qpart121(f)minus qpart212(f) + x1part1212(f) part2121(x2f) = minuspart212(f) + part121(f) + x3part2121(f)

part21212(x2f) = qpart2121(f) + qminus1x2part21212(f) part12121(x2f) = minuspart1212(f) + qx3part12121(f)

We continue the above calculations to find that

part121212(x2f) = qpart12121(f) + part21212(f) + x2part121212(f)

part212121(x2f) = minusqpart12121(f)minus part21212(f) + x2part121212(f)

and the braid relation for x2f follows from the inductive hypothesis

6 A Diagrammatic Approach to q-Symmetric Polynomials

61 Introduction to a q-Bialgebra

In the previous section we answered Question 51 in an algebraic way by defining q-analogs of theclassical elementary and complete symmetric functions In this section we generalize the diagrammaticmethod used in [8] in order to study this question from the perspective of bialgebras

Let NΛq be a free associative Z-graded C-algebra with generators hm for m ge 0 We define h0 = 1and hm = 0 for m lt 0 and let q isin Ctimes The homogenous part of NΛq of degree ` has a basis hααkwhere

hα = hα1 middot middot middothαz for a composition α = (α1 αz) of `

17

Define a multiplication for homogenous x and y on NΛqotimes2 as follows where deg(x) denotes the degreeof x

(w otimes x)(y otimes z) = qdeg(x)deg(y)(wy otimes xz)

We can make NΛq into a q-bialgebra by letting the comultiplication on generators be

∆(hn) =

nsumk=0

hk otimes hnminusk

and by letting the counit be ε(x) = 0 if x is homogenous and deg(x)gt 0We can impose through the braiding structure that

∆(hahb) =asumj=0

bsumk=0

(hj otimes haminusj)(hk otimes hbminusk) =asumj=0

bsumk=0

qk(aminusj)(hjhk otimes haminusjhbminusk)

For any partitions λ and micro of n consider the set of double cosets of subroups Sλ and Smicro of Sn SλSnSmicroFor every C in this set let wC be the minimal length representative of C and let `(wC) be the length ofthis minimal length representative We will now attribute a bilinear form to NΛq

(hλ hmicro) =sum

CisinSλSnSmicro

q`(wC)

This bilinear form admits a diagrammatic description Let hn be an orange platform with n non-intersecting strands coming out of it When computing (hλ hmicro) with `(λ) = z and `(micro) = y drawz orange platforms at the top of the diagram representing λ1 λ2middot middot middot λz Draw y orange platforms atthe bottom of the diagram representative of micro1 micro2middot middot middot microy We require that |λ| = |micro| so that the topplatforms and bottom platforms have the same number of strands

Consider the example (h121 h22) In the following diagram snippets of the strands from each plat-form are shown

Every strand must start at one platform at the top and end on another platform at the bottom Nostrands that have originated from one platform may intersect The strands themselves have no criticalpoints with respect to the height function no two strands ever intersect more than once and there are notriple-intersections where three strands are concurrent Diagrams are considered up to isotopy Withoutany restrictions there would be n such diagrams if |λ| = n since there would be no limitations onthe ordering of the strands However due to the above rules there are only 4 possible diagrams in thecomputation of (h121 h22) shown below

Define(hλ hmicro) =

sumall diagrams D representing (hλhmicro)

q number of of crossings in D (61)

In the above example (h121 h22) = 1 + 2q2 + q3

18

We can extend the bilinear form toNΛqotimes2 by stating that any diagram in which strands from distincttensor factors intersect contributes 0 to the bilinear form

(w otimes x y otimes z) = (w y)(x z)

Let I be the radical of the bilinear form inNΛq In [8] the authors prove for any q that multiplicationand comultiplication are adjoint In other words for all xy1 y2 in NΛq

(y1 otimes y2∆(x)) = (y1y2 x) (62)

62 The Elementary q-Symmetric Functions

We now use the bilinear form of q-symmetric functions to study one of their important bases the ele-mentary q-symmetric functions

Define elements ek isin NΛq by ek = 0 for k lt 0 e0 = 1 and

ksumi=0

(minus1)iq(i2)eihkminusi = 0 for k ge 1 (63)

Equivalently leten = qminus(n2)

sumαn

(minus1)`(α)minusnhα (64)

Lemma 61

1 The coproduct of an elementary function is given by ∆(en) =

nsumk=0

ek otimes enminusk

2 If λ n then (hλ en) =

1 if λ = (1 1)

0 otherwise

Proof We begin by demonstrating (2) from which (1) will follow To show (2) it suffices to show that

(hmx en) =

(x enminus1) if m = 10 otherwise

We will utilize strong induction on n in order to find (hmx ekhnminusk) The base cases n = 0 1 are easyto show There are two cases to consider by the inductive hypothesis applied to k lt n Either there isa strand connecting hm and ek or there is not Just as we used an orange platform to denote hn wewill use a blue platform to denote ek The rules of the diagrammatic notation are the same for the blueplatforms as they are for the orange platforms

k

m

nminus k minusm

k nminus k

mlowast lowast lowast lowast lowast

x

If there is not a strand connecting hm and ek the configuration contributes qkm(x ekhnminuskminusm)

19

k minus 1

mminus 1

nminus k minusm + 1

k nminus k

mlowast lowast lowast lowast lowast

x

If a stand connects hm and ek this configuration contributes q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) We have thusshown that (hmx ekhnminusk) = qkm(x ekhnminuskminusm) + q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) Now we are equippedto consider (hmx ek)

(minus1)n+1q(n2)(hmx en) =

nminus1sumk=0

(minus1)kq(k2)(hmx ekhnminusk)

=nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus1sumk=0

(minus1)kq(k2)+(mminus1)(kminus1)(x ekminus1hnminuskminusm+1)

=

nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus2sumk=0

(minus1)k+1q(k+12 )+(mminus1)(k)(x ekhnminuskminusm)

= (minus1)nminus1q(nminus12 )+nm(x enminus1h1minusm)

Corresponding terms from the two sums cancel in pairs since q(k2)+km = q(

k+12 )+k(mminus1) leaving only the

k = nminus 1 term in the first sum The second statement of the lemma thus followsWe will now use (2) to prove (1) This follows from equation 62

(∆(ek) hλ otimes hmicro) = (ek hλhmicro) =

1 λ = (1`) micro = (1p) `+ p = k

0 otherwise

We now calculate the sign incurred when strands connect two blue (ek) platforms

(minus1)n+1q(n2)(en en) =

nminus1sumk=0

(minus1)kq(k2)(en ekhnminusk)

= (minus1)nminus1q(nminus12 )(en enminus1h1)

= (minus1)nminus1q(nminus12 )(∆(en) enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )

nsumk=0

(ek otimes enminusk enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )(enminus1 enminus1)

One may solve this recursion to find that (en en) = qminus(n2)Here the second equality follows from not-ing that at most one strand can connect hnminusk and en (so that k = n minus 1) the third equality followsfrom adjointness and the fourth and fifth equalities follow from the diagrammatic considerations of theprevious lemma

To summarize the diagrammatics of the bilinear form thus developed

20

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

Remark 415 In the even case let X be a Euclidean vector space with dimension n and let C[X] be thealgebra of C-valued functions on X Then this result about sl2 plays a major role in the study of higherdifferential operators on C[X] This is because the representation theory of sl2 allows for the reductionof degree to the second order [13] As a result our results in this section should correspondingly have arole in further studying differential operators in the odd case

5 q-nilHecke Algebras

Until now we have been concerned with the odd symmetric polynomials in variables x1 x2 xnwhere xixj = (minus1)xjxi for 1 le i 6= j le n This immediately suggests the question what if one replacesthe minus1 by any constant q isin Ctimes Specifically we ask the following questions

Question 51 Is it possible to study q-symmetric polynomials for which xixj = qxjxi when i gt j

Question 52 Are there q-analogs of evenodd divided difference operators and nilHecke algebras Sofar such structures are known only for the even case (q = 1) and the odd case (q = minus1)

In this section we answer both questions in the affirmativeWe work in the Z-graded q-braided setting throughout LetC be a commutative ring and let q isin Ctimes

be a unit If VW are graded C-modules and v isin V w isin W are homogeneous the braiding is theldquoq-twistrdquo

τq V otimesW rarrW otimes Vv otimes w 7rarr q|v||w|w otimes v

(51)

where | middot | is the degree function By q-algebra we mean an algebra object in the category of gradedC-modules equipped with this braided monoidal structure likewise for q-bialgebras q-Hopf algebrasand so forth

Remark 53 Note that the q-twist described above is its own inverse only when q2 = 1 which correlatesto the even and odd cases When q2 6= 1 the corresponding theory becomes more complex There-fore the q-nilHecke algebras that we introduce later in this section are nontrivial generalizations of thepreviously studied even and odd nilHecke algebras

Definition 54 The q-algebra P qn is defined to be

P qn = C〈x1 xn〉(xjxi minus qxixj = 0 if i lt j) (52)

where |xi| = 1 for i = 1 n

Note that P qn sim= otimesni=1Pq1 There are two interesting subalgebras of P qn that can be thought of as q-

analogs of the symmetric polynomials Define the k-th elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middotxin

and define the k-th twisted elementary q-symmetric polynomial to be

ek(x1 xn) =sum

1lei1ltltiklenxi1 middot middot middot xin

where xj = qjminus1xj

Definition 55 The q-algebra of q-symmetric polynomials in n variables denoted Λqn is the subalgebra ofP qn generated by e1 en Likewise for the twisted q-symmetric polynomials Λqn and e1 en

The type A braid group on n strands acts on P qn by setting

15

1 σi(xj) = qxi+1 if j = i

2 σi(xj) = qminus1xi if j = i+ 1

3 σi(xj) = qxj if j gt i+ 1

4 σi(xj) = qminus1xj if j lt i

and extending multiplicatively

Definition 56 For i = 1 nminus 1 the i-th q-divided difference operator parti is the linear operator P qn rarr P qndefined by parti(xi) = q parti(xi+1) = minus1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + σi(f)parti(g) (53)

for all functions f g isin P qn We call Equation 53 the q-Leibniz rule

Lemma 57 For every i and every j lt k parti(xkxj minus qxjxk) = 0

Proof Since parti(xj) = 0 for j gt i+1 one may reduce the lemma to having to prove that part1(x2x1minusqx1x2) =0 part1(x3x1 minus qx1x3) = 0 and part1(x3x2 minus qx2x3) = 0 These statements follow from the q-Leibniz rule

Therefore parti is a well-defined operator on P qn

Lemma 58 The following relations hold

parti(xki ) =

kminus1sumj=0

qjkminus2jminusj2+kxjix

kminus1minusji+1

parti(xki+1) = minus

kminus1sumj=0

qminusjxjixkminus1minusji+1

Proof We induct on k The base case (k = 1) follows from the definition of the parti and the powers of qarise mostly from xni+1x

mi = qmnxmi x

ni+1 for all mn isin Z+

Our q-divided difference operators also annihilate the twisted elementary q-symmetric polynomialsjust as the even divided difference operators annihilate the elementary symmetric functions

Lemma 59 For every i = 1 nminus 1 and every k parti(ek) = 0 Hence Λqn sube⋂nminus1i=1 ker(parti)

Proof We can express ek as

ek =sum|J|=kii+1isinJ

xJ +sum|J|=kminus1ii+1isinJ

qf(Jik)xJ(xi + qxi+1) +sum|J|=kminus2ii+1isinJ

qg(Jik)xJxixi+1

for certain Z-valued functions f g The result then follows from parti(xi + qxi+1) = parti(xixi+1) = 0 and theq-Leibniz rule

Having discussed q-divided difference operators we can now construct algebras for every q 6=0 1minus1 that have many similarities to the even and odd nilHecke algebras For every such q we de-fine a q-nilHecke algebra generated by xi and parti for 1 le i le n subject to the relations found in thefollowing two lemmas (510 and 511)

Lemma 510 The following relations hold among the operators parti and xi (left multiplication by xi)

1 part2i = 0

2 partjparti minus qpartipartj = 0 for j gt i+ 1

3 xjxi = qxixj for i lt j

4 partixj minus qxjparti = 0 for j gt i+ 1

5 qpartixj minus xjparti = 0 for j lt i

6 partixi minus qxi+1parti = q

7 xiparti minus qpartixi+1 = q

16

Proof To show that part2i = 0 note that we can reduce to i = 1 and proceed by induction Since parti(1) = 0the base case follows Suppose that part2i (f) = 0 Then note that

part21(x1f) = part1(qf + qx2part1(f)) = qpart1(f)minus qpart1(f) + x1part21(f) = 0

part21(x2f) = part1(minusf + qminus1x1part1(f)) = minuspart1(f) + part1(f) + x2part21(f) = 0

part21(x3f) = part1(qx3part1(f)) = q2x3part21(f) = 0

which completes the proof of the first statement in the lemmaStatement 3 follows by definition Statements 4 5 6 and 7 follow from a suitable application of the

q-Leibniz rule Statement 2 follows from an inductive argument We can reduce to i = 1 and j = 3Suppose that partjparti = qpartipartj if j gt i+ 1 Then

part3part1(x1f)minus qpart1part3(x1f) = (qpart3(f) + x2part3part1(f))minus q(part3(f) + x2part1part3(f)) = 0

part3part1(x2f)minus qpart1part3(x2f) = (minuspart3(f) + qminus2x1part3part1(f))minus q(minusqminus1part3(f) + qminus2x1part1part3(f)) = 0

part3part1(x3f)minus qpart1part3(x3f) = (q2part1(f) + q2x4part3part1(f))minus q(qpart1(f) + q2x4part1part3(f)) = 0

part3part1(x4f)minus qpart1part3(x4f) = (minusqpart1(f) + x3part3part1(f))minus q(minuspart1(f) + x3part1part3(f)) = 0

part3part1(x5f)minus qpart1part3(x5f) = q2x5part3part1(f)minus q(q2x5part1part3(f)) = 0

thereby completing the induction

Lemma 511 partiparti+1partiparti+1partiparti+1 + parti+1partiparti+1partiparti+1parti = 0

Proof This result follows from an inductive argument we reduce to i = 1 and assume that the braidrelation holds true for some function f Then we check that the braid relation is true for x1f x2f x3f and x4f (since the behavior of xjf for j ge 4 is the same as that of x4f ) For brevity we will show theargument for x2f only

part1part2(x2f) = qpart1(f) + q2x3part1part2(f) part2part1(x2f) = minuspart2(f) + qminus2x1part2part1(f)

part212(x2f) = qpart2part1(f)minus q2part1part2(f) + qx2part212(f) qpart121(x2f) = minusqpart1part2(f) + part2part1(f) + part121(f)

part1212(x2f) = qpart121(f)minus qpart212(f) + x1part1212(f) part2121(x2f) = minuspart212(f) + part121(f) + x3part2121(f)

part21212(x2f) = qpart2121(f) + qminus1x2part21212(f) part12121(x2f) = minuspart1212(f) + qx3part12121(f)

We continue the above calculations to find that

part121212(x2f) = qpart12121(f) + part21212(f) + x2part121212(f)

part212121(x2f) = minusqpart12121(f)minus part21212(f) + x2part121212(f)

and the braid relation for x2f follows from the inductive hypothesis

6 A Diagrammatic Approach to q-Symmetric Polynomials

61 Introduction to a q-Bialgebra

In the previous section we answered Question 51 in an algebraic way by defining q-analogs of theclassical elementary and complete symmetric functions In this section we generalize the diagrammaticmethod used in [8] in order to study this question from the perspective of bialgebras

Let NΛq be a free associative Z-graded C-algebra with generators hm for m ge 0 We define h0 = 1and hm = 0 for m lt 0 and let q isin Ctimes The homogenous part of NΛq of degree ` has a basis hααkwhere

hα = hα1 middot middot middothαz for a composition α = (α1 αz) of `

17

Define a multiplication for homogenous x and y on NΛqotimes2 as follows where deg(x) denotes the degreeof x

(w otimes x)(y otimes z) = qdeg(x)deg(y)(wy otimes xz)

We can make NΛq into a q-bialgebra by letting the comultiplication on generators be

∆(hn) =

nsumk=0

hk otimes hnminusk

and by letting the counit be ε(x) = 0 if x is homogenous and deg(x)gt 0We can impose through the braiding structure that

∆(hahb) =asumj=0

bsumk=0

(hj otimes haminusj)(hk otimes hbminusk) =asumj=0

bsumk=0

qk(aminusj)(hjhk otimes haminusjhbminusk)

For any partitions λ and micro of n consider the set of double cosets of subroups Sλ and Smicro of Sn SλSnSmicroFor every C in this set let wC be the minimal length representative of C and let `(wC) be the length ofthis minimal length representative We will now attribute a bilinear form to NΛq

(hλ hmicro) =sum

CisinSλSnSmicro

q`(wC)

This bilinear form admits a diagrammatic description Let hn be an orange platform with n non-intersecting strands coming out of it When computing (hλ hmicro) with `(λ) = z and `(micro) = y drawz orange platforms at the top of the diagram representing λ1 λ2middot middot middot λz Draw y orange platforms atthe bottom of the diagram representative of micro1 micro2middot middot middot microy We require that |λ| = |micro| so that the topplatforms and bottom platforms have the same number of strands

Consider the example (h121 h22) In the following diagram snippets of the strands from each plat-form are shown

Every strand must start at one platform at the top and end on another platform at the bottom Nostrands that have originated from one platform may intersect The strands themselves have no criticalpoints with respect to the height function no two strands ever intersect more than once and there are notriple-intersections where three strands are concurrent Diagrams are considered up to isotopy Withoutany restrictions there would be n such diagrams if |λ| = n since there would be no limitations onthe ordering of the strands However due to the above rules there are only 4 possible diagrams in thecomputation of (h121 h22) shown below

Define(hλ hmicro) =

sumall diagrams D representing (hλhmicro)

q number of of crossings in D (61)

In the above example (h121 h22) = 1 + 2q2 + q3

18

We can extend the bilinear form toNΛqotimes2 by stating that any diagram in which strands from distincttensor factors intersect contributes 0 to the bilinear form

(w otimes x y otimes z) = (w y)(x z)

Let I be the radical of the bilinear form inNΛq In [8] the authors prove for any q that multiplicationand comultiplication are adjoint In other words for all xy1 y2 in NΛq

(y1 otimes y2∆(x)) = (y1y2 x) (62)

62 The Elementary q-Symmetric Functions

We now use the bilinear form of q-symmetric functions to study one of their important bases the ele-mentary q-symmetric functions

Define elements ek isin NΛq by ek = 0 for k lt 0 e0 = 1 and

ksumi=0

(minus1)iq(i2)eihkminusi = 0 for k ge 1 (63)

Equivalently leten = qminus(n2)

sumαn

(minus1)`(α)minusnhα (64)

Lemma 61

1 The coproduct of an elementary function is given by ∆(en) =

nsumk=0

ek otimes enminusk

2 If λ n then (hλ en) =

1 if λ = (1 1)

0 otherwise

Proof We begin by demonstrating (2) from which (1) will follow To show (2) it suffices to show that

(hmx en) =

(x enminus1) if m = 10 otherwise

We will utilize strong induction on n in order to find (hmx ekhnminusk) The base cases n = 0 1 are easyto show There are two cases to consider by the inductive hypothesis applied to k lt n Either there isa strand connecting hm and ek or there is not Just as we used an orange platform to denote hn wewill use a blue platform to denote ek The rules of the diagrammatic notation are the same for the blueplatforms as they are for the orange platforms

k

m

nminus k minusm

k nminus k

mlowast lowast lowast lowast lowast

x

If there is not a strand connecting hm and ek the configuration contributes qkm(x ekhnminuskminusm)

19

k minus 1

mminus 1

nminus k minusm + 1

k nminus k

mlowast lowast lowast lowast lowast

x

If a stand connects hm and ek this configuration contributes q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) We have thusshown that (hmx ekhnminusk) = qkm(x ekhnminuskminusm) + q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) Now we are equippedto consider (hmx ek)

(minus1)n+1q(n2)(hmx en) =

nminus1sumk=0

(minus1)kq(k2)(hmx ekhnminusk)

=nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus1sumk=0

(minus1)kq(k2)+(mminus1)(kminus1)(x ekminus1hnminuskminusm+1)

=

nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus2sumk=0

(minus1)k+1q(k+12 )+(mminus1)(k)(x ekhnminuskminusm)

= (minus1)nminus1q(nminus12 )+nm(x enminus1h1minusm)

Corresponding terms from the two sums cancel in pairs since q(k2)+km = q(

k+12 )+k(mminus1) leaving only the

k = nminus 1 term in the first sum The second statement of the lemma thus followsWe will now use (2) to prove (1) This follows from equation 62

(∆(ek) hλ otimes hmicro) = (ek hλhmicro) =

1 λ = (1`) micro = (1p) `+ p = k

0 otherwise

We now calculate the sign incurred when strands connect two blue (ek) platforms

(minus1)n+1q(n2)(en en) =

nminus1sumk=0

(minus1)kq(k2)(en ekhnminusk)

= (minus1)nminus1q(nminus12 )(en enminus1h1)

= (minus1)nminus1q(nminus12 )(∆(en) enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )

nsumk=0

(ek otimes enminusk enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )(enminus1 enminus1)

One may solve this recursion to find that (en en) = qminus(n2)Here the second equality follows from not-ing that at most one strand can connect hnminusk and en (so that k = n minus 1) the third equality followsfrom adjointness and the fourth and fifth equalities follow from the diagrammatic considerations of theprevious lemma

To summarize the diagrammatics of the bilinear form thus developed

20

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

1 σi(xj) = qxi+1 if j = i

2 σi(xj) = qminus1xi if j = i+ 1

3 σi(xj) = qxj if j gt i+ 1

4 σi(xj) = qminus1xj if j lt i

and extending multiplicatively

Definition 56 For i = 1 nminus 1 the i-th q-divided difference operator parti is the linear operator P qn rarr P qndefined by parti(xi) = q parti(xi+1) = minus1 parti(xj) = 0 for j 6= i i+ 1 and

parti(fg) = parti(f)g + σi(f)parti(g) (53)

for all functions f g isin P qn We call Equation 53 the q-Leibniz rule

Lemma 57 For every i and every j lt k parti(xkxj minus qxjxk) = 0

Proof Since parti(xj) = 0 for j gt i+1 one may reduce the lemma to having to prove that part1(x2x1minusqx1x2) =0 part1(x3x1 minus qx1x3) = 0 and part1(x3x2 minus qx2x3) = 0 These statements follow from the q-Leibniz rule

Therefore parti is a well-defined operator on P qn

Lemma 58 The following relations hold

parti(xki ) =

kminus1sumj=0

qjkminus2jminusj2+kxjix

kminus1minusji+1

parti(xki+1) = minus

kminus1sumj=0

qminusjxjixkminus1minusji+1

Proof We induct on k The base case (k = 1) follows from the definition of the parti and the powers of qarise mostly from xni+1x

mi = qmnxmi x

ni+1 for all mn isin Z+

Our q-divided difference operators also annihilate the twisted elementary q-symmetric polynomialsjust as the even divided difference operators annihilate the elementary symmetric functions

Lemma 59 For every i = 1 nminus 1 and every k parti(ek) = 0 Hence Λqn sube⋂nminus1i=1 ker(parti)

Proof We can express ek as

ek =sum|J|=kii+1isinJ

xJ +sum|J|=kminus1ii+1isinJ

qf(Jik)xJ(xi + qxi+1) +sum|J|=kminus2ii+1isinJ

qg(Jik)xJxixi+1

for certain Z-valued functions f g The result then follows from parti(xi + qxi+1) = parti(xixi+1) = 0 and theq-Leibniz rule

Having discussed q-divided difference operators we can now construct algebras for every q 6=0 1minus1 that have many similarities to the even and odd nilHecke algebras For every such q we de-fine a q-nilHecke algebra generated by xi and parti for 1 le i le n subject to the relations found in thefollowing two lemmas (510 and 511)

Lemma 510 The following relations hold among the operators parti and xi (left multiplication by xi)

1 part2i = 0

2 partjparti minus qpartipartj = 0 for j gt i+ 1

3 xjxi = qxixj for i lt j

4 partixj minus qxjparti = 0 for j gt i+ 1

5 qpartixj minus xjparti = 0 for j lt i

6 partixi minus qxi+1parti = q

7 xiparti minus qpartixi+1 = q

16

Proof To show that part2i = 0 note that we can reduce to i = 1 and proceed by induction Since parti(1) = 0the base case follows Suppose that part2i (f) = 0 Then note that

part21(x1f) = part1(qf + qx2part1(f)) = qpart1(f)minus qpart1(f) + x1part21(f) = 0

part21(x2f) = part1(minusf + qminus1x1part1(f)) = minuspart1(f) + part1(f) + x2part21(f) = 0

part21(x3f) = part1(qx3part1(f)) = q2x3part21(f) = 0

which completes the proof of the first statement in the lemmaStatement 3 follows by definition Statements 4 5 6 and 7 follow from a suitable application of the

q-Leibniz rule Statement 2 follows from an inductive argument We can reduce to i = 1 and j = 3Suppose that partjparti = qpartipartj if j gt i+ 1 Then

part3part1(x1f)minus qpart1part3(x1f) = (qpart3(f) + x2part3part1(f))minus q(part3(f) + x2part1part3(f)) = 0

part3part1(x2f)minus qpart1part3(x2f) = (minuspart3(f) + qminus2x1part3part1(f))minus q(minusqminus1part3(f) + qminus2x1part1part3(f)) = 0

part3part1(x3f)minus qpart1part3(x3f) = (q2part1(f) + q2x4part3part1(f))minus q(qpart1(f) + q2x4part1part3(f)) = 0

part3part1(x4f)minus qpart1part3(x4f) = (minusqpart1(f) + x3part3part1(f))minus q(minuspart1(f) + x3part1part3(f)) = 0

part3part1(x5f)minus qpart1part3(x5f) = q2x5part3part1(f)minus q(q2x5part1part3(f)) = 0

thereby completing the induction

Lemma 511 partiparti+1partiparti+1partiparti+1 + parti+1partiparti+1partiparti+1parti = 0

Proof This result follows from an inductive argument we reduce to i = 1 and assume that the braidrelation holds true for some function f Then we check that the braid relation is true for x1f x2f x3f and x4f (since the behavior of xjf for j ge 4 is the same as that of x4f ) For brevity we will show theargument for x2f only

part1part2(x2f) = qpart1(f) + q2x3part1part2(f) part2part1(x2f) = minuspart2(f) + qminus2x1part2part1(f)

part212(x2f) = qpart2part1(f)minus q2part1part2(f) + qx2part212(f) qpart121(x2f) = minusqpart1part2(f) + part2part1(f) + part121(f)

part1212(x2f) = qpart121(f)minus qpart212(f) + x1part1212(f) part2121(x2f) = minuspart212(f) + part121(f) + x3part2121(f)

part21212(x2f) = qpart2121(f) + qminus1x2part21212(f) part12121(x2f) = minuspart1212(f) + qx3part12121(f)

We continue the above calculations to find that

part121212(x2f) = qpart12121(f) + part21212(f) + x2part121212(f)

part212121(x2f) = minusqpart12121(f)minus part21212(f) + x2part121212(f)

and the braid relation for x2f follows from the inductive hypothesis

6 A Diagrammatic Approach to q-Symmetric Polynomials

61 Introduction to a q-Bialgebra

In the previous section we answered Question 51 in an algebraic way by defining q-analogs of theclassical elementary and complete symmetric functions In this section we generalize the diagrammaticmethod used in [8] in order to study this question from the perspective of bialgebras

Let NΛq be a free associative Z-graded C-algebra with generators hm for m ge 0 We define h0 = 1and hm = 0 for m lt 0 and let q isin Ctimes The homogenous part of NΛq of degree ` has a basis hααkwhere

hα = hα1 middot middot middothαz for a composition α = (α1 αz) of `

17

Define a multiplication for homogenous x and y on NΛqotimes2 as follows where deg(x) denotes the degreeof x

(w otimes x)(y otimes z) = qdeg(x)deg(y)(wy otimes xz)

We can make NΛq into a q-bialgebra by letting the comultiplication on generators be

∆(hn) =

nsumk=0

hk otimes hnminusk

and by letting the counit be ε(x) = 0 if x is homogenous and deg(x)gt 0We can impose through the braiding structure that

∆(hahb) =asumj=0

bsumk=0

(hj otimes haminusj)(hk otimes hbminusk) =asumj=0

bsumk=0

qk(aminusj)(hjhk otimes haminusjhbminusk)

For any partitions λ and micro of n consider the set of double cosets of subroups Sλ and Smicro of Sn SλSnSmicroFor every C in this set let wC be the minimal length representative of C and let `(wC) be the length ofthis minimal length representative We will now attribute a bilinear form to NΛq

(hλ hmicro) =sum

CisinSλSnSmicro

q`(wC)

This bilinear form admits a diagrammatic description Let hn be an orange platform with n non-intersecting strands coming out of it When computing (hλ hmicro) with `(λ) = z and `(micro) = y drawz orange platforms at the top of the diagram representing λ1 λ2middot middot middot λz Draw y orange platforms atthe bottom of the diagram representative of micro1 micro2middot middot middot microy We require that |λ| = |micro| so that the topplatforms and bottom platforms have the same number of strands

Consider the example (h121 h22) In the following diagram snippets of the strands from each plat-form are shown

Every strand must start at one platform at the top and end on another platform at the bottom Nostrands that have originated from one platform may intersect The strands themselves have no criticalpoints with respect to the height function no two strands ever intersect more than once and there are notriple-intersections where three strands are concurrent Diagrams are considered up to isotopy Withoutany restrictions there would be n such diagrams if |λ| = n since there would be no limitations onthe ordering of the strands However due to the above rules there are only 4 possible diagrams in thecomputation of (h121 h22) shown below

Define(hλ hmicro) =

sumall diagrams D representing (hλhmicro)

q number of of crossings in D (61)

In the above example (h121 h22) = 1 + 2q2 + q3

18

We can extend the bilinear form toNΛqotimes2 by stating that any diagram in which strands from distincttensor factors intersect contributes 0 to the bilinear form

(w otimes x y otimes z) = (w y)(x z)

Let I be the radical of the bilinear form inNΛq In [8] the authors prove for any q that multiplicationand comultiplication are adjoint In other words for all xy1 y2 in NΛq

(y1 otimes y2∆(x)) = (y1y2 x) (62)

62 The Elementary q-Symmetric Functions

We now use the bilinear form of q-symmetric functions to study one of their important bases the ele-mentary q-symmetric functions

Define elements ek isin NΛq by ek = 0 for k lt 0 e0 = 1 and

ksumi=0

(minus1)iq(i2)eihkminusi = 0 for k ge 1 (63)

Equivalently leten = qminus(n2)

sumαn

(minus1)`(α)minusnhα (64)

Lemma 61

1 The coproduct of an elementary function is given by ∆(en) =

nsumk=0

ek otimes enminusk

2 If λ n then (hλ en) =

1 if λ = (1 1)

0 otherwise

Proof We begin by demonstrating (2) from which (1) will follow To show (2) it suffices to show that

(hmx en) =

(x enminus1) if m = 10 otherwise

We will utilize strong induction on n in order to find (hmx ekhnminusk) The base cases n = 0 1 are easyto show There are two cases to consider by the inductive hypothesis applied to k lt n Either there isa strand connecting hm and ek or there is not Just as we used an orange platform to denote hn wewill use a blue platform to denote ek The rules of the diagrammatic notation are the same for the blueplatforms as they are for the orange platforms

k

m

nminus k minusm

k nminus k

mlowast lowast lowast lowast lowast

x

If there is not a strand connecting hm and ek the configuration contributes qkm(x ekhnminuskminusm)

19

k minus 1

mminus 1

nminus k minusm + 1

k nminus k

mlowast lowast lowast lowast lowast

x

If a stand connects hm and ek this configuration contributes q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) We have thusshown that (hmx ekhnminusk) = qkm(x ekhnminuskminusm) + q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) Now we are equippedto consider (hmx ek)

(minus1)n+1q(n2)(hmx en) =

nminus1sumk=0

(minus1)kq(k2)(hmx ekhnminusk)

=nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus1sumk=0

(minus1)kq(k2)+(mminus1)(kminus1)(x ekminus1hnminuskminusm+1)

=

nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus2sumk=0

(minus1)k+1q(k+12 )+(mminus1)(k)(x ekhnminuskminusm)

= (minus1)nminus1q(nminus12 )+nm(x enminus1h1minusm)

Corresponding terms from the two sums cancel in pairs since q(k2)+km = q(

k+12 )+k(mminus1) leaving only the

k = nminus 1 term in the first sum The second statement of the lemma thus followsWe will now use (2) to prove (1) This follows from equation 62

(∆(ek) hλ otimes hmicro) = (ek hλhmicro) =

1 λ = (1`) micro = (1p) `+ p = k

0 otherwise

We now calculate the sign incurred when strands connect two blue (ek) platforms

(minus1)n+1q(n2)(en en) =

nminus1sumk=0

(minus1)kq(k2)(en ekhnminusk)

= (minus1)nminus1q(nminus12 )(en enminus1h1)

= (minus1)nminus1q(nminus12 )(∆(en) enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )

nsumk=0

(ek otimes enminusk enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )(enminus1 enminus1)

One may solve this recursion to find that (en en) = qminus(n2)Here the second equality follows from not-ing that at most one strand can connect hnminusk and en (so that k = n minus 1) the third equality followsfrom adjointness and the fourth and fifth equalities follow from the diagrammatic considerations of theprevious lemma

To summarize the diagrammatics of the bilinear form thus developed

20

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

Proof To show that part2i = 0 note that we can reduce to i = 1 and proceed by induction Since parti(1) = 0the base case follows Suppose that part2i (f) = 0 Then note that

part21(x1f) = part1(qf + qx2part1(f)) = qpart1(f)minus qpart1(f) + x1part21(f) = 0

part21(x2f) = part1(minusf + qminus1x1part1(f)) = minuspart1(f) + part1(f) + x2part21(f) = 0

part21(x3f) = part1(qx3part1(f)) = q2x3part21(f) = 0

which completes the proof of the first statement in the lemmaStatement 3 follows by definition Statements 4 5 6 and 7 follow from a suitable application of the

q-Leibniz rule Statement 2 follows from an inductive argument We can reduce to i = 1 and j = 3Suppose that partjparti = qpartipartj if j gt i+ 1 Then

part3part1(x1f)minus qpart1part3(x1f) = (qpart3(f) + x2part3part1(f))minus q(part3(f) + x2part1part3(f)) = 0

part3part1(x2f)minus qpart1part3(x2f) = (minuspart3(f) + qminus2x1part3part1(f))minus q(minusqminus1part3(f) + qminus2x1part1part3(f)) = 0

part3part1(x3f)minus qpart1part3(x3f) = (q2part1(f) + q2x4part3part1(f))minus q(qpart1(f) + q2x4part1part3(f)) = 0

part3part1(x4f)minus qpart1part3(x4f) = (minusqpart1(f) + x3part3part1(f))minus q(minuspart1(f) + x3part1part3(f)) = 0

part3part1(x5f)minus qpart1part3(x5f) = q2x5part3part1(f)minus q(q2x5part1part3(f)) = 0

thereby completing the induction

Lemma 511 partiparti+1partiparti+1partiparti+1 + parti+1partiparti+1partiparti+1parti = 0

Proof This result follows from an inductive argument we reduce to i = 1 and assume that the braidrelation holds true for some function f Then we check that the braid relation is true for x1f x2f x3f and x4f (since the behavior of xjf for j ge 4 is the same as that of x4f ) For brevity we will show theargument for x2f only

part1part2(x2f) = qpart1(f) + q2x3part1part2(f) part2part1(x2f) = minuspart2(f) + qminus2x1part2part1(f)

part212(x2f) = qpart2part1(f)minus q2part1part2(f) + qx2part212(f) qpart121(x2f) = minusqpart1part2(f) + part2part1(f) + part121(f)

part1212(x2f) = qpart121(f)minus qpart212(f) + x1part1212(f) part2121(x2f) = minuspart212(f) + part121(f) + x3part2121(f)

part21212(x2f) = qpart2121(f) + qminus1x2part21212(f) part12121(x2f) = minuspart1212(f) + qx3part12121(f)

We continue the above calculations to find that

part121212(x2f) = qpart12121(f) + part21212(f) + x2part121212(f)

part212121(x2f) = minusqpart12121(f)minus part21212(f) + x2part121212(f)

and the braid relation for x2f follows from the inductive hypothesis

6 A Diagrammatic Approach to q-Symmetric Polynomials

61 Introduction to a q-Bialgebra

In the previous section we answered Question 51 in an algebraic way by defining q-analogs of theclassical elementary and complete symmetric functions In this section we generalize the diagrammaticmethod used in [8] in order to study this question from the perspective of bialgebras

Let NΛq be a free associative Z-graded C-algebra with generators hm for m ge 0 We define h0 = 1and hm = 0 for m lt 0 and let q isin Ctimes The homogenous part of NΛq of degree ` has a basis hααkwhere

hα = hα1 middot middot middothαz for a composition α = (α1 αz) of `

17

Define a multiplication for homogenous x and y on NΛqotimes2 as follows where deg(x) denotes the degreeof x

(w otimes x)(y otimes z) = qdeg(x)deg(y)(wy otimes xz)

We can make NΛq into a q-bialgebra by letting the comultiplication on generators be

∆(hn) =

nsumk=0

hk otimes hnminusk

and by letting the counit be ε(x) = 0 if x is homogenous and deg(x)gt 0We can impose through the braiding structure that

∆(hahb) =asumj=0

bsumk=0

(hj otimes haminusj)(hk otimes hbminusk) =asumj=0

bsumk=0

qk(aminusj)(hjhk otimes haminusjhbminusk)

For any partitions λ and micro of n consider the set of double cosets of subroups Sλ and Smicro of Sn SλSnSmicroFor every C in this set let wC be the minimal length representative of C and let `(wC) be the length ofthis minimal length representative We will now attribute a bilinear form to NΛq

(hλ hmicro) =sum

CisinSλSnSmicro

q`(wC)

This bilinear form admits a diagrammatic description Let hn be an orange platform with n non-intersecting strands coming out of it When computing (hλ hmicro) with `(λ) = z and `(micro) = y drawz orange platforms at the top of the diagram representing λ1 λ2middot middot middot λz Draw y orange platforms atthe bottom of the diagram representative of micro1 micro2middot middot middot microy We require that |λ| = |micro| so that the topplatforms and bottom platforms have the same number of strands

Consider the example (h121 h22) In the following diagram snippets of the strands from each plat-form are shown

Every strand must start at one platform at the top and end on another platform at the bottom Nostrands that have originated from one platform may intersect The strands themselves have no criticalpoints with respect to the height function no two strands ever intersect more than once and there are notriple-intersections where three strands are concurrent Diagrams are considered up to isotopy Withoutany restrictions there would be n such diagrams if |λ| = n since there would be no limitations onthe ordering of the strands However due to the above rules there are only 4 possible diagrams in thecomputation of (h121 h22) shown below

Define(hλ hmicro) =

sumall diagrams D representing (hλhmicro)

q number of of crossings in D (61)

In the above example (h121 h22) = 1 + 2q2 + q3

18

We can extend the bilinear form toNΛqotimes2 by stating that any diagram in which strands from distincttensor factors intersect contributes 0 to the bilinear form

(w otimes x y otimes z) = (w y)(x z)

Let I be the radical of the bilinear form inNΛq In [8] the authors prove for any q that multiplicationand comultiplication are adjoint In other words for all xy1 y2 in NΛq

(y1 otimes y2∆(x)) = (y1y2 x) (62)

62 The Elementary q-Symmetric Functions

We now use the bilinear form of q-symmetric functions to study one of their important bases the ele-mentary q-symmetric functions

Define elements ek isin NΛq by ek = 0 for k lt 0 e0 = 1 and

ksumi=0

(minus1)iq(i2)eihkminusi = 0 for k ge 1 (63)

Equivalently leten = qminus(n2)

sumαn

(minus1)`(α)minusnhα (64)

Lemma 61

1 The coproduct of an elementary function is given by ∆(en) =

nsumk=0

ek otimes enminusk

2 If λ n then (hλ en) =

1 if λ = (1 1)

0 otherwise

Proof We begin by demonstrating (2) from which (1) will follow To show (2) it suffices to show that

(hmx en) =

(x enminus1) if m = 10 otherwise

We will utilize strong induction on n in order to find (hmx ekhnminusk) The base cases n = 0 1 are easyto show There are two cases to consider by the inductive hypothesis applied to k lt n Either there isa strand connecting hm and ek or there is not Just as we used an orange platform to denote hn wewill use a blue platform to denote ek The rules of the diagrammatic notation are the same for the blueplatforms as they are for the orange platforms

k

m

nminus k minusm

k nminus k

mlowast lowast lowast lowast lowast

x

If there is not a strand connecting hm and ek the configuration contributes qkm(x ekhnminuskminusm)

19

k minus 1

mminus 1

nminus k minusm + 1

k nminus k

mlowast lowast lowast lowast lowast

x

If a stand connects hm and ek this configuration contributes q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) We have thusshown that (hmx ekhnminusk) = qkm(x ekhnminuskminusm) + q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) Now we are equippedto consider (hmx ek)

(minus1)n+1q(n2)(hmx en) =

nminus1sumk=0

(minus1)kq(k2)(hmx ekhnminusk)

=nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus1sumk=0

(minus1)kq(k2)+(mminus1)(kminus1)(x ekminus1hnminuskminusm+1)

=

nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus2sumk=0

(minus1)k+1q(k+12 )+(mminus1)(k)(x ekhnminuskminusm)

= (minus1)nminus1q(nminus12 )+nm(x enminus1h1minusm)

Corresponding terms from the two sums cancel in pairs since q(k2)+km = q(

k+12 )+k(mminus1) leaving only the

k = nminus 1 term in the first sum The second statement of the lemma thus followsWe will now use (2) to prove (1) This follows from equation 62

(∆(ek) hλ otimes hmicro) = (ek hλhmicro) =

1 λ = (1`) micro = (1p) `+ p = k

0 otherwise

We now calculate the sign incurred when strands connect two blue (ek) platforms

(minus1)n+1q(n2)(en en) =

nminus1sumk=0

(minus1)kq(k2)(en ekhnminusk)

= (minus1)nminus1q(nminus12 )(en enminus1h1)

= (minus1)nminus1q(nminus12 )(∆(en) enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )

nsumk=0

(ek otimes enminusk enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )(enminus1 enminus1)

One may solve this recursion to find that (en en) = qminus(n2)Here the second equality follows from not-ing that at most one strand can connect hnminusk and en (so that k = n minus 1) the third equality followsfrom adjointness and the fourth and fifth equalities follow from the diagrammatic considerations of theprevious lemma

To summarize the diagrammatics of the bilinear form thus developed

20

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

Define a multiplication for homogenous x and y on NΛqotimes2 as follows where deg(x) denotes the degreeof x

(w otimes x)(y otimes z) = qdeg(x)deg(y)(wy otimes xz)

We can make NΛq into a q-bialgebra by letting the comultiplication on generators be

∆(hn) =

nsumk=0

hk otimes hnminusk

and by letting the counit be ε(x) = 0 if x is homogenous and deg(x)gt 0We can impose through the braiding structure that

∆(hahb) =asumj=0

bsumk=0

(hj otimes haminusj)(hk otimes hbminusk) =asumj=0

bsumk=0

qk(aminusj)(hjhk otimes haminusjhbminusk)

For any partitions λ and micro of n consider the set of double cosets of subroups Sλ and Smicro of Sn SλSnSmicroFor every C in this set let wC be the minimal length representative of C and let `(wC) be the length ofthis minimal length representative We will now attribute a bilinear form to NΛq

(hλ hmicro) =sum

CisinSλSnSmicro

q`(wC)

This bilinear form admits a diagrammatic description Let hn be an orange platform with n non-intersecting strands coming out of it When computing (hλ hmicro) with `(λ) = z and `(micro) = y drawz orange platforms at the top of the diagram representing λ1 λ2middot middot middot λz Draw y orange platforms atthe bottom of the diagram representative of micro1 micro2middot middot middot microy We require that |λ| = |micro| so that the topplatforms and bottom platforms have the same number of strands

Consider the example (h121 h22) In the following diagram snippets of the strands from each plat-form are shown

Every strand must start at one platform at the top and end on another platform at the bottom Nostrands that have originated from one platform may intersect The strands themselves have no criticalpoints with respect to the height function no two strands ever intersect more than once and there are notriple-intersections where three strands are concurrent Diagrams are considered up to isotopy Withoutany restrictions there would be n such diagrams if |λ| = n since there would be no limitations onthe ordering of the strands However due to the above rules there are only 4 possible diagrams in thecomputation of (h121 h22) shown below

Define(hλ hmicro) =

sumall diagrams D representing (hλhmicro)

q number of of crossings in D (61)

In the above example (h121 h22) = 1 + 2q2 + q3

18

We can extend the bilinear form toNΛqotimes2 by stating that any diagram in which strands from distincttensor factors intersect contributes 0 to the bilinear form

(w otimes x y otimes z) = (w y)(x z)

Let I be the radical of the bilinear form inNΛq In [8] the authors prove for any q that multiplicationand comultiplication are adjoint In other words for all xy1 y2 in NΛq

(y1 otimes y2∆(x)) = (y1y2 x) (62)

62 The Elementary q-Symmetric Functions

We now use the bilinear form of q-symmetric functions to study one of their important bases the ele-mentary q-symmetric functions

Define elements ek isin NΛq by ek = 0 for k lt 0 e0 = 1 and

ksumi=0

(minus1)iq(i2)eihkminusi = 0 for k ge 1 (63)

Equivalently leten = qminus(n2)

sumαn

(minus1)`(α)minusnhα (64)

Lemma 61

1 The coproduct of an elementary function is given by ∆(en) =

nsumk=0

ek otimes enminusk

2 If λ n then (hλ en) =

1 if λ = (1 1)

0 otherwise

Proof We begin by demonstrating (2) from which (1) will follow To show (2) it suffices to show that

(hmx en) =

(x enminus1) if m = 10 otherwise

We will utilize strong induction on n in order to find (hmx ekhnminusk) The base cases n = 0 1 are easyto show There are two cases to consider by the inductive hypothesis applied to k lt n Either there isa strand connecting hm and ek or there is not Just as we used an orange platform to denote hn wewill use a blue platform to denote ek The rules of the diagrammatic notation are the same for the blueplatforms as they are for the orange platforms

k

m

nminus k minusm

k nminus k

mlowast lowast lowast lowast lowast

x

If there is not a strand connecting hm and ek the configuration contributes qkm(x ekhnminuskminusm)

19

k minus 1

mminus 1

nminus k minusm + 1

k nminus k

mlowast lowast lowast lowast lowast

x

If a stand connects hm and ek this configuration contributes q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) We have thusshown that (hmx ekhnminusk) = qkm(x ekhnminuskminusm) + q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) Now we are equippedto consider (hmx ek)

(minus1)n+1q(n2)(hmx en) =

nminus1sumk=0

(minus1)kq(k2)(hmx ekhnminusk)

=nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus1sumk=0

(minus1)kq(k2)+(mminus1)(kminus1)(x ekminus1hnminuskminusm+1)

=

nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus2sumk=0

(minus1)k+1q(k+12 )+(mminus1)(k)(x ekhnminuskminusm)

= (minus1)nminus1q(nminus12 )+nm(x enminus1h1minusm)

Corresponding terms from the two sums cancel in pairs since q(k2)+km = q(

k+12 )+k(mminus1) leaving only the

k = nminus 1 term in the first sum The second statement of the lemma thus followsWe will now use (2) to prove (1) This follows from equation 62

(∆(ek) hλ otimes hmicro) = (ek hλhmicro) =

1 λ = (1`) micro = (1p) `+ p = k

0 otherwise

We now calculate the sign incurred when strands connect two blue (ek) platforms

(minus1)n+1q(n2)(en en) =

nminus1sumk=0

(minus1)kq(k2)(en ekhnminusk)

= (minus1)nminus1q(nminus12 )(en enminus1h1)

= (minus1)nminus1q(nminus12 )(∆(en) enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )

nsumk=0

(ek otimes enminusk enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )(enminus1 enminus1)

One may solve this recursion to find that (en en) = qminus(n2)Here the second equality follows from not-ing that at most one strand can connect hnminusk and en (so that k = n minus 1) the third equality followsfrom adjointness and the fourth and fifth equalities follow from the diagrammatic considerations of theprevious lemma

To summarize the diagrammatics of the bilinear form thus developed

20

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

We can extend the bilinear form toNΛqotimes2 by stating that any diagram in which strands from distincttensor factors intersect contributes 0 to the bilinear form

(w otimes x y otimes z) = (w y)(x z)

Let I be the radical of the bilinear form inNΛq In [8] the authors prove for any q that multiplicationand comultiplication are adjoint In other words for all xy1 y2 in NΛq

(y1 otimes y2∆(x)) = (y1y2 x) (62)

62 The Elementary q-Symmetric Functions

We now use the bilinear form of q-symmetric functions to study one of their important bases the ele-mentary q-symmetric functions

Define elements ek isin NΛq by ek = 0 for k lt 0 e0 = 1 and

ksumi=0

(minus1)iq(i2)eihkminusi = 0 for k ge 1 (63)

Equivalently leten = qminus(n2)

sumαn

(minus1)`(α)minusnhα (64)

Lemma 61

1 The coproduct of an elementary function is given by ∆(en) =

nsumk=0

ek otimes enminusk

2 If λ n then (hλ en) =

1 if λ = (1 1)

0 otherwise

Proof We begin by demonstrating (2) from which (1) will follow To show (2) it suffices to show that

(hmx en) =

(x enminus1) if m = 10 otherwise

We will utilize strong induction on n in order to find (hmx ekhnminusk) The base cases n = 0 1 are easyto show There are two cases to consider by the inductive hypothesis applied to k lt n Either there isa strand connecting hm and ek or there is not Just as we used an orange platform to denote hn wewill use a blue platform to denote ek The rules of the diagrammatic notation are the same for the blueplatforms as they are for the orange platforms

k

m

nminus k minusm

k nminus k

mlowast lowast lowast lowast lowast

x

If there is not a strand connecting hm and ek the configuration contributes qkm(x ekhnminuskminusm)

19

k minus 1

mminus 1

nminus k minusm + 1

k nminus k

mlowast lowast lowast lowast lowast

x

If a stand connects hm and ek this configuration contributes q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) We have thusshown that (hmx ekhnminusk) = qkm(x ekhnminuskminusm) + q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) Now we are equippedto consider (hmx ek)

(minus1)n+1q(n2)(hmx en) =

nminus1sumk=0

(minus1)kq(k2)(hmx ekhnminusk)

=nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus1sumk=0

(minus1)kq(k2)+(mminus1)(kminus1)(x ekminus1hnminuskminusm+1)

=

nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus2sumk=0

(minus1)k+1q(k+12 )+(mminus1)(k)(x ekhnminuskminusm)

= (minus1)nminus1q(nminus12 )+nm(x enminus1h1minusm)

Corresponding terms from the two sums cancel in pairs since q(k2)+km = q(

k+12 )+k(mminus1) leaving only the

k = nminus 1 term in the first sum The second statement of the lemma thus followsWe will now use (2) to prove (1) This follows from equation 62

(∆(ek) hλ otimes hmicro) = (ek hλhmicro) =

1 λ = (1`) micro = (1p) `+ p = k

0 otherwise

We now calculate the sign incurred when strands connect two blue (ek) platforms

(minus1)n+1q(n2)(en en) =

nminus1sumk=0

(minus1)kq(k2)(en ekhnminusk)

= (minus1)nminus1q(nminus12 )(en enminus1h1)

= (minus1)nminus1q(nminus12 )(∆(en) enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )

nsumk=0

(ek otimes enminusk enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )(enminus1 enminus1)

One may solve this recursion to find that (en en) = qminus(n2)Here the second equality follows from not-ing that at most one strand can connect hnminusk and en (so that k = n minus 1) the third equality followsfrom adjointness and the fourth and fifth equalities follow from the diagrammatic considerations of theprevious lemma

To summarize the diagrammatics of the bilinear form thus developed

20

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

k minus 1

mminus 1

nminus k minusm + 1

k nminus k

mlowast lowast lowast lowast lowast

x

If a stand connects hm and ek this configuration contributes q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) We have thusshown that (hmx ekhnminusk) = qkm(x ekhnminuskminusm) + q(kminus1)(mminus1)(x ekminus1hnminuskminusm+1) Now we are equippedto consider (hmx ek)

(minus1)n+1q(n2)(hmx en) =

nminus1sumk=0

(minus1)kq(k2)(hmx ekhnminusk)

=nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus1sumk=0

(minus1)kq(k2)+(mminus1)(kminus1)(x ekminus1hnminuskminusm+1)

=

nminus1sumk=0

(minus1)kq(k2)+km(x ekhnminuskminusm) +

nminus2sumk=0

(minus1)k+1q(k+12 )+(mminus1)(k)(x ekhnminuskminusm)

= (minus1)nminus1q(nminus12 )+nm(x enminus1h1minusm)

Corresponding terms from the two sums cancel in pairs since q(k2)+km = q(

k+12 )+k(mminus1) leaving only the

k = nminus 1 term in the first sum The second statement of the lemma thus followsWe will now use (2) to prove (1) This follows from equation 62

(∆(ek) hλ otimes hmicro) = (ek hλhmicro) =

1 λ = (1`) micro = (1p) `+ p = k

0 otherwise

We now calculate the sign incurred when strands connect two blue (ek) platforms

(minus1)n+1q(n2)(en en) =

nminus1sumk=0

(minus1)kq(k2)(en ekhnminusk)

= (minus1)nminus1q(nminus12 )(en enminus1h1)

= (minus1)nminus1q(nminus12 )(∆(en) enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )

nsumk=0

(ek otimes enminusk enminus1 otimes h1)

= (minus1)nminus1q(nminus12 )(enminus1 enminus1)

One may solve this recursion to find that (en en) = qminus(n2)Here the second equality follows from not-ing that at most one strand can connect hnminusk and en (so that k = n minus 1) the third equality followsfrom adjointness and the fourth and fifth equalities follow from the diagrammatic considerations of theprevious lemma

To summarize the diagrammatics of the bilinear form thus developed

20

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

1 For each crossing there is a factor of q in the bilinear form

2 If two blue platforms are connected by n strands there is a factor of qminus(n2)

3 At most one strand can connect a blue platform to an orange one

63 Relations Between Elementary q-Symmetric Polynomials

In this subsection we apply the diagrammatic method in order to study relations between q-elementarysymmetric polynomials

Define Symq sim= NΛqR where R is the radical of our bilinear form

Lemma 62 If qn = 1 then hn1 is in the center of NΛq

Proof First suppose q is a primitive nth root of unity Construct all ordered k + 1-tuples of nonnegativeintegers that sum to nminus k Let Rnminuskk+1 be the set of all such k + 1-tuples For any tuple (a1 a2 middot middot middot ak+1)let |(a1 a2 middot middot middot ak+1)| be the sum of the entries of the tuple

For these tuples (a1 a2 middot middot middot ak+1) define the map f as follows

f(a1 a2 middot middot middot ak+1) = (ka1 (k minus 1)a2 (k minus 2)a3 middot middot middot ak 0)

DefineP (n k) =

sumRnminuskk+1

q|f(a1a2middotmiddotmiddot ak+1)|

Example 63P (7 2) = 1 + q + 2q2 + 2q3 + 3q4 + 3q5 + 3q6 + 2q7 + 2q8 + q9 + q10

m

x

Consider the above diagram representative of (hn1hm ekx) In the diagram n = 7 and m = 3 Thethree strands from e3 rdquosplitrdquo the seven h1rsquos into groups of 1 2 1 and 0 This is a 3 + 1-tuple that sums to7minus3 = nminusk = 4 Numbering the h1rsquos from left to right note that the first h1 contributes qk intersectionsthe third and fourth h1rsquos contribute qkminus1 intersections and so on In general the diagrams in which nostrand connects hm and ek contribute P (n k)(hnminusk1 hm x) to (hn1hm ekx)

mminus 1

x

If a strand connects ek to hm then it intersects the other nminus (kminus 1) strands connecting some h1 to xcontributing a factor of qnminusk+1The other intersections contribute P (n k minus 1) Putting this case and theprevious case together we obtain that

(hn1hm ekx) = P (n k)(hnminusk1 hm x) + qnminusk+1P (n k minus 1)(hnminusk+11 hmminus1 x) (65)

21

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

m

x

mminus 1

x

Similarly the above two diagrams show that

(hmhn1 ekx) = qmkP (n k)(hmh

nminusk1 x) + q(mminus1)(kminus1)P (n k minus 1)(hmminus1h

nminusk+11 x) (66)

Now consider the case when k = n+ 1 In this case there is only one diagram for the bilinear formand it can be shown that

(hn1hm en+1x) = (hmminus1 x)

(hmhn1 en+1x) = qn(mminus1)(hmminus1 x)

which are equal since qn = 1 Now if k le n we claim that P (n k) = 0 for all n 6= k This follows fromthe fact that qn = 1 that qnminus`) 6= 1 for ` isin (1 2 3 middot middot middot nminus 1) and the fact that

P (n k) =

(n

k

)q

The above statement follows from a bijection establishing P (n k) as the Gaussian binomial coeffi-cient

(nk

)q It is known that the coefficient of qj in

(nk

)q

is the number of partitions of j into k or fewerparts with each part less than or equal to k P (n k) yields the same result since f takes every k+1-tupleto a k + 1-tuple with last term 0 Each term must be less than or equal to n minus k since we have imposedthat the sum of all the terms is nminus k

We substitute P (n k) = 0 in (65) and (66) to find that both products (hn1hm ekx) and (hmhn1 ekx)

are 0 unless n = k or n = k minus 1 (already addressed) If n = k then(hn1hm enx) = (hm x) + qP (n nminus 1)(h1hmminus1 x)

(hmhn1 enx) = qnm(hm x) + q(mminus1)(nminus1)P (n nminus 1)(hmminus1h1 x)

Since qmn = 1 and P (n nminus 1) = 0 the above two expressions are equal We therefore have the desiredresult when q is a primitive root of unity By using some basic number theory and the recursive propertyof the Gaussian polynomials that (

n

k

)q

= qk(nminus 1

k

)q

+

(nminus 1

k minus 1

)q

one may extend the result to any root of unity

22

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

64 Insertion

In this subsection we develop the novel idea of insertion as a method for developing further relationsin NΛq Note from the previous arguments in this section that many diagrammatic relations betweenelementary symmetric functions involve evaluating the bilinear form (hλ ekx) for some λ k and x isinNΛq The insertion method aids in the general computation of this bilinear form

Let λ and micro be compositions such that λ = (λ1 λ2 middot middot middot λz) and micro = (micro1 micro2 middot middot middot microz) The length of λand micro which will be denoted by `(λ) and `(micro) is z Define |λ| = λ1 + λ2 + middot middot middot + λz Let σk` be a binarysequence of 0rsquos and 1rsquos with k total elements ` of which are 1 Let Ok` be the set of all σk` for given k and` The size of the set Ok` is

(k`

)

Define subtraction and multiplication of compositions in a component-wise manner

λminus micro = (λ1 minus micro1 λ2 minus micro2 middot middot middot λz minus microz)

λmicro = (λ1micro1 λ2micro2 middot middot middot λzmicroz)

Let Tmn be the composition with m elements all of which are n Let λGk = (λk+1 λk+2 middot middot middot λz) and letλLk = (λ1 λ2 middot middot middot λk) Further let r(λ) denote the composition (λ1 λ1+λ2 λ1+λ2+λ3 middot middot middot λ1+middot middot middot+λz)Define (hλ ekx)hmicro to be the result when computing (hλ ekx) but with hmicro appended to the beginning ofhα all bilinear forms (hα x) We call this process insertion

Example 64(h2h3 e1x)h1 = (h1h1h3 x) + q2(h1h2h2 x)

We now show some applications of insertion The first is a result that simplifies the computation ofa specific bilinear form

Lemma 65 The equation (hnhλ ekx) = q(kminus1)(nminus1)(hλ ekminus1x)hnminus1 + qkn(hλ ekx)hn holds

Proof We utilize casework and the diagrammatic approach There are two cases either there exists astrand connecting hn and ek or there is not

x

If there exists a strand connecting hn to ek then summing across all possible diagrams we obtain(hλ ekminus1x)hnminus1 The insertion of hnminus1 is due to the fact that nminus 1 strands from hn intersect x and mustbe accounted for when summing However each of the n minus 1 strands from the hn platform intersectseach of the k minus 1 strands from ek to hλ This case contributes q(kminus1)(nminus1)(hλ ekminus1x)hnminus1

x

If no strand connects hn to ek then summing across all possible diagrams we obtain (hλ ekx)hn Theinsertion of hn is due to the fact that n strands from hn intersect x which must be accounted for in thesummation However each of the n strands from the hn platform intersects each of the k strands fromek to hλ so this case contributes qkn(hλ ekx)hn

These are the only two possible cases and putting the two cases together yields the desired result

23

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

Also note that

(hn ekx) =

(hn x) if k = 0(hnminus1 x) if k = 10 if k lt 0 or k gt 1

since at most one strand can connect hn and ekWe can now compute the general bilinear form (hλ ekx) thereby facilitating the discovery of further

relations between elementary symmetric functions

Lemma 66 We have that

(hλ ekx) =

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

Proof We induct on m If m = 1 then the proposition becomes

(hλ ekx) =1suml=0

sumO1l

q|(λ1minusσ1l )(kminusr(σ

1l ))|(hλG1

ekminuslx)hλL1 minusσ1

l

(67)

which reduces to Proposition 12Now assume that the result holds for m Then

(hλ ekx) =msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|(hλGm ekminuslx)h

λLmminusσml

=msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminuslminus1)(λm+1minus1)(hλGm+1

ekminuslminus1x)hλLmminusσm

lhλm+1minus1

+

msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

We therefore have

(hλ ekx) =m+1suml=1

sumOml

q|(λLmminusσmlminus1)(T

mk minusr(σ

mlminus1))|q(kminusl)(λm+1minus1)(hλGm+1

ekminuslx)hλLmminusσm

lminus1hλm+1minus1

(68)

+msuml=0

sumOml

q|(λLmminusσml )(Tmk minusr(σ

ml ))|q(kminusl)(λm+1)(hλGm+1

ekminuslx)hλLmminusσm

lhλm+1

(69)

Let aml denote a composition in Oml that ends in a 0 Let bml denote a composition in Oml that ends ina 1 Let Aml and Bm

l be the set of all aml and bml respectively Now consider the terms indexed only by1 le l le m

msuml=1

sumBm+1l

q|(λLm+1minusb

m+1l )(Tm+1

k minusr(bm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusb

m+1l

+

msuml=1

sumAm+1l

q|(λLm+1minusa

m+1l )(Tm+1

k minusr(am+1l ))|(hλGm+1

ekminuslx)hλLm+1minusa

m+1l

=m+1suml=0

sumOm+1l

q|(λLm+1minusσ

m+1l )(Tm+1

k minusr(σm+1l ))|(hλGm+1

ekminuslx)hλLm+1minusσ

m+1l

24

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

The terms indexed by 1 le l le m match their corresponding terms in the Proposition It remains toconsider the cases l = 0 and l = m For l = 0 note that there does not exist a bm0 and for l = m + 1note that there does not exist an amm From here it is easy to see that these terms satisfy the propositionas well (the l = 0 term can be found in the second sum of (68) and the l = m + 1 term can be found inthe first term of (68))

Therefore an explicit formula for the bilinear form can be given by

(hλ ekx) =

0 if k ge z + 1

q|(λLkminus1minusσ

kminus1kminus1)(T

kminus1k minusr(σkminus1

kminus1))|(hλLkminus1minusσkminus1kminus1

hλkminus1) if k = z

sumOzminus1kminus1

q|(λLzminus1minusσ

zminus1kminus1)(T

zminus1k minusr(σzminus1

kminus1))|(hλLzminus1minusσzminus1kminus1

hλzminus1)

+sumOzminus1k

q|(λLzminus1minusσ

zminus1k )(T zminus1

k minusr(σzminus1k ))|(hλLzminus1minusσ

zminus1k

hλk) if k lt z

7 Conclusion and Further Research

Through this work we have contributed towards the program of oddification by studying propertiesof odd Dunkl operators in relation to diverse ideas in mathematics namely we connected odd Dunkloperators to odd divided difference operators the classical Yang-Baxter equation and the important Liealgebra sl2 We used inductive arguments and introduced refinements of the odd divided differenceoperators and the odd Dunkl operators in order to prove our main results By discovering odd versionsof the Dunkl Laplacian and sl2-triples which play important roles in the representation theory of evensymmetric polynomials we have strengthened the odd theory and provided new areas of investigationfor future researchers

In Section 4 we gave an action of sl2 on skew polynomials through a variant of the Khongsap-Wangodd Dunkl operator In the future we will try to describe the weight spaces and isotypic decompositionof this representation We could also apply our results by studying higher degree differential operatorsin the odd case since the representation theory of sl2 allows us to conveniently reduce degree to secondorder [13]

Ellis one of the authors who introduced the odd nilHecke algebra asked if there were odd analogsof other symmetric polynomials such as Jack polynomials or Macdonald polynomials Here we outlinea procedure for answering his question and making progress towards finding odd Jack polynomialsWe first introduce the odd Cherednik operators

Yi = minusαxiηi +sumklti

sik minus (nminus 1) (71)

Applying arguments similar to those used by Khongsap and Wang in [18] we can find that

1 YiYj = YjYi

2 siYi = Yi+1si minus 1

3 siYi+1 = Yisi + 1

4 siYj = Yjsi for j 6= i i+ 1

The next step would be to find a scalar product for which the odd Cherednik operators are self-adjoint One can then define the odd Jack polynomials as eigenfunctions of the odd Cherednik opera-tors and study their properties as in [20] Since the odd Cherednik operators are closely related to the

25

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

odd Dunkl operators and the rik we introduced in Section 2 the work in this paper would contributesignificantly towards the study of odd Jack polynomials

Factorization entails yet another problem of interest in the odd theory For example one can use themethod of undetermined coefficients to show that for odd n

xn1 minus xn2 = (x1 + ax2)

nminus1sumk=0

vkakxnminus1minusk1 xk2

where vnminus1an = minus1 and vk is defined as follows

vk =

1 k equiv 0 3 (mod 4)

minus1 k equiv 1 2 (mod 4)

Such identities arise in subtle ways in the action of operators on P and the study of these kinds ofnoncommutative factorizations have separate combinatorial interests as well

In Section 5 we introduced q-nilHecke algebras for all q 6= 0 1minus1 It would be interesting to study ifthe q-nilHecke algebras categorify an interesting Lie theoretic algebra and whether they can be used toconstruct invariants of links or other geometric structures One could also begin a diagrammatic studyof the q-nilHecke algebras as in [9]

In the same section we defined elementary q-symmetric functions which brings up the problem offinding relations between these generators and further studying the structure of Λqn We introduced amethod for solving this problem using diagrams in Section 6 and found some of these relations How-ever the remaining relations between the ei are much more complex than their even or odd counterpartsand merit further study When q3 = 1 for example the following degree 6 relation holds

v1 = e11211 + e12111 + e21111

v2 = e1122 minus 2e1221 + 3e2112 + e2211

v3 = 2e1131 minus 2e114 + 2e1311 minus 2e141 + 3e222 + 2e1113 minus 2e411

v1 + q2v2 + qv3 = 0

where eλ = eλ1 eλk for λ = (λ1 λ2 λk)We also conjecture that our results and especially the definition of odd Cherednik operators have

connections to recent geometric work of Braden Licata Proudfoot and Webster who have constructedcategory O for certain Cherednik algebras As a result we believe that the ideas in this paper willfurther develop the program of oddification and also create a more thorough understanding of higherrepresentation theoretic structures

8 Acknowledgements

I would like to thank my mentor Alexander Ellis for introducing me to his own research in representa-tion theory and helping me obtain a more intuitive understanding of the odd construction In additionProfessor Pavel Etingof Professor Tanya Khovanova and Dr Ben Elias thoroughly edited this paper andprovided encouragement I also express gratitude to MIT PRIMES USA for giving me the opportunityto conduct this research

References

[1] S Axler P Gorkin and K Voss The Dirichlet problem on quadratic surfaces Math Comp 73 (2004)637651

26

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

[2] T H Baker and P J Forrester Isomorphisms of type A affine Hecke algebras and multivariableorthogonal polynomials arxivorgpdfq-alg9710036pdf

[3] W Baratta and P J Forrester Jack polynomial fractional quantum Hall states and their generaliza-tions arxivorgabs10072692

[4] L Crane I Frenkel Four-dimensional topological quantum field theory Hopf categories and thecanonical bases J Math Phys 35 51365154 (1994) arxivorgabshep-th9405183

[5] Z Dancso A Licata Odd Khovanov Homology for Hyperplane Arrangementsarxivorgabs12052784

[6] H De Bie B Orsted P Somberg and V Souccek Dunkl operators and a family of realizations ofosp1|2 Preprint arxivorgabs09114725

[7] C F Dunkl Differential-difference operators associated to reection groups Trans Amer Math Soc311 (1989) 167183

[8] A P Ellis and M Khovanov The Hopf algebra of odd symmetric functions Advances in Mathemat-ics 231(2) 965-999 2012 arxivorgabs11075610

[9] A P Ellis M Khovanov and A Lauda The odd nilHecke algebra and its diagrammatics Interna-tional Mathematics Review Notices 2012 arXivmathQA11111320

[10] A P Ellis and A Lauda An odd categorification of quantum sl2 arxivorgabs13077816

[11] P Etingof Lectures on Calogero-Moser systems arxivorgabsmath0606233v3 Preprint 2006

[12] I Gelfand D Krob A Lascoux B Leclerc V Retakh and J-Y Thibon Noncommutative symmetricfunctions Advances in Mathematics 112218348 1995 arxivorgabshep-th9407124

[13] G J Heckman A remark on the Dunkl differential-difference operators Barker W Sally P (eds)Harmonic analysis on reductive groups Progress in Math 101181191 Basel Birkhauser Verlag 1991

[14] S J Kang M Kashiwara and S J Oh Supercategorication of quantum Kac-Moody algebras II2013 arxivorgabs13031916

[15] B Kostant and S Kumar The nil Hecke ring and cohomology of gp for a Kac-Moody group gProceedings of the National Academy of Sciences of the USA 83(6)15431545 1986

[16] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras I The classical affinetype Transf Groups 13389412 2008 arxivorgabs07040201

[17] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras II The rational doubleafne type Pacic J Math 23873103 2008 arxivorgabs07105877

[18] T Khongsap and W Wang Hecke-Clifford algebras and spin Hecke algebras IV Odd double affinetype SIGMA 5 2009 arxivorgabs08102068

[19] M Khovanov A categorification of the Jones polynomial arxivorgabsmath9908171

[20] F Knop and S Sahi A recursion and a combinatorial formula for Jack polynomialsarxivorgabsq-alg9610016

[21] A Lauda and H Russell Oddification of the cohomology of type A Springer varietiesarxivorgabs12030797

27

[22] P Ozsvath J Rasmussen and Z Szabo Odd Khovanov homology arxivorgabs07104300

[23] F Plomp Dunkl operators and Fischer decompositions igitur-archivelibraryuunl

[24] M Rosler Dunkl Operators Theory and Applications arxivorgabsmath0210366

[25] H J Schmidt and J Schnack Symmetric polynomials in physics arxivorgpdfcond-mat0209397pdf

[26] A Shumakovitch Patterns in odd Khovanov homology arxivorgabs11015607

[27] R P Stanley Enumerative Combinatorics vol 2 Cambridge University Press Cambridge UK1999

[28] E Witten Khovanov Homology And Gauge Theory arxivorgabs11083103

28