\1

CHAPTER II

The Inductive Groupoid G(6 n )

It follows from theorems from Chapter I that the semigroup 6 n is an idem­

potent generated regular semigroup. Henc the inductive groupoid G(6 n ) of

6 n may be constructed from the biordered set En = E(6 n ) by the procedur

described in Section 1.3. V.:e proceed to apply these to G(6 n ) which will lead

to some interesting consequences for the semigroup 6 n.

1. THE BIORDERED SET AND GREEN'S RELATIONS ON 6 n

\\"e know that 6 n is regular. Hence En = E(6n) is a regular biordered

set. Now elements of En are idempotent endomorphisms of V and e E En

if and only if 91(e) EB ~(e) = V. Since e is singular, we have lJ1(e) =/:- 0

(i.e.) lJ1(e) =/:- {O}; since no confusion is likely, in the sequel, we shall use the

shorter notation). Conversely, if U ffi ltV = V with U =/:- 0, there is a uniqu

e E En with U = lJ1(e) and TrV = ~(e). We denote this unique idempotent

by e(U, ltV). Therefore, the biorder structure of En can be described in terms

of these direct sum decompositions. In particular the quasio.der relations w T

and wi on En can be obtained as follows.

PROPOSITIO\" 1 For t. f E En) e:.J f if and only iflJ1(J) ~ lJ1(e) and ewlJ

zf and only ifr.R(e) ~ r.R(J).

1 II THE INDUCTIVE GROUPOID G(6 11 )

PROOF ewT J =} fe = e.

But 91(/) ~ 91(/e) = 91(e).

Conyersely 91(/) ~ 91( e) =} u(1 - f) E 1J1(e) for ev ry u E V. lIenee

u(1 - J)e = 0 for every u E 1 =} (1 - f)e = 0 =} e = fe. That is ewT f.

Again ew l f =} ef = e

Now 91(e) = 91(eJ) ~ 91(/).

Conyersely, 91(e) ~ 91(/) =} ue E 91(/) for every u E V =} uef =

ue for every u E V =* ef = e. That is ew l f. 0

Since w = wi n wT, for e,f E En ewf if and only if 91(e) ~ 91(1) and

lJ1(e) ~ 91(/). The basic product in En is the composition of endomorphisms

restricted to DEn. A more general construction of biordered sets from ompl­

mented modular lattices are in [20].

Recall that in a semigroup S we define the relations. R, £ and :J by aRb

ifandonlyifaSU{a} = bSU{b}. a£bifandonlyifSaU{a} = SbU{b} and a:Jh

if and only if aSUSaUSaSU {a} = bSUSbUSbSU{b}. Also, we define V to be

the equivalence generated by R U £. It can be show·n that V = R 0 £ = £ a R.

Finally we define H = R n £. These are equi valences on S called the G l' n s

relations [1,12]. The £-class (R-class, :J-class, H-class, V-class) containing

an element a of a semigroup S is denoted by La(Ra, la, JIa, Da).

\Ne will now characterize the Green's relations on 6(V). Since 6(V)

is a regular subsemigroup of L(V), two elements of 6(V) are R or £ rclatcd

in 6 n if and only if the corresponding relations holds in L( V). Hence first

we characterize the Green's relations in L(V). For that we n ed the following

result.

Let S be a semigroup. There is a partial order ~ defined on the set. of

11.1 THE BIORDERED SET AND GHEEN'S RELATIONS 0:-; 6 n

all 'R -class S/R [£- las 5/£] as follO\\"5

R t :::; Rs <===> tESS' <===? t = su for sam u E 5'

L1 :::; Ls <===> t E 5' s <===? t = us for some u E 5'

where

19

{

55'

5 U {I}

if 5 contains an identity element

otherwis \\"h re sl = 1 = Is V s E 5

Since 5/R is in one-to-one correspondence with the set of right ideals, this

defines a partial order on he set of all right ideals of 5 which is clearly the in­

lusion. Similarly, the relation:::; defined on 5/£ induces the inclusion relation

on the set of all left ideals.

LE1L\IA 2 Let s, t E L(II). Then

(1) Rt :::; Rs <===? 1J1(t) 2 1J1(s) .

(2) Lt:::;L s <===> !.R(t)~!.R(s).

PROOF (1) Rt :::; Rs => t = su for some u E L(V)'. Now x E lJ1(s) ==}

xs = a ==} xsu = (xs)u = O. That is. xt = 0 ==} x E lJ1(t). Therefore,

1J1(s) ~ 1J1(-t).

Conversely, suppose that 1J1(s) ~ 1J1(t). Define l arbitrarily on the com­

plement of !.R(s) in II and for x E !.R(s) letting y E V be such that ys = x. De­

fine xl = yt. The fact that lJ1(s) ~ 1J1(t) ensures that l is \\' 11 defined. Cl arly

l E UV) and ysl = xl = yi for every y E V. showing that t = sl ==} R t :::; R s .

20 II THE INDUCTIVE GROUPOID G( 6 n )

(2) Lt ~ L. =? t = us for some u E L(V)'. ow 91(t) = 91(llS) ~ 91( ).

That is, 91(t) ~ 91(s).

Conversely, let 91( t) ~ 91(s). For each x E. V, we may choose one

y E V such that y = xt and declare .,d = y. It follows that l E L(V) and

xls = ys = xt for every x E V. Hence ls = t =? L t ~ Ls . 0

TIIEORE~I 3 Let s, t E L(V)) then we have the following

(1) tLs if and only ifl]{(t) = l]{(s).

(2) tRs if and only if'Jl(t) = 'Jl(s).

(3) tHs if and only ifl]{(t) = l]{(s) and 'Jl(t) = 'Jl(s).

(4) tDs if and only if rank t = ranks.

(5) J = D in L(V).

PROOF t£s {=} L t = Ls and tRs {=} Rt = Rs . Hence statements (1)

and (2) of this theorem are consequence of the above lemma. Also tHs if and

only if tRs and t£s. Hence we get (3).

(4) Assume that tDs. Then there exists l E L(V) such that tLl and lRs.

By part (1) t£l =? l]{(t) = l]{(l) and hence rank t = rank l. Again by part

(1) lRs => 'Jl(l) = 'Jl(s). Hence land s have the same nullity and hence the

same rank. Therefore, t and s have the same rank.

Conversely, assume that s, t E L(V) have the same rank. I-renee l]{(s) and

l]{(t) have the same dimension. So there exists an isomorphism cP: l]{(t) -t l]{(s).

Defi.ne u = to. Then l]{(u) = l]{(s) and 'Jl(u) = 'Jl(t) since 0 is an isomorphism.

Hence [rom parts (1) and (2) we get uLs and uRt ;. s£uRt => sVt.

11.1 THE BIOR.DERED ET AND GR.EE:-I'S I1ELATIONS ON 6 n 21

(5) first of all 1) ~ :J in general. :\ow 1) = £ 0 R = R 0 £ is the smallest

equi\'alence relation £ U R containing both £ and R. vVe defin a:Jb to mean

S'a ' = S'bS'. Hence we get £ ~ :J and R ~ :J. Therefore: 1) ~ :J.

To prove the other inequality. we first show that for s, t E L(V), s E it if

and only if rank s :S rank t. If E it, then s = (t17 for some ~ and "I E L(V)'.

Now

Also '7 induces a linear transformation of 91(t) onto 91(try) and so dim91(t) 2

dim 91(t1]). Hence ranks:S rank(t1]):S rankt.

Conversely, let rank :S rank t. Th n dim 91(s) :S dim 91( t). Hence th re

xists a surjective linear map "I: 91ft) -t 91(s). therefore 91(t1]) = 91( ). II·nce

[77£S. :\ow t1] E tL(V) ==? tryIp') ~ tL(V) ==? i/7] ~ it. Sin e t1]Ls,

L(V)t1] = L(V)S.

Therefore,

Thus s E it if and only if rank s :S rank t. Hence s:J t if and only if rank t =

ranks. So by part (4) s:Jt if and only if s1)t. Hence in L(V),:J = 1). 0

Thus we have the following characterizations of Green's relations on 6 n .

THEOn.E~I 4 For j, 9 E 6 n we have the following

(a) j£g ~ 91(/) = 91(g).

(b) jRg ~ 'J'C(/) = l)1(g).

(c) jHg ~ fJ1(J) = 91(g) and 1)1(J) = l)1(g).

(d) jVg ~ rank(J) = rank(g).

(e) V=:J.

22 II THE Il'-:OUCTIVE GROUPOID G(bn )

\i\"e observ that the characterizations of the biorder structure on En

immediately leads to a charact rization of the Green's relations L, Rand 1i

given in (a). (b) and (c) above.

DEFI TITION 1 A regular semigroup S is completely semisimple if no two

distinct V-related idempotents are comparable by the natural partial order.

PROPOSITION 5 b n is completely semisimple.

PROOF Let e, f E En with eVf and e ~ f. Then dim 1J1( e) = dim 1J1(1) and

dim91(e) = dim91(1). Also e ~ f implies ef = fe = e. Therefore, 91(e) ~

91(1) since 91( e) and 91(1) are fini te dimensional it follows that 91( e) = 91(1).

So eLf by Theorem 4 and fe = f. Thus e = f. This proves the result. 0

2. THE PROPER SET OF COMMUTATIVE CYCLES IN 6 n

\Ve shall say that E-cycle I in Q(En ) is commutative in b n (or in G(bn ))

if it is c6 n -commutative. We denote the set of all commutative E-cycles in

6 n by r n. Then r n is a closed proper set of E-cycles in Q(En). ~ow for any

c = c(eoel'" en) E Q(En),

lS a singular endomorphism of the vector space V; since 6 n is idempotent

generated, every element of 6 n is of this form by Theorem 1.2. Note that, if c

is reduced, then we have

eOe2 ... en, if eOLel) en-l Ren

ele3'" en, if eoRel, en-1Re nTc =

ifeOe2"' en, eOLel) en-lLe n

ele3'" en, if eoRe]) en-lLe n

11.2 TilE Pn.OPER SET OF COI'.!/'.IUTATIVE YCLES IN 6 n 23

v\ e shall say that Te has th normal JOl'ln if the first case abo,' holds. B)

introducing trivial edges into c, if necessary we can always ensure that th

first case above holds. To avoid ambigui ty, we shall adopt the convention that,

unless otherwise indicated,' the expression Te is in the normal form. This in

particular means that the number n of '"ertices of c is odd and the number of

edges is even. Moreover, Te = Tel if and only if c(ee/l ee)CC(Jel fel )c,-l Ern.

Jotice that an E-cycle I is commutati,"e in 6 n if and only if T-y = e-y. Thus

rn = {T I is an E-cycle and T-y = e-y}.

\Ne therefore, have a mapping T: 9(En ) -t 6 n which. in view of the isomor­

phism betvveen S(Grn ) = S(G(6 n )) and 6 n is equivalent to the map llVrn

defined by Equation vVrn = ernPr;; where ern is th surjective evaluation on

Gr n and pr;; denotes the canonical mapping of Grn onto Grn/p(Grn ). Next

lemma characterizes commutative E-squares.

LEl\IMA 6 For e, f E En) the following statements are equivalent.

(a) There exists g, h E En such that 8 = (% J) is a commutative E-square.

(b) 91(J) ~ 91(e) @ N where N = 91(e) n 1J1(J).

(c) (e-J) 2 191(e)=O.

PROOF (a) ===? (b): Since 8 is an E-square, either eRg or e£g; to fix

notation we may assume that eRg. Since 8 is commutative we have efe = e

and gh = e. Therefore, the map g: V -t vg is a lin ar isomorphism of 91(e) onto

91(1) and the map h: w -t wh is its inverse. For, vg = vg, vgh = vgh == ve = v.

It is easy to see that for each v E R(e), vg == v +n where n E lJ1(e). Similarly,

for 1e E 91(1), wh = w + m for m E 91(1). Hence for ach w E 91(J),

there is a unique v E 91( e). n E 91(e) and m E 1J1(J) such that w = v + nand

24 II THE INDUCTIVE GROUPOID G(6 n )

v+n+m = v. This implies that n+m = 0 so that n = -m E 91(e)n91(f) = N.

Therefore, (b) holds.

(b) ===? (c): Let B f = {Wi: i = 1,2, ... ,r} be a basis of fJt(f). It follows

from (b) that for each Wi there is a unique Vi E fJt( e) and ni E N such that

Wi = vi + ni. Then Be = {Vi: 1 ::; i ::; r} is a linearly independent set. For, if

L aivi = 0, then L ai(wi - ni) = 0 and so L aiWi = L aini· This impliesIii i

that L aiWi is a vector in fJt(f) n N ~ fJt(f) n 91(f) = {O} and so L aiWi = O.. .I I

Since B f is a basis, we have ai = 0 for all i. Also dim Be = r. Hence Be

is a ba~is of fJt(e). Now for any i, vi(e - I) = Vi - Wi = -ni and since

ni E N = 91(e) n 91(f), it follows that vi(e - 1)2 = (-ni)(e - I) = O. Since

Be is a basis of fJt( e), (c) holds.

(c) ===? (a): Let, as above, Be = {Vi: 1 ::; i ::; r} be a basis of fJt(e) and

let Wi = vii for each i. Then there exists vi E R(e) and ni E 91(e) such that

vii = Wi = vi +ni. For Vi E Be, we have from (c),

since vile = vi, it follows that Wi = vi+ni for any i. Since f is an idempotent,

we have nif = wiI-vii = 0 and so ni E 91(e)n91(f) = N. Now if L aiWi = 0,

then we get using the equality Wi = ni +Vi that L aiVi = - L aini E N, weI I

get ai = 0 for all i. Thus B f = {Wi: 1 ::; i ::; r} is a basis of fJt(f). Also

fJt(f) EB 91(e) = V. For if, u E fJt(f) n 91(e) then u = ~ aiwi for some ai E f{.I

Hence u = "L aivil and also we have ue = 0 = L aivile. This implies. "I I

that L aivi = O. Hence ai = 0 for all i. Thus we have u = O. HenceI

fJt(f) n 91(e) = {O}. So there exists an idempotent 9 wi th 91(g) = 91(e) and

fJt(g) = fJt(f). Hence by Theorem 4, we have eRg£f. Now for each i, we

have observed that ni E N from which it follows that Wi( e - 1)2 = O. Thus

11.2 THE PROPER SET OF COMMUTATIVE CYCLES IN 6 n 25

( - 1)2 I 91(J) = 0 and so, as above there xists an idempotent h with

e£hRf· Then by Clifford-Miller Theorem (see [1]) since 9 E Lf n Re the

product fe E Hit. Since vile = Wje = Vi for all i, we conclude that fe = h.

So efe = eh = e. Hence (a) holds. o

For each a E 6 n, let a* = 1- a. Clearly, for any a E 6 n , a* E En if and

only if a E En. Further, for e E En, 91(e*) = 91(e) and 91(e*) = l]1(e). Hence

o= (~ j) is an E-square if and only if 0* = (;: j: ) is an E-square. Notice

that (e - 1)2 = (e* - f*)2. Since, for e E En, 91(e) EEl 91(e) = V, (e - 1)2 = 0

if and only if (e - 1)2 I 91(e) = 0 and (e - 1)2 I 91(e) = O. The following

observation is an immediate consequence of the lemma above.

COROLLARY 7 Let 0 = (~ j) be an E-square in Q(En). Then 0 and 0* are

commutative E-squares in 6 n if and only if (e - 1)2 = O. 0

Since rn is a closed and proper set of E-cycles, by (PI) we have fa C

fa ~ rn. This implies that every singular E-square is commutative in 6 n .

The next lemma shows that every E-square in r n belongs to fa.

LDI:YfA 8 Let 0 = ({ r.) be an E-square in Q(En). Then 0 is commutat£ve

in 6 n if and only if it is degenerate or 0 is both row and column singular.

PROOF In view of the observation above, it is sufficient to show that if 0

is commutative and not degenerate, then 0 is both row and column singular.

Suppose that 0 is commutative. Then by Lemma 6, we have

91(k) ~ 91(1) EEl N where N = 91(J) n 91(k)

If 91(1) = 91(k), we have fRk by Lemma 4(b) and then g = k and f = h. So

o is degenerate. Similarly, if N = 0, then by (*), 91(k) = 91(1) and so k£f

26 II THE INDUCTIVE GROUPOID G(6 n )

by Theorem 4(b). In this case f = 9 and h = k and so 8 is again d gen rate.

So we may, in the following assume that N is a proper non-zero subspace of

1J1(J) and 1J1(k ).

Let N' be a complement of N in 1J1(J). Then by hypothesis, N' i O. Let

e = e(i\f fJ{(J)ffiN'). By Proposition 1, g,k E wT(e). Also, if {Ui: 1 ~ i ~ r} is

a basis of fJ{(J) = fJ{( h) by (*) there is a basis {Wi: 1 ~ i ~ r} of fJ{( k) = fJ{(g)

such that

Wi = Ui + ni where ni E N for all 1 ~ i ~ r

Since 1J1(J) = lJ1(g), we have Vig = Wig = wi for all i. Therefore, Vige

Wie = Vi for all i. Since, clearly, lJ1(ge) = lJ1(g) = 1J1(J), we have ge = f.Similarly Vik = (Vi + ni)k = Wik = wi so that vike = Wie = Vi· Also

1J1( ke) = 1J1( k) = 1J1(h) so that ke = h. Thus 8 = U: 1) and hence 8 is row

singular.

Suppose that e' = e(N';fJ{(J) ffi N). From (*) it follows that 91(k) ~

fJ{(e'). Also fJ{(J) = fJ{(h) ~ fJ{(e'). Hence by Proposition 1, k, h E wl(e').

Then e'h£h£ f and so by Theorem 4(a) fJ{( e' h) = fJ{(J). Since 1J1(J) = N ffi

N', if n E 1J1(J), then ne' E N and so ne'h = O. Hence 1J1(J) ~ 1J1(e' h).

Since dim 1J1(J) = dim 1J1(e' h) it follows that 1J1(J) = 1J1(e' h) and so f = e' h.

Similarly, 9 = e'k and so 8 = (e~h e; ). Hence 8 is column singular. 0

In [18] it is shown how a partial order can be defined on a regular semi­

group. The relevant result is given below. If S is semigroup for T ~ S we

write E(T) to denote the set of idempotents for T ~ S.

PROPOSITION 9 Let S be a regular semigroup and x, yES. Then the

following are equivalent.

II.2 THE PROPER SET OF COMMUTATIVE CYCLES IN 6 n 27

(1) xS ~ yS and there exists f E E(Rx ) with x = fy·

(2) For each f E E(Ry) there exists e E E(Rx ) with ewf and x = ey.

(3) For each f E E(Ly) there exists e E E(L x ) with ewf and x = yeo 0

For proof see Proposi tions 1.1 and 1.3 of [18]. If x, y satisfy any (and so

all) of the above conditions then we write x :::; y. In [18] it is shown that this

defines a partial order on S and is named the natural partial order on S.

PROPOSITION 10 If,:::; " with respect to the partial order in the ordered

groupoid 9(En) then T-y :::; T-y' with respect to the natural partial order on the

semigroup 6 n. lVloreover) if, :::; " in 9(En) and if,' is commutative so is ".

PROOF Let, = (ho, hI, ... ,hn) and " = (eo, el"" ,en) be in 9(En) such

that T-y and T-y' are in the normal form. Now,:::;,' implies howeo and

, = ho*,' by Equation (2) in §1.2. Then

Therefore, T-y = hoTy' where howeoRT-y' since T-y' is in the normal form. Hence

by Proposition 9 T-y :::; T-y' with respect to the natural partial order on 6 n.

" commutative means en = eo and T-y' = eo· Hence hn = ho and

T-y = hoeo = ho, which implies that, is commutative. Hence the last statement

follows. 0

\Ve have the following characterizations of commutative E-cycles in 6 n .

28 II THE INDUCTIVE GROUPOID G(6 n )

PROPOSITION 11 An E-square 0 is commutative in 6 n if and only if 0 is

singular. lHoreover) if n 2 2) an E-cycle I ba Ed at e is commutative if and

only if I E r T and satisfies the condition that whenever 0 is an E-square based

at e'we with To ::; T'Y) then 0 is commutative.

PROOF The first statement follows from Lemma 8. Now by definition, the

E-cycle I based at e is commutative if and only if T'Y = e. Hence for an E­

cycle 0 based at e'we, To ::; T'Y implies that To = e' and so 0 is commutative.

Conversely, assume that I satisfies the gi"en conditions. Since I EfT,

TAl is a scalar transformation, say T'Y = Ae. Then 91(T'Y) = 91( e) and 91(T'Y) =91(e). Let a =1= v E 91 (T'Y) and let N' be a complement of (v) containing

91(T'Y)' Define T' = e'T'Y where e' = e(N', (v)). Then 91(e') = (v) ~ 91(e)

which implies e' wi e and 91(e') = N' 2 91(e) which implies e' wT e. Hence

e' w e. Then T' = e'T'Y = e' Ae = Ae' e = Ae' and since n 2 2. rank T' = 1 ::;

dim N' = nullity T'. Therefore, T' = Ae' is a sclar matrix having rank::; nullity

and hence by Proposition 4.8 T' = To for some E-square 0 based at e'. Now

T' = e'T'Y where e' w eRT'Y' Hence by Proposition 2.8 T' ::; T'Y in the natural

partial order in 6 n . Then by the hypothesis, 0 must be commutative which

implies that A = 1. Therefore, T'Y = e. This proves that I is commutative. 0

The proposi tion above shows that the proper set of E-cycles f n (the set

of those E-cycles which are commutative in 6 n ) is, in some sense 'generated"

by the proper set fa of singular E-squares. Hence it is natural to ask whether

f n is th.e closure of fa in the sense of §6 [19]. The question is equivalent to :

is it true that all commutative E-cycles in 6 n can be obtained as combination

of commutative squares. We therefore, formulate the following.

Is it true that 6 n is the free idempotent generated regular semigroup

generated by En·

11.2 THE PROPER SET OF COMMUTATIVE CYCLES IN en 29

The answer is 1 o. en is not the free idempotent generated regular

semigroup generated by En. We can show it by an example.

Consider the V-class of rank(n - 1) idempotents in en' vVe know that

there are so many commutative E-cycles in that particular V-class. In the

free idempotent generated regular semigroup generated by En, the E-cycles

are all generated by singular E-squares. But there are no singular E-squares

of rank(n - 1) except the identity square. Hence there are no commutative·

E-cycles of rank(n - 1) in the free idempotent generated regular semigroup

generated by En except the identity squares. Hence en is not isomorphic to

the free idempotent generated regular semigroup generated by En.

vVe will give an example of a commutative E-cycle in the V-class of

rank(n - 1) idempotents in en.

EXAMPLE 1

Consider the semigroup of all singular 3 x 3 matrices over the R. We construct

a commutative E-cycle of length 6 in the V-class of rank 2 idempotents in

this particular semigroup. The matrices eo, el, e2, e3, e4, es given below are all

idempotents of rank 2.

r~a

~J [~0

~J [i0

!Jeo = 1 el = 1 e2 = 0

a 1 a

r~a

~J [~a

!J [i0

]e3 = a e4 = 0 . es = 1a 1 0

Let UI,U2,U3 denote the vectors (100)(0,1,0) and (0,0,1) respectively." ,

30

Then

~(eo) = (UI uz)

~(el) = (UI uz)

~(ez) = (UI' U3)

91(e3) = (UI' U3)

~(e4) = (UI' uz +U3)

91(es) = (UI uz +U3)

II THE I:-<DUCTIVE GROUPOID G(6 n )

lJ1(eo) = (U3)

lJ1(el) = (uz - U3)

1J1( ez) = (U2 - U3)

lJ1(e3) = (uz)

1J1( e4) = (uz)

lJ1(es) = (U3)

Then by Theorem eo£el Rez£e3Re4£es Reo· Hence I = c( eo el ez e3, e4 es, eo)

is an E-cycle based at eo. Moreover T-y = eo le2 e3 ol eSeo = eo· Hence I is a

commutati\'e E-cycle of length 6 based at eo in the V-class of rank 2 id m-

potents in 63·