Abstract. In this paper, we study the properties of the inversion statistic and the Fibonaccimajor index, Fibmaj, as defined on standard Fibonacci tableaux. We prove that these twostatistics are symmetric and log-concave over all standard Fibonacci tableaux of a given shapeμ and provide two combinatorial proofs of the symmetry result, one a direct bijection on theset of tableaux and the other utilizing 0, 1-fillings of a staircase shape. We conjecture that theinversion and Fibmaj statistics are log-concave over all standard Fibonacci tableaux of a givensize n. In addition, we show a well-known bijection between standard Fibonacci tableauxof size n and involutions in Sn which takes the Fibmaj statistic to a new statistic called thesubmajor index on involutions.
Keywords: tableaux, Fibonacci tableaux, involutions, statistics, major index, log-concave, in-versions
1. Introduction
In 1975, Richard Stanley defined the Fibonacci lattice [9] which gave rise to a newkind of tableaux called Fibonacci tableaux. In [6], the second author extended thewell-known definitions of the inversion statistic and the major index for permutationsto an inversion statistic and Fibonacci major index for standard Fibonacci tableaux.MacMahon [7] proved a famous theorem stating that the inversion statistic is equidis-tributed with the major index on permutations in Sn, i.e.,
∑π∈Sn
qmaj(π) = ∑π∈Sn
qinv(π). (1.1)
The second author gave a similar result for the inversion statistic and the Fibmajstatistic for standard Fibonacci tableaux in [6]:
∑T
qinv(T ) = ∑T
qFibmaj(T),
604 N. Cameron and K. Killpatrick
where the sum is over all standard Fibonacci tableaux of a given shape μ , althoughthis result was previously proven in the context of a certain set of permutations byBjorner and Wachs [2]. In this paper, we prove that the inversion statistic and theFibmaj statistic are both symmetric over all standard Fibonacci tableaux of size n.
In addition to the symmetry property of the inversion and Fibmaj statistic, westudy a second property of these statistics called log-concavity. A sequence a0, a1, . . . ,ak of real numbers is said to be log-concave if a2
i ≥ ai−1ai+1 for all 1 ≤ i ≤ k−1. Apolynomial is said to be log-concave if its coefficients form a log-concave sequence.A related concept is that of unimodality. A sequence a0, a1, . . . , ak is called unimodalif for some 0 ≤ i ≤ k we have a0 ≤ a1 ≤ ·· · ≤ ai ≥ ai+1 ≥ ·· · ≥ ak. A polynomial issaid to be unimodal if its coefficients form a unimodal sequence.
Log-concave and unimodal sequences show up often in many different areas ofcomputer science and mathematics, including combinatorics, algebra, probability,statistics, and geometry. The study of these types of sequences and polynomials hasa rich history and in 1989 Stanley [10] gave a nice survey of the variety of methodsthat are useful for showing that a sequence is log-concave or unimodal, since provingsuch properties for a sequence can be quite a challenging and difficult task. This sur-vey was updated by Brenti [5] in 1994 due to a number of new techniques that hadbeen developed by that time.
In this paper, we prove that the inversion statistic and the Fibmaj statistic arelog-concave over all standard Fibonacci tableaux of a given shape μ . As a corollary,we obtain the result that these statistics are also unimodal over all standard Fibonaccitableaux of a given shape μ . In addition, we conjecture that the Fibmaj statistic is log-concave over all standard Fibonacci tableaux of a given size n and discuss progressmade towards the proof of this conjecture and related results.
Finally, we define an interesting new statistic called the submajor index on in-volutions and show that a certain bijection between standard Fibonacci tableaux andinvolutions takes the Fibmaj statistic to the submajor index.
2. Background and Definitions
2.1. Permutation Statistics
Much work has been done to study permutation statistics on permutations in thesymmetric group Sn. Given a permutation σ = σ1σ2 · · ·σn ∈ Sn where σi = σ(i), wedefine an inversion to be a pair (i, j) such that i < j and σi > σ j. Then the inversionstatistic, inv(σ), is the total number of inversions in σ , i.e.,
inv(σ) = ∑i< j
σi>σ j
1.
For example, for σ = 3 2 8 5 7 4 6 1 9 , inv(σ) = 15. We define the descent setof σ , Des σ , as the set of all i such that σi > σi+1. Similarly, we define the inversedescent set of σ , Ides σ , as the set of all i such that i + 1 is to the left of i in σ =σ1σ2 · · ·σn. In addition, define the excedance set of σ , Exc σ , as the set of all i suchthat σi > i.
Symmetry and Log-Concavity on Fibonacci Tableaux 605
The major index of a permutation σ , written maj(σ), is
maj(σ) = ∑i∈Des σ
i,
and the corresponding inverse major index is
imaj(σ) = ∑i∈Ides σ
i.
Since Ides σ = Des σ−1, we have imaj(σ) = maj(σ−1).
In 1915, MacMahon proved the following important theorem:
Theorem 2.1. (MacMahon, 1915) [7]
∑π∈Sn
qinv(π) = ∑π∈Sn
qmaj(π).
MacMahon’s theorem has played a central role in many areas of enumerativecombinatorics and has been widely studied, refined, and generalized.
2.2. Symmetry, Log-Concavity and Unimodality of Sequences
A sequence a0, a1, . . . , ak of real numbers is called symmetric if ai = ak−i for0 ≤ i ≤ k. A polynomial is called symmetric if the sequence of its coefficients isa symmetric sequence. As defined in Section 1, a sequence a0, a1, . . . ,ak of realnumbers is said to be log-concave if a2
i ≥ ai−1ai+1 for all 1 ≤ i ≤ k−1 and a polyno-mial is said to be log-concave if its coefficients form a log-concave sequence. Sim-ilarly, a sequence a0, a1, . . . , ak is called unimodal if for some 0 ≤ i ≤ k we havea0 ≤ a1 ≤ ·· · ≤ ai ≥ ai+1 ≥ ·· · ≥ ak and a polynomial is said to be unimodal if its co-efficients form a unimodal sequence. Probably the best known example of a sequencethat is symmetric, unimodal, and log-concave is the nth row of Pascal’s triangle, i.e.,the sequence
(n0
),(n
1
), . . . ,
(nn
).
As far back as 1839, Rodriguez [8] gave the generating function for inversions onpermutations in Sn:
It is a classic result of log-concavity (see [4] for a proof) that the product of twolog-concave functions is log-concave. Thus one can see that the inversion statisticis log-concave over Sn. In 2005, Bona [3] finally gave a combinatorial proof of thisresult, which we present in Section 6.
2.3. Pattern Avoidance in Permutations
Let Sn be the symmetric group of all permutations of the set [n] = {1, 2, . . . , n}and suppose π = π1π2 · · ·πn and σ = σ1σ2 · · ·σn are two permutations in Sn. We saythat π is order isomorphic to σ if πi < π j if and only if σi < σ j. For any π ∈ Sn andσ ∈ Sk for k ≤ n, we say that π contains a copy of σ if π has a subsequence that is
606 N. Cameron and K. Killpatrick
order isomorphic to σ . If π contains no subsequence order isomorphic to σ , then wesay π avoids σ . This is the notion of classical pattern avoidance and we write σ withdashes between each of the elements σ1, σ2, . . . , σk to indicate that these elementsneed not be adjacent in π .
In a generalized permutation pattern (see [1] for a complete introduction), we mayrequire some of the letters in σ to be adjacent which will be indicated by not writingdashes between these letters in σ . For example, a 3−1−2 pattern in π would be anysubsequence πi, π j, πk for i < j < k with π j < πk < πi. A 31−2 pattern in π wouldbe any subsequence πi, πi+1, π j for i+ 1 < j with πi+1 < π j < πi.
2.4. Fibonacci Tableaux
Fibonacci tableaux arise from the Fibonacci lattice first defined in the followingmanner by Stanley [9] in 1975. Let A = {11, 12, . . . , 1r, 2} and let A∗ be the set of allfinite words a1a2 · · ·ak of elements of A (including the empty word).
Definition 2.2. The Fibonacci differential poset Z(r) has as its elements the set ofwords in A∗. For w ∈ Z(r), we say z is covered by w (i.e., z � w) in Z(r) if either
(1) z is obtained from w by changing a 2 to 1k for some k if the only letters to the leftof this 2 are also 2’s, or
(2) z is obtained from w by deleting the leftmost 1k for any type k.
We use the term shape to refer to an element in Z(1). The Ferrers diagram for anelement μ ∈ Z(1) is formed by replacing a 1 with a single dot and a 2 with two dots.The size of the Fibonacci shape is the sum of the 1’s and 2’s that make up the shape.For example, the Ferrers diagram for the Fibonacci shape μ = 122121 looks like
• • •
• • • • • •
and has size 9.
Definition 2.3. Given a Fibonacci shape μ of size n and its corresponding Ferrersdiagram, a partially standard Fibonacci tableau is a filling of the Ferrers diagramwith positive integers such that the bottom row decreases from left to right and everycolumn decreases from bottom to top. A standard Fibonacci tableau is a partiallystandard Fibonacci tableau using the integers 1 through n.
An example of a standard Fibonacci tableau of shape μ = 122121 is
3 4 29 8 7 6 5 1
Definition 2.4. We define the column reading word, wc(T ), for a standard Fi-bonacci tableau T by reading the columns from bottom to top, right to left.
For the tableau above, wc(T ) = 152674839.Due to the restrictions on the standard Fibonacci tableaux (i.e., columns must
decrease from bottom to top and the bottom row must decrease from left to right),we obtain an immediate bijection between the set of column reading words for stan-dard Fibonacci tableaux and the set of permutations which avoid both the generalizedpatterns 312 and 321.
Symmetry and Log-Concavity on Fibonacci Tableaux 607
3. Statistics on Fibonacci Tableaux
Definition 3.1. Given a standard Fibonacci tableau T , we say i is an element of thedescent set of T , written Des(T ), if i + 1 appears below or in a column to the rightof i in the tableau T .
This means that if i is in the descent set of T then i+1 is to the left of i in wc(T ).Thus Des(T ) = Ides(wc(T )).
Definition 3.2. Let T be a standard Fibonacci tableau and let wc(T ) be the columnreading word of T . We define the inversion statistic of T , inv(T ), as
inv(T ) = inv(wc(T )).
Similarly, we will define the inverse major index of T , imaj(T ), as
imaj(T ) = imaj(wc(T )).
Since imaj(wc(T )) = ∑i∈Ides(wc(T )) i = ∑i∈Des(T) i, then we can define imaj(T ) =
∑i∈Des(T) i. Since we will only be using the latter statistic in the context of Fibonaccitableaux, we will refer to this statistic as the Fibonacci major index of T , i.e.,
Fibmaj(T ) = ∑i∈Des(T)
i.
Definition 3.3. Define T minμ as the standard Fibonacci tableau obtained by filling
the columns of the Ferrers diagram for the shape μ with the numbers 1 through nconsecutively from right to left, top to bottom. Additionally, let n j denote the numberin the bottom of the jth column of height 2 (reading right to left).
For example, if μ = 122121, then
T min122121 =
7 5 2
9 8 6 4 3 1
and n1 = 3, n2 = 6, and n3 = 8.Note that T min
μ has the minimum number of inversions for any standard Fibonaccitableau of shape μ .
A Fibonacci shape μ of size n can be described as the shape of the Fibonaccitableau uniquely determined by a sequence n1, n2, . . . , nk of elements that are in thefirst row and in a column of size 2 in the tableau T min
μ . For example, the sequence(3, 5, 9) definitively describes the tableau
T min211221 =
8 4 2
9 7 6 5 3 1
of size 9 and thus gives the Fibonacci shape μ = 211221.
608 N. Cameron and K. Killpatrick
Definition 3.4. Let μ be a Fibonacci shape. Define
mμ = ∑i
ni,
where the ni’s correspond to T minμ .
For example, m122121 = 3 + 6 + 8 = 17.
Definition 3.5. Define T maxμ as the standard Fibonacci tableau obtained by filling the
top most row of μ with the numbers 1 through k from left to right and the bottom rowwith the remaining numbers k + 1 to n from right to left.
For example, for μ = 122121, then
T max122121 =
1 2 3
9 8 7 6 5 4.
Note that T maxμ has the maximum number of inversions for any standard Fibonacci
tableau of shape μ .
Lemma 3.6. For any Fibonacci shape μ ,
mμ = inv(T min
μ)+ inv
(T max
μ).
Proof. Let μ be a Fibonacci shape and consider the kth column of height two, wherenk = j. Then in T min
μ , j − 1 is the element on top of j and j − 1 induces a singleinversion (with j) in the corresponding column reading word. Now suppose that inT max
μ , l is the element in the top row of the kth column of height 2. Since the elementj below l is larger than l and every element in any column to the right of l is smallerthan j, in the corresponding column reading word l forms inversions with each ofthese j−1 elements. Thus the contribution to inv of this column in T min
μ is 1 and inT max
μ is j−1, giving a total contribution of j.
4. Symmetry of the Fibmaj Statistic
In [6], the author shows that
∑T
qFibmaj(T ) = ∑T
qinv(T),
where the sum is over all standard Fibonacci tableaux T of a given shape μ , althoughthe result itself had been proven by Bjorner and Wachs [2] for a set of permutationsthat are in one-to-one correspondence with standard Fibonacci tableaux (the authormade the connections to Fibonacci tableaux). This implies the additional result thatthe Fibmaj statistic is equidistributed with the inversion statistic over all standardFibonacci tableaux T of size n.
Letf (μ) = ∑
Tqinv(T ) = ∑
TqFibmaj(T),
where the sum is over all standard Fibonacci tableaux T of a given shape μ . Then wehave the following result:
Symmetry and Log-Concavity on Fibonacci Tableaux 609
Lemma 4.1. The function f satisfies the recurrence
f (ε) = 1, where ε denotes the empty shape,
f (1μ) = f (μ),
f (2μ) =(
q + q2 + · · ·+ q|μ|+1)
f (μ),
where μ is a Fibonacci shape.
Proof. The base cases are trivial to check. For any standard Fibonacci tableau of shape1μ , n is placed in the single box in the leftmost column and so in the column readingword for the tableau, n is the last number in the word. Thus this n does not contributeto the inversion statistic nor does it change the inversion statistic that is computed onthe numbers 1 through n− 1 that are in the tableau of shape μ , thus f (1μ) = f (μ).For any standard Fibonacci tableau of shape 2μ , n is placed in the bottom cell ofthe leftmost column. We will let x denote the number that is placed above n in theleftmost column. The column reading word for 2μ is then the column reading wordfor μ , call it wμ , followed by n x. Since n is the largest number in the word, it doesnot form inversions with any of the numbers in wμ , so inv(2μ) = inv(wμ) + (thenumber of inversions that x forms with wμ n). Since x ranges from n−1 to 1, x formsanywhere from 1 to (|μ |+ 1) inversions. Thus f (2μ) =
(q + q2 + · · ·+ q|μ|+1
)f (μ).
If we define f (μ) = Pμ(q), then we have the following symmetry result:
Theorem 4.2. The inversion statistic and the Fibmaj statistic are symmetric over allstandard Fibonacci tableaux of a given shape μ , i.e.,
Pμ(q) = qmμ Pμ(1/q).
Proof. With the result from Lemma 3.6 and the use of the recurrence in Lemma 4.1,the proof follows immediately from induction on the size of μ .
5. Two Combinatorial Proofs of the Symmetry of the Inversion Statistic for Fi-
bonacci Tableaux
Although the recurrence gives us an immediate algebraic proof, we now also presenttwo different combinatorial proofs, one utilizing a direct method for taking a Fi-bonacci tableau of shape μ with an inversion number of k and turning it into a Fi-bonacci tableau of shape μ with an inversion number of mμ −k and the other makinguse of the symmetry of 0, 1-fillings of a staircase shape of size n−1.
5.1. A Direct Bijection on Fibonacci Tableaux
Definition 5.1. Let T be a partially standard Fibonacci tableau of shape μ . Let a jdenote the jth smallest element in T . We define g(T ) as follows:
610 N. Cameron and K. Killpatrick
(1) If μ = 1 or μ = 2, then g(T ) = T .(2) Suppose that μ = 1ν with ν �= /0 and let the size of μ be n. Then, T = anT ′ where
T ′ is a partially standard tableau of shape ν . Let g(T ) = ang(T ′).
(3) Suppose μ = 2ν with ν �= /0 and let the size of μ be n. Then T =akan T ′ where
T ′ is a partially standard tableau of shape ν . Note that T ′ contains the ele-
ments a1, a2, . . . , ak−1, ak+1, . . . , an−1. In this case we define g(T ) =an−kan g(T ′)
where T ′ is the partially standard tableau of shape ν that contains the elementsa1, a2, . . . , an−k−1, an−k+1, . . . , an−1 in the same relative position as the elementsa1, a2, . . . , ak−1, ak+1, . . . , an−1 appeared in T ′.
For example, suppose
T =3 4 2
9 8 7 6 5 1.
Then,
g(T ) =9 g
(3 4 28 7 6 5 1
)
=5
9 8 g
(3 27 6 4 1
)
=5 3
9 8 7 g
(2
6 4 1
)
=5 3
9 8 7 6 g
(24 1
)
=5 3 1
9 8 7 6 4 g
(2
)
=5 3 1
9 8 7 6 4 2.
Theorem 5.2. The function g is an involution on the set of standard Fibonacci tableauof shape μ . Furthermore, if T is a standard Fibonacci tableau of shape μ theninv(g(T )) = mμ − inv(T ).
Proof. That the function g is an involution on the set of standard Fibonacci tableauof shape μ is immediate from the definition of g. We will prove that g has the statedeffect on the inversion number by induction on the number of columns in T :
Base case: Let T be the standard Fibonacci tableau 1 of shape μ = 1. Then inv(T ) = 0and m1 = 0. In this case, g(T ) = T and inv(g(T )) = 0−0 = 0. Let T be the standard
Fibonacci tableau12 of shape μ = 2. Then inv(T ) = 1 and m2 = 2. In this case,
g(T ) = T and inv(g(T )) = 2−1 = 1.
Inductive step: Assume that for T ′, a standard Fibonacci tableau of shape ν , whereν has k columns, if inv(T ′) = l, then inv(g(T ′)) = mν − l. Now let T be a standard
Symmetry and Log-Concavity on Fibonacci Tableaux 611
Fibonacci tableau of shape μ and size n where μ = 1ν or μ = 2ν , i.e., μ has k + 1columns.
(1) If μ = 1ν then T = nT ′ where T ′ has shape ν and g(T ) = ng(T ′). Note thatwc(T ) = wc(T ′)n. Since n is the largest element in T , we have that inv(T ) =inv(wc(T )) = inv(wc(T ′)n) = inv(wc(T ′)) = inv(T ′). By the same reasoning,inv(g(T )) = inv(g(T ′)). Also, since the columns of height 2 in T are in the samepositions as the columns of height 2 in T ′ then we have mμ = mν . Supposeinv(T ) = l. Then inv(T ′) = l and, by the induction hypothesis, inv(g(T ′)) =mν − l. Hence, inv(g(T )) = inv(g(T ′)) = mν − l = mμ − l = mμ − inv(T ).
(2) If μ = 2ν then T =kn T ′ where T ′ has shape ν and g(T ) =
n−kn g(T ′). In this case,
mμ = n+mν . Also, we have wc(T ) = wc(T ′) n k and wc(g(T )) = wc(g(T ′)) n n−k. This implies that inv(T ) = inv(wc(T )) = inv(wc(T ′))+ (n− k) = inv(T ′)+(n − k) and inv(g(T )) = inv(wc(g(T ))) = inv(wc(g(T ′))) + (n − (n − k)) =inv(g(T ′)) + k. Now, making use of inv(T ) = inv(T ′) + (n− k), inv(g(T )) =inv(g(T ′))+ k, and the induction hypothesis, we have
inv(g(T )) = inv(g(T ′))+ k
= mν − inv(T ′)+ k
= mμ −n− inv(T ′)+ k
= mμ − (n− k)− inv(T ′)
= mμ − (n− k)− (inv(T )− (n− k))
= mμ − inv(T ).
In the previous example, with
T =3 4 2
9 8 7 6 5 1
and
g(T ) =5 3 1
9 8 7 6 4 2,
we have inv(T ) = inv(wc(T )) = inv(152674839)= 9 and inv(g(T )) = inv(wc(g(T )))= inv(241673859)= 8. Since m122121 = 17, we have inv(g(T )) = m122121 − inv(T ).
This gives a combinatorial proof that the inversion statistic is symmetric over allstandard Fibonacci tableaux of a given shape μ . Utilizing this result and the bijec-tion given in [6], one can obtain a combinatorial proof that the Fibmaj statistic issymmetric over all standard Fibonacci tableaux of a given shape μ .
5.2. A Symmetry Proof Using 0, 1-Fillings
Our second proof of this result utilizes a bijection between Fibonacci tableaux and0, 1-fillings of staircase shapes. The advantage of the second proof is that the natural
612 N. Cameron and K. Killpatrick
symmetry of 0, 1-fillings of staircase shapes will lead immediately to the result thatthe inversion statistic is symmetric.
We say λ = (λ1, λ2, . . . , λk) is a partition of n if λ1 ≥ λ2 ≥ ·· · ≥ λk > 0 and∑k
i=1 λi = n. A partition is described pictorially by its Ferrers diagram, an array of ncells into k left-justified rows with row i containing λi cells for 1 ≤ i ≤ k. A staircaseshape of size n is the shape of the partition λ = (n, n−1, . . . , 1). We label the cells ofa shape as (i, j) with i, the row label, increasing from left to right and j, the columnlabel, increasing from top to bottom. The cells of a staircase shape can be split into ndiagonals {dk}k=1,2,...,n where dk = {cells (i, j) | i+ j = k + 1}.
A 0, 1-filling of a partition shape is a filling of each cell of the shape either by a 0or a 1.
Definition 5.3. Let μ be a Fibonacci shape of size n and, as in Definition 3.3, let{ni}i=1,..., j denote the sequence of elements that appear in the first row and in acolumn of height 2 in T min
μ (reading right to left). A 0, 1-filling of the staircase shapeof size n− 1 is said to be μ-admissible if there is exactly one cell filled by 1 in thediagonal dni−1 for each 1 ≤ i ≤ j and all other diagonals are filled by 0.
For example, let μ = 211221 be a Fibonacci shape with sequence {ni}i=1,..., j =(9, 5, 3) then a μ-admissible 0, 1-filling of the staircase shape of size 8 is
Note that for a given μ there are many μ-admissible 0, 1-fillings of the corre-sponding staircase shape.
Lemma 5.4. There is a bijection between μ-admissible 0, 1-fillings of a staircaseshape of size n−1 and standard Fibonacci tableaux of shape μ and size n.
Proof. In the case where n = 1, the empty 0, 1-filling of size 0 corresponds to theFibonacci tableau T = 1.
Suppose n > 1. Let S be a μ-admissible 0, 1-filling of a staircase shape of size n−1. If there is no 1 on the longest diagonal, dn−1, then μ = 1ν and the correspondingFibonacci tableau is T = nT ′, where T ′ is the Fibonacci tableau of shape ν describedby the 0, 1-filling of the first n−2 diagonals of S. Suppose there is a 1 on the longest
diagonal dn−1 in row i, then μ = 2ν and T =n− i
n T ′, where T ′ is the Fibonacci tableauof shape ν described by the 0, 1-filling of the first n− 3 diagonals of S, relabeled inthe natural way to contain the numbers 1, 2, . . . , n− i−1, n− i+ 1, . . . , n−1.
For the reverse direction, let T be a Fibonacci tableau of shape μ . Consider thejth (reading right to left) column of height two in T . Let k j be the number at thetop of this column of height two. Suppose there are i j numbers less than k j in any
Symmetry and Log-Concavity on Fibonacci Tableaux 613
column to the right of the column containing k j. Then place a 1 in row n j − (i j + 1)of diagonal dn j−1 in a staircase shape of size n−1. Do this for all such j and then fillin the rest of the staircase shape with 0’s. Since for all j, n j > n j−1 +1, any diagonalcontaining a 1 will have a preceding diagonal containing all 0’s, thus the 0, 1-fillingobtained in this manner is μ-admissible.
Lemma 5.5. Given a standard Fibonacci tableau T of shape μ and size n, the inver-sion statistic for T can be determined by the sum of the row indices of cells filled by1 in the corresponding 0, 1-filling of the staircase shape of size n−1.
Proof. This statement is easy to prove by induction and the base cases are left tothe reader as they are straightforward. Suppose n > 1. Let S be a μ-admissible 0, 1-filling of a staircase shape of size n−1. If there is no 1 on the longest diagonal, dn−1,then μ = 1ν and the corresponding Fibonacci tableau is T = nT ′, where T ′ is theFibonacci tableau of shape ν described by the 0, 1-filling of the first n−2 diagonalsof S. Since n is at the end of the column reading word for T , then inv(T ) = inv(T ′).
If there is a 1 on the longest diagonal dn−1 in row i, then μ = 2ν and T =n− i
n T ′,where T ′ is the Fibonacci tableau of shape ν described by the 0, 1-filling of the firstn−3 diagonals of S, relabeled in the natural way to contain the numbers 1, 2, . . . , n−i− 1, n− i + 1, . . . , n− 1. The column reading word for T is equal to the columnreading word for T ′ followed by n and then n− i. Then inv(T ) is equal to inv(T ′)plus the number of inversions that n− i makes with those elements in T ′, which is i.Thus if there is a 1 in row i in T , the contribution to the number of inversions in T isalso i.
Now given a Fibonacci shape μ , let φ be a function on μ-admissible 0, 1-fillingsof the staircase shape of size n−1 that reflects a filling across the diagonal i = j.
Theorem 5.6. The function φ is an involution on the set of μ-admissible 0, 1-fillingsof the staircase shape of size n−1. Furthermore, if S is a μ-admissible 0, 1-filling ofthe staircase shape of size n−1 and T is the standard Fibonacci tableau of shape μthat corresponds to S, then inv(T ) = mμ − inv(T ) where T is the standard Fibonaccitableau of shape μ that corresponds to φ(S).
Proof. Since the shape μ is determined from the 0, 1-filling by which diagonals con-tain 1’s, the shape is clearly preserved by the function φ . In addition, the conditionsfor being a μ-admissible 0, 1-filling are also clearly preserved under the function φ ,so φ is an involution on the set of μ-admissible 0, 1-fillings of the staircase shape ofsize n−1.
By Lemma 5.5 we have that the sum of the indices of the rows containing 1’s in aμ-admissible 0, 1-filling S of a staircase shape of size n−1 is equal to the inversionstatistic for the corresponding standard Fibonacci tableau T of shape μ . If thereis a 1 in row i j on diagonal dn j−1, then in φ(S) there is a 1 in row n j − i j on thesame diagonal dn j−1. Then inv(T ) = i1 + i2 + · · ·+ ik (note that these i j’s need notbe distinct). Then in φ(S) the row indices of cells filled by 1 are n1 − i1, n2 − i2, . . . ,nk − ik, and inv(T ) = (n1 − i1) + (n2 − i2) + · · ·+ (nk − ik) = (n1 + n2 + · · ·+ nk)− (i1 + i2 + · · ·+ ik) = mμ − inv(T ).
614 N. Cameron and K. Killpatrick
6. Log-Concavity of the Fibmaj Statistic
Let f (q) = anqn + an−1qn−1 + · · ·+ a1q + a0 be a function with non-negative integercoefficients. Recall that such a function is called log-concave if ak+1ak−1 ≤ a2
k for1 ≤ k ≤ n− 1 and unimodal if there exists a k such that ai−1 ≤ ai for 1 ≤ i ≤ k anda j ≥ a j+1 for k ≤ j ≤ n− 1. It is an easy exercise to verify that the condition forlog-concavity is equivalent to akam ≤ ak+iam−i for 0 ≤ i ≤ m− k, by noting that thelog-concavity of f (q) means ak+1/ak is a weakly decreasing sequence. It is knownthat for finite positive sequences, log-concavity implies unimodality [4].
As in the previous section, we denote by f (μ) the distribution on all standardFibonacci tableaux of a give shape μ of the statistics inv or Fibmaj counted by q.That is,
f (μ) = ∑T
qinv(T ) = ∑T
qFibmaj(T),
where the sum is over all standard Fibonacci tableaux T of a given shape μ . Then wehave the following theorem:
Theorem 6.1. The inversion statistic and the Fibmaj statistic are log-concave overall standard Fibonacci tableaux of a given shape μ .
Proof. By applying induction and the fact that the product of two log-concave func-tions is also log-concave [10] to the given recurrence for f (μ), we obtain our theorem.
Corollary 6.2. The inversion statistic and the Fibmaj statistic are unimodal over allstandard Fibonacci tableaux of a given shape μ .
Proof. From the recurrence in Lemma 4.1, it is clear that the coefficients of f (μ)form a finite positive sequence and thus since f (μ) is log-concave, it is unimodal.
Making further use of the recurrence given in Lemma 4.1, we can obtain a moregeneral recurrence for the inversion statistic and the Fibmaj statistic on all standardFibonacci tableaux of size n. Let
F(n) = ∑T
qinv(T ) = ∑T
qFibmaj(T ),
where the sum is over all standard Fibonacci tableaux of a given size n. Then sincethe recurrence in Lemma 4.1 relies only on the size |μ | of the remaining shape, wecan generalize to the following recurrence for F(n):
This leads us to the following conjecture, which has been tested by Maple for alln up to 35.
Conjecture 6.4. The inversion statistic and the Fibmaj statistic are log-concave overall standard Fibonacci tableaux of a given size n.
Symmetry and Log-Concavity on Fibonacci Tableaux 615
In 2005, Bona [3] gave the first (and only, to this point) combinatorial proof thatthe generating function for inversions on permutations in Sn is log-concave. We out-line his proof here since it is helpful to see what happens when we apply his proofto the set of permutations that are column reading words for Fibonacci tableaux (i.e.,permutations that are 312- and 321-avoiding).
For his combinatorial proof, Bona proves by induction that there exists an injec-tion fn,k,k+2 from Ik × Ik+2 → Ik+1 × Ik+1 where Im is the set of permutations in Snwith m inversions. The interested reader can check Bona’s original paper for the basecases, we will present the inductive step.
Let (p, q) ∈ Ik × Ik+2 with p = p1 p2 · · · pn and q = q1q2 ·qn. Then:
(1) If p1 < n and q1 > 1, increase p1 by one and decrease the entry in p that was onelarger than p1 by one. Call the new permutation p′. Similarly, decrease q1 byone and increase the entry of q that was one larger than q1 by one. Call this newpermutation q′. Note that p′ starts with an entry larger than 1 and q′ starts withan entry smaller than n. Let fn,k,k+2(p, q) = (p′, q′).
(2) If p1 = n or q1 = 1, remove these entries to obtain permutations p∗ and q∗.Perform the natural relabeling on the entries in p∗ and q∗ to obtain permutationsin Sn−1. By induction (there are some technicalities to check here), apply theinjection to the pair (p∗, q∗) to obtain the pair (p, q). Prepend p by p1 and qby q1 and increase the entries larger than or equal to the prepended entry by 1.Finally, set fn,k,k+2(p, q) = (q1q, p1 p). Note that either q1q starts with 1 or p1 pstarts with n, thus we have obtained a different pair than the result of part one.
Why doesn’t this injection work to give us a combinatorial proof that the inversionstatistic is log-concave for standard Fibonacci tableaux of size n? Suppose in p wehave p1 = m, p2 = a, and p3 = m + 1 or p1 = m, p2 = m + 1, and p3 = a for some1 ≤ m ≤ n− 1 and a < m and suppose q1 �= 1. Then we are in case (1) and p′ =m + 1 a m · · · or p′ = m + 1 m a · · · , respectively. Unfortunately, this gives us a p′
that is not 312-avoiding (in the first case) or not 321-avoiding (in the second case) andthus is not a column reading word for a Fibonacci tableau. We hope that some kindof swapping similar to case (2) in the Bona algorithm may work since q′ will neverbegin with q′1 = m, q′2 = a, and q′3 = m+ 1 or q′1 = m, q′2 = m+ 1, and q′3 = a sincethis implies q began with a 312 or a 321 pattern. At this point, we have not found aswapping algorithm for these special cases that works to give an injection.
This work does lead to the larger question of the log-concavity of the inversionstatistic over different subsets of pattern avoiding or generalized pattern avoidingpermutations in Sn (i.e., other than the subset of permutations that are both 312- and321-avoiding).
7. A Bijection with Involutions
Recall that an involution is a permutation consisting solely of one and two cycles. Thefollowing function φ is a well-known bijection between standard Fibonacci tableauxof size n and involutions in Sn. Let T be a standard Fibonacci tableau. To form aninvolution φ(T ) = π in Sn, let the columns of height two of T be the two cycles in π
616 N. Cameron and K. Killpatrick
and the columns of height one of T be the one cycles in π . For example, let
T =3 4 2
9 8 7 6 5 1.
Then φ(T ) is the involution π = 1 5 8 7 2 6 4 3 9 . Note that this gives animmediate bijection between involutions and column reading words for Fibonaccitableaux, which ultimately gives a bijection between involutions and permutationsthat are both 312 and 321 avoiding.
One might now ask the question about what happens to the inversion and Fibmajstatistics under this bijection. There does not seem to be a clear image for the inver-sion statistic that is easily obtainable from the involution. For the Fibmaj statistic,however, the story is different.
Definition 7.1. Given a permutation σ ∈ Sn, define the submajor index as
submaj(σ) = ∑i∈(Des σ ∩ Exc σ)
i.
For the example above, Des π = {3, 4, 6, 7} and Exc π = {2, 3, 4}, thus we havesubmaj(π) = 3 + 4 = 7.
Lemma 7.2. Let T be a standard Fibonacci tableau of size n and let φ(T ) = π . Thenthe set of elements in the top row of a column of height two in T is exactly Exc π .
Proof. If k is in a column of height one in T , then πk = k so k is not in Exc π . Ifjk
forms a column of height two in T then π j = k and πk = j. Since j < k, πk < k, so kis not in Exc π . Also, π j > j, so j ∈ Exc π .
Theorem 7.3. Let T be a standard Fibonacci tableau of size n. Then
Fibmaj(T ) = submaj(φ(T )).
Proof. To begin, consider Fibmaj(T ) = ∑i∈Des(T ) i. Since the bottom row of T de-creases from left to right and the columns in T decrease from bottom to top, noelement in the bottom row of T can be in Des(T ). Thus if i ∈ Des(T ), then i is in thetop row of a column of height two in T , which means i ∈ Exc φ(T ) by Lemma 4.1.Now let φ(T ) = π and for i in the top row of a column of height two in T , considerthe following three possible relative positions of i and i+ 1 in T .
(1) Suppose i+ 1 is in a column to the left of i in T , so i is not in Des(T ). Since thebottom row of T must decrease, i + 1 must also be in the top row of a column
of height two in T . Suppose the columns containing i + 1 and i arei+ 1
m andik,
respectively. Then k < m and we have πi = k and πi+1 = m so πi < πi+1. Thus iis not in Des π .
(2) Suppose i + 1 is in a column to the right of i in T , so i ∈ Des(T ), and letik be
the column containing i in T so πi = k. If i+ 1 is in a column of height one thenπi+1 = i + 1 and since i + 1 is to the right of the column containing i we have
Symmetry and Log-Concavity on Fibonacci Tableaux 617
k > i+ 1. Thus πi > πi+1 so i ∈ Des π . If i+ 1 is in a column of height two then
either the column looks likei+ 1
j orj
i+ 1. In both cases, πi+1 = j and since thebottom row of T decreases and the columns of T decrease from bottom to top wehave k > j, thus πi > πi+1, so i ∈ Des π .
(3) Suppose i is in a column with i+1, i.e.,i
i+ 1, in T . In this case, i ∈ Des(T ). Thenπi = i+ 1 and πi+1 = i, thus i ∈ Des π .
Thus if i ∈ Des(T ), then i ∈ (Des π ∩ Exc π).On the other hand, suppose i ∈ (Des π ∩ Exc π). Since i ∈ Exc π , we know
from Lemma 4.1 that i is in the top row of a column of height two in T , i.e.,ik. Since
i ∈ Des π , πi > πi+1. Suppose i+1 is in a column to the left of the column containingi in T . Since the bottom row of T decreases i+ 1 must be in the top row of a column
of height two in T , i.e.,i+ 1
m with m > k. Then πi+1 = m > k = πi which contradictsthe fact that i ∈ Des π . Thus, i+1 is either in the same column as i or in a column tothe right of i in T , that is, i ∈ Des(T ).
Since the Fibmaj statistic on Fibonacci tableaux has now been shown to be equiv-alent to the submaj statistic on involutions we have as a consequence of Theorems4.2, 6.1, and 7.3, that the submajor index is symmetric, log-concave, and unimodalover the appropriate subset of involutions in Sn. It remains an open problem to deter-mine if there is a larger subset of Sn on which the submajor index has some or all ofthese properties. If our conjecture about the inversion statistic being log-concave onall Fibonacci tableaux of a given size n turns out to be true, it would imply that thesubmajor index is log-concave over all involutions of size n.
Acknowledgments. The authors would like to thank the referees for many helpful commentsto improve the quality of the paper, and in particular, for suggesting the interesting connectionto 0, 1-fillings of a staircase shape.
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