is an embedding if r is chosen big enough. □ Corollary 1.9 Let rj = Aa M n (A^)/M J _i —> Mt(K) be the homomorphism found above. If S C Mn(K) is a semigroup, then
cl(#5)) - #cl(S)). Proof. Let T 3 ?](S) be a 7r-regular subsemigroup of Mt(K). Then V — T fl rj(Mn(K)) is also 7r-regular because so is n(Mn(K)) (see the proof of Corollary 1.5). Clearly V D rj(S). Let n{a) eT\ae Mn(K). Then, for some k > 1, n(a)k lies in a subgroup H of T7. So rj(a)krj(x) = 77(e) for some x,e G M n (A r ) such that 77(e) is the identity of H and 77(x) G fl". By Lemma 1.8, 77(e) = 77(e2) implies that e2 = Ae for an 5-th root of unity A. Hence, replacing e by e s , we may assume that e = e 2 . Let G be the maximal subgroup of Mn(K) containing e. Since rj(x) =
CHAPTER
12
1. GENERAL
TECHNIQUES
rj(exe), we come to xs — (exe)s £ eMn(K)e. Similarly, aks G eMn(K)e. Now rj(aksxs) = 77(e) implies that (aksxs)s — e. Therefore, the inverse xs(aksxs)s~1 of aks in G is contained in 7?_1(T1/). We have thus shown that n~l{T') C Mn(K) is a 7r-regular semigroup which contains 5. Therefore cl(5) C 7 / - 1 (T / ), so 77(01(5)) C rfTf-^V) C V C T. It follows that r](c\(S)) C 01(77(5)). Since 77(01(5)) is 7r-regular and contains 77(5), we also have 77(cl(5)) 5 cKrl{S)). This completes the proof. □
1.4
Combinatorial techniques
For a given semigroup 5 C Mn(K) the nonempty intersections 5 fl D with maximal subgroups of Mn(K) will play a crucial role in the approach developed in Chapter 3. Here, we present a combinatorial technique which indicates the role of these intersections. The first result of this type was obtained in [43]. We follow the improved approach of [116], (compare also [70], Chapter 10, and so called repetitive mappings [73] in connection with Lemma 1.11 below). L e m m a 1.10 The following conditions are equivalent for a matrix a G Mn(K) 1. a G D for a maximal subgroup D of
Mn(K),
2. rank(a 2 ) = rank(a), 3. Kn is the direct sum of the subspaces ker(a) and Im(a), 4. the minimal polynomial Xa{x) of a is not divisible by x2, 5. a is conjugate to a matrix I
for some g £ GLj(K)
and
some j , 0 < j < n. Proof. If a £ D, then a2 G D, so that rank(a 2 ) = rank(a). The latter implies that ker(a) H Im(a) = 0. Hence 1) implies 2) and 2) implies 3). If 3) holds, then the restriction of a to Im(a) is an automorphism. Changing the basis of Kn we derive 5). Assume that 5) holds. If a — 0 or rank(a) = n, 4) is clear. Otherwise, Xa(x) = x-xg(x), and x 5 (0) / 0. Hence 4) follows. If 4) holds, then the Jordan form of a in Mn(K) is
1.4.
COMBINATORIAL
TECHNIQUES
13
as in 5). Therefore a G G' for a maximal subgroup G' of Mn(K). Since Mn(K) fl G' is a subgroup of Mn(K), 1) follows. This proves the lemma.
□
The basic combinatorial lemma reads as follows. L e m m a 1.11 Let X be a nonempty set, n, £;0, fc1?... , kn natural num bers, and r : X* —> { 0 , 1 , . .. , n} a mapping such that r(tuv) < r(u) /or a// £,u,t> in the free monoid X* generated by X. Then, for every w G X* of length at least k0 • • • kn in X there exists /, 0 < / < n, such that w has a factor of the form Wi • • • w^ with all Wi G X* \ {1}, and r(wi • • • Wj) — I for all i,j with 1 < i < j < k\. Proof. Let r
n
=
^n?
r
n— 1 — knkn—1-)
• • • ■> ^0
=
^n^n—1 ' ' ' ^0-
Assume first that w G X* is of length at least kn and r(w) = n. Then iu has a factor of the form Xi • • • x^.n with all x2- G X. Hence, for all i,j r(xi • • • Xj) > r(w) = n implies that r(x2- • • • Xj) = n and the assertion follows. Thus assume that I < n and the assertion is true for all elements x G X* with r{x) > I. Let w G X* be of length r/ = fc/H+i a n d r(u;) = /. Then w has a factor wx • • • iu^ with every w;2- of length rl+1. If /' = r(w t ) > / for some z, then the assertion follows by induction, applied to W{ because r/ + i > r//. Otherwise, r(wi) < I for every i. Therefore, for all z, j < ki with i < j we have / = r(w) < r{wi • • • Wj) < r(wi) < /, which implies that r{w{ • • • Wj) — I. This completes the proof. □ The following result shows that sufficiently long products of ele ments of Mn(K) contain subproducts which lie in maximal subgroups of Mn(K). T h e o r e m 1.12 There exists a natural number N — N(n) such that for all elements a i , . . . , a # of Mn{K) there exist i < j , 1 < z, j < TV, with a{- • • aj G D for a maximal subgroup D of Mn(K).
CHAPTER
14
1. GENERAL
TECHNIQUES
Proof. Let k0 = 1, h = (j) Q + 1 for it = 1,2,... , n - 1, and fc knn = = 1. Let ww == aQlx •■• ■• a■Na. N.WeWeapply applyt ht hpreceding t precedinglemma lemma Define T NV == kk00 ■■ ■■ ■■ kknn.. Let rank(u). It follows that there with the function r defined denned by r(u) = = rank(t;). exists Z < n such that w = uuu>i factorization ofofwwasasaaword word • •• •■wwklvklv isisaafactorization Wl ■ in aw, and rank(u; • •i) •3) ==l for I forrvery rveryy y1 1<<<<j j<,-■■• u;,- ■ • u;, • ^lielin lies in i naximam i naximam subgroup of Mn{K). Let V^ = ker(«,,■) ker(«;,•) and W, W{ = lm(w I m ( ^t)) for every i. Since r a n kk ((TWO ++ 11 ) = r a n k (^^, ) , we must have V- D Wi ^ ++1x = 0. On the other hand Vj nn^i^O ^i,i- 0 for i < j , because by Lemma 1.10 rank(u>,- • ■ • •■ww 3)3) >> For rvery yubspace e of V put Z' r a n kk(t(K^ • • • •3f). Z' ' z, Zx A • • • A zr G ^ +1 , / ±0 0 and A r V, y , where ^z i, ,. ..... ,,zzrr is a fixed basis of Z. Then F/ VJ' A W{ Vj (]), and the latter is the dimension \A' A W/ = 0 for i < j . Since Jfe, k, > (?), of A'(K it follows that there exists j , 1I < j < kh such that W\ = Al(Kn), aiW{ ctj^Wj.-i for for rome romeaak k G€ A'. K. Therefore .Terefore aiW> + •■■ •.. ++a^Wj-
^ i A ^ ' = E < i
A
^ '
= 0.
a contradiction. This completes the proof. □ R e m a r k The above proof shows also that, if . ,wr G Mn(K) are ifwwu...,w u . .rG matrices of rank k for r > (j) and such that rank(u;i = k, then r a n k ^ •••r) Wi-'-Wj ii sontained di n maximal lubgroup po Mn(K) for rome ej with 1 < i < j < r. We shall see in Section 2.2 that it is enough to assume r >
(l)here. We shall need another result on factors of sufficiently long words in a free semigroup. First, we state the following well-known observation. L e m m a 1.13 Let P be a subset of a finitely generated free semigroup Z on a finite set X. Then the following conditions are equivalent 1. there exists r > 1 such every z G Z which is a word of length > r has a factor in P, 2. every infinite word in the generators from X has a factor in P.
1.4.
COMBINATORIAL
TECHNIQUES
15
Proof. Clearly 2) is a consequence of 1). So, suppose 1) does not hold. Then there exists an infinite set F C Z of words with no factors in P. Then F has an infinite subset F\ all of whose elements start with the same generator xi G X. Next, there exists an infinite subset F2 of Fi whose all elements have equal initial factors of length two, say xix2, where x2 € X. Continuing this way, we come to an infinite word x — Xix2 • ■ - which has no factors in P, because each finite factor of x is a factor of a word from F. The result follows. □ The above lemma is often used in the context of the following exten sion of van der Waerden's theorem on arithmetical progressions (cf.[73]), obtained in [11], see also [73], Problem 4.1.1, [70], Lemma 10.1.6. T h e o r e m 1.14 Let a : N —y X be a function into a finite set X. Then there exist p > 1 and an element x G X such that for every q>\ there exist ii < i2 < • • • < iq with a(i\) = • • • = a(iq) = x and i3+i — i3 < p for j = 1 , . . . , g - 1. Our next result comes from [53]. We give a proof only in the case where S is a group, the case which will be needed later. Corollary 1.15 Let x — X\X2 • • • be an infinite word in elements of a free semigroup Z on a set X. Assume that (j) : Z —> S is a homomorphism into a finite semigroup S. Then there exists an idempotent e G S and p > 1 such that for every q > 1 there are q consecutive factors r/i,. . . , uq in x of length < p with c/)(ui) = e for i = 1,. .. , q. Proof. Let S be a group. Consider the sequence yn — x\ • ■ • xn for n — 1 , 2 , . . . . Let a : N —> S be defined by a(n) — 4>(yn)- From Theorem 1.14 it follows that there exists p > 1 and e G S such that for every q > 1 there exist ix < i2 < • • • < z g + i such that ^>(yz>) = e and ij+i — ij < p for j = 1,. . . , q. Since S is a group, the consecutive factors = 1 , . . . , g, are all equal to the identity of S. □ Uj = cj)(xlj+i • • • xtj+l),j The last combinatorial technique discussed in this section is con cerned with linear recursions. Here we follow [33]. Let u i , u 2 , • • • be a sequence of elements of a field K. Suppose this sequence satisfies a linear recurrence of order r w m + r = a!U m + r _! + a 2 u m + r _2 H
h arum
16
CHAPTER
1. GENERAL
TECHNIQUES
for m = 0 , 1 , 2 , . . . , where a2- G K are constants. Let k(x) = 1 — a\x — a2x2 — • • • — arxr G A'[x], The formal power series g(x) — YALQ uix% £ K[[x]] is called the generating function for this sequence. Then C(x) — g(x)k(x) is a polynomial of degree at most r — 1. By the characteristic polynomial of the recurrence we mean f(x) — xr — aixr~1 — • • • — ar G K[x]. We can assume that ar ^ 0, since otherwise this recurrence is of smaller order. Write f(x) = (x - a i ) e i • • • (x - as)e% e H
\- es = r
where ct{ G K. The rational function g(x) = C(x)/k(x) may be written in terms of partial fractions g(x) = DJUi E;=i A « / ( l ~ otkx)1. Let F denote the ring of integers if ck(K) — 0 and let F be the prime subfield of K if ch(/\) > 0. Every (1 — akx)1 is invertible in F(c*A:)[[z]] with inverse E~=o ( " ^ K ^ Then Pk(m) equal to a polynomial in ra.
= £ £ , / ^ ( " t V O is on F
T h e o r e m 1.16 Let um G K, m = 0 , 1 , . . . , be a recurrent sequence of order r. Then for all m > 0 we have um = J2t=i Phi™*)**™, where every Pk{x) £ F(a.k)[x\ is a polynomial of degree at most e^ — 1. Conversely, assume that the elements of a sequence um G K, m > 0, satisfy um — YX-\ ^fc(m)a5T f° r some polynomials Pk(x) G K[x] and some c*k G K. One can proceed as in [33], but in the reverse order. First, since every polynomial h(x) G F[x] satisfies h(m) = J2ili[m^2i ) for some fixed 7; G K we see that the generating function for this sequence can be written in the form
*(*) = £ EE/»K( ro+ _ i r 1 W m m=0k = li=l
\
l
l
)
for some scalars flki G K and some natural numbers e^. This implies that 5 ( x ) = E L i E i = i ( / W ( ! - a**)1')- Therefore ^(x)fc(x) = C(z) for a polynomial C(x), where fc(z) = FLU — o^z) 6 *. Finally, it follows that the sequence um satisfies a linear recurrence of order r = EJUi ek with characteristic polynomial f(x) — xrk(l/x).
1.4.
COMBINATORIAL
TECHNIQUES
17
The above theorem applies also to every sequence um = az^b, m = 0 , 1 , . . . , where a, 6, z G Mn(K). Namely, the minimal polynomial for the matrix z yields a linear recurrence zm+r = a 1 z m + r _ 1 + a2zm+r~2 + • • • + arzm, at- G K, of order r < n, which leads to um+r
- a1um+r_1 + a 2 w m + r _ 2 H
h a r w m for m = 0 , 1 , . . . .
Therefore, if um = (i^j )*j=i>---»rc' then for every z, j = 1 , . . . , n we have u ij — Z)JUi Pikj{m)aT f° r some polynomials P-j(m) and some a*. G A". We conclude with another useful observation on sequences of the form azmb for a,b,z G Mn(K). L e m m a 1.17 Let a,b,z G Mn(K). Assume that z G D for a maximal subgroup D of Mn(K). If rank(az m 6) < j for some j < n and every m — 1, 2 , . . . , then rank(ae6) < j for the idempotent e G D. Proof. From Lemma 1.6 we know that A^(azmb)) = 0 for ra > 1. Moreover, AJ(e) is the identity of the subgroup AJ(D) of M/n\(K) and AJ(z) G AJ(D). Let r be the rank of AJ(e). Then X0AJ(e)+\1A:i{z) + • • • + A r (A J (2r)) r = 0 for some A, G K such that A0 ^ 0. Multiplying by A ; (a) on the left and by A3(b) on the right we come to AJ(aeb) = AJ(a)AJ(e)AJ(b) = 0. This implies that rank(aefr) < j , as desired. □
Chapter 2 Full linear monoid In this chapter we present the basic information on the structure of the full linear monoid Mn(K) over a field K. This includes a discus sion of the ideal structure and the Rees presentations of the principal factors of Mn(K), a useful analogue of the Bruhat decomposition for GLn(K), and the semigroup interpretation of the exterior power map pings on Mn{K). Certain auxiliary results on nilsubsemigroups and nil subfactors of Mn(K) are given in Section 2. Their connections with the properties of sets of idempotents are presented. An analogue of the classical combinatorial approach to GLn(K), leading in particular to a Tits system on Mn(K), is presented in Section 3. The case of a finite field K — Fq is treated in detail in Section 4, with an emphasis on the representation theory of Mn(Fq). If not stated otherwise, elements of Mn(K) will be identified with endomorphisms of Kn (written as column vectors), acting on Kn by left multiplication.
2.1
Structure
We start with a description of J, 7£, and £-classes of L e m m a 2.1 Green's relations on Mn(K)
Mn(K).
are given by
1. ajb if and only if GLn(K)bGLn(K) = GLn{K)aGLn(K), is also equivalent to rank(a) = rank(6), 19
which
20
CHAPTER
2. FULL LINEAR
MONOID
2. aRh if and only if aGLn(K) to Im(a) — Im(6),
— bGLn[K\
which is also equivalent
3. aCb if and only if GLn(K)a to ker(a) = ker(6).
— GLn(K)b,
which is also equivalent
Proof. Let a, 6 G Mn(K) be matrices of rank j . Since aKn,bKn are isomorphic as K-spaces, there exists an invertible matrix g G Mn(K) such that gaKn = bKn. Let e 1 ? . . . , en be a basis of Kn such that fe(ei),. .. )b(ej) is a basis of bKn. Choose /,- G Kn such that ga(fi) = b(e{) for i — 1 , . . . , j . For any basis / i , . . . , / n of Kn we have ga/i = 6, where /i G GLn(K) is such that /i(e,-) = fi for every z. Hence b G GLn(K)aGLn(K). Therefore GLn(K)aGLn(K) = GLn{K)bGLn(K). Since the latter implies that aj^fr, which in turn implies that the ranks of a and 6 are equal, assertion 1) follows. Assume that aKn = bKn. A reasoning as above, with g = 1, shows that ah = bfor some /i G GLn(K). Consequently aGLn(K) = bGLn(K). On the other hand, if aMn(K) = bMn(K), then Im(a) = aMn(K)Kn = bMn(K)Kn = Im(6), which proves 2). Assume that ker(a) = ker(6). Choose a basis u i , . . . , un of A"n such that Uj+i,... , u n is a basis of ker(a). Then 6 ( ^ i ) , . . . ,b(uj) are linearly independent, so there exists g G GLn(K) such that g{b(u%)) — aiui) for z = 1 , . . . , j . Therefore #6 = a. It follows that GLn(K)a = GLn(K)b. On the other hand, if Mn(K)a = Mn(K)b, then 6(ker(a)) = 0. Since rank(a) = rank(fr) and ker(a) C ker(6), we must have ker(a) = ker(6). This implies that 3) holds. D It is now clear that every 7^-class R of Mn(K) contains an idempotent, namely any projection e onto the common image of the endomorphisms from R. Moreover, conjugating by an element g G GLn(K), we can assume that e is of the form e =
. Thus, R consists of all
matrices of rank equal to the rank of e which are of the form
a b 0 0
Similarly, the transpose Rl of R is the £-class of e. Recall that a matrix a G Mn(K) of rank j < n is said to be in reduced row elementary form if the following conditions are satisfied: 1. the leading coefficient of every nonzero row of a is 1,
2.1.
STRUCTURE
21
2. for every i < j , the i + 1-th row of a has more leading zero coeffi cients than the z-th row, 3. the leading coefficient of a nonzero row is the only nonzero coeffi cient in its column. Let Yj be the set of all matrices of rank j which are in the re duced row elementary form. Put Xj = Yj, the transpose of Yj. As sume that x,x' G Xj. Suppose that Im(x) = Im(x'). By induction on j we show that x = x'. Let t>i,... , vj and i ^ , . . . ,Wj be the subse quent nonzero columns of x,x' respectively. If j = 1, then the as sertion is clear because Linx(^i) = LinA'(^i). Assume that j > 1. It is easy to see that Linx(^i, • • • ,Vj) = L i n ^ ( ^ i , ■ • - ,Wj) implies that Vk G Lin/f (iu2, • • • ,Wj) for k = 2 , . . . , j . This and a symmetric argu ment show that Linx(^2 5 • • • ? vj) — Lin^(u?2,... , Wj). Deleting the first row and the first column of the considered matrices we see that the induction hypothesis implies that Vk = Wk for k = 2 , . . . ,j. Moreover j
Vi = ^2,a{Wi for some c^ G K. 2=1
The rows of £ , # ' containing the leading l's of the columns Vi,W{ have all other entries equal to 0. Comparing the coordinates of the vectors in the displayed equality, corresponding to these rows, we come to a2 = 0 , . . . ,e*j = 0. Consequently a x = 1 and vx — wx. Therefore x = x\ as desired. It is well known that every matrix a of rank j is column equivalent to a matrix x G Xj (that is, a can be brought to x by a sequence of elementary column operations). Moreover, if two matrices a, b are column equivalent, then a = bg for some g G GLn(K), so Im(a) = Im(6). It follows that Xj is a set of representatives of the set of 7£-classes of Mn(K) consisting of matrices of rank j . Similarly, Yj can be interpreted as the set of ^-classes. L e m m a 2.2 Let 0 < j < n. If e G Yj is the diagonal idempotent, then let G3 be the group of units of the monoid eMn(K)e. Then Gj ~ GLj(K), each xgy, where x G Xj,y G Y3,g G Gj, has rank j and every matrix of rank j has a unique presentation in this form.
22
CHAPTER
2. FULL LINEAR
MONOID
Proof. Assume that a £ Mn(K) has rank j . Let b £ Mn(K) be an endomorphism of rank j such that ker(6) = ker(a) and Im(6) = eKn. Then b = eb and there exists y £ Y3 such that y — gb for some g £ G3 (apply Gauss elimination). Clearly, ker(y) = ker(a). Therefore, there exists c £ Mn(K) such that cy = a. Then a = cgb = (ce)gb (elimination on columns). Choose h £ Gj such that ceh £ Xj. Then a — (ceh)(h'g)b, where h! denotes the inverse of h in Gj. Therefore a is of the desired form. For every xgy there exist u,v £ Mn(K) such that ux — e — yv (because xCe and yTZe). Hence rank(;rgy) > ia,nk(uxgyv) — rank(g) = j , so that rank(xgy) = j . Assume now that xgy — x1 g'y' for some x' £ Xj,y' £ Yj,g' £ Gj. Since rank(x^y) = rank(x) = j , we must have \m(xgy) = Im(a:). Hence Im(x) = Im(x'), which implies that x = x'. A dual argument shows that y — y'. Now x(g — g')y — 0 implies that g — g1 — e(g — g')e = ux(g — g')yv — 0, so that g = g' and the result follows. □ Let Mj = {a £ M n ( / \ ) | rank(a) < j } for j = 0 , 1 , . . . ,n. We are now able to describe the semigroup structure of Mn(K). T h e o r e m 2.3 The sets 0 = M 0 C Ml C • • • C Mn = Mn(K) are the only ideals of the monoid Mn(K). Each Rees factor Mj/Mj-i is isomorphic to the completely 0-simple semigroup M(GLJ(K),XJ,YJ,QJ), where the matrix Q3 — (qyx) Z5 defined for x £ X3,y £ Yj by qyx = yx if yx is of rank j and 6 otherwise. Proof. From Lemma 2.1 we know that M3 are the only ideals of Mn(K). Lemma 2.2 implies that (^,x,y) H* xgy determines an isomorphism of the semigroup of matrix type M(GLj(K),Xj,Yj,Qj) onto Mj/Mj-i because (xgy) ■ (x'g'y1) = x(gyx'g')y in M3jM3-\ if rank(yx') = j and it is zero otherwise. Finally, by Lemma 2.1, each nonzero 7£-class (£class, respectively) of the latter semigroup contains an idempotent a projection on the subspace of Kn that is the common image of all transformations in this 7^-class (respectively, a projection whose kernel is the kernel of all transformations in this £-class). Therefore, this is a completely 0-simple semigroup. □ The ideal M\ ~ Mi/Mo of Mn(K) admits also another description. Let X be a set of vectors in Kn representing all different 1-dimensional
2.1.
STRUCTURE STRUCTURE
23 23
subspaces. Let Y be a set of representatives of 1-dimensional subspaces n of the space dual to K . It is asy to check that Mx ~ M(K*, X, Y, Q), where the yx-entry of Q is the 'scalar product' (y,x), [29]. Every principal factor M3/M3^-X will be approached via its egg-box pattern arising from Theorem 2.3. The following observation will be often used for M equal to one of these factors. P r o p o s i t i o n 2.4 Let S be a completely 0-simple subsemigroup of a completely 0-simple semigroup M. IfS,Ss is one of the Green relations K,£,%, on M, and on 5, respectively, then we have S\ H,£,%, SlsS = SsProof. Assume aKsb for some a,6 a, 6 £ 5. Clearly, aKb. aTlb. On the other hand, if aKb, then choose e = e2 £ 5 such that aS = eS. Then aKe, so that bKe. bile. Thus b = eb G eS and consequently 6 £ a5. Similarly Similarly a £ bS, which implies that aSsb. A symmetric argument works for the the £-classes. □ The Rees presentation of M3/M3^ in Theorem 2.3 is given in terms terms of subspaces of Kn. Sometimes, the following group theoretic description description is more useful, especially because it reflects the representation theory theory of the monoid Mn{K) and allows to develop combinatorics of M Mnn{K) (K) in the case of a finite base field K. For an idempotent e of rank j lei let P, = Pj = {a £ GLn(K) {K) | ae = eae], eae}, Pf = {a £ GLn{K) | ea = eae}, lei let Uj, Uf be the unipotent radicals of P3, Pf, respectively, and H3 = {a ££ GLn(K) | (1 - e)a = a ( l - e) = 1 - e},#* e}, H] = = {a £ GLn(K) | ea ea = = e}, L3 = {a £ GL„(A:) GLn{K) | ea = ae}. If e = ( J n ae = = ae}.
thes ' then these
subgroups of GLn{K) (AT) are of the form
t
0
';=(;:).'H::).M;;)- Hi ') ^ - ( ;
0
/ ) ^ - ( ( : ) . ^ ( ( ; ) - -
Note that Pj is a semidirect product of L3 and U3}. It is easy to see that Pf P3 coincides with the set of nonsingular nonsingula matrices a such that rank(eae) = j . (If a £ GLn{K) (/l) satisfies the latter latte Pf condition, row elementary operations allow us to find a matrix g £ P]
24
CHAPTER
2. FULL LINEAR
MONOID
such that ga G Pj, so that a G P~Pj. The converse is clear.) Moreover P~ Pj = U~LjUj, so that every matrix z G Pf Pj can be written in the form z = uAu with u £ Uj7u_ e U~,1 £ Lj. This presentation is unique. In fact, if vlw = /' for some v G U~7w G [/,-,/ G Lj, then t?; = l'w~l — w'V for some w' G f/j. It follows easily that / = /', whence v — w' = 1 and consequently w — 1. Let GLn(K)IPj, GLn(K)/P~ denote the sets of left, respectively right, coset representatives of the respective subgroups of GLn(K). We are ready to give a description of the factors Mj/MJ_1 entirely in terms of the structure of the group GLn(K). P r o p o s i t i o n 2.5 Let j < n. Then MjjM3_\ is isomorphic to the semi group M(Hj,GLn(K)lP3,GLn(K)lP~,Nj), where the matrix N0 — (nyx) is defined for x G GLn(K)/P3,y G GLn(K)/P~ by: nyx = h l if yx G U~lUj for some h G HjJ G Lj such that h~ l G # * , and nyx = 9 ifyx iP~Pr Proof. GLn(K) acts transitively on the set of 7£-classes of Mn(K) con tained in Mj \ Mj_l by left multiplication. The stabilizer of this action is the group Pj. Therefore, the set of nonzero 7^-classes of Mj/Mj_x can be identified with a set of left coset representatives of Pj in GLn(K). A similar argument, applied to the set of £-classes under right multiplica tion by GLn(K), allows us to identify it with GLn(K)/P~. We have shown that every %-class contained in Mj \ Mj_i is of the form xPjeP'y for some x G GLn(K)/PJ:y G GLn(K)/Pf. Hence, it is equal to xeHjey. Therefore Mj \ Mj-i = GLn(K)eGLn(K) — UxyxeHjey, where the summation runs over all possible coset repre sentatives x,y. It follows that : M{E3,GLn{K)IP3,GLn{K)jp-,N3)
—► M, \ M ^
given by (g, x, y) i-» xegey is a bijection. We know that (xegey)(xfeg'ey') is of rank j if and only if rank(eyx'e) = j , which is equivalent to yx' G Pf Pj. Moreover, in this case eyx'e — enyx>e and so (xegey)[x'eg
ey) =
x{egnyxlge)y
This means that our map is an isomorphism. □
2.1.
STRUCTURE
25
Define R = {a = (atJ) G Mn(K)\ al3 G {0,1} and a has at most one nonzero entry in each row and in each column }. It is clear that R is the semigroup of all one-to-one partial transformations of the set { 1 , . . . , n). R is called the symmetric inverse monoid. It is the semigroup analogue of the symmetric group. The structure of R is particularly nice. L e m m a 2.6 The sets R0 C Rx C • ■ • C Rn = R, R3■. = R n M,, are the only ideals of R. The Rees factors R3/ R3_x are completely 0simple semigroups with presentations M(Wj, (n), rM,7j), where W3 is the group of j x j permutation matrices and I3 is the identity matrix. Moreover, Green's relations on R are the restrictions of the correspond ing relations on Mn(K) and alZb if and only if aWn = bWn, aCb if and only if Wna = Wnb, ajb if and only if WnaWn = WnbWn for a,b G R. Proof. Let a G R3 \ Rj-i> Permuting rows and columns of a we can bring this matrix to the form e =
, where / is the j x j identity
matrix. Therefore a = (ue)(ev) for some u,v G Wn. Adjusting the order of the nonzero columns of ue and of the nonzero rows of ev we come to a = xgy for some x G X3 fl R,y G Y3 H R and g G G3 fl R. It follows that Rj \ Rj^ = (Xj H R)(G3 H R)(Y3 H R). The above shows also that R3/R3_i has no nonzero ideals, so it is completely 0-simple. It is clear that G3C\R~ W3 and \X3 H R\ = \Y3 n R\ = (j) (the number of choices for the columns in which the nonzero entries of matrices m Y3 C\ R are located). It is easy to see that either yx — e or rank(y:r) < j for x G Xj fl R,y G Y3; fl R. The proof of Theorem 2.3 implies that Rj/R3-\ has the desired presentation. By Proposition 2.4 Green's relations on R are induced from Mn(K). They come from the action of Wn because a permutation of rows (of columns, respectively) brings any a G R to a uniquely determined diagonal idempotent. □ Let B C GLn(K) be the group of upper triangular matrices. B is a Borel subgroup of GLn(K). The corresponding Weyl group is W — Wn - the group of permutation matrices. The following analogue of the Bruhat decomposition for GLn(K) was established in [115]. In this context R is called the Renner monoid of Mn(K). T h e o r e m 2.7 Mn(K) = {JaeRBaB and if BaB = Ba'B, then a = a'. Ifae Mn(K), then we have GLn(K)aGLn{K) = GLn{K)eGLn(K) = UaeWeW BaB, where e = e2 G R is such that rank(a) = rank(e).
26
CHAPTER
2. FULL LINEAR
MONOID
Proof. For i , j G { 1 , . . - ,n} and t G K let et-j be the corresponding matrix unit and x%3(i) — I + teij. Let a = {ai3) G Mn(K). The operation a —t Xij(t)a adds t times row j to row i, while a —»■ axl3(t) acts similarly on columns of a. We will use X{j(t) for z < j only because in this case Xij(t) G B. Thus, we allow addition of rows from below to above and addition of columns from left to right. If all entries of the first column of a are zero then we go to the second column. If the first column has a nonzero entry, then let j \ be the largest integer such that ajli ^ 0. Using the ( i l l ) entry of a we eliminate the remaining nonzero entries in row ji and in column 1. Thus, we find u,v G B such that a' = uav has these properties. Multiplying by a diagonal matrix we can assume that ajji = 1. If all entries of the second column are zero, move to the third column. Otherwise, let j 2 be largest with a^ 2 / 0. Then j 2 ^ j \ - Using the (j 2 2) entry of a' we eliminate the nonzero entries in rows jx, j 2 a n d columns 1,2, except perhaps for the entries ( j i l ) , (j 2 2). Again, (j 2 2) can be chosen 1 if it is nonzero. Proceeding in this way, we come to an element of R. This proves the first assertion. Suppose now that Ba'B = BaB for some
AJ : Mn(K)
—> Mrn\(K)
satisfies
2.2.
IDEMPOTENTS
the following
AND NIL
SEMIGROUPS
27
conditions
1. if a,b G Mn(K) \ Mj_i, then we have aSb in Mn(K) if and only if AJ(a)SAj(b) in Mrn\(K) for each of Green's relations S =
2. ifMj/Mj-i is identified with M{GJ,XJ,Y3,QJ), then A3(M3) can be identified with M{K*,X'j,Yj,Q'j), where X'2 = Aj{Xj),Yj = J = k (Yj)i
then clearly Aj(a)CAj(b)
in M,n\(K).
On the
other hand, if the latter holds, then rank(a) = rank(6) by Lemma 1.6. Suppose that Mn(K)a ^ Mn(K)b. Then ker(a) ^ ker(6). Hence there exists s G Mj such that rank(as) = j and rank(&5) < j . Thus A-7'(as) 7^ 0, while A-7 (6s) = 0. Lemma 2.1 implies that A-7 (a) and AJ:(6) are not in the same £-class of M/n\(K). A dual argument works for the relation 72., so that 1) follows. It is clear that A^(Mj) is a completely 0-simple semigroup over K* with the sets of rows and columns indexed by X'^Y-, respectively. If xgy <E Mj \ M3_i, where x G X^y G Y^g G Gj, then from Lemma 1.6 we know that AJ(xgy) — AJ(x) det(g)A3(y). Therefore 2) follows. □
2.2
Idempotents and nil semigroups
We start by introducing the notion of a triangular set of idempotents. This notion arises from a technical condition, which is used in Chap ter 3 in the proof of the structural theorem for arbitrary subsemigroups of Mn(K) (Theorem 3.5). Apparently, it is also connected with nilpotency of nil semigroups. Recall that a semigroup with zero 9 is a nil semigroup if for every s G S there exists k > 1 with sk — 6. If Sk — {0} for some &;, then S is called power nilpotent, or just nilpotent. (The latter notion will have a more general meaning, for historical reasons, in Chapter 5 only.) The key features of triangular sets of idempotents show up already in the context of subsemigroups of completely 0-simple semigroups. The defining condition was first considered in [127]. We follow [119], where it was introduced in full generality and used to study the more general case of skew linear semigroups, see Chapter 9.
28
CHAPTER
2. FULL LINEAR
MONOID
First, we need some preparatory results. L e m m a 2.9 Let T = (V, E) be an oriented graph, where V is the set of vertices and E C V x V is the set of edges. Then the following conditions are equivalent: 1. for any n > 1 and any sequence v1 ? . . . , vn of vertices of T there exists i G { 1 , . . . ,n} such that ( ^ , ^ + i ) G E, with indices taken modulo n, 2. there exists a linear order ^ on V such that (v,w) G E v ■< w.
whenever
Proof. Assume that Y satisfies condition 1). By transfinite induction, we shall define a sequence of subgraphs Ta of T with the same set ofvertices and the set of edges Ea C E such that: (i) each graph Ta — (V, Ea) satisfies condition 1), (ii) Ep C Ea for a < /?, (iii) (v,v) G Ea for every v G V, (iv) the last graph Fao does not contain double edges (there are no v,w G V with (v,w) G Eao and (w,v) G Eao). Put Ei = E. Then for any v G V we have (u, u) G -Si (because condition i) may be applied to the sequence v). Assume that Ep has been defined for j3 < a, where a = a' + 1. Let r;,u; be distinct vertices of V such that (v,w),(w,v) G 2?a/. Suppose that none of the graphs (V, Ea> \ {(?;, tu)}), (V, £ a / \ {(iu, t>)}) satisfies i). There exist u 1? . . . , t>n G V and i ^ , . . . , w m G V such that (t^,^-+i) ^ an( £<*' \ {{v,w)} l ( W J , W J + I ) ^ Eai \ {(w,v)} for z = 1 , . . . , n and j = 1 , . . . ,m. Assume that there are two equal vertices in the sequence v i , . . . ,vn. Thus there exist i, j G { 1 , . . . , n } , i < j , such that ?;,• = Uj and V{, U{+i,... , Uj-i are all different (z ^ j — 1 because otherwise V{ — uz-+i, so (iii) implies that (i;t-,i>t-+1) G i£a/ \ {(^,VK)}, contradicting the choice of t>!,... , vn). Then the sequence u l 5 . . . , u n may be replaced by V{,... , Vj-i- Therefore, we may assume that u 1 ? . . . , t>n are all different, and similarly that i u i , . . . , ium are different. Since (V, Ea>) satisfies i), there exists i0 G { 1 , . . . , n} such that (vl0,Vi0+i) G i£ a /. But (i^ 0 ,i>; 0+1 ) ^ jBa/ \ {(u,w)}, so that ( ^ 0 , ^ 0 + i ) = (v,w). Similarly, there exists j 0 G
2.2.
IDEMPOTENTS
AND NIL
{ 1 , . . . , m } such that (wJO,wJO+1) (*)
vuv2,...
SEMIGROUPS = (w,v).
,vio-i,vio
= w30+u... = w = vt0+u...
29
Consider the sequence:
= v ,wrn,wl,w2,...
,wJ0_uwJ0
,vn
Suppose that (vt,vl+l) G Ea, for some i G {z0 + 1 , . . . ,n, i , . . . , i0 - 1}. Since (vi,vi+i) 0 Ea, \ {{v,w)}, we get {vt,vl+1) = {v,w). This implies that Vi = vl0,i ^ ilQ, contradicting the fact that vu . . . , vn are different. Hence (vi,vi+i) g Ea> for every i G {z0 + l , . . . , n , l , . . . , z - l } . Similarly, we prove that (^-, w j + 1 ) 0 £ a , for every j G { j 0 + 1, • • • , m, 1 , . . . , j 0 1}. This shows that i) is not satisfied for the sequence (*), a contradic tion. Therefore one of the graphs (V, Ea> \ {(v, w)}), (V, Ea> \ {(w, v)}), say the former one satisfies i). Define Ea — Ea> \ {(v,w)}. It is obvious that (i),(ii),(iii) are satisfied by Ea. If a is a limit ordinal, define Ea — p[p
CHAPTER
30
2. FULL LINEAR
MONOID
A linearly ordered set T of nonzero idempotents of a semigroup S with zero 9 is called a triangular set of idempotents if e -< f implies
that ef = 0foi ej
eT.
Recall that, in a semigroup S = M(G, X, Y, P) of matrix type over a group G, the equality si • • • sn = 9 for some S{ G 5, n > 1, implies that there exists j < n such that SjSj+i = 9. This will be frequently used without further reference. L e m m a 2.10 Let M = M(G,X, Y, P) be a semigroup of matrix type. Assume that every triangular set of idempotents in M has at most n elements. If S is a nil subsemigroup of M, then Sn+2 = 9, and Sn = 9 whenever M is a completely 0-simple semigroup. Proof. Assume that S is not nilpotent of index n + 2, (respectively, of index n if M is completely 0-simple). Then there exist s 0 , • • • , ^ n + i G S (respectively S i , . . . , s n G S), such that so • ■ • s n + i ^ 9 (respectively s i ''' sn i=- #)• Thus S{Si+i ^ 9 for every i G { 0 , . . . , n } , (i G { 1 , . . . , n — 1} respectively). By the multiplication rule in M there exist idempo tents e,-,z = 0 , 1 , . . . , n (i = 1,2,... , n — 1,) such that siei = s2-, e 2 s l + i = s z + i. If M is completely 0-simple, then we also have idempotents e0> en such that Si = e0Si and sn = snen. Since S is a nil semigroup, for every z, j with 1 < i < j < n we have (s z s; + 1 • • • Sj)2 = 9. Then one of the elements Si-st'+i> • • • ,Sj-\sj,sjSi is equal to 9. But s^s/c+i ^9 for fc = 1, 2 , . . . , n — 1. Therefore SjS; = # for i < j . Let z, j be such that 0 < z < j < ™- We have 5jejet-5,-+i = SjSi+i = #. Since also Sjej ^ 9 and e,-3t-+i ^ #, it follows that eje2- = 9. We have proved that e 0 , . . . , e n form a triangular set of idempotents in M. This is impossible by the hypothesis on M. Therefore Sn+2 = 9 (Sn = 9 respectively). □ R e m a r k A converse of this lemma is also true. Assume that T C M = M.{G,X,Y,P) is a triangular set of idempotents such that \T\ = n < oo. Let S = {s G M\ se = s, fs = s for some e, / G T, e ^ / } . We will first prove that 5 is a subsemigroup of M. Let s,£ G 5. By definition there exist e ! , e 2 , / i , / 2 £ ^ s u c n that e i ^ / i 5 e 2 ^ /2 a n d 5 = s e i , s = / i 5 , t = te 2 ,t = / 2 t . Consider the following two cases: i) ex -< f2 Then st = (sei)(f2t) = s9t = 9. Therefore st G S. ii) h ■< ei
2.2.
IDEMPOTENTS
AND NIL
SEMIGROUPS
31
Assume that st 7^ 9. Then st = ste2,st = fxst and e2 ^ f2 r< ei ^ / i . This means that st G S. Next, we will prove the following useful fact: (*)
If st = gstg ^ 9ior someg G T, 5,t G 5, t h e n s = gsg&ndt — gtg.
Let 5,t G 5 and let e i , e 2 , / i , / 2 G T be defined as above. Since st 7^ 0, case ii) holds. We know that st = s£e2 and st = gstg. It follows easily that e2Cg in M. Thus e2 = g (e2 j^ g implies e2g = 9 or ge 2 = #, a contradiction). Similarly fi = g. By case ii) we have e2 ■< f2 -< ex -< fx. Since also e2 = g = fu this yields e2 = f2 = ex — fx— g. Thus 5 = gsg and t = gtg, as desired. Let TV = {s G M\s = se,s = / s for some e , / G T, e -< / } U {#}. S\N is composed of maximal subgroups of M intersecting T. By (*) N is an ideal of S. Moreover TV is a nil semigroup (because s2 — (se)(fs) = s(ef)s = s6s = 9). We will show that TV is not nilpotent of index < n. Indeed, there exist e x , . . . , e n G T with ex -< e2 - < • • • - < en. Choose st G iV as follows: S{ = S{ti,Si — et-_|_iS{ for z = 1 , . . . , n — 1. If i > 2, then we have SiSi-i — (s2-et-)(ez-st-_i) 7^ # since s2-ez- 7^ 6 and et-st-_i 7^ #. Thus also s n _ i s n _ 2 • • ■ 5i 7^ 9, which shows that A^n_1 7^ 9. We note that S \ N a disjoint union of n groups which are jT-classes of S. It is clear that S is a 7r-regular semigroup. The above shows that if there exists a triangular set of idempotents of cardinality n in M, then there exists a 7r-regular semigroup S C M which has n ^-classes containing idempotents and a nilideal N which is not nilpotent of index < n. We now investigate the notion of a triangular set of idempotents in Mn(K). The following key observation is based on an idea similar to that used in the proof of Theorem 1.12. In view of Corollary 2.12 it also leads to an improvement of the bound obtained there. L e m m a 2.11 Let T C Mn(K) be a set of elements of Mj\M3_x whose images in Mj/Mj-i form a triangular set of idempotents. Then \T\ < &
)
■
Proof. First assume that T C M x is a triangular set of idempotents of Mn(K). Suppose that eu . . . , e n + 1 G T are such that e{e3 = 0 if i < j .
CHAPTER
32
2. FULL LINEAR
MONOID
Let 0 / u t G Im(e,-) C Kn. Since U i , . . . ,u n +i a**e linearly dependent, there exist A; and ak+\,... , a n + i £ /i such that vk = a^+i^fc+i H
h
ttn+iVi-
Thus ekVk = J2?i=k+iaiekvi = 0 (note that ekV{ — 0 for i > k). This contradicts the fact that ekVk = vk ^ 0, establishing the assertion in the case j — 1. Assume now that T C Mj \ Mj_x has cardinality exceeding t — vyj and satisfies the conditions of the lemma. Apply the exterior power map A-7. From Lemma 1.6 it follows that A J (T) inherits the assumptions on T. Moreover, it has the same cardinality as T and consists of matrices of rank one in Mt(K). This contradicts the first paragraph of the proof, completing the argument in the general case. □ Corollary 2.12 Assume
that S i , . . . ,s/n\ £ Mn(K)
are matrices of
rank j such that rank(sx • • • s/n\) = j . Then there exist i < k such that Si ■ • • sk lies in a maximal subgroup of
Mn(K).
Proof. The proof of Lemma 2.10 shows that one can construct a tri angular set of idempotents in Mj/M3-i of cardinality (n) + 1 if the assertion does not hold. This contradicts Lemma 2.11. □ Corollary 2.13 Let T C Mn(K) be a set of idempotents of rank j . Assume that ( e x / ) 2 £ Mj_i for every x £ (T) 1 and every e,feT with e / / . Then there exists a linear order < on T such that ef £ Mj_i whenever e -< f. In particular, \T\ < ( n ) . Proof. Let V = T, and E C V x V be defined by the rule (e, / ) £ E if and only if ef £ Mj_i or e = f. We will prove that T — (V, E) satisfies condition 1) of Lemma 2.9. Let e x , . . . , e n £ E and ex ^ en. Then, by the hypothesis (ei(e 2 e 3 • • • e n _x)e n ) 2 £ Mj_i. Hence, there exists i such that (eiei+i) £ E (with indices taken modulo n). If ex = e n , then (e n ,ei) £ £ . By Lemma 2.9 there exists a linear order on V satisfying the desired condition. From Lemma 2.11 it follows that \T\ < ( n ) . □ Now, we turn to nil semigroups arising from P r o p o s i t i o n 2.14 Let S C Mn(K)
Mn(K).
be a semigroup.
Then
2.2.
IDEMPOTENTS
AND NIL
SEMIGROUPS
33
1. If S has a zero <9, then S C (/ - 9)Mn(K)(I - 6) + 6 ~ M , ( / i ) , where t — n — rank(#). Moreover, if S is a nil semigroup, then
2. If J is an ideal of S and S \ J does not intersect maximal sub groups of M n ( / \ ) , then S/J is nilpotent of index not exceeding m — \Yk=i [kj- In particular, this holds if S is i\-regular and S/J is a nil semigroup. Proof. 1) Since 9s = s9 = 6 for every 5 e S, s = (I-6)s(I-6) + 0 and the first assertion follows via the map s \-t (I — 6)s(I — 6). To prove the second, it is then enough to consider the case where 6 — 0 and t — n. We will use standard facts about irreducible semigroups, presented in Section 4.1. For another proof we refer to [24], 17.19. Conjugating in Mn(K) we can assume that S is in a block triangular form of Propo sition 4.4 with irreducible or zero diagonal blocks T^\i = 1 , . . . , r. It is enough to show that there are no irreducible blocks, because in this case S is upper triangular with zero diagonal and Sn — 0 follows. If some TW is irreducible, then the identity e of the /i-linear span of T^ can be written in the form J2i=i aisi f° r some S{ € T^ and OL{ G K (see Lemma 4.1). But the trace of each S{ is zero because S is nil. Therefore tr(e) = 0, a contradiction. Hence 1) follows. 2) Consider the chain 5 0 C 5i C ••■ C S n , where Sj = S D M3. If Sj 7^ 0 and a i , . . . , a/ n \ G Sj, then from Corollary 2.12 it follows that n ai • • • asn\ G J U Mj_i. Hence (Sj)
C J U 5j_i. This easily implies that
m
S
c J. Assume that S is 7r-regular and S/J is nil. Then every intersection of S with a maximal subgroup of Mn(K) is contained in J by Corollary 1.5, so the above applies. □ It is clear that the last assertion of 2) is not true without the as sumption that S is 7r-regular. In fact, S = Z C Q with the ideal / = U ^ x pl.S, where pi,p2, • • • are the subsequent prime numbers, is a counterexample. We conclude with the following direct consequence of Lemma 2.10 and Lemma 2.11, or of Proposition 2.14 and Lemma 1.6.
CHAPTER 2. FULL LINEAR
34 Corollary 2.15 Assume factor Mj/Mj-u Mj/M^u
2.3
that N is a ml subsemigroup
Jj < n, of the monoid Mn(K).
MONOID
of a principal
Then N® M " ) = 9. 0.
Combinatorics on Mn(K).
In this section we present more detailed information on the structure of the full linear monoid Mn{K). We concentrate on the combinator ial aspects of this monoid. This is justified by the role of the analo gous results for the full linear group GLn(K), cf. [13],[18]. As before, Mj,j = 0 , 1 , . . . , n, will denote the ideals of Mn(K). Also, 5, R stand for the Borel subgroup consisting of upper triangular matrices in GLn(K) and for the Renner monoid, respectively. By T, U C B we mean the subgroups of diagonal matrices and of unipotent matrices, respectively. To get a deeper insight into the combinatorics of Mn{K) one needs, as in the case of GLn(K) (and also of more general classes of groups cf.[13],[18]), a notion of the length function. Our aim is to prove several technical results on the nature of this function, culminating in Propo sition 2.27, and then to use it in Theorem 2.30. The latter shows that Theorem 2.7 leads to a nice analogue of the powerful Tits system for GLn(K). In our presentation we closely follow Solomon, [129], whose work is in part modelled after the classical papers of Chevalley and Iwahori, [14],[41]. Note that another approach to the length function was taken by Renner in [115]. Let W C Mn{K) be the group of permutation matrices and let S — { ( 1 , 2 ) , ( 2 , 3 ) , . . . ,(n - 1,n)} be the set of its 'simple reflections' - a distinguished set of generators of W. By EtJ we denote the matrix unit with 1 in position (i,j) and 0 elsewhere. If r £ { 1 , . . . , n } , then every element aeRr = Rf)(Mr\ Mr_i) is of the form a = ££= 1 Eikjk, where 1(a) = {i1?... ,ir},J(a) = {ju... , j r } are subsets of { 1 , . . . , n } of cardinality r. Write ika = jk and ajk = ik. Thus
^T Eitia = a = J2 Eajtj. «'6/(a) ie/(a)
(2.1) (2.1)
j€J{a) j€J(a)
Note that i .->• ia,j >-► aj are biiecttve maps ffom 1(a) to J(a), and from J(a) to /(a), respectively. If w £ W, then I(w) = { 1 , . . . , n } = J(w) and wi = iw'1 for all i £ { 1 , . . . , n } . Let x* denote the transpose of
2.3. COMBINATORICS
ON
Mn(K)
35
x G Mn(K). Since £*• = EJt, we have 1(a) = J(a*) and J ( a ) = /(a*). Also (z'a)a* = z for i G / ( a ) and a*(aj) = j for j G J ( a ) . The group WK x W acts on i? by (w, v)a — wav~l for a G i? and it;, i> G VT.
(2.2)
Define a graph with vertex set J?r as follows. Two vertices a, 6 are adjacent if there exists s G S such that a = sb or there exists s G 5 with a — bs. The graph is connected because 5 generates W and /T is a W x W-orbit of i? by Lemma 2.6. For a,b e Rr define the distance <£(a, 6) from a to 6 £(a, 6) = min{/(iu) + /(K/) | w, w' G VK and a = w W }
(2.3)
where Z(w) is the length of w in the generators from S. We will define 1(a) = J(a, z) for a suitably chosen z G Rr. Let m = £12 + £23 + • • • + £n-i,n.
(2.4)
Then mr
= mn~T
= £ i ) n _ r + i + E2,n-r+2
H
h £r,n
(2.5)
is a matrix of rank r. We define the length of a G Rr by the rule 1(a) — mm{l(w)
+ l(w')\ w,w' G W and a = wmrw'}.
(2.6)
Then rar is the only element in Rr of length zero. Moreover, \l(sa) — 1(a)| < 1 and \l(as) — l(a)\ < 1 for every a G R and 5 G S. Our first aim is to give a combinatorial description of 1(a) for a G R. This will be accomplished in Proposition 2.27. Let A = {(i,j)\l
(2.7)
and A + = {(*, j) G A| i < j } , A " - {(z, j ) G A| z > j}.
(2.8)
W acts on A by w(ij) = (wi,wj) for iy G W. For s e S which is the transposition of /c and k + 1 let a s = (A;,fc+ 1) G A + . Then S
(A+\{aJ) = A+\{a.}.
(2.9)
CHAPTER 2. FULL LINEAR MONOID
36 IfweW,
then let
tf'(iu) = {a G A + | w~la G A + } , V"(w) = {a G A + | w _ 1 a G A " | 2 . 1 0 ) Thus A + = V'(w)U1$"(w)
(2.11)
where the symbol U is used to emphasize that the union is disjoint. Note that (z, j) G ty"(w) if and only if (j, z) is an inversion of the permutation fc H-> tufc of the set { 1 , . . . , n } . Thus n(w) = |^/"(u;)|, where n(u;) denotes the number of inversions of w. From (2.9) it follows that the function w— i »■ ty"(w) satisfies the 'cocycle condition' V"(sw) = s$"{w) U {a,} if as G V\w) q"(w) = sV"(sw) U {as} ifas
G#"H
(2-12)
where the unions are disjoint and thus / n{w) + l if a , E t f ' H /91ON n(siu) = S ) ( T -r ^ lTf/// \ (2.13) v J 1 n(u;) — 1 it a s G W (it;) Our aim is to prove several formulas analogous to (2.13) with W replaced by R and to use them to derive a formula for 1(a), a G i? r , in terms of A. For a subset A C { 1 , . . . , n } define the following partition of A Aoo(A) = {(z,j) eA\i#AJ#A}
(2.14)
Aoi(A) = { ( i , j ) G A | z ^ A , j G A 1 0 (A) = {(z,j) e A | z G
An{A) =
A}
A,j#A}
{(ij)eA\ieAjeA}
If x,y G {0,1} and a e R, define subsets * ^ ( a ) and ^ ^ ( a ) of A by * x y ( a ) = Axy(I(a)),
$xy{a)
= Axy(J(a)).
(2.15)
For any subset T of A we write T + = T fl A + . Next, define *'(«) = { ( M ) e ^ ( a ) | (ta, ja) € A + } $'(a) = {(*,» G $ + ( a ) [ ( a t , a j ) G A + } *"(«) = { ( M ' ) e * i 1 ( a ) | ( a t , a i ) e A - }
(2.16)
2.3. COMBINATORICS
ON
Mn(K)
37
Since J(a) = I (a*), we have **„(*) =
tf^O,
*+(a) = *+(a*)
(2.17)
for x , y G {0,1}. Also $'(a) = tf'(a*), $"(a) = * > * ) ■
(2.18)
There is a duality between the pairs J, 0 and / , $ . Namely, to each '$-statement' concerning left multiplication a \-± sa there corresponds a dual '^-statement' concerning right multiplication a \-^ as, which may be deduced from it if we replace a by a* and use the fact that (sa)* = a*5. For example (2.12) yields $"(ws) = s&'{w) U {as} if as G $'(w) $"(u,) - 5 $ ,, (iw5) U { a j if as G $ " H L e m m a 2.16 The map (z, j ) i-> (ja,ia)
(2.19)
is bijective from \P"(a) to
#"(a*)Proof. za > ja. so that tf'V),
Suppose that (z, j) G \P"(a). Then i G / ( a ) , j G J(a),z < j and Thus j a G J(a),za G J(a),ja < ia and {ja)a* = j > i — (za)a* (ja,za) G \P"(a*'). Replacing a by a* we see that, if (i\jr) G then ( j V , i V ) G *"(a**) = tf"(a). D
Define a function n : R —> N by n(a) = |tf"(a)|. If a = UJ G 1^ this agrees with n(u;) defined earlier. It follows from Lemma 2.16 that n(a*) = n(a). If a = rar, then / ( a ) = { 1 , . . . , r } and za = z + n - r for so i/>"(a) is empty and thus n(a) = 0. If w e W and n(tu) w = 1. It is not true that if a G Rr and n(a) = 0, then a overcome this difficulty we introduce another function m : For A C { 1 , . . . , n } define moi(A) - | A + ( A ) | and m 10 (A) = |A+(A)|.
(2.20) i = = R
G /(a), 0, then mr. To —> N .
(2.21)
38
CHAPTER
2. FULL LINEAR
MONOID
L e m m a 2.17 If / / A C { 1 , . . . ,,n} n } /»as has cardinality r, then mcn(A) m = ££ (( ** -- 1) 1) -01(A) = keA keA
^^ -- ^^ **
•MA) = £ < » - * ) - « ^
keA
Proof. Write A = {kt....
, kr}, where kY < ■ ■ ■ < kr. For 1 < q < r let
Aq01(A) = {(i,j)
e A+,(A)\ j = kq}.
Since Aq01(A)
=
{(l,kq),(2,kq),..,,(kq-l,kq)} \{(A;i, A:g), (/c2, fcg),... , (/cg_i, &g)}
we have | A ^ ( A ) | = (kq --l)-(q1) - (q - 1). Since A 0 1 (A) = UJ = 1 A ^ ( A ) is a disjoint union, the first formula follows. The second formula is proved similarly. □ If a E R define m01(a) = m = mw(J(a)) moi(/(a)) = | ^ ( a ) | , m10 (a) = 01(I(a)) 1 0(a)
= |$+(a)|(2.22)
and put m(a)=moi(a)+m m(a) = m01{a) lo(a). + m10{a). From the definitions and (2.17) it follows that, if r G Rr, then = £ m(a) = i€/(a) i€l(a)
(i - 1) + £
(n - jJ)) - r^( r - 1).
(2.23)
jeJ(o) jeJ{a)
Since J(a) y(a) = I(a*), /(a*), it follows from (2.17) and (2.22) that m001i(a) (a) mw10(a*) (a*) = = r(n -- r). Similarly, since 1(a) = J(a*) we have m10(a) m„i(a*) m = r(n - r). Thus, if r €e Rr, then 01(a*) = m(a)+m(a*) 2r(n-r). m(a) + m(a*) = 2r(n-r).
+ +
(2.24)
Define p : i? fl —► N by p(a) = m(a) + n(a). We will prove in Proposi tion 2.27 that p(a) = 1(a).
2.3. COMBINATORICS
ON
Mn(K)
39
L e m m a 2.18 If a G i? r , then p(a) > 0 with equality if and only if a = mr. Proof. We have already noticed that both \I>oi(mr) a n d $^ 0 (m r ) are empty. So is $ " ( m r ) . Thus p(mr) = 0. Suppose now that a G Rr is such that p(a) — 0. Then m(a) = 0 = n(a). The former equality shows that l^ r oi( a )l — 0 = |^io( a )l- Since 1(a) and */(a) have cardinality r, it follows from Lemma 2.17 that 1(a) = { 1 , . .. , r } and J(a) — {n — r-\-1,.. . , n } . Since | ^ ; / ( a ) | = n(a) = 0 we have ia < ja for all 1 < i < j < r. Since ia G J(a) = {n — r + 1,. . . , n} and the map i \—> ia is bijective from 1(a) to J ( a ) , we must have za = n — r + i for 1 < i < r. Thus a — mr. D In view of (2.14),(2.15) and (2.16) each a G R determines the fol lowing partition of A + A + = * + (a) U tf + (a) U tf+(a) U #'(a) U
tf"(a).
(2.25)
This replaces the two part partition (2.11) corresponding to an element w G W. We need analogues of (2.13) for the sets in this partition. These will be proved in Lemma 2.22. If w G VK, then I(wa) = wl(a) and I(aw) = 1(a)- It follows that if w G W and x,y £ {0,1}, then $xy(wa)
— w^xy( a) and tyxy(aw) — ^!xy(a).
(2.26)
L e m m a 2.19 Suppose that x,y G {0,1}, a G i? and s G S*. TTien we
i. 5 ( * + ( a ) \ { a J } ) = * + ( 5 a ) \ { a J , 2. s(tt"(a) \ {as}) = V'(sa) \
{as}.
Proof. Since ^tyia) \ ias} — ( A + \ {as}) H tyxy(a), the first assertion follows from (2.9) and (2.26). Suppose that (z, j) G * " ( a ) \ { a j . Then z G 1(a), j G /(a),z < j and ia > j a . Thus sz G I(sa),sj G /(5a) and sz < s j because (z,j) ^ a s . Since (si)(sa) = (is)(sa) = ia > ja = (js)(sa) = (sj)(sa), it follows that (sz,sj) G V(sa). Thus s(V(a) \ {as}) ^ ^"(sa) \ {as}. To get the reverse inclusion replace a by sa. □ L e m m a 2.20 Suppose that x,y G {0,1}, a G i? ana7 5 G 5. T/ien
CHAPTER
40
2. FULL LINEAR
MONOID
1. as G tyXy{a) if and only if as G S&yx(sa), 2. as G $'(a) if and only if as G
W(sa).
Proof. To prove the first assertion suppose for example that x and 6 = 1 . Write as = (A;,k + 1), where 1 < k < n — 1. It follows (2.26) that we have the following equivalences: as G \Poi( a ) ^ k ^ and k + 1 G I{a) <£> sk G / ( a ) and s(fc + 1) £ / ( a ) <£> k G / ( s a ) A; + 1 0 / ( s a ) <& as e \p 10 (sa). The proof of the second assertion is similar. □
= 0 from 1(a) and
L e m m a 2.21 For any a G R,s G 5 we have: 1. If as G \Poo(a) ^ e n
5a
= a-
2. If as G \Poi(a) #&en *J 1 (a) = 5 * J 1 ( 5 a ) U { a J * + ( 5 a ) = 5*S,(a) U { a s } tf"(sa) = sV'(a) 3. If as G *io(a) ^ e n tffo(a) = 5 * + ( 5 a ) U { a j *+(5a) = s*+(a)U{as} #"(sa) = s # " ( a ) ^. 7 / a a G * n ( a ) tfien tt+oM = stf+(a) V"(sa) = stf"( a ) U {«.} »/a 5 G tf'(<0 tf"(a) = sV"(sa) U { a j z / a s G tt"(a) Proof. Write a s = (fc,fc+ 1), where 1 < fc < n-1. To prove 1) suppose that as G *oo(a). Then k £ 1(a) and k + 1 0 / ( a ) . Thus sz = z for all i G / ( a ) . Since sE{j = Esi^ for all z, j G { 1 , . . . , n } , we have 5a = a. This proves 1). We will deduce 2),3),4) from Lemma 2.19, Lemma 2.20, and the fact that the union (2.25) is disjoint. Note that the unions in 2),3),4) are disjoint because sas G A~. To prove 2) suppose that as G ^oi(a). Then ot-s G \Pio(sa) by Lemma 2.20. Thus as £ $10(a) and thus as g SI?01(sa).
2.3. COMBINATORICS
ON
Mn{K)
41
It follows from Lemma 2.19 that s(*j}i(a) \ {as}) = *5i(sa) \ { a s } = tfJi(sa) and tfj"0(sa)\{aj = s(^+(.sa) \ {a s }) = s#f 0 (sa). This proves the first two assertions of 2). Since as G ^oi(«), we have as £ $ n ( a ) and thus a s £ * n ( s a ) . Therefore a s £ * " ( a ) and a s £ * " ( s a ) . Now the remaining assertion of 2) follows from Lemma 2.19. To prove 3) suppose that as G *io(a). Then as G \Poi(sa) by Lemma 2.20. Thus we may apply 2) with sa in place of s. This proves 3). To prove 4) suppose that <xs G * n ( a ) . Then as g \P 0 i(a) and as $ *io(a), so as £ \P 0 i(sa) and a s & ^io(sa) by Lemma 2.20. Now the first two assertions of 4) follow from Lemma 2.19. If as G #'( a )> then as 0 \P"(a) and also as G V(sa) by Lemma 2.20. Now the third assertion of 4) follows from Lemma 2.19. If as G ^ " ( a ) , then as G ty'(sa) so the last assertion of 4) follows from the third by replacing a by sa. □ L e m m a 2.22 For any a G R,s G S we have 1. If as G ^oo(a) then sa = a. 2. If as G ^oi(<^) then ra(sa) = ra(a) — 1 and n(sa) — n(a). 3. If as G tyio(a) then m(sa) = m{a) + 1 and n(sa) = n(a). 4- If OLS G \Pn(a) then m(sa) — ra(a) and n sa
^ >-
\
n
(a)-l
ifa8eV"{a)
Proof. It follows from (2.26) that m10(sa) = |*+ 0 (a*s)| = |*f 0 (a*)| = m 1 0 (a). Thus we may replace m by ra0i in each of 2)-4). Now the assertions follow at once from Lemma 2.21. □ Corollary 2.23 / / a G R and s G 5, £/ien sa = a or p(sa) = p(a) ± 1. Note that the assertions in Lemma 2.22, which compare m(sa) with 772(a), may be expressed in a single formula: if x,y G {0,1}, then OLS £ ^a;y(a) implies that m(sa) — m(a) — x — y.
(2.27)
Recall that $xy(a) = ^ x y ( a * ) . Since 5a* = (as)* andrank(a) = rank(as), it follows from Lemma 2.22 that, if x,y G {0,1}, then ots G $:ry(a) implies that m(as) - ra(a) = y - x.
(2.28)
CHAPTER
42
2. FULL LINEAR
MONOID
Note that n(a*) = n(a) by (2.20). Also $'(a) = *'(a*) and $"(a) = *"(a*) by (2.18). Thus, the analogue of Lemma 2.22 for right multipli cation is: L e m m a 2.24 For any a G R,s G S we have: 1. If as G $oo(a) ^ n as = a. U. / f a s G $oi( a ) ^ e n m(as) — 771(a) — 1 ana7 n(as) = 77(a). 3. 7 / a s G $io( a ) ^ e n m(as) — m(a) + 1 ana7 77(05) = 77(a). 4. If OLS G $ n ( a ) i7ien m(as) = 777(a) ana7
n l a 5 j
~ \ n(a)-l
i/a8G$"(a)
Corollary 2.25 / / a G i? ana7 5 G 5, t/ien as = 5 or p(as) = p{a) ± 1. L e m m a 2.26 If a G Rr and a ^ m r , then there exists s G S such that p(sa) = p(a) — 1 or p(as) = p(a) — 1. Proof. Suppose first that 1(a) ^ { 1 , . . . , r } . Write / ( a ) = { z 1 ? . . . , z r }, where i1 < ■ • • < ir. Then either (i) il > 1 or (ii) there exists g G {2,. . . , r } such that zg — zg_i > 1. If (i) occurs, let k = i x — 1. If (ii) occurs, let k = iq — 1. Then k 0 / ( a ) and k + 1 G / ( a ) so (&:, A; + 1) G *oi(a). Define s G 5 by as — (k, k+l). From 2) in Lemma 2.22 it follows that m(sa) = 777(a) — 1 and n(sa) = 77(a). Hence p(sa) = 79(a) — 1. Thus we may assume that 1(a) — { 1 , . . . , r } . If J(a) ■=£ {n — r + 1 , . . . , n} it follows by a similar argument using 3) of Lemma 2.24 that there exists 5 G S such that p(as) = p(a) — 1. Thus we may assume that 1(a) = { 1 , . . . , r } and J(a) = {n - r + 1 , . . . , n } . Then a = £-=i Eiiia, where { l a , . . . ,ra) — {n — r + 1 , . . . ,77}. Since a / mr there exists p G { 1 , . . . , r - 1} such that pa > (p + l ) a . Thus (p,p + 1) G ^ " ( a ) . Define 5 G S by a s = (p,p + 1). It follows from 4) in Lemma 2.22 that
p(sa) — p(a) — 1. □ P r o p o s i t i o n 2.27 Ifa G Rr then 1(a) = 777(a) + n(a).
2.3. COMBINATORICS
ON
Mn{K)
43
Proof. By induction on 1(a) we first show that p(a) < 1(a). Let a = wmrw\ where l(w) + l(wf) = 1(a). If 1(a) = 0, then w = 1 = wf so that a — mr. From Lemma 2.18 it follows that p(a) = 0. Suppose that 1(a) > 0. Then l(w) > 0 or l(w') > 0. Assume that l(w) > 0 (the other case goes similarly). Write w — sw"', where 5 G 5 , w " G W^ and l(w") — l(w) — 1. Let b = sa = w"mrw'. Then 1(b) < 1(a). The induction hypothesis implies in view of Corollary 2.23 that p(a) < p(b) + 1 < 1(b) + 1 < 1(a), as claimed. Next, we prove that 1(a) < p(a) by induction on p(a). If p(a) = 0, then a — mr by Lemma 2.18, so 1(a) = 0. Suppose that p(a) > 0. Then a ^ rar, so Lemma 2.26 implies that there exists s £ S such that p(sa) < p(a) or p(as) < p(a). If p(sa) < p(a), then by induction 1(a) < l(sa) + 1 < p(sa) + 1 < p(a). The argument in the latter case is similar. This completes the proof. □ Corollary 2.28 Suppose that a 6 R and s £ S. If l(sa) = 1(a) then sa = a. If I (as) = 1(a) then as = a. In view of Proposition 2.27, the precise circumstances in which l(sa) — 1(a) + 1 and l(sa) — 1(a) — 1 are given by Lemma 2.22. Simi larly, the circumstances in which I (as) = 1(a) + 1 and I (as) — 1(a) — 1 are given by Lemma 2.24. Our next aim is to prove the multiplication formula for the orbits BaB,a £ i?, which exploits ormation. If (z,j) £ A, then let x{j(t) = 1 + tEtJ for t £ K. Then XZJ = {xij(t)\t £ K} is the corresponding 'root subgroup'. We recall some basic facts about these subgroups, see [13]. A subset T of A is said to be closed if it has the following property (hj)
e r,(j,fc) £ T,z 7^ k imply that (i,fc) £ T
(2.29)
If T C A + let Ur be the subgroup of U generated by the Xtj with (z, j) £ T. If T is a closed subset of A + , then every u £ Ur may be uniquely written in the form
"= II xi&a)
( 2 - 3 °)
(t,j)er
where ti3 £ K and the product is taken in any fixed order. If A + = r ' U r / ; , where T\ T" are closed subsets of A + , then U = Ur>Ur» and Ur H Ur> = 1.
(2.31)
CHAPTER
44
2. FULL LINEAR
MONOID
Let w G W. Then $'(w),$"(w) are closed subsets of A + . We define subgroups U'w,Uu of U by U'w = U$i(w),U^ = c T ^ ) . Since A+ = $'(w) U $"(u;), we have t ^
= £/ = U:U'W and C/: (1 U1: = 1.
(2.32)
Every element in .BtfB may be written in the form bwu"', where b £ B and u" G £7" are uniquely determined. We use the partition (2.25) to define subgroups Ufa,U" for a e R. Namely , let Q\a) = *&(a)U*+1(a)U&{a),
0"(a) = $ + ( a ) U
(2.33)
Note that for a = iu G W we have &(a) = $'(w) and 0"(a) = $"(w). L e m m a 2.29 If a G i?, £/ierz 0 ' ( a ) , 0 " ( a ) are closed subsets of A + anc/ A+ = 0 ' ( a ) U 0 " ( a ) . Proof. Suppose that (z',j) G 0'(a) and (j,A:) G 0'(a),z ^ fc. If (ij) G $J 0 (a) U $Ji(a), then z £ J ( a ) so (z,fc) G $ J 0 ( a ) u $ o i O ) Q ©'(a) because i ^ k. Suppose that (z, j ) G $'(a). Then i G J(a)->j G J ( a ) and az < a j . Since j G i) £ $"( a )- If & £ «/(«) then, since z ^ fc, we have (z,fc) G $io(«) C 0 ' ( a ) . If k G J ( a ) , then (j, A;) G $"(a) so a j > ak. Thus az > aA; so (i,k) G $"(a) C 0"(a). Thus 0"(a) is closed. The assertion A + = 0'(a) U 0"(a) follows from (2.25) with a* in place of a. □ Define subgroups [7^, [7" of [7 by ^ = J/e»(„)andC^' = C/e»(a)-
(2.34)
From (2.31) it follows that U'JJ: = U = U:U'a and U'a n ^
= 1.
(2.35)
If a = iy G Wj then [7^, [7^' have the earlier meaning and (2.35) agrees with (2.32). We will need the fact that, if ij G { 1 , . . . , n } , then 13
[0
otherwise
^ '
'
2.3. COMBINATORICS
ON Mn(K)
45
and
■Mo*" ~
w
It follows that Xij(t)a = a if j 0 / ( a )
(2.38)
ax{j(t) — a if z ^ J ( a ) and x%3(t)a = axiai3a(i) if z,j G /(a) axl3(t) = x at - ia j(t)a if i , j G J ( a ) .
(2.39)
We are ready to prove the formula for the product of B x 5-orbits BaB, a G R. The obtained trichotomy replaces the duality of the clas sical case of GLn(K), [40], § 29. Recall that BsB • 5u;i? is equal to BswB if Z(stu) > Z(iu), or to # s w i ? U B w B , if Z(siu) < l(w). T h e o r e m 2.30 For every a G R and every have ( BaB if BsB • BaB = I BsaB if [ BsaB U BaB if
simple reflection s G S we l(sa) = 1(a) l(sa) = 1(a) + 1 l(sa) = 1(a) - 1
Proof. We will show that ( BaB if as G *oo(a) BsB ■ BaB = i 5sa5 if a s G *io(a) U tf'(a) ( R s a # U BaB if a s G *oi(a) U #"(a) Then the assertion will follow from the behaviour of the functions ra(a), n(a) under left multiplication a H->- sa, determined in Lemma 2.22, and from the fact that 1(a) = m(a) + 72(a), established in Proposition 2.27. If BsaB — BaB then sa — a by Theorem 2.7. From Lemma 2.22 it follows that as G ^00(a). Thus as G \J/oi(#) U ty"(a) imphes that BsB i- BsaB. The left-hand side may be replaced by sBa and equality may be replaced by inclusion provided that we show for as G ^01(a) U \I/"(a) that sBa meets both BsaB and BaB.
CHAPTER
46
2. FULL LINEAR
MONOID
Write U = U'aU'J. We have tf"(5) = {as} and V(s) = A+ \ { a 5 } . From (2.9) it follows that sV'(s) = tf'(5)> so that sU'3s = U's. Let k be such that a s = (k,k + 1). Then £/" = A ^ + i . Hence 5B = sTU = T s C O T = TsU'assXkM1 C BsXkM1. If a s G *ooO) U *io(a), then fc + 1 0 / ( a ) , so from (2.38) it follows that Xktk+ia — a- Then sBa C fisa C BsaB. If a s G ^oo( a ) 5 then sa = a by Lemma 2.21. Hence sBa C ,Ba5. If as G ^oi(a), then we must show that sxkik+i(t)a £ i?si? U BsaB. This is clear for t = 0. If £ zfi 0, let /i G GLn(K) be the diagonal matrix with entries — t - 1 , t in positions fc,fc + 1, and the other diagonal entries equal to 1. It is easy to check that sxk,k+i(t) =
hxk,k+i{-t)xk+i,k(r1)'
Since as G \Poi (&), we have fc ^ ^(&)5 which in view of (2.38) implies that Xk+i,k{t~~1)a = a - Therefore sxk,k+i(t)a = hxk}k+i(—t)a £ BaB, as desired. Suppose a s G ^ 11(a). Then &;,£; + 1 G / ( a ) and (2.39) implies that xk,k+i(t)a — aXka,(k+i)a(t)- If « s £ ^'( a )> then ka < (k + l)a. Hence sxk,k+i(t)a> G B 5 a R If a s G * " ( a ) , then a s G W(sa) by Lemma 2.20. Thus, replacing a by 5a we come to sBsa C BaB. Since s B s C BUBsB by the remark preceding the theorem, it follows that 5 5 a — sBssa C ( 5 U BsB)sa C BsaB U 5 a 5 . This completes the proof of the theorem.
a For the rest of this section we assume that K = ¥q is a finite field with q elements. Our aim is to derive basic numerical information re lated to the presented structural approach. P r o p o s i t i o n 2.31 For every n > 1 we have \GLn(Fq)\
= (q-
1 ) V ( - 1 ) / 2 n ( l +
Let n3 = I G M F J K ^ - ^ I G L ^ F J U G L ^ ^ F , ) ! ) - 1 , for j = 1,... , „ . Then the semigroup Mj/Mj_i has rij nonzero TZ-classes, rij C-classes and \MJ\M]_l\ = n)\GL](¥q)\.
nonzero
Proof. It is clear that the order of GLn(Fq) is the number of ordered bases of the F g -space F™ (to a given g G GLn(Fq) associate the row
2.3. COMBINATORICS
ON Mn{K)
47
vectors of g). The first vector can be chosen in qn — 1 ways. Then there are qn — q choices for the second vector, qn — q2 for the third and finally qn — qn~l choices for the last vector. Therefore
IGLB(F,)I
= n V - **■)=nV iitf -1) i=0
2= 1
i=l
= ( 9 - i ) v ( n - 1 ) / 2 n ( i + ? + ••• +
From Proposition 2.5 we know that the number of nonzero 7^-classes of MJ/MJ_I is equal to the index of the subgroup Pj = {a £ GLn(Fq)\ ae = eae} in GLn(Fq),
where e = e2 = I
J has rank j . Since the order of
Pj is equal to qJ(n~J}\GLj(Fq)\\GLn_3(Fq)\, the assertion on the number of 7£-classes follows. A similar argument works for the ^-classes of Mj/Mj_i. In view of Theorem 2.3 this proves the formula for \MJ\MJ_1\. D In order to compute \BaB\ in terms of 1(a) we need another technical result. L e m m a 2.32 If a £ R, then BaB — BaU". Moreover, ifbxaux = b2au2 for some 61,62 £ B and u\,u2 £ U", then u\ — u2 and b\a = 62a. Proof. First we show that if a £ R and u £ £/, then au £ [/a if and only if u £ C/^.
(2.40)
Suppose that u £ U'a. Take T = ©'(a) in (2.30) and write u — Tixij(tij)i where the order of the factors is chosen so that the terms with (z, j) £ $oo(°0 U $01 (a) appear on the left. From (2.38) we know that axi3(t) — Xij(t) for (z,j) £ $oo( a ) ^ $01 ( a )- Thus au — a[]a;,-j(t,j), where the product is over (z, j) £ $'(a). If (z, j ) £ $'(a), then z £ J(a)J £ J ( a ) and az < a j . Then a j £ 1(a) and (aj)a = j . It follows from (2.32) that if t £ 7\, then ax{j(t) = xa{j(t) = xai^aj(t)a £ Ua. Therefore au £ U'al which proves the first implication. Conversely, suppose that u £ U and au £ Ua. We can write w = u'u"', where u £ U'a,u" £ £/"• Then aw' £ [/a by the first part of the argument. Thus au" £ (7a. If (z, j ) £ $J 0 (a), t h e n *.-j(*K = a * b y (2.38). Write
CHAPTER 2. FULL LINEAR MONOID
48
u" = yz, where y G C/$//(a) and z is a product of factors X{j(t) with (z,j) G $^ 0 ( a )- Then za* = a* so aya* = aim* G f/aa*. Since aa* is a diagonal idempotent, it follows that ay a* is upper triangular. If Y is a closed subset of A + and v G £/r? then it follows by induction on the number of factors Xij(t) of t; which are different from 1 that we may write
f = l+
£
^^'
(2-41)
W)er for suitable ^- G K. Apply (2.41) with T = $"(a) and v — y. Since ja* = a j for j G
]P
tijEaiyCLJ.
(t,j)€S"(a)
Since aya* — aa* is upper triangular and ai > aj for (z, j ) G $"(«), it follows that Uj = 0 for all (z,j) G $"(a). Thus y = 1 and z = u" G U". Now apply (2.41) with Y = $to(a) and v — z. Write
z = l+
£
t^.
W)e*f0(a) The indices j which occur here are not in J(a). On the other hand, the elements of Ua are K-linear combinations of elements E^ with j G J{a). Thus Uj — 0 for all (z,j) G $i~0(a)> so that z = 1. Therefore u" = y^ = 1 and u = u' €U'a. This completes the proof of (2.40). Since aT — Ta, it follows from (2.35) and (2.40) that BaB = BaTU = BaU'JJ'Z Q BaU'J C B a R Hence £ a £ = BaU'J. Suppose that bxaui = b2au2 for some &i,62 £ B and u x , u 2 G £/"• Then au2u^ G 5 a . From (2.40) it follows that u2u^1 G U'a. Since C/^ ^ U" = 1, this means that wi = u 2 . Therefore b\a — b2a, as desired.
□
Corollary 2.33 / / a G R C M n (F g ) and rank(a) = r, tfien | £ a f i | =
(g-l)V ( r _ 1 ) / y ( a ) .
2.4. COMPLETE 2.4. COMPLETE REDUCIBILITY REDUCIBILITY
OF OF MM n{F n{F q)q)
49 49
Proof. Let a = £*=i £ L i Eikik Ei
Thus Thus
fc=l k=l \
\Ba\ = (g-iygEie/C)**'- 1 ).
From Lemma 2.17 it follows that moi(a) a _— 1i r\r0 ' "r(( rr_- 1l )) //22 m I| £5 aa| I _= (Co 0l(a)
Choose T = 0"(a) = $f 0 (a) U $"(a) in (2.30) From (2.22) we know |$f„(a)| (a). In view of (2.18) and (2.20) we have |$"(a)| = that |$f 1 0(a). o (<0| = m10 = n(a). n{a). Now the uniqueness in (2.30) implies that |*"(a*)| = n(a*) = \JJ"\ — oomimio(ll) °('1)+n( \U"\ +n(<Jo)' From Lemma 2.32 it follows that a ) ) mmi \BaB\ = \Ba\\U':\ \BaB\ \Ba\\U':\ ==( (g g- - l)-" l)-"99-(-))/V -(-))/V, l,la )) ++ i o( o(a a)+ )+n n((a a).).
Since /(a) 1(a) = = m(a) + n(a) by Proposition 2.27, the assertion follows. □
2.4
Complete reducibility of Mn(Fq).
We turn to the complete reducibility problem for the monoid Mn(Fq) over a field K. For K = C, the complex numbers, this problem was approached in a series of papers by Edigarjan and Faddeev, cf.[22], and independently by Munn (unpublished). Munn verified the assertion for n < 4. The solution was claimed by Faddeev in [23], but a complete proof has never been given. In [98], Putcha and the author proved much more generally that the complex semigroup algebra of a finite monoid of Lie type M is semisimple (see Chapter 8). Kovacs then obtained an elementary proof for the special case M = M n (Fg), [66]. To present the proof of Kovacs we need an auxiliary notion. A ma trix c G Mn(Fq) is called a semiidempotent if cr = = c r + 1 for some r > 1. By the rank sequence a(c) of a semiidempotent c we mean the sequence r a n k ( c ) , r a n k ( c 2 ) , . . . ,rank(c n ). Consider the Jordan form of c. It is clear that rank(c') - rank(c rank(c*', ++11 ) is the number of niipotent Jordan blocks of
50
CHAPTER
2. FULL LINEAR
MONOID
degree greater than i and the nonnilpotent Jordan blocks are identity matrices. It follows that the rank sequence determines the Jordan form of a semiidempotent c. Since the Jordan form of a semiidempotent is a matrix with entries in {0,1} C F g , this implies that two semiidempo tent s have the same rank sequence if and only if they are conjugate in Mn(F,). If a,b,c G Mj \ Mj-i are such that ca = 6, then ca = b for the unique isomorphism c : Im(a) —> Im(6). We can treat c as a one-to-one partial map on F™ (that is, a map from a subspace of F™ into F™) and define partial maps c 2 , c 3 , . . . in the usual way. It is clear that c is a semiidempotent partial map (that is c7* = Zf*1 as partial maps for some r > 1) whenever so is c. By the rank sequence of c we then mean the sequence dim(Im(c*)),z = 1 , . . . ,n. We are ready for the main result of this section. Here, Q3 and rij,j = 1 , . . . n, have the same meaning as in Proposition 2.31. T h e o r e m 2.34 Assume that K is a field such that ch(/\) does not divide q. Then every sandwich matrix Q3,j = 1,. .. , n, is invertible in the matrix ring Mn.(K[GLj(Fq)]) and
tfo[Mn(FJ]~eMny(tf[G^(F,)]). t= l
In particular, K[Mn(Fq)] is a semisimple does not divide the order of GLn(Fq).
algebra if and only if ch(K)
Proof. We will show that every contracted semigroup algebra A3 — K0[Mj/Mj-i],j — 1,. . . ,ra — l, has a left identity. Then the assertion will follow via Maschke's theorem. In fact, the existence of a left identity easily implies that x \-+ xoQj yields an isomorphism of Aj, viewed as the corresponding Munn algebra, onto M n .(AT[F q ]), see Proposition 4.13. (In particular, Q3 is also right invertible.) Fix some j < n. Consider an element of Aj which is of the form = / Z)c ka(c)c5 where ka^ G K and the summation runs over the set of semiidempotents c in Mj \Mj-\. We claim that the coefficients ka^ can be chosen so that, whenever a,b £ Mj \ Mj_i, the sum of the ka^ over the c with ca = b is 1 if a = b and it is 0 otherwise. Then fx — x in Aj for every x G Mj/Mj_i, so that / is a left identity of Aj, as desired.
2.4. COMPLETE
REDUCIBILITY
OF
Mn(Fq)
51
To prove the claim, note that the above condition leads to a system of linear equations in unknowns fcT. Formally, this system consists of one equation for each pair a, 6. Assume that the set {c\ ca = b} is nonempty (otherwise all coefficients in the respective equation vanish). We will show that 1. the coefficient of kT in the corresponding equation depends on a, b only via the rank sequence cr(d(a,b)), where d(a,b) : Im(a) —y Im(6) is the unique linear map such that d(a, b)a = 6; so that equa tions corresponding to pairs a, 6 with a common cr(d(a,6)) are all the same. (Note that the coefficient of kT is the number of semiidempotents of rank j with rank sequence r that are extensions of d(a,b) to endomorphisms of F™.) 2. this coefficient is 0 unless r majorizes cr(d(a,b)) and it is a power of q if r = cr(of(a, b)). This will imply that, when the equations and the unknowns are listed according to the lexicographic order on the set of the relevant sequences, the system is triangular with all diagonal coefficients being powers of q. Since the latter are units in K by the hypothesis, the system has a solution in K, which means that Aj has an identity, as desired. To prove 1),2) fix some a, b £ M3 \ Mj-i and assume that ca — b for a semiidempotent c of rank j . Note that ker(a) = ker(fe). Denote by c the one-to-one map Im(a) —> Im(6) that is the restriction of c. Since c is uniquely determined by the pair a, 6, we must have c = d(a,b). It follows that the rank sequence for c majorizes the rank sequence for d(a, b) (because c1 is a restriction of c1 for i = 1 , 2 , . . . ) . Put d = d(a,b). Let d : Im(6) —> Im(a) be the isomorphism such that GW, dd are are identity maps on Im(6), Im(a), respectively. It is clear that, for i > 1, the composition of partial maps ddl+1 is the restriction of dl to the domain of dl+l. Assume that dl(x) = ddl(y) for some x,y that lie in the domains of the respective partial maps. Then dl(y) = ddd\y) = ddl{x) e Im(d z + 1 ), so that dd\y) G I m ( d > + 1 ) . It follows that Im((f) H \m{ddl) = Im(dV + 1 )
(2.42)
for i > 1, because the converse inclusion is clear. Next, we show that Im((f) = lm{ddt+l)
© (ker(e) 0 Im(
(2.43)
CHAPTER
52
2. FULL LINEAR
MONOID
whenever e is an extension of d to a semiidempotent with rank sequence equal to that of d. The two summands on the right-hand side lie in Im(a) and ker(e), respectively, so they form a direct sum (because Im(a) © ker(e) = F™). By (2.42) this sum is contained in lm(dl). Since lm(dl) C IrruV) and a(d) = (r(e), it follows that I m ( ^ ) = lm{el). But dim(ker(e) nlm(e 2 )) = dim(Im(eO)-dim(Im(e z + 1 )) and dim(Im(d>' + 1 )) = dim(Im(rf' +1 )) imply then that the dimension of this sum is equal to that of Im(ef). This proves equality (2.43). Conversely, assume that U is a complement of Im(a) in F™ such that lm(dl) = lm{ddt+1)
© (U H Im(flP)) for z = 1,2,...
(2.44)
Let eu be the unique endomorphism of F™ that annihilates U and acts on Im(a) as d. Then cc/(Im(cf)) - eu{lm{ddl+l))
= Im(euddi+1)
= Im(dddi+1)
= Im(cf + 1 ).
Since Im(eu) = Im(d), this implies by induction that Im(e^) = Im(cf) for i > 1. Therefore, the rank sequences of ej/ and d are equal. Now, d is the restriction of the semiidempotent c, so there exists k > I such that Im(e^) = Im(dk) C Im(a) and e/* = G^"1-1. Since eu,d are equal on Im(a), it follows that e^ +1 = e\j. Therefore eu is a semiidempotent. We have shown that extensions of d to semiidempotents with the same rank sequence as d are in one-to-one correspondence with comple ments U of Im(a) satisfying (2.44) (because we must have e = e ker ( e )). Thus, to prove 2), it is enough to show that Im(a) has a complement U satisfying (2.44) and the number of such complements is a power of q. Let k > 1 be such that d is the identity map on lm(dk). If U is one of the desired complements of Im(a), then it gives rise to a descending chain of subspaces Ui = UCi Im(
(2.45)
that satisfy the following conditions: tf* = E/jb+i = ■ • • = 0, Ul/Ul+1 is a complement to (Ut+i + in lm{dl)/Ul+1
(2.46) lm(ddl+1))/Ul+1
for i = 1,2,... ,
(2.47)
2A. COMPLETE
REDUCIBILITY
OF
Mn(Fq)
U/Ui is a complement to (Ux + lm(a))/Ui
53 in F " / t / i .
(2.48)
Conversely, assume that t/, L^, C/ 2 ,... is a descending chain of subspaces of F™ that satisfies these three conditions. We will show that then U is a complement of Im(a) and (2.44),(2.45) are satisfied. First, by induction we show that U{ C\ lm(ddl) = 0 for i = k, k — 1 , . . . , 1. In fact, for i — k the assertion is clear. If k > i > 0, then Uinlm(d
=
U% 0 lm{ddl) n Im(cf)
(2.49)
because by (2.47) U% C Im(
£/ t -nIm(
=
t/ 1 - + inlm(dcf + 1 ) by (2.47)
=
0 by the inductive hypothesis.
In particular U% H Im(^>' + 1 ) = 0. Since (2.47) yields Im(rf') = Ut + Im(ddl+1), it follows that this is a direct sum. But Im(a) = Im(drf), so in view of (2.48) we also get U H Im(a) - U H (C/j + Im(a)) 0 Im(a) = C/j fl Im(a) = 0. Hence U is a complement of Im(a). Finally, U C\ {U{ + Im(ddl+l)) = Ui + (Un Im(ddl+1)) by the modular law, and the second component on the right-hand side is contained in U D Im(a) = 0, so that (2.44),(2.45) follow. This shows that, if one considers (2.46) as a definition, next chooses t/fc_i,. . . , U\ in that order subject only to (2.47) and chooses U subject to (2.48), then the U so obtained is a complement of Im(a) that satis fies (2.44), and that each such complement is obtained this way. The number of complements of a subspace of an F g -space is a power of q. Hence, the total number of choices in the above construction also is a power of q. This completes the proof of 2). It remains to prove that 1) holds. As above, suppose that ca = b for some a,6 G Mj \ Mj_i and a semiidempotent c £ Mj. Let d — d(a,b). Constructing the chain of subspaces f/^-i, • • • , [ / i , [ / w e see from (2.47) that Uk-X ©Im() = I m ^ ' 1 ) . Since dim(Im(d>')) = dim(Im(cP')) for each z, it follows that the number of possible choices for Uk-\ depends on dim(lm(dk)) and dim(Im(d*- 1 )) only. Similarly, by (2.47) we see that the number of possible choices for [/,- (with £ 4 _ i , . . . , {/,•+1 cho sen already) depends on the dimensions of Im(dl)/Ui+i and of (C/i+i +
CHAPTER
54
2. FULL LINEAR
MONOID
Im(ddt+l))/Ui+i. The latter subspace is isomorphic to Im(ddl+1) be cause, as seen in (2.49), we must have U3 0 Im(dd^) = 0 for j — i + 1 once this has been verified for j = fc, k — 1 , . . . , i + 2. Hence, by down ward induction it follows that this number depends only on the rank sequence of d. The same applies to the number of possible choices for U (with £ 4 - i , . . . jU\ constructed already), which in view of (2.48) de pends only on dim(C/i) and dim(Im(a)) = dim(Im(d)). Therefore, the number of extensions of d to semiidempotents with the same rank se quence as that of d depends only on this rank sequence. It follows that equations corresponding to pairs a, b with the same rank sequence are identical, as desired. This completes the proof of the theorem. □ Once we know that Mn(Fq) is completely reducible over K, we can construct the complete set of nonequivalent irreducible represen tations. From Theorem 2.34 and its proof it follows that they are de termined by the maps x —> (j)(xf3), where f3 <E JKO [Mj] is the iden tity modulo Ko[Mj-i] and <j> is a homomorphism of Ko[Mj/Mj-i] ~ Mnj(K[GLj(Fq)]) onto a simple algebra (hence, (j) is determined by an irreducible representation of GLj(Fq)). However, there is a direct way to construct all irreducible representations, avoiding the computation of fj (which is Qj1 when Mj/Mj-i is identified with the completely 0simple semigroup M(GLj(Fq),Xj,Yj,Qj)). This follows from our last result in this section and from the description of complex irreducible representations of the group G f L n (F g ), cf.[30]. T h e o r e m 2.35 Let : G3 ~ GL3(Fq) —> End(V), j < n, be an ir reducible complex representation. Let W — V ®c X, where X is the C-space with basis X3 and let e = I j . Define $ : Mn{Fq)
J be the idempotent
—+ End(W), for a e M n ( F 9 ) , veV,xGX3
of rank by
4>(a)(v ® x) — (j>(g)(v) ® x if rank(ae) = j where ax = xfg for some x' G Xj,g
G Gj, and
(f)(a) = 0 i / rank(ae) < j . Then is an irreducible representation of Mn(Fq) and it agrees with (j) when restricted to V ® eC ~ V. Moreover, every irreducible complex representation of Mn(Fq) is equivalent to a representation of this type.
2.4. COMPLETE
REDUCIBILITY
OF
Mn(Fq)
55
Proof. We have seen in Theorem 2.3 that every ax of rank j can be uniquely written in the form x'gy' for some x' G X3,y' G Yj,g G Gj. Since axCe, it follows also that y' = e, so indeed ax = x'g. Assume that ax = yg, by = z/i for some a, 6 G M n ( F g ) , x, y, z G X, and g,h E Gj. Then 6ax = 6yg = zhg, and hence ^(&a)(ua:)
=
(f)(hg)(v) ® z = cj)(h)((j)(g)(v)) ® z
=
^{b){^{g)(v)
® y) = 0 ( 6 ) $ ( a ) ( u <8> x))
which implies that is a representation. It is clear that (f)(h)(v ® e) = h(v) ®efov v eV,h £ G3. Note that C[M n (F 9 )] acts on C[G3] X by: ,
.
v
^
y
f a'a (g) x' where ax — x'q' if rank(ax) = j [ 0 if rank(ax) < j
for a, g, x and x', g' as above. Since C0[M3/M3_i] has an identity element / by Theorem 2.34, it is easy to see that f(g ® x) = g ® x. But <> / arises from the composition of this action with the natural homomorphism C[Gj] ® X —> End(V) ® X. Therefore, ^ ( / ) is the identity map. It follows that for any nonzero submodule W of W we have ftM3)W ^ 0. Hence ~cj>(eMj)W ^ 0 because M3eM3 = Af,-. Since ~4>(ea)(v ® x) = ^ ( e a i ) ( u ) ® e or it is 0, we have ^ ( e M ^ W 7 C V ® e. The fact that 1/ is an irreducible Gj-module implies that V ® e C W7. Then W = VK because (x)(V ® e) = V ® x for every x G X3. This means that W is an irreducible C[M n (F g )]-module. According to Theorem 2.34, irreducible representations of Mn(Fq) are in one-to-one correspondence with the irreducible representations of the groups GL3(Fq),j = 1,. .. , n. To complete the proof it is then enough to show that any two of the representations , ft constructed above are not equivalent. This is clear if ,' are representations of GLj(Fq),GLk(Fq), respectively, where j ^ A;, because 0(M 2 ) = 0 for i < j . If j = /c, this is a consequence of the fact that the restriction of (j) to Gj agrees with <j>. □ Let B C GLn{Fq) be the Borel subgroup considered in Section 2.1. Let 6 = eB = IB]'1 J^beB & £ C[GLn(Fq)]. It is known that the algebra eC[GLn(Fq)]e, called the Hecke algebra of B, is isomorphic to the group ring C[W] of the symmetric group W, and it controls the decomposition
56
CHAPTER
2. FULL LINEAR
MONOID
of the permutation representation of G on GjB [18]. In a similar way, using Theorem 2.34 and the material of Section 2.3, it is shown in [129] that we have cC[M n (F,)]e ~ C[R] and the former has a strong flavour of the Hecke algebra, controlling representation theory for GLn(Fq). We note that modular representations of Mn(Fq) have also been studied. In particular, it was shown in [28] that Fp[Mn(Fp)] is not an algebra of finite representation type, that is, it has infinitely many nonequivalent indecomposable representations. Modular representations af forded by the space of homogeneous polynomials were further studied in [68]. Irreducible modular representations of F p [M n (F p )] have been described, and extensively used in a series of papers on algebraic topol ogy, cf.[34]. Irreducible modular representations of finite monoids of Lie type (see Chapter 8) were considered in [111], [114].
Chapter 3 Structure of linear semigroups In this chapter we present the general structure theorem for arbitrary subsemigroups S of Mn(K). The starting idea is to study the ideal chain determined on S by the ideals M0 C M1 C • • • C Mn = Mn(K) and to see how S fills the egg-box pattern at every level Mj \ Mj_i. The theorem associates with S a collection of finitely many (at most 2 n ) linear groups Ga and as many sandwich matrices Pa over these groups. Up to conjugation, Ga are all groups generated by the nonempty inter sections S f! D with maximal subgroups D of Mn(K). This description of S shows a strong analogy with Wedderburn's structure theorem for finite dimensional algebras. The philosophy is then to study S via the 'associated linear groups' Ga and the group action they determine on cl(5).
3.1
Uniform semigroups and associated groups
Our first aim is to introduce a class of subsemigroups of completely 0-simple semigroups that will be crucial for the approach presented in this chapter. P r o p o s i t i o n 3.1 Let U be a subsemigroup of a completely 0-simple semigroup S with zero 0, which intersects nontrivially every nonzero Hclass of S. Fix a maximal subgroup D of S. Let G denote the subgroup 57
58
CHAPTER
3. STRUCTURE
OF LINEAR
SEMIGROUPS
of D generated by U C\ D. Then there exists a subsemigroup I of U and a Rees matrix presentation S = A4(D, X, Y, Q) of S such that I = -M(T, X, Y, Q) is a semigroup of matrix type over the subsemigroup T = U2 C\ D of D and I = A4(G,X,Y,Q) is the smallest completely O-simple subsemigroup of S containing U. Moreover, the construction of I does not depend on the choice of D, in particular for any other maximal subgroup D' of S the subgroup of Df generated by U 0 D' is equal to I fl D', so that it is isomorphic to G. Proof. Let X,Y be the sets of nonzero 7£,£-classes of S respectively, and let x G X,y G Y be such that D = S^j \ { I defined by
= =
<£((%'*'> M ' ) ) = siytuXJSi>ytfuXJ> =
siytqjid,uxjl (t>({t,i,j))(j){{t',i,j'))
Suppose that siytuX3- — skyquxi for some z,A: G X,jJ G Y, and £,q G T°. Since 5^j is a maximal subgroup of S and siy G S ^ , t G 5 ^ | , i ^ G S ^ , it follows that t — 9 whenever SiytuXJ is the zero of S. Similarly, q = 9 in this case. Otherwise, we must have i — k,j = I because siytuxj lies in S$ fl S^y Since S is completely O-simple, there exist a, 6 G S such that asiy,uxjb G T. The fact that (asiy)t(uxjb) — a(siy)q(uxjb) implies now that t — q. Therefore extends to an isomorphism M(D,X, Y, Q) —> S. Put I = Utj 5^Gu^- U {9} C 5. Since siyGuXJskyGuxi
C siyGuxi U {6>},
it follows that 7 is a subsemigroup of 5. But
/ n sgj = s ^ G ^ u {0} - G u {0}
3.1. UNIFORM
SEMIGROUPS
59
because sxy,uxy G T. It follows easily that / is a completely 0-simple semigroup that can be identified with M(G, X, y, Q). Let z G U. Then there exist c G S(x),d G S ^ such that czd ^ 0. These c, e/ can be chosen so that c,d £ I. Therefore czd G T. Now, if c = (g,x,j),d = (/i,fc,y) £ - M ( T , X , y , Q ) and * - (/,z,/), then czd = (gq3ljqlkh,x,y). Since g,h,qji,qik G T, it follows that / G G. Hence z £ I. Therefore U C I. Then T C [/ D D C G, so that G is the group generated by t / f l D , as well. It is clear that / is the smallest completely 0simple subsemigroup of S that contains / , so it is the smallest one that contains U. In particular, / does not depend on the choice of the maximal subgroup D. Now, if D' is another maximal subgroup of S, then by the above, the semigroups / ' , / coincide, so that U fl D' generates a maximal subgroup of / . This completes the proof of the proposition. □ Let S = Ai(D,X,Y,P) be a completely 0-simple semigroup. A subsemigroup U of S will be called a uniform subsemigroup of S if U intersects each nonzero H-class of a completely 0-simple subsemigroup S' = M(D,X',Y',P'), where X' C X,Y' C Y and P' is the corre sponding Yf x X'-submatrix of P. From the above result it follows that 5 has the smallest completely 0-simple subsemigroup U containing U. This U will be called the completely 0-simple closure of U in S. It can be viewed as a 'group approximation' of U. The common (up to iso morphism) group G generated by the intersections U fl D with maximal subgroups D of S' will be called the group associated to the uniform subsemigroup U of S. (Note that this definition does not imply that 6 G U. In fact, this is not so if 9 ^ [/, or equivalent ly if all entries of P' are nonzero. However, including 9 in U is sometimes convenient and produces no ambiguity.) Uniform semigroups were introduced in [88]. A special class of such semigroups already appeared in the study of irreducible linear semi groups, [80], [106], (however, these were only certain subsemigroups of Mn(K) and not of an arbitrary Mj/Mj-\.) R e m a r k We note that a semigroup U that is a uniform subsemigroup of two completely 0-simple semigroups Si, S2 with nonisomorphic com pletely 0-simple closures t/i, C/2 can be constructed. First, every almost poly cyclic group G (a finite extension of a poly cyclic group), which is
60
CHAPTER
3. STRUCTURE
OF LINEAR
SEMIGROUPS
not almost nilpotent, has a free noncommutative subsemigroup X, cf. Chapter 6. Therefore, the subgroup of G generated by X and the free group of rank equal to the rank of X both can be considered as com pletely 0-simple closures of X. Here is another construction, in which the groups generated by the nonempty intersections of U with the maximal subgroups of £/i, U2 are isomorphic. Let Si = A4(G, 1,2, P ) , where G is the free group on generators x,y and the sandwich matrix P is given by pu = l,£>2i = 1LetU = V1UV2withV1 = {{g^^)\ge(x,y)lx}^dV2 = {{g,l,2)\ge (x,y)1y}. It is easy to see that U is a cancellative subsemigroup of Si and Si is its completely 0-simple closure. On the other hand, U is isomorphic to the free semigroup U' = (x,y) C S2 = G via the map (#, l , i ) H^ g. Clearly, S2 is the completely 0-simple closure of U' in S2. However, such a situation cannot occur if every cancellative subsemigroup T of U satisfies the Ore condition: xT H yT ^ 0, Tx D Ty ^ 0 for every x,y G T. Namely, assume that a,6 G U are such that aRh in S\. Since U is a uniform subsemigroup of Si, there exist x,y G U such that ax, by £ D for a maximal subgroup D of Si. But U 0 D is an Ore semigroup, so axs = byt for some s,t G U fl D. It follows that aTZb in Si if and only if au — bw for some u, ID G (7. Since the same applies to S 2 , the 7?.-class partition of U coming from Si is the same as that coming from S2. A symmetric argument works for the £-class partitions. Hence, if Di is a maximal subgroup of Si nontrivially intersected by [/, then U f) Di = U D D2 for a maximal subgroup D2 of S2. The maximal subgroups G i , G 2 of UllU2 are the classical groups of quotients of the respective intersections, because they are generated by these intersections by Proposition 3.1. It follows that Gi ~ G2 for an isomorphism which is the identity on U D G\. We may write Ut = M(GUX,Y, Pi),<7 2 = M{G2,X,Y, P 2 ). Since U is a uniform subsemigroup of Uu and of [/2, the sandwich matrices P i , P 2 can be chosen as in the proof of Proposition 3.1. In particular, they are determined by [/, so we can take P1 = P 2 . It follows that Ux ~ U2. It can be verified that, in the above case, the semigroup U = U\ is the semigroup of quotients of U in the sense studied in [25] and [26]. Namely, every 5 G U can be written as s = ux~l = y~lv for some u,v G U and some x G U fl D,y G U Pi D\ where D, D' are maximal subgroups of
3.1. UNIFORM
SEMIGROUPS
61
U and the inverses are taken in the respective groups. Moreover, semi groups U of this type admit an abstract characterization, not referring to U. We note that, if a semigroup U has no free noncommutative subsemigroups, then every its cancellative subsemigroup satisfies the Ore condition. We continue with some examples of uniform semigroups. E x a m p l e s . 1. A subsemigroup of a group; the multiplicative semi group of a subring of a division algebra. 2. More generally, let S be the multiplicative semigroup of a prime right Goldie ring 72, cf. [36]. Then S has an ideal I that is a uniform subsemi group of some J\A(GLj(F),X, Y, P ) , where F is a division algebra such that the classical quotient ring of R is isomorphic to Mn(F). The same can be said about the image of a semigroup T under a homomorphism of its semigroup algebra K[S] onto a prime right Goldie ring R. This is a consequence of the structure theorem presented in the next section, or more precisely, of its extension discussed in Chapter 9. 3. Let S = M.{G, 2,2,7), where G is the free nonabelian group on x,y and I is the 2 x 2-identity matrix. Then S can be considered as a subsemigroup of M2(K[G]). Let U be the subsemigroup of S of the form U =
(xy y~\xY
(Xyy
\
y-\xYyj
(meaning that the only nonzero entry of every nonzero matrix of U lies in the corresponding indicated subset of G.) It is clear that U is a uniform subsemigroup of S and the associated group is the cyclic infinite group. Moreover, the proof of Proposition 3.1 shows that U ~ A 4 ( ( : E ) , 2 , 2,1). Note that the 'entries' of U do not generate a cyclic infinite group. 4. Let S = M n (Z) C M n ( Q ) . Since for every a G M n ( Q ) there exists z G Z such that za G 5, it follows that S intersects all H-classes of M n ( Q ) . Therefore, for every j , (S H Mj)/(S H Mj-i) is a uniform subsemigroup of Mj/Mj-i. 5. Let X be a finitely generated free semigroup and w G X a prim itive word in X (this means that w is not a power of its proper subword). Let Xw = {a G X\a is not a subword of any wn,n > 1}. The Rees factor X/Xw has an ideal Iw with finite complement and such that Iw is a uniform subsemigroup of an inverse semigroup with fi-
62
CHAPTER
3. STRUCTURE
OF LINEAR
SEMIGROUPS
nitely many 'H-classes. In fact, there is a natural embedding of Iw = {a £ X \ Xw\ a has w as a subword } U {8} into the semigroup S = Ai((x),l(w),l(w),P), where l(w) is the length of w, (x)1 is an infinite cyclic monoid and P is the matrix with diagonal ( 1 , 1 , . . . , l , x ) and zeros off the diagonal, cf. [87], Chapter 24. We shall often use the fact that for certain classes of semigroups 'uniform' implies 'completely 0-simple'. Corollary 3.2 Let U be a uniform subsemigroup of a completely 0simple semigroup S. If a power of every a £ U is a von Neumann regular element of £/, then U is completely 0-simple. In particular, this applies if U is n-regular. Proof. Let U be the completely 0-simple closure of U in S. Choose a maximal subgroup D of U and let e = e2 £ D. If a £ U C\ D, then the hypothesis implies that a3k is regular in U for some k > 1. Hence there exists x £ U such that a3h = a3hxa3k. Let b = axa. Then a3k = a3h~1ba3k~1 and we must have b £ U 0 D. Cancellativity in D implies that ba3h~2 = a3k~2b — e. Therefore, e £ S and a is invertible in S H D. It follows that S H D is a group. Since S Pi D generates D (by Proposition 3.1), we have D C S. It follows easily that U — U. □ As we shall see below, more can be said in the case where U is a uniform subsemigroup of some principal factor MJ/MJ_1 of the monoid Mn(K). L e m m a 3.3 Let S C Mn(K) be a semigroup. Assume that for some maximal subgroups Di,D2 of Mn(K), consisting of matrices of the same rank, there exist a, b £ S such that a(SC[Di)b C SC\D2. Then the groups G\,G2 generated by SC\Di, SC\D2 in D 1 ,Z? 2 , respectively, are conjugate in Mn(K). Proof. Let j be the rank of matrices in Di,D2. Let e, / be the identities of Z)i, D2 respectively. We can replace a by any ax, where x £ S Pi Di, and 6 by any y6, where y £ S n.D\. This allows to assume that a, 6 £ Mj \ Afj_i, and consequently a £ fMn(K)e,b £ eMn(K)f. Let T C Mj/Mj-i be the union of the 7^-classes of Mj \ Mj_i intersected by the set (S H £>i) U (S H D 2 ) U {a, 6}. Then T U {(9} is a completely 0-simple
3.2.
STRUCTURE
THEOREM
63
subsemigroup of M2jM3-\ and SC\T determines modulo Mj-i a uniform subsemigroup U of T U {9}. Moreover, by Proposition 3.1, G i , G 2 are maximal subgroups of the completely 0-simple closure of U in Mj/Mj-i. In particular, bG2a = G±. Since ba G S Pi Gi, replacing a by some a z , z G Gi, we can assume that ba — e. Then ab = aeb — a&a& G G 2 , so that ab = / . Since the idempotents e, / have the same rank, there exists g G GLn(K) such that g e ^ - 1 = / . Let CGLn(K){e) be the centralizer of e in GLn[K). Choose h G sC GLn (*r)(e) 0{xe
GLn(K)\x-la
G GJ.
(It is easy to see that this intersection is not empty. In fact, conjugating in Mn(K)
we can assume that e = I
I . Then g~la G
Mn(K)e.
Since fgTZfTZa by Lemma 1.1 and g~l acts on the set of 7^-classes of Mn(K) by left multiplication, we also have g~1aTZg~1 fgTZe. Hence g~la G eMn(K).
Therefore g~xa G Dx — I
V G GGLn(/i)(e) = I
n
) such that y~1g~1a
. Hence, there exists G G\ and we can put
h — gy.) Now h~la G Gi and h~labh = ^ " V / i = flf"1/? = e G Gi. But bhCfhCe because 6 £ / and h acts on the set of ^-classes of Mn(K) by right multiplication. Since also bh G e M n ( / i ) , it follows that bh G Z?!. The displayed formula implies that bh G G\. Therefore h~lG2h — h~1abG2abh — GibG2aGi = Gi, as desired. □ The following is an immediate consequence of Lemma 3.3 and of the definition of a uniform subsemigroup. Corollary 3.4 Let U be a uniform subsemigroup of some MjjM3-\. Assume that U intersects maximal subgroups D\,D2 of Mj/Mj-i. If G i , G 2 are the groups generated by U 0 D1:U fl D 25 then G i , G 2 are conjugate in Mn(K).
3.2
Structure theorem
Our aim in this section is to prove the main structure theorem for ar bitrary linear semigroups. We also present the resulting philosophy of
64
CHAPTER
3. STRUCTURE
OF LINEAR
SEMIGROUPS
studying S C Mn{K) via associated groups, the corresponding sand wich matrices and group actions. This is a natural extension of the structural approach to finite semigroups - via completely 0-simple prin cipal factors, whose role is now played by the uniform components of S. The following theorem is due to the author, [88], and extends the result known before for 7r-regular linear semigroups, cf. [87], Theorem 3.10. We say that a subset A of a semigroup T is a 0-disjoint union of certain A a , a G Aj if A = [ja Aa and f]a Aa is either empty or it is the zero of T. T h e o r e m 3.5 Let S C Mn(K)
be a semigroup. Define the sets
Sj = {a G 51 rank(a) < j}
Tj
=
{a £ Sj\ S1aS1does
not intersect maximal subgroups of Mn(K)
contained in Mj \
Mj-i}.
Then So C 7\ C S, C T2 C • • • C 5 n _ ! = Tn C Sn = S are ideals of S (if nonempty). 1. for every j we have TjJ nilpotent ideal of S/Sj-i,
Moreover C 5 j _ i , so that Nj = Tj/Sj-i
is a
2. (Sj \ Tj) U {9} C Mj/Mj-i is a 0-disjoint union of uniform subJ semigroups U[ \... ,f/W, n3 < ^ of MJ/MJ_1 that intersect different TZ- and different C-classes of Mj/Mj-i, moreover N3 does not intersect H-classes of Mj/Mj-i intersected by S3 \ Tj, 3.
UWUP C N3 for every
k ^ z; moreover lj\^Nj, NjU^ C Nj and U[j)N3U[3) = 9 in MjjMj_x. In particular, u\3) can be considered as ideals of S/Tj.
Proof. It is clear that every Sj,Tj
is an ideal of 5, if nonempty. Fur-
ther, Tj/Sj-i C M2jM2-X consists of nilpotents. Thus T>3) C S3^ by Corollary 2.15. We then are left with proving assertions 2),3) on the Rees factors Sj/Tj.
3.2. STRUCTURE
THEOREM
65
Let T = 5 i / 5 i _ i and put J = M3IM3.UN = T3/S3^. By U,K,C we denote Green's relations on J and by 9 the zero element of J. Suppose that a eT\N. Then xay G C for some x,y G ( 5 / S j - i ) 1 , where C is the union of all T D D, I) - a nonzero maximal subgroup of J. The elements x,y can be chosen from T (replace x,y by xd, dy for any d G T fl D). Since xay is not nilpotent in T, it follows that ayx G C. Therefore az, za G C for z = yx G T. Define a relation ~ on C by a ~ b if there exist u,v eT
such that
uvHa.vuHb
It is clear that ~ is reflexive and symmetric. Assume that a ~ b, b ~ c for some a,b,c e C. Then we also have u'v'Hb, v'u"Hc for some u', t>' G T. Hence u'TZvlZb and uCv'Cb. Since ui> ^ 0, we have uu' ^ 9 because u"R,v. Similarly, v'v! ^ 9 implies v'v ^ 0. Then also {v'v){uu') ^ 9 and (uu')(v'v) / 9 (note that pqr — 9, p,q,r G J, implies that pq = 9 or gr = 0). Since v'vuu"Hv'u"Hc and uu'v'v'Huv'Ha, this implies that a ~ c. This shows that ~ is an equivalence relation on C. Denote by C^i G / , the equivalence classes of ~ on C. It is clear that different classes do not intersect the same C- or 7^-classes of J. Let D{ — {a G T\ there exists t G T such that a£,£a G C;}. Assume that a G T \ N. The first paragraph of the proof shows that az,za G C for some z £ T. Hence az ~ za, so there exists i G / such that ae D{. Therefore T\N = \JieI D{. It is clear that D% DDk = 9 for i 7^ k because C;, Ck intersect different 71- and ^-classes of J. Let z G Di^u G -DA- Then :cz, za; G Ct-, wy, yu G C^ for some x, y G T. In particular zxz ^ 9 and uyw ^ 9. If xzwy ^ A", then xzuy G ,Dm for some m £ I. Clearly m — i because xzTZxzuy and m — k because uyCxzuy. This contradicts the choice of i ^ k and shows that xzuy G A/". Then (zxz)(uyu) G A", and zxzHz^uyuHu. Since zuHzxzuyu, from the definition of A" it follows that zu G N. This proves that DZDA: ^ A/". Let jBt- = {a G J | aT^x, a £ y for some x,y G D,-}U{0}. First, we claim that D2- intersects all nonzero ^-classes of J contained in E%. Choose x,y G Di. We know that there exist u,z eT such that xu,ux,yz,zy G C2. The fact that ux ~ yz implies that there exist v,v' eT with vv'Hyz and v'vliux. In particular yzKv.uxCv. Then clearly yRv^xCv. Simi larly, since xw ~ zy, there exist w,w' e T with ww'Hxu and w'wWzy. Then xu7Zw,zyCw, so that xTZw^yCw. It is easy to see that v,w e D%. (For example, vv'Hyz, yz G Ct- imply that W G Ct-, so r; ^ N and
66
CHAPTER
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OF LINEAR
SEMIGROUPS
hence v G Dk for some k G I. But vTZy implies then that k = i.) This establishes the claim. Let a G Ei,a ^ 6. Then aTZx for some x G A - Hence there exists t e T such that xt G C2. Then xt G A C £ 2 . Since xtRa, it follows that the 7£-class of a in J contains a maximal subgroup which is also contained in E{. The same applies to the £-class of a. Therefore, E{ is a completely 0-simple subsemigroup of J. We have also shown that JJy' — Dt U {6} is a uniform subsemigroup of E^ and hence of J. Note also that TC\Ei = D{. In fact, we have seen already that every a G TC\Ei cannot lie in Dk for k ^ i. Hence a G D%; U N. Since D{ intersects all Hclasses of E{, there exists b G D{ such that ab lies in a maximal subgroup of E{. Therefore a £ N, and so a G A , as claimed. The inclusions NU$j) C N,U^3)N C TV are clear because TV is an ideal of T. Every nonzero element a&, for a G U\3' ,b G £^ , is in the 7£-class of a and in the £-class of 6. Hence U^U^ C N if i ^ k. If su £ N for some s G S/Sj-i^u G C^ , then the fact that suCu implies that su G U± . This and a symmetric argument show that U\3' is an ideal of S modulo Tj. Suppose that U^NU^ ^ 6 for some i G / . Then there exist a,b e D{ such that axb ^ 0 for some x G TV. Now axb G TV because TV is an ideal in T. But axb1Za,axb£b, so that ax6 G JE; and hence axb G A This contradicts the fact that TV D A — 0, completing the proof of 3). Let x G C,-, y £ Ck for some z 7^ &. If z G T is such that (xzy)2 / 0, then xzy G C. Moreover xzyTZx and xzyCy. Therefore xzy G Ct- fl C^, a contradiction. This means that {xzy)2 — 9. It follows that (egf)2 = 6 for the idempotents e, / G J such that eWx, fHy and every g G J which is 7^-related in J to some element z G T. From Corollary 2.13 we then see that there are at most fnJ ^-classes on C. It follows that | / | < ( n J. This completes the proof of the theorem. □ Using the notation of the proof, define a relation p on C by: apb if there exist a 1 ? . . . , a m G C, m. > 1, such that ax = a , a m = 6, and for every z = 1 , . . . , m — 1, the elements ai^a^i are ^-related or Crelated in J. Clearly, this is an equivalence relation on C. Moreover, if a, 6 G C and aTZb (respectively a£6), then abW)^baHa (respectively aVHa, baHb), so that a ~ fr. This implies that the relation p is contained in ~ . The last paragraph of the proof shows in fact that there are at
3.2.
STRUCTURE
THEOREM
67
most f n j p-classes on C. The structure of S is therefore described in terms of the ideal chain S = Sn 3 Sn-i 5 • • • 5 So, Sj — S H Mj (note that we may have S{ = 0 for some i). The Rees factors Sj/Sj-i C Mj/Mj_i represent the 'layers' of S and are approached via the egg-box pattern on the completely 0-simple semigroup Mj/Mj_\. Recall that we adopt the con vention Sj/Sj-i — Sj if Sj-i — 0. Each nonempty Sj_i, j > 1, is an ideal in Sj and the Rees factor S(j) = Sj/Sj-i is a 0-disjoint union \JQUa\J N of the maximal nil ideal TV of S(j) and of certain semigroups Ua^a G A, intersecting different TZand different £-classes of Mj/Mj_i. In particular, UaUp C Ar for a ^ f3. If the minimal rank of matrices in S is j > 1, so that Sj-\ — 0, then we must have Sj = UaiA — {a} and TV = 0. Every Ua is an 'order' in a unique completely 0-simple subsemigroup Ua of Mj/Mj-\. That is, Ua intersects all nonzero 'H-classes of Ua and every maximal subgroup H of Ua is generated by Ua fl H. These Ua are uniquely determined by the above conditions. All Ua coming from all possible layers Sj/Sj-i are called the uniform components of S. With a slight abuse of lan guage, the inverse image of Ua \ {0} in the corresponding Sj \ Sj-i (0 denoting the zero of Sj/Sj-i here) will sometimes also be called a uni form component of S, depending on whether we view the matrices in Mj \ Mj_i as elements of Mj/Mj_x o r Mn(K). If a uniform component Ua of S, treated as a subset of S, is an ideal of S or Ua U {0} is an ideal of S, then we call Ua an ideal uniform component of S. In this case Ua, originally defined as a subset of Mj/Mj_i, will be identified with (Pa \ {0}) U {0} C Mn(K) (that is, the zero of Ua is replaced by the zero matrix). Also, if 0 G S, this allows to include the zero matrix into f/a, so that the latter is indeed an ideal of S. By the nilpotent components of S we mean the maximal nil ideals Nj of S(j) (or, again, the inverse images in S of the sets of nonzero elements of Aj, if convenient), for all j . The maximal subgroups of Ua are called the groups associated to S. In particular, the groups associated to S that come from the same Ua are conjugate by Corollary 3.4. Hence, Ua can be viewed as a 'group approximation' of Ua and it is called the completely 0-simple closure of
u a. Thus, to every S C Mn(K) there corresponds a collection of at most 2 n associated linear groups (these are linear groups generated by
68
CHAPTER
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OF LINEAR
SEMIGROUPS
the nonempty intersections S C\ D with the maximal subgroups D of Mn(K)) and sandwich matrices over these groups (in fact, over S f) D) arising from Proposition 3.1. R e m a r k i) If S C Mn(K) has a zero element 0, then Proposition 2.14 allows us to identify 9 with the zero matrix by passing to S ~ eSe C eMn(K)e ~ Mt(K) for e = 1 - 9 and t = n - rank(0). If we first conjugate 0 to a diagonal idempotent, then from the description of the egg-box patterns on Mn(K)1 eMn(K)e it is clear that this identification does not affect the structure of S coming from Theorem 3.5. ii) We note that uniform components can be described in terms of the rank of certain matrices. Namely, a £ S is in a uniform component of 5, if rank(a) = rank(xay) = rank((zay) 2 ) for some x,y £ S. From the above proof it follows easily that the uniform components of S con sisting of matrices of rank j are the equivalence classes of the relation: a <-> b if aHxby for some x,y £ Sj\ £ j - i , defined on every nonempty T = Sj\ Tj. As a consequence of Proposition 3.6 we shall see that this relation coincides with the one given by the following condition: there exist x,y £ Sj \ Sj_i such that x1Za,xCb,y7Zb,yCa. iii) The proof also shows that any subsemigroup T of a completely 0-simple semigroup J is a 0-disjoint union \JaeA Ua U N for a nil of index two (but not necessarily nilpotent) ideal TV of T and uniform subsemigroups Ua of T (A can be an infinite set, though) that satisfy the remaining assertions of the theorem. These Ua are called uniform com ponents of T. iv) There is a striking analogy of the assertion of Theorem 3.5 with that of Wedderburn's theorem on the structure of finite dimensional algebras. Namely, Nj plays the role of the nilpotent radical, while the uniform components U\3' behave like the simple blocks of the algebra modulo the radical. (More exactly, they correspond to 'orders' in sim ple algebras.) The connection with irreducible representations will be discussed in Chapter 4. v) It is easy to see that any homomorphic image and any ideal of a 7r-regular semigroup is also 7r-regular. Therefore, if S C Mn(K) is 7r-regular (for example periodic, Zariski closed or regular), then from Corollary 3.2 it follows that every uniform component of S is completely 0-simple. vi) If S is finitely generated, then S C Mn(R) for a finitely generated
3.2.
STRUCTURE
THEOREM
69
domain R C K. Therefore, S is a subdirect product of finite subsemigroups S(3,f3 G B, which are the images of S in the algebras Mn(Kp) for certain finite fields Kp such that R is a subdirect product of Kp,(5 G B. The important point here is that the numerical invariants coming from Theorem 3.5 are for all Sp bounded by a function of n. More can be said on the egg-box pattern of the nilpotent radicals Nj of the factors Sj/Tj C M2jM2_Y arising from Theorem 3.5. A part of the available information is given below. P r o p o s i t i o n 3.6 Let S be a subsemigroup of a completely 0-simple semigroup M. Assume that S = Ux U • • • U Ur U N where U% are the uniform components of S and N is the maximal nil ideal of S. Then there exist partitions X = Xx U • • • U Xr+UY = Y0 U • • • U Yr of the sets X,Y of 7Z- and C-classes of M respectively, such that for Mtj = {a G M\ allx, aCy for some x G A,-, y G Y3} and NtJ =
NnM,3,
we have 1. N = Nx U N2 U N3 U N4 U {#}, where Nx = Ui<j 1, then Nl3Nkl = e ifi
^1 N2
Ny
N4
^3
70
CHAPTER
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OF LINEAR
SEMIGROUPS
Proof. Let X{, Y{ be the sets of nonzero 1Z, ^-classes respectively, of M that intersect U{. Put Xr+1 =X\ [fi=i xu Yo = Y \ \Ji=1 Yx. Let z,j G { 1 , . . . ,7*}. Assume that a G U{,b G f/j,z ^ j , are such that there exists z G S with a7£z and bCz. Since Ui,U3 are uniform subsemigroups of M, there exist c G f/t-,d G C/j such that c1Za,dCb and c2 ^ 6 ^ d2. Then cz ^ 0 and zcl ^ 0, so that czd ^ <9. If z G C/tis such that x £ c and y G t/j satisfies ylld, then xzy G UizUj satisfies xzylZx, xzyCy. Since from Theorem 3.5 we know that {/t-, C/j intersect all W-classes of the completely 0-simple semigroups Mzz-, Mjj respectively, it follows that UizUj C AT;J intersects all ^-classes of M contained in M{3. Suppose additionally that aCw, blZw for some w G S. Then a symmetric argument shows that N intersects all ^-classes of M contained in M3{. Hence there exists v G S such that vTZd and vCc. Then vzl~Ld: zvT~Lc^ which implies that c ~ of, where ~ is the relation used in the proof of Theorem 3.5. This contradicts the fact that i 7^ j . We have shown that Nji = 0 whenever NXJ ^ 0. Assume that Ni3 7^ 0 and Njk 7^ 0. We have seen that xzy ^ 6 for some z G Ni3,x G c^-,y G Uj. Also, we can choose w G A^-^ with wlZy. Then xzw ^ 0. Hence xzw G A/,-*., so Aft-fc 7^ 0. This completes the proof of 2). Suppose that for some z l 5 . . . ,z m G { 1 , . . . , r } , ra > 2, we have Nt'mt'i 7^ 0 and Nijij+1 7^ 0 for j = 1 , . . . , m — 1. The above shows that ^imim-i 7^ 0- Since we also have ATt-m_lim / 0, this is a contradiction. Consider the graph V = (V, £ ) , where V = { 1 , . . . , r } and (z, j ) G £ if A^-j = 0. From the above it follows that there exists i with (z,j) G £ for every j . Therefore, an induction applied to the graph F' = (V\Ef), where V = V \ {1} and E' = E n (V x V ) , allows to show that there exists a reordering of the set { 1 , . . . , r } such that (z, j ) G £ for z < j . This means that A^- = 0 whenever i < j . If NijNki 7^ 0 for some j > 1,/c < r, then C/j[4 7^ 0, so that j > k. But j < ij < k imply then that k < i and / < j>. This completes the proof. □ The structural approach resulting from Theorem 3.5 raises the ques tion of the status of arbitrary cancellative subsemigroups of the given S C Mn(K) as well as of the independence of the structural description of S of the embedding of S into a full matrix monoid Mm(F) for a field F and m > 1.
3.2. STRUCTURE
THEOREM
71
First, as seen in Remark after Proposition 3.1, two isomorphic free noncommutative semigroups X,X' C GLn(K) can have nonisomorphic completely 0-simple closures. If G is a free nonabelian subgroup of GLj(K),j > 1, (see Chapter 6) then the semigroup M3/Mj_i^ de fined for M n ( / \ ) , contains an isomorphic copy of the semigroup S\ = wA/f(G, 1,2, P) described in this remark. Therefore, the free semigroup (x,y) has two faithful representations into Mj/Mj-i with nonisomorphic completely 0-simple closures. There is also a further reason for the dependence of the decomposition of S on the matrix representation. For example, if J is a completely prime ideal of S C GLn(K) (that is, S\J is a subsemigroup of 5 ) , consider the embedding cj) : S —> M 2 n ( / i ) given by 5 4
T
j for s G S \ J and s ^
n
for s £ J.
Clearly, S and (f>{S) intersect different number of layers Mj \ M3_\ of Mn(K), M2n(K) respectively, so their decompositions coming from The orem 3.5 are different. So, speaking about uniform components of S we shall always have in mind the given embedding S C Mn(K). However, the set of the groups associated to a linear semigroup S has in many cases 'maximal members' which are determined uniquely up to isomorphism - they do not depend on the matrix presentation of S. P r o p o s i t i o n 3.7 Let S C Mn(K) be a semigroup with no free noncommutative subsemigroups contained in maximal subgroups of Mn(K). Assume that T is a cancellative subsemigroup of S and j the minimal rank of matrices in T. Then T has no noncommutative free subsemi groups and T has a group of quotients H which is isomorphic to the group gp(T D D) for any maximal subgroup D of Mn(K) consisting of matrices of rank j and intersecting T. In particular, if cj) : S —> Mrn(K') is an embedding, for a field K' and some m > 1, and G is a group asso ciated to (f)(S), then G is isomorphic to a subgroup of a group associated to S. Proof. For e = e2 G D we put / = T 0 eMn(K). Then I is a right ideal of T and T fl D is a left ideal of / . Since a one-sided ideal of a free noncommutative semigroup contains a free noncommutative semigroup, it follows that T has no such subsemigroups. Therefore T satisfies the Ore condition. Therefore, within the group H of right quotients of T
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CHAPTER 3. STRUCTURE OF LINEAR SEMIGROUPS
(which is also its group of left quotients) we have H = TT-1
D IT1
D IT(IT)-1
= ITT~lrl
= IHT1
= H.
Further
g p ( r n D ) - (T n D){T n D)-1 D i(T n D)(i(T n D))-1 = /gp(rnD)/- 1 = ir1 = H. Finally, assume that E is a maximal subgroup of Mm(K') such that U = (S) n E ^ 0. Let G = gp{U) C M m ( / i / ) . Then T = (j>~\U) is a cancellative subsemigroup of S and therefore TT'1 ~ gp(T fl D) for a maximal subgroup D of Mn(K). Since U ~ T, £? embeds into gp(TflD). Now gp(T f l J ) ) C gp(S 0 D) and the result follows. D The conditions of Theorem 3.5 imply that the ideal chain ^ C Ti C Si C • • • C Sn = S can be refined in the following way. Corollary 3.8 Let S C Mn(K). / l C /
Then there exists an ideal chain 2
C - C /
r
= 5
such that r < 2 n + n — 1 and /j and each Rees factor Ik/Ik-i Z5 ci ther nilpotent of index at most fnJ for some j < n or it is a uniform subsemigroup of some Mj/Mj-i. A chain of this type will be called a structural chain of S. (Note that it may be not unique because of the possible permutation of the order of uniform components coming from the same rank in Mn(K).) The following corollary generalizes the result on chains of principal ideals of 7r-regular semigroups [87], Theorem 3.13, improving also the bounds given there. Corollary 3.9 Let S C Mn(K)
be a semigroup.
1. Assume that v\ • • • vq £ Sj\Sj-i for some j £ { 1 , . .. , n } and some V{ £ S j , where q = 2f n J. Then there exists a uniform component U of S contained in Sj \ Sj-i such that vi^Vk+i £ U for some k. In particular v^Vk+i € U.
3.2. STRUCTURE
THEOREM
73
2. Let t = 2 n n r = i (?)- Assume that axSl D a2Sl D • • • D atSl ^ 0 for some az- G S. Put a$ — 1. TTien £/iere exists a uniform compo nent U of S such that a^ = a z u, am — a{Uw for some i < k < m and u,w G U such that uTZuw in Mn(K). 3. If S is n-regular, then the length of every chain of nonzero prin cipal right (two-sided) ideals of S is less than t (t2 respectively). Proof. Let S^) = S3j S3-\. Let W be a uniform component of S^y To prove 1) we can assume that W ^ Syj. Then from Proposition 3.6 it follows that W(S{J)Y(S{J) \ W){SU)yW = 9 in Su). Suppose 1) does not hold. Then each W contains at most one element of the set V = {i>i,.. . ,vq}. But Nj is nilpotent of index < q/2 and WW C N3 for every uniform component W ^ W of S3. Therefore v\ ■ • • vq must be zero in 5^-), a contradiction. Thus, 1) follows. 2) By induction on n — j we show that: if a^S*1 D • • • D avSl for some a2- G S \ S^-i, and p > rJ_1 = 2n'J+l U?=j (?), then the assertion of 2) is valid for this chain. For j = 1 this will prove 2). If j — n, then 5 \ 5j_i C GLn(K) is a uniform component of 5. Thus, the assertion is trivial in this case. Assume that j < n. As before, let q — 2f n J. Then r2_x = qr3. Sup pose that p > rj_i. Let vr G S be such that a r + 1 = a r v r + 1 , r = 1,. .. ,j9 — 1. Put t^ = a^ If iur = vr • • • v r + r —i ^ 5^, for some r G { 1 , . . . , (g — l ) r j + 1}, then the chain vrSl D • • • D vr • • • r; r + r i _iS' 1 satisfies the induction hypothesis. Hence, we can find elements u, w in a uniform component U of S such that uTZuw in Mn(K) and either vr ■ • • vk = vr ■ • • V{U, vr - • • vm — vr • • • i;z-?iitf for some i < k < m with z > r, or v r • • • Vk = u and vr • • • vm = uw for some k < m,k > r. Then dk = aiU,am — aiuw, or ak = a r _ i u , a m = ar-iuw, and we are done. Thus, we can assume that every iu r , r = 1 , . . . , (q — \)r3 + 1, lies in S3. But and aqrj = Wiivrj+i • • • W(g_i)r>+i G 5j \ Sj-i- Hence, 1) applied to this product shows that u — Wirj+1,w — W(i+iyj+1,uw all belong to a uniform component U of 5 for some / G { 0 , . . . , g — 2}. Then ulZuw in Mn(K). Since a(j+i) ri = CLirju^{i+2)rj = aWjuw, this completes the inductive proof of 2). 3) If S is 7r-regular, then every uniform component U of S is com pletely 0-simple by Corollary 3.2. Suppose that axSl D - - D atS1 ^ 0
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for some az- G S. The elements u, w found in 2) must satisfy uUl = uwU1. Hence uSl = uwS1 and so a^S1 = amSl, a contradiction. There fore, the assertion on chains of principal right ideals follows. Let r = t2. Suppose that there exist 6 1 ? . . . ,6 r G 5 such that S^fr^ 1 D ••• D S1brS1 7^ 0. Then x^y,- = &,-+i for some xt-,y2- G S'1. Put x 0 ,yo — 1 and define U+i = &i2/o • • • 2/i, Zi+i = Xi- ■ • x0bi
forz = 0 , . . . , r - l . T h e n t i S 1 D •■• D t r 5 ! and S 1 ^ 2 . . . D S 1 ^ . T = { t 1 5 1 , . . . , t r 5 1 } , Z = { 5 1 z i , . . . ,Slzr}. We know that |T| < * |Z| < t. Let VK = { 1 , . . . , r } . Define a function : W ■—> T 0 ( 0 = (tiS1,S1Z{). Since r = £2, there exist a,/? G VK, a / /?, such 0 ( a ) = 0(/3). Hence t^S*1 = tpS1, S1za = S1zp, and so S1baSl
= =
5 x x a _ i •••z 0 6 1 yo--'2/a-i£' 1 = ^^yo'-'Va-iS1 1 l 1 S zpy0--ya_1S = 5 x /? _ 1 •••xo&iyo---2/a-i5 rl
=
S ^ a ^ - i ' • ' XotaS1
=
Slx(3_l--x0bly()--yp_lSl
— S1Xp_i
Let and x Z by that
• • • XotpS1
=
Slb(3Sl.
This contradicts the supposition and proves 3). □ Our basic approach, resulting from the structure theorem, is to study S C Mn(K) via its uniform components. The aim here is to reduce problems on S to questions on the associated linear groups and the properties of the corresponding sandwich matrices. This approoach can be successful once we are able to transfer the properties from the cancellative subsemigroups of type SC\D, D a maximal subgroup of Mn(K), to those of the associated groups gp(5 fl D). However, to understand the global structure of S one often needs also to handle the relation ship between different uniform components. It seems that the action of the associated groups on the uniform components of S that lie below is crucial here. This approach may be described as follows. L e m m a 3.10 Let D be a maximal subgroup of Mn(K) and e — e2 G D. Consider the monoid M(e) = eMn(K)e ~ M ran k( e )(^0- If U is a uniform component of S C Mn(K) intersecting M(e), then UC\M(e) is a uniform component of S D M(e). If additionally S is n-regular, then S fl D is a group and it acts by conjugation on U fl M(e).
3.2. STRUCTURE
THEOREM
75
Proof. Let j be the rank of matrices in U. Since M(e) = eMn(K) fl M n ( / i ) e , from Lemma 1.1 it follows that M(e) cuts a 'rectangle' from Mj\Mj-i (viewed via its egg-box pattern). Fix s o m e / = f2 G Mj\Mj_i such that TV = / M , flC/fl M(e) 7^ 0. We view V = U U {9} as a subsemigroup of M2jM3_x. Then TV U {0} can be considered as a right ideal in (17 n M(e)) U {0} C M3jM3_x. Suppose that N2 = 9 in V. Let g £ D. Then Af[/# \ {0} is contained in a single 7^-class of Mj/Mj-i and so NVg C TV U Nj, where Nj is the nilradical of Sj/Sj-\. Hence 2 (NVg)N = 9 and consequently (gNV) = 0. Since NV is a right ideal of V, it intersects all ^-classes of Mj/Mj-i that come from the elements of UnfMj. Hence gNV and gV intersect the same ^-classes. Therefore
(gvy = e. If VgV =fi 6, then it is a nonzero ideal of V, so it intersects all %classes intersected by V. Therefore g(VgV) ^ 9 (because g does not annihilate TV), which is impossible. Hence VgV = 9. Since g G S fl D is arbitrary, Lemma 1.17 implies that VeV = 9. But then VN — VeN C VeV — 9. Since V is a uniform subsemigroup of Mj/Mj-i and N C V, it follows that A^ = 0, a contradiction. This means that we cannot have TV2 = 0. Therefore, (U H M(e)) U {9} has no 'rows' with zero multiplication. A similar argument works for the 'columns', hence {U Pi M(e)) U {0} is a uniform subsemigroup of MjjMj-\. Every element a G (S f) M(e)) \U which is 71- or £-related to an element in U fl M(e) must be in the nilpotent radical Nj of Sj/Sj-\. Hence, a lies in the nilpotent radical of (SC\ M(e))j/(S Pi M(e))j_i. This implies that U C\ M(e) is a uniform component of STl M(e). The remaining assertion is clear. □ We conclude with some simple examples. In particular they show that, in general, there is no obvious connection between the sandwich matrices and the associated groups of S and those of its closure cl(5). E x a m p l e s . 1. Let S C Mn(K) be the semigroup of all diagonal ma trices. Then 5 is a semilattice of groups (each of type (K*)j,j < n). S is regular, and at every rank j it has exactly \jj groups (^uniform components of S.) This shows that the bound on the number of uniform components in Theorem 3.5 is sharp. 2. Let S be the semigroup of all upper triangular matrices in Mn(K). S is 7r-regular by Corollary 1.5. We use the notation of Theorem 3.5.
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CHAPTER 3. STRUCTURE OF LINEAR SEMIGROUPS
It is easy to check that N3 \ {9} consists of rank j matrices that have < j nonzero diagonal entries (applying the Jordan form of any matrix a of this type we see that a power of a has rank < j , so N3 indeed is a nil ideal of S3^-3^.) Also, S3/T3 = u[j) U • • • U U& for ccmplltely 00 simple semigroups Ut with maximal subgroups isomorphic to the group of upper triangular matrices in GL3{K). Each u\j) \{0} consists of matrices with a given pattern of zero entries on the diagonal {u\j) \ {6} is a subsemigroup and U^U^ = 9 for k / z, so these indeed are the uniform components of S.) It is easily that this also leads to a decomposition of S as a semilattice of nilpotent extensions of completely 0-simple semigroups U^. 3. Let S C Mn(K) be'the set of all monomial matrices. That is, a G 5 if each row (and each column) of a has at most one nonzero entry. It is easy to check that S is a regular semigroup whose only ideals are S3 = Sf) Mj, j = 0 , 1 , . . . , n. Moreover, S./S^ is isomorphic to a completely 0-simple semigroup M(G3, ( j j , ( " ) , / j ) , where G3 is the group of all monomial matrices in GL3(K) and I3 is the identity matrix. This can be compared to the monoid R used in Section 2.1. 4. Let S C Mn{K) be a Zariski closed connected monoid with group of units G. Then S is ^-regular and the uniform components of S are of the form GaG, where rank(a) = rank(a 2 ), cf.[108]. 5. Let S- = ( s , e ) C M 2 ( Q ) , where
5=
( o « ) ' e = 0 o)
for some a > 0. Let also S' = (s"\e). First, we claim that the set U of rank one matrices of S is a uniform component of S and S = UU{s) (so it is a union of its two uniform components.) Clearly, U = (e, esn, sne; n = 1 , 2 , . . . ) and U has no nilpotents. Since snesk £ U for n, k > 0, U intersects each W-class of M 2 (Q) which is ^-related to some sne and ^-related to some es^. Therefore U is a uniform component of S and the associated group is isomorphic to the subgroup of Q* generated by the nonzero entries of the matrices in (es n e; n — 0 , 1 , . . . ) . It is clear that cl(5) = cl(S"). However, the entries of the first row of each matrix es~n have opposite signs, while the entries of every t G S are nonnegative. It follows that S" intersects some £-classes of M 2 (Q) that do not intersect S. Consequently, cl(5) intersects more ^-classes of M 2 (Q) than S.
3.3.
CLOSURE
77
Let D be the maximal subgroup of M 2 (Q) containing e. The (1,1)entry of the matrix es~ne is equal to (a - n)a~n~l, so that it is negative for some n > 1. Therefore the groups G, G' generated by S 0 £>, S" fl D, respectively, do not satisfy G' C G. It follows that the maximal subgroup of cl(S) containing e is bigger than G.
3.3
Closure
Our mam objective in this section is to study the closure cl(S') of a semigroup S C Mn(K) and to discuss connections between the structure of S and of cl(5). When studying the structure of S it is often convenient to replace S by its linear homomorphic image S' in such a way that the given uniform component U of S becomes an ideal of Sf. The following lemma, though rather technical, is essential for our approach. It allows us, in particular, to consider a linear semigroup with an ideal uniform component / as a subsemigroup of a linear semigroup in which / is an ideal. L e m m a 3.11 Assume that U is a uniform component of a semigroup S C Mn(K). Let U be the completely 0-simple closure of U in the cor responding Mj/Mj-i and G a maximal subgroup of U. Then 1. S and the semigroup So generated by S U G intersect the same 7i-classes of Mn(K) contained in Mn(K) \ Mj_i, 2. U is a uniform component of SG. Therefore U is an ideal of the Rees factor S = SQ/J, where J
— {a
rank(a) < j or rank((6ac) 2 ) < j
for all 6,c G SG}In particular, if U is an ideal uniform component of 5, then S = SUU = SG. Proof. Since GUI generates £/, it is clear that U is contained in a uniform component V of SG. Let a E U be nonzero (the set of nonzero elements of U is identified with a subset of SG) and let s e S. Then, since U is a uniform subsemigroup of (7, there exists a' G U such that
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SEMIGROUPS
a"Ua in Mn(K). If rank(as) = j , then we have aMn(K) = asMn(K) and Mn(K)af = Mn{K)a. Hence rank(a's) = j and asT-ia's. This means that the %-class of as in Mn(K) contains an element of S. The same holds for sa. A similar argument shows that, if z G SQ is of rank j and is in the 'H-class of an element of 5*, then, for any t G S U G, each of the elements £z, zt also is in the H-class of Mn(K) intersected by 5, whenever it is of rank j . Therefore, an induction on the length of the elements of SQ as words in SUG allows to show that S and So intersect the same ^-classes of Mn(K) consisting of matrices of rank j . In view of the structure theorem (see also Proposition 3.6) this implies that U intersects the same ^-classes of Mn(K) as V. Assume that as is in the H-class of an element of U for some a G U and 5 G 5. Choose e G U such that as — ase. Since UU = {/, we can write e = xy for some x G £/, y G U. Then asxKa! sx for an element a1 £ U such that a'l-La. The structure theorem applied to 5 implies that a'sz G 6^ and, since sa: and a'sx are in the same £-class of Mn(K), we must also have sx G U. Hence as — asxy G UUU C [/. This shows that {75 f! V C U because V, 17 intersect the same H-classes of Mn(K). Similarly one gets SU D V C (7. Again by an induction argument, this implies that V = f/ is a uniform component of So- Since the matrices of rank j in SGGSQ are contained in U U A/", where A is the nilpotent component of SG consisting of matrices of rank j (see Proposition 3.6), U is an ideal of S. Finally, assume that U is an ideal uniform component of S. Suppose that a G U and s G S are such that as G J. As above, there exist v G U,u G £/ such that uv £ U satisfies ui>a = a. Choose If u' G t/ with u'l-iu. Then w a s G J implies that u'vas G J C\ U. Therefore u'vas — 0. It follows that as — 0 because u'vaCa. Similarly, sa G J implies as — 0. This proves that S U U is a semigroup and [/ is its ideal. Consequently SG = £ W . Moreover J = {0} or J = 0, so that SG = 5. This completes the proof. □ If SG ^ Mn(K) is the semigroup constructed above, then 7 can be considered as a subsemigroup of SG/{SG H M ^ ) C Mn(K)/MJ_1. In particular, Lemma 1.8 often allows to study I C »?G/(S'G H M J _ I ) as linear semigroups. A further step in this direction will be made in Lemma 3.26. We are now ready to get some insight into the construction of c\(S).
3.3.
CLOSURE
79
R e m a r k Let M( 0 ), • •. , M( n j be the sets of matrices in Mn(K) of ranks 0 , . . . , n respectively. Put S^) — S Pi M(z). Let j be maximal with 5(j) 7^ 0. Write 5^°^ = S. Consider the nonempty intersections S Pi D with maximal subgroups D of Mn(K) whose elements are of rank j . We define S^ = (5,11(5 fl D ) " 1 ) , where ( 5 n D)~l denotes the set of inverses in D of elements of Sn D and the summation runs over all such D. Next, construct S^2' D S^1' proceeding in the same way with respect to S^ and the set 5/ _^ of matrices of rank j — 1 in 5 ^ . After j steps we reach a subsemigroup S^ of Mn(K). It is clear that, in each step r > 1, S g - S^1] for fc > j - r + 2. M oreover, from Lemma 3.11 it follows that S
(j-r + l) = (S(j-r + l)i (A(j-r + l))~ ) ° ^ ( j - r + l ) ,
where A^V denotes the set of 'group elements' in S^ Pi M^y (Note also that, in each step r, it is enough to invert the elements of S C\ D for only finitely many maximal subgroups D in M^_ r + 1 ) - one for each uniform component of S^ coming from this level.) This easily implies that S^ = cl(5). Moreover, each element of S^ is a word w(s\,... ,5 g ) in some S i , . . . , sq G S^-1' that allows local inverses of those S{ that lie in A(Zr+l\- Therefore, each z G cl(5) is an iterated word z — Wj(wj]1(. ..)>••• iwj-\
(• • •)) °f this type.
Next we show that cl(5) is determined by the closures of finitely generated subsemigroups of S. L e m m a 3.12 Let S C Mn(K)
be a semigroup.
Then
1. cl(5) = {jTc\(T), where the union runs over all finitely generated subsemigroups T of 5, 2. for every finitely generated subsemigroup R of cl(S) there exists a finitely generated subsemigroup T of S such that R C cl(T). Proof. Let A = \JT cl(T). If Si, s2 G A, then the definition of A implies that there exist finitely generated subsemigroups T\,T2 of S such that st G cl(Tt-) for i = 1,2. Clearly, Sls2 G cl((T 1 ,T 2 )). Since (TUT2) C S is finitely generated, it follows that s1s2 G A. Hence A is a semigroup.
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Moreover, cl(Ti) is 7r-regular, so s™ = s™ts™ for some t G cl(T) and m > 1. Therefore, A is 7r-regular. Since 5 C A C 01(5), it follows that A = cl(5). Assume that R — (i>i,... , Ufc) for some u i , . . . , ^ £ cl(5). From 1) it follows that ^- G cl(K) for a finitely generated subsemigroup V{ of 5 , z = 1 , . . . , k. Then 7? C cl((V 1 ,... , 14)) and ( V I , . . . , Vk) C S is finitely generated. The assertion follows. □ The following result explains the basic relation between the compo nents of S and the components of its closure. T h e o r e m 3.13 Let U be a uniform component of S C Mn(K) a nilpotent component of S. Then
and N
1. N = N' C) S = N" n S for some nilpotent components N', N" of the Zariski closure 5, and of cl(S) respectively, 2. U = U' fl S — U" Pi S for some uniform components 5 , cl(S') respectively.
£/', U" of
Proof. We view A'' as a subset of the appropriate Mj \ Mj-\. If z G A/", then (zz) 2 G Mj_ x for every x £ S. Since Mj_x is a closed subset of Mn(K), it follows that (zx) 2 G Mj_! for every x G 5. Thus, z 5 is a right ideal of S which is nil modulo Mj_x. Therefore z G N', where Nf is the nilpotent component of S consisting of matrices of rank j . Hence N C N'. Clearly, N' H 5 is a nil ideal of S modulo M3_x. Hence N'ClS CN, and so N = N'H S. Since S C cl(5), we have d ( S ) = 5. The above applied to cl(5) yields AT' Pi cl(S') = AT", where N" is the nilpotent component of cl(5) consisting of matrices of rank j . Hence AT n s = TV' n s = N. Assume that U\, U2 are uniform components of S consisting of ma trices of rank j . Let AT C Mj \ Mj_i be the nilpotent component of S. From Theorem 3.5 it follows that U\ C U[,U2 Q U'2 for some uniform components U[ of S. Also, if U\ ^ U2 and a G f/i, 6 G £^2, then ax6 G AT, so that (axby)2 G Mj_i, for all x,y £ S. As above, this implies that (axby)2 G Mj_i for all z,y G S. Hence a, 6 are not in the same uniform component of S (and hence of cl(S')) so that U[ 7^ U2. Now U\ C. U[C\S and it follows that the latter does not intersect uniform components of S other than U\. Moreover U[C\S does not intersect N = N'DS because
3.3.
CLOSURE
81
U[nN' = 0. Hence Ui = U[ D S. Similarly, since Ul C U" for a uniform component U[' of cl(5) and cl(5) = 5, we get Ux C U[' = cl(5) n V for a uniform component V of S. Hence U\ QV,Ui C U[ imply that V — U[. This completes the proof. □ The following example shows that cl(5) may have more uniform components than S. E x a m p l e Let S = (x, e) C M 3 (Q), where / 1 1 0 \ / 1 0 0 \ a : = 1 0 0 , e = 0 0 0 . \ 0 0 1/ \ 0 0 1 / Then ex~1e =
/ 0 0 0 \ 0 0 0 6 c l ( 5 ) . The (1,1) entry of xn is positive, so
\ 0 0 1I n
that ex e is a matrix of rank 2 for every n > 1. The same is true of every a G ( e x n , x n e , exne\ n > 1). This easily implies that the set of elements of rank 2 in S forms a uniform component U of S and S — U U (x). Therefore cl(S) has more components than S because ex~1e lies in a uniform component of 01(5). As noted before, the uniform components of cl(5) can intersect more 1Z- (and also more C-) classes of Mn(K) than those of S. However, this does not happen in the following special case. P r o p o s i t i o n 3.14 Let U be a uniform component of a semigroup Mn(K) such that U intersects finitely many IZ-classes of Mn(K). the uniform components U',U" of the Zariski closure 5 , and of respectively, containing U intersect the same IZ-classes of Mn(K)
S C Then c\(S) as U.
Proof. Let R be the union of the 7^-classes of Mn(K) intersecting U. Then R = \Ji=1e{Mn(K) \ M,_i for some e% = e2t e Mn(K),r > 1, where j is the rank of matrices in U. Each elMn(K) is a closed subset of Mn(K). Hence Z = R U Mj-i is closed. Choose u G U. If s G S, then either su G Mj-i or suCu. In the latter case su G U U AT, where A^ is the nilpotent component of S consisting of matrices of rank j . Hence (sut)2 G Z for every s,t G S, so it follows that we also have (sut)2 G Z for every s,t G S.
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Let W = {w G U'\ wCu for some u G U} \ R. Suppose that W ^ 0. Choose w G W. Since [/ is a uniform component of 5, we can find u £ U such that w £ u and u 2 G (/. Hence wuKw in Mn(K). The first paragraph of the proof shows that (wut)2 G Z for every t G 5. But IOU G £/' and £/' is a uniform component of S. Hence t can be chosen so that wut G D for a maximal subgroup D of Mn(K) \ Mj-i. So (wut)2 G D. Now (wut)2 £ Z = RU Mj_i implies that wut G i?. On the other hand, since wutlZw and w ^ R/it follows that wut ^ i2, a contradiction. This shows that W — 0, proving the assertion on U'. Since [/" C {/', this completes the proof. D Uniform components of the type considered above have also the fol lowing important property. P r o p o s i t i o n 3.15 Assume that a uniform component U of a semigroup S C Mn(K) intersects finitely many TZ-classes of Mn(K). Then for every maximal subgroup G of c\(S) intersecting U we have G = gp(S Pi G). Proof. We will show that, for every e x t e n s i o n ^ C ^ = (S", (S'nD)-1), for a maximal subgroup D of Mn(K), used in the construction of cl(*S) (see the remark preceding Lemma 3.12), we have gp(S' C\G) — gp(SfD D G). Since cl(5) is obtained from S in finitely many steps of this type, the assertion will follow. On the other hand, if V is the uniform component of some S'D C cl(S) and U C V, then the pair V, S'D inherits the hypothesis on £/, S by Proposition 3.14. Therefore, it is enough to prove this for S' = S. Let V be the uniform component of cl(S) containing SdG. Let W be the union of 'H-classes of Mn(K) intersecting U. From Proposition 3.14 we know that V intersects the same ^-classes of Mn(K) as U. It is enough to show that SD H W C U because gp(S C\ G) is a maximal subgroup of U and S D H G C SD H W. Consider an arbitrary element x G S D fl W. Then x = S161S2... bk-\Sk for the inverses b{ of some elements of S Pi £) and some st- G S U {1}. Clearly, we may assume that k > 2. There exist u,w € U such that uxw G W. If we show that uxw G £/, then it is easy to see that also x G £/, as desired. Therefore, replacing si by usi and s^ by s^u;, we can assume that si,Sfc G C7 and ^i = 5i6i . . . 5fc_i,£2 = ^fc-SA; are contained in V. Hence x2 G W (use the definition of VK and the fact that x2Csk and sk G C/).
3.3.
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83
First, consider the element x2. Write b = bk,s = sk. Let b is the inverse in D of an element a G 5 n D. It is clear that the group H = g p ( 5 n JD) acts transitively on the set of 7^-classes of Mn(K) intersected by Hs by left multiplication. In view of Theorem 3.5 this implies that Hs C V. Therefore, as £ V. Since as £ -£, we have as £ 5 H V C [/, see Theorem 3.13. Since bs £ W, we must have bs = / 6 s for an idempotent / £ £/. Now as = {a2f)(bs) £ U C V, which implies in particular that a2feV. Hence a 2 / £ V fl S G = f/ by Lemma 3.11. Therefore as £ [/,6s £ W imply in view of the displayed equality that bs £ U. In particular, this shows that x2 £ £/. Let F be the maximal subgroup of U containing / . Since fx2 = x2, we have z = (xiv)(zx 2 ) £ W, where v £ (7 Pi F and z £ F is the inverse of v in F. We know that zx2 £ £/ because x2 £ [/. Moreover a^iu = Sx&i • • • bk_2(sk_iv) £ VK has a shorter presentation than x. Therefore, an induction allows us to assume that X\V £ U. Then x = (xiv)(zx2) £ [/, as desired. This proves our claim on SD, completing the proof of the proposition. □ Even more can be said in case S is finitely generated. P r o p o s i t i o n 3.16 Assume that U is a uniform component of a semi group S C Mn(K) such that U intersects finitely many TZ-classes of Mn(K). If S is finitely generated, then the group associated to S that comes from U also is finitely generated. Proof. Choose e = e2 £ D for a maximal subgroup D of Mn(K) that intersects U. If rank(e) = j , then the image I of C/U {0} in MjjM2_x is a uniform subsemigroup. Put / ' = Ze, where I is the completely 0-simple closure of I in M2jM2-x and let G denote the subgroup of D generated by U H D = S H D. Here K0[I] is an ideal of K0[SG/T] for an ideal T of the semigroup SQ constructed in Lemma 3.11 (see Corollary 3.8). G = ele \ {9} acts on /' by right multiplication. Hence K0[I'] is a right K[G]-module. Note that it is a free K[G]-module of finite rank, say r, by the hypothesis. Indeed, picking one element from every nonzero 'H-class of I intersected by V7, we get a basis eu ... , er of this module. We can choose ei = e. Now, each s £ S acts by left multiplication on KQ[I'], SO it determines an endomorphism of KQ[I']. Hence, we get a homomorphism
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cf> : K0[S] —> Mr(K[G]), which maps S into column monomial matrices Mr(G U {0}) over G°. By the hypothesis (S) = (<£(si),... ,<(>&)) for some Si G S. Let H C G be the group generated by the nonzero entries of all matrices (s,). It is clear that S C\ D C H. Since G is generated by S H D, it follows that G = # is a finitely generated group. □
3.4
0-simple semigroups
From the point of view of the ideal structure of a semigroup S the knowledge of 0-simple semigroups that can arise as principal factors of S is crucial. The aim of this section is to discuss this problem for linear semigroups S. We focus on the nature of 0-simple linear semigroups, showing in particular that they are uniform semigroups of certain special type, the study of which reduces in some sense to simple subsemigroups of linear groups. We start with a very useful, but not widely known, result of Jones, [50]. Recall that a 0-simple semigroup with a nonprimitive idempotent contains the bicyclic semigroup, that is the semigroup given by the pre sentation B = (p, q),pq = 1. Moreover, a 0-simple 7r-regular semigroup must be completely 0-simple, [15], Theorem 2.54 and Theorem 2.55. P r o p o s i t i o n 3.17 Assume that S is a 0-simple semigroup nonzero idempotents. Then
with no
1. The bicyclic semigroup is a homomorphic image of a subsemigroup
ofS. 2. If the relation 1Z is trivial on 5, then S contains a free mutative subsemigroup.
noncom-
Proof. If S has a zero element, then it is denoted by 9. Suppose that x — xy for some nonzero x,y £ S. 0-simplicity of S implies that there exist <s,t G S such that y — sxt. Hence x = xy = xsxt and sx — (sx)2t, where sx ^ 9. Therefore (sx)lZ(sx)2 in S. First, consider the case where TZ is trivial on S. The above shows that x ^ xy for every 9 ^ x,y G S. There exists b £ S such that b2 ^ 9 (otherwise S is nil, so it is 7r-regular and hence completely 0simple, which is not possible because S is nil). Now b = ab2c for some
3.4.
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nonzero a,c G S. We claim that the elements ab,a2b generate a free subsemigroup of S. Suppose that u,v are distinct (nonempty) words in ab,a2b that are equal in S. Let w be the longest common initial segment of u, v as words in ab1a2b (w can be the empty word). Then u = wui,v — wvi for some words ui,vi. Note that ui,Vi are not empty. (Otherwise w — wv\ or w — wui, respectively, in 5, which is not possible because 11 is trivial on S.) Let for example U\ — (a2b)u2 and V\ — (ab)v2, where u2,v2 may now be empty. For any i > 1 inductively we show that b = ab2c = (ab)(ab2c)c = (ab)2bc2 = • • • =
(ab)lbc\
Next, b = ab2c — a(ab2c)bc = (a2b)b(cbc), so that by induction ab = a{ab2c) = {a2b)bc = {a2b)2b{cbc)c = • • • = (a26)16(c6c)1'-1c. Multiplying ux = (a2b)u2 on the right by elements of the tye bc\b(cbc)lc several times (depending on whether the resulting element ends with a power of ab or a power of a2b) we can find, in view of the two displayed identities, an element x £ (a,6, c) such that U\X = ab. Then wab — wu\X — wv\X — (wab)v2x. This contradicts the hypothesis on 5, and establishes 2). To prove 1) it is now enough to consider the case where 1Z is nontrivial on S. Then xy — x ^ 6 for some x,y G S, soby the first paragraph of the proof there exist a, b E S\ {0} such that a — a2b. We will show that T — (a, b) maps onto the bicyclic semigroup B — (p,q\pq = 1). Using the relation a2b — a we can write each t G T in the form t — v(ab)rnan^ where m > 0, n > 0 and v G V = (6, (a6)6, ( a 6 ) 2 6 , . . . ) x . Put 5 = a&. Let Z(u) denote the length of t; in V. Then a5 = a implies that s for all v G V \ {1} and also (*)
a^+'v
= a, al^+l(vsman)
= an+l for all v G V.
We claim that, for any two elements vsrnan,v'slaJ vsman
G T as above
= f / s V implies that /(u) = l(v'),n
—j.
Assume for example that l(v) > l(vf). By (*) applied twice a n+i = a /M+i( U 5 m a nj = a'H+i
(,/5 v )
= a'^-'^a^ 1 .
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Since a is not nilpotent (because 9 ^ a = a2b = • • • = akabk for all k > 1) and S has no nonzero idempotents, it follows that n — j — l{v) — l(v'). Suppose that l(v) > l(v'). Since al^v = s and akbk = s forfc> 1, we get al^v\vsman)bn = (al^v)sm+1 = sm+2 and al^v\v'sla3)bn
=
a^-^ia^v'yiaW^-i l{v) l{v,) l+2 n J
=
a
-
= an-Jbn-J
s b-
= s.
Then sm+2 = 5, hence a power of s is a nonzero idempotent, a contra diction. Thus l(y) < l(v'), so that l(v) = l(v') and n — j , as claimed. Now, the formula (f)(vsman) = ql^pn defines a surjection (/> : T —> B. We shall repeatedly use the equalities as — a, al^v>v — s for v ^ 1. It is easy to see that l(wsmv) = l(w) + l(v) for v, w G V, v ^ 1. Let v' £ V and z,j > 0. Write z = vsrnanv'sla3\x = vsman,y = l 3 m+1, n+J v's a . Assume that t>' = 1. Then z — vs a if n = 0 and z = l v n+3 ifn ^ o. Hence, in both cases <j>(z) = p i )q = pl^qnp°q3 = vsman+3 (x)(y). Hence, assume that u' =^ 1. First consider the case where /(V) > n. Then z = vsmanv'sla3 = vsm+lwsla3 for some w G V with /(w) = Z(i>') - n if n / 0, and z = vsmv,sla3 ifrc= 0. Hence, in both cases <£(*) = g ' H + ' W - ' y = JMpnqKv')pi = (x)0(y). Finally, if /(>') < n, then z = v ^ a V s V = vsman-l^si+laj = vsman-l(v')+j^
Jjence5 ^
=
ql(v)pn-l(v')+j
=
^ ( 3 ) ^ ) Jn this Case, too.
Thus, is the desired homomorphism. □ Corollary 3.18 Let J be a 0-simple principal factor of a semigroup S C Mn(K). If S has no free noncommutative subsemigroups or J con tains a nonzero idempotent, then J is a completely 0-simple semigroup. Proof. Suppose that the relation 1Z is nontrivial on J. Then x = xy G J for some nonzero x, y G J. Since J is 0-simple and it is of the form A/B for some ideals B C A of J, it follows that the set A \ B of nonzero elements of J can be considered as a subset of Mj/Mj—i for some j . Then x = xy ^ 6 in Mj/Mj-i leads to y — y2 by Lemma 1.1. Thus, in view of Proposition 3.17, either of the hypotheses implies that J has a nonzero idempotent. But S cannot have infinite chains of idempotents.
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Therefore J has a primitive idempotent, so it is completely 0-simple. □
We now concentrate on 0-simple linear semigroups. Several ideas and examples presented below come from [126], where simple semigroups of matrices were considered. L e m m a 3.19 Let S C Mn[K) be a nontrivial 0-simple semigroup. Then there exists j such that the nonzero elements of S have rank j and the image S of S in M2jM2_\ is a uniform subsemigroup of M3jM2_x isomorphic to S. Moreover, for every maximal subgroup D of Mn(K), S n D is a simple subsemigroup of D whenever it is nonempty. Proof. The first assertion is an immediate consequence of the structure theorem. We only have to identify the zero of S (if it has one) with the zero of the corresponding Mj/Mj-\. Let a G S D D. If b G S Pi D, then b — xo?y for some x,y G S because Sa3S = S. Thus, xaTZb and ayCb in Mn(K), which implies that xa,ay G S 0 D. This means that (S H D)a(S f! D) — S Pi D, as desired. □ If | 5 | ^ 1, then define S as the image of S C Mn{K)mMn{K)IM^u where j is the minimal rank of a nonzero element of S. (Note that, if S has a zero, it is not necessarily the zero matrix.) T h e o r e m 3.20 The following conditions are equivalent for any semigroup SCMn{K),\S\^l, 1. S is 0-simple, 2. S = S2;, S is a uniform subsemigroup of some Mj/Mj-i and Sf)D is simple for every maximal subgroup D of Mn(K) intersecting £*, 3. S is a uniform subsemigroup of some Mj/Mj-i, S fl D is simple for every maximal subgroup D of Mn(K) intersecting S and S C\ H1H2 = {Sn ffi)(5 H H2) for every U-classes HUH2 of Mn{K) intersecting 5, 4- S\ {0} = IJ y(S fl D)x for a maximal subgroup D of Mn(K) such that SC\D is simple, where the summation runs over x G XdS, y G Y fl S for some subsets X, Y of the 1Z-, respectively C-class of the idempotent e G D in Mn(K).
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Proof. Assume that S is 0-simple. By Lemma 3.19 S is a uniform subsemigroup of Mj/Mj-i for some j . Let s G H\,t G H2 for W-classes Hi,H2 of Mn(K) not contained in Mj-\. Let S^s\S(t) D e the intersec tions of 5 with the £-class of s, respectively 7£-class of t in Mj/Mj-i. Then S^S(t) is an ideal of 5 . If it is nonzero, we must have S = S^S^t)What contributes to the 7{-class of st is (S H Hi)(S 0 # 2 ) - Hence 3) is a consequence of 1). If S is uniform, then for every a G S there exist b,c £ S such that bcHa in Mn(K). Therefore, 2) follows from 3). If 2) holds, then it is easy to see that every nonzero ideal I of S contains the set \J S fl D where the summation runs over all maximal subgroups of Mn(K). From Proposition 2.14 it follows that S/1 it must be nilpotent. Since S = S2, this means that / = 5, which establishes ! ) ■
Assume that 4) holds. The definition of a uniform semigroup is then satisfied for 5 , viewed as a subsemigroup of Mj/Mj_i,j — rank(e). This and the fact that S C\ D is simple imply that every nonzero ideal of S contains S 0 D. Therefore it contains 5, hence S is 0-simple. If S is 0-simple and s G 5 f l D , then, using the above notation, we come to S = ( S « S { s ) ) ( S ^ S { s ) ) = S « ( S n D)S(s). Hence, 4) is a consequence of 1). This completes the proof. □ Corollary 3.21 Assume that S C Mn(K) is a 0-simple semigroup and j is the common rank of nonzero elements of S. If Mj/Mj-i is identified with M(GLj(K),X, Y, P ) , then for every subsets X\ Y' of X, Y respec tively, the corresponding subsemigroup S P\ A4(GLj(K), X', Y', P') C Mj/Mj-i also is 0-simple whenever it is a uniform subsemigroup of
Proof. This is a direct consequence of condition 3) in Theorem 3.20.
□ Assume that condition 4) of the theorem is satisfied. Then, conju gating e = e2 G D to the appropriate diagonal form, we may define T = {t G GLj(K)\
( Q J
G ^ n D } . Then the nonzero elements of
S are of the form u c
0\ ( h 0\ ( v d\ _ ( uhv 0 ) \ 0 0 J \ 0 0 ) ~ \ chv
uhd \ _ / g chd ) ~ \ag
gb \ agb ) '
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where h,u,v G T ~ S D D, g = u/w,a = cw _ 1 ,6 = u _ 1 d, and a, 6 are rectangular matrices of the appropriate sizes. Denote the displayed matrix by s(g,a,b). For fixed a, b let Ta& = {g G T| s(g,a,b) G 5 } . Let j = rank(e). Without loss of generality, assume that S has a zero 8. So, there exist sets A, 5 of rectangular (n — j) x j , j x (n — j) matrices respectively, and a collection of subsets Tab C GL3(K),a G A, 6 G B, such that 5 = {s(g,a,b)\
aeA,beB,ge
Tab} U {0}
where Tab(I + ba')Ta>h> C Ta6/ if rank(J + 6a') = j with e = I
. Moreover T(a&) = U^afc, where the summation runs
over all a, 6 with s(g,a,b) in a given 'H-class of M n ( / i ) , is a simple subsemigroup of GLj(K) if rank(7 + ba) = j . Using the notation of the proof we see also that, if S C Mn(K) is 0-simple, then S(t)S^ = 5 fl D for every s,t E S such that s7le,tCe, because (SOD)2 C 5 ( s ' 5 ( i ) . The proof of Proposition 3.1 shows, m view of condition 4) that S — \Ja Sa for a collection of (0-simple) semigroups of matrix type Ia ~ Ai(S D D,X, Y, Pa). However, as shown in Exam ple 2 below, S is not necessarily isomorphic to a semigroup of the latter form. We continue with some examples of 0-simple linear semigroups that are not completely 0-simple. E x a m p l e s . 1. It is well known and easy to check that
*={(; j)|o''eRt}'*={(!!')"Ac£R+} are simple subsemigroups of G L 2 ( R ) . 2. Let C/ a b 0 \ (a b+ c 0 \ ] S= < 0 c 0 , 0 c 0 | a,b,c G R + \ . [\a i 0/ \0 0 0/ J It is easy to see that S = (SnDi)L\{SnD2)
for two maximal subgroups
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D DuUDD 2 2 of M 3 ( R ) . These groups are determined by the idempotents / 1 0 0 \ / 1 0 0 \ ci = Ij 0 1 0 I , e 2 = I 0 1 0 . 0 o 0 // \\ 0o o 0 o 0 /y \\ 1i o Theorem 3.20 implies that 5 S is 0-simple (use condition 2)). The completely 0-simple closure of S in M2/M1x is isomorphic to M(G,2,1, PP), ), where G is the group of 2 x 2 upper triangular matrices with positive diagonal entries and pn = p12 = = 1. However, the simple subsemigroups
^^M(2 - { ( o 6 r)}" ) } ' sS nn H(»<)} H(o<)} are not isomorphic. 3. Let H,G be subgroups of a group D. Assume that GxH is a subsemigroup of D for some x £ D. Then xHGx C GxH. Hence, for any gGG,hGH there exist gug2 £ G,huh2 G H such that {GxH)gxh{GxH)
= G{xHgx)hGxH
= Ggl{xh1hGx)H
D
D Ggi{g2xh2)H
G^xh^nGxH G^xh^hGxH = GxH.
Therefore 5 = GxH is a simple semigroup. This allows to construct simple subsemigroups in GL2{K) (K) that are not groups. For example, if G = H is the group of diagonal matrices in SL 5 L22(R) ( R ) and a; is a transvection, it is easy to check that S is of this type. The theorem above reduces the description of 0-simple subsemigroups of Mn(K) {K) to simple subsemigroups of GL,(K) GL3{K) for j < n. However, not many examples of the latter type have been constructed. An example of a simple irreducible subsemigroup of GL2{K) that is not a group will be presented in Section 4.1. It is a subsemigroup of a finitely generated group. However, if S itself is finitely generated, examples of this type cannot be constructed. P r o p o s i t i o n 3.22 A finitely generated 0-simple semigroup S C Mnn{K) (K) is completely 0-simple. ,sm) embeds into M,jM for some Proof. We know that S = (s (si,... Mj/Mj-i u... 3^ j < n. It is clear that the completely 0-simple closure of S in M.jM^ M^M^
3.5.
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intersects finitely many ^-classes of the latter semigroup. Choose t0 G A — { 5 1 , . . . , 5 m } , t 0 ^ 0. Since S2 = 5, we have t 0 = tiiii for some Ui G 5 and £x G A. Next £1 = t2u2 and similarly ^ = £2-+1ut-+1 for all i > 1, where i^- G S, £, G A. There exist j < k such that tj = £*.. Therefore £*. = tj = t^u^ • • • Uj+2uj+i- This implies that e = u^ • • • u J + 1 is a nonzero idempotent, cf. Lemma 1.1. The assertion follows because e G 5 must be a primitive idempotent. □
3.5
Some reduction techniques
The aim of this section is to establish some technical results that pro vide useful reductions when the structure theorem is applied. They often allow to simplify the structural chains of S and the structure of the groups associated to S. One can then approach many problems by induction on the length of a structural chain or on certain numerical invariants of the associated groups. We start with a result saying that every uniform component of S is, in a rather strong sense, controlled by any of its intersections with maximal subgroups of Mn(K). L e m m a 3.23 Let U be a uniform component of a semigroup S C Mn(K). Fix any maximal subgroup D of Mn(K) that intersects U. Then there exists a finite set Z C U such that for every a G U there exist x,y G Z with xay G S 0 D. Proof. We view U as a subset of Mj/Mj-i for some j < n. From Lemma 1.6 we know that V — AJ(U) U {0} is a subsemigroup of the full linear monoid of dimension t — (n) over K. Moreover, every nonzero matrix in V has rank 1 and V is a uniform subsemigroup of the com pletely 0-simple subsemigroup Mi of Mt(K). Let W = H Wu C Kl where Wu — ker(u) and the intersection runs over all u G V,n / 0. There exist iy 1 ? ... , wr G V, r < t, such that W = WW1 fl • • • fl WWr. Suppose that W{V = 0 for some v G V, v 7^ 0, and every i = 1 , . . . , r. Then lm(v) C ker(iut-), so that lm(v) C WWI fl • • • fl WWr. Thus lm(v) C Wu for every 0 ^ u G V. This means that Vv = 0, contradicting the fact that V is a uniform subsemigroup of M\. Hence, for each 0 7^ v G V there exists k such that WkV G V \ {0}. A dual argument allows to find elements 2 1 , . . . ,zq,q < t, such that
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for every 0 ^ v G V we have vz% G V \ {0} for some z. Therefore w ^ ^ ^ 0. Let B C [/ be a finite set that contains an inverse image of {w\,. . . , wr, zi,.. . , zq}. From Lemma 1.6 it follows that, if a G £7, then there exist x,y e B such that xay G C/. Let T be the union of all %-classes H of Mn(K) such that for every h G / / there exist 6 G B,c £ B with c7^/i and 6£/i. For every 'H-class H of M n (A r ) contained in T, there exist x', y' G C/ such that xf(UP\H)y' C U 0 D (because (7 U {#} is a uniform subsemigroup of Mj/Mj-i). Since there are finitely many such ^-classes and xay G £/, the result follows.
□
If U is an ideal uniform component of 5, then the sandwich matrix of U actually comes from its finite submatrix in a very simple way. L e m m a 3.24 Assume that U C Mj \ Mj_Y is a uniform component of a semigroup S C Mn(K) such that C/U{0} is a subsemigroup of 5 U { 0 } . Let D be a maximal subgroup of Mn{K) intersecting U and e — e2 G D,rank(e) = j . Let X C U 0 Mn(K)e be a set of representatives of the nonzero H-classes of Mn(K) intersected by U and Y C U fl eMn{K) a set of representatives of the nonzero C-classes of Mn(K) intersected by U. Let Xf C X,Yf C Y be bases of the subspaces of Mn(K) spanned by the sets X,Y, respectively. Denote by C the set of T-L-classes of Mn(K) that are TZ-related to some x G X' and C-related to some y G Y'. Then T — (JHEC{S H H) U {6} C Mj/Mj_i is a uniform subsemigroup of Mj/MJ_1 contained in U U {9}. Moreover \X\ < jn, \Y\ < jn, and the sandwich matrix P = (pyx)xex,yeY of U can be chosen so that for every x G X and y G Y we have
_ J ylP'x yx
) 9
if it is in D otherwise
where P' — (pyx)xeX',y£Y' and u^v are the coordinate vectors of u,v in the bases X\Y', respectively. Proof. Suppose that aT = 0 in Mn(K) for some a £T\ {9}. For every u G U there exists q G X such that qRu. Hence q — J2i ^i%i, where A,- G K and x% G X' C T. It follows that aq = 0, so that au = 0. Therefore aU = 0. This contradicts the fact that UU{9} is a uniform subsemigroup of M3jM3_x. A dual argument shows that Ta ^ 0 in M n (A r ) for (9 ^ a G
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93
T. This implies that each 'row' (and each 'column') of T intersects a maximal subgroup of M2jM3_x. Therefore, T is a uniform subsemigroup of M3/M3.l. Let x G X,y G Y Then x = £t-at-zt-,T/ = T,kPkyk for some at,flk G 7\ and x{ G X\yk G Y'. Hence yx = Y^,i,kaiPkykXi. Since yx, and every y*;X2-, is either the zero matrix or it is in D, it is enough to define Pyx = yx for x G X,y G Y, (see the proof of Proposition 3.1 for the way a sandwich matrix of U can be determined). It is clear that the cardinality of X, and of Y, does not exceed jn. □ From Lemma 1.8 we know that every 5 / ( 5 fl Mj) is an 'almost lin ear' semigroup and it is linear whenever 5 is finitely generated. This allows us often to reduce the study of U to the case where the matrices in U have the least rank among all nonzero matrices in 5. In particular, Lemma 3.24 implies that the sandwich matrix of every uniform compo nent of 5 is determined by its finite submatrix. The next reduction step is given below. It allows to 'cut off' the nilpotent component from the given layer of 5 and also to 'separate' the uniform components of this layer. L e m m a 3.25 Let K be a finitely generated field. Assume that S C Mn(K) is a semigroup and J is the ideal of matrices in S of the least nonzero rank, and zero if it is in 5. Let C/i,. . . , Ur be the uniform com ponents of J and N the maximal nilpotent ideal of J. Then there ex ist congruences pi,... ,p r on S such that the natural homomorphism 7r : 5 —> S/pi x • • • x S/pr x S/J satisfies the following conditions: 1. the kernel of the induced homomorphism Ko[S] —> the contracted semigroup algebras is nilpotent,
KO[TT(S)}
of
2. 7r(5) is a linear semigroup such that the ideal 7r( J) consists of all matrices in 7r(5) of minimal nonzero rank, and zero if it is in 7r(5), and 7r(J) has no nonzero nilpotent ideals, 3. each S/pj x S/J is a linear semigroup such that Uj/pj (identified with Uj/pj x {6}) is its ideal uniform component containing all matrices of the least nonzero rank, 4. the homomorphism 5 —> S/pj is one-to-one when restricted to every intersection UjPiD with a maximal subgroup D of Mn(K). In particular, the groups associated to Uj and Uj/pj are isomorphic.
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Proof. We may assume that 0 G 5, adjoining it to 5, if necessary. We shall also assume that J(~)GLn(K) — 0, because otherwise the assertions are clear with 7r the identity map. Let Ui be the smallest completely 0-simple subsemigroup of Mn(K) containing [/,-. From Lemma 3.11 it follows that S C S ' C Mn{K) for Sf = {S \ J) U J ' with J ' = U1 U • • • U Ur U AT' and N' the nil radical of J 7 . Clearly, Ui are the uniform components of J'. Since, under any homomorphism of 5', the image of £/,- is a completely 0-simple closure of the image of £/,-, it follows that it is enough to prove all assertions for Sf. Therefore, we shall assume that S = S". Let I3 = {x G KiS^U^xS1^ = 0}, where K{S} denotes the subalgebra spanned in Mn(K) by S. Define congruences pj,j = 1 , . . . , r, on 5 by apjb if a — b £ Ij for a,b £ S. Since 5*/^ embeds into the finite dimensional algebra K{S}/Ij, it is a linear semigroup. From Theorem 3.5 we know that UjNUj = 0 and UjUkUj = 0 for j ^ k. Hence the homomorphism TT3 : S —> Sjp3 maps J \ Uj to the zero of S/pj. Suppose that s,t G J are such that TTJ(S) = 7Tj(t) is nonzero. Then s,t £ U3 and K0[UjSl](s — t)K0[SlUj] = 0 in
Ko[S\. By Lemma 1.8 S/pi x • • • x 5//or x S/J is a linear semigroup because J — SnMk-i for some k < n. Moreover, choosing for S/J in Lemma 1.8 r big enough, we can ensure that every two matrices of different ranks are mapped to matrices of different ranks under the homomorphisms S —> S/pj x S/J and under IT. In particular, TT(J),TT3(J) X {6} con tain all matrices of minimal nonzero rank in 7r(5) and 7Tj(5) X 5 / J , respectively. But TTJ(J) = U3/p3. Therefore, conditions 2) and 3) are satisfied. Put p — f)r3=1pj. It is clear that n(S) can be identified with S/p\ where sp't if and only \i s — t OT s^t ^ J and spt. Moreover, the kernel L of A'ot*?] —> j^o[7r(5')] is the subspace of A"0[Sr] spanned by the set {s — t\sp't}. Assume that Sip'U for some Si,U £ J,i = 1,2,3. Let z = (si — ti)(s2 — ^2)(53 — ^3)- If a i l Si,U are in Uj for some j , then we have seen that z = 0 in A'o[S]- Otherwise z G Ko[N] because U3Uk Q N for j 7^ k. Hence L3 C K0[N]. Since A^n = 0, L is nilpotent. This shows that 1) holds. It is clear that p3 is trivial on cancellative subsemigroups of Uj, so
3.5.
SOME REDUCTION
TECHNIQUES
95
4) follows. □ Using the notation of the lemma above, let J3 — ( J \ [ / J ) U { 0 } . Clearly n factors through the composition of the natural homomorphisms S —> S/N —+ SIJx x • • • x S/Jr. Uj can be considered as an ideal of S/Jj. In order to make the latter a linear semigroup we had to factor out the congruence coming from the middle annihilator of Uj. The point is that the resulting kernel is rather small. Our next reduction procedure is concerned with homomorphic im ages of the groups associated to S. It extends a classical result on linear groups. The proof heavily depends on the techniques developed in this case [83], [134]. It is based on a refinement of the proof of [134], The orem 6.4, see Theorem 6.1. Recall that an element g of a subgroup G of Mn(K) is called unipotent if all its eigenvalues A £ K lie in the set {0,1} and g is semisimple if it is conjugate in Mn(K) to a diagonal matrix. P r o p o s i t i o n 3.26 Let T — S D D ^ 0 for a semigroup S C Mn(K) and a maximal subgroup D of Mn(K). Assume that G C D is the group generated by T and H is a normal subgroup of G which is Zariski closed in G. Let U be the uniform component of S containing T and SG = (S,G) C Mn(K). Then there exists a homomorphism : SQ —> Mr(K),r > 1, such that 1. 4>(T) C D' for a maximal subgroup D' of
Mr(K),
2. the subgroup G' of D' generated by 4>(T) contains D' C\ 4>(S) and is isomorphic to G/H' for some H' C 77, 3. f = f2 £ {H) consists of scalar matrices in fMr(K)f, J^, J = (j)(U) is an ideal uniform component of 4>(S) and G' is a maximal subgroup of the completely 0-simple closure J of J in Mr(I<). Moreover, the image of any unipotent (respectively, semisimple) element of G is a unipotent (semisimple) element of(G). In particular, if H is unipotent, then (/>(H) = f and (j>(G) ~ G/H.
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SEMIGROUPS
Proof. Lemma 3.11 implies that it is enough to find a homomorphism (j> such the corresponding assertions hold for SG,G,U in place of £, T, U. In fact, if <j) : SG —> Mr(K) is such a homomorphism, then D'C\(j)(S) C D' H (SG) = D'H <j>(e)(SG)(e) = D' n {G) = G", w h e r e e = e2 G G.
Since (U) intersects all ^-classes of the completely 0-simple ideal (f>(U) of (/)(SG)I all assertions will follow. Therefore, we consider the case S = SG only. In particular T = G. From Lemma 3.25 it follows that, passing to a homomorphic image of 5, we can assume that U is an ideal uniform component of S. We will find a homomorphism (/> : S —> Mr(K),r > 1, such that {U). (with zero, if it is in 4>{S)) is a completely 0-simple ideal of (e) — <\>(eUe) — 4>(G) is a m a x i m a l s u b g r o u p of (S) (or
(G) U {0} if 0 G (f>(S)). Let D' be the maximal subgroup of
Mr(K)
c o n t a i n i n g <£(e). T h e n , again D' H (S) = D' H (e)(S)(e) = (G).
Hence, assertions 1) - 4) will follow. To construct we shall proceed along the lines of the proof of The orem 6.4 in [134]. We may assume that e = I
J . Let rank(e) = k.
Let KY,KX denote the polynomial rings in indeterminates Xij,i,j = 1 , . . . , n; X{j, i,j — 1 , . . . , /c, respectively, that are the coordinate rings of Mn(K) and eMn(K)e. Let J be the annihilator ideal of H in KX, m the maximal total degree of any polynomial from a fixed finite gen erating set for J, Am the subspace of polynomials of degree < m in KY, Jm = J f) Am. Then each s G S acts on Am via the map g —y gs, defined by gs(Y) — g(sY). Here Y denotes the collection of the appro priate indeterminates written in the form of an n x n matrix and sY cor responds to the product of s and Y, hence the substitution of J2k sikXkj in place of X{j. This yields a homomorphism of S into E n d ^ A m , induc ing on Am a structure of a K[S'j-submodule of KY. We may identify S with its image in End#-(A m ). Let E(Am) be the exterior algebra of Am and let u : End# Am —> E n d ^ E(Am) be the homomorphism induced by the exterior power, see Section 1.3. Now Tm = Am D KX is a A"[G]-submodule of Am and E n d x Tm is mapped by a into End# E(Tm) C End# E(Am). Let P i , . . . , Pt be a basis of Jm over K and w = Pi A- • -APt G i?(A m ). Let VK = E S G 5 c r ( 5 ) ( w ; ) ^ £ E(Am). It is a A'[S]-module with a K[G}submodule WG = ^2geG a{9)(w)KThis determines a representation
3.5. SOME REDUCTION
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97
End A- W such that a\G : G —> EndA- WG. As in the proof of Theorem 6.4 in [134] we see that H is the normalizer of Jm in G. From the proof of Theorem 6.3 in [134] we know that crf(H) coincides with the set of scalar matrices in crf(G) C fMr(K)f, where End# W is identified with Mr(K),r — dimA- W, and / = cr'(e) = I
n
n
j . Indeed, if h G H, g G G, then by Lemma 1.6
^ ( / ^ ' ( ^ H = a\a)a\g^ha){w) ,
1
a {g){det{g- hg)w)
= ,
= det(/i)a ( 5 r)(^),
where det stands for the appropriate 'local determinant' of the elements of H as matrices in EndK{Am). So, cr'(h) acts as the scalar det(h) on W. Moreover, the kernel of this representation restricted to G is contained
mH. Finally, in each of the steps in the proof, unipotent (semisimple) matrices of G are mapped to unipotent (semisimple) matrices of the respective image of G. Therefore, the result follows. □ The following example shows that, in general, one cannot accom plish the isomorphism (G) = G/H in the above reduction step. E x a m p l e Let S C M2(Q) be the multiplicative semigroup of all matri ces of rank at most one. S is completely 0-simple with a Rees presenta tion S — ^(Q^TX^Y^P)^ where Y is the set of all reduced row echelon forms of nonzero matrices of 5, X is the transpose of Y, and the sand wich matrix P = (pyx)yeY,xex is such that pyx is equal to the (1, l)-entry of the matrix yx. Consider the sequence yn of elements of Y with the first row ( l , n ) , n = 1,2,... , and the sequence of elements xn of X with the first column (1, — l / n ) f , n = 1, 2 , . . . . It is clear that the submatrix Q z= (qij)ij>i of P determined by these sequences satisfies qa = 0 and qij ^z 0 for i ^ j . Let (/> : Q* —> {1} be the trivial homomorphism. Let (P) = Q v ) , where p — 0 if p = 0 and p — \ otherwise. It is easy to see that the determinant of every submatrix Qn = (%-)l-|j=i,...>n? n > 1, is nonzero. Let S' = M({1},X, Y, <j)(P))- The above implies that the contracted semigroup ring Qo[5"] contains subalgebras isomorphic to the matrix algebras M n ( Q ) for all n > 1 (see Proposition 4.13). Let e G X f) Y be the unique idempotent. Then G — eSe \ {0} is a maximal subgroup of 5, G ~ Q*. Suppose that n : S —> Mt(Q)
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SEMIGROUPS
is a homomorphism such that n(G) is trivial. Then n factors through S*. This is a contradiction because n determines a homomorphism of algebras Q0[S] —^ Mt(Q). We conclude with a result which shows that the completely 0-simple closure U of a uniform component U of S can be viewed, in some cases, as a localization. We say that the sandwich matrix P of U is invertible over the group ring K[G] if U ~ M(G,r,r,P),r < oo, and P has an inverse in the matrix ring Mr(K[G]). Clearly, a Rees matrix presentation of U is chosen here, but it is well known that, if we also have U ~ M(G,r,r, P ' ) , then P' is invertible over K[G] (see Section 1.1). Note also that KQ[U] ~ M(K[G],r,r,P) (the Munn algebra over K[G]) and the latter is isomorphic to Mr(K[G]) via the map a i-> a o P, where o denotes the usual matrix product, as explained in Section 4.2. L e m m a 3.27 Assume that U is a uniform subsemigroup of its com pletely 0-simple closure U such that the sandwich matrix is invertible over K[G] and U f) G satisfies the right Ore condition for every maxi mal subgroup G of U. Then KQ\U\ is a right localization of K0[U]. Proof. Write U = M{G,j,r,P), with P invertible in Mr(K[G\). Re call that the element of U with g G G at the (z,j)-entry, and zeroes elsewhere, is denoted by (#, i , j ) . Let Uij be the set of nonzero elements of U that are in row i and column j . Put U^ = U^ H U. We use the fact that Uji — UjiU^1, where Un is a group from the first column. Indeed, pu 7^ 0, and UnU~xl — Un by the hypothesis, so that (*)
UfiUa1 2 UflUHU?
D UfiU* = Un
In particular, for every element y £ U3i there exists u G Un such that yu G C/ji, because e = e2 G Un is a right identity for Uj\. View A'o[C/] as a subset of Mr(K[G]), see Section 1.1, (but they have different multiplication of course). Define the subset C — {P~l o s\ s G A} of Mr(K[G]), where A is the set of diagonal matrices with entries in G. Clearly C consists of invertible elements in K0[U]. Let Cf consist of those elements of C that lie in /\o[f/] (when treated as elements of K0[U]). It is enough to show that for every matrix z G A"o[C^] there exists c = P _ 1 o s G C' such that zc G A"o[C^]- But zc — zoPoc — zos. So we need to find s such that z o s and P _ 1 o s are in K0[U}. Hence,
3.5. SOME REDUCTION
TECHNIQUES
99
for each 1 < q < r and for the finitely many elements from the support of the q-th. columns of P - 1 and z (we treat them as elements of A'o[f/]) it is enough to find t — sq = (g, q,q) £ U such that t multiplies these elements (on the right) into U. For simplicity assume that q — 1. So, given a finite collection of elements Xk — ( ^ , 4 , 1 ) in £/, we need an element t = (g, 1,1) £ U\\ such that Xk o t = (h^g^ik-, 1) £ U for every k. But U\\Un C /7n imphes that replacing x^ by xj. — Xk o a for any fixed a £ C/n it is enough to find an element u £ Un such that x'^u £ Uik\ for all A:. Existence of such u follows from (*) above ((*) allows first to find u for the first of the elements x'k, and then adjust it step by step by right multiplication by elements of Un so that it works for all x'k - like in the process of finding a common denominator for finitely many fractions). This proves the lemma. □
Chapter 4 Irreducible semigroups A semigroup S C Mn(K) is irreducible if S ^ {0} and the column vector space Kn has no proper S'-invariant subspaces, where 5 acts on Kn by left multiplication. In other words, Kn is an irreducible K{S}module. As in the case of linear groups, this class has certain special properties and it provides a general technique for dealing with arbitrary semigroups of matrices. In this chapter we first discuss general results on irreducible semigroups. Then, the theory of irreducible representa tions of a semigroup S C Mn(K) is set up in terms of the structure of S. As an intermediate step, we describe all irreducible representations of a completely 0-simple semigroup. Finally, as an application, results on triangularizability of linear semigroups are obtained. The basic in formation on the properties of irreducible linear groups can be found in [83], [134].
4.1
Structure
If K is algebraically closed then, by Burnside's theorem, S is irreducible if and only if it is absolutely irreducible, that is K{S} = Mn(K). For an arbitrary field K, the following result is an easy consequence of the standard facts on the structure of finite dimensional algebras. L e m m a 4.1 Assume that S C Mn(K) is an irreducible semigroup. Then K{S} is a simple algebra and 1 G K{S}. Moreover, if S has a zero element, then either it is the zero matrix or S — {1} and n — 1. 101
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Proof. K{S} is a semisimple algebra because otherwise the Jacobson radical J of K{S}, being nilpotent, yields a proper S-invariant subspace JKn of Kn. If K{S} has a central idempotent e ^ 0, then eKn is an invariant subspace. Therefore e = 1, so that K{S} must be simple. If S has a zero element #, then, in view of Proposition 2.14, we must have 9 = 0 or 0 = 1. The assertion follows. □ For any subset A C Mn(K) by R(A) we denote the row space of A, that is, the space spanned by the rows of all matrices in A. C(A) stands for the column space of A. L e m m a 4.2 If S C Mn(K) is irreducible, then S satisfies any of the following equivalent conditions 1. dinix R(S) = dim^ C(S) = n, 2. K{S}Kn
= Kn and if Sv = 0 for some v £ Kn, then v = 0,
3. if Sx = 0 or xS = 0 for some x £ Mn(K),
then x = 0.
Proof. Assume that 1) holds. Let e i , . . . , en be the standard basis of Kn. Then K{S}Kn = AT{5}(Et e.-AT) = £.■ #{S}e t - = C(S) = A' n . Moreover, Su = 0 implies that, for every w £ R(S), the scalar product of w and u is zero, so we must have v = 0. Hence 2) is a consequence of 1). If 2) holds and xS = 0 for some x £ Af n (A r ), then 0 = xAT{S}/! 71 = x / \ n , so that x = 0. Assume that 3) holds. If C(S) ^ Kn, then it is easily seen that there exists x £ Mn(K),x ^ 0, such that xS* = 0, a contradiction. This and a symmetric argument show that 1) holds. Finally, if S is irreducible, then 2) is satisfied because K{S}Kn and {v £ Kn | Sv — 0} are S'-invariant subspaces. □ Irreducibility of S carries heavy consequences for the structure of the 'bottom layer' of S (that is, of the ideal consisting of matrices of the least nonzero rank and the zero matrix, if it is in S). Recall that S C Mn(K) is completely reducible if Kn is a direct sum of S'-invariant subspaces V{ on which S acts irreducibly. That is, 0 ^ SV{ C V{ and V{ has no proper S-invariant subspaces. As before, Mj stands for the ideal of matrices of rank at most j in Mn(K). P r o p o s i t i o n 4.3 Assume that S C Mn(K) is a semigroup. the minimal rank of nonzero matrices in S. Then
Let j be
4.1.
STRUCTURE
103
l.ifS is completely reducible, then Sj = SnMj is a 0-disjoint union of ideal uniform components of S. Moreover, for every maximal subgroup D of Mn(K) intersecting Sj \ {0}, D H Sj is completely reducible as a subsemigroup of eMn(K)e ~ Mj(K), where e = e2eD. 2. if S is irreducible, then Sj is the only ideal uniform component of S and D fl Sj is an irreducible subsemigroup of eMn(K)e. Proof. 1) Let A' n = Vi © • • • © Vr for irreducible AT{5}-submodules V%. If Af is a nilpotent ideal of 5, then each NV{ is 5-invariant, so it must be trivial. Hence, 5 has no nonzero nilpotent ideals. Therefore, the first assertion follows from the structure theorem in Section 3.2. We have eKn = eVi-\ h eVr. Choose a G D H Sj. Then there exists b G D such that ba = e = ba. If ev\ + • • • + evr = 0 for some vt G K', then avi + • • • + avr = 0. Since aV{ C V^-, the subspaces aV{ form a direct sum and we have av{ = 0 for every i. Then ev{ = bavi — 0. Hence eKn — eVi © • • • © eVr. Since aV{ C eVi and rank(a) = rank(e), we have aVi © • • • © aVr = eVl © • • • © eVr. Consequently eVt = aVt C Vt for every i. Choose 0 7^ w G eV{. From Corollary 1.5 we know that gp(5j r\D) = c\(Sj H D) C A^{^ H £>}. Moreover, Lemma 3.11 implies that eSe C c l ( ^ n D ) . Now K{S3^D}w D eK{S}ew = eK{S}w - eV;because IU G K* a n d the latter is an irreducible A"{S'}-module. This proves that each eV- is an irreducible K{Sj fl D}-module. 2) Suppose f/, V are two different ideal uniform components of S. Then UV C [/ and [71/ C V. Since [/, V do not intersect nontrivally, we must have UV = 0. But VKn is an 5-invariant subspace of Kn. This contradicts Lemma 4.2. Therefore, the hypothesis and 1) imply that U — Sj indeed is the unique ideal uniform component of S. An argument as in 1) shows that Kn is an irreducible K{S D D}-module.
□
The key reduction step reads as follows. P r o p o s i t i o n 4.4 Let S C Mn(K) be a semigroup. Then there exist natural numbers n0 — 0 , n i , . . . , nr and diagonal idempotents eu . . . , er such that the nonzero entries of et- are in columns n z _i + 1 , . . . , nt-, respectively, n = n\ + • • • + n r , and S is conjugate to a semigroup T C Yli<j e%Mn(K)ej such that for every i either e{Tex = 0 or et-Tet-
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is an irreducible subsemigroup of etMn(A")ez- ~ Mni(K). block upper triangular (T^ 0
TC
\
0
* T^
* *
* *
\
0
TM
)
and such that every nonzero projection T^ block is irreducible.
SEMIGROUPS That is, T is
= e{Te{ ofT onto a diagonal
Proof. If S is irreducible, then r — \ and S — T = T^1'. Otherwise, there exists a proper 5-invariant subspace V C Kn. Choose a chain of S'-invariant subspaces 0 ^ V\ C V2 C Vr = V of maximal length. Then there exists a basis vi,... , vn of 7\ n such that f i , . . . , v^ is a basis of V{ for i = 1 , . . . , r and some k{. Conjugating in Mn(K) we can bring S to a block upper triangular form with the sizes of diagonal blocks n x = kx and rii = fcz- — fc2_! for i — 2 , . . . , r. The choice of the chain V{ easily implies that the block diagonal projections are irreducible whenever nonzero. □ Corollary 4.5 We have K{S} = K{Ux} + • • ■ + K{Um} + J(K{S}), where <J(K{S}) denotes the Jacobson radical of the algebra K{S} and f / i , . . . , Um are uniform components of S. Moreover, if S is in the block triangular form of Proposition 4-4 an^ n Z5 ^e corresponding block di agonal projection, then J(K{S}) = ker(7r) Pi K{S}. Proof. From Corollary 3.8 we know that R = K{S} has an ideal chain 0 = i? 0 C R\ C • • • C Rt = R such that every i? 2 /R{-\ is either nilpotent or it is the image in R/Ri_i of a subalgebra Ax — K{U{} for a uniform component U{ of S. We put A{ — 0 for factors of the former type. By induction on t we show that R = Ai -\ \- At + J(R) for any algebra R of this type. If t = 1 the assertion is clear. It is easy to see that R' — R/Ri satisfies the induction hypothesis with respect to the chain R[ — Ri/Ri,i = 1 , . . . t, and the subalgebras A[ defined as the images in B! of A{. Therefore R' = A[ + • • • + A!t + J(R). Since Rx C J(R) (if it is nilpotent) or Ri = Ai, we come to R = A\ + • • • + At + J(R). Assume that S is in the block triangular form of Proposition 4.4. Let 1 T^ ' C eiMn(K)e{ be an irreducible block diagonal projection of S. Then
4.1.
STRUCTURE
105
K{T^} is a simple algebra, so the algebra ir(K{S}) is a semisimple. Since ker(7r) is nilpotent we get J{K{S}) — ker(7r) H K{S}. □ It is clear that every completely reducible semigroup S C Mn(K) is conjugate to a block diagonal semigroup with irreducible diagonal blocks. Moreover, the algebra K{S} is semisimple. The above form of S fits well into our structural approach. L e m m a 4.6 Let S C Mn(K) be in the block upper triangular form of Proposition 4-4- Let n : S —> Mn(K) be the corresponding block diagonal projection. Then 1. rank(7r(a)) = rank(a) for every element a £ S which belongs to a uniform component of 5, 2. cl(7r(5)) = 7r(cl(5')) and for every e — e2 £ cl(5) there exists g £ GLn(K) Pi [J2i<j eiMn(K)ej) such that 7r(e) = g~1eg. Proof. Let R = E,-<j e,-M n (A r )e i , Z = YJl=i^Mn{K)el and let : R —> Z be the natural projection. Clearly, n — (j)\S- R is 7r-regular by Proposition 1.3, so that cl(5) C R. Let a £ R. Recall that rank (a) is the dimension of the /i-column space of a. Since cf> is a linear map, it is easy to see that columns z l 7 . . . , z^ of a are independent whenever columns Zi,. .. , i\k of {a) are indepen dent. Therefore rank(a) > rank((a)). Assume that / = f2 £ R. It is well known that every two de compositions of 1 into a sum of minimal orthogonal idempotents of R are conjugate, cf.[24], 18.23. Therefore there exists x £ R f) GLn(K) such that x~lfx is diagonal. Then (j)(x~1(j)(f)x) = (j){x~l)cj){f)(t){x) = <j)(x~lfx) = x~lfx, so the foregoing implies that rank(x _ 1 (/)(/)x) > _1 r a n k ( x / x ) . Hence rank(<^>(/)) > r a n k ( / ) and the equality follows. Since rank(<^>(/)) = r a n k ( x _ 1 / x ) , these two idempotents are conjugate in the semisimple algebra Z C R. Hence / , (/) are conjugate in i?, which proves the second part of 2). If a is an element of a uniform component U of 5, then there exists x £ S such that ax £ U and axTif for an idempotent / £ Mn(K). Then / £ cl(S') C R. Clearly, (j)(ax)T-L(f)(f)^ so these matrices have the same rank. Since we know that r a n k ( / ) = rank((/>(/)), it follows that rank((/)(ax)) = rank(<^(/)) = r a n k ( / ) = rank(ax). Hence rank(a) >
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rank((a)) > rank((/>(ax)) = rank(a^) = rank(a). Therefore rank(a) = rank((cl(S)) is 7r-regular as a homomorphic image of cl(5). Since it contains T = (/>(£), we have (cl(5)) D cl(T). On the other hand, V = {a G R \ 4>(a) G cl(T)} is a 7r-regular subsemigroup containing S. In fact, let v G V. We know that v n lies in a subgroup H of /?, so there exists u £ H such that vnuvn = i;71. Then (#) is a subgroup of Z and c/>(u) is the inverse of cj){vn) in (j)(H). Since (V) = cl(T) is 7r-regular, and <j)(vn) G 4>(V), it follows that <£(u) is the inverse of <j)(v)n in a maximal subgroup of (V) D <£(cl(S)). This implies that cl(<£(S)) = <£(cl(S)), which completes the proof. □ The following lemma is crucial for applications of the structure the orem in the context of reducibility and triangularizability of semigroups of matrices. L e m m a 4.7 Assume that S C Mn(K) is in the block upper triangular form of Proposition 4-4- As above, write R — J2i<3 eiMn(K)ej and e Z = J2i iMn(K)ei. If fa : R —y Z{ — ezZe2- is the natural projection on an irreducible diagonal block, then 1. there exists a uniform component T of S such that t(T), or i(T) U {0} ifO G (f>i{S), is an ideal of (f>i(S) and it intersects the same Ti-classes of Mn(K) as a uniform component V of i(S), 2. if D is a maximal subgroup of Mn(K) intersecting 4>i(T) and e = e2 G D, then gp(&(T) H D) = gp(V H D) and ^ ( V ) n D is irreducible as a subsemigroup of eMn(K)e ~ A^rank(e)(^)5 3. there exists a maximal subgroup D' of Mn(K) such that 2-(T D D') C D and gp(&(T n D')) = gp(^(T) n D). Proof. We may assume that 0 G 5, adjoining the zero matrix to 5, if necessary. Suppose that the image in Z t of every uniform component of S is zero. Then i(S) is a nil semigroup, so it must be nilpotent, which contradicts its irreducibility. Hence we can choose a uniform component T of S consisting of matrices of minimal possible rank j such that i(T) ^ 0. Then <^t-(5j_i) = 0 because the choice of T implies
4.1.
STRUCTURE
107
that (j)i(Sj-i) is a nil ideal of i(S). We know that the j - t h layer of S is of the form S(j) — U\ U • • • U Uq U N, where U{ are uniform components of S and N is the nilradical of S(j). Say T = Ux. Clearly &(N) = 0. It follows that i(T) U{0} = 4>t(Sj) is an ideal of (j)i(S). Consequently, it is irreducible (as a subsemigroup of Z t = e{Mn{K)et- ~ M3(K)). Since zpreserves the relations 7£ and £ , (j>i{T) intersects all nonzero 7^-classes of a completely 0-simple subsemigroup of the corresponding Rees factor (R H Mk)/(R H Mk-i) C Mk/Mk-i (k = the rank of matrices in &(T)). Hence, the structure theorem applied to (f>i(S) easily implies that 4>i(T) is contained in a uniform component V of (f)i(S). Since i(T) U {0} is an ideal of V U {0}, V&{T)V C i(T) is irreducible in Z{. From Proposition 4.3 it follows that 4>i(T f! D) is irreducible as a subsemigroup of eMn(K)e. Since <j>i(T) fl Z) is an ideal in Vfl Z), it is easy to see that gp(,(T) C\ D) = gp(V H D). (In fact, c//i G g p ( i ( T ) n Z>) for g G V fl Z), /z G (j>i(T)C\D, implies that g G gp(t(T)nZ)), so that the nontrivial inclusion follows). This completes the proof of 2). If a G T is such that i(a) G Z>, then z(a2) G Z). Hence rank(a) = rank(a 2 ) because <^(<Sj_i) = 0. Therefore a G D' for a maximal sub group D'ofMn{K). Clearly ,(# HZ>') C Z>. Since {TnDf)T(TnD') C (T H Z>') U S'j-i, it follows that
(/)^(TnD/)((/>^(T)nZ))(/>^(^nZ},) c <j>t((TnD')T{TnDf))nD c '). Hence t-(T) f l D C gp(')). On the other hand, it is clear that we also have {(T H D') C <^(T) n Z>. Therefore 3) follows. □ Clearly, the simplest class of uniform semigroups consists of subsemigroups of GLn(K). We continue with an example of a simple subsemigroup of GLn(K) which is not a group (and so it is not completely 0-simple), but it is irreducible. It is due to Kelarev, [60], and answers the question asked in [106]. E x a m p l e Let F be a free nonabelian group with free generators x,y. It is well known that F contains a free subgroup G of infinite rank with free generators # i , # 2 ? . . . . We define a chain of subsemigroups of
108
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G inductively by n+1 S^ = (*!), S& = <*!), 5< 5<"+1») =
(xn+^S^xZUG^) {xn+uS^x-l.G^)
for n > 1, where G<"> denotes the subgroup generated by 5 K Clearly 5W = (S^x-^x^ C 5< n+1 ) and G ^ is the free group generated by xu... , x „ . Put 5 = (Jn>iS ( n ) - If s,t G 5, then ssx^t' i ^1 r 1 G 5, where n > 1 is such that s,t G 5 n . Therefore s = K " ^ * - 1 ) ^ ^ G StS. 5*5. This shows that 5 is a simple semigroup. By induction on n we prove that 5<"> does not contain the identity e of G. This is clear for S^ = (Xl). Assume that e g 5 for some n > 1. Every element tt G € 5< n+1 ) has a unique presentation of shortest length (n) t = txt2 ...tm with U G € G G<") U (x > 11 n+u n+ i, X and all y,- are in (x n + 1 ) U S W W Ji ^; ^G ' ' ' "' ' . By induction on r we first show that:
ijuMusw, ii) if some t{ G (x >, then i = m or t
n n+1 t+1 G 5< >. ii) if some t{ G (xn+1), then i = m or tt+1 G 5. If r = 1, then f = i * + 1 for some fc > 1 nor i € S^x^G^. Since e g 5 W , in the latter case we have tx G S^ \t2 = x ^ , and m = 2 or e g 5("), in the latter case we have i x G 5<">,i2 = * ^ i , and m = 2 or m = 3. Hence, i) and ii) are satisfied. m = 3. Hence, i) and ii) are satisfied. Let r > 1. Let y2 • • ■ y r = *i • • • t'p be the shortest presentation with Let r > 1. Let y2 ■ ■ ■ yr = t[ • ■ • ^ be the shortest presentation with t't G G^ U {xn+uxZl+l). By the induction hypothesis (condition i) ap t\ G G<"> U ( i n + i . i ^ x ) . By the induction hypothesis (condition i) ap plied to t[ ■ ■ ■ t'p) we have t[ G (x n+ i> U S™. plied to ii • ■ ■ g we have kt[ G ( z n + i ) U 5. n Assume first that Vl = x n+1,k > 1. If t\ G 5< n>, then tx = y 1 > m = Assume first that yi = z£ +1 ,fc > 1. If t\ G 5< >, then h = yum = p + 1, and ti+1 = ^ for i = 2 , . . . ,p. The induction hypothesis applied p + 1, and i ! + i = t'i for i = 2 , . . . ,p. The induction hypothesis applied to y2 • • • yp implies that ii) holds. Hence, consider the case t\ G (x n + 1 ). to y2 ■ ■ ■ yp implies that ii) holds. Hence, consider the case t[ G (xn+1). It follows that U = yit\,m = p, and tt = t\ for i = 2 , . . . ,p. Therefore, It follows that h = ylt'l,m = p, and tx = t\ for i = 2 , . . . ,p. Therefore, ii) holds in this case, too. It is clear that 1) is satisfied. ii) holds in this case, too. It is clear that 1) is satisfied. Hence, assume that yi = sx'^g for some s G 5 and t = 5 ^ • • • <;, or ii) applied to y2 ■ ■ ■ yr) implies that either t'3 G 5<»> and t = st'3--- fp, or t'2 G 5<") and i = st' ■ ■■t' respectively. Hence h = st' or h = st' is t2 G( n5<) n> and i = st'22 ■ ■ ■ptfp, respectively. Hence h = 3st'3 or tx = 2st'2 is in 5 ( n ) . This contradicts our supposition and proves that i) holds for t. in 5 . This contradicts our supposition and proves that i) holds for t. Since t = sx'^gt^ ■ ■ • t', we must have t = st'3---t' or t = st2 ■ ■ ■ t', so Since t = sxl\xgt\ ■ ■ ■ t', we must have t = st'3---t' or t = st'2 ■ ■ ■ t', so it is easy to see that iifalso is satisfied. This completes the proof of the it is easy to see that iifalso is satisfied. This completes the proof of the inductive claim. inductive claim. From From i) i) applied applied to to tt = = ee it it now now follows follows that that S 5nn+ +11 does does not not contain contain the identity of G. In fact, otherwise e = t ---t implies m = 1, 1, while while 1 m the identity of G. In fact, otherwise e = tl---tm implies m =
4.1.
STRUCTURE
109
e i S H Therefore S = (jn>1 S™ is not a group. Now, consider the following faithful representation of F into 2 x 2 matrices over rationals, [57], Theorem 14.2.1,
*-*(o
i ) '
^ ( 2
i)"
We claim that the image of S in M 2 (Q) is an absolutely irreducible semigroup. If the subalgebra A = K{S} spanned by S in M 2 (Q) has dimension < 3, then the group of units U(A) of A is solvable. But G^ C [/(A), a contradiction because G^ is free nonabelian. We continue with two results on finiteness of irreducible semigroups. P r o p o s i t i o n 4.8 Assume that S C Mn(K) is irreducible. Let j be the minimal nonzero rank of matrices in S. If S Pi D is finite for every maximal subgroup D of Mn(K) consisting of matrices of rank j , then S is finite. Proof. Let / = {a G S | rank(a) < j}. By Proposition 4.3, / is an ideal uniform component of S and it is irreducible. The hypothesis implies that / fl H = S PI H is finite for every 'H-class of Mn(K) consisting of matrices of rank j . More precisely, / is a completely 0-simple semigroup over a finite group G. By Lemma 4.1 Y^=\ Kai — 1 f° r some k > 1, Az G K and a2- G I. Clearly, these can be chosen so that k < n2. If x G 5, then x = (Y^ \tai)x(J2 ^iai) = YL h^3ixa>y i
i
i,j
But e^Sa,/ C i/jjU{0}, where Hij is the intersection of / with an W-class of Mn{K). It follows that | 5 | < (|G| + l) 7 1 '. □ P r o p o s i t i o n 4.9 Let 5 C Mn(K) be an irreducible semigroup over an algebraically closed field K. If the set {tr(s) | s G 5 } is finite, then S is finite. Proof. Since A" = ~K, irreducibility of S implies that K{S] = Mn(K). Hence, there exist ai,... , an2 G S which form a basis of Mn(K). Let A = { A i , . . . , \k] Q K be the set of all traces of the matrices in S. Then t r ( a x x ) , . . . ,tr(a n 2x) is an n 2 -tuple of elements of the set A. We will
CHAPTER
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show that the system of linear equations tr(aix) = a n , . . . , tr(a„2a;) = a n 2, a,- € K, has at most one solution in Mn(K). This will prove that | 5 | < kn . It is enough to consider the corresponding homogeneous system tr(a z x) = 0, i' = 1,. .. , n 2 . By the choice of a2 and linearity of the trace, tr(arc) = 0 for every a £ Mn(K) whenever x is a solution of the latter system. Since the trace is a non-degenerate bilinear form, we must have x = 0. □ We note that the above is no longer true if K is not algebraically closed. However, it remains true under certain assumptions on the al gebra K{S}, which are satisfied in particular if ch(K) — 0. We give a simplified version of the approach presented in [85]. Recall that a finite dimensional simple algebra R is separable over K if the centre of R is a separable field extension of K. If A is a simple subalgebra of Mn(K) and A ~ Mk(D) for a division algebra Z), then n = kr[D : K] for some r > 1. (It can be shown that r is the dimension of the centralizer of A in Mn(K) over its centre, cf. [44], Theorem VI.4.2). P r o p o s i t i o n 4.10 Let S C Mn(K) be an irreducible semigroup. As sume that the simple algebra K{S} — Mk(D) is separable over K and ch(K) does not divide r = n/(k[D : K]). If {tr(s) | s 6 S} is a finite set, then S is finite. Proof. Separability of K{S} implies that R = K{S} ®K K 1S a semisimple TT-algebra, cf.[21], Chapter 6. But 1 G 1<{S} C Mn(K) and R can be identified with K{S}. Every a £ K{S} can be viewed as a map Dh —> Dh. But D, as a A'-space, can be identified with K^D:K^. Hence, Dh ~ KkiD^ C Kk[D:K]. Let tiR(a) be the trace of a e R as a linear map K embedding
'
Mk(D)
—> K
~ K{S}
' . Since rk[D : A"] = n, we have a natural
A
Mk[D:K]{K)
A
Mn(K),
which is constant on K. Here \b maps any matrix c to a block diagonal matrix with r diagonal blocks, each equal to c. Moreover tr('0(/)(a)) = r • trfi(a) for a G K{S}. By the Noether - Skolem theorem [36], The orem 4.3.1, there exists g £ GLn(K) such that ^(j){a) — g~1ag for a £ K{S}. Then tr(a) = r • trR(a) for every a £ K{S}. Since ch(A')
4.2.
MUNN ALGEBRAS
AND
REPRESENTATIONS
111
does not divide r by the hypothesis, it follows that {tr#(a) | a £ S} is a finite set. Suppose tin(ax) — 0 for some a £ R and every x £ R. Choose x so that ax is an idempotent of rank one in a simple block of R. Then trji(ax) — 1. Therefore, tr# is a non-degenerate bilinear form on R. An argument as in Proposition 4.9 allows to prove that S is finite. □
4.2
M u n n algebras and representations
Our aim in this section is to describe irreducible representations of com pletely 0-simple semigroups. We introduce an important class of semi group algebras arising from completely 0-simple semigroups. They are crucial for investigating local properties of arbitrary semigroup algebras. They first appeared in the independent work of Munn and Ponizovskii, cf. [15], providing one of the main tools in the representation theory of semigroups. Our presentation is in the spirit of [79], [87]. For another approach to irreducible representations and many further results in this direction we refer to [15],[80],[106]. Let R be an associative K-algebra. Let X, Y be nonempty sets and P = (pyx) a generalised Y x X - matrix with pyx £ R. Consider the set A4(i2, X, Y, P) of all generalised X x Y - matrices over R with finitely many nonzero entries. For any A — (axy),B = (bxy) £ M.(R, X,Y, P) addition and multiplication are defined as follows A + B — (cxy) where cxy — axy + bxy for x £ X, y £ Y, AB — A o P o 5 , where o stands for the usual product of matrices. Also let a A = (aaxy) for a £ K. Then R = A4(R, X, Y, P) subject to these operations becomes an associative Kalgebra, called an algebra of matrix type over R. If not stated otherwise, we consider only the case where R has an identity and every row and every column of P contains a unit of R. Then R is referred to as a Munn algebra over R. The crucial motivating example comes from the following observa tion. L e m m a 4.11 Let G be a group, X, Y - nonempty sets and P - an Y X X-matrix over a group with zero G U {9}. Then the contracted semigroup algebra Ko[M(G,X, Y, P)] of the semigroup of matrix type A ^ ( G , X , Y , P) is naturally isomorphic to the algebra of matrix type M(K[G],X,Y,P) over the group algebra K[G\.
CHAPTER
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Proof. Define a mapping : M(G,X,Y,P) —> M(K[G],X,Y,P) by ((#,x',y')) ^ (axy), where axy = g if x = x',y = y' and a ^ = 0 oth erwise. From the definitions it follows that is a homomorphism into the multiplicative semigroup of the algebra M(K[G],X, Y, P). Since the images of the nonzero elements of M(G,X,Y, P) are linearly indepen dent, it is clear that the extension of (j> to a homomorphism of K-algebras K0[M(G,X,Y,P)] —► M(K[G],X,Y,P) is an isomorphism. □ We will use the above lemma by identifying the contracted semigroup algebra of a completely 0-simple semigroup with an appropriate Munn algebra. The following notation is motivated by our convention on completely 0-simple semigroups, see Section 1.1. A matrix (axy) G R such that ^x'y1
—
' 5 0>xy
= 0 for x 7^ x\y
^ y' is denoted also by (r,x\y').
For
any x G X,y G Y we put kfy = {(r,a;,y) | r G i?}. If X' C X,Y' C Y are nonempty subsets, then R/X,[ ~ ^xtX'^eY' R(x)- Thus R/X,{ ~ M{R,X\Y',PyX'), where PY>X> is the Y' x i ' - s u b m a t r i x of P. Let <£ : i? —> R' be a homomorphism of algebras. Then <j)(P) stands for the Y x X-matrix {(f>{pyx))- Further, <£:£—►£' =
M{R',X,Y,(j)(P))
denotes the induced homomorphism, that is, (f>(A) = (<^(axy)), where A = (axy) G R. If T is a subset of it!, then, for any sets X, Y, an X x Y-matrix A over i? is said to lie over T if all entries of A are in T. We put M{T,X,Y,P) = {A G £ | Alies overT}. Thus, it clear that ker(0) = A4 (ker (), X , Y , P ) and <£(£) = A4((#),X, Y, (P)). We start with an important auxiliary result providing necessary and sufficient conditions for a Munn algebra to have an identity. The fol lowing well known linear algebra argument will be used, cf.[15], Theo rem 5.11. L e m m a 4.12 Let R be a finite dimensional K-algebra and let P be an m x n-matrix over R for some integers m,n > 1. If m > n, then Q o P = 0 for a nonzero n x m-matrix Q over R. If n > m, then P o Q' — 0 for a nonzero n x m-matrix Q' over R. P r o p o s i t i o n 4.13 Let R be an algebra with a nonzero finite dimen sional homomorphic image. Then the following conditions are equiva lent for any Munn algebra R = M(R,X,Y,P)
4.2.
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AND REPRESENTATIONS
113
1. the algebra R has an identity, 2. X, Y are finite sets of the same cardinality and P is invertible as a matrix in M\x\(R). Moreover, if 1),2) hold, then R ~
M\X\(R).
Proof. Assume that R has an identity E. Then there exist finite subsets U C X, Z C Y such that E G R^}y Let x £ X. We know that for any yeY (*)
( l , a , y ) = E(l,x,y)
= EoPo
(l,x,y)
e
R$].
This implies that x G £/ and so X = [/ is a finite set. Moreover, from (*) it follows that the x-ih column of the matrix E o P consists of zeros except the (x,x)-th entry which is equal to 1. Therefore, E o P is the X x X identity matrix. Similarly, the fact that E is a right identity of R implies that Y = Z is a finite set and the Y x Y matrix P o E is the identity matrix. To establish 2) it is enough to show that \X\ = \Y\ since then £ is the inverse of P in the algebra M\x\(R). Let : i? —» R! be a homomorphism onto a nonzero finite dimensional algebra and let : R —> R! = M(Rf,X,Y,(/)(P)) be the induced homomorphism. Then R! has an identity {E) o (P) o A = A o (P) o (/>(£) for all A e B! and from Lemma 4.12 it follows that \X\ — \Y\. This proves 2). Let if; : R —> M\X\(R) be given by %j)(A) = A o P . For any A, B £ i?,A G A", we have il>{AB)
= I/J(AOPOB)
i(;(A + B) = {A + B)oP
=
AoPoBoP
= (AoP)
= i/;(A)il>(B),
+ (BoP)
0(AA) = (XA) oP = \(AoP)
= 0(A) + 0 ( 5 ) ,
= A0(A).
If 2) holds, then I/J is an isomorphism because P is invertible in This proves 1), as well as the remaining assertion. □
M\X\(R)-
Corollary 4.14 Let S = M(G,X,Y,P) be a semigroup of matrix type over a group G. Then the following conditions are equivalent 1. the algebra K0[S] has an identity,
114
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2. X, Y are finite sets of the same cardinality and P is an invertible matrix in M\x\{K[G\). Moreover, if 1),2) hold, then KQ[S] ~ M\x\(K[G\), and S is a com pletely O-simple semigroup. If S is an inverse semigroup, then 1),2) are equivalent to the fact that S has finitely many idempotents. Proof. Since the augmentation homomorphism maps K[G] onto K, Proposition 4.13, in view of Lemma 4.11 establishes the equivalence of 1) and 2), and the fact that K0[S] ~ M\X\(K[G]). Moreover, the invertibility of P implies that P has no zero columns or rows, so that S is completely O-simple. Assume now that S is an inverse semigroup. It is well known that S ~ M(G,X,X,Q), where Q is the identity X x X-matrix, cf.[15], Theorem 3.9. Hence, by Lemma 1.1, S has finitely many idempotents if and only if X is a finite set. This completes the proof. □ Thus, in the case described in Corollary 4.14, irreducible representa tions of S are in one-to-one correspondence with irreducible representa tions of the maximal subgroup G of S. An important example of such a situation was presented in Theorem 2.34. Our main aim in this section is to prove an extension of this result to the case of arbitrary completely O-simple semigroups. Recall that, if T C R is a nonempty subset of an algebra i?, then by /^(T),r j R(T I ) we mean the left, respectively right, annihilator of T in R. We write l(T),r(T), if unambiguous. L e m m a 4.15 For any x £ X,y £ F we have l(R) = {A £ R\Ao P = 0} = lfi(R(y)). If the rows of P are left R-independent as elements of the left R-module RY, then l(R) — 0. Proof. Assume that AR^ = 0 for some A £ R, y £ Y. Then A o P o ft(y) _ 0, so AoPo(r, x, y) = 0 for any r £ /?, x £ X. Hence r annihilates, on the right, all columns of the matrix A o P. Since r = 1 £ 7?, we get A o P = 0. Hence, the first assertion follows. If A o P = 0 for some 0 / A — (axy) £ R, then for any z £ X, Az o P = 0 where Az = (bxy) with bxy = axy if x — z and bxy = 0 if x / z. It follows that a nontrivial left i?-combination of rows of P is zero. □
4.2.
MUNN ALGEBRAS
AND
REPRESENTATIONS
115
Let Row(P) C RY be the left P-submodule generated by the rows of the matrix P. In other words Row(P) = Y,yeY RPy, where Py is the y-th row of P. Moreover, for any set Z, denote by Mrzow(R) the algebra of all Z x Z-matrices over R with finitely many nonzero rows subject to the natural addition and multiplication. The following is a direct consequence of the definitions and of Lemma 4.15. L e m m a 4.16 The rule (A) = A o P defines a homomorphism algebras (/> : R —> Mrx0W(R) such that 1. 4>{R) is the subalgebra of Mx°w(R) rows of which lie in Row(P),
of K-
consisting of all matrices the
2. ker() = /(P). The submodule Col(P) of the right P-module Rx is defined dually to Row(P), that is, Col(P) is the submodule of Rx generated by the columns P x , x G X, of P. The following result shows that /(P) is de termined by some specific subsets of P , which are also essential when representing the algebra R modulo r(R). L e m m a 4.17 Let R = M.(R,X,Y,P) be a Munn algebra over an al gebra P , and let Z C X be a subset such that the right R-submodules Col(P) = Zxex PXR andJ2xezPxR of Rx coincide. Then R = r(R) + R^andl(R) = lk(R^)). Proof. Let (r, x, y) G P . From the hypothesis it follows that the column Px of P may be written as Ylk=i PXkrk for some n > l , r 1 ? . . . , rn G R,xu... ,xn G Z. Define A G P by A = E L i ( r ^ ^ ^ ) - ( r ^ ^ ) - J t i s easy to see that P o A = 0, so A G r ( P ) . Since X^Ui(r/cr> ^ y) € P(^), we get (r, x,y) G r ( P ) + ^?m\ which proves the first equality. Now
KR) = ln{r(R) + Rfz]) = lfi(r(R))nla(R$) = Rnlk(R$) n
= ^(P(%}).
It is clear that the right-left symmetric analogues of Lemma 4.15, Lemma 4.16, and Lemma 4.17 may be proved. For this, one has to interchange the roles of the left and right annihilators, the roles of the mappings A —> A o P, A —> P o A, and those of the P-modules Row(P),Col(P).
CHAPTER
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For any ideal J of R we denote by B ( J ) , or BR(J) if ambiguous, the set { A G P | P o A o P lies over J}. It is called the basic ideal of R determined by J. The ideal B(0) may be characterized through some annihilators arising from R. L e m m a 4.18 Let R be an algebra of matrix type over an algebra R with identity. Then B(0) is the middle annihilator of R and R/B(0) ~ (R/l(R))/r(R/l(R))
~
(R/r(R))/l(R/r(R)).
P r o o f . Let A G R. Then A lies in the kernel of the natural homomorphism P —► (R/l{R))/r(R/l(R)) if and only if RA G l(R). The latter is equivalent to the fact that RAR = 0. Since R has an identity, this happens if and only i f P o A o P = 0, that is A G B(0). The second isomorphism is established similarly. □ We will describe a connection between the classes of modules over R and P , which exploits the basic ideals of P introduced above. Let V be a left P-module. Then there is a natural induced M^0W(R)module structure on the direct sum Vx, written as column Jf-tuples. Thus, Vx may be regarded as an P-module through the homomorphism of algebras 0 : R —> Mrxow(R) defined by (A) = A o P. This module will be denoted by V(P). In other words, the action of P on V(P) is given by Av = A o P o v for v G V(P),A G P . Let VQ(P) = {v G V(P) | Rv = 0}. Then V0{P) is an P-submodule of V(P) and VQ(P) = {v G V{P) | A o P o v = Ofor all A G P } = {^ G V ( P ) | P o u - 0} because TR(R) — 0. L e m m a 4.19 Let V be a left R-module. 1. znnR(V(P)/V0(P))
Then
= £(annR(V)),
2. ifV is an irreducible R-module, then V(P)/Vo(P) R-module,
is an irreducible
3. the rule W *-> W(P) defines an embedding of the lattice of Rsubmodules ofV into the lattice of R-submodules ofV(P). P r o o f . 1) We have a n n ^ ( y ( P ) / y 0 ( P ) ) = {A G P | AV{P) C V0{P)} = {AeR\AoPove V 0 (P)for allu G V(P)} = {AeR\PoAoPo
4.2.
MUNN ALGEBRAS
AND
REPRESENTATIONS
117
v = Ofor allu G V ( P ) } = { i G f i | P o A o Plies over ann H (\/)} = £(ann*(V)). 2) Let v — (vx)xeX G V"(P) \ Vb(P). Then P o v is a nonzero element of Vx, so there exists y G K such that the y-th entry 2: of P o 1? is nonzero. Let u — (ux) G V(P). Since V is an irreducible P-module, for any x G X there exists rx £ R such that r^z = iij.. Let A G P be defined as the matrix with zeros everywhere except the y-th column, where it has rx at the (2, y)-th entry, for all x G X. It is easy to see that u = Ao P o v — Av ^ Rv. This proves 2). 3) is straightforward. □ We note that 23^(0)3 = 0, so BR is contained in the annihilator of any irreducible P-module V. Hence V is an irreducible P/23(0)-module. If T is a ring with a nonzero idempotent e and I f is a left T-module, then eW is a left eTe-module which carries some important properties of W. In particular eW is irreducible or faithful whenever so is W. In the special case of Munn algebras, we derive the following result. L e m m a 4.20 Let V be a left R/B(0)-module. If pyx is a unit o / P , for some y G Y,x G X, then E = (p'^x^y) is an idempotent of R and EV has a natural structure of a left R-module. Moreover 1. EV is an irreducible R-module module, 2. EV is a faithful R-module ifV
if V is an irreducible
is a faithful
R/B(0)-
R/B(0)-module.
Proof. We may treat V as a left P-module with 23(0)V = 0. It is clear that E2 = E and P ~ ERE via the map r i-> (rp~^x, y). Hence EV C V may be regarded as an P-module. Thus 1) is a direct consequence of the foregoing remark. If ann^(V) = 23(0), and (EAE)EV — 0 for some Ae R, then EAE G 5(0). Thus EAE = E o ( P o EAE 0 P ) o E = 0, which proves 2). □ We now consider a special case which is of interest when looking at finite dimensional representations of Munn algebras. Assume that P = M(R^X^Y^P) is a Munn algebra over a simple Artinian algebra P . Let P = Mr(D) for some r > 1 and a division algebra D. IfT,Z are nonempty sets and A = (atz) is a T x Z-matrix over P , then we write A for the (T.r) x (Z.r)-matrix obtained from A by erasing the matrix brackets of all entries atz of A. Here T.r, Z.r denote the disjoint unions
CHAPTER
118
4. IRREDUCIBLE
SEMIGROUPS
of r copies of sets of cardinality | P | , \Z\ respectively. Similarly, treating elements of (Dr)T as {1} x T-matrices over D r , an element v G D T r is associated with any element v G ( D r ) T . With this notation we have the following result. L e m m a 4.21 The algebras M{Mr{D),X,Y, are isomorphic.
P) and
M{D,X.r,Y.r,P)
Proof. Let A,B G M{Mr(D),X, Y, P ) . It is easy to see that AoB = A o P o B — AoPoB = AB. Since A \-> A is a linear map, it is an isomorphism of algebras. D We define the rank of an Y x X-matrix P over M r ( D ) , and write r a n k ( P ) , as dim^ (/>p((D r ) y ), where (j)P is the homomorphism of left D-spaces (Dr)Y —► {Dr)x determined by P. So <j>P{y) = (V o P)\ with w* denoting the transpose of w. If r = 1 and X = Y is finite, then our notation for the rank agrees with that used before for matrices over a field. Note that, under the natural isomorphisms of D-spaces (Dr)Y ~ DYr,(Dr)x ~ DXr, 4>P corresponds to the isomorphism 4>p : Yr Xr D —y D given by p(w) = (w* o P)K Thus, dirn^ # ( D y r ) = dim£>(p(Z)r)r). Since the former is equal to dim^ Row(P), as a direct consequence we get the following corollary. Corollary 4.22 rank(P) = rank(P) = dimz)Row(P). Our next auxiliary result is well known and easy to prove. L e m m a 4.23 Let P be an Y x X-matrix Then dim£> Col(P)
=
over a division algebra D.
sup{A; | P has an invertible k x k submatrix}
— dim£> Row(P). This shows that rank(P) may be equivalently defined as the dimen sion of the image of (Dr)x under the homomorphism of right D-spaces defined dually to p by the action of P on the left. P r o p o s i t i o n 4.24 Let D = A4(D,X,Y,P) be a Munn algebra over a division algebra D. Assume that rank(P) = t < oo. Then D/B(0) ~ Mt(D).
4.2.
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AND REPRESENTATIONS
119
Proof. Since rank(P) = dim£> Col(P), there exist xu ... ,xt 6 X such that the corresponding columns PXl,... , PXt of P are right DXl independent and Col(P) = £ | = 1 P D. Put Z = { z i , . . . , xt}, and let : bfz] —> Mrzow(D) be the homomorphism defined by (A) = A o PYZ. Clearly, the latter algebra can be identified with Mt(D). In view of Lemma 4.23 the matrix PyZ n a s t rows which are left D-independent. Thus, from Lemma 4.16 it follows that (D\zl) has dimension t2 as a left vector space over D. Therefore (j> maps D\zl hence again Lemma 4.16 yields (*)
bfz]ll(D%])
^ (D§) ~ MT(D)
onto MZ0W(D),
~
and
Mt(D).
On the other hand, from Lemma 4.15 (cf. the remark following it) it follows that r(D) = rb(D{$), and consequently r(b)C\Dfz\ = r(D^) ~
~ ry\
~
\z) + r ( D ) , we come to oince by Lemma 4.17 D = D)J D/r(D)
= (D$
+ r(b))/r(D)
~ D$/(r(D)
n £,%>) =
^ M ^ g ) .
Thus, from Lemma 4.18 it follows that D/BD(0)
~ (D/r(D))/l(D/r(D))
= (^ ( % ) /r( J Dg))//(D ( %Vr( J D ( %>)).
Applying Lemma 4.18 once again with respect to the algebra D)zl we get
D/BD(0) ~ (^>//(^))/r(^V'(^(% } ))The latter algebra is in view of (*) isomorphic to Mt{D). This completes the proof of the proposition. □ Corollary 4.25 Let R = M(R, X, Y, P) be a Munn algebra over a sim ple Artinian algebra R = Mr(D),r > 1, for a division algebra D. Then R/B(0) ~ Mt(D) provided that t = rank(P) < oo. Proof. By Lemma 4.21 R ~ M(D,I.r,M.r,P). Thus Proposition 4.24 and Corollary 4.22 imply that R/B(0) ~ MTank(P)(D) = Mt(D). D We are now ready for the main result of this section.
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T h e o r e m 4.26 Let S — A4(G,X,Y, P) be a completely 0-simple semi group. Assume that : G —> Mr(K) is an irreducible representation of G such that rank((/>(P)) — t< oo. Then the induced map <j> M{G,X,Y,P)
—+R = M{Mr{K),X,Y,(P))
—► R/B(0)
~
Mt(K)
is an irreducible representation of S. Moreover, every irreducible rep resentation of S arises in this way and two such representations are equivalent if and only if they are equivalent on G. Proof. V = Kr is an irreducible K[G]-module. So V((P))/Vo((P)) is an irreducible A = M((j)(K[G)), X ,Y,
V0(4>(P)) = {ve V((P)) 14>{P) o v = o}, the action of A extends to the action of R = M(Mr(K),X, Y, (/>(P)) on V((P))/Vo({P)). This module has dimension rank((P)) by Corol lary 4.22. Its annihilator is # R ( 0 ) by Lemma 4.19. This implies that the resulting representation of S is accomplished by the natural homomorphism <j>. Let ^ : M(K[G],X,Y,P) —y Mt(K) be the extension of an irre ducible representation of S to its Munn algebra. Then ip factors through the natural map M{K[G],X, Y, P) —± C = M{i>{K[G}),X,Y^{P)). From Lemma 4.20 we know that ip determines an irreducible represen tation %j)\G of G. So ij)(K[G]) ~ Mk(D) for a division algebra D by Lemma 4.1. Suppose the rank of tp(P) over D is not finite. Then Proposition 4.13 implies, in view of Lemma 4.23, that for every n > 1 C contains a subalgebra of matrix type isomorphic to M/ cn (D). Hence, the inverse image of such an algebra in M(K[G],X,Y,P) intersects S nontrivially and lies in the kernel of ?/>. This is a contradiction since S is completely 0-simple. Therefore tp(P) has a finite rank over D. Since BQ(0) is nilpotent, Corollary 4.25 implies that C has only one irreducible module. So, this module must be induced from ip\o in the way described in the first paragraph of the proof. It follows that ip is equivalent to the representation for = I/J\Q. This completes the proof of the theorem. □
4.3.
4.3
IRREDUCIBLE
REPRESENTATIONS
121
Irreducible representations
It is well known that irreducible representations of a finite semigroup come from irreducible representations of the completely 0-simple prin cipal factors of S. We have seen in Section 4.2 that every irreducible representation of a principal factor is determined by an irreducible rep resentation of its maximal subgroup. A beautiful instance of this phe nomenon is given in Theorem 2.35. More generally, it turns out that irreducible representations of a linear semigroup S come from the rep resentations of the uniform components U of S. However, there is no natural one-to-one correspondence with irreducible representations of S n D for a maximal subgroup D of the completely 0-simple closure U. The situation, in general, is much more complicated than that in the classical case. One of the reasons being that representations of S D D do not always extend to representations of the group g p ( 5 Pi D) C D. We start with a technical result of independent interest. L e m m a 4.27 The relation -< defined on the set of uniform components of a semigroup S C Mn(K) by U -< V if SVSOU ^ 0 is a partial order. Proof. It is clear that -<< is reflexive. If U -< V and V -< U, then £/, V consist of matrices of the same rank. Therefore, the structure theorem (Theorem 3.5) implies that we must have U — V. Assume that U -< V and V X W. Consider the ideal SWS of S. Then the Zariski closure SWS is an ideal of S and it intersects V. Since S is 7r-regular, from Remark iv) after Theorem 3.5 it follows that SWS contains every uniform component of S which it intersects nontrivially. Since V is contained in a uniform component of S by Proposition 3.13, we must have V C WVS. Suppose that SWSMJ = 0. Then U{SWS)U C M^u where j is the common rank of matrices in U. Since M3_x is a closed set, it follows that USWSU C M j - i . B u t U -< V implies that SVSnU ^ 0, so that 0 ^ U(SVS)U H U C USWSU. This contradiction shows that U -< W, whence ~< is transitive. □ Recall that, for convenience sake, a uniform component of a semi group S C Mn(K) is sometimes viewed as a subsemigroup of the cor responding principal factor M2jM2-X of Mn(K), and sometimes as a subset of Mj \ Mj_i. T h e o r e m 4.28 Let S C Mn(K)
be a semigroup.
Then
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1. If (j> : S —> Mr(L) is an irreducible representation over a field L, then there exists a unique uniform component U of S such that (f)(1) = 0 for I = {x G S\ SxS 0 U = 0}. Moreover <j>(U) is an ideal of (j)(S). Hence factors through S/I and it determines an irreducible representation of U. 2. Conversely, every irreducible representation n : U —> Mr(L) of a uniform component U of S determines a unique irreducible repre sentation : S —> Mr(L) such that cjyp — 77. Proof. 1) We may assume that 0 G S. Choose a minimal, with respect to the order -<:, uniform component U of S which is not contained in the ideal / ' = {x G S \ 4>(x) = 0} of S. Such a component exists because oth erwise 4>{S) is a nil semigroup, hence it is nilpotent by Proposition 2.14, which contradicts its irreducibility. We claim that U is uniquely deter mined. Suppose that U' is another component with this property. Then from Theorem 3.5 it follows that SUS \ U and SU'S \ U' are ideals of S and they do not intersect uniform components of S that are not con tained in I'. Therefore, (f)(SUS \ U) is a nil ideal of c^(S'), so it must be zero because <j>(S) is irreducible. Hence (/>(SUS) — 4>(U), and similarly <j>(SU'S) = (U'). Now ^(SUS^SU'S) = cj>{UU') = 0 because UU' does not intersect {/, U'. Irreducibility of <j)(S) implies that (j)(SUS) = 0 or <j>(SU'S) = 0. But this contradicts the choice of U and [/', proving our claim. It is clear that / is an ideal of S. If / 2 ^'> then as above we see that / intersects a uniform component of S. This leads to a uniform component U' ^ U minimal over /', which is not possible. Hence / C /'. Since 4>{U) is an ideal of (5), it is irreducible. Hence 1) follows. 2) From Chapter 3 we know that U can be viewed as an ideal of a Rees factor S/J. If rj : U —> Mn(L) is an irreducible representation, then by Lemma 4.1 there exists e G L[U] such that 17(e) = 1. Let (j>(x) = 77(ex) for x G L[S], where x G L[S/J] denotes the image of x under the map L[S] —> L[S/J]. It is clear rj(ex) = n(ex)r](e) = rj(exe) = n(xe). Hence 4>(xy) — n(exye) — rj(ex)rj(ey) — (j)(x)(f)(y), so that (f> determines a representation of S which coincides with 77 on U. If (/>' : S —> Mn(L) is another representation with this property, then '{x)ri{e) = (//(z)'(e) = '(xe) = r7(xe) = (x). The result follows. □
4.3.
IRREDUCIBLE
REPRESENTATIONS
123
R e m a r k Let S C Mn(K) be a 7r-regular semigroup. The above theorem leads in view of the material in Section 4.2 to the following classification of irreducible representations. Let Q be the set of maximal subgroups of 5, one from each regular J'-class of S. For G G £?, let VQ be the set of all nonequivalent irreducible F[G]-modules. If F is a field, then the irreducible F[S]-modules W are in one-to-one correspondence (up to equivalence) with the ordered pairs (G, V), where G £ Q and V G VGIf W is given, then there is a unique ^7-class J of S such that JW ^ 0 and IV = 0, where I = {x G 5 | J £ SxS}. Moreover, the dimension of VK is equal to rank(<^(P)) for a sandwich matrix corresponding to J. It is clear that, in general, a representation <j> of a uniform semi group U with a completely 0-simple closure U does not extend to a representation of U. Namely, if <j>(a) — 0 for some a G U H D and a maximal subgroup D of [/, then we would have (j) — 0. However, even if <j>[a) ^ 0 for all a ^ 0,a G C/, extending may not be pos sible. Indeed, let S C Mn{K),n > 1, be an irreducible semigroup consisting of matrices of the same rank j < n. For example, consider S = {a = (dij) G M n ( R ) | azj > 0,rank(a) = 1}. Let X be a free semi group such that there exists an onto homomorphism (j) : X —> S. Then (/> does not extend to a representation '(G) C GLn(K), which contradicts the choice of j . Clearly, S can be replaced by a finitely generated irreducible subsemigroup T. Then U — X is a finitely generated free semigroup, so it can be viewed as a linear semigroup U C GL2(Q) with U — gp(U) C GL2(Q) (see Example in Section 4.1). For the above reasons we do not get a natural correspondence be tween irreducible representations of U and of U H D for a maximal subgroup D of U. Our next aim is to give an abstract characterization of the bottom layer of an irreducible semigroup, proved in [106]. We say that rows y, y' of the sandwich matrix P = (pyx) of a completely 0-simple semigroup S = M(G,X,Y,P) are similar if there exists g G G such that pyx — gpy'x for every x G X. Similarity of columns x , x ' is defined by the condition pyx — pyx>g for some g G G and every y G Y.
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SEMIGROUPS
P r o p o s i t i o n 4.29 Let U be a uniform subsemigroup of a completely 0-simple semigroup S with completely 0-simple closure U C S with a maximal subgroup G. Then U has an irreducible representation <j> : U —> Mn(K) extending to a faithful representation of U if and only if the following conditions are satisfied 1. the sandwich matrix of U — M(G,X,Y,P) (under some Rees matrix presentation) has no similar rows and no similar columns, 2. there exist r > 1 and a faithful irreducible representation rj : G —> GLr(K) such that the matrix (j)(P) (viewed as a matrix over K with \Y\r rows and \X\r columns) has rank n. Moreover, ifUHG is a right Ore semigroup, then it is enough to assume in 2) that a faithful representation rj : U D G —> GLr(K) is given. Proof. If 1),2) hold, then the extension of rj to a homomorphism of Munn algebras R = M(K[G\,X,Y,P) —> M(MrjK),X,Y,(/){P)) —► Mn{K) yields an irreducible representation of [/, see Theorem 4.26. Suppose that <j)(a) — cj)(b) for some a, 6 G U. From the construction of (j> we know that (a — b) G # R ( 0 ) . Hence U(a — b)U = 0. Choose an idempotent e G U such that ea — a. Then (a — eb)oPoU = (a — eb)U = 0. Therefore (a — eb) o P = 0. In view of 2) this implies that we must have a%eb. Clearly a — eb in this case. Hence aCb. Similarly one shows that aTZb. Then aHb, and P o (a — b) o P = 0 imply that a = b. Therefore <j> is faithful. Assume now that : U —y Mn(K) is a faithful irreducible represen tation. Then 4>(U) is an irreducible, completely 0-simple subsemigroup of Mn(K). From Theorem 4.26 we know that rank(<£(P)) = n and the restriction of (j> to G is an irreducible representation. Suppose P has two similar rows. Say, pyx = gpy,x for some y, yl G Y, some g G G and for every x G X. Fix some z £X,h G G. Let a = {h,z,y),b = (hg,z,yf). Then ac = be for every c G U. This is impossible by Lemma 4.2, unless y = y'. A symmetric argument works for the columns of P. Finally, it is easy to see that G = (U Ci G)(U Pi G)'1 implies that the unique extension of a faithful representation of U D G to G is also faithful. The result follows. □ We conclude with a result on a representation of an irreducible semigroup 5 it terms of the translational hull ft(S). Let A, 0 be the
4.3.
IRREDUCIBLE
REPRESENTATIONS
125
sets of left, respectively right translations of a semigroup S. That is, all A G A,p G 0 are maps S —> S satisfying \(xy) = \(x)y and p(xy) = xp(y) for every x,y G 5. Recall that the translational hull fi(5) of S is the set of all pairs {(A,/9)|A G A,/9G 0,xA(y) = /9(x)yforx,y G 5 } with operation (A,p) • (A',p') = (A'A,/>//), [15]. Let 7(5) be the idealizer in Mn(K) of a semigroup S C Mn(7\T). That is, 7(5) — {a e Mn(K) \aS,Sa C 5 } . Clearly, every element a G 7(5) determines an element u(a) = (A a ,p a ) G 0 ( 5 ) , which represents the left and right multiplications \a,pa by a. This semigroup is of interest because of the following result obtained in [106]. P r o p o s i t i o n 4.30 Assume group. Then
that S C Mn(K)
is an irreducible
semi
1. if S is the bottom layer of a semigroup T C M n (7i"), then T C 7(5); moreover the bottom layer of 7(5) is equal to S if 5 zs completely 0-simple, 2. the natural map u : 7(5) —> $7(5) is an
isomorphism.
Proof. 1) Since 5 is an ideal of T, the inclusion T C 7(5) is clear. So, assume that 5 is completely 0-simple. Let a G I(S) be of rank equal to the common rank of nonzero matrices in 5. Since 5 is irreducible, from Lemma 4.1 we know that 1 G 7i{5}. Hence aS ^ 0. Therefore aS C 5 implies that a7£6 in Mn(K) for some b £ S. Similarly, a £ c for some c G 5. Since 5 is completely 0-simple, it follows that a%d for some d G 5. Then aS Q S implies that a e S. 2) Let CJ(S) = (As,y9s) for 5 G / ( 5 ) . Suppose that (A5,/os) = (^t,Pt) for some s,t e S. Then (5 - t ) 5 = 0. Since 1 G 7f{5}, we must have s — t = 0. This means that LJ\S is an embedding. Thus, S : u;(5) —> 7(5) C Mn(K) given by £(a;(,s)) = 5 is an irreducible representation of u(S). We have AAS = AA(s) and psp(x) = p(x)s = #A(s) whenever (A,/>) G 0 ( 5 ) and x G S . Therefore (*)
(Aa, />S)(A, p) = (AA(5), ^ A(s) ) G CJ(5).
Similarly (\,p)(\s,ps) G C J ( 5 ) , so that LJ(S) is an ideal of 0 ( 5 ) . Hence, 5 extends to a representation 5 : 0 ( 5 ) —> Mn(K). Then 5 is an
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ideal of 5(Q,(S)) because S(UJ(S)) = S. Consequently 5(Q,(S)) C I(S). Let y G /(S r ). Then for every x G 5 we have yx,xy G 5 and yx = 5(o;(yx)) = £(u;(y))£(u;(>)) = 5(u(y))x. Hence (y - 6(u(y)))S = 0, so that y = £(cj(y)). Therefore /(AS) = S(Ct(S)). It remains to show that 5 is injective. Suppose that 5(a) = 5(b) for some a, 6 G fi(S'). Then i ( u a ) = ( J ( M ) = 5(ub) — 5(ub) and i(aw) = S(au) = <£(6u) = 5(bu) for every u G UJ(S). This implies that ua = ub and au = bu for every w G w(S). If a = (A,/>), 6 = (A/,/?/), then in view of (*) this means that \\(x) — \\'{x) f° r every x G S. A symmetric argument shows that PP(X) — Pp'(x)- So (A(x) — A'(x))5 = 0 = S(p(x) — p'(x)), which implies that A = A' and p = p1'. So a = 6, as desired. The result follows. □
4.4
Triangularizable semigroups
Our next aim is to describe triangularizable semigroups in terms of the structure theorem of Chapter 3. The obtained conditions reduce the problem to the case of groups of matrices. It follows also that the global requirements concerning all elements of S can be replaced by 'local' conditions concerning the sets of elements of S of the same rank. The triangularizability problem for a subsemigroup S of the monoid Mn(K) has been considered by several authors, with a special emphasis on the case of the field of complex numbers, [56],[71],[112],[113],[133]. Here 5* is called triangularizable if there exists g G GLn(K) such that the semigroup g~~1Sg is upper triangular. Several sufficient conditions are known. In particular, in the case of characteristic zero or > n / 2 , a semigroup with constant trace must be triangularizable, [56],[133]. Permutability of the trace is also sufficient, [113]. The effect of certain spectral conditions on triangularizability was studied in [71],[112]. Note that, in the case of groups of matrices, several necessary and sufficient conditions are known over fields K of 'good' characteristic, cf. [71], [113]. Since we only set up the problem in terms of the structural approach, our results do not require any restriction on the field K. Recall that every maximal subgroup D of Mn(K) consists of units of the monoid eMn(K)e for some e — e2 G Mn(K), so it can be identified with GLt(K), where t = rank(e). Since every idempotent of Mn(K) is conjugate to a diagonal idempotent, triangularizability of gp(S Pi D) in Mn(K) is equivalent to its triangularizability in GLt(K).
4.4.
TRIANGULARIZABLE
SEMIGROUPS
127
If T is a uniform component of 5, then let ET denote the set of all idempotents that are the identities of maximal subgroups of Mn(K) intersecting T. In other words, e == e2 £ Mn(K) is an element of ET if and only if there exists t £ T such that £ = ete and rank(t) = rank(e). Let (ET) denote the semigroup generated by ET. S C Mn(K) is called unipotent if the eigenvalues of every s £ S lie in the set {0,1}. Some sufficient conditions for the triangularizability of unipotent semigroups were obtained in [56],[112]. These results will be improved in Corollary 4.33. We are now ready for our main result. T h e o r e m 4.31 The following conditions are equivalent for a semigroup S C Mn(K) 1. S is
triangularizable,
2. every uniform component T of S is a subsemigroup of 5, every (ET) is a unipotent semigroup and every nonempty intersection S fl D with a maximal subgroup D of Mn(K) is triangularizable, 3. for every uniform component T of S (ET) is unipotent, for every fl,6 G S, if rank(a) = rank(6) = rank(afr) = rank((a&) 2 ), then rank(a 2 ) = rank(a), and the groups associated to S are triangu larizable. Proof. Conjugating in Mn(K) we can assume that S is in the block up per triangular form of Proposition 4.4. We use the notation of the proof of this proposition. Let n : R —> Z = Y7i=i eiMn(K)e{ be the corre sponding block diagonal projection. Assume first that 1) holds. Then Z consists of diagonal matrices. Let a, 6 £ S be such that rank(a) = rank(fr) = rank(afr) = rank((a6) 2 ). From Theorem 3.5 it follows that a, 6, ah lie in a uniform component of S. Therefore, from Lemma 4.6 we know that 7r(a), 7r(6), 7r(a6) have rank equal to the rank of a. Since these are diagonal matrices, we also have rank(7r(a 2 )) = rank(7r(a)). There fore rank(a 2 ) > rank(7r(a 2 )) = rank(7r(a)) = rank(a), which implies that rank(a 2 ) = rank(a). Since every nonempty intersection S fl D with a maximal subgroup D of Mn(K) is upper triangular, so is the group gp(SC\D). (ET) is unipotent because ET is a triangular set of unipotent matrices. This proves 3). Assume that 3) holds. Every uniform component T of S inter sects all nonzero %-classes of a completely 0-simple semigroup T C
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4. IRREDUCIBLE
SEMIGROUPS
Mj/Mj-uj > 1. We know that f is of the form f ~ M{G,X,Y,P). Let a G T. Since aT contains a nonzero idempotent e G T, there ex ists b G T such that ahHe. Hence rank(afr) = rank((afr) 2 ), and clearly rank(a) = rank(afr) = rank(6). The assumption implies that a2 G T. This shows that all entries of the sandwich matrix P are nonzero. Con sequently T, viewed as a subset of 5, is a subsemigroup of S. Therefore 2) is a consequence of 3). Finally, assume that 2) holds. Let 7r2- be the projection onto a diago nal block Zi ~ Mm(K) such that 7rz(S') is irreducible as a subsemigroup of Z{. We aim to show that ra = 1. By Lemma 4.7 there exists a uniform component T of S such that 7rt-(T) U {0} is an ideal of 7rt-(5) that inter sects the same ^-classes of Mn(K) as a uniform component V of 7^(5). Moreover, by the hypothesis 7rt-(T) is a subsemigroup of iri(S). Since 7r,(T) U {0} is an ideal of ^ ( 5 ) , 7rt-(T) is irreducible in Z 2 . Let D, D', e be chosen as in Lemma 4.7. Then gp(7r2(T) C\ D) = gp(7r,-(T fl /)')) is an irreducible subgroup of the corresponding matrix ring eMn(K)e. Because T C\ D' is triangularizable, 7rz(T H Z)') is triangularizable and consequently gp(7r z (TnD / )) is triangularizable. Therefore we must have rank(e) = 1. This implies that 7rt-(T) C Mi, the ideal of matrices of rank at most 1 in Mn(K). Hence 7rt-(T) U {0} meets the same H-classes of Mn(K) as a completely 0-simple semigroup T' ~ Ai(G, X, Y, P ) , where G is a subgroup of the multiplicative group K* of the field K. But T is a subsemigroup of 5, so it is contained in a union of maximal subgroups of Mn(K) and ET,T must meet the same 'H-classes of Mn(K). Hence TT^ET) meets all ^-classes of Mn(K) that are intersected by 7Tt(T) (in particular 7r2(T) is contained in a union of maximal subgroups of Mn(K) consisting of matrices of rank 1). Also, since ni((ET)) is unipotent by 2), it must consist of idempotents. This means that 7rz((.ET}) = TT^ET) is a completely simple subsemigroup of X" over the trivial subgroup of G. Hence, if a, 6 G TT^ET) are in the same 7^-class of Mn(K), then as = s = bs for every s G 7r z (£' T ). This implies that at = bt for every t G 7rf-(T). Therefore (a - b)7Tt(T) = 0. But TT2(T) is irreducible. It fol lows that a = 6, and hence | X | = 1. Similarly, one shows that \Y\ — 1. Therefore iTi(ET) is a singleton, and consequently m = 1. This shows that 1) holds, completing the proof of the theorem. □ We note that, in conditions 2),3) of Theorem 4.31, the assumption that all semigroups (ET) are unipotent can be replaced by the require-
4.4.
TRIANGULARIZABLE
SEMIGROUPS
129
ment that they are triangularizable (clearly, triangularizability of the sets ET is sufficient). This is an easy consequence of the proof. Corollary 4.32 A semigroup S C Mn(K) is triangularizable if and only if its all uniform components are triangularizable, or equivalently, each of the nonempty sets S3 \ S3_uj — 1 , . . . , n, where S3 = S C\ Mj, is triangularizable. Proof. If T C Sj \ S3__i is a uniform component of 5 , then the semi group (T) generated by T is contained in T U Sj-i. Therefore, T is its uniform component. If every T is triangularizable, then so is (T) and Theorem 4.31 applied to (T) implies that condition 2) of this theorem is satisfied. The assertion follows. □ Corollary 4.32 shows that certain known sufficient conditions, for example permutability of the trace in the case of fields of 'good' charac teristic, [113], Theorem 1, can be weakened by requiring only that they hold for each of the sets Sj \ 5j-iThe theorem and the corollary remain valid in the more general con text of subsemigroups of Mn(F), where F is a division algebra. (Here, a G Mn(F) is called unipotent if a — e is nilpotent for the idempotent e G Mn(F) such that anl~Le in Mn(F)). This is a consequence of our proof and of the fact that a similar structure theorem holds in this case. (The only difference, not essential for the proof, is that the number of uniform components may be infinite and the nilradicals TV of Sj \ Sj-i are nil modulo Sj-i but do not satisfy N^J) C SJ-I in general. However, if S is triangular, then there are at most 2 n uniform components and each TV is nilpotent modulo the corresponding Sj_i, see Section 9.1.) If S is triangular, then a simple description of the uniform compo nents of S can be given. Namely, every component T is the collection of matrices in S with a fixed location of nonzero diagonal entries and such that rank(a) = rank(7r(a)), where n is the diagonal projection. This is an easy consequence of the structure theorem from Section 3.2. Corollary 4.33 Assume that S C Mn(K) is a unipotent semigroup. Then S is triangularizable if and only if the uniform components of S are subsemigroups of S. Proof. The necessity follows from Theorem 4.31. Since unipotent groups are triangularizable, the sufficiency follows as in the proof of
130
CHAPTER
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SEMIGROUPS
the implication 2) = > 1) in Theorem 4.31 (the proof works also if one assumes that 7rt-(T) is unipotent in place of the hypothesis that TT^ET)) is unipotent). □
Chapter 5 Identities In this chapter we study the influence of identities of certain types on the structure and properties of a linear semigroup. Identities in the coordinate ring K[xij\i,j = 1 , . . . ,n] of Mn(K), semigroup identities for S C Mn(K), and also polynomial identities of the semigroup algebra K[S] are considered. In what follows, if M is a class of groups, then we say that a group H is an almost Af-group, if H has a normal subgroup of finite index, which is an A/'-group. If every finitely generated subgroup of a group G is in AA, then G is called a locally A^-group.
5.1
Semigroup identities
Let / be an element of the polynomial ring K\t\,... , t m ] , where m > 1. If A is a subset of Mn(K): then we say that the identity / = 0 holds in A if for every matrices a^ = (a\j ) G A, k = 1 , . . . , m, we have / ( a i , . . . ,am) = 0inMn(/iQ. L e m m a 5.1 Assume that the identity f = 0 /io/ds m a subset A C Mn(,ftr), w/iere / G l f [ £ i , . . . , t m ] . T/ien / = 0 /io/ds m the Zariski closure A. Proof. For every i = 1 , . . . ,ra define A,
=
{ a € Mn(A0|/(ai,... ,am) = 0 for all dj G A, j < z, and all a*. G A, A: > i}. 131
132
CHAPTER
5.
IDENTITIES
It is clear that each A; is a closed subset of Mn(K) and A C Ai C A. It follows that Ax = A. Next, if At = A for some i > 1, then A C Al+1 C A. Therefore we must have A z+ i = A. By induction it follows that A m = A. This means that / = 0 holds in A. □ Let S be a semigroup. By an identity in S we mean an identity of the form v(xi,... , xm) = w(a?i,... , x m ) , where v, iu are any two differ ent words in a collection of indeterminates x\,... , z m . In particular, if 5 is a group, then we shall consider only identities of this type, for em phasis saying that S satisfies a semigroup identity. It is well known that existence of such an identity in S implies that there exists an indentity in two variables that is satisfied in S. This is an easy consequence of the fact that the free semigroup of finite rank ra, ra > 2, embeds in the free semigroup (x,y). As an example, consider the class of nilpotent semigroups introduced via a semigroup identity, independently in [77] and [86], see also [87]. Let x, y, ui, ?i2,. •. be elements of a semigroup S. Consider the sequences of elements defined inductively as follows £o = X, yo = y
We will sometimes write xm(x,y, ^ i , . . . , u m ) , y m ( x , y , U i , . . . , u m ) for ^m 7ym- S is a nilpotent semigroup if there exists ra > 1 such that x m — y-m f° r all x, y, u l 5 . . . , um E S. We will then say that the identity Xm = Ym holds in S. The minimal n with this property is called the nilpotency class of S and S is called m-nilpotent. If S is a group, this agrees with the classical definition. More generally, if S is cancellative, then S is ra-nilpotent if and only if it has a group of quotients that is ra-nilpotent, [77],[86],[87]. This class will be studied in detail in the next section. Corollary 5.2 Assume that a semigroup S C Mn(K) tity. Then S also satisfies this identity.
satisfies an iden
Proof. Let v(x\,... ,xm) — w(x\,... , x m ) be an identity satisfied in S. Lemma 5.1 applies to show that the latter identity is satisfied for all matrices in S. D
5.1.
SEMIGROUP
IDENTITIES
133
Next, we show that satisfying an identity is a local property of a semigroup S C Mn(K). L e m m a 5.3 The following conditions are equivalent for a semigroup S C Mn(K) 1. S satisfies an identity, 2. every nonempty intersection S Pi D with a maximal subgroup D of Mn(K) satisfies an identity, 3. every group associated to S satisfies an identity. Proof. Clearly, 2) is a consequence of 1). From Corollary 5.2 it follows, in view of Corollary 1.5, that 2) implies 3). Assume that 3) holds. First, we show that each factor T = Sj/Sj-i arising from Theorem 3.5 satisfies an identity. Let s,t E T. Then either stss £ D for a maximal subgroup D C Mj \ M3_i or it is zero in T. In the former case, s £ D. Let v(x,y) — w(x,y) be an identity satisfied in gp(5 Pi D) C D. Hence, in both cases we have v(stss,s) — w(stss,s) in T. It is clear that the words v(xxyx,x),w(xyxx,x) are different, so v(xxyx1x) = w(xyxx^x) is an identity in T. Therefore, it is enough to show that, if R and R/1 satisfy some identities for an ideal / of a semigroup /?, then R also satisfies an identity. Thus, assume fi(x,y) = gi(x,y) is an identity in R/I and f2(x,y) = g2(x,y) is an identity in / . We can assume that f2lg2 are words in x,y of equal length, replacing them by 72^2,^2/2, if necessary. If 5, t £ R and fi(s,t) — gi(s,t) £ R \ / , then the choice of / 2 ,g 2 implies that / 2 ( / i ( M ) 5 0 i ( M ) ) = 92(fi{s,t),gi(s,t)). It is easy to see that the two respective words in s,t are different, so f2{fi(x,y),gi(x,y)) = g2(fi(x,y),gi{x,y)) is an identity in R. □ We will need the following fundamental result on linear groups, re ferred to as Tits alternative. The proof can be found in the original paper [132] or in [83], see also [78], Appendix B. T h e o r e m 5.4 Let G C GLn(K) be a subgroup. Then either G has a solvable normal subgroup N such that G/N is periodic or G has a free nonabelian subgroup. If ch(K) — 0 or K is finitely generated, then either G is almost solvable or it has a free nonabelian subgroup.
CHAPTER
134
5.
IDENTITIES
The following theorem is due to Rosenblatt [118]. It is an extension of results on solvable groups of polynomial growth due to Milnor and Wolf, see [67]. T h e o r e m 5.5 Let G be a finitely generated solvable group. Then either G is almost nilpotent or it has a free noncommutative subsemigroup. We shall only prove this theorem in the special case where G is a linear group. To this end we need the following result. L e m m a 5.6 Let A be an abelian normal subgroup of a group G. Assume that G has no free noncommutative sub semi groups. For every g £ G and a £ A, let ak = gkag~k for k £ Z. Then there exists a natural number z and integers g 1 ? .. . ,qz £ {0,1, —1}, not all equal 0 such that a\Y • • • a\z — 1. Moreover, the group generated by the set \a\\ k £ Z} is finitely generated. Proof. Clearly, we may assume that a ^ 1. The hypothesis implies that for every x,y £ G there exist two different words w(x,y),v(x,y) in x, y such that w(x, y) = v(x, y) in G. Let w(x, y) — xniymi ■ • • xnsyms and v(x^y) — xklylx • • • xktylt where all exponents are nonnegative and ni^m^kjjj > 0 except possibly ni,ki,ms,lt. Cancellation on the left allows us to assume that nx > 0 and k\ — 0. Next, replacing w(x,y) by w(x,y)v(x,y) and v(x,y) by v(x,y)w(x,y), we can assume that the words w,v have equal length. Applying this to the elements g,ga we get a relation of the form 9ni(ga)^g^(ga)^
• ■■gn-{ga)m- = (gaf g^gaf
g** ■ ■•9kt(ga)1'
where £* =1 (n t - + mt) = Y?j=i(kj + If) with kY = 0. Since gk(ga)1 ak+iak+2 • ■ • a,k+ig , this relation can be rewritten as
=
( a n i + 1 a n i + 2 • • • a r i ) ( a r i + n 2 + 1 a r i + n 2 + 2 ■ • • «r2) * ' ' ( a r s _i+n s +i * ' ' ars)gTs = («/c1+i«A:1+2 • •' api)(aPl+k2+1aPl+k2+2
• ■ • aV2) • ■ • (aPt__l+kt+l
•••
aPt)gv\
where pj — k\ + h + • • • + &j + lj for j = 1, 2,. .. , £ ad r, = n\ + mi + • ■ • + nx + mi for i = 1,2,... , s. Then r s = p t , so the factors gr%gPt can be cancelled on both sides. Since A is abelian, 0 = k\ ^ ni and the sequences of indices appearing on both sides of this equality are
5.1.
SEMIGROUP
IDENTITIES
135
increasing, we get a relation of the desired form: a\l ■ • • aqzz = 1, with z — rs and q\ ^ 0. Choose z as small as possible. Then qz ^ 0. Clearly, z > 2 because otherwise ax = 1, and hence a — 1. We have shown that az is a word (allowing inverses) in a i , . . . , a^-i. Conjugating by g we see that a 2 + 1 is a word in a 2 , . . . , a z , and therefore a word in a,\ . . . , a 2 _i. An easy induction implies that every a*., k > 1, is a word in a i , . . . a 2 _i. Similarly, it follows that a\ is a word in a 2 , . . . , a z , so that a 0 is a word in a i , . . . , a z _i. Then a_i is a word in a 0 , . . . , o.z-i, and hence in a i , . . . ,az-\. Again, by induction it follows that every a -ki k > 1, is a word in a 1 ? . . . , a 2 _i. This means that the latter elements generate the group generated by the set {a^\ k 6 Z } . □ Proof of Theorem 5.5 for G C GLn(K). Suppose that G has no free noncommutative subsemigroups. From Malcev's theorem we know that G has a normal subgroup T of finite index which is triangularizable in Mn{K), see [134], Theorem 3.6. Then T is finitely generated, so replacing G by T and K by K, we can as sume that G C GLn(K) is upper triangular. We proceed by induction on n, the case n = 1 being clear. Let e i , . .. , en be the diagonal idempotents of rank one and let / be the identity matrix. Then the induction hypothesis implies that the projections G\ = (/ — ei)G(I — ex) and G 0, then it must be finite. Then CiC"1 is a root of unity, a contradiction. Therefore ch(K) = 0. Let L be a finitely generated subfield of K such that G C GLn(L). We can assume that L is a subfield of the field C of complex numbers. It has been shown that there exist r > 1 such that blx £ J2rj=i t^xZ for every i > 1. This implies that 6 is integral over Z.
CHAPTERS.
136
IDENTITIES
Then |6| ^ 1, since every element of C which is integral over Z and has absolute value 1 is a root of unity, see [8], p. 104-105. Now, replacing c by c1 or by c~l for i big enough, we may assume that |6| > 2. From Lemma 5.6 we know that for a natural number z and some integers 9i> • • • ? Qz € {0,1, —1} with qz ^ 0, we have a?1 • • • aqz* = I. Therefore qib + q2b2 + • • • + qzbz = 0. This contradicts the fact that |fe| > 2. Hence, suppose that Cic" 1 is a root of unity for every c G G. The set A of roots of unity in a finitely generated field L C K such that G C GLn(L) is finite by Lemma 1.7. It follows easily that F — {c G G| CiC'1 = 1} is a subgroup of finite index in G. It is clear that TV lies in the centre of F. But F/N C G/iV, so it is almost nilpotent. This implies that F is almost nilpotent. Therefore G is almost nilpotent. This completes the proof of the theorem. □ Recall that, if every finitely generated subgroup of a group G is almost solvable, then we say that G is locally almost solvable. It is well known that a linear group of this type has a solvable normal subgroup H such that G/H is a periodic linear group, see [134], Theorem 10.8. In particular, G is a periodic extension of a solvable group. Similarly, every locally almost nilpotent group G C GLn(K) is a periodic extension of a nilpotent group. This will be used in the following result, coming from [5]. The proof requires standard facts on the structure of algebraic groups.. All of these results are stated in Section 6.1. P r o p o s i t i o n 5.7 Let K be an algebraically closed field which is not the closure of a finite field. Assume that G is a closed connected subgroup of GLn(K). Then the following conditions are equivalent 1. G satisfies a semigroup
identity,
2. G is nilpotent, 3. G has no free noncommutative
sub semigroups.
Proof. As mentioned after Lemma 5.1, nilpotent groups satisfy semi group identities of some special type (see also Section 5.3). Hence 1) is a consequence of 2). Since 1) implies 3), it is enough to prove the implication 3) =^ 2). Thus, assume that G has no free noncommutative subsemigroups. Then G has no free nonabelian subgroups, so from The orem 5.4 it follows that G has a solvable normal subgroup H such that
5.1.
SEMIGROUP
IDENTITIES
137
G/H is periodic. Let R be the solvable radical of G (the largest closed connected solvable normal subgroup). Suppose that H ^ G. Then G/R is a semisimple algebraic group and H/(H f) R) is its solvable normal subgroup. The identity component of the Zariski closure of H/(H H R) is a closed connected solvable normal subgroup of G/R, so it must be trivial. Therefore H/(H D R) is finite. It follows that G/R is periodic. In particular, a maximal torus of G/R is periodic. This is impossible by the hypothesis on K. Hence we must have H = G and G is solvable. Now, Theorem 5.5 implies in view of the hypothesis that G is locally almost nilpotent. Therefore G has a nilpotent normal subgroup TV such that G/N is periodic. Replacing TV by the identity component of its closure we may assume that N is closed and connected. On the other hand, since G is connected, it is well known that the set Gu of unipotent elements of G forms a closed nilpotent subgroup of G and G is a direct product of Gu and a maximal torus T of G, see Theorem 6.3. Then GUN is a normal subgroup of G which is also closed, see Corollary 6.5. The algebraic group G/GUN is a periodic torus, hence it is trivial by the hypothesis on K. The sets 7VS, Nu of semisimple, respectively unipotent elements of N are subgroups of TV and N — Ns x Nu by Theorem 6.3. Since N is a normal subgroup of G, Ns is a normal subgroup of G. But G — GUN — GUNS implies then that G = Ns x Gu is nilpotent. This completes the proof. □ It is clear that the assertion of Proposition 5.7 does not hold for K = F p . Namely, consider G — GLn(Fp). Corollary 5.8 Let S C Mn(K) tions are equivalent 1. S satisfies an
be a semigroup.
The following
condi
identity,
2. every group associated to S is almost
nilpotent.
Proof. Extending K, if necessary, we may assume that K is alge braically closed and it is not the closure of a finite field. If 1) holds, then by Corollary 5.2 5 satisfies an identity. From Corollary 1.5 it follows that every group H associated to S is contained in a max imal subgroup G of ~S. But G is the group of matrices of maximal rank in eMn(K)e f) 5 , where e = e2 G G (see Remark iv) before
CHAPTER
138
5.
IDENTITIES
Proposition 3.6). Therefore G is a closed subgroup of a maximal sub group of Mn(K). Hence Proposition 5.7 implies that G is almost nilpotent. Therefore H is almost nilpotent and 2) follows. On the other hand, every almost nilpotent group satisfies an identity of the form xm(xk,yk,u\,... ,ukm) = yrn{xk,yk,uk,... ,ukm), where xm,yrn are the words defined after Lemma 5.1. Therefore, the implication 2) => 1) follows from Lemma 5.3. □ We note that the results of this section will be strengthened in Chap ter 6.
5.2
Almost unipotent semigroups
Our aim in this section is to describe two classes of semigroups S C Mn(K) defined by identities over the coordinate ring K[%ij\i,3=\,...,nJ a s considered in Lemma 5.1, of some special types. We follow [93] and [100]. ^ Let K be the algebraic closure of the field K. An element s G Mn{K) is called unipotent if the eigenvalues of the matrix s G Mn(K) lie in the set {0,1}. If there exists k > 1 such that sk is unipotent, then s is called almost unipotent. A semigroup S C Mn(K) is almost unipotent if every group asso ciated to S is an almost unipotent group. Clearly, this condition is stronger than that considered in Corollary 5.8. Hence, every semigroup of this type satisfies an identity. This property can be described by a number of equivalent conditions. P r o p o s i t i o n 5.9 The following conditions are equivalent for a linear semigroup S C Mn(K) 1. S is almost
unipotent,
2. for some m > 1 the identity xm(I — x m ) n = 0 holds for x G 5, where I denotes the identity matrix. 3. the set of eigenvalues of elements of S is finite, 4- S is conjugate to a subsemigroup of Mn(K) that is block upper triangular with finite projections onto the diagonal blocks.
5.2. ALMOST
UNIPOTENT
SEMIGROUPS
139
Moreover, if K is finitely generated, these conditions are satisfied if and only if. 5. every element s 6 S is almost
unipotent.
Proof. Assume that 1) holds. Since S has at most 2n uniform compo nents, Corollary 3.4 implies that there exists m > 1 such that for each s e S the matrix sm lies in a subgroup Ds of Mn(K) and is unipotent. Let e = e2 G Ds. Conjugating e to a diagonal idempotent we see that (/ - sm)n = I-e. Since sm = sme, the identity of 2) follows. Therefore 1) implies 2). It is clear that 3) is a consequence of 2). Assume that 3) holds. By Lemma 4.4 S is conjugate to a block upper triangular subsemigroup T of Mn(K) whose projections T i ? i , . . . , Tr>r, r < n, onto the diagonal blocks are absolutely irreducible or zero. From Proposition 4.9 it follows that each Ti>t- must be finite if 3) holds. Hence 3) implies 4). Assume that 4) holds. Let G be the group generated by T Pi D for a maximal subgroup D of Mn{K). The block diagonal projections (Tr\D)ij, z = 1 , . . . , r, of TOD are finite subsemigroups of the respective projections G{j of G. Since (T H D)i^ generates G2)Z-, we must have (TCiD)ij — G{}{. Hence each G;jt- is finite and G must be a finite extension of a unipotent group. This implies that 1) is satisfied. If s £ Mn(K) is almost unipotent, then the nonzero eigenvalues of s are roots of unity. It is known that there are finitely many roots of unity that are roots of polynomials of degree n over a given finitely generated field, see Lemma 1.7. Therefore, for such a field 7\, 5) is equivalent to 3). This completes the proof. □ Note that condition 2) of the above proposition implies that the Zariski closure S of S also is an almost unipotent semigroup, because it satisfies the same identities as S by Lemma 5.1. R e m a r k Let Z l 5 . . . , Zr be the /i-subalgebras of Mn(K) corresponding to the diagonal blocks determined by T ^ i , . . . , Tr>r of the above proof. That is, Zi is spanned by TM- if Tz-jt- is irreducible and Z{ is the ap propriate 1-dimensional diagonal block if Tt-jt- is a zero semigroup. The block diagonal projection ix : T —> Z = Z\ x . . . x Zr is a homomorphism that gives a Z-gradation on T. The components of idempotents
140
CHAPTER
5.
IDENTITIES
_1 Z, are subsemigroups of T. T(e)=n) = 1^(e),e=e ( e ) ,2 e = e2 G GZ, T. Since each e can be diagonalized by conjugating within Z, T(ej is conjugate to a semigroup V such that each projection VJt- is a diagonal idempotent in Z,. Therefore, V is an upper triangular semigroup which is unipotent. Let A be the semigroup of all upper triangular matrices with diagonal equal to the diagonal of e. Then J = {a G A\ rank(a) = rank(e)} is an ideal of A. From Theorem 3.5 it follows that J is a uniform component of A. If D is the maximal subgroup of Mn{K) containing e, then JflDisa group. Hence J is a completely simple semigroup, which is the principal factor of e in A (see Proposition 3.1). Thus, from Theorem 3.5 it follows that / = J D V is a uniform subsemigroup of J, whose associated group is unipotent. Therefore T(e) has an ideal 7(e), which is its uniform component, whose associated group is unipotent and such that (T(e))* C I(e) for some Jjfc > 1. All this should be compared to the semilattice decomposition of upper triangular subsemigroups of Mnn(K), [108], Corollary 1.16, Theorem 5.12, Corollary 6.32.
Assume now that S = (au... ,at)C Mn{K) is an almost unipotent semigroup that is in the block upper triangular form of 4) in Proposition 5.9. Let e3 be the identity of the diagonal block Z3,j = 1 , . . . ,r. For k < j write x^J = e^e^ ekaie3, so that a{ = £fc £1^=1 ^=1 k<' ^ . . We define the sets C{ti = e;oe,-, C^i+i = CiiXi)i+iCi+1
= Ct,i+iCi+1,j U . . . U Cij-tCj-u
U
ClXijC^.
It is clear that each Cit3 is a finite set (induction). Let D{j,i < j , be the fi-submodule of the ring Mn{K) generated by dj, Ct,J, where R = Z if ch(A') = 0 and R is the prime subfield of K otherwise. L e m m a 5.10 With the above notation we have 1. IfsGS,
then elS se3 2 G D.j for %e
i<j.
2. Ifs = ahil...alm se3 is a sum of G { 1 , . . . ,t}, and i < j , then et%se, im,iq at most hj-i(rn) , where h is a polynomial h^t(m) elements of the set Cl3 k l3> in m of degree k, for k = l , . . , r c - l!
5.2. ALMOST
UNIPOTENT
SEMIGROUPS
141
Proof. 1) follows from the block multiplication of the matrices in S. Namely, for s G 5, by induction on the length m of s in the generators a i , . . . , at of 5, we show that, if z < j , then e ^ e , G A,z+i A-+1.J + • • - + A . i - i ^ j - i . j + CliXijCjj
C DM.
For m = 1 the assertion is clear. Let g G { 1 , . . . ,£}. Then i e t -(sa g )ej = 5^(ei5e/)(e/a g ej). From the definitions it follows easily that CPiPCPik C CPik for p < A;, and also Cv,kXk,r Q Cv,kCk,r Q Cp}r, for p < k < r. Since by the induction hypothesis (eisei)(eiaqej)
G (A'^+iA+i,/ + • • • + A , / - i A - i , / +
C^X^iCl^X^^
the inductive claim follows. This implies that 1) holds. The above block multiplication pattern easily implies that to get the desired bound one can take h0(m) = 1, /i^(l) — 1 for k — 1 , . . . , n — 1, and inductively hk(m) = ^(ra- — 1) + hk-i(m — 1) + . . . + /i 0 (ra — 1) for k > 0,ra > 1. By induction on k we show that hk(m) is a polynomial in m of degree A;. For k = 0 this is clear. Let /c > 0. The induction hypothesis implies, in view of the definition, that hk(m) — hk(m — 1) is a polynomial of degree k — 1. It is well known, and easy to check, that in such a case hk(m) is a polynomial of degree &, see [67], Lemma 1.5.
□
Corollary 5.11 If ch(K) > 0, then every almost unipotent semigroup S C Mn(K) is periodic. Moreover, if it is finitely generated, it must be
finite. Proof. If a G 5, then there exists k > 1 such that ak G D for a maximal subgroup D of Mn(K) and a^ is unipotent. Therefore afcr is the identity of Z), where r is a power of the characteristic of J\, so S is periodic. Assume that S is finitely generated. Since that sets dj are finite, Lemma 5.10 implies that S must be finite. Note that the latter follows also from the more general assertion of Theorem 7.3. □
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P r o p o s i t i o n 5.12 Let S C Mn(K) be a finitely generated almost unipotent semigroup in the block upper triangular form of 4) in Proposi tion 5.9. If Cij^Dij have the earlier meaning, then 1. the set E = {a £ Mn(K)\ eiaei £ C^-, eiae3 £ Dij for i < j} is a n-regular almost unipotent semigroup containing S. 2. the groups associated to S are finitely
generated.
Proof. We have seen that E is a semigroup containing S. From Propo sition 5.9 it follows that E is almost unipotent. Let D be a maximal subgroup of Mn(K) such that E Pi D ^ 0. We know that the group G generated by E H D has a subgroup V of finite index that is unipotent. The finiteness of the index easily implies that E C\ V generates V as a group, see [87], Lemma 7.4. Let u £ Ef)V. Then (u — e)n = 0, where e is the identity of V. Let R = Z if ch(K) = 0 and R = Fp if ch(/\) = p > 0. Hence e is an .R-combination of u , . . . , un. If u~l denotes the inverse of u in Z), then we get eiu~le3 £ Yl?=i ezulejR C DZjJ for i < j . Since every Ct',i = ez'5'e2 is a finite semigroup, the projection elGel is a finite group, so we also have ez-u-1e2- £ CZ)2-. Therefore u _ 1 £ £7, which shows that y = g p ( £ : n l / ) C £ . Since £ n D generates G and [G : V] < oo, it follows that G C E. This proves 1). From Lemma 4.6 it follows that, conjugating in the Ji-subalgebra of Mn(K) that is block upper triangular with the block pattern of 5, we can assume that e is a diagonal idempotent. We can also assume that S = E because each group of the form G as above is a subgroup of E and a subgroup of a finitely generated almost nilpotent group is finitely generated. If exeex — 0 then G C (/ — e 1 ) M n ( / i ) ( J — e ^ , so that one can proceed by induction on the number r of diagonal blocks. (G is finite if r = 1.) This allows to assume that e\eei 7^ 0. Similarly, we can assume that ereer ^ 0. It is enough to show that the block upper triangular unipotent subgroup W — {g £ G\ eigei = eieei for i — 1 , . . . , r } of G is finitely generated (note that [G : W] < co.) Let F = e + eDire. Since Dir is a finitely generated /^-module, the same is true of eDi ) r e, whence F is a finitely generated group. Therefore FC\W also is finitely generated. On the other hand, induction shows that W/(FP\ W) is finitely generated, because it is isomorphic to a subgroup of Gi x G 2 , where G, = (I - ex)G{I - e x ), G2 = (I - er)G{I - er) are
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the corresponding projections of G. Hence G is finitely generated, as desired. □ Our next aim is to discuss a more general class of semigroups, those that are 'almost solvable'. It turns out that this class also is determined by identities over K[xij]i,j=i,... of some special type. P r o p o s i t i o n 5.13 Assume that the associated groups of a linear semi group S C Mn(K) are almost solvable. Then the maximal subgroups of S are almost solvable. Proof. Let C/j, Wj,j — 1, 2 . . . , be words in #, y defined inductively as follows Ux = x, Wi= y, UJ+l = Uf, WJ+1 = UJWJUJ for odd j , U3+l = WJUJWJ, W3+l = Wf for even j . We claim that, ifx,y £ Mn(K), then u = U2n+i,w = W2n+i lie in a max imal subgroup D of Mn(K). Since rank([/ t - +1 ) < rank([/,), rank(U l+1 ) < rank(W / z ), and rank(M/ 2+1 ) < rank(H / 2 ),rank(M / t + i) < rank(f/,-), it is clear that either f/2n+i = W2n+i = 0 or there exists i < 2n — 1 such that rank(C/t-) = rank(W^) = rank(C/l-+i) = rank(W 2 + i). In view of the defini tion, this implies that f/;+i, Wi+i lie in a maximal subgroup of Mn(K), and consequently Uj,Wj lie in this subgroup provided that j > i. This proves the claim. By the hypothesis and Malcev's theorem, [134], The orem 3.6, the group H generated by S fl D is almost triangularizable. From Corollary 3.4, we know that there are finitely many conjugacy classes of linear groups arising from S in this way. Therefore, there exists N > I such that if x, y £ 5 , then [c(ab— ba))n = 0 for every a,b,c £
(uN,wN).
If a, 6, c are fixed words in x, y and £, y run over 5 , then this is a poly nomial identity which is satisfied in S. Therefore, by Lemma 5.1 this identity must be also satisfied for every x,y £ S. Let G be a maximal subgroup of S and let e = e2 £ G. Suppose that x , y £ G generate a free nonabelian subgroup. Then uN\wN also generate a free nonabelian subgroup X (as, clearly, these two words in x,y do not commute, and every subgroup of a free group is free). The above implies that the com mutator ideal J of the /i-subalgebra A generated by uN, wN in Mn(K)
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has a basis consisting of nilpotents. It is well known that J must be then nilpotent (for example, use the fact that the trace of every a G J is zero and apply Lemma 4.1 and Proposition 4.4). From Corollary 1.5 it follows that e £ A is the identity of A and X lies in the group of units A* of A. However, the nilpotency of J easily implies that A* is solvable, a contradiction. In view of Theorem 5.4 this shows that G has a solv able normal subgroup H such that G/H is periodic. Since K can be chosen nonalgebraic over its prime subfield, the connected component Gc of G is solvable (otherwise, the semisimple group GC/1Z(GC), with 1Z(GC) denoting the radical of G c , would have a periodic torus, which contradicts the choice of K). Hence G is almost solvable, as desired. □
We shall see in Chapter 6 that, if a linear semigroup S has no free noncommutative subsemigroups and ch(K) — 0 or K is finitely gener ated, then the groups associated to S are almost solvable. Hence, the maximal subgroups of S are almost solvable by Proposition 5.13. Let G = G L n ( F p ) , n > 1, where F p is a proper subfield of K. Then G is periodic, but the Zariski closure G' of G in GLn(K) is not a periodic extension of a solvable group. Namely, G' contains the group B' C GLn(K) of upper triangular matrices, because B' is the closure of B' fl G L n ( F p ) , so G' = GLn(K) by Theorem 2.7. Therefore, 'almost solvable' cannot be replaced by ca periodic extension of a solvable group' in the statement of Proposition 5.13.
5.3
Nilpotent semigroups
In this section we discuss nilpotent semigroups, as defined in Section 5.1 via an identity Xm = Ym for some m > 1. Thus, the reader should be warned that the term 'nilpotent' used here is more general than that in the foregoing chapters, where it was used for a semigroup S with zero 9 such that Sm — 6 for some ra > 1. (To make a distinction, the semigroups of the latter type will be sometimes called power nilpotent.) It is known that every nilpotent cancellative semigroup S has a group of classical quotients G which is nilpotent and of the same nilpotency class as S. Groups, and linear groups in particular, satisfying certain related semigroup identities, introduced in [124], have been recently studied in [6], [105], [123]. In particular, finitely generated residually finite groups
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satisfying a semigroup identity must be almost nilpotent, [123]. On the other hand, the results of Section 5.1, which will be strengthened in Chapter 6, show that a linear semigroup S C Mn(K) satisfying any identity has a strong flavour of nilpotency. A natural question that arises here is to decide which of these semi groups are nilpotent. Because of the powerful classical theory of nilpo tent linear groups, cf.[134], one can also ask whether such semigroups can be approached via group theoretical methods. We note that the very special case of nilpotent connected algebraic monoids has been recently considered in [39]. The aim of this section is to develop a structural approach to nilpo tent semigroups of matrices. First, we find a structural characterization of such semigroups. Next, we show that the smallest 7r-regular subsemigroup cl(S') C Mn(K) containing S is very close to S. Namely, cl(S') is also nilpotent, 5, cl(5) intersect the same maximal subgroups of Mn(K) (even more: S intersects all %-classes of cl(5) contained in regular ^/-classes of cl(S')) and H — gp(S fl H) for every maximal sub group H of c\(S). Moreover, there are at most 2 n maximal subgroups of Mn(K) intersected by S. These are very special properties, which allow in view of the general structure theorem discussed in Section 3.2, to apply group theoretical tools in the study of an arbitrary nilpotent semigroup S C Mn(K). The results of this section come from [94]. Recall that for a subset A of a semigroup S we denote by E(A) the set of idempotents of A. First, we collect a few simple observations on nilpotent semigroups. L e m m a 5.14 1. Let J be an ideal of a semigroup S with zero 0. If there exists k > 1 such that (aS)k = 0 for every a £ J, and S/ J is nilpotent, then S is nilpotent. 2. if S is a uniform subsemigroup of a completely 0-simple semigroup M and S is the completely 0-simple closure of S in M, then S is nilpotent if and only if S is a Brandt semigroup over a nilpotent group, that is S ~ M.(G, Z, Z, / ) for a nonempty set Z, a nilpotent group G and the Z x Z identity matrix I. 3. if S C Mn(K)
is nilpotent, then cl(5) C S are nilpotent.
Proof. 1) Assume that S/J is m-nilpotent. So, if x,y,ui1u2,... then either xm = ym or x m , y m £ J. If q > 1, then the word
£ 5, Xm+q
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contains 2q 1 copies of the word X^ separated by some u;, Y{. Therefore, in the latter case xm+q,ym+q = 9 whenever 2q~l > k. This proves 1). 2) If ef = fje = e , / ^ e for some e , / G £ ( S ) \ {#}, then, for every ra > 1, x m ( e , / , e , . . . , e ) , y m ( e , / , e , . . . , e) are not in the same Kclass of 5 . Therefore, choosing s,t £ S such that sT/e and f H / , we see that x m ( s , £ , s , . . . ,5) and y m (s, /, s , . . . , s) are not in the same 'H-class, hence they are not equal. This implies that each 7£-class of 5* has only one idempotent if S is nilpotent. A similar argument works for the Cclasses. By [77], every maximal subgroup G of S is nilpotent because it is generated by the nilpotent semigroup G D S. It is well known that S is of the desired form in this case, [15]. Conversely, assume that S is a Brandt semigroup over a nilpotent group G. Let r = max(m,2), where m is the nilpotency class of G. Choose x, y, u 1 ? . . . , ur G S. If xuiy = 0 or yuxx — 9, then x2 = 0 = y2. Suppose that xuiy,yuxx are nonzero. Write x — (g,w,z),y — (h,w',z') for g,h G G and w,w',z,z' E Z. The assumptions imply that w = w' and z = z'. Hence, we must have xtT~Lyi for every i > 1. If xr ^ ^, it follows also that all wt- are in the same 'H-class H of S - the only one that satisfies x i / y 7^ ^- Therefore w = t {fit z,w) for some /z- G G. Since G is m-nilpotent, this easily leads to xr — yr (because this equality reduces to the corresponding equality in G with g, h in place of x, y and /,- in place of U{). It follows that S is nilpotent of class at most max(m,2). This proves 2). 3) From Corollary 1.5 we know that cl(5) C S. The latter semigroup satisfies the same identities as S by Corollary 5.2. □ Since Mn{K) has no infinite chains of idempotents, the principal factors of every regular semigroup S C Mn{K) are completely 0-simple (or completely simple). Hence, if S is also nilpotent, then Lemma 5.14 implies that S is inverse, so that E{S) is a commutative semigroup. However, if S is not regular, then E{S) need not be a subsemigroup of S. For example, 5 = { ( \
\ ) >{\
\ ) \ \
J ) >0> = < ^ ) >
^
M2{K) is nilpotent by assertion 1) of Lemma 5.14. If / is a principal right ideal of a subsemigroup S of MnjM3_x such that / C Sj/Sj-i and / is nil, then it is easy to see that I2 — 9. In this context assertion 1) of Lemma 5.14 (with k = 2) will be later applied. Note that it was mistakenly stated in [69], Proposition 2.3, that a nilpotent regular semigroup must be a semilattice of nilpotent groups
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(though, a slightly different definition was used there). Our first aim is to look at the closure of a nilpotent semigroup S. We will prove that cl(5) is very close to S. In fact, Theorem 5.15 shows that they are as close as possible in terms of the uniform components and the associated groups. Namely, every uniform component of c\(S) is the completely 0-simple closure of a uniform component of S. T h e o r e m 5.15 Assume that the completely 0-simple closure of every uniform component of a semigroup S C Mn(K) is a Brandt semigroup. Then 1. every uniform component of c\(S) intersects the same of Mn(K) as a uniform component of S,
H-classes
2. for every maximal subgroup G of c\(S) we have G — g p ( 5 f) G), 3. S intersects at most 2n maximal subgroups of
Mn(K).
This applies in particular to every nilpotent semigroup S C
Mn(K).
Proof. Let Ej be the set of nonzero idempotents of all Brandt semi groups mentioned above and contained in Mj/M3_i. From Proposi tion 3.6 we know that Ej is a triangular set of idempotents. Hence \Ej\ < ( n ) by Lemma 2.11. This implies that 3) holds. Now, assertions 1),2) follow from Proposition 3.15, Proposition 3.14 and its left-right dual. The last assertion is a direct consequence of Lemma 5.14. □ Now we will focus on irreducible nilpotent semigroups. We say that a G Mn(K) is a monomial matrix if each row and each column of a con tains at most one nonzero entry. S C Mn(K) is a monomial semigroup if S is conjugate to a semigroup consisting of monomial matrices. More generally, if 1 = e\ + . . . + er for some mutually orthogonal idempotents e2- £ Mn(K), and for every i there exists at most one j with e{sej ^ 0, and for every j there exists at most one i with e{Stj ^ 0, then we say that 5 is a block monomial matrix with respect to e 1 ? . . . , er. This allows us to define block monomial semigroups S C Mn(K). P r o p o s i t i o n 5.16 Assume that S C Mn(K) semigroup. Then
is an irreducible
nilpotent
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1. cl(5) is block monomial with respect to idempotents e i , . . . , er such that E(cl(U)) \ {0} = {ei,. . . , er} for an ideal U of S. Moreover, if D is an algebraically closed field, then S is monomial. 2. S and cl(S) intersect the same H-classes of
Mn(K).
3. E(cl(S)) is a commutative subsemigroup ofc\(S) and it contains at most frJ idempotents of rank qn/r for every q £ { 1 , . . . ,r},butno nonzero idempotents of other ranks. Consequently \E(cl(S))\ < 2 r and S intersects at most 2r maximal subgroups of Mn(K). Proof. First note that the set U of matrices of the least nonzero rank in S (with zero if it is in S) is an ideal of S that is a uniform component of 5, see Proposition 4.3. Clearly, U also is irreducible. U can be viewed as a subsemigroup of Mn(K) and the proof of assertion 2) of Lemma 5.14 shows that it is a Brandt semigroup or a group (if 0 £ S). In the latter case, the irreducibility of U implies that U C GLn(K), so that S C GLn(K). Hence, adjoining 0 to S if necessary, we may assume that U C U = c\(U) C Mn(K) is such that H = g p ( # H U) for every maximal subgroup H of cl(C/), U intersects all 'H-classes of cl(£/), and cl(U) ~ M{G,Z,Z,I) for a set Z and a nilpotent group G. Since the idempotents of cl(S) are orthogonal in Mn(K), we have \Z\ — r < n. Let e = ei + . . . + e r be the sum of all nonzero idempotents of c\(U). Since U is irreducible, cl(U) also is irreducible. This implies that e = 1 because c\(U) — ec\(U)e. Hence 1 = ex + . . . + er is a decomposition into a sum of orthogonal idempotents. Conjugating in Mn(K) we can assume that e i , . . . , er are diagonal. Let 5 G S. Since S' = S U c\(U) is a subsemigroup of Mn(K) by Lemma 3.11, et-s,,set- G cl((7) for every z. If e^6 ^ 0, then e ^ = eise3 for a unique j . Therefore, in the block decomposition s — Y?i j=\ eisej, for every i there exists at most one j such that e^se^ / 0. Similarly, for every j there exists at most one i with ez-5ej ^ 0. That is, s is block monomial with respect to e i , . . . , e r . If additionally Z) is an algebraically closed field, then the irreducibility of c\(U) implies by Proposition 4.3 that e2-cl(f/)e2- is irreducible as a subsemigroup of ezMn(K)ei — Mt(K),t = rank(e ? ). But it is a group with zero, so it is monomial by [134], Theo rem 1.14. Since e i , . . . , er are diagonal, after an appropriate conjugation in M n (A"), all diagonal blocks e; cl([/)e; consist of monomial matrices. Assume that x = e^xe^ £ cl((7) is nonzero. Let ej; = e^e^e,- be the
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149
matrix whose only nonzero block is the identity t x t matrix. Then eji G cl(f7) and 0 ^ xe3{ G cl(C7). This implies that x is a monomial matrix. Hence, cl((7) is monomial. Let A be the set of all matrices a G Mn(K) that are block mono mial with respect to e i , . . . , er and with ezaej G e» cl(f/)ej for every z, j . Denote by B C GLt(K) the group corresponding to the maximal subgroup e 1 cl([/)e 1 \ {0} of c\(U). Clearly S C A and k = gt, for <7 = 1 , . . . , r, are the only ranks of nonzero matrices of A. It is easy to see that {An Mk)/{An Mk_x) ~ M{Gq, Q , Q , / ) where G9 is isomorphic to the group of block monomial matrices of rank qt contained in (ex + . . . + eq)Mn(K)(ei + ... + eq) and with nonzero blocks contained in B. Therefore A is 7r-regular (even regular) and so c\(S) C A. This completes the proof of 1). Moreover, A = RM, where M is the group of invertible matrices in the monoid Y7i=i ^i^ei and R is the corresponding inverse monoid (isomorphic to the full symmetric inverse submonoid in Mr(K)). Let tp : A —> R be the natural homomorphism. In other words, ip(a) is the element of R having the same location of nonzero blocks as a and all nonzero blocks equal to the identity t x t matrix. It is easy to see that the inverse image in A of any maximal subgroup H' of R is a subgroup H of A (in fact, H is the set of all matrices a G A with the location of zero blocks identical to the location of zero blocks a matrix a' G H'). In particular, every maximal subgroup of A is the inverse image of a maximal subgroup of R. For the same reason, every %-class of A is the collection of all matrices of A with the patterns of nonzero blocks of some fixed types. Since
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corresponds to the group of diagonal invertible matrices. R e m a r k Recall that S C Mn(K) is completely reducible if Kn is a direct sum of irreducible K{S} - modules. In this case S is conjugate to a block diagonal semigroup S' C Mn(K) whose diagonal blocks are irreducible. Proposition 5.16 implies that Sf is block monomial with respect to the set of idempotents coming from all irreducible blocks. It can be checked that the proof of the second part of Proposition 5.16 extends to this case, showing that 5, cl(5) intersect the same %-classes of Mn(K). If a G Mn(K) is a monomial matrix, then a n! is diagonal. This is clear for a G GLn(K). If rank(a) < n, then b = an G G for a maximal subgroup G of Mn(K). Then W[ is diagonal, where j = rank(6) (compare Lemma 2.6). It follows that indeed a n! is diagonal. Corollary 5.17 If S C Mn(K) is a nilpotent semigroup, then S is a triangularizable - by - (periodic of bounded index) subsemigroup of Mn(K). Proof. S is conjugate to a semigroup S' C Mn(K) which is block up per triangular and whose diagonal blocks are irreducible or zero. By Proposition 5.16 1) we may assume that S consists of monomial matri ces. Therefore, for any s G S, every projection of snl onto a diagonal block is a diagonal matrix. Hence, the assertion follows. □ Let T be a uniform component of a semigroup S C Mn(K) such that the completely 0-simple closure T of T is a Brandt semigroup with finitely many idempotents. Let e l 5 . . . , er be the idempotents of the maximal subgroups i ^ , . . . , Hr of Mn{K) intersected by T and k the rank of matrices in T. We know that the semigroup ST — (S, gp(STl H)) intersects the same 'H-classes of M& \ Mk-\ as 5*, and T is a uniform component of ST-> see Lemma 3.11 and the remark following it. Then ST = {S \ Sk-i) U T has a natural semigroup structure and it is a Rees factor of ST- Moreover T is an ideal of ST- It is well known that the sum of idempotents of T is the identity of the contracted semigroup ring Ko[T] and Ko[T] ~ Mr(K[G]) for the maximal subgroup G of T, see Section 4.2. Consider the mapping « T : K[ST] —> Ko[T] de fined for s G ST by ax(s) — e^s, where e^ = ex + . . . + er G A ' [ 5 T ]
5.3.
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and a is the image of a G K[ST] in K0[ST]- It is a homomorphism because ~e~r is central in A ^ S r ] - It is easy to see that ax(s) has a mono mial pattern with respect to eT,... ,e7, so that the nonzero entries of a?(s) as an r x r matrix determine a 'quasi permutation' of the set A = { 1 , . . . , r } (a map a : A —> AU{0} that is one-to-one on cr _1 (A)). We say that ar(S) C Ko[T] has no inversions if for every s,t G S and e t -,ej,e p ,e g G E(T) \ {9} with et- ^ ej,e p ^ e g , at least one of the ele ments aT(eqsej),aT(eqtei),aT{eptej),aT(epsei) is zero (or equivalently, one of the matrices eqsej,eqtei,eptej,epsei has rank < k). Recall that we often identify a uniform component T of S with the subset of the corresponding Sk \ Sk-i- We are now ready for a structural description of nilpotent semigroups. T h e o r e m 5.18 The following conditions are equivalent for a semigroup S C Mn(K) 1. S is nilpotent, 2. the groups associated to S are nilpotent and for every x, y,u,z G 5, if x,y,xuy,yux,xzx,yzy lie in the same uniform component of S, then xl-Ly in M n ( / \ ) , 3. the groups associated to S are nilpotent, the completely 0-simple closure of every uniform component T of S is a Brandt semigroup with finitely many idempotents and CYT(S) has no inversions. Moreover, if k denotes the maximum of the nilpotency classes of the groups associated to 5, then there exists a function / ( n , k) such that the nilpotency class of S is bounded by f(n,k). Proof. Assume that S is nilpotent and there exists a uniform compo nent T of S containing x,y,xuy,yux,xzx,yzy for some x,y,u,z G S. Let u2m = z and u 2 m _i = u for m — 1, 2 , . . . . Since in a completely 0-simple semigroup ab ^ 0,bc ^ 6 imply abc ^ 0, it follows that £m = xm(x,y,uu... ,um) G T and ym = ym(x,y,uu... ,um) G T. Therefore xmT-Lx and ymT-ty if m is even. But the identity Xm = Ym holds in S for some m. Hence xl-iy in Mn(K). Since H — g p ( 5 fl H) for every group H associated to 5, H must be nilpotent, [77]. Therefore 1) implies 2).
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Assume that 2) holds. Suppose first that a uniform component T of S intersects two different maximal subgroups H, G of Mn(K) that are in the same 7^-class of Mn(K). Let x,u,z G S H H,y G S D G. Then xuy.yzy G G,yux,xzx G JJ. This contradicts 2). A similar argument shows that T does not intersect different £-related maximal subgroups of Mn(K). Thus, in view of Theorem 5.15, the second part of 3) is satisfied. Suppose that a ^ ( 5 ) has an inversion. Let { e l 5 . . . , e r } be the idempotent s of the maximal subgroups of Mn(K) intersecting T and let I = rank(ei). Then there exist i , j , p , g , i ^ j,p / q, such that rank(e g sej) = rank(e9£et-) = rank(e p te J ) = rank(epse2-) = / for some s,t e S. Choose x,y eT such that x = etxep,y = e3yeq. Then ysy = yeqse3y G T,ytx = yeqte{x G T,xty = xevte3y G T,xsx — xevse{x G T. This contradicts 2). Therefore 3) follows. Assume that 3) holds. To prove 1) we may assume that 0 G 5, adjoining 0 to S if necessary. From the general structure theorem de scribed in Section 3.2 we know that S has an ideal chain S = T\ D . .. D T m , m < 2 n + n, such that T m and each factor Ti/Ti+i is either a nonzero nil subsemigroup of a completely 0-simple semigroup or it is a uniform component of S with a completely 0-simple closure that is a Brandt semigroup over a nilpotent group and with finitely many idempotents. We show by induction on j that every factor S/Tj is nilpotent (under the remaining assumption of 3)). Since the linearity of S/Tj will not be needed, we simply assume that S/Tm is nilpotent (this takes care of the case m = 1 as well). If Tm is a nil semigroup, then assertion 1) of Lemma 5.14 applies with k = 2, showing that S is nilpotent. Thus, assume that T = Tm is a uniform component of S. As mentioned above, it is known that Sf = SUT has a semigroup structure extending that of S. Then K0[T] ~ Mr(K[G]),r > 1, is a ring with an identity element and it is an ideal of Ko[S']. Therefore it is a ring direct summand of /iro[5'/] and the centre of Ko[T] is contained in the centre of K0[S']. Identifying K0[f] with Mr(K[G]), we see that the set Z of invertible matrices in the centre of Mr(K[G]) lies in the centre of the algebra / l o ^ ] - Let p be the relation on S' defined by spu if and only if s G uZj 5, u G T, or s = u. It is easy to see that p is a congruence on S". Moreover, an easy calculation shows that 5 ' is nilpotent of class at most t + 1 if S'/p is nilpotent of class t. Let Z = {gl \ g G G} C Mr(K[G]), where / is the identity matrix. Let p be the congruence defined by spu if
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SEMIGROUPS
s G Zu,s, u G T, or s = u. Since G is g-nilpotent for some q and Z con tains the scalar matrices over the centre of G, an induction on q shows that S' is nilpotent of class at most t + q if Sf/p is nilpotent of class t — q. The location of nonzero blocks in every a^(,s), 5 G *?, is preserved under the map S* —> S'/p, so the additional assumption of 3) carries over to S/p. Therefore, we will further assume that G is trivial, so that the H-classes of T are singletons (hence T — T) and Ko[T] — Mr(K). It is easy to see that S embeds into the direct product K0[S/T] 0 J\ 0 [T] via the map s i-> (s',es), where s' is the image of s in K0[S/T] and e is the identity of ATop1]. By the induction hypothesis, S/T is nilpotent, say it satisfies the identity Xp = Yp. Therefore, we only need to show that eS C Ko[T] is nilpotent. Choose z, y, i^, u 2 , . . . G eS. If x v>Vv & eT — T C A"0[T], then we have xp = yp because eS/eT is a homomorphic image of S/T. Hence, we only need to consider the case where xv,yv G T. Now, if Xpl-Lyp in T — A^(l, r , r , / ) , then they are equal. Thus, sup pose that Xp,yv are not in the same 7^-class. If xp+2 — 0 or y p + 2 — $5 then xp+3 — 9 ~ yp+3. Hence, assume that xp+2 7^ $ / Up+2- Thus, aubzbua, buazaub are nonzero for a — xp,b = yv G T and some u, z G eS*. If a7£6 in T then u(bzbua)1Zu(azaub) because the left multiplication by u permutes the 7^-classes of T. Since aubzbua, buazaub are nonzero and T is a Brandt semigroup, it follows that aCb. Hence oH6, a contra diction. Thus a, b are not in the same 7^-class. Similarly one shows that they are not in the same £-class. Hence a = e z ae g ,6 = ejbep for some e z -,ej,e p ,e g G E(T) with i ^ j,p ^ q- Since aza,bzb,aub,bua are nonzero, this contradicts 3) (u,z would give an inversion). This shows that the identity Xp+3 — Yp+3 holds in 5, so that 5 is nilpotent. The existence of a function / ( n , k) is an easy consequence of the proof. □ Example where
Let S = E3UG3C
a=
M3(K)
and S' = E3 U {a,6} C M 3 (A'),
/ 0 0 1 \ / 0 1 0 \ 0 1 0 , 6 = 0 0 1 , \ 0 0 0/ \ 0 0 0 /
i? 3 is the semigroup of 3 x 3 matrix units with zero and G3 is the sub group of order 3 in GL3(K) consisting of permutation matrices. Then S is nilpotent, but Sf is not nilpotent, though both 'layers' S2/ S[, S[ of
154
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S' are nilpotent. Corollary 5.19 Let S C Mn(K) be a semigroup such that every non empty intersection SC\G with a maximal subgroup G of Mn(K) is nilpo tent. Then S is nilpotent if and only if every finitely generated subsemigroup of S is nilpotent. Proof. Let P be a finitely generated subsemigroup of S. Every group associated to P is a subgroup of a group H associated to S and H — g p ( 5 f! H) is nilpotent by the hypothesis and [77]. From Theorem 5.18 it follows that P is nilpotent of class at most /(n,fc), where k is the maximal nilpotency class of a group associated to S. Therefore S also satisfies the identity Xf(n^) — Yf(n,k)- D Corollary 5.20 Assume that every uniform component of S is con tained in a subgroup of Mn(K). Then S is nilpotent if and only if all uniform components of S are nilpotent. The assumption is satisfied in particular if S is n-regular and E(S) is contained in the centre of S. Proof. The first assertion is a direct consequence of condition 3) in the above theorem because, with the notation used there, T \ {0} is a group, so E(T) \ {9} is a singleton, in this case. If S is 7r-regular with central idempotents, then every regular J^-class of S is a group, so the assertion follows. □ This result covers the case of nilpotent connected algebraic monoids M, recently considered in [39]. Note that for such monoids the Zariski closure in Mn(K) of the group of units G is equal to M. Therefore, in view of Lemma 5.14, the definition of nilpotency used in [39] (the nilpotency of the group G) is equivalent to our definition. Theorem 5.18 can be used to show that the class of semigroups considered in [45] (where U i , u 2 , . - . used in our definition are allowed to be chosen from S1) coincides with the class of nilpotent semigroups, when restricted to subsemigroups of Mn(K). Corollary 5.21 If S C Mn(K)
is nilpotent, then S1 is nilpotent.
Proof. We show that condition 2) of Theorem 5.18 holds for S1 if it holds for S. Assume that x, y, xuy, yux, xzx,yzy £ T for some x, y,u,z G
5.3.
NILPOTENT
SEMIGROUPS
155
S1 and a uniform component T of S1. Suppose that z , y are not in the same %-class of Mn(K). Then T C M n _i, so that x,y G S and T is a uniform component of S. We use the fact that the completely 0-simple closure of T is a Brandt semigroup. This easily implies that either u E S or z G S. First assume that z — 1. Then x = e2xez-,y = e^7/e^ for some e2-,ej G E(T). By the supposition z ^ j . The quasi permutation corre sponding to C£T(U) switches z,j because ax{eiuej) ^ 0,aj(ejue z ) 7^ 0 (since xuy,yux G T). Therefore, the quasi permutation corresponding to ax(u2) fixes i and j . This means that e2u2ez'He2-, hence xu2xl-Lx, and similarly yu2yl~Ly. Thus, we can replace z = 1 by u2 G 5, coming to a contradiction with the assumption on S. Thus, assume that u — 1. Then xy,yx G T implies that xy,yx,yxuyx,xyuxy,yxzxy,xyzyx G T and zy = e,-a;et-, yx = ejyej for some e,-, ej G E(T). Moreover xy, yx are not in the same 'H-class because of the supposition on x, y. As above, we can re place u by z2 G 5, coming to xy,yx,yxz2yx,xyz2xy,yxzxy,xyzyx G T. This again contradicts the assumption on S. Thus, the assertion follows.
□
We note that the nilpotency class of S1 can exceed that of S. For example, this is so for the Brandt semigroup 5 = . M ( { l } , n , n , / ) , n > 1, over the trivial group, which is 2-nilpotent. Using the functions QT, we can also extend the comment following Lemma 5.14 as follows. P r o p o s i t i o n 5.22 Assume that S C Mn(K) Then (E(c\(S))) is a finite semigroup.
is a nilpotent
semigroup.
Proof. E(cl(S)) is finite by Theorem 5.15. Let F = (E(c\(S))). Choose a uniform component T of cl(S'). Then the semigroup ax(F) is generated by idempotents a y ( e ) , e G E(cl(S)). But ax(cl(S')) has a block pattern that is monomial with respect to the decomposition e = ei + ... + er of the identity e of A'olT]? where E{T) — {ei,. .. , e r } . Therefore, the elements of OLT{E{C\{S))) are block diagonal idempotents. Since each nonzero block is an invertible matrix, this implies that these idempo tents commute, so that aT(F) is a finite semigroup. Moreover, T fl F embeds into (XT(F). The ideal chain used for S in the proof of the im plication 3) => 1) in Theorem 5.18 yields now an ideal chain of F, whose factors are either finite or nil subsemigroups of completely 0-simple semi groups. Hence, each factor is locally finite, so F must be locally finite,
CHAPTER
156
5.
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see [87], Proposition 2.2 or Theorem 7.3. Since it is finitely generated, it must be finite. □ Our last aim is to show that the class of semigroups S C Mn(K) whose all 'layers' Sj/Sj-i are nilpotent is not very far from the class of nilpotent semigroups. Namely, such a semigroup is periodic modulo a nilpotent subsemigroup of a very special type, described in Corol lary 5.20. T h e o r e m 5.23 Assume that the groups associated to S C Mn(K) are nilpotent and the completely 0-simple closure of every uniform compo nent of S is a Brandt semigroup. Let M be the set of all x £ S such that for every uniform component T of S and every f £ E(T) \ {0} either xf = fxf or xf £ T. Then 1. M is a nilpotent
semigroup,
2. S is periodic of bounded index modulo M, 3. cl(S),S,M
intersect the same maximal subgroups of
Moreover, if S is TT-regular (regular respectively), (regular).
Mn(K).
then M is TT-regular
Proof. If T is a uniform component of 5, then let E(f) \ {6} — { e 1 ? . . . , e r } , see Theorem 5.15. Let x £ M. The definition of M is equivalent to saying that OLT{X) £ e\Te\ + . . . + erTer C A'o|T] (for all T). In other words, a?(x) has a block diagonal form with respect to the blocks arising from the decomposition 1 = t\ + . .. + er £ /io|T]. Since a? is a homomorphism, it follows that M is a subsemigroup of S. The argument above shows also that M 0 T is contained in the union of maximal subgroups of Mn(K) intersecting T. Therefore, every uniform component of M must be contained in a maximal subgroup of Mn(K). From Corollary 5.20 it follows that M is nilpotent. It is clear that, if 5 £ S and r! divides q, then ^ ( s ) 9 is block diago nal. Since r < |J5(5)|, this implies that S is periodic of bounded index modulo M. Consequently, 5, M intersect the same maximal subgroups of Mn(K). Hence 3) follows from Theorem 5.15. Finally, assume that S = cl(5). Let a £ H fl M for a maximal subgroup H of cl(5). Let b £ H be such that ab = e = e2 £ H. It
5A.
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ALGEBRAS
157
is easy to see that, for every uniform component T of 5, ax(b) has a block diagonal form because ax (a) has such a form. Therefore, b G M. This implies that M = cl(M). A similar argument also shows that M is regular whenever S is regular. Namely, if aba = a,bab = b for some a G M, 6 G 5, then it follows easily that every ay(6) also is block diagonal. This completes the proof. □ If S C Mn(K) is periodic of bounded index modulo a nilpotent subsemigroup, then S satisfies an identity of the form Xm(xk,y\ul...
,ukJ = Ym(xk,yk,ul...
,ukJ.
As in the proof of Lemma 5.14 this implies that the completely 0-simple closures of uniform components of S are Brandt semigroups. The groups associated to S must be almost nilpotent by Corollary 5.8. In [45] the structure of semigroup algebras of nilpotent semigroups was studied, in particular via prime Goldie homomorphic images, lead ing naturally to nilpotent subsemigroups of the matrix monoids Mn(F) over division rings F. The results of this section go over to this more general setting, though the proofs require more work because we are unable to use S and to immediately derive the nilpotency of cl(5), [94].
5.4
P I algebras
In this section we study conditions on a linear semigroup S C Mn(K) which ensure that the semigroup algebra F[S] over a field F satisfies a polynomial identity. In fact, we find necessary and sufficient conditions in terms of the sandwich matrices of the uniform components of S and also we prove that they are equivalent to a combinatorial property for S. The results come from [97]. Throughout this section, our basic reference to PI - algebras is [117]. We will also need certain results on semigroup algebras [87]. A semigroup S has the property Vm,m > 2, if for every a i , . . . , am G S we have ax • • • am — a a (i) • • • CLa(m) f° r a nontrivial permutation a. S has the permutation property V if S satisfies Vm for some m. Our starting point is the following observation.
CHAPTER
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5.
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L e m m a 5.24 Assume that S is a semigroup such that the semigroup algebra F[S] over a field F satisfies a polynomial identity of degree m. Then S satisfies Vm. Proof. It is well known that F[S] satisfies a multilinear identity of degree m, [117]. Therefore, there exist a i , . . . , a m £ F such that Xi'--xm = J2
We note that the converse to Lemma 5.24 is not true in general, even for group algebras, due to the following fundamental result, see [102]. P r o p o s i t i o n 5.26 The group algebra F[G] over a field F satisfies a polynomial identity if and only if one of the following holds 1. ch(F) = 0 and G is almost abelian, 2. ch(F) = p > 0 and G has a normal subgroup H of finite index such that [H, H] is a finite p-group.
We will focus on linear semigroups with permutation property. L e m m a 5.27 If S C Mn(K) its Zariski closure S.
is a semigroup having Vm, then so does
5.4.
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ALGEBRAS
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Proof. We proceed as in the proof of Lemma 5.1. For every nontrivial permutation a of the set { 1 , . . . , m } and every i = 1 , . . . , ra, put Xi{c)
=
{a>% G S\ ax • • • am = aCT(1) • • • a a ( m ) for all ctj G S,j < z, and all a*. G 5, fc > z}
Then every Y; = ILXJ X{(a) 1S a closed subset of S. Since 5 C Yx C 5 , it follows that Yx = S. If Y{ = 5 for some z, then 5 C Y^-+1 C S. Hence Yi+i — S. Therefore, by induction Ym — S. This means that 5 has Vm. D We shall need some results on closed subsemigroups S of Mn(K). Re call that by Theorem 1.4 such a semigroup is 7r-regular and the uniform components of S coincide with regular ,7-classes of 5, see Remark iv) preceding Proposition 3.6. In other words, they determine completely 0-simple principal factors of S. For more background in this area we refer to [108]. L e m m a 5.28 Let S C Mn(K) irreducible components. Then
be a Zariski closed semigroup with m
1. if G is a maximal subgroup of S with the identity component G c ,
then \G/GC\ < m, 2. if J is a regular J-class of 5, then E(J) with at most m2 irreducible components.
is a closed subset of S
Proof. 1) Let / be the identity of G. Then fSf is the image of S under the morphism x \-> fxf. Hence fSf has at most m irreducible components. Since G is an open subset of fSf, the same is true of G. 2) Let k be the rank of matrices in J. If a G S, then define 5(a) as the sum of all products of k eigenvalues of a. Since 5(a) is a coefficient of the characteristic polynomial of a, it follows that 5 : S —> K is a morphism. Clearly, the set Z = {x G S\ x2 = x,rank(x) < k} is closed. But X — {x G X\ 5(x) = 1} is then also closed. Note that X coincides with the set of idempotents of rank k in S. Let U\,... , Ut be the uniform components of S consisting of matrices of rank k. We know that each of them is a regular J^-class of S and one of them is equal to J. Then X = Xi U • • • U Xu where X{ = Ut H X, so one of the X% coincides with E(J). In order to show that Xi are closed it is enough to
CHAPTER
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5.
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prove that every Y{ — (Jj Xj \ Xt is closed. Let Vi = (SnMk)\Ui. From Theorem 3.5 we know that Vz = {x e S D Mk\ {xUif G M*_i}. Since Mk-i is closed, it follows that V{ is closed. But V{ D X = Y{. Since X is closed, so is Y{. It follows that E(J) is closed, as desired. Let e G E(J) and let G be the maximal subgroup of S containing e. Conjugating in Mn(K)
we may assume that e = I
I . Define
[/ = { ( x , y ) | x , y G S,eyxe G G}. £/ is a nonempty subset of the linear closed semigroup S x S C M 2n (A"). If 9(exe) denotes the determinant of exe treated as a rC x k matrix, then (x,y) \-> (9{eyxe)) l is a morphism on U. Thus (x,y) H* (eyxe)~l is also a morphism on [/ (where the inverse is taken in G). Therefore the map tp : (x,y) \-> x(eyxe)~1y is a morphism from U into E(S). Since eyi\)[x^y)xe — eyxe G G, it follows that i/j(x,y) G E(J). On the other hand, if / G E(J), then xey = / for some x,y £ S. Let ex = eyxe. Then ex G J f l eS'e = G is an idempotent. Hence e\ — e. So (x,y) G £/ and xj;(x,y) = f. We have shown that ip is a morphism of £/ onto E(J). Since S x S has m 2 irreducible components by the hypothesis, and [/ is an open subset, U has at most m 2 irreducible components. Hence E(J), being an image of £/, has at most m 2 irreducible components. D The following 'zigzag' lemma is very useful, see [108]. L e m m a 5.29 Let S C Mn(K) be a Zariski closed connected semigroup. Assume that SeS — SfS for some idempotents e , / G S. Then there exist e1,e2, fi, f2 G E(S) such that elZeiCf{lZf and e£e 2 7^/ 2 ^/? where 7Z,£ stand for Green's relations in S. Proof. Let k = rank(e). We claim that there exists ex G E(S) such that tR,ex and exfjf in 5. Suppose otherwise. Let H,H' be the 'H-classes in S of e , / respectively. Then eSe \ H = eSe Pi Mk-\ is a closed set. Similarly, fSf \ H' is closed. By the hypothesis there exist x,y G S such that xey — f. Since eS is closed, the set X = {a £ eS\ fxaf G fSf \ H'} is closed. Also Y — {a G S\ ae G e5e \ 77} is closed. Since fxeyf = / , we have ey ^ X. Clearly e ^ Y. Since 5 is connected, so is eS. It follows that eS %. X U Y. So there exists a £ eS such that a ^ I U K Then ea = a^fxaf G i / ' and ae G //". Hence there exists z G 5 such that zae = e. Then za2 — zaea — ea = a. Therefore Sa2S — SaS, so a lies in a maximal subgroup of S. Let ex G £ ( 5 )
5.4.
PI
ALGEBRAS
161
be the identity of this group. Then eKex in 5 because ea = a. Hence
SeJS
= SafS
D SfxafS
= SfS, and consequently SeJS
= 5/5,
which contradicts our supposition and proves our claim. Thus, there exists ex G E(S) such that eReY and e1fjf in 5. Since the principal factor of 5 containing ex,f,e1f is completely 0-simple, it follows that there exists an idempotent / i G £ ( 5 ) such that e i £ / i and fiTZf in 5. This proves the first assertion. The proof of the second assertion is similar. □ Recall that a rectangular band B is a direct product of a right zero semigroup and a left zero semigroup. In particular, B satisfies the identity xyzw = xzyw. L e m m a 5.30 Let 5 C Mn(K) property V. Then
be a Zariski closed semigroup with the
1. if G is a maximal subgroup of 5, then the identity component Gc of G is abelian, 2. if J is a regular J-class of 5, then every irreducible component A of E(J) is a rectangular band. Proof. Gc satisfies Vm, and hence by Lemma 5.25 it has a normal subgroup H of finite index such that the commutator subgroup [i7, H] is finite. It is well known that Gc has no closed subgroups of finite index, see [40], Proposition 7.3. Hence H is dense in Gc. On the other hand, we have a morphism (/> : Gc x Gc —> Gc given by (f)(x,y) = xyx1y~1. Now Gc x Gc is irreducible, H x H is dense in Gc x Gc and c/>(H x H) is finite, so (j){Gc x Gc) = i. Now Ufc>i ^ i s a semigroup with closure T. Hence T is a connected semigroup satisfying Vm. Now the irreducible set T x • • • x T (ra times) is a finite union of the closed sets Ba = {(au...
,am)\at
G T,a1 • • • am = a a(1) • ■ -a a (m)},
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Hence T x • • • x T = Ba for some a 7^ 1. Thus, T satisfies a fixed permutation identity. It is known that this implies that, for some k > 1, Tk satisfies the identity xyzw = xzyw, see [87], Lemma 21.5. It follows that E(T) is a subsemigroup of T. Let J i , . . . , Jt be the regular Jclasses of T. If e , / G J& for some fc, then by the connectedness of T there exist e i , / i G E(Jk) such that eTZeiCfilZf in T, see Lemma 5.29. Therefore ee\ = e i , e i / i = ex and / 1 / = / . Consequently eeifif G */&• Thus, the above identity yields e / ( e i / i ) / = e(e1fi)f G Jfc. It follows that e / G Jfc. This implies that each E(Jk) is a rectangular band. Also, by Lemma 5.28, each E(J^) is closed and irreducible. Since A C £J(T) = Ufc E(Jk), we see that A C E(Jr) C £ ( J ) for some r. So A = # ( J r ) is a rectangular band. D Corollary 5.31 Assume that S C Mn(K) erty. Then S satisfies an identity.
has the permutation
prop
Proof. This is a direct consequence of Lemma 5.27, Lemma 5.30 and Lemma 5.3. □ L e m m a 5.32 Let S = AA(G,X,Y, P) be a completely 0-simple group and E C S, E ^ 0, a rectangular band. Then 1. for some X' C X,Y' C Y and the Y' x Xf submatnx is the idempotent set of M{G,X'\Y'',(3),
semi
Q of P, E
2. any two rows of Q are similar, that is, for every y,y' G Y' there exists g G G such that pyx — gpy>x for all x G X'. Proof. Let X' = {x G X\ (g,x,y) G E for some g G G, y G ^ j and y = {y G y | ( # , £ , y ) G £ for some g £ G,x £ X}. Then the first assertion is clear. Assume that y,y' G Y',x' G X' and g = PyX'Py'x> G G. If x G X 7 , then e = (p~^x,y),f = (p~,x,, x'', y') G E. Since e/e = e, we obtain PyxPyx'Py'l'Py'xPyx = Pyx' Hence pyx = gpy,x, as desired. □ R e m a r k Let S C Mn(K) be a 7r-regular semigroup with the permuta tion property. Assume that T is a completely 0-simple principal factor of S. Since T is contained in a principal factor of 5 , by Lemma 5.30 E(T) \ {6} is a union of finitely many rectangular bands Eu... , Er.
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ALGEBRAS
163
We can then refine this union to write £ a s a disjoint union of finitely many rectangular bands. This is a consequence of the fact that E{ D Ej is a rectangular band and E{ \ [E{ D Ej) is a disjoint union of at most three rectangular bands. If T = M(G,X, Y, P ) , then it follows that X, Y can be partitioned as X = Xx U • • • U Xp, Y = Yx U • • • U Ys so that any two rows of the submatrix Pij of P corresponding to the semigroup T^ = M(G, Xj, Yi, Pij) C T are similar. Thus T is a 0-disjoint union of the semigroups T^ and each Txj is either a null semigroup or Tij \ {6} is a direct product of the almost abelian group G and a rectangular band.
Let F be a field. If T is a completely simple semigroup over an almost abelian group G such that E(T) is a rectangular band, then it is well known that T is a direct product G x E{T). Therefore, F[T] ~ F[G] ® F F[E(T)}. Now, F[G] is a PI - algebra by Proposition 5.26 and F[E(T)] also is a PI - algebra since E(T) satisfies the multilinear identity xyzw — xzyw. Hence, by [117], Theorem 6.1.1, F[T] is a PI algebra. If M is a completely 0-simple principal factor of a Zariski closed semi group S C Mn(K) satisfying V, then by Lemma 5.28 and Lemma 5.30 the set of non-nilpotents of S can be covered by at most m 2 completely simple semigroups of the above type. This will allow us to prove that F[M] is a PI - algebra. The proof of the main result of this section is based on this idea. In view of Lemma 5.27 the following theorem characterizes arbitrary linear semigroups whose semigroup algebras are PL T h e o r e m 5.33 Let S C Mn(K) be a n-regular semigroup. following conditions are equivalent 1. S has the permutation
Then the
property,
2. E(S) is a finite union of rectangular bands and every subgroup of S is almost abelian,
maximal
3. F[S] satisfies a polynomial identity for a field F, 4. F[S] satisfies a polynomial identity for every field F. Moreover, if these conditions hold and J(F[S\) denotes the Jacobson radical of F[S], then J{F[S\) is nilpotent of index at most 3 71 " 1 rij=i (j)
164
CHAPTERS.
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and F[S]/J(F[S]) satisfies an identity of degree 2ra(2 m — 1) + 2 , where m is the number of irreducible components of the Zariski closure S. Proof. That 1) implies 2) follows from Lemma 5.27, Lemma 5.28 and Lemma 5.30. By Lemma 5.24 1) is a consequence of 3). Since 3) follows from 4), we only need to prove that 2) implies 4). It is enough to consider the case where ch(F) = 0. In fact, an identity of the algebra Q[S] yields an identity in F p [5], and therefore in every F[S]. If J is an ideal of 5, then F0[S/J] ~ F[S]/F[J]. Moreover, the class of PI - algebras is closed under ideal extensions. Hence, by Theorem 3.5, it suffices to show that F[V] is a PI - algebra for every completely 0-simple principal factor V
of S. By the hypothesis V ~ -M(G, X, Y, P) for an almost abelian group G and E(V) \ {8} = Ex U • • • U Er, where each E{ is a rectangular band. We can construct subsemigroups Vj — A4(G,Xj,Yj,Pj) of V, where Xj C X, Yj C y, with E(Vj) — Ej,j — 1 , . . . , r. Since V is completely 0-simple, we have X = X\ U • • • U Xr. Let Z 1 ? . . . , Zt be all nonempty subsets of X, which are minimal in the Boolean algebra generated by X\,... , Xr. Then X — Z\ U • • • U Zt. Clearly t < 2r — 1 and this is a disjoint union. Consider the homomorphism <j> : F0[V] —> Mx(F[G]) given by a t-> a o P. Here F0[V] is identified with the Munn algebra M{F[G\,X,Y,P) and MX(F[G]) is the algebra of X x X matrices over F[G] with finitely many nonzero rows. From Lemma 4.16 we know that the kernel of <j> is equal to the left annihilator of F 0 [V]. Moreover (/>(F0[V]) coincides with the set of all elements of Mx{F[G]) whose rows lie in the left i^[G]-submodule of F[G]X generated by the rows of P. Let M C F[G}X be the submodule generated by the set of all projections of the rows of P onto F[G}ZJ C F[G}X, j = 1 , . . . ,t. Consider the subalgebra A of MX(F[G\) consisting of all matrices whose rows lie in M. For each j choose a row z3 with a nonzero projection onto F[G)Zj. Let Z = {ZJ\ j = 1,. .. ,£}. From Lemma 5.32 we know that any two nonzero projections onto a given F[G]Zj are similar. Since every row of P is the sum of its projections, it follows that M is a left P[G]-module generated by Z and (F0[V]) C A. Consider the Munn algebra R = M(F[G],X,Z,Q), where for every z G Z the z-th. row of Q is equal to z. Let ip : R —> Mx{F[G]) be given by a \-> a o Q. Clearly, (F0[V]) C A = i/>(R). In particular, we can identify (f)(F0[V]) with a subalgebra of R/l(R), where l(R) is the
5.4. PI
ALGEBRAS
165
left annihilator of R. On the other hand, we have a homomorphism R —> Mz(F[G]) ~ Mt(F[G]), determined by a H- Qoa. Its image is isomorphic to R/r(R), where r(R) is the right annihilator of R. Moreover, G has an abelian normal subgroup H of index q < oo. Hence F[G] embeds into Mq(F[H]). Therefore R/r(R) embeds into Mqt(F[H}). Consider the maps F0[V] —> R/l(R) —> R/(r(R) + 1{R)). By the Amitsur - Levitzki theorem, see [117], it follows that R/(r(R) + l(R)) (being an image of R/r(R)) sat isfies the standard identity S2qt(xi1. .. ,x2qt) — 0 of degree 2qt. Hence (F0[V]) C R/l(R) satisfies the identity x0S2qt(xi:. • • ,x2qt) = 0. Since ker((/>)F0[V] = 0, this implies that Fo[V] satisfies the identity X
0^2qt{Xli
• • • 5 X2qt)X2qt+l
= 0.
Therefore F[S] is a PI - algebra, as desired. This completes the proof of the equivalence of conditions 1) - 4). From Lemma 5.28 and Lemma 5.30 it follows that above we can take H = Gc and q < m,r < m 2 , so that t < 2 m ' - 1. Let S3,T3 be as in Theorem 3.5. Each Sj/Tj is a 0-disjoint union of completely 0-simple ideals. Hence, the algebra Fo[Sj/Tj] is a direct product of the corresponding contracted semigroup algebras. If G is a maximal subgroup of Sj/Tj, then G embeds into GLj(K). Since ch(F) = 0, it is well known that the algebra F[G] is semisimple (for example, use [102], Lemma 7.1.5, Proposition 7.4.3, the fact that finitely generated linear groups are residually finite and Maschke's theorem). Therefore the Jacobson radical J(F[Sj/Tj]) is nilpotent of index at most 3, [87], Corollary 5.18. (In fact, this is also a consequence of Theorem 4.26. Namely, since all primitive homomorphic images of F[Sj/Tj] are finite dimensional simple F-algebras, because it is a PI - algebra, one only needs to look at the intersection of the kernels of all homomorphisms corresponding to finite dimensional irreducible representations of Sj/Tj (n) over F). Since T>jJ C Sj-U it follows that J{F[S])r = 0, where r = Let R be a simple algebra which is a homomorphic image of F[S]. As in Theorem 4.28 we see that R is an image of the semigroup algebra F0[V] of a completely 0-simple principal factor V of S. We have checked above that F0[V] satisfies the identity x 0 S 2 g t(^i, • • • ,x2qt)x2qt+i = 0. Hence, it satisfies x 0 S p ( x 1 , . . . , x p ) x p + 1 = 0, with p — 2m(2 m — 1).
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IDENTITIES
Since F[S] is PI, F[S]/J(F[S]) is a subdirect product of simple algebras, [117]. Therefore it satisfies the same identity. This completes the proof of the theorem. □ We conclude with the following extension of Theorem 5.33. The proof requires standard facts about Noetherian and PI - algebras, [117], results on Noetherian semigroup algebras, [87], and a representation theorem from [2]. Corollary 5.34 Assume that S is a semigroup such that F[S] is a right Noetherian algebra. If S has the permutation property, then F[S] embeds into a matrix algebra over a field K D F. In particular, S is a linear semigroup and F[S] is a PI - algebra. Proof. Let Q be a prime ideal of F[S] and let SQ denote the image of S in F[S]/Q. Then F[S]/Q embeds into Mn(E) for a division al gebra E over F. We shall see in Section 9.2 that SQ has an ideal U which is a uniform subsemigroup of a completely 0-simple semigroup V C Mn(E). We may assume that V is the completely 0-simple closure of U. Then H = GnU ^ 0 for a maximal subgroup G of V. Since F[S] is right Noetherian, from [87], Lemma 7.21, it follows that the group ring F[HH~l] is Noetherian. Therefore G — HH~1 is a finitely generated group and Proposition 5.26 and Lemma 5.25 imply that F[G] is a PI - algebra. Since F[SQ] is right Noetherian and U intersects every 7iclass of V, it follows that the number t of 7^-classes of V is finite. This implies that F0[V] is a PI - algebra (as in the proof of Theorem 5.33 we see that F0[V] modulo its left annihilator embeds into the algebra Mt(F[G})). Therefore F[U] is a PI - algebra. Since the image of F[U] in F[S]/Q is a nonzero ideal and F[S]/Q is prime and right Noetherian, the latter also is a PI - algebra, see [117]. This implies that F[S] mod ulo its prime radical satisfies a polynomial identity. Since the radical is nilpotent, F[S] also satisfies an identity. Finally, right Noetherian semi group algebras satisfying a polynomial identity are finitely generated, [87], Theorem 19.14. Therefore, from [2] it follows that F[S] embeds into a matrix ring over a field extension of F. □ Further results on PI - algebras of linear semigroups will follow from the material presented in Chapter 7. This is due to the well-known fact that finitely generated PI - algebras have polynomial growth, [67], Corollary 10.7, or [87], Theorem 19.4.
Chapter 6 Generalised Tits alternative Tits alternative asserts that a finitely generated linear group G either is almost solvable or it contains a free nonabelian subgroup, Theorem 5.4. This, together with Theorem 5.5, implies that G is almost nilpotent or it contains a free noncommutative subsemigroup. Moreover, by Propo sition 5.7, any linear group (not necessarily finitely generated) is almost nilpotent if and only if it satisfies a semigroup identity. On the other hand, in view of our structural approach, it seems natural to study the status of cancellative subsemigroups of a linear semigroup 5, and in particular to ask whether a generalisation of Tits alternative holds for S. In this chapter we show that a finitely generated linear semigroup T C GLn(K) with no free noncommutative subsemigroups generates an almost nilpotent subgroup of GLn(K). This extends the results on finitely generated linear groups and finitely generated solvable groups mentioned above. We call this result the 'generalised Tits alternative', we extend it to an arbitrary linear semigroup S C Mn(K) and obtain consequences for the structure of the Zariski and 7r-regular closures of such an S. Our approach comes from [100] and is based on a refinement of arguments of Tits and Rosenblatt, [132],[118]. We refer to [7] and [40] for the general material on algebraic groups, extensively used in this section. Some of the key auxiliary results are stated in the first section. A complete presentation of the ideas of Tits can be found in [83] and [78]. 167
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6.1
6. GENERALISED
TITS
ALTERNATIVE
Auxiliary results
In this section we state several classical results, used in the proof of the result of Tits [132] and required for the refinement of his argument presented in the next section. Let G be a linear algebraic group. In other words, G is a closed subgroup of GLn(K) for a field K and some n>\. For the proof of the following result we refer to [134], Theorem 6.4. T h e o r e m 6.1 If H is a closed normal subgroup of a linear algebraic group G, then G/H is a linear algebraic group. The connected component Gc of the identity is a normal subgroup of finite index in G. It is a connected group, and it has no nontrivial closed subgroups of finite index. Thus, we will further assume that G is connected. Recall that G has the largest solvable normal subgroup TZ(G). This is a closed subgroup and it is called the radical of G. The group G is semisimple if 71(G) is trivial. Recall also that a matrix a £ Mn(K) is called semisimple if it is diagonalizable in Mn(K). An element g £ G is semisimple if g is a semisimple matrix. Note that this does not depend on the chosen rational linear representation of the group G. By Gs we denote the set of all semisimple elements of G. A torus of G is a closed connected diagonalizable subgroup of G. A torus of an algebraic linear group G is conjugate to the product (A"*)r of r copies of the multiplicative group of the field K, for some r > 0. The next classical result is extracted from [40], Theorem 22.2, Corol lary 26.2, Theorem 28.5. T h e o r e m 6.2 Assume that G is a semisimple algebraically closed field. Then
algebraic group over an
1. for every closed normal subgroup H of G, the algebraic group GjH is semisimple, 2. the commutator
subgroup [G?, G] coincides with G,
3. Gs = (JgeG 9~lTg for any maximal torus T of G and Gs a nonempty open subset of G.
contains
6.1. AUXILIARY
RESULTS
169
The second extreme is the class of solvable connected groups. Let Gu denote the set of unipotent elements of G. Recall that the unipotent radical of a linear group G is the largest unipotent normal subgroup of G. For the following result we refer to [40], §§19.2, 19.3, 21.4. T h e o r e m 6.3 Assume that G is a connected algebraic group over an algebraically closed field. Then 1. if G is solvable, then G is triangularizable and G is a semidirect product of Gu and a maximal torus T of G] moreover Gu is the unipotent radical of G and it is a closed subgroup, 2. G is nilpotent if and only ifGs is a subgroup. In this case Gs is a closed subgroup and G = Gs X G n , 3. if G has only one maximal torus, then G is nilpotent. An intersection of a closed subset of a topological space V with an open subset of V is called locally closed. A union of finitely many locally closed subsets is called a constructible subset of V, see [40], Chapter 4.4. In particular, a constructible set contains a dense open subset of its closure. L e m m a 6.4 Let <j> : G —> H be a morphism of linear algebraic groups. If A C G is a constructible set, then 4>{A) is a constructible subset of H. In particular, 4>(A) contains a nonempty open subset of G. So U C H for a dense open subset U of the closure IT. If h G if, then hU~l, [/are dense open subsets of IT, so they intersect nontrivially. Therefore H — UU. As H is a subgroup, we come to H = H. O It is clear that a nonempty open subset of a connected group G is dense in G. Hence, so it the intersection of any two such subsets. We will need the following extension of this fact.
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L e m m a 6.6 Assume that G is a connected group and A, B are dense subsets of G such that A contains a nonempty open subset V of G. Then AH B is a dense subset of G. Proof. Since B is dense in G, we have V H B ^ 9. This set is open in B, so it must be dense in B. Therefore it is dense in G and the assertion follows. □ A field K is called locally compact if there is a norm | | on K such that the normed space A", | | is locally compact. The field R of real numbers, the field Q p of p-adic numbers and the field Fp((t)) of power series over the finite field F p are of this type. Here these fields are considered with their standard norms. These are: the absolute value on R , the p-adic norm on Q p , and for the latter field, the t-adic norm = defined by \Y^ra^l\ ^~T', if a r G Z is such that ar ^ 0. Moreover, if K is a finite extension of one of these fields, then there is a unique extension of the norm to K and K is locally compact with respect to this norm. This, and the following converse result, may be found in [10], §9. T h e o r e m 6.7 Every locally compact field is a finite extension of one of the fields R , Q P , F P ( ( * ) ) . In particular, this implies that locally compact fields are perfect. Our next auxiliary result turns out to be very useful. Its proof does not require Theorem 6.7, since it leads directly to the three classes of locally compact fields mentioned above, see [132], [83], 53.1.1. L e m m a 6.8 Let a be an element of a finitely generated field K which is not a root of unity. Then there exists an embedding a : K —> L, for a locally compact field L with a norm | |, such that \o~(a)\ j^ 1. The space Kn and its projective space P are endowed with the topology induced from the field K, that is, the product topology on Kn and the quotient topology on P coming from the natural mapping 7r : Kn \ {0} —> P. It is well known that P is compact in this topology. Let x = (rci,... ,rr n ) be an affine coordinate system on Kn. We define dx : Kn x Kn —> R by dx(a,b) - sup \xi(a) - xt-(6)|.
6.1. AUXILIARY
RESULTS
171
The equation y0 = 0 defines a hyperplane H in P. Then x{ — yQlyi, for z = l , . . . , n, is an affine coordinate system on P and Dx = P \ H is the domain of this system. A distance function d : P x P —> R + on the projective space P is called admissible if it defines a metric compatible with the topology on P and if, for every affine coordinate system x in P and every compact subset C of Dx there exist ra, M > 0 such that mdx\Cxc
< d\cxc <
Mdx\CxC.
It is known that an admissible distance function exists on P, see [132], 3.3, [83], 53.2. For a linear map g : P —> P we denote by || g || the norm of g with respect to the metric d. That is, u p
% ( q ) , # ) )
d(a,b) where the supremum runs over all a, 6 £ P such that a ^ b. It is known that we have \\ g \\ < oo, [132], Lemma 3.5, [83], 53.2. Any semigroup S C GLn(K) acts on the projective space P of Kn. This is accomplished via the natural map GLn(K) —> PGLn(K): onto the projective linear group. For a linear map g : P —> P, let g be a representative of g in GLn(K). Let x{x) — IHLil^ ~ \ ' ) ^ ^ M ^ e the characteristic polynomial of g. Put fi = {i | |At-| = sup{|Aj| 11 < j < n } } , where | | stands also for the extension of the norm to K. Let xi(x) = Uien(x - A 0 a n d X^{x) = Ui^{x - A,-)- Clearly x = X1X2. Since the extension of | | to K is unique, every A"-automorphism of K preserves the norm. Therefore, the coefficients of Xi{x) a r e invari ant under all such automorphisms. Since K is a perfect field, this im plies that X15X2 G K[x\. Applying the Jordan form of s we also get ker(xi(#)) © ker(x2(^)) = Kn- Define A(g),A'(g) as the subspaces of P that correspond to ker(xi(^)),ker(x2(^)), respectively. In particular A(g) H A'(g) = 0. Now, for any s £ GLn(K) we denote by A(s),A'(s) the spaces A(g),A'(g), where g is the image of 5 in PGLn(K). With this notation we have the following important technical result. L e m m a 6.9 ([132], L e m m a 3.8) Let g : P —> P be a linear map, C C P a compact subset.
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1. Assume that g is semisimple, A(g) is a point and C Pi A'(g) = 0. Then \\ gn\c || —>n^oo 0 and for every neighbourhood X of A(g) there exists N > 1 such that gk{C) C X for k > N. 2. Assume that g(C) C Int(C), the interior of C in P, and || g\c \\ < 1. Then A(g) is a point contained in Int(C).
6,2
Free subsemigroups
Recall that a cancellative semigroup T is almost nilpotent if it has a group of quotients that is a finite extension of a nilpotent group. This notion can also be given an intrinsic definition in terms of subsemigroups of finite index and semigroup identities introduced in Section 5.1. We say that a subsemigroup T of S is a subsemigroup of finite index if there exists a finite set Z C S such that for every s G S there exists z G Z with sz G T. L e m m a 6.10 The following conditions are equivalent for a cancellative semigroup S 1. S has a nilpotent subsemigroup T of finite
index,
2. S has an almost nilpotent group of quotients. Proof. Assume that 1) holds and a finite set Z C S is chosen for T. If 5 , l G 5 , then there exist x,y G Z such that sx,ty G T. Now sxTDtyT ^ 0, because T is nilpotent, so S has a group of right quotients G. Let H be the subgroup generated by T. Then H is nilpotent (see Chapter 5) and it follows that [G : H] < \Z\. Since S has no free noncommutative subsemigroups, G is a (right and left) group of quotients of G. Conversely, if H is a nilpotent normal subgroup of finite index in the group G of quotients of 5, then S intersects all cosets of H in G. Therefore, we may take as Z any set of coset representatives contained
in S. □ The main result of this chapter reads as follows. T h e o r e m 6.11 Let S C Mn(K) following conditions
be a linear semigroup.
Consider the
6.2. FREE SUBSEMIGROUPS
173
1. S has no free noncommutative
sub semigroups,
2. the associated linear groups are almost
nilpotent,
3. every cancellative subsemigroup of S is almost 4- S satisfies a semigroup
nilpotent,
identity.
Then the following implications hold: 2) <& 3) <$ 4) => 1). Moreover, if the field K is finitely generated, then 1) => 2). Proof. The equivalence of conditions 2) and 4) was established in Corollary 5.8. Proposition 3.7 implies that 3) is a consequence of 2). Clearly, 2) is a consequence of 3) and 1) is a consequence of 4). It remains to prove that, if 1) holds and K is finitely generated, then 2) is satisfied. Replacing S by its intersections with maximal subgroups of Mn(K), it is enough to consider the case where S C GLn(K), n > 1. The proof will be completed by establishing the following assertion: (*) If S C GLn(K) and K is finitely generated, then either S has a free noncommutative subsemigroup or S is almost nilpotent. Let G C GLn(K) be the subgroup generated by S. First consider the case where G is almost solvable. We extend the proof of Theorem 5.5, showing that the hypothesis on the nonexistence of free subsemigroups in S is sufficient there. Using the reduction argu ment of that proof, and the notation used there, we may assume that G is upper triangular and there exists a normal subgroup H of finite index in G such that H/N is nilpotent, where N C H is the subgroup consisting of unipotent matrices with all non-diagonal entries, except possibly the ( l , n ) - t h entry, equal to zero. Replacing S by S C\ H we may further assume that G/N is nilpotent, of nilpotency class r, say. Let x,y e S. Then xr(x,y,Ui,... ,ur)N — yr(x,y,Ui,... ,wr)7V for the Malcev words xr, yr in #, y and any fixed U i , . . . , ur £ G. Since G is up per triangular, the definition of x r , y r imphes that xriyr have the same diagonal, say ( c l 7 . .. , c n ) . First, consider the case where C\C~l is a root of unity or xr — yr (for every choice of x, y, u l 5 . . . , ur £ S). Since K is
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ALTERNATIVE
finitely generated, there exists k > 1 (independent of Ci,c n ) such that c\ = c£. It follows easily that xr(x,y,uu...
,ur)kyr(x,y,ui,...
,ur)h
= y r ( x , 2 / , u i , . . . , u r ) * x r ( x , i / , u i , . . . ,1/,.)* k
because x^y may differ only in the ( l , n ) - t h entry, but in this case their (1,1) and (n, n) entries are all equal. Hence S satisfies a nontrivial identity, so its Zariski closure Sf in GLn(K) also satisfies this identity by Corollary 5.2. Since Sf is a group by Theorem 1.4 and Corollary 1.5, G C S' and so Proposition 5.7 implies that S',G are almost nilpotent. So suppose that CiC~l is not a root of unity for some x, y, u l 5 . . . , ur £ S and s = xr(x, y, u l 5 . . . , ur) ^ t — y r (x, y, u 1 ? . .. , ur). Let T be the subgroup generated by s,£. We know that t = sa for some a G N. Then we get the conclusions of Lemma 5.6 for the pair 5, a (the argument used there requires only the fact that (s,sa) is not a free noncommutative semigroup). As in the proof of Theorem 5.5, this leads to a contra diction. This completes the argument in the case where G is almost solvable. Assume now that G is not almost solvable. We will use the Zariski topology in the K-algebraic group GLn(F), where F is a sufficiently big algebraically closed extension of K, and the topology induced to certain subsets of GLn(F). It is well known that the F-closure S of S in GLn(F) is an algebraic K-group. Let S denote the connected component of 5 . Since S has finite index in 5, the group S is generated by S Pi S . But S H S is a closed group, so it must be equal to S . If S H S contains a free noncommutative subsemigroup, then we are done. If SOS is almost nilpotent, then so is S = S D S by Corollary 5.2 and Corollary 5.8, and hence S and S are almost nilpotent. Therefore, it is enough to establish assertion (*) for the semigroup S n S . But this semigroup is dense in the connected A'-group S . This allows us to assume that S is a connected group. Consider the natural AT-homomorphism <j) : S —> S/TZ(S), where T^{S) is the solvable radical of S. Since G C S is not solvable, it follows that S 7^ 1Z(S). Hence 4>(S) is dense in the nontrivial semisimple Kgroup S/1Z(S) and it is contained in the set of A'-rational points of this group. Clearly, it is enough to show that (S) has a free noncommuta tive subsemigroup. Thus, we can assume that S is nontrivial, connected and semisimple.
6.2. FREE
SUBSEMIGROUPS
175
From Lemma 1.7 we know that there are finitely many roots of unity that satisfy a polynomial of degree n over K. Hence, the degrees of torsion semisimple elements of 5 divide a natural number N. Let B — {g G S \ gN = l j . B i s a closed subset and B ^ 5 because otherwise 5 is periodic of bounded index, and so almost nilpotent by Corollary 5.8, which contradicts the assumption. From Theorem 6.2 it follows that the set A = {g G 5 | g is semisimple} contains a nonempty open subset of 5 . Hence, by Lemma 6.6, there exists s G 5 Pi ( 5 \ B) Pi A. Then s is a semisimple element of infinite order. Extending K we may assume that the eigenvalues of s lie in K. One of the eigenvalues, say A, is not a root of unity. From Lemma 6.8 we know that there exists a locally compact field K' with a norm | | and a homomorphism a : K —> K' such that |cr(A)| ^ 1. We will further assume that K is locally compact and |A| ^ 1. The unique extension of | | to an algebraic extension of K will also be denoted by | |. Let r be the number of the eigenvalues of s of maximal norm. Since the commutator subgroup [5,5] coincides with 5 by Theorem 6.2, we must have det(s) = 1. Hence r < n and the maximal norm eigenvalues of s is not equal to 1. Replacing 5, 5 by their images under the exterior power homomorphism g \-t A r (g), we can assume in view of Lemma 1.6 that r = 1. (Note that all homomorphic images of 5, A r (5) in particular, are semisimple by Theorem 6.2.) Passing to a finite field extension of K (it is also locally compact), we can assume that 5 has a block triangular form with absolutely irreducible diagonal blocks. Replacing 5 by the block containing an eigenvector corresponding to the eigenvalue A, we can assume that 5 is absolutely irreducible. Clearly, 5 is nontrivial because |A| ^ 1. From Section 6.1 we know that there exists a metric d : P x P —> R + inducing a topology that agrees with that coming from the field K. Moreover, on open subsets of P that can be identified with 7 i n _ 1 , d is equivalent in the usual sense with the metrics introduced via the cartesian metrics on Kn~l. For a linear map g : P —> P the norm \\g\\ of g with respect to the metric d satisfies || g || < oo, see Section 6.1. Also, A(s), A'(s) have their earlier meaning. L e m m a 6.12 Let S C GLn(K) be a semigroup such that 5 is a con nected group. If there exists a semisimple element s G 5 such that A(s) is a point, then the set Z = {sf G 5 | A(s') is a point } is dense in 5 .
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ALTERNATIVE
Proof. Choosing an appropriate irreducible component of the set s N = {sk | k G N } we can assume that there exists an infinite subset M C N such that sM is K-irreducible. Let U = {ue
S\uA(s)<£A'(s)}.
Then U is open in S. It is easy to see that sA(s) — A(s). Since A(s) D A'(s) = 0, it follows that s G U, so UPiS / 0. Let u G £/fl5. Since A'(s) is closed in the topology of P, there exists a neighbourhood X of the point A(s) in P such that uX D A'(s) = 0, where X denotes the closure of X in the topology of P. From the first assertion of Lemma 6.9 applied to C — uX it follows that there exists Nu > 1 such that sk(uX) C X and || ^ 1 % || < 1 for k > Nu (note that || sku\^ \\ < \\ sk\u^ || || u ||). Take any k > Nu. The second assertion of Lemma 6.9 applied to g — sku and C — X implies that A(sku) is a point. Hence sku G Z for u € UC\S and k> Nu. Now the set sMn{*|*>Ntt} i s d e n s e i n SM because the latter is irreducible. Hence sMu C Z. Let ra0 G M. We have shown that 5 mo ((7 n S ) C Z . Consequently *™°I77TS C Z. Since I T n 5 = 5 (see Lemma 6.6), we come to S — s m ° S C Z. This shows that Z is dense in S. D The element 5 G 5 found earlier has precisely one eigenvalue of maximal norm, so that A(s) is a point. By Lemma 6.12 Z is dense in S. Let T C 5 be a maximal torus of 5. Then T diagonalizes in a basis ^ i , . . . , v n of Fn. By Aj(t) we denote the eigenvalue of t G T corresponding to ^-. Define [7^- = {t G T | A;(£) 7^ Aj(t)}. The sets U{j are open in T and at least one of these sets is nonempty, because otherwise T would be the only maximal torus of 5 , which in view of Theorem 6.3 leads to a contradiction. Let t0 G C\Ux^Ui3. We may assume that u 1 } . . . , vr is the set of simple eigenvectors of t0 (that is, eigenvectors corresponding to simple eigenvalues of t 0 ). The choice of t 0 implies that each simple eigenvector v of an element t G T lies in { v l 5 . . . , vr}, up to a scalar multiple. Let w G Kn C Fn be a vector corresponding to the point A(s), and W C F n a hyperplane corresponding to A'(s). Then ^ is a simple eigenvector of 5. Define L[s)
— {g G 5 I if w' is a simple eigenvector of g, W its g — invariant complement, then w (jt W and w' ^
W}.
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SUBSEMIGROUPS
177
We will show that L(s) contains an open subset of S. Let V
=
{g € S\gv{ £W,w £ g Lm(vu... for i — 1 , . . . , r } .
,Vi-Uvi+u...
,vn)
Then V is an open subset of S as an intersection of the open subsets {g £ S\gvt £ W} and {g £ ~S\g~lw £ L i n ( ^ , . . . ,Vl_uvt+u... ,un}, for i = 1 , . . . , r. Hence V is nonempty because otherwise one of these sets is empty (because S is connected), which would contradict absolute irreducibility of S. Let v be a simple eigenvector of gtg~l, where g £ V, £ £ T. Then g - 1 ?; is a simple eigenvector oft. Hence g~lv £ Lin(^) for some i £ { 1 , . . . , r}. Therefore v £ L'm(gvi) and t; ^ W by the definition of V. A similar argument shows that g L i n ( u i , . . . , ^ _ i , i> l + i,... , u n ) is a g£g _1 -invariant complement of v and it does not contain w. The definition of L(s) implies now that gtg~l £ L(s). Let (f> : S x T —> S be given by the formula (#, t) = gtg~l for (g,t) £ 5 x T. We have shown that 0(V x T) C L(s). But c/>(V x T) is a constructible set as an image of V x T. Hence it contains an open subset of cf)(V x T), see Lemma 6.4. Since V is dense in 5, we have <^(V x T) = (^(iS x T) = 5 by Theorem 6.2. Therefore, L(s) contains an open subset of 5 , as claimed. The set C of elements g £ 5 such that if is not an eigenvector of g is open and nonempty because S is absolutely irreducible. Hence we may choose an element t £ Z fl L(s) 0 {g £ S | g is semisimple } fl C. In particular, A(t) is a point. Since w is not an eigenvector of t by the definition of C, it follows that A(£) ^ A(s). Also A(s) £ A'(t) and A(t) ^ A'(5) by the definition of L(s). The following lemma shows that there exists n0 > 1 such that s n °,£ n ° generate a free noncommutative semigroup, completing the proof of the theorem. L e m m a 6.13 Assume that s,t £ GLn(K) are semisimple. Assume also that A(s),A(t) are points such that A(s) ^ A(t),A(s) £ A!(t) and A(t) 0 A'(s). Then there exists n0 > 1 such that sn°,tn° generate a free noncommutative semigroup. Proof. There exist neighbourhoods Us,Ut of the points A(s),A(t) respectively, such that Us H Ut = 0, Us H (A'(s) U A'{t)) = 0 and Ut fl (A'(s) U Af(t)) = 0, (as before, X denotes the closure of X in P ) . Choose p#Ua\JUt\JA'{s)\JA'(t).
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TITS
ALTERNATIVE
From Lemma 6.9, applied to X — Us and C = {p} U Us U Ut, it follows that there exists n0 > 1 such that sk({p} UUsUUt) C t/ s . Similarly, ad justing n0 if necessary, we get tk({p}UUsUUt) C [/t for k > n 0 . Suppose that snoaitnoa2 ■■■ = tnoblsnob2 ••• for some a i , a 2 , . . . , 6i,6 2 , - - - € N . An easy induction shows that sn°aitn°a2 ■ • • (p) £ f/s and tno&i5n0&2 • • • ( » £ C/t, contradicting the fact that UsfMJt — 0. Similarly, p ^ UsUUt implies that a nontrivial word in s n o , t no cannot be the identity of GLn(K). The assertion follows. □ R e m a r k The argument of Tits can also be modified to obtain an alter native proof of the almost solvable case in the above theorem. Indeed, it is enough to consider the situation where K is finitely generated and S C GLn(K) is such that the group S is connected and solvable. Then, by Theorem 6.3, 5 is triangularizable in GLn(K') for a finite field exten sion K' of K. Hence, extending K we can assume that S is indecompos able and S is contained in the group of upper triangular matrices over K. For an element g £ S let Ai(g),... , Xn(g) be the subsequent diagonal entries of g. Define : S —> GLn(F) by 4>(g) — g\\(g)~l. Then is a /i-homomorphism of algebraic groups. If (f)(S) is almost nilpotent, then S is almost nilpotent because the kernel of <j> is central. So, passing to 4>(S), we can assume that Ai(g) = 1 for all g £ S. If for every s £ S each Xi(s) is a root of unity, then {A2-(s) | s £ 5, i — 1 , . . . , n} is a finite set. In this case the group generated by S is almost unipotent, and so we are done. Thus, assume that Az(s) is not a root of unity for some s £ S and some i. As in the proof of the semisimple case, Lemma 6.8 allows us to assume that K is locally compact and |At-(s)| / 1. Replacing s by s~l, if necessary, we may assume that |Az-(s)| > 1. S acts on the projective space P of Kn. Since |Ax(<s)| = 1 and |A t (s)| ^ 1, we have A(s) ^ P. Let W, W C Kn be the linear spaces corresponding to A(s),A'(s) re spectively. Define WF(s) = L'mF(W) and W'F(s) = LmF(W). Assume first that WF(s) is S'-invariant. Then, extending K, we can assume that W — L i n / ^ e i , . . . ,er\ W = L i n x ( e r + i , . . . , e n ) , where ex is the z-th vector of the standard basis of 7i n , and S is upper block triangular with respect to the decomposition Fn = WF(s) © W'F(s). Since W,W are 5-invariant, the matrix s is block diagonal subject to this decomposition. Consider the transpose ST C S . Suppose that M ^ F ( ^ T ) (defined similarly as WF(s) above) is S -invariant. Note that WF(sT) — ker(xi(5) T ), where Xi{x) ls the polynomial used in the defi-
6.2. FREE
SUBSEMIGROUPS
179
nition of A(s). Hence it is easy to see that WF(sT) = WF(s) is a direct summand of the S -module F n , contradicting the indecomposability of S. Therefore, replacing S by ST, if necessary, we can assume that WF(s) is not ^-invariant. We apply the exterior power A r to S. Since r = dimF WF(s), we come to an element A r (s) G A r (£) which has only one simple eigenvalue of maximal norm (see Section 1.3). Therefore A(Ar(s)) is a point. This allows us to assume that s G S is such that A(s) is a point, 5 is trian gular with Xx(g) - 1 for g G 5 , and WF(s) is not 5-invariant. Choose a subset M C N , | M | = oo, such that the set sM is 7\-irreducible. Let Xi(s) be the (unique, because A(s) is a point) eigenvalue of s of maximal norm. Then i > 1. It is easy to see that for u G 5 there exists Nu > 1 such that, for A; > A^, A^us*) is an eigenvalue of usk of maximal norm. Let Z = {sf G S | Xi(s') is an eigenvalue of s' of maximal norm}. Then usk £ Z for u £ S and k > Nu. An argument as that at the end of the proof of Lemma 6.12 allows us to show that Z is a dense subset of S. Let t G Z fl {g G 5 | WKF(S) is not g -invariant}. (The latter of the inter sected sets is open in S and nonempty by the choice of 5, so that the intersection is nonempty.) We have chosen elements s,£ G S such that Xi(s),X{(t) are eigenvalues of s,£, respectively, of maximal norms and A(s) ^ A(t). Replacing s,t by their restrictions to L i n ( e i , . . . , et-) we can assume that n — i. Then A'(s) = A'(t) and L i n ( e i , . . . , e n _i) is the corresponding subspace of Kn. The elements s,t satisfy the hypotheses of Lemma 6.13, except for being semisimple. However, as mentioned in [132], the assertions of Lemma 6.9 (and hence also of Lemma 6.12 and Lemma 6.13) are valid without this restriction. Therefore, (s,t) contains a free noncommutative subsemigroup. Now we can summarize the information on the associated groups of a semigroup S C Mn(K) with no free noncommutative subsemigroups. Corollary 6.14 Assume that S C Mn(K) has no free noncommuta tive subsemigroups. Then every group G associated to S has a nilpotent normal subgroup H such that G/H is periodic. Moreover, G is locally almost nilpotent and it is almost solvable if ch(K) — 0 and almost nilpo tent if K is finitely generated. Proof. Recall from Section 5.1 that a locally almost nilpotent linear group G has a normal subgroup H with a periodic quotient G/H. G
180
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TITS
ALTERNATIVE
is almost solvable if ch(K) = 0, by Theorem 5.4. Thus, the assertion follows from Theorem 6.11. □ If R is a finitely generated domain, then every locally almost nilpotent subgroup of GLn{R) is almost nilpotent, cf.[134], Theorem 10.14. The following extension of this result is an immediate consequence of Theorem 6.11. C o r o l l a r y 6.15 Assume that G C GLn(K) is a linear group over a finitely generated field K. If G is locally almost nilpotent, then it is almost nilpotent. We note that, in contrast to the case of linear groups, finitely gen erated (noncancellative) linear semigroups satisfying the conditions of the theorem need not have polynomial growth, even if the associated linear groups are finitely generated. This topic will be discussed in the next chapter. Let S C Mn(K). The groups associated to the Zariski closure S of S in Mn(K) are just the maximal subgroups of 5 , see Section 3.2. Thus, S has a better structure than S. From Corollary 6.14 we know that, if S has no free noncommutative subsemigroups, then the groups associated to S are periodic extensions of nilpotent groups. Moreover, Theorem 6.11 implies that, if S is finitely generated, then S has no free noncommutative subsemigroups. The latter is no longer true if S is not finitely generated. For example, the group G of upper triangular matrices in GLn[K), n > 1, K $£ F p , with periodic diagonal is a solvable group which is a periodic extension of a nilpotent group, but its closure ( = the group of all upper triangular matrices) is not of this type. Simply, the Zariski closure of a semigroup S C Mn(K) is not controlled by the Zariski closures of the finitely generated subsemigroups of S. However, we have seen in Section 3.3 that this is the case for the 7r-regular closure
cl(5) of S. C o r o l l a r y 6.16 Let S C Mn(K) be a semigroup. If S has no free noncommutative subsemigroups, then cl(5) has no free noncommutative subsemigroups . Proof. This is a direct consequence of Theorem 6.11 and Lemma 3.12.
□
6.2. FREE
SUBSEMIGROUPS
181
Note that Corollary 3.18 yields an important information on the ideal structure of a linear semigroup S with no free noncommutative subsemigroups. Namely, 0-simple principal factors of S are completely 0-simple. If S C Mn(K) is a connected algebraic semigroup (that is, Zariski closed and irreducible as an algebraic variety), then the maximal sub groups of S are connected algebraic groups, see [108]. If K <2 F p , then from Theorem 6.11 and from Proposition 5.7 it follows that S has no free noncommutative subsemigroups if and only if the maximal subgroups of S are nilpotent, which is also equivalent to the fact that S satisfies an identity. Finally, we note that some other variants of Tits alternative have been studied for groups of units of integral group rings of finite groups, [35], for skew fields,[74],[75], and domains,[64].
Chapter 7 Growth It was shown in Chapter 6 that a finitely generated linear semigroup S C Mn(K) with no free noncommutative subsemigroups 'locally' is almost nilpotent in the sense that each nonempty intersection S fl D with a maximal subgroup D of Mn(K) generates an almost nilpotent subgroup of D. However, in contrast to the case of linear groups, the growth of such a semigroup S often is not polynomial. The aim of this chapter is to look for a structural description of linear semigroups of polynomial growth. To this end we first present certain important special classes of semigroups of polynomial growth, proving in particular that unipotent semigroups are of this type. Then we discuss the second extreme case of 'almost semisimple semigroups', and we explain why semigroups of these two types should be crucial for the structure of arbitrary linear semigroups of polynomial growth. Also the class of semigroups of linear growth and its connection with a combinatorial property, called repetitivity, are studied. Techniques resulting from the general structure theorem for linear semigroups, obtained in Chapter 3, are essential for our approach. The polynomiality of the growth of S turns out to depend not only on the properties of the uniform components of S but also on the interaction between them. We refer to [67], [81] for the general theory of growth of associative systems. 183
184
7.1
CHAPTER
7.
GROWTH
Burnside theorem and some examples
Let S be a finitely generated semigroup. Choose a generating set V = { a l 5 . . . ,«t}- By V m , r a > 1, we denote the set of all elements of S which are of the form vx • • • vm for V{ G V. Then the cardinality dy(m) of the set V U V2 U • • • U V m , m > 1, is called the growth function of S corresponding to the generating set V. If dy(m) is bounded by a polynomial in ra, then we say that S has polynomial growth. It turns out that this is independent of the choice of the generating set V. The minimal degree of such a polynomial is called the degree of growth of S. Moreover, the Gelfand - Kirillov dimension of S is defined as GK(S) = limsuplog m dy(m). In other words, GK(S') < oo if and only if S has polynomial growth. If S is not finitely generated, then GK(S') is defined as the supremum of GK(T) where T runs over the set of finitely generated subsemigroups of S. More generally, growth and Gelfand - Kirillov dimension can be defined for any A'-algebra A, by considering the /\-dimension of the linear span of VU • • -U V m for a generating set V of A. Clearly, GK(S') = GK(Ar[5f]) for any semigroup S. It is known that GK(A) can be equal to 0,1 any real number > 2 or oo. Dimensions 0 and 1 are distinguished also from the structural point of view. Namely, GK(A) = 0 if and only if A is locally finite, that is, every finitely generated subalgebra of A is finite dimensional. Also, algebras of dimension 1 are very special. If GK(A) = 1 for a finitely generated K-algebra A, then the prime radical 13(A) of A is nilpotent and A/B(A) is a finite module over its centre, which is finitely generated, [128]. In particular, A is a PI - algebra. The following fundamental result due to Gromov,[32], and Bass,[4], is the starting point for studying the growth problem for linear semi groups. T h e o r e m 7.1 Let G be a finitely generated group. Then G has polyno mial growth if and only if G is almost nilpotent. Moreover, in this case GK(S) = J2iirii where r\- is the torsion - free rank of the i-th factor of the lower central series of a nilpotent subgroup of finite index in G. In view of Theorem 7.1 and of the results of Chapter 3, in case of a semigroup S C Mn(K) one needs in the first place to consider
7.1. BURNSIDE
THEOREM
AND SOME EXAMPLES
185
the relation between GK(S) and GK(G) for a cancellative semigroup S generating a group G. This was done by Grigorchuk, [31], cf. [87]. T h e o r e m 7.2 Let S be a cancellative semigroup. Then GK(S') is fi nite if and only if S has a group of quotients G and GK(G) is finite. Moreover, in this case GK(5) = GK(G). The next step level of the groups affects the growth first proved in [82]
is to see how the 'local' information (that on the associated to a given linear semigroup S C Mn(K)) of S. The first step looks rather promising. It was through quite a different approach.
T h e o r e m 7.3 The following conditions are equivalent for a semigroup S C Mn(K) 1. S is locally finite, 2. S is periodic, 3. GK(5) = 0, 4- GK(STl G) = 0 for every maximal subgroup G of Mn(K) ially intersecting S.
nontriv-
Proof. We know that 1),3) are equivalent. Clearly, 2) is a consequence of 1). If S is periodic, then so is every nonempty SC\G. So Sf)G is a pe riodic linear group, and therefore it is locally finite, [17], Theorem 36.2. Hence 2) implies 4). It remains to show that 1) is a consequence of 4). This is an easy consequence of the structure theorem. Namely, it is easy to see that a completely 0-simple semigroup over a locally finite group is locally finite, [87], Lemma 2.3. Also, if TN = 9 for a semigroup T, then T is locally finite. Next, if / is an ideal of a semigroup T and / , T/I are locally finite, then T is locally finite, cf.[87], Proposition 2.2. But S has an ideal chain S = Ii D • • - D It such that every factor T of this chain is either completely 0-simple or it satisfies TN — 9 for some TV > 1, Corollary 3.8. Therefore, the assertion follows by induction on t. D It turns out that one cannot expect the equivalence of 3) and 4) in general. Polynomial growth of every group associated to a given S C Mn(K) does not even imply that S has polynomial growth.
186
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E x a m p l e There exist linear semigroups of subexponential growth with all associated groups isomorphic to the infinite cyclic group. Namely, let Z be the ring of integers, M 3 (Z) the ring of 3 x 3 matrices over Z. Denote by S the subsemigroup generated in M 3 (Z) by the matrices / 1 0 0 \ e= 0 0 0 \ 0 0 1 /
and
5
/ 1 1 1 \ 0 2 1 . = \ \ 00 00 11 /
We claim that the growth of S is not polynomial. For this, first note / 1 On K \ that gn=\ 0 T an , where a0 = 0,b 0,6oo = = 0, and for n > 1 \ 0 0 1 / an
= = = =
a„_! + 2n_1 1 + 2 + --- + 2 " - 1 2n-l
and K
= = -=
6„_1 + an_ 1 + l a n _i + a n _ 2 + al+bl+n-l Ya2 + ( 2 " - 1 - 1) + 2 " - 2 + . . . + 2 + l = 2 ' l - l
Consider the elements t = egn>egn> ■ ■ ■ egn> G G S. The (1,3)-entry of t is equaljo 6 ni +• • • + &„.. Let A m = {6 ni + - • -+6„ -+bn.\s | nn,i ++ -- • -+nss <m; s > 1} and A m = {6ni + --- + 6„ 3 |n 1 + --- + n s + 5 < m ; 5 > 1}. Then A m C A 2 m because if m + • • • + ns < m, then s < m. The maximal number of the set Am is equal to bm. Thus, the sets Am,Am + bm+1, ■■■ ,Am + b2m are disjoint (because max(A m + bm+k) = bm + bm+k < bm+k+1) and of the same cardinality. Since all of them lie in A 3 m , we come to |A IA33m| ~| >3" 1 - 1 1Agm-i ! > • • • > 3 m - x 3 m - 2 • • • 3° = S"1T'"11^-^ -1)/22.. Therefore m |A | > |A 3 m| > s™^-!)/2. |A 22 .. 3m 3m | > |A 3 m| > 3 (m-!)/2.
Hence Hence
W 1C)
\A
l>W
3 mU-iV2
^ AOg3m+i 06 g2-3m l ^ 2 - 3 m | d-
1C)
n
__ m(m - 1) — — — 2(m +—1)— .
It follows that the function / ( m ) = \{s G S\ 5 | s is a word of length at most m in e,g}
7.1. BURNSIDE
THEOREM
AND SOME EXAMPLES
187
is not bounded by a polynomial, as desired. We note that |\A Ann\ | < Ek
0 0 0 ] |i e zJ.
Finally, S has no elements of rank one. Therefore all the associated groups of S are cyclic infinite. Before studying the structure of arbitrary linear semigroups of poly nomial growth, we show that certain classes of semigroups considered in Chapter 5 are of this type. T h e o r e m 7.4 Let S C Mn(K) be a finitely generated semigroup permutational property. Then S has polynomial growth.
with
Proof. It is known that a finitely generated PI - algebra has finite Gelfand-Kirillov dimension, [67], Corollary 10.7. Hence, the assertion follows from Theorem 5.33. Let us note that a purely combinatorial proof, not referring to PI - algebras, can be derived from Shirshov's height theorem, [87], Theorem 19.4. □ Assume now that S = (au... ,at) C Mn(K) is an almost unipotent semigroup that is in the block upper triangular form of Proposition 5.9. As in Section 5.2, let e3 be the identity of the diagonal block Zjtj = 1 , . . . , r, and a?,... , a« the diagonal blocks of a,-, that is a? = W edaj{e,r Write 4i g5 = efcflie,-, so that a{ = Efcj=i, Efcj=i,k<j
C,C,-::i+i i+i = c7,C^,,^w,i+iG;+i,,+i ),-.X;),--|.iG,-+lii+1
CHAPTER
188
7.
GROWTH
where Xii3 = {x\3] \ k = 1 , . . . , * } , and inductively, for i < j - 1, Cij = Citi+iCi+itj U . . . U dj-iCj-ij
U
CijiXijChJ.
Let A',j, * < i ? be the i?-submodule of the ring Mn(K) generated by CXy3, where R = Z if ch(K) = 0 and R is the prime subfield of K otherwise. From Lemma 5.10 we know that, if s — ail . . . airn^i3 £ { 1 , . . . ,£}, and i < j , then e{se3 is a sum of at most hj-i(m) elements of the set Ct-fj, where hk is a polynomial in m of degree fc, for k — 1,. . . , n — 1. Corollary 7.5 Le£ 5 C Mn(K) be a finitely generated almost semigroup. Then S has polynomial growth.
unipotent
Proof. If s — a{x . . . aim = ^ ^{se3- 6 5*, then C{sc3 is an element of length < hj-i(m) in the generators dj of the (additive) abelian group Dij. Therefore the claim follows. □ Note that, if ch(/\) > 0, then a finitely generated almost unipotent semigroup S C Mn(K) is finite because it is periodic, Theorem 7.3. Each Dij is then finite. Our last class of examples of semigroups of polynomial growth re quires preparatory results on uniform semigroups. L e m m a 7.6 Assume that J = M(G:X,Y,P) is a completely 0-simple ideal of a semigroup S and U is a uniform subsemigroup of J such that J is the completely 0-simple closure of U in J and Q = U U (S \ J) is a subsemigroup of S. Then 1. GK(S) < 2sup(GK(5/J),GK(G r )), provided that at least one of the sets X, Y is finite, 2. GK{S) = GK{Q) = s u p ( G K ( S / J ) , G K ( G ) ) if X,Y
are finite.
Proof. We consider the case where r = \X\ < oo only, because the proof of assertion 1) in case \Y\ < oo is symmetric. Since G is a group generated by a subsemigroup of Q, by Theorem 7.2 we have GK(G) < GK(Q). Also, Q/U = S/J is a homomorphic image of Q, hence G K ( 5 / J ) < GK(Q). It remains to show that GK(5) < A s u p ( G K ( 5 / J ) , G K ( G ) ) ,
7.1. BURNSIDE
THEOREM
AND SOME EXAMPLES
189
where A = 1 if | y | < oo and A = 2 otherwise. Let e = e2 G J, e ^ 9. Choose nonzero elements cx — e, e 2 , . . . , e r , such that ez- G J and e z £e for all z, but e,-, ej are not in the same 7£-class of J if i ^ j . As in Proposition 3.16 (see also Section 4.2), we have a homomorphism : MS) —► M r (tf[G]) whose kernel TV is the left annihilator of A"o[^]- Consider ^ : K0[S] —► M r (tf[G]) © /i 0 [5/J] given by ^ ( x ) = 0(x) + x, where x is the image of x under the natural map K0[S] —> K0[S/J]. Let A = NnK0[J}. Then A2 - 0 and ker(^) = A. Introduce a congruence ~ on 5 by the rule s ~ t <=> s = t or s,t G J and (5 — £)J = 0. (In other words, (s — t) G N, so 5 / ~ can be identified with ip(S).) Thus ip factors through K0[S/ ~] and hence the square of the kernel of the map K0[S] —> K0[S/ ~] is zero. From [67], Corollary 5.10, it follows that GK(K0[S]) < 2 GK(K0[S/ - ] ) • But GK(K0[S/
- ] ) < sup(GK(M r (AT[G])),GK(5/J)).
Since we know that GK(K[G}) = GK(M r (A r [G])), [67], Proposition 5.5, it follows that GK(S/ ~ ) < GK(G) and GK{S/ ~ ) < s u p ( G K ( G ) , G K ( 5 / J ) ) . This proves assertion 1). Now, assume that | y | is finite. Let s G J- If t ~ s, then tx = sx for every x £ J easily implies that t1Zs. Moreover, every 'H-class of J that is 7^-related to 5 contains at most one element t of this type. Therefore, we get the following relation between the growth functions ds and dS/~ for S and S/ ~ subject to the generators Si,. .. , sm of S and (si),. . . , (sm) of S/ ~ , respectively
d 5 H
190
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Corollary 7.7 Assume that U is a uniform subsemigroup of a com pletely 0-simple semigroup J with closure U. Then GK(U) — GK((7) = GK(T) for any intersection T of U with a maximal subgroup of U. Proof. We may assume that J = U. Let J = M(G,X,Y,P). Let H be the maximal subgroup of J containing T. Then H ~ G is the group generated by T. It is enough to show that G K ( J ) < GK(T). So, choose any finitely generated subsemigroup / of J. Then / is con tained in a completely 0-simple semigroup J' = M.{G, Xf,Yf, P ' ) , where X' C. X,Yr C. Y are finite subsets. Therefore, replacing J by J', we may assume that the sets X, Y are finite. The assertion follows from Lemma 7.6. □ Note that the group ring C — K[H] is a subalgebra of A — Ko[U] and it is a 'corner' in A in the sense that CAC C C. It is known that GK(AC 2 A) = GK(C) in this case, cf. [1]. Since UH2U = £/, in view of Theorem 7.1 we also come to GK(T) = GK(H) = GK(t/). The following result, relying on the lemma above, extends the ideas of Theorem 7.3 to certain classes of linear semigroups of particular in terest. T h e o r e m 7.8 Assume that the maximal subgroups of Mn(K) intersect ing a finitely generated semigroup S C Mn(K) are contained in finitely many TZ-classes of Mn(K). Then the following conditions are equivalent 1. S has polynomial
growth,
2. every nonempty intersection S C\G with a maximal subgroup G of Mn(K) has finite G elf and - Kirillov dimension, 3. S has no free noncommutative
sub semigroups,
4- every group associated to S is finitely generated and almost nilpotent. Moreover, in this case GK(5) is bounded by a function f(n,r), r = maxT{GK(T)} for T running over the set of all nonempty sections s n G.
where inter
7.1. B URNSIDE
THEOREM
AND SOME EXAMPLES
191
Proof. It is clear that 1) => 2) and 2) => 3), cf. Proposition 3.7. The implication 3) ^ 4) follows Theorem 6.11 and from Proposition 3.16. Thus, assume that 4) holds. From Corollary 3.8 we know that S has an ideal chain S = h D • • • D Ik such that, if T3 — Ij/Ij+i for j = 1 , . . . , A: — 1, and Tk = Ik, then every T{ either is a uniform subsemigroup of a completely 0-simple semigroup Jt C M r / M r _ i for some r, or it is power nilpotent of index not exceeding (n) for some q < n, and k < 2n + n. Moreover, in the former case the maximal subgroup of J{ is isomorphic to the group of quotients of every maximal nonzero cancellative subsemigroup of T2 (note that 4) implies that these semigroups have groups of quotients). The hypothesis implies that J2 has finitely many 7£-classes. From Lemma 3.11 we know that J2 U (S \ 7Z) has a natural semigroup structure (as a Rees factor of a semigroup containing S) and contains S//t-+i as a subsemigroup. By induction on k we show
that GK(5) < oo. If k — 1, then S is power nilpotent or uniform. In the former case 1) is obviously satisfied, while in the latter case 1) follows from Corollary 7.7. Assume now that k > 1. By the induction hypothesis G K ( 5 / 4 _ i ) < oo because it has a shorter ideal chain with the required properties. If S is power nilpotent, then the fact that K0[S/Ik-i] = Ko[S]/Ko[h-i] implies in view of [67], Corollary 5.10, that GK(5) < r GK(5//fc_i), where r is the nilpotency index of Ik_x. Otherwise Ik_x is uniform, so GK(S') < oo by Lemma 7.6. The remaining assertion is a direct consequence of the proof. □ For the basic facts on rings with Krull dimension, we refer to [81]. Corollary 7.9 Assume that S C Mn(K) is (Malcev) nilpotent or the algebra K[S] has right Krull dimension. Then S has polynomial growth if and only if every nonempty intersection SC\G, for a maximal subgroup G of Mn(K), has finite Gelfand - Kirillov dimension. Proof. Let R be an 7^-class of Mn(K) such that S fl R is contained in a uniform component U of S. We know that there exists an ideal / of S such that U is an ideal of S/I. Then K0[R fl 5] is a right ideal of K0[S/I]. If K[S] has right Krull dimension, the same is true of K0[S/I}. In particular, K0[S/I] has finite right Goldie dimension. It follows that U intersects finitely many 7^-classes of Mn(K). If S nilpotent, then the
192
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latter follows from Theorem 5.18. Thus, the assertion is a consequence of Theorem 7.8. □ We conclude with a general observation on linear semigroups of poly nomial growth. Recall that the unipotent radical of a linear group G is the largest normal subgroup of G consisting of unipotent matrices. P r o p o s i t i o n 7.10 Assume that S C Mn(K) semigroup of polynomial growth. Then
is a finitely
generated
1. S satisfies an identity, 2. there exists a finitely generated subring R C K such that cl(S) C Mn(R), 3. groups associated to S are almost nilpotent and finitely generated modulo unipotent radical. Proof. 1) is a direct consequence of Theorem 6.11. Let G be the subgroup of Mn(K) generated by a nonempty intersec tion S fl D with a maximal subgroup D of Mn(K). From Theorem 7.1 we know that GK(G) < oo and G is almost nilpotent by Theorem 6.11. Let H be a nilpotent normal subgroup of G of finite index. By induction on the nilpotency class of H we show that G has a finitely generated subgroup TV such that a power of every g E G lies in N. Let A be a finitely generated subgroup of Z(H) of maximal torsion-free rank. If H is abelian, then put TV = A. Now # , N have the same rank, so that H is torsion modulo N and the claim follows. Otherwise, by the induction hypothesis H/Z(H) has a finitely generated subgroup B/Z(H) with the desired property. Let N be the group generated by 6 1 : . . . , bu A, where &i,... , bt, Z(H) generate B. It is clear that for every h € H there exists k > 1 with hk e N. Thus, the claim follows. Since G is the group generated by S D Z), it follows that TV = g p ( 5 n A/"), cf.[87], Lemma 7.5. Hence, we can find generators u 1 ? . . . , Uk of the group iV such that the inverses Vj of Uj in G lie in S O D. Then the semigroup C/ = ( S T | J D , U 1 , . . . , u k ) contains N. This easily implies that G C U C Mn(R), where R C K is the subring generated by the entries of (finitely many) generators of 5, and the entries of ui,... , Uk. Now, in view of the remark following Corollary 3.11, cl(S') can be constructed from S by a finite chain of extensions, each obtained from
7.2. LINEAR
GROWTH
AND
REPETITIVITY
193
the previously constructed T D S hy inverting the elements of T n M for a group M associated to T. The above shows that we only need to invert finitely many elements in each step. Therefore 2) follows. Let U be the unipotent radical of G. From [134], Theorem 3.6, we know that G has a normal subgroup F of finite index which is triangularizable in Mn{K). So u~lFu is triangular for some u G GKn(K) and u~lFu C Mn{Ri) for the subring Rx of K generated by R,u and i r 1 . Therefore F/U ~ (u-lFu)l(vTlUu) embeds into {R\)m for the group of units R* of i? l 5 where m is the rank of matrices in G. Since Ri is finitely generated, R{ is a finitely generated group, cf. [58], Corol lary 4.10. Therefore assertion 3) follows. □
7.2
Linear growth and repetitivity
In view of Theorem 7.1 and of the recurrent behaviour of sequences of matrices (see Theorem 1.16 and the comment following it), one might expect that GK(S') either is a natural number or it is infinite, for every S C Mn(K). It seems that, at this stage, one cannot hope for a descrip tion of semigroups S C Mn(K) of a given natural Gelfand - Kirillov dimension r. However, this is possible in the distinguished case r — 1, in which we say that S has linear growth. In this section we give such a criterion and compare it to a combinatorial property for S. Let 5 be a semigroup, and let s = s l 5 . . . , sm be a sequence of elements of S. A fc-factorization of 5 is a sequence t — £ l 5 . . . ,£*., where the tj's are the values of k consecutive factors of 5, that is, tj = 5^.5^+1 • • • s i i + 1 _i for j = 1 , . . . , k and some 1 < zx < i2 < • • ■ < U+i < m + 1. We say that t is a power fc-factorization if t\ — • • • = t^. A semigroup S is said to be repetitive if and only if, for each finite subset X of S and every integer k > 0, there exists a positive integer L = L(S,X,k) such that every sequence «s l 5 ... , sL of elements of X has a power fc-factorization. Repetitive semigroups were introduced by Justin in [51]. He then developed the theory in a series of papers. For example, a well-known corollary to Ramsey's theorem implies that every finite semigroup is repetitive (cf. [73, §4.1]). Our first main result (Theorem 7.17) describes all repetitive linear semigroups, generalizing [55] and [63]. It allows us to construct exam-
194
CHAPTER
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pies of finitely generated repetitive linear semigroups of subexponential growth, and also of polynomial but not linear growth. Our second main result (Theorem 7.20) describes all semigroups S C Mn(K) of GelfandKirillov dimension one. Such semigroups turn out to be repetitive. Fur ther, we show that if we impose an additional structural restriction on a semigroup S C Mn(K), then repetitivity becomes equivalent to sublinear growth (Corollary 7.22). Clearly, the class of repetitive semigroups is closed for subsemigroups and for homomorphic images. Also, a semigroup is repetitive if and only if all its finitely generated subsemigroups are repetitive. L e m m a 7.11 Let I be an ideal of a semigroup S. Then S is repetitive if and only if S/I and I are repetitive. Proof. Choose a finite subset X of S. For any k > 1, let m = L(S/I,X,k). Next, put Y = Xm n 7,n = L(I,Y,k) and L = mn. We claim that L satisfies the definition of L(S,X,k). Consider a sequence s = S i , . . . , SL of elements of X and its factorization u = u\,... , u n , where U{ = 5(i_i) m +i • ■ • «szm for i — 1 , . . . , n. If i ^ , . . . , un £ / , then u has a power ^-factorization by the choice of n. Otherwise, U{ (£ I for some i and therefore U{ has a power fc-factorization by the choice of m. The latter is a power fc-factorization of s as well. Hence S is repetitive.
□
L e m m a 7.12 Let U be a uniform subsemigroup of a completely 0simple semigroup M. Then U is repetitive if and only ifUf)G is repetitive for every maximal subgroup G of M. Proof. The necessity is clear. Thus, assume that every U H G is repet itive. Since repetitivity of U can be verified on finitely generated subsemigroups of [/, it is enough to consider the case where the sets of 7Zand ^-classes of M are finite. Then U = Vi U ■ • ■ U 14, where each Vi is the intersection of U with a principal left ideal of M. Moreover, each Vi = W{1 U • • • U Wim, where Wzj is the intersection of V{ with a principal right ideal of M. Hence Wi3 \ {6} either is equal to some U f! G or Wf- = 0. So Wij is repetitive. Therefore, it is enough to show that a semigroup S which is a finite union of its repetitive right (left) ideals i?2-,z = 1,. . . , r, is itself repetitive. By induction, it is enough to consider the case r = 2. By symmetry, we consider the right ideals only.
7.2. LINEAR
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195
Let I be a finite subset of U. Put X{ = X H R{,i = 1,2. Let m = L(R2,X2,k) and Yi = Xi U X i X m . Since Yi is contained in the repetitive right ideal i?i, the number L(R\,Yi,k) can be defined and we put n — m(L(Ri,Yi,k) + 1). We claim that n can serve as L(U,X,k). Let s = s l 5 . . . , s n be a sequence of elements of X. Let 5z-n . . . ,Sinii < • • • < z/, be the subsequence of all elements of 5 that are in X\. If s has m consecutive elements which all belong to X*i, then the factor formed by these elements has a power A:-factorization by the choice of m. Otherwise, every factor of s of length m contains an ele ment of Xi. Then zx < m and i j + 1 < ij + m for all j — 1 , . . . , / — 1. Hence n < (/ + \)m and by the choice of n we get / > L(i?i, Yi, &). Since 7?i is a right ideal, all elements Uj = 5Zj • • • <SzJ+1-i of the factorization u — u\ • • • u/_i belong to Yi. Therefore 5 has a power A;-factorization because / — 1 > L(R\,Yi,k). The assertion follows. D L e m m a 7.13 Assume that H is a subgroup of finite index in a group G. If H is repetitive, then G is repetitive. Proof. Since H contains a normal subgroup of G of finite index, we may assume that H is a-normal subgroup of G. Let I b e a finite subset of G and k > 1. Consider an infinite word X\X2 .. . with X{ G X. From Corollary 1.15 it follows that there exists p > 1 such that for every q > 1 there exist %i < i2 < • • • < i 9 +i such that z J + 1 — ij < p for j — 1 , . . . , g and the consecutive factors Uj = xt-.+1 • • • x 2j+1 , j = 1 , . . . ,g, are in i7. Now, let A be the set of all elements of G which are words of length at most p in the elements of X. Choose q — L(H,A,k). By the choice of q we know that ux • • • uq has a factor (as a word in U{) which is a kpower. Hence has such a factor. The assertion follows from Lemma 1.13. □ The next two fundamental lemmas are due to Justin [53], [54]. L e m m a 7.14 If a commutative semigroup S is repetitive, then S does not contain free commutative subsemigroups of rank two. Proof. Let X — (x^y)1 be the free monoid on x,y. Let A be the commutative free monoid on x, y and a : X —> A the natural map. The idea is to construct arbitrarily long words w in x,y such that a(w) £ A has no power 5-factorization.
CHAPTER
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For a word x\ • • • xn G X we denote by w[z,i] the factor X{ • • • Xj, where z < j . Consider the homomorphism : X —> A given by (x) = xzzzy, (y) =
xyyyy.
Let g £ X and / = (#). Suppose that a ( / ) contains a 5-power, say, /1/2/3/4/5 with fu = f[iu + l,z u +i] are such that a ( / i ) = • • • = a(/5). Let zu = 5rw + j u , where 0 < j u < 5, and let zu be the ru + 1-th letter of g, for u = 1 , . . . ,6. Write Su = (zu) and W u = Su[l,ju] (with W n = 1 if j u — 0). Consider the homomorphism rj : X —> Z 3 given by rj(x) = 1,77(2/) = 2. Define d u = 77(WU). Now iu+1 - iu — ju+i - ju (mod 5) for z = 1,2,3,4,5, and z n + i - iu is equal to the length of fu. Since all fu have equal length (as they are equal modulo a ) , we come to ju+2 ~ ju+i = iu+i - Ju (mod 5) for u = 1,2, 3,4. Suppose one of i u + 1 — juiu = 1,2,3,4, 5, is nonzero. Then it is easy to see that j \ — i 6 , a contradiction. Therefore j u = j for some j . Since all a ( / u ) are equal, also all rj(fu) must be equal. Therefore V{f[l^u]) + rj(fu) = */(/[l,iu+i]) implies that ry(/[l,z w ])-r/(/[l,z n + 1 ]) G Z 3 does not depend on u. Since r](j)(x) = 0 and r](j>{y) — 0, it follows that 77(/[l,z u ]) = rj(wu) = du. This leads to du+2 — du+i = <^u+i — du (mod 3) for u = 1, 2,3,4. Since rj(xx) ^ rj(xy) and rj(xxx) ^ rj(xyy), it follows that either all 2^ are equal or i 7^ 2,3. So, either j = 0,i = l , i = 4, or all z u are equal. In all these cases, if u G { 1 , . . . , 6}, we put i'u = iu + t with t = —j if j ^ 4, or £ = 1 if i = 4, and we define fu — f[i'u + 1, i'u+1]. It is easy to see that in every case /{f^f3/4/5 ls a 5-power modulo a. Moreover, since i'u = 0 (mod 5), we have ,fu = <j){gu), where gu are consecutive factors of g. But ( a ( x ) , a ( y ) ) C A is a free commutative subsemigroup, so a<j)(w) determines uniquely a(w) for any w G X 1 . Hence, the factor g i ^ ^ S ^ s of g is a 5-power modulo a. Now, put hi — x and / i n + 1 = (/zn) for n > 1. We have shown that every a(hn) G A has no power 5-factorization. The assertion follows. D The next result includes the fact that the infinite cyclic group is repetitive, which is a generalization of the van der Waerden theorem on arithmetic progressions (see [73]).
7.2. LINEAR
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197
L e m m a 7.15 Every almost cyclic group is repetitive. Proof. By Lemma 7.13 it is enough to consider the case of the infinite cyclic group Z. Let T C Z be a finite subset and k any natural number. Put m = sup(T) and m' = inf(T). Let X be a free semigroup of rank \T\ and (j> : X —> Z a homomorphism which is one-to-one on the generators of X. Assume first that T consists of positive integers. Let Y be the free semigroup on yu . . . ,y m . Define a homomorphism i\) : X —> Y by ip(x) = yt- • • y1 if x G X and (f)(x) = t. It is clear that cf>(f) is the length of ^ ( / ) , for every / G X. Let w G X, with ip(w) = z1 • • • zu z3 G y, 1 < j < t. For 1 < j < t let w3 denote the shortest initial factor of w such that <j)(wj) > j . If Zj = yp, we have (j)(w3) = j +p-
1.
Applying van der Waerden theorem ([73]) to the word ip(w) G Y we see that there exists an integer V = V(m,k + 1) > 1 such that for t > V there exist jx < ... < j ^ + i , some r > 1 and some yp such that Js+i ~ Js=r z js ~ VP
1 <s
Next, we define factors hs of w by W
js+i
= w
jshs
f° r 1 < 5 < fc.
From the displayed equalities it follows that (hs) = ( J s + 1 + p - 1) - (js + P - 1) = r. Hence, w has k consecutive factors hs whose images under <j> are equal if t > V. Since t > m'\w\, the latter holds if the length of w exceeds m'
Now, consider the general case, where A C Z is any finite subset. Consider an infinite word S in the generators of X. By Lemma 1.13 it is enough to show that it has k consecutive factors with equal images under cf>. Write S — fiSi, where /t- is the initial factor of S of length z, for i > 1. Assume first that (fi) are bounded. Then there are infinitely many indices ix < i2 < • ■ • such that <^(/,--) = (fij+1) for
CHAPTER
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all j > 1. If the words h3 are defined by fi3hj — ftj+l,j = 1,2,... , then we have 4>(hj) = 0. Since hj are consecutive factors of 5, we are finished. Therefore, suppose that (/;) are not bounded. Then S can be factorized as S = g\g
where gj G X is such that 1 < (f)(gj) < m = sup (A). In fact, it is enough to put S\ = S and Sj = gjSj+i where gj is the shortest initial factor of Sj such that (#j) > 0. Now, let Z be the free semigroup on the set {21,22,... } and rj : Z —> X be defined by rj(zi) — gi for i > 1. Then the homomorphism i/> = rj : Z —> Z satisfies tp(zi) > 1. Therefore, by the first part of the proof, the infinite word Ziz2z3 • • • has k consecutive factors u 1 ? . . . , uj. with equal images under I/J. Then /i2- = rj(vi), i = 1 , . . . , fc, are consecutive factors of 5 with equal images under (j). This completes the proof of the lemma. □ Recall that a semigroup S is divided by a semigroup T if T is a homomorphic image of a subsemigroup of S. L e m m a 7.16 Let G be an almost nilpotent group, and let S be a finitely generated subsemigroup of G. If S is not divided by a free commutative semigroup of rank 2, then the group of quotients SS~l is almost cyclic. Proof. Since SS~l is also almost nilpotent, we may assume that G — SS~l, so G is finitely generated. Let N be a nilpotent normal subgroup of finite index in G. Then [87], Lemma 7.5, implies that ( 5 f) N)(S Pi TV)-1 = A/", because [G : N] < 00. Also, N is finitely generated. Denote by n the nilpotency class of N. We proceed by induction on n. If n = 1, then Af is an abelian group. By the hypothesis, S has no free commutative subsemigroups of rank two. Therefore S P\ N is of torsion-free rank < 1. By [87], Proposition 23.1, rk(N) - r k ( 5 n N). A finitely generated abelian group of torsion-free rank < 1 is almost cyclic. Hence G is almost cyclic. Suppose that n > 1. Consider N/Z(N), where Z(N) is the center of N. Since the nilpotency class of N/Z(N) is less than n, the factor N/Z(N) is almost cyclic. Choose a normal subgroup A of N such that Z(N) C A, N/A is finite and A/Z(N) is cyclic. In particular, the index of A in G is finite. There exists an element x m A such that A is generated by x and Z(N). Hence A is abelian. As above it follows that A is almost cyclic. Therefore G is almost cyclic. □
7.2. LINEAR
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199
Now we can state the first main theorem of this section, [62]. T h e o r e m 7.17 The following semigroup S
conditions
are equivalent for a linear
1. S is repetitive, 2. S is not divided by a free commutative
semigroup of rank 2,
3. all the associated groups are locally almost cyclic. Proof. Let S C Mn(K) be a semigroup which is not divided by a free commutative semigroup of rank 2. Take any maximal subgroup D of Mn(K) intersecting 5, and consider the associated linear group G = gp(D f! S) C D. Certainly S does not contain free noncommutative subsemigroups of rank two. If we look at any finitely generated cancellative subsemigroup T of G, then it follows from the generalised Tits alternative, Theorem 6.11, that the group TT~l is almost nilpotent. Hence it is almost cyclic by Lemma 7.16. Since every subgroup H of G generated by a finite number of elements hi = Si^f 1 ,... , h^ — s^1^ where s l 5 £ l 5 . . . , s^, t^ £ DOS/is contained in the group of quotients of the cancellative subsemigroup of S generated by S i , £ i , . . . , 5 ^ , ^ £ 5 , we see that G is locally almost cyclic. Suppose now that S is a linear semigroup such that all associated groups of S are locally almost cyclic. By Corollary 3.8 5 possesses a finite ideal chain such that each factor is nilpotent or is contained in a completely 0-simple semigroup H with maximal subgroup isomorphic to a group associated to S. Clearly, all nilpotent factors are repetitive. So, consider a factor of the latter type. Every almost cyclic group is repetitive by Lemma 7.15. Therefore all associated linear groups of S are repetitive. From Lemma 7.12 it follows that the corresponding semigroup H is repetitive, too. Since the class of repetitive semigroups is closed for ideal extensions by Lemma 7.11, we see that S is repetitive. Finally, if S is repetitive, then S is not divided by the free commu tative semigroup of rank 2 by Lemma 7.14. This completes the proof of the theorem. □ Note that the 'if part of Theorem 7.17 can be also proved via com binatorial methods developed in [42].
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There exist repetitive linear semigroups whose growth is not even polynomial. Namely, the semigroup S C M 3 (Z), considered in Section 7.1, generated by the matrices / 1 0 0 \ / 1 1 1 \ e = 0 0 0 a n d l f = 0 2 1 Vo o I / v o o I / is of this type. In fact, S is repetitive by Theorem 7.17. On the other hand, there exist repetitive semigroups S C with polynomial but nonlinear growth.
Mn(K)
E x a m p l e Denote by 5 the subsemigroup generated in M 2 (Z) by the elements
a =
(
0 0)'
6 =
( o
°)
e =
( 0
0)'
where a, (3 E Z are relatively prime. It is readily verified that the elements ane6m, for m,n = 1,2,... , are all different. Therefore GK(5) > 2. Since S = {anebm\ m,n > 0 } U { a n , 6 n | n > 1}, it is easy to see that in fact GK(5) = 2. Note that S is contained in a union of four ^-classes of M 2 ( Q ) . All the associated groups are cyclic infinite, so that S is repetitive. The next two examples show that it is impossible to replace 'locally almost cyclic' by 'almost cyclic' in Theorem 7.17, even if the whole semigroup is finitely finitely generated. generated. semigroup is E x a m p l e There exist finitely generated repetitive linear semigroups whose associated groups are not almost cyclic. Namely, denote by S the subsemigroup generated in M 3 (Z) by
e=
/ 1 0 0 \ 0 0 0 \ 0 0 1 /
and g =
/ 1 1 0 1/2 \ 0 0
1 \ 1 . 1 /
7.2. LINEAR
GROWTH
1 an Then g = | 0 1/2" 0 0 n
AND
REPETITIVITY
201
bn an | , where for n > 0 1 = = =
a ^ j + 1/2"- 1 1 + 1/2 + - •• + 1/2"- 1 2-1/2"-1
and K
= b.,n-1 + O-n-l + 1 = a n _! + a n _ 2 -\ \- a2 + ai+bi+n - 1 = (2 - 1/2"- 2 ) + (2 - 1/2"- 3 ) + • • • + (2 - 1) + 1 + (n - 1) = 2 ( n - l ) + n - ( l + l / 2 + --- + l / 2 " - 2 ) = 3 n - 2 - ( 2 - l / 2 n - 2 ) = 3 n - 4 + l/2"-2.
Denote by D the maximal subgroup of Ms(Q) containing e. Put G (D n S)(D n S)'1. Then
egne eDPiS
C
Clearly, G is a torsion-free abelian group. Consider the associated linear groups of S. One of them is the cyclic group generated hy g. It contains all elements of rank three. Further, for any n, there exists k such that kbn is an integer. Therefore
W
e) =
/ 1 0 1 \ 0 0 0 . It follows that \0 0 1/ the set of all matrices of rank two in S forms a uniform component of S with associated group G being abelian and a periodic extension of a cyclic group. Thus S is repetitive by Theorem 7.17. If G were finitely generated, then it would be cyclic, because it is a torsion-free abelian group, which is a periodic extension of a cyclic
belongs to the cyclic group generated by
CHAPTER
202
7.
GROWTH
group. But {3n - 4 + l / 2 n _ 2 | n > 1} C Q is not contained in any set of the form aZ, for a € Q. Hence G is not cyclic, and so it is not finitely generated. A similar example can also be given in the case of positive charac teristic. E x a m p l e Let p be a prime, F p a finite field of p elements, and let S be the subsemigroup generated in M 3 (F p (x)) by
e = | 0
1 0 0 \ 0 0 a n d
0 0 1 /
/ 1 1 0 0 = 0 x 1
Vo o i
As above we can verify that, up to isomorphism, the associated groups of S are the cyclic group generated by g and a p-group G consisting of matrices of the form 1 0 * ' 0 0 0 0 0 1 Therefore S is repetitive. However, it is easy to see that G is not finite, hence not almost cyclic. L e m m a 7.18 Let G be a linear group, and let G = TT~l group T. Then the following are equivalent:
for a semi
1. G is repetitive, 2. GK(G) < 1, 3. for every x,y £ T there exist integers i,j such that (z,j) ^ (0,0) and x% — y-7. Proof. The equivalence of 1) and 2) follows from Theorem 7.17 and Theorem 7.1. If G is a group with GK(G) < 1, then for every x,y G G there exist ij,k,l > 1 with x{yj = xky£ and (z,j) ^ (£;,£). Therefore 3) is a consequence of 2). Assume that 3) holds. To prove 1) it is enough to consider the case where G is finitely generated. Condition 3) clearly implies that
7.2. LINEAR
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AND
REPETITIVITY
203
T is not divided by the free commutative semigroup of rank 2. From Theorem 6.11 it follows that G is almost nilpotent. By Lemma 7.16 G is almost cyclic. Therefore Lemma 7.15 implies that G is repetitive. □ We will use a special property of semigroups of linear growth, [67], the proof of Lemma 2.4, or [52]. If S = ( s i , . . . , s*.), then let a : X —y S be the homomorphism of the free semigroup on x l 5 . . . , Xk extending the map Si; >-)> X{, i = 1 , . . . , k. We equip X with the order -< defined by v -< w if \v\ < \w\ or |r;| = |w| and v precedes w lexico graphically. For each s G *?, a~1(s) contains a unique word v which is minimal with respect to ^< . Then a(v), viewed as a word in s i , . . . ,5^, is called the normal form of s. P r o p o s i t i o n 7.19 The following conditions are equivalent for a finitely generated semigroup S = ( s i , . . . , s^) 1. S has linear growth, 2. there exists a finite set F C Sl such that S C A, where A = {uzlv I u,v,z G F, z 7^ 1,£ > 1}. Moreover, if s — 5Z1 • • • st-m G 5 is the normal form of 5, then some uzlv is a factorization of s i\ ' ' ' sim-> for u-> v-> z and ^ as above. In particular, m > /. Now we are ready to describe all semigroups S C Mn(K) with linear growth. This involves a simple extension of condition 3) of Lemma 7.18. The result comes from [62]. T h e o r e m 7.20 Let S C Mn(K) conditions are equivalent:
be a semigroup.
Then the following
1. GK{S) < I, 2. for every x,y,z G S, there exist positive integers i,j,k,£ xlyzJ = xkyze and (z, j) ^ (k,£).
such that
If, in addition, S is finitely generated, then the groups associated to S are almost cyclic.
CHAPTER
204
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Proof. Since the set of words {xlyzJ | z, j > 0} in the alphabet x,y,z has n(n + l ) / 2 words of length < n, it is clear that 2) is a consequence of 1). Now assume that S satisfies condition 2). It is enough to consider the case where S is generated by a finite number of elements, say Si,. . . , SkWe can assume that S is a semigroup containing the zero matrix, since otherwise we could adjoin it to S. We know that S has an ideal chain 0 = To ^ h ^ * * * ^ h — 5*, where each Ij+i/Ij is a subsemigroup of a principal factor M{/Mi_i of Mn(K) and it is either nilpotent or isomorphic to a uniform component of S. By induction on m we shall show that GK(S/It-m) < 1 for m = 0 , 1 , . . . , f. This will prove 1). The case m = 0 is clear. Suppose that GK(£y7j + 1 ) < 1 for some j<£. C a s e 1. Ij+i/Ij can be identified with a uniform component U of S. As in the proof of Theorem 7.8 S/ Ij can be considered as a subsemigroup of a disjoint union S = (S\ Ij+i) U [/, where U is a completely 0-simple ideal of S and U intersects all nonzero W-classes of U. Given that GK(S/IJ+i) < 1, from Proposition 7.19 it follows that there exists a finite set F C S1 such that S \ 7 j + 1 C A, where A = {uz£v | u,z,v G F,H > 1}. Moreover, if s = 5Z1 • • • slrn G 5 \ Ij+i, then s = uz£v is a word of length < m in s 1 ? . . . , s*.. Suppose that £7 = M ( G , X , y , P ) for infinite sets X, Y. Take any element s £ (7, say 5 = 521 • • • s tm 7^ 0. Denote by 6 the shortest terminal factor of 5 that lies in U. Then s = ab and 6 = s?-tc for some £ and some c £ (S \ Ij+i)1- Hence b £ sZt A Therefore all possible b belong to the set k
A' = |J 5f-A = {uV)V I u W GF',I> 1} for a finite set F' C 5 1 . Moreover 6 = Situzlv with u,z,v G F and £ < m. Note that 5 and b generate the same left ideals in U. In particular, there exists TV such that the elements of U of length at most m lie in at most Nm £-classes of U. Similarly, the shortest initial factor e of s that is in U comes from a set A" of the same type (defined by a finite set F" C 5 1 ), and s,e generate the same right ideals in U. This means that each of the sets X, Y is a union of finitely many subsets, each corresponding to one of the sets {uzlv \ £ > 1} as u, z, v
7.2. LINEAR
GROWTH
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REPETITIVITY
205
range over F ' , F" respectively. The above shows also that every %-class of U is ^-related to an element of B = {uzlv
\u,z,ve
E,£>
1}
for the finite set E = F U F' U F" C S\ and is ^-related to an element of this set. From Lemma 3.24 we know that there exists a finite subset Q of U such that for every t G U we have qtq' ^ 0 for some q,<( G Q. Clearly, tlltq' in £/ in this case. Since E and Q are finite, but the set X of 7^-classes of U is assumed to be infinite, it follows that there exists a G Q such that {uz£v \ £ > l}a intersects infinitely many 1Z~ classes of U for some fixed u,z,v G E. Similarly, since Y is infinite, b{cx£d | £ > 1} intersects infinitely many £-classes of U for some 6 G Q and some c,d,x G F . Since there exists / G U such that a / 6 j^ 6 m U (because U intersects all nonzero %-classes of U)1 it follows that {uzivafbcxkd
\k,£>l}
intersects infinitely many ^-classes and infinitely many 7?,-classes of U. The same must be true of the set {zzyxk \ k,£ > 1}, where y — vafbc. Note that zlyxkKz£ y and z£yxkCyxk in U whenever z£yxk is nonzero. l h £ k If z yx = 0 for some £, fc, then z 'yx ' — 0 for all £' > £,k' > fc, which contradicts that {z£ yxk \ £',k' > 0} intersects infinitely many 1Z- and ^-classes of U. Hence z£yxk ^ 0 for every £, k. The cyclic semigroup generated by z acts by left multiplication on the set of 7£-classes of U. Hence z£y1Zzpy if and only if p = £. Similarly, yxkCyxr if and only if k — r. Therefore all elements z£yxk, for k,£ = 1 , 2 , . . . , are different. This contradicts condition 2), proving that either \X\ < oo or \Y\ < oo. Lemma 7.18 and Proposition 3.16 show in view of Theorem 7.17 that G is almost cyclic. Our next aim is to prove that G must be finite if one of the sets X and Y is infinite. For example, assume \X\ — oo. Suppose also that G is infinite. Then there exists a nonperiodic element x G U Pi H for a maximal subgroup H of U. Again, we see that there exist u, z, v G S/Ij, a G £/, such that \uzlva \ £ > 1} intersects infinitely many 7£-classes of U. Then the set \zlva \ £ > 1} also has this property. Choose / G U so that afx j^ 0 m U. As above, it follows that zivafxk,zpvafxr are not in the same 7^-class of U if p ^ £. Therefore all elements z£vafxk,
CHAPTER
206
7.
GROWTH
for k,£ > 1, are different. This contradicts condition 2) and proves the claim. Assume first that U = M(G,X,Y,P), where |G| < oo and \X\ < oo (the case where \Y\ < oo goes similarly). Then U = U, because H C\ S = H for any maximal subgroup H of U (since H D S generates H). Let T be any £-class of U. Then \T\ < \X\ -JG| < oo. As noted above, the number of £-classes of U containing elements of length < m in s i , . . . ,5fc is at most Nm. Therefore there are at most \T\Nm elements of U that are words of length < m. Also, by the induction hypothesis, there is a linear bound on the number of elements of length < m in S \ IJ+\. It follows that GK(S/Ij) < 1, as desired. Consider finally the second case where U — M(G, X, y, P) and X, Y are finite. By Theorem 7.17 and Lemma 7.18, since G is almost cyclic, and from the induction hypothesis and from Lemma 7.6 it follows that GK(5) < GK(5) = s u p ( G K ( S / / i + i ) , G K ( G ) ) < 1, proving the inductive claim again. C a s e 2. We consider the case that N = Sj+i/Sj is nilpotent, say N* = 9 where we may assume that t > 2. Consider any finite subset X ofS1. By 2), (3K(X)
> l)(Vx,y^,u; G
1
X){3iJ,k,£)
jL (k,l)<mdxlywzJ
=
xkywz*.
Define X' = X U{uzpvw,uvzpw
| u,z,v,w
G X,p=
1,...
,K(X)},
which is also finite. Let m be a positive integer. Using the technique of Case 1 (which relies on Proposition 7.19) there is a finite set F C S1 such that if w is a word in the generators of length < m, which, when evaluated in S/Sj, does not lie in AT, or lies in N but has no proper subword which evaluates to an element of A/", then w lies in Ai = {uzlv
| w, z, v G F, 1 < £ < m } .
Put F1 = F and, for i = 2 , . . . , t - 1, F{ = {Fi^Y
and A{ = {uzev \ u,z, v G F n 1 < £ < m}
7.2. LINEAR
GROWTH
AND
207
REPETITIVITY
Note that Ax C Al+i for i = 1 , . . . ,£ - 2. Let 5 = 5Zl • • • S{ ,, where m' < m. We prove that s G At-\. If s G S\Sj+i then 5 G Ai C A*_i and we are done. Suppose that s G N. Clearly s can be written in the form s = xxyi where Xi is the shortest initial factor of s which lies in N. If yx = 1 then s = x^. Consider the case yi ^ 1. If yi 0 TV then put x2 — y\. If yi G AT then y\ — x2y2 where x2 is the shortest initial factor of y\ which lies in N. Continuing in this way we obtain a decomposition s = x\ - • • xr for 1 < r < t where, for i — 1 , . . . , r, no proper initial segment of Xi lies in N. Hence x l 5 . .. , xr G Ax. We prove by induction on i = 1,. .. , r that xx • • • X{ G A{. The induction begins since xx G Ax. Let z > 1. Suppose, as inductive hypothesis, that xi • • • X{_i G At-_i. Also x2- G Ai C A 2 _i. Hence, for some ui,u2,Zi,z2,v\,v2 G i ^ - i , and some f 1,-^2 such that 1 <£u£2 < m, and X{ — u2z22v2.
x\ • • • X{-i — u\z^v\ Thus Xi • • • X{ —
UiZ1lViU2Z2l2V2.
We may suppose that £\ and £2 have been chosen so that (£\,£2) is minimal with respect to the lexicographic order such that the previous equation holds for some Ui,Zi,Vi,u2,z2,v2 G Fz-_i. But z\ViU2Z2
—
z v u
i l 2zl
for some z , j , k,£ such that 1 < i,j,k,£ < K(Fi-i) the lexicographic order. If £\,£2 > K(Fi-i) then X\ • ' ' Xi — U\Z^
u\U2Z2
V2 — U\ZY
and (i,j) < {k,£) in
VIU2Z2
V2
which is a contradiction, since (£1-(k-i),l2-(l-j))<(£1,£2) with respect to the lexicographic order. Thus £\ < K(Fi-i) or £2 < AT(Ft-_i). If £1 < # ( # - 1 ) , then u1z[1v1u2,z2,v2 G (F,_ 1 ) / = F 2 , so £ 2 x i - - - x z G At-. If £2 < K(Fi-i), then uuzuvlu2z 2 v2 G ( F t - i ) ' = ^i,
208
CHAPTER
7.
GROWTH
so again x\ • • ■ X{ £ A{. It follows by induction that x\ • • • X{ £ A{ for i — 1 , . . . , r. In particular, s £ A r C A*_i. It follows now that S/Sj has at most |A t _i| < ra|Ft_i|3 elements that are words in the generators s 1 ? . . . ,sk of length at most m. But SyS; = TV U S\SJ+1 and GK(5/5^+1) < 1 by an inductive hypothesis. Therefore GK(S/Sj) < 1, which completes the inductive step. Thus 1) holds. Finally, note that in the course of the proof we have shown that the associated groups are almost cyclic whenever S is finitely generated and satisfies 2). □ We note that the assertion of Theorem 7.20 is not valid for arbitrary semigroups. For example, consider any finitely generated periodic group which is not almost nilpotent. The following corollaries compare the class of repetitive semigroups with the class of semigroups of Gelfand-Kirillov dimension at most one. Corollary 7.21 Every semigroup S C Mn(K) dimension one is repetitive.
of Gelfand - Kirillov
Proof. The associated linear groups have Gelfand - Kirillov dimension at most one by Theorem 7.2. Hence they are locally almost cyclic by Theorem 7.1. Therefore Theorem 7.17 implies that S is repetitive. □ Recall that a linear semigroup S C Mn(K) is the union of its uniform components if and only if, for every a £ 5 , there exist x,y £ S such that the rank of xay is equal to the rank of a, and xay lies in a maximal subgroup of Mn(K). For example, regular semigroups are of this type. The proof of Theorem 7.20 shows that uniform components U of a finitely generated semigroup S C Mn(K) of polynomial growth can be of two types only. Either U is a completely 0-simple semigroup over a finite group and U has finitely many 7^-classes or finitely many £-classes, or U intersects finitely many ^-classes of Mn(K) and the associated group is infinite. So, for S equal to the union of its uniform components, this can be rephrased as: (*) S is contained in the union of a finite number of TZ- and £-classes of Mn(K) and if S D H is infinite for a maximal subgroup H of Mn(K), then S intersects finitely many Hclasses of Mn(K) that are TZ- or £-related to H.
7.3. ALMOST
SEMISIMPLE
SEMIGROUPS
209
Corollary 7.22 Assume that S C Mn(K) is a finitely generated semi group that is the union of its uniform components. Then the following conditions are equivalent 1. GK(5) < 1, 2. S is repetitive and S satisfies condition (*), 3. the groups associated to S are almost cyclic and S satisfies condi tion (*). Proof. The proof of Theorem 7.20 actually shows that 1) and 3) are equivalent. The equivalence of 2) and 3) follows from Theorem 7.17 and Proposition 3.16. □ Note that in the second example after Theorem 7.17 S is repetitive and intersects finitely many ^-classes of M 2 (Q), but GK(S') > 1. Thus Corollary 7.22 does not extend to arbitrary semigroups of matrices.
7.3
Almost semisimple semigroups
The first aim of this section is to show that the growth function of a finitely generated linear semigroup S C Mn{K) is controlled by its behaviour on finitely many cancellative subsemigroups of S. From Sec tion 7.1 we know that, if the growth of S is polynomial, then every cancellative subsemigroup T of S has a group of quotients G C Mn(K) that is almost nilpotent and of finite rank. We prove that the latter condition, strengthened by the hypothesis that the unipotent radical of every such G is finite, is sufficient for S to have polynomial growth. Moreover, the degree of growth of S is then bounded by a polynomial / ( n , d) in n and the maximal degree d of growth of cancellative T C S. Let S = ( s i , . . . , 5 r ) be a semigroup. By \s\ mean the minimal length of s £ S in the generators s l 5 . . . , sr. If A C S is a subset, then by /^(ra) we denote the cardinality of the set of elements of S that can be written as words of length at most m in s l 5 . . . , sr. In particular, ds(m) = / s ( r a ) is the growth function corresponding to this set of generators. Let J be an ideal of a finitely generated semigroup S = (su . . . , sr) and J C I for a completely 0-simple semigroup / . Let h(m) denote the number of %-classes of / that contain elements of S that are words
210
CHAPTER
7.
GROWTH
of length < m in s i , . . . , s r . We call h[m) the growth function for the number of %-classes of I with respect to J and the generators S i , . . . , sr. L e m m a 7.23 Let I be a completely 0-simple ideal of a semigroup U and S = ( s i , . . . , s r ) a subsemigroup of U. Let J = S f) I and suppose that S/J has polynomial growth of degree d. Then the growth function of the number of H-classes of I with respect to J and the generators Si,. . . , sr is bounded by a polynomial of degree not exceeding Ad. Proof. Let Dn = {w = Si1 - - - S{k G J\k < n, proper subwords of w are not in J}. Then \Dn\ < r(ds/j(n)) since every w G Dn is of the form w — S{Z^ where i G { 1 , . . . , r} and z (£ J. Let 0 7^ w = SiY • • • S{k G J. Then w = Wiv, where w\ — s2l • • • sH, v — 5 n+i ' ' " sik a n ( ^ ^ i 1S t n e shortest initial factor of w that lies in J. Write w — x\y\, where yx is the shortest terminal factor of Wi (treated as the word s^ • • • st-t) that lies in J. Note that v,X\ can be empty words. Then w = XiyiV, so that wTZxiyi in / . Moreover, x\ (£ J and y\ G Dn. Hence, such an element w can lie in at most ds/j(n)\Dn\ 7^-classes of / . A symmetric argument shows that w can lie in at most ds/j(n)\Dn\ £-classes of / . Hence, h(n) < ds/j{n)2\Dn\2 < r2ds/j(n)A. The assertion follows. D Corollary 7.24 Let I be a completely 0-simple ideal of a semigroup U and S = ( s i , . . . , sr) C U,J = Snl. If S/J has polynomial growth and for every H-class H in I we have fsnH(m) ^ cmd for some c,d > 0, independent of m and H, then S has polynomial growth. Proof. fj(m) < cmdh(m) has polynomial growth by Lemma 7.23. Hence, the assertion follows from the fact that d(rn) < fj(m) + ds/j(m).
a Our first aim is to show that the growth of a linear semigroup S is in some sense determined by the cancellative subsemigroups of S. T h e o r e m 7.25 Let S — ( s ^ . . . , s r ) C Mn(K) be a semigroup. For each uniform factor Ii/Ii+i of a structural chain S = IQ D I\ D • ■ • D h of S choose a cancellative subsemigroup T that is of the form S 0 H for a maximal subgroup of Mn(K). Then the following conditions are equivalent
7.3. ALMOST
SEMISIMPLE
1. S has polynomial 2-
SEMIGROUPS
211
growth,
has polynomial growth for every cancellative subsemigroup T of S which is of the form T = S P\ H for a maximal subgroup
/T("I)
H ofMn(K), 3- frX™) has polynomial growth for every Tt-. Moreover, in this case, GK(5) < g{n,d), where g is a function of n and the maximal degree d of polynomials bounding the functions fo^m). Proof. Proof. The implications 1) =» 2) and 2) => 3) are obvious. Thus, assume that 3) holds. We will inductively show that GK(5/J t -) < oo. If S/h is nilpotent, then it is finite and G K ( 5 / / i ) = 0. If S/h is uniform, then from Corollary 7.7 it follows that G K ( 5 / / i ) = GK(7\). Suppose we know that GK(S/I{) < oo. Again consider two cases. If Ulk+i is nilpotent, then (Ii/Ii+1)k = 0, where k < (n\ j denoting the common rank of matrices in Ii\Ii+1. Hence GK(5// t - + i) < k GK(S/I{) < cx3 by [67], Corollary 5.10. Assume now that Ii/Ii+i is uniform. From Lemma 3.24 we know that for every %-class H in Mn(K) such that T — Hfl (Ii \ I{+i) 7^ 0 there exist x , y G /,- \ /t-+i (chosen from a finite set Z) such that 0 ^ xTy C T{. Since z \-t xzy, z G T, is a one-to-one mapping, it follows that / r ( m ) ^ / r t ( m + ^0> where A^ = 2{max|u;| \w £ Z}. Hence frim) is bounded by a polynomial independent of the choice of H. Corollary 7.24 implies that GK(S/I{+i) < oo. This establishes 1). Existence of the function g follows from the above proof and the fact that every structural chain must satisfy t < 2 n + 1 . □ As noted in examples in Section 7.1, there exist finitely generated semigroups S C Mn(K) whose all maximal cancellative subsemigroups have finite Gelfand - Kirillov dimension, but the growth of S is not polynomially bounded. From Proposition 7.10 we know that, whenever a finitely generated S C Mn(K) has polynomial growth, then every group associated to S is almost nilpotent and is finitely generated modulo its unipotent radical. Our main goal is to show that the problem (as that in the examples mentioned above) can only come from the unipotent radical of these groups. We will prove that for a class of S C Mn(K), the polynomiality
CHAPTER
212
7.
GROWTH
of the growth of S can be decided by measuring the Gelfand - Kirillov dimension of all (in fact, of finitely many) such H. We will need some preparatory lemmas. 1 0 0 0
L e m m a 7.26 Let s l 5 . . . , s r G M n ( / \ ) , e idempotent
of rank one, and let 0 7^ es zi
ij G {1, •. • , r } , m > 1, £ G K.
■ S:
e
G Mn(K) x 0 0 0
be the
for some
Then
1. If A is the set of entries of the matrices s i , . .. s r , then x is a sum of nm~l elements of the form ax • • • a m , a2- G A. 5. Le£ E be a subfield of K with [K : E] — k < 00, arcrf : Mn(K) —> Mnk(E) the embedding coming from the regular repre sentation K —> Mk(E). If B is the set of entries of the matrices mv(a) for a G A with minimal valuation and, if 3) holds, then \v[x)\ < mN for some N dependent on A,C and I only. Proof. 1) is clear. 2) follows from 1) and the fact that x can be viewed as a scalar k x k submatrix of Mnk(E). 3) By the hypothesis x~l — ai(x) • • • 07(2;), for cr,- G G(K/E). Thus, l x~ is the nonzero entry of the product of eaj(sil) ■ • • ~cf3(slrn)e, j = 1 , . . . , /. So, the assertion follows as in 1). 4) By 1) we have v(x) > min{i;(ai • • • am)} > mv(a). But 3) implies that v(x~l) > mlv(c) for some c G C. Hence 4) follows. □
7.3. ALMOST
SEMISIMPLE
SEMIGROUPS
213
L e m m a 7.27 Let X = {xtJ),i = = 1 , . . . ,n, j = 1 , . . . ,t, 6e a matrix of rank t with real coefficients. Then there exists a G R such that, if \a H■ < M Ka;ii ---■+■ a+t Xatixtlt|\< M l^n K^ni + /or some au...
,auM
\-atxnt\
< M
G R, then |a3| < aM /or j = 1 , . . . ,t.
Proof. If t = 1, then |on f c l| < M for every k and xn / 0 for some i because X has rank 1. Hence, we can put a = frn^M. Assume that t > > 1. Let for example xn be a nonzero element of the first column of X. For j = 2 , . . . , n we have |a2(xj2 - x12xnx1)
+ ■■■ h a*(xJt -
< \alXjl + ••• + atxjt| + {a.xnxj^
1 )] xx^x^ ltXjlx^)\
+ ••• +
a^ux,^]
<M + M\Xjlxrf\. This yields a system of inequalities with an (n - 1) x (t - 1) matrix Y that has rank t - 1. (We performed elementary row operations on X eliminating the nonzero entries below in). This allows us to complete the proof by induction on t. □ L e m m a 7.28 Le£ L D D = F(tu... ,tu))D D F be field extensions such that [L : D] = p < oo and tu... ,tn zs a transcendence basis of D over F. LetS=(su... ,sr) C Mn(L), and a G A u t ( F ) . Assume £hat e G Mn(L) is a diagonal idempotent idem-potent of rank 1 and es2l • ■ • •lme eas the only nonzero entry x < GE F. F. There There exists exists afield afield isomorphism isomorphism r™ r™LL —> —> L'L' such that the only nonzero entry of ef(sn) ■ • ••(s, m )e G € Mn{U) is equal to a{x), where T : Mn(L) C Mn{V) is the natural"'extension of r. Proof. Extend u to D, sending each *,- to tf,-, —► u and then to a' : MP{D) —>■ MV{D). Let L be embedded into Mp(D) via the regular representation. Put L' = a'{L) and extend r = a\L to r : Mn{L) (L) —>■ M„(L')Mn(L'). All these mappings agree with a when restricted to F. Hence, if / denotes the identity matrix, then er(Sli)
■ ■ -T(sim)e
= r{estl • ■ ■ sime) = T(e(xl)e)
has the nonzero entry equal to a(x).
□
= ef(xl)e
=
ea(x)e,
CHAPTER
214
7.
GROWTH
P r o p o s i t i o n 7.29 Let S — ( 5 1 , . . . , s r ) C M(K) be a finitely generated semigroup. Let e G Mn(K) be a diagonal idempotent of rank one and z i , . . . ,zt free generators of a subgroup H of K*. Suppose that A is a subset of S such that for each a — Sj1- • • Sjm G A with 0 / s = eae^jk G { l , . . . , r } , r a > l , the nonzero entry of the matrix s is of the form x = z% \ ' " * z%t for some ij G Z. Then there exists a constant M (independent of a,m and j i , . . . ,jm) such that \ij\ < Mm for j = 1,. . . ,£. Proof. We shall proceed by induction on t. However, the case t — \ will be dealt with later. First, we show how to proceed with the induction step. Let L be a finitely generated subfield of K such that S C Mn(L) and let A"0 C / \ o ( ^ i , . . . ,t s ) C L for the prime subfield K0 of K and a transcendence basis £ 1 , . . . ,t s of L over K0. It is known that this can be chosen so that E = A"o(^i, • • • ,t s ) C L is a separable field extension, cf.[135], Theorems 11.13.30 and 11.13.31, so that extending L we can assume that E C L is a finite Galois extension. Let TV = NL/E be the corresponding norm. Then for G — G(L/E) we have l[a(x)
= N(x) =
N(zY)---N(zl<)
a(EG
and clearly N(x),N(z{) G E. The only nonzero entry of the matrix e s ^( ji)'' '&{sjm)e 1S equal to cr(x), where a is the natural extension of a to an automorphism of Mn(L). We consider two cases. Case 1. NL/E(x) is not a root of unity. Let for example N(zi),... , N(zp), 1 < p < t, be a maximal subset of {N(z{)\ i = 1 , . . . , t} that freely generates a subgroup of L*. If p < t, then there exists k G N such that N(zj)k = N(zi)aiJ ■ - - N(zp)api for j = p + 1 , . . . ,£ and some a%3 G Z. Hence, N(x)k = N{z1)h • • • N(zv)h*, where 6g = kiq + z p+1 a g?p+1 + • • ■ + itaq,t for g = 1 , . . . ,p. Moreover, the matrix in eMn(L)e with the only nonzero entry equal to N(x)k is a word of length (m + 2)k\G\ in the generators of the semigroup S' = <e,
7.3. ALMOST
SEMISIMPLE
215
SEMIGROUPS
Ml(a') < M(m + 2)k\G\,q = 1 , . . . ,p, for a constant M, where /(a') is the minimal length of a' in the generators of 5'. Note that M is independent of (m + 2)Jb|G|, and so of m because k depends on zu ... , zt only, and M is independent of the choice of a € Aaxid j u . . . ,jm. Thus, \b.\ < M'm for q = 1,...p, where M ' = 3Mfc|G|. Now xkz7bl
=y\2---y\',
where y3 = z) for j = 2 , . . . ,p, and y, = ^*zf° I j for j = p + 1 , . . . ,t. But y 2 , . . . , yt are free generators of a subgroup of L*. Hence the above equality fulfills the hypotheses of the proposition with respect to the semigroup S" = {S, u, v), where u,v e Mn(L) are such that the nonzero entries of u = eue, v = eve are zu z7l respectively, and the subset A" = {(eae)V>|a G A,b, > 0} U {(eae)kubl\a £ AM > 0}. Consequently, the induction hypothesis implies in view of |6X| < M'm that \ij\ < M"((m + A)k + \bi\) < (5k + M')M"m
for j = 2 , . . . .t
for some M" (independent of a e A,m,Ju... ,jm) because xkz7bl is the nonzero entry of ewe for a word w of length (m + 4)k+ |&i | in S". In view of the bound |6X| < M'm, this completes the inductive argument (establishing a linear bound on all ^ | ) . Assume now that p = t. Then we replace the elements x,zu... ,zt by N(x),N(Zl),... ,N(zt) G E. Thus, passing to the semigroup S' and its subset {eax{a)ea2(a)e ■ ■ ■ ea]G]{a)e\ a € A} we can assume that x, zu ... , zt e E. If A'0 = E, and ch(A') = p > 0, all z{ G F p have finite order, a contradiction. Otherwise, we treat E as the field of quotients of the polynomial ring R = Fp[tu... ,ts] or of R = Z[iu ... ,Q. Let q be a prime element of R and S the q-adic valuation on E. Write c, = 5(zt) for i = 1 , . . . ,t. From Lemma 7.26 2) it follows that a: is a sum of (n/e)" 1 " 1 elements of the form bx ■ ■ ■ bm with 6,- chosen from a finite set B C E independent of x. Writing bt = fig-1 for some fug G R, we see that x = fg~m where / is the sum of all / i • • ■ fm. If q is a polynomial of a nonzero degree, then it follows that 8(x) does not exceed the maximal degree of / x ■ ■ ■ fm. If q is a constant, then ch(A') = 0 and 5(x) does not exceed the absolute value of any nonzero coefficient in the polyno mial / . Therefore 6(x) < (nk)"1-^"1 for some h G R independent of x. Hence, we always have |c 1 ? 1 + • • • + ctit\ < Nm for some AT (dependent Oil 5]_, . . . , 5 r only).
CHAPTER
216
7.
GROWTH
Suppose ci = • • • = ct — 0 for every such q. Then zu... ,zt lie in the group of units of R, cf.[135], Corollary VI.10.3. Since the latter is finite, this is a contradiction. Thus, for some such 5 one of the c,-, say cu is nonzero. Now, x = z\l (z2z^C2)i2 • • • (ztz^Ct)lt, where kx = CiH + c2i2 + • • • + ctiu a n d Vi = ZiziCi,i = 2 , . . . ,t, are free generators of a subgroup of L*. Then
and the left side is the nonzero entry of ewe for a word w of length < m-\-Nm in s i , . . . , s r , u, t?, the latter two defined as above. Applying the induction hypothesis to the semigroup (S,u,v) and its subset consisting of these elements iu, we see that \ij\ < M(N + l ) m for j = 2 , . . . , t, for some M (independent of ra). Since |cxZi + • • • + c t z t | < Nm, and cx 7^ 0, we also get a desired linear bound on |zx|. This completes the inductive argument in Case 1. Case 2. NL/E{X) is a root of unity. As above, passing to xk = (z\)n • • • (z^)u for some k > 1, we can assume that NL/E(X) = 1. Then Lemma 7.26 4) shows that \v(x)\ < Nm for some A^ (independent of m) and every valuation with value group Z of L. If for some such v one of v ( z i ) , . . . ,v(zt) is nonzero, then we can decrease t as in Case 1, performing the inductive step. Therefore, we need to consider only the case where v(zi) — ••• = v(zt) — 0 for every such v. It is well known that this implies that each Z{,z~x is integral over C — Z or C = F p , depending on the characteristic of K. (In fact, if a G L is not integral, then v(a) ^ 0 for the a~ 1 C[a~ 1 ]-adic valuation v on C [ a - 1 ] . Then v can be extended to a valuation on L that is again discrete of rank one, cf.[135], pp.51,52 and p.85). If C = F p , we have (zi) Q Fp[zi] is finite, contradicting the assumption on z{. Thus, assume that ch(A') = 0. Let F be a finite Galois extension of Q containing z i , . . . , 2 t . Extending L we can assume that F C L. We know that the integral closure B of Z in F contains the group H generated by zu... ,zt. Let u — ux + u2, where a{ G G ( F / Q ) , z = 1 , . . . ,u 1 ? are such that (Ti(F) C R and G{,(J\ G G ( F / Q ) , Z = Wl + 1 , . . . ,u, are the conjugate pairs of the remaining automorphisms of F. It is known that there exists k > 1 such that the u x t matrix
(lnk#)lki
7.3. ALMOST
SEMISIMPLE
SEMIGROUPS
217
has rank t, cf.[83], 47.2. Using the presentation xk = (z*)1'1 • • • (^)t'« and the fact that zf,... ,z\ freely generate a subgroup of L*, as be fore, we can reduce our problem to the case where k = 1. Now, for every a = <jt- G G ( F / Q ) , we have |cr(x)| = H ^ ) * 1 • • • cr(^) u |. Moreover W(x)\ < am and |cr(x) _ 1 | < bm for some a,6 £ R independent of m, by Lemma 7.26 1),3) and Lemma 7.28. Therefore | In \a(x)\\ < Mm for some M independent of m. Hence liilnlor^i)! + . . . +
Ztln|a(^)||
<
Mm.
Lemma 7.27 yields \i3\ < Nm for j = 1 , . . . , t , and some N independent of m. This completes the inductive argument in Case 2. It remains to check the validity of the assertion in case t = 1, that is x — z%i . However, it is clear from the above reasoning that it takes care of this case, too. This completes the proof of the proposition. □ We are now in a position to prove the first main result of this section. T h e o r e m 7.30 Let S — (s1:... , s r ) C Mn{K) be a finitely generated semigroup such that every nonempty intersection T = S fl F with a maximal subgroup F of Mn(K) has a group of quotients G C F that is almost nilpotent and finitely generated modulo its unipotent radical U. Then the number of cosets in G/U that contain elements of T that are words of length < m in S i , . . . , sr is bounded by a polynomial in m, depending only on the torsion-free rank of G/U and on the rank of matrices in T. Proof. From [134], Theorem 3.6, we know that the group G of quo tients of T has a nilpotent normal subgroup D of finite index that is triangularizable in fMn(K)f = M r a n k ( / ) ( / i ) , where / = f2 diag(D). Then diag(D) has a subgroup B of finite index with all rank one diagonal projections being torsionfree subgroups of K* (the group D/(D H U) ~ diag(D) C (/i*) rank (/) is finitely generated since it has finite index in the finitely generated group G/U). Thus, there is a finite set Z C G such that for every g e G there exists z G Z with zg G D and diag(zg) G B. Let
218
CHAPTER
7.
GROWTH
e < / be a diagonal idempotent of rank one. From Proposition 7.29 it follows that for each nonzero a = ezs^ • • • S{ke,k < m,z G Z, such that diag(zsil • • • S{k) G 5 , the power of each generator of the projec tion eBe of B appearing in the presentation of a has absolute value < N(m + 1), where TV is independent of m and Z i , . . . , ifc. Hence, the set {ezse\ z G Z^diag(zs) G B, length of 5 < TO} has at most (2N(m + 1) + 1)* elements, where t is the rank of eBe. Since this argu ment works for every diagonal of rank one e < / , we get a polynomial bound on qm — \{diag(zsil • • • Sik) G B\z G Z,k < m}\. Now, the num ber of cosets in G/U that contain elements of T that are words of length < TO in S i , . . . , sr does not exceed qm\Z\. This proves the theorem. □ For any semigroup S we define the rank of S by the formula rk(S') = sup{t| S has a free commutative subsemigroup on t free generators}. It is clear that, if S C Mn(K), then rk(5) = r k(T) for a cancellative subsemigroup T of S contained in a maximal subgroup of Mn(K), see Proposition 3.7. If T has a group of quotients G which has a finite nor mal subgroup A such that G/A is almost abelian, then rk(T) = rk(G) and the latter is equal to the Gelfand-Kirillov dimension of G, [87], Proposition 23.2. In particular, this applies to the case described in Theorem 7.31. (Note that groups associated to a semigroup S satis fying conditions of this theorem are in fact almost abelian and finitely generated, cf. Lemma 5.25 and Proposition 7.10). The above result leads naturally to a condition under which we are able to prove the finiteness of GK(5). Namely, assume that there exists q G N such that l^foi*)] < Q for every t G T, where fa : T —> G/U is the natural homomorphism. Suppose that a i , . . . , ar G G are distinct elements that lie in the same coset of U. Since G is the group of quotients of T,ai — WiV~l for some W{,v G T. Then W{ = a^v G T are in the same coset of (7, so that r < q. It follows that the above condition is equivalent to saying that U is a finite group. (So, U must be trivial if ch(/i) = 0.) In this case, Theorem 7.30 yields a polynomial bound on / ^ ( m ) . T h e o r e m 7.31 Let S C Mn(K) be a finitely generated semigroup of finite rank r such that every group G associated to S is almost nilpotent, (equivalently, S has finite rank and satisfies an identity). If each G has finite unipotent radical, then S has polynomial growth of degree bounded by / ( n , r ) , where f is a function of n and r.
7.3. ALMOST
SEMISIMPLE
SEMIGROUPS
219
Proof. We know that G has a normal subgroup D of finite index with finite unipotent radical [/, and D/U embeds into (L*)k for a finitely generated subfield L C K and some k < n. Since ik(G) — rk(Z)/C7), [87], Proposition 23.2, it follows that D/U is finitely generated. In view of Theorem 7.25 we need to show that frirn) is polynomially bounded for every cancellative subsemigroup of S of the form T = S Pi G. This is a direct consequence of Theorem 7.30. The existence of the function / ( n , r ) follows from the proof. □ R e m a r k 1) From the remark preceding Theorem 7.31 and the proof it follows that S C Mn(K) has polynomial growth if and only if the asso ciated groups G = TT~l are almost nilpotent and for each t £ T that is a word of length < m in the generators of S we have l ^ 1 (f>x{t)\ < q(m) for a fixed polynomial q. In fact, in this case / r ( m ) a ls° is polynomially bounded. 2) The hypotheses of the theorem amount to say that each G is an almost diagonalizable group (so, our S C Mn(K) can be called 'almost semisimple') which is also finitely generated. If K is of characteristic zero, then the hypotheses of the theorem can be reformulated, due to the following observation. P r o p o s i t i o n 7.32 Assume that ch(A') = 0. Then S C Mn(K) is al most semisimple if and only if the groups associated to S are almost nilpotent and a power of every element s £ S is diagonalizable. Proof. The necessity is clear. So, suppose that D is a nilpotent sub group of finite index in a group G associated to S. If Du, Ds denote the sets of unipotent and diagonalizable elements of D respectively, then DU,DS are subgroups and (DU,DS) — Du x J9S, [134], Theorem 7.11. If a power of every s £ S is diagonalizable, then S 0 D C Ds because ch(/\) = 0. Hence D = {S n D)(S H D)~l C D8, so D is diagonalizable.
□
We conclude with the case of 2 x 2 matrices, in which the description can be simplified. Corollary 7.33 Let S C M2(K) be a finitely Then the following conditions are equivalent
generated
semigroup.
CHAPTER
220 1. S has polynomial
7.
GROWTH
growth,
2. S satisfies an identity and has finite rank, 3. each cancellative subsemigroup of S embeds into a finitely ated linear group of polynomial growth,
gener
4- S fl GL2(K) has an almost nilpotent group of quotients, if non empty, and every nonempty intersection S H H with a maximal subgroup H ^ GL2(K) of M2(K) lies in a finitely generated sub group of H. Proof. Clearly, S fl GLn{K) is finitely generated (if nonempty) when ever so is S'. If 1) holds, then groups associated to S are finitely generated modulo unipotent radical by Proposition 7.10. So all such groups are fi nitely generated. Hence 3) follows via Proposition 3.7 and Theorem 7.1. A nilpotent subgroup of GL2(K) is almost abelian, cf.[134], Chap ter 7. Hence 2) is a consequence of 3), cf. Lemma 5.3. A subgroup of finite rank of the multiplicative group of a finitely generated field is finitely generated by Lemma 1.7. Hence, 4) follows from 2). The implication 4) =>- 1) is a consequence of the proof of Theo rem 7.31 because the unipotent radical of every S fl H is trivial. □ We conclude this chapter with some remarks concerning the general case. From Proposition 7.10 we know that the groups associated to a linear semigroup S are almost nilpotent and finitely generated modulo the unipotent radical, whenever S has polynomial growth. The basic idea of the approach relies on the fact that every almost nilpotent lin ear group is isomorphic to a subdirect product of an almost unipotent linear group and an almost diagonalizable linear group. The interaction between the 'group elements' z of S (that is, elements z G S fl D for a maximal subgroup D of Mn(K)) with the 'unipotent parts' of the uniform components U of S that lie below z decides whether S is of polynomial growth. It was shown in the course of the proof of Theo rem 7.30 that the problem does not come from the 'semisimple part' of the group associated to U. We introduce a formalism that allows us to study S via its cor responding 'decompositions'. Let D b e a maximal subgroup of Mn(K)
7.3. ALMOST
SEMISIMPLE
SEMIGROUPS
221
such that SC\D generates an almost nilpotent group G and lies in a uni form component U of S. Let T C Mn(K) be the semigroup generated by SUG. Denote by Su the Rees factor of S with respect to the ideal Sj\U, where j is the rank of matrices in U and Sj = {x G S | rank (a;) < j}. From the structure theorem and from Lemma 3.11, it follows that the layers of T are of the form T^T^ = UUU^U.. .Ut//U N', where U is the completely 0-simple closure of U in Mj/Mj_i, U = U[, U^. • • ,U't are the uniform components of T and N' is the nilradical of Tj/Tj-i. More over, the semigroup Sg — T/(Tj\U) is a disjoint union of S\Sj = T\Tj and f/, so it contains SuT modulo Tj_i and N' is a subdirect product of T/(Tj \ U-),i — 1 , . . . ,£. So, by [67], Corollary 5.10, GK(S') < oo if and only if each of S/(Sj \ Ui) has polynomial growth. So, we consider Su only. Since S is finitely generated, it is easy to see that T C Mn(L) for a finitely generated subfield L of K, see Proposition 7.10. Let ~ . ,f>v denote the congruence Ann(U)
°
{(x,y) G S0 x % | x,y G C/, t/(x - y)t/ = 0}. The construction of Lemma 3.25 allows us to find a linear representation £ of T such that £(T \ Sg) = 0 and £ restricted to Sjj is determined by ~Anntu\ ■ So, the kernel of the map K0[T] —> K0[£(T)] is nilpotent, hence again by [67], Corollary 5.10, we may assume that Sg is linear and all matrices of the least nonzero rank in SQ are in U. L e m m a 7.34 Assume that G is an almost nilpotent group associated to a semigroup S C Mn(K). Then there exist normal subgroups GU,GS of G such that Gu is unipotent, Gs is diagonalizable, and G has an almost diagonalizable rational linear representation G^ and an almost unipotent rational linear representation G^ with G^ ~ G/GU^G^ ~
GIG,. Proof. We can assume that K is algebraically closed. It is well known that the Zariski closure F of G also is almost nilpotent, so that the connected component Fc of F must be nilpotent (use [134], Lemma 5.3, Theorem 5.11 and Exercise 7.2). Let Gu = G C\ Fu where Fu is the unipotent radical of F c , and let Gs be the maximal closed diagonalizable normal subgroup of G, cf. [134], Theorem 14.22. Then the assertion follows from [134], Theorem 6.4, Lemma 6.5 and Lemma 6.6. □
CHAPTER
222
7.
GROWTH
So G is a subdirect product of an almost diagonalizable linear group G^ and an almost unipotent linear group G^u\ The map G —> G^ has a natural extension to U = M(G,X,Y,P)
—► U^
=
M(G/GS,X,Y,PU).
It is easy to see that we get an extension to a homomorphism SQ —> S^
= {SQ \U)U
constructs U^ Sk
]
C/(n), which is identity on SQ \ U. Similarly, one
— M(G/GU1X,Y,
Ps) and a homomorphism SQ —>
= (S \ U) U f/W. We approach Su via the embedding Su —>
Sffi x Su\ where S^\S^ homomorphisms of SQ.
are the images of Su under the respective
Let (f)s be the linear representation of SQ extending G —> G/Gu and one-to-one on Sfr\U, (f>s(Sfj) x S~
which arises from Lemma 3.26. Let ip : SQ —>
be the resulting map. If
xHy in U and consequently x — y, because the kernel of <j)s\G is equal to Gu and Gu trivially intersects Gs. So, (f)s factors through S~ and SQ embeds into (J)S(SQ) X S~ . Now, s{U) is an ideal which is a uniform component whose associated group is also rationally isomorphic to G^s\ Trying to prove that GK(S') < oo by induction on the number of uniform components, we may assume that GK(Su/U) < oo. Since Su/U ~ S{J]/U{s) ~ s(su)/Mu) a n d t h e g r o u P associated to (f>s{U) is almost diagonalizable, Theorem 7.25 and Theorem 7.30 imply that GK(S(S[/)) < oo. So, it remains to see when S)j has polynomial growth, because this is now equivalent to GK(Su) < oo, see [67], Propo sition 3.12 and Proposition 3.13. It was claimed in [92], [93], that roughly speaking, a finitely gener ated linear semigroup S has polynomial growth if and only if the groups associated to S are finitely generated and almost nilpotent and for every uniform component U of S the semigroup Su modulo ~Ann(u(u)) is almost unipotent (and that the latter condition can be dropped if ch(K) > 0). An equivalent condition was also claimed in terms of lin ear recurrencies azma, m = 1,2,... , for a, z G S. However, there is a gap in the proof, resulting from the fact that in general S^ does not allow a reduction to the linear case, see the example after Proposition 3.26.
7.3. ALMOST
SEMISIMPLE
SEMIGROUPS
223
In case SJJ' can be replaced by a linear semigroup u(Su)i as Sy was replaced by 4>s(Su) above, one shows that the problem reduces to the case of subsemigroups of the form (a,z), where a £ U,z £ Su, and rank(2r2) = rank(z) > rank(a). So, it is not surprising that the recur rent properties of the sequence azrna, m — 1,2,... , should be decisive, because the group H associated to (a,z) and containing a is generated by nonzero elements of this type and by powers of a, (see also Theo rem 7.25). We briefly discuss the case where ch(K) > 0, since the conjectured result seems quite readable in this case. Namely, one expects S^ to be finite modulo ~Ann{uM) (f° r this, it is enough to check that the factor semigroup is periodic). This in turn implies that the completely 0-simple semigroup C/(u)/ ~Ann(u{u)) is finitely generated (cf. [87], Lemma 2.1), so it has finitely many H-classes. Then its maximal subgroup G^ is finitely generated, whence G is finitely generated as we assume that G is finitely generated modulo its unipotent radical. Consequently, the unipotent radical of G is finite. This is exactly what is desired, in view of Theorem 7.31. Let a, z be as above. Assume that a £ G. Fix some p, q £ { 1 , . . . , n} and consider the sequence um = uffl of (p, q)-th entries of azma,m = 1 , 2 , . . . . Matrices azma satisfy a linear recurrence coming from the char acteristic polynomial of z. So
«m = EPi(m)A7l t=l
for some polynomials P t and Az £ K, see Section 1.4. Extending the methods of [93] one can show that the group H C G associated to (a, z) and containing a is finitely generated. So, in view of Theorem 7.31, we see that GK((a, z)) < oo exactly when H is finitely generated (note that (a, z) has only one uniform component consisting of matrices of ranks equal to rank(a)). However, for lack of linear methods in S)j\ we are unable to extend this to SuOne might expect that the polynomiality of the growth of S can always be verified on 2-generated subsemigroups (a, z) of our special type, provided we know that the associated linear groups G are almost nilpotent and finitely generated modulo the unipotent radical. It seems also that the growth functions of linear semigroups have a rather regular
224
CHAPTER
7.
GROWTH
behaviour. In particular, one might expect that the Gelfand-Kirillov dimension of S C Mn(K) is an integer whenever it is finite.
Chapter 8 Monoids of Lie type In this chapter we discuss an exceptional class of finite monoids. They are finite analogues of linear algebraic monoids, in a way that finite groups of Lie type are analogues of algebraic groups. Their theory was created as an abstraction of the theory of linear algebraic monoids and developed in a series of papers by Putcha and Renner. The full linear monoid M n ( F g ) , discussed in Chapter 2, is the basic example. Actually, among the topics presented before, only this and the material of Chapter 4 are relevant for our aims here. After giving the necessary background on groups of Lie type, we state the definition and the main properties of such monoids. A universal object in the class of monoids built on a given group of Lie type, as Mn(Fq) is built on G L n ( F g ) , is then described. Exceptional properties relating irreducible representations of monoids of Lie type are presented in the last section. Throughout this chapter F will stand for a field.
8.1
Definition and examples
Let G be a finite group. We say that G has a Tits system (or admits a J9A/"-pair) if there are subgroups B,N of G which generate G and such that T = B H N is a normal subgroup of N and W = N/T (called the Weyl group) has a generating set S (referred to as a set of Coxeter generators for G) consisting of order two elements and satisfying sBa C BsaB U BaB sBs ^ B
for all aeW,s for all s <E S 225
GS
226
CHAPTER
8. MONOIDS
OF LIE
TYPE
Here, abusing notation, we write Ba for the coset Bna,na G AT, such that a = Tnai with a similar convention for the cosets aB. If a G W7, then the length /(a) is defined as the minimal length of a in the generators from S. We put Z(l) = 0. It turns out that W has a unique element a^ of maximal length. Then do is of order two and aQSaQ = S. Define B~ = a0Ba0. If / C 5, then by Wj we denote the subgroup of W generated by I. Here W$ — {1}. Then P / = BWiB and Pf — B~WiB~ are subgroups of G which are their own normalizers. Moreover, these are the only subgroups containing P , B~ respectively. The axioms imply that G — Uaew BaB: a disjoint union. This is called the Bruhat decomposition of G. Since B~ = a0Ba0, we also have G = (J aG w B~aB, a disjoint union. The conjugates of B are called the Borel subgroups, the conjugates of P/ are the parabolic subgroups of G. If x G G, then x~1Pjx^x~1Pfx are called opposite parabolic subgroups. Hence, if P, P~ are opposite parabolic subgroups, then x~lP~x,x G P, is the set of all opposites of P and there exists g £ G such that g~lPg — Pug-lP~g = Pf for some / C S. If IJ< C 5, then P 7 n PK = PInK. Speaking about a group of Lie type, we will assume that G admits a split i?TV-pair satisfying some commutator relations, see [13], page 61. In other words, G satisfies some extra conditions, which in particular are satisfied by any group Ha of constants of a connected reductive group H C GLn{Fq) with respect to the Frobenius map a defined by a((aij)) — (CL-J). We note that groups of this type have played a crucial role in the classification of finite simple groups. Then there is a normal subgroup U of B such that B = f/T, U Pi T = {1}. So B~ = C/"T, where U~ = a0Ua0. Moreover, if L 7 = PT D Pf for / C 5, then there exist normal subgroups Uj in P j and Uf in Pf such that UIf]PI = {l},C/f H Pf = {1}, and P 7 = LjU^Pf = LiUf. This is called the Levi decomposition of P/, Pf. Also [//, C/^- are called the unipotent radicals of P j , Pf and Lj is called a Levi factor of Pt. More generally, if P = x _ 1 P / x , P ~ = x~~1Pfx are any opposite parabolic subgroups of G, and L = P fl P ~ , then P = LRU(P), P~ — LRU{P~), where P U ( P ) = x~xUix, RU(P~) = x~lUfx. The unipotent radical RU(P) does not depend on the choice of P ~ , while any two Levi factors of P are conjugate by an element of RU(P). U is generated by certain subgroups Xa,a G 3>+, called the root subgroups. In fact, U is the product of all Xa in any order. Similarly, U~ is the product of some root subgroups Xa^ a G $~ = — $ + . Moreover, T
8.1.
DEFINITION
AND
EXAMPLES
227
normalizes the root subgroups Xa,a G $ = $ + U $ ~ , and VK permutes the subgroups I a , a G $ . If / C S, then Lj is generated by T and the AVs such that Xa,X-a are both contained in Pj. Then [// is the product, in any order, of the Xa, a G $ + , such that X_a £ PIm Similarly, Uj is the product, in any order, of the X_a with Xa C [//. Each Li is a group of Lie type with a J5TV-pair B f] Lj,N D Lj. Moreover (B fl Li)Ui = B and ( £ ~ n Li)Uj = 5 " . All this can be found in [13]. If s G 5, a G W, then /(as) = /(a) ± 1. Moreover, we have the follow ing multiplication rule for the components of the Bruhat decomposition BsB
BaB = { R **f_ _ jj i{' f l ) = \\a\ + } [ BsaB U i^ai* 11 /(sa) = /(a) — 1
Applying the map g h-» g _ 1 we also get (BaB)(BsB) or (BaB)(BsB) = BasB.
(8.1) v
y
— BasB U J?ai?
E x a m p l e Let G = GLn(Fq). Then we define B as the group of upper triangular matrices and T as the group of diagonal matrices. Next, W consists of permutation matrices and N = F*W. Then S — { 5 1 ? . . . , 5 n _ i } , where 5,- is the simple reflection interchanging z,z + 1. Now a 0 = Eni + En-it2 + ' •' + £i,n with E{j standing for the matrix with 1 in the (i,j) position and zeros elsewhere. Then B~ is the group of lower triangular matrices, and U C JB, f/~ C B~ are the subgroups of all unipotent matrices. If / C 5, then P j , Pf consist of the correspond ing block upper, respectively lower, triangular matrices. Now Lj is the group of block diagonal matrices. $ + is in one-to-one correspondence with ordered pairs (i,j) with 1 < i < j < n. $~ is in one-to-one corre spondence with the pairs (i,j) with 1 < j < i < n. The root subgroups are of the form X{j = {I + aEij \a G F g } . We will need the following technical result, which is a special case of standard general facts on the intersections of parabolic subgroups, see [13], Proposition 2.8.6, Theorem 2.8.7, Proposition 2.8.9. It follows via the root subgroup calculus and it is stated in the form needed later. Note that a~1Lia,a~1Uia,a~lUfa make sense for a G W, I C 5, because T normalizes L/,/7/ and Uj. L e m m a 8.1 Let G be a group of Lie type, a £W 1. Pi 0 a~lUja
= (L/ H a~lUja)(C/7
n
and I,K
a'Wfa),
C S. Then
CHAPTER
228
8. MONOIDS
OF LIE
TYPE
2. aPia~l fl Li is a parabolic subgroup of Li with unipotent aUia~l fl L j ,
radical
3. PK H Pf = LInK{LK
fl E/f),
H UT)(UK fl P f ) and C/7nK =
UK(LK
^. t/AT C 7, ifeen C/7 C UK, PK = ( P * H L / X P ^ n t / / ) = ( P * n L j ) E / j and PA' H Lj is a parabolic subgroup of Li with Levi decomposition p K n L / = ([/KnL/)LK. Let M be a finite monoid with group of units G. If X C M, by £ ( X ) we denote the set of idempotents of X. Write E = £ ( M ) . If e € £ , then we define P(e) = {x 6 G | xe = exe}, [/(e) = {x G G | xe = e},
P~(e) = {z € G | ex = exe}, U~{e) = {x G G | ex = e},
L(e) = {x e G\xe = ex}. It is clear that these are subgroups of G. (For P ( e ) , P ~ ( e ) we use the finiteness of G here.) M is called a monoid of Lie type on the group G if M is a finite regular monoid with zero such that the following two conditions are satisfied 1. the ^-classes of M are of the form GaG for a G M, 2. for all e G £ , P ( e ) , P ~ ( e ) are opposite parabolic subgroups of G and Ru(P{e)) C £/(e), i*u(P"(e)) C [/"(e). Our definition differs from that originally given in [109], but as it will be seen in Corollary 8.6, the two definitions are equivalent. L e m m a 8.2 Let M be a monoid of Lie type on a group G.IfeE E(M), then the H-class of e in M is equal to P(e)eP~(e) = eL(e) = L(e)e and L(e) = P(e) Pi P~{e) is the common Levi factor of P ( e ) , P ~ ( e ) . Moreover, the 1Z- and C-classes of e in M are eG,Ge respectively. In particular, M is unit regular, that is, M = E(M)G = GE(M). Proof. If xeyKe for some x,y G G, then xeHe and eyTie because the principal factor of e in M is completely 0-simple. Therefore exe = xe and eye = ey, so we get x G P{e),y G P~{e). It is clear that L(e) — P(e) H P'{e). Since Ru(P(e))e = e and e i ^ ( P - ( e ) ) = e, it follows that
8.2.
UNIVERSAL
MONOIDS
OF LIE
TYPE
229
xey £ L(e)eL(e). But le — ele = el for every / £ L(e) implies that L(e)e = eL{e). Let J = GeG. Now, eG = (eL(e))G = (eGe n J ) G = e J f l J, and similarly Ge = JeDJ. The remaining assertion is now clear.
□ It is easy to see that Mn(Fq) is a monoid of Lie type on the group GLn(Fq), see Chapter 2. Other important examples arise from the following more general construction. E x a m p l e Let M C Mn(Fq) be a Zariski closed connected submonoid with zero such that the group of units G of M is a reductive group. That is, RU(G) = {1}. Assume that a surjective morphism a : M —> M is given. If Ma = {x £ M | cr(x) — x] is a finite monoid, we call it a finite reductive monoid. It is known that Ma is a monoid of Lie type, cf. [111]. For example, if a is the Frobenius map cr((a2j)) = [a]-) and M = M n ( F g ) , then Ma — Mn(Fq). An excellent introduction to the theory of monoids of this type can be found in [130]. Examples of another type will be given in the next two sections.
8.2
Universal monoids of Lie type
The aim of this section is to introduce, for a given group of Lie type G, a monoid M = Ai( G ) on G which has certain universal property. Namely, the group G with a principal factor of any other monoid of Lie type on G form a monoid which is a homomorphic image of the corresponding monoid arising from M.. Therefore, the local properties of M can often be deduced by studying M. Moreover, the intersection properties of structural subgroups of G are beautifully reflected in M, and consequently the representation theory of M. can be set up in terms of G, as we will see in the next section. Suppose that uv — z for some u £ Uf,v £ Ui,z £ Lj. Then u = zv~l £ Uj Pi Pr = Uj fl Li = {1}, so also z = 1 = v. Since Pf Pi = Uj LIUI and Li normalizes [//, this easily implies that we have a natural map Pf Pi —> L/, defined by ulv H> / for u £ Uf, / £ Luv £ [//. We extend this map to a map r\i : G —> Li U {6} defining rji(x) = 0 for x ^ Pf Pi- Since L/ normalizes Ui,Uf, it follows also that rji(ax) = rji(a)rji(x),
rji(xb) = rji(x)rji(b)
(8.2)
CHAPTER
230
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TYPE
for a £ Pj ,b E Pi,x £ G. Hence, rji restricted to P/, P 7 is a homomorphism and rji(ax) = arji(x),
r)i(xa) = rji(x)a
(8.3)
for a G Lj,x G G. The following result comes from [111]. T h e o r e m 8.3 Let G be a group of Lie type with Coxeter generators S. We start with idempotents ej, I C S. Let M. = U/C5 Jl U {^}> a disjoint union, where Jj — GerG/ = , with relation = given by if x~lx G P/, y i y - 1 G P f and rj^x^x)
xejy = x1eIy1
=
n^y'1)
for x,Xi,y,yi G G. If a — xejy G Ji^b — se^t G J x , £^en define f xleInKmt
if ys G UjlmUKi
1 0
t/
I € £/, m G LA-,
ystPfPK
Then M. is a monoid of Lie type on G. Proof. Since 77/ is a homomorphism when is clear that = is an equivalence relation. multiplication on A4 is well-defined. Let a and b = se^t^bi — Sie^t\ G JK t>e such that
restricted to Pj or P7~, it First, we check that the — xejy^ai = x^eiy^ G Ji ax = a , ^ = b. Then
x~lx G P/, yiy" 1 G P f , ^ / ( x ^ x ) =
rjiiyiV'1),
s~'s G P K , M " 1 G P ^ , W ^ i " ^ ) = r ^ i * " 1 ) Now yisi — (yiy~l)(ys)(s~1si) and it follows that ys G Pf PK if and only if yxsi G PJPK- Let ys,y1s1 G Pf^W = LSjLiLKl]K. Then there exist /,/i G Li,m,mi G L# s u c n that y 5 £ UjlmUK and y ^ G U^limiUK' Hence a6 = xlemx'^t^ibi = ^l/ie/nA'^i^i- Since x]" 1 ^ G Pr, we get Jy 1 X £ Uirji(x lx). Since Uj is a normal subgroup of P/, / - ^ - ^ Z G Ufa1 Tufa1 x)l (8.4)
= Wr^/Cyiy-1)/ - ^/(/r'yiy" 1 /) because Z,^ G Lj (see (8.3)). Next z
=
l l
\ viy~l1
= 'r1(yi5i)5^15(y5)_1/
= Uj m^KS^1
sUKm~lUj
—
UjmiS^lsrrrlUKUj
8.2.
UNIVERSAL
since m^s^1s
MONOIDS
OF LIE
TYPE
231
G PR- SO, for some Ui,u2 G f/f and v G UK we have z = uirriis^1
sm~lvu2.
Let z\ — mxs{xsvrrxv. Now z G Pf, because y i y - 1 G P f and l,lx G £ / . Hence zx = Uizu2 implies that zx G Pf and 7y/(z) = Vi{zi)- Since «sj" 5 G PR: and m G LA', we have zx — mxr\K{ys'{1s)mTlw for some 1 5 772 1 w G UK- Let z2 = m^A^si" " ) " - Then z2 G L A and zx = 22w G P A n P f = LInK(LK
n C/f )(£/* 0 P f )
by Lemma 8.1. Since L/ n K ^ ^ K and P A = LKUK is a semidirect product, it follows that z2 G LInK(LK fl Uj),w G t/# 0 P f = ({/# H L 7 )(C/A' H UJ). Hence, in view of (8.2), *7/(*) = ^/(^O = Vi{z2w) = m{z2)rii{w), 7/7(z2) G L/n/c, 7//(w) G C/A' H L/. In particular, 77/(2:) G P/ fl P A = P/nA- So, by (8.4) Z- 1 *- 1 */ G C//77/(^) C C//P/nx C P/nA-.
(8.5)
Moreover, since U\ U C/A ^.UI C\ UK and 77/(ty) G UK, rUnKiK1^1*1)
= TllnKilliz))
= r]InK{rji{z2)Vl{w))
=
llnhivii^)),
which in view of f/f C f/f,A- is equal to Vlnhi^)
= VinKimxriKis^s)™-1).
(8.6)
Similarly to (8.5) one shows that rriitit~1m~l G PfnK- Now, since M - 1 G P f and m G LA', we get raiM-1^-1 G mirjK(tit~1)m~lUf. 1 -1 Since f/f C f/f,^ and rji(si s) = ^ / ( M ) , in view of (8.6) we have VlnK(rnlt1t~1m~~1)
= 1
r)InK{rnir)K{tit~l)m~l) 1 1
= ^/nA^i^SiS- )™- ) = ^m^r^r ^)It follows that the multiplication on .M is indeed well-defined. Next, we prove associativity. Let a = xeiy G J/, fr = seA'^ G J f , c = ne^u G JNJ f° r some I,K,N C 5 and x,y,s,t,u,v G G. Put i7 =
CHAPTER
232
8. MONOIDS
OF LIE
I D K D N. Suppose that (ab)c ^ 9. Then ys G Uj hhUx li G L1J2 G £ K , and ab = xliemxhtSince (a6)c ^ 0,
TYPE
for some
l2tu G UjnKl3UUN for some fc G Lmx^U € ^AT- SO (a6)c = xlilseahvNow t/^,^ = Ux(Uf fl L # ) by Lemma 8.1. Hence iu G l2lUjnKhhUN
= l2lUK{Ujt\LK)hhUN
=
U^\UjfMK)hUUN.
So, for some z G t/J" D L # , tu G C^K^ ZI3I4UN and 6c = sl^zlseKnNhv.
(8.7)
Now
{ys)l^zl3 G {Ujhl2UK)l^zh = UjhUKzk = UjhzhUK (because 2 , I 3 6 ^ x ) , which is equal to Ujhzl^hhUK
=
UjhhUK,
the latter since z G C/j~, /1 G L/. Since [/# C UKHN view of (8.7) we see that a(bc) = xlilsenhv
an
d M3 G L j , in
= (ab)c.
Similarly, a(bc) ^ 9 implies that a(6c) = (a6)c. Therefore M. is a monoid with group of units G = JsIt remains to show that M. satisfies the conditions defining monoids of Lie type. Every J'-class J of M contains some Ge/G, / C S. It is also clear that MejM\GeiG consists of non-generators of the ideal MejM. Therefore J = GejG. Since M is finite and every J'-class contains an idempotent, AA is regular. Let x G P{ei)- Then xej — e/xej. So ejxei ^ 9,x G PfPi. Thus xei = ejxei = rji(x)ei. Hence x~lrji(x) G Pi. Since rji(x) G L/, we have x G P/. Therefore P(e 7 ) C P 7 . Clearly Lj C P ( e / ) . If u G [//, then
8.2.
UNIVERSAL
MONOIDS
OF LIE
TYPE
233
77/(11) = 1. So uei = ej and UI C P(e 7 ). Hence P 7 = L/C// C P(e 7 ) - P 7 . Similarly P " ( e j ) = Pf and e/t/f = {e/}. We claim that every idempotent / G Ge/G is conjugate let / = xe 7 y. Then xejyxejy — xe 7 y implies that e/yxe/ = yx G P / "P / ,77 / (yx) = 1. Then yx G {/ft//. So t ^ y = vx~l u G Uj ,v G C/j and / = xe/y = xv~leju~ly
—
P ( e 7 ) . So to ej. So, e 7 . Hence for some
(xv~l)ei(vx~l),
so indeed / = g~1eig for some g £ G. It follows easily that P ( / ) = g-lP{eI)g,p-{f) = g^P'ie^g and U(f) = g~lU{eI)g,U-{f) = g-lU(eI)g. Therefore, the defining condi tions checked above for the idempotent ej also hold for / . This completes the proof of the theorem. □ R e m a r k We have seen in the proof of the theorem that, for every J C 5 , one has P ( e / ) = Pj and Uj C £/(ej) in the monoid .M. If xej — ej for some x G G, then x E Pi and 77/(2:) = 1. Hence x G £//. Therefore we get U(ei) = Ui. Similarly P~(e/) = P 7 ~,t/~(e/) = f/7. Also e/x = xe/ if and only if x G Pj H P/~ = L j . Hence L(ej) = Lj and it is isomorphic to the maximal subgroup eLj of M.. Choose any subset I C S. Consider J? = J j U {0} C A4. Then j £ is a subsemigroup of M and it is isomorphic to a principal factor of A4. So, it must be completely 0-simple. We will describe a Rees matrix presentation of Jj. By the above remark, every maximal subgroup of Jj is isomorphic to ejjjej \ {9} ~ Lj. Write GjPj for a complete set of representatives a 1 ? . . . , at of cosets gPi,g G G. Similarly, let G/Pf = {&!,... , 6 t } be the representatives of all Pfg,g G G. The sandwich matrix is defined by Qj = (rj^bjdi)). In other words, we get J ° ~ M(Li,G/Pi,G/Pf ,Qi). Here, for xe/y G J/ we write x = a 2 /iUi,y = U2/2&J f ° r some z,j and some / i , / 2 G £7,^1 G ^7/,u 2 G C/f. So, x e /2/ *s identified with (/1/2, at-, 6j). This can be compared with Proposition 2.5. Note that { J j | I C S} is the set of nonzero ^-classes of M and that JPK = JinK f o r a n y A K ^ S- N o w > l e t ^ / = G U J / U {9}. Then A4 7 is itself a monoid of Lie type on G. L e m m a 8.4 If I C 5, £/ien e/A^e/ zs isomorphic monoid of Lie type on the group Lj.
to the
universal
234
CHAPTER
8. MONOIDS
OF LIE
TYPE
Proof. It is easy to see that ei(GeKG)ei = LiemK'Li U {9} for every K C S. So eiMei — \JKCI LiejcLi U {#}. Let K C J. As noted above, P(ejc) — PK a n d L(ex) — LK in M. Hence, by Lemma 8.1 {x G Lj | xeK = e^-ze^} = ^A' H L J is a parabolic subgroup of Li with unipotent radical UK H L; = {X G i ; | z e # = eK}. Since a symmetric assertion holds for P ^ , f/^, the result follows easily. □ For an J'-class of a monoid M of Lie type on G the set J U {#}, with multiplication a • b = ab if ab G J and a • b — 6 otherwise, becomes a completely 0-simple semigroup isomorphic to a principal factor of M. Then M(J) = G U J U {#}, with obvious multiplication, becomes a monoid of Lie type. This is called a local monoid of M. The universal property of M has now the following form. Corollary 8.5 Let M be a monoid of Lie type on the group G. Assume that J = GeG is an J-class of M with e G E(M) and P(e) conjugate to Pi for some I C S. Then there exists an onto homomorphism : Mi —► M(J) such that f = {ei) satisfies P(f) = PhP~(f) = Pf and (j) is the identity map on G. Proof. We know that g~1P(e)g = PI,g~1P~(e)g = Pf for some g G G. Let / = g~leg. Then P(f) = Pr and P'(f) = Pf, and hence C/ 7 / = / = fUf. Assume that xeiy — sett in GeiG for some x , y , s , £ G G. Then s~xx G Pi,ty~l G Pf and ^ / ( s - ^ ) = ni(ty~l). So 5 _ 1 x G lU^ty'1 G [/}"/ for some / G L/. Since £/// = / = / J / 7 m ^ 5 w e Se^ s~lxf — '/ and / £ y - 1 = / / . Now x / y = s/£ because / / = / / by Lemma 8.2. Hence, xejy y-> xfy determines a map Mi —> M(J), which is identity on G. To check that this is a homomorphism, note that xfysft = xfrji(ys)ft for every x, y, 5, t G G because Uif — f — fUf. □ As an application, we derive an important technical result concern ing the set E(M) of idempotents of any monoid of Lie type M. We will use the fact that PiUfUi = PiPfPi = G for I C S. For this, it is enough to know that BB~ B = G. By induction on the length of a G W we first check that BaB n B " / 0 . This is clear if 1(a) = 0 since B fl B~ = T. Let a = sc, where c G W, s G 5 are such that 1(a) > 1(c). Then BsBBcB = 5 a 5 by (8.1). Since £ c £ n 5 " ^ 0 ^ £ s £ n B " by the induction hypothesis, we get BaB fl B~ ^ 0, as desired. From the Bruhat decomposition it now follows that G = BB~ B.
8.2.
UNIVERSAL
MONOIDS
OF LIE
TYPE
235
Corollary 8.6 Let M be a monoid of Lie type on the group G. Then 1. if e , / £ E(M),
then ejf
2. if e , / £ E(M) and ej'/, eKei,eiCe2,f1le2,
if and only if g~x fg = e for some g £ G, £/ierz £/iere ezzsJ ei,e 2 £ E(M)
such that
Proof. Because of Corollary 8.5 it is enough to check that the assertions hold for the universal monoid M. on G. For 1) this was done in the proof of Theorem 8.3. To prove 2), conjugating, we may assume that e = ej for some I C. S. Moreover, / = x~xe\x for some x £ G. Note that x £ G — PjPfUi. Hence, we may choose y £ Pf Pjx f) Uj. Then there exists z £ Pjx such that y £ Pf z. Now y £ [// implies that ex = ey is an idempotent. Clearly eflZe. Moreover zx~l — q £ Pj implies that the idempotent e2 = z~lez = x~lq~1ez satisfies e2lZf. Similarly, yz~l £ Pf implies that e\Ct2. □ Finally, we state without proof an analogue of Theorem 2.7 in our more general context. As in the case of GLn(K) C Mn(K), it is a basis for developing a combinatorial approach to any monoid M of Lie type with group of units G. By U — U(M) we denote the set of ^/-classes of M. We have seen above that for every J i , J 2 € U(M) there exists J £ U(M) such that JXJ2 C J U {<9}. Then we write J = Jx A J 2 . In this way U(M) becomes a lattice. From the multiplication rule in Ai it follows easily that e^e/ = exm for K, I C. S. So, this lattice can be represented by {ej \ I C S} U {#}. Let J be a ^-class of M. Then J = GeC? for some e £ E(M) and P ( e ) , P ~ ( e ) are opposite parabolic subgroups of G. So there exists a: £ G such that x _ 1 P ( e ) x = P j , x _ 1 P ~ ( e ) x = P/~ for some I C. S. Then e j = x _ 1 e x £ J is an idempotent such that P(ej) = Pi and P~(ej) = P 7 _ . We put A(Af) = {ej \ J £ £V(M)} and £ 0 - UneTV n " 1 A(M)n. T h e o r e m 8.7 Assume that M is a monoid of Lie type. Define R — (N,A(M)) C M. Then 1. A(M) is a cross section lattice of idempotents of M, £/m£ is, for all J\,J2 £ U(M) we have eJxej2 — ej2ejx = eJz for some J3 £ U(M), 2. R = NEo is an inverse monoid with group of units N and E(R)
=
236
CHAPTER
3. = defined by ex = e'y if e = e',xy
8. MONOIDS x
OF LIE
TYPE
G T, is a congruence on R,
J. R' = R/ = is an inverse monoid with group of units W and E(R') ~ F 0 , 5. BuB = BvB if and only if u = u, /or u,v e R, and M = UueR BuB, R' = IJeGA(M) We'W, where e' denotes the image of e in R'.
8.3
Connections with group representations
Foundations of the representation theory of monoids of Lie type are presented in this section. First, we show that the complex algebra C[M] is semisimple. This extends Theorem 2.34. Secondly, we prove that all irreducible modular representations of M restrict to irreducible representations of the group G. These are exceptional properties within the class of all finite regular monoids. The results come from [98] and
[in]. We shall consider representations (j) : M —> Mn(F), over a field F, such that (f)(6) = 0. Therefore, irreducible representations of M over F correspond to irreducible F 0 [M]-modules. If V is an F-space, then GL(V) denotes the group of F-automorphisms of V. We first explain the role of subgroups of the type P(e) for the representation theory of the group G. L e m m a 8.8 Let M be a finite monoid with zero and group of units G. Let e G E(M). Suppose J — GeG is a J-class of M such that J° — J U {9} is an ideal of M. Let V be an irreducible F0[M]-module, for a field F, such that eV ^ 0. Let P = {a G G\ae = eae} and (j) : P —> GL(eV),xl> : G —> GL(V) denote the associated representations. Suppose the sandwich matrix Q of J° (under a Rees presentation of J°) istxt and invertible. Then ip is equivalent to the induced representation
8.3.
CONNECTIONS
WITH GROUP REPRESENTATIONS
237
Hence i — j . It follows that k < t. Let x G G. Then xP = g{P for some i. Since Pe — ePe, we get xeV — xPeV = giPeV — gzeV. So W = J2i 9ieV i s invariant under J ° . Since J ° is an ideal of M, and V is an irreducible C 0 [M]-module, W — V. Hence dim(V) < k • dim(eV), so that k = t and V = gxeV © • • • © gteV. Let g £ G. For every z there exists j such that ggiP = ^jP. So 9jlQ9i £ ^ a n d gJ1ggi(eV) = eK This means that V is the F[G]module induced from the F[P]-module eV, see [17], §12D. The assertion follows. D The first main result of this section, originally proved in [98], reads as follows. T h e o r e m 8.9 Let M be a finite monoid of Lie type. Then C[M) is a semisimple algebra. According to the results of Sections 4.2 and 4.3, every irreducible complex representation 0 of M comes from a representation of a max imal subgroup H of M. We need to show that every
CHAPTER
238
8. MONOIDS
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TYPE
In other words, x is a left identity of the algebra F0[eZ] such that it annihilates on the left the remaining 7£-classes of Z. Choose nonzero idempotents e = e\,... , et from all distinct nonzero 7^-classes of Z. By the hypothesis, there exist automorphisms c^- of Z such that crt(e) = e2-,z = 1 , . . . , £. Since every cr2- permutes the nonzero 7^-classes of Z, it is clear that every (Ji(x) is a left identity of F 0 [e;Z], which annihilates e^-Z, j 7^ z, on the left. Here a; are treated as the induced automorphisms of F 0 [Z]. It follows easily that 5D*=i &i(x) is a left identity of F0[Z]. Thus, Q is left invertible, and since F0[Z] is finite dimensional over F, we see that Q is invertible in Mt(F[H]). D Recall that, if Q = (qij) is a matrix over H°, for a group # , and (j) is a representation of iJ, then we let (j>{Q) = (0(%)). Corollary 8.11 Assume that for every irreducible complex representa tion (f) of a maximal subgroup H of a finite completely 0-simple semi group Z = M(H,t,t,Q) we have ( A 1 ? . . . ,At)(Q) = ( 1 , 0 , . . . ,0) for some matrices Ax,... , At G Mt(C[H]). Assume also that for every nonzero e,f G E(Z) there exists a G Aut(Z) such that a(e) — f. Then Q is invertible in Mt(C[H]) and C[Z] is a semisimple algebra. Proof. The algebra C[H] is semisimple by Maschke's theorem. Since the assumption allows us to solve the equation O n , . . . , x t ) Q = ( l , 0 , . . . ,0) in every projection onto a simple component of this algebra, the hy potheses of the foregoing lemma are satisfied. The result follows. □ We will need another technical result on groups of Lie type. L e m m a 8.12 Let I C 5, a i , a 2 G W be such that a\ is of minimal length in the coset Wiax. If /(ai) > l(a2) and ax ^ a 2 , then axB H P f P 7 a 2 = 0. Proof. Since B D LhN D Lj form a BTV-pair for L 7 , by the Bruhat decomposition for Lj we have PfPj
= UJLIUI
= UJ(B-
n L/)W Z (£ n L/)J7j -
B-Wrf.
8.3.
CONNECTIONS
WITH GROUP REPRESENTATIONS
239
For any a G W, s G 5, by (8.1) we get B~aBs
— a0Ba0aBs
C a0Ba0aB
U a0Ba0asB
= B~aB U
B~asB.
Let a 2 = si • • • 5t, where st- G 5 and /(a 2 ) = £. Then it follows that P f P 7 a 2 = P " W 7 P 5 l • • ■ st C
|J
P " W^ 2 1 • • ■ 5 l V fl.
*i<"-<«V
Suppose that axB D Pf Pia2 ^ 0. By the Bruhat decomposition the union G — \Jaew B~aB is disjoint. Hence ax G Wjs^ • • • sir for some ii < • • • < ir. Since ax is of minimal length in Wja1 and l(ax) > l(a2) = t, we must have r —t and ax = a 2 . This contradiction completes the proof.
□
A representation (/> of L = Lj is called cuspidal if for every parabolic subgroup P of L with unipotent radical V we have I ^ G y <^(u) = 0. The study of all irreducible complex representations of L reduces, in some sense, to the case of cuspidal representations. Namely, if is an irreducible representation of L, then there exist K C / , K ^ / , and an irreducible cuspidal representation n of L^ such that, if n is the trivial extension of n to PK H L (that is, W(u) = 1 for u G C/^ Pi L), then 0 is a component of the induced representation ip = ftp nLi see [13], Chapter 9. This often allows an inductive approach to representations of G. The following lemma, being the starting point of such an induction, is the first step in the proof of Theorem 8.9. L e m m a 8.13 Let (j> : Li —> GLn{C) be an irreducible cuspidal repre sentation. Then there exist A i , . . . , At G M n ( C ) such that (Au...
,At)4>(QI) = (In,0,...
,0),
where Qj is the t x t sandwich matrix corresponding to the principal factor J ° = GejG U {6} of M and In is the identity t x t matrix. P r o o f . Write L = Li,P = Pi, P~ = P z ". We start by carefully choosing the coset representatives of G/P and G/P~. We know that any coset of Wi in W has a unique element of minimal length, [13], Proposition 2.3.3. Let these right coset representatives be ai = 1, a 2 , . . . , as. Then, by the Bruhat decomposition,
G = B~WB = (J B-Wia{B = \J P~axB. i
i
(8.8)
CHAPTER
240
8. MONOIDS
OF LIE
TYPE
Thus G = U» P~cb{B. We claim that this union is disjoint. Suppose that P~ciiB — P~a,jB for some i,j. Assume for example that /(a z ) > l(a,j). Since d{B intersects P~'a^ we see by Lemma 8.12 that az = a J7 as desired. Also by the the Bruhat decomposition G = BWB
= | J Ba;1 WTB = \J Ba~lP. i
(8.9)
i
Thus G = Ui Ba~lP. We claim that this union is also disjoint. Suppose that Ba~lP = BafP. Then PatB = Pa3B. Again, if l(a{) > l\a3), then by Lemma 8.12 we get a7- = a,j. For i = 1 , . . . , s, let V- = U H a^U^i,
Yt = UD a~lUjau
Zx = [/ n a" 1 !^-.
Since U is the product of the positive root subgroups in any order, we see that U = Y{ZXY{. Hence P~axB = p-axU
= P-aM,
Ba~lP
= UaJlP
= Yxa~l P
(8.10)
for i = 1 , . . . , 5 , because a; normalizes T and t/jT, L C P ~ , [//, L C P. Let a; = Trii^rii G AT. Suppose that t>i,t>2 £ K" are such that P~n{V\ — P~riiV2. So n ^ i t ^ n " 1 G P ~ . But it is easy to see that niV\v2ln~l G [//. Since P ~ f)Ui = L C\Ui = {1}, we get ^i?;^ 1 = 1 and vi = v2. Hence, by (8.8),(8.10), riiV,v G VJ,z = 1 , . . . , s, is a set of distinct right coset representatives for P~ in G. Suppose y x ,y2 € K' a r e such that y 1 n ~ 1 P = y2n~lP. Then riiyi1y2n~1 G P. But riiyiy2ln~l G E/f- Hence, similarly, yiy^ 1 = 1 and yx = y2. Thus, by (8.9),(8.10), yn~l,y e ^ , i = l , . . . , s , is a set of distinct left coset representatives of P in £?. Therefore, as noted after Theorem 8.3, the entries of the matrix Qj correspond to the pairs (n,t;, y n " 1 ) , where v G VJ,y G Yj,i,j = 1 , . . . , 5. If x G G, for simplicity let x = r]i(x). Then the entry in the position (riiV^ynJ1) is Let N\y{L) be the normalizer of L in VK. We next arrange the a's so that di = 1 , . . . , a r G Nw{L) and a*. ^ Nw{L) for fc = r + 1 , . . . , s. Further, let /(a x ) < l(a2) < • • • < /(a r ). Since [7j~ C C/~, we have
n = c/ni/7- = {i}. Thus, we are trying to find AniV that
G M n ( C ) for t; G VJ, i = 1 , . . . , s, such
^ A n t ^ ( r v * 7 ) = /n TliV
(8.11)
8.3.
CONNECTIONS
WITH GROUP REPRESENTATIONS
241
and Y
An An^{n tV<j>(n tvynf) %vynf)
= 0 for y G Yh j = 2 , . .2,...,s. . ,s. (8.12)
Let 1 < j < r. Then at{ G NW(L). Thus, for any root subgroup Xa of Ui, a{Xaar [//, Xaa-1l is either contained in Ui £/, or £/f. t/f. Since £/ [//7 is the product of its root subgroups in any order, it follows that Ui = Y,Vi ,r. YiVi for i = l , . . . ,r.
(8.13)
Now, for y G Y, v G V V5, GG G we we have have by by (8.2) (8.2) t, xx G
n^x n ^
1 = (n,-yn,(mynT'Ywx) )(nit;x)
= ^y^ruvx = ^ W W
= ^pxx. .
(8.14)
From (8.13) and (8.14) it follows that for all x G € G Y
4>(n^ux) ==\Y\\Yt\Y Y 4>{nm), {n&x) 4>{w*), i i= =l ,1. ,....,. r ,. r.
u£Uj U£UT
(8.15) (8.15)
vEVl
1 1 Now let j > r + + 1. By Lemma 8.1 R = ^PaJ a^PaJ nD L is a parabolic 11 subgroup of L with unipotent radical V = ajU^J ajU^' fl~lL. ~l I . Suppose Suppose PP == L. L. Then aj1!^ !^- C P. Since j > r + + 1, a53 <£ G- NN SinceLLisisgenerated generated W(L). W(L). Since by T and root subgroups, there is a root subgroup Xa C L such that 1 a ljXaa a^3 - £^ L. I . So aJ aj1lX *^aaj C tf Ui. L,a?X-aaj C f/f, tff, 7. Hence X _ a C L^fX.^ 11 contradicting the assumption a" aj !^^ C P. We have shown that R ^R^L. L. Since £ is cuspidal, it follows that
Y Y
4>(a) (a) == 0.°-
(8.16) (8.16)
a6V
V,ii < r. Since nj1™, Ux, we have by (8.3) Let a G V, ™,- G J//, Y
1 l 4>{niunj ^{niunj )^))4)(a) = = ^Y
u£Uj u&Jj
4>(niUnjla) 4>{niun~
u£Uj
Hence, considering such equalities for all a G V, by (8.16) we get £
^(^J
1
) = 0.
(8.17)
CHAPTER
242
8. MONOIDS
OF LIE
TYPE
Now let w G U. Then w — Iv for some / G L,v G {//, because U C P. Since ra, G N G (L), we have by (8.17), (8.3) y ^ (f)(riiuwnjl)
— ^2
= (j)(niln~l) J^
^{niulvnj1)
(/>(nz(l-1ul)vnj1)
uGt//
= (f)(riiln~l) ^2 (niunjl) = 0. uGt/ 7
Thus, for w £ U, J2 (t>{nxuwn-1) = 0,
z < r, j > r + 1.
(8.18)
From general considerations it follows that, if a solution to (8.11), (8.12) exists, it must be such that AUxV = Ani for v G K', 2■ — 15 • • • , r*. So at this time let AniV = Ani for u G K , i = l , . . . ,r. We further let An%v — 0 for z > r + l,v G VJ- Then, by (8.15) applied to x = yn~l and (8.18) applied to w = y, we see that (8.12) is automatically valid for j > r + 1, y G l j . Hence (8.11), (8.12) now become by (8.15)
E ^ n , £
t=l
E | ^ 7 T A n t £ (ftniuynj1) i=i lr*l tier// Now for j < r,a,j G NW(L). Y;- = U^a-lUja3
(8.19)
nGf/7
= 0, y G Yy, j = 2,. . . , r. (8.20)
Hence a~lUja3{M
C PCla^Ufaj
= {1}. So, by Lemma 8.1
= (L^afUja^U^afUja3)
C [//.
Thus (8.20) simplifies to r
X E T^n i = l \1^\
t
£
(KriiunJ1) = 0,
j = 2 , . . . , r.
(8.21)
uEUj
Similarly, if i < r, then aiU^'1 Thus
H L = {1}. Hence axUia~x C [//"[//.
n ^ ^ " 1 = 1 for u G f//, i = 1 , . . . , r.
(8.22)
8.3.
CONNECTIONS
WITH GROUP REPRESENTATIONS
243
Furthermore, by Lemma 8.12 munj1
= 0 for 1 < j < i < r, u € Uj.
(8.23)
By (8.22), (8.23) we see that (8.19), (8.21) finally become \Ui\Ani
\TJ
\
W\An3 r
i-i
(8.24)
i
+ £ TylAm r
\ j\
= In
t=i l *l
E H^unJ1)
= 0 for j = 2 , . . . , r.
u€t/7
(8.25)
This is a triangular system of equations in Ani,... , AUr, which can now be solved inductively. This completes the proof of the lemma. □ P r o o f of T h e o r e m 8.9. From Proposition 4.13 and Maschke's theo rem it follows that the assertion is equivalent to the invertibility of the sandwich matrices of all principal factors of M. So, by Corollary 8.5, it is enough to prove that C[A4/] is semisimple, where Aii — GUGerGU{9}, for any I C S,I ^ S. We prove this by induction on | / | . If | / | = 0, then every irreducible representation of Lj = B is cuspidal. Hence, the as sertion follows from Lemma 8.13 and Corollary 8.11. So, assume that | / | > 0. Write e — ej. Let Qi be the sandwich matrix of J j = GeGU {9} and L = LT — eL. By Corollary 8.11 and Lemma 8.13 it is enough to show that <j)(Qi) is invertible for any noncuspidal irreducible repre sentation of L. Let cf) be such a representation. As explained before Lemma 8.13, there exist K C I,K ^ / , and an irreducible cuspidal representation n of LK such that if W is the trivial extension of ir to Z = PK fl L (that is, W(u) = 1 for u G UK H L), then
C P,
PK
C P",
LA- C
L, UJ C £/£, C/7 C [/*.
Let / = e * and MK = GU GfG U {0}. Let MKj = M i f U A4/ with the G's identified and with the zeros identified. So MKj
- G U GeG U GfG U {0}
CHAPTER
244
8. MONOIDS
OF LIE
TYPE
can be considered as a submonoid of a Rees quotient of M. Namely, for a — xey G GeG, b — sft G GfG we have ah={
x
vi{ys)ft
1 6
if
yseP-P
otherwise
and 1 0 otherwise Clearly MK,I is a monoid of Lie type. By Lemma 8.4 e M ^ / e is a monoid of Lie type on the group L, in fact eMe ~ (M.(L))K- Since / < e, / G CTWA',/ 6 - Since I ^ S, C[eM.xje] is semisimple by the induction hy pothesis. Now GfGU {6} is the unique minimal nonzero ideal of A4K,IHence, by Theorem 4.26 and Theorem 4.28, there exists an irreducible C 0 [A^x,/]-module V such that the C[Lx]-module fV corresponds to the representation n. Now eV is an irreducible CQ[eA4xjej-module. By Lemma 8.8 the C[L]-module eV corresponds to the representation ip and dim(eV) = d i m ( / V ) • — ^ —
(8.26)
Also by the induction hypothesis, the sandwich matrix of GfG U {9} is invertible. Hence, by Lemma 8.8 dim(V) = d i m ( / V ) • j^r
(8.27)
\"K\
Since PK C Pj = P and [// C f/A- C PA-, by Lemma 8.1 we have PK = (PK 0 L)(PK n Ui) = (PK n L)£/, and
I^A-I = |(Pic n L)U!\ = |i>* n L\ • |[/ 7 |. Hence, by (8.26), (8.27) dim(V) = dim(eV) • - ^ -
=
dim(e
y) . M .
(8.28)
Since {GeGf C GeG U {#}, V is also a Co^ 0 ]-module. We claim that it is completely reducible. Let A be the image in End(V) of the algebra C 0 [
8.3.
CONNECTIONS
WITH GROUP REPRESENTATIONS
245
MR,i which does not annihilate V, it follows that V is irreducible over C 0 [ J £ ] . Hence End(V) is the the image of C 0 [J&]. By Theorem 4.26 and Lemma 4.18 the radical of the algebra C 0 [«//], viewed as a Munn algebra, consists of elements x such that Qj o x o Qj = 0. In other words {GeG)x(GeG) = 0 in C 0 [J?], hence also in C 0 [ . M A V ] - Since (GeG)(GfG)
= (GfG){GeG)
=
GfGU{9},
we see that (GfG)x(GfG) = 0. Hence, the image of x in End(V) must be zero. This means that the radical of C 0 [Jj] maps to zero, so A is a semisimple algebra. Now, from (GeG)(GfG) = GfG U {9} it follows that A contains the identity of End(V). This shows that V is indeed a completely reducible C 0 [Jj]-module. Applying the remark after Theorem 4.28 to the irreducible components of this module, we come to dim(V) = rank(^((5/)). The sandwich matrix Qr of J ° = GeGU {6} is t x t, where t = | G | / | P | . Thus if>(Qi), as a matrix over C, is t' x t\ where t' = t • dim(eV). Hence (8.28) implies that ip(Qi) is invertible. This completes the proof of the theorem. □ As mentioned before, irreducible representations of G are classified, via cuspidal representations, according to which parabolic subgroup they come from, [13], Chapter 9. Irreducible representations of M. are classified according to which principal factor of M. they come from, see Section 4.3. The following result provides a connection. Corollary 8.14 Let M be a monoid of Lie type on the group G and e G E(M). Let n be an irreducible complex representation of the group L(e) containing U(e)PiU~(e) in its kernel. LetW be the trivial extension ofn to P(e) (that is, W(Ru(P(e))) = 1) and (j) the representation induced from P(e) to G. Then (j> extends to an irreducible representation of M. Moreover, every complex representation of G extending to an irreducible representation of M arises in this way. Proof. Let J = GeG U {9} be the corresponding principal factor of M. The sandwich matrix of J is invertible by Theorem 8.9. If L is the maximal subgroup of J containing e, then L = L(e)e = eL(e) by Lemma 8.2. So, if TT satisfies the conditions of the corollary, then it factors through L, because [/(e) D U~(e) = {x G L(e) | ex — e}. Hence, it determines an irreducible representation of J. Via the natural map
246
CHAPTER
8. MONOIDS
OF LIE
TYPE
M —> J, this leads to a representation fi of M. From Corollary 8.8 it follows that fio — 4>-> a s desired. The second assertion is also an easy consequence of Theorem 8.9 and Corollary 8.8, applied to an idempotent e £ M which is minimal with respect to having nonzero image under the considered representation of M. □ An alternative proof of Theorem 8.9 was obtained in [120]. We note that a classification of irreducible complex representations of M can be given, as in Theorem 2.35, see also Section 4.3. However, the explicit formulas for the identities of the algebras of the principal factors of M are not known. Next, we discuss irreducible modular representations of M.. We say that a prime number p is the natural characteristic of the group G if U is a p-group. For example, this is the case if G = Ha for a Frobenius map a on a linear reductive group H C GLn(Fp). By a modular representation of a monoid M of Lie type on G we then mean a representation M —> Af n (F p ). The sandwich matrices of the principal factors of M are not invertible in the corresponding matrix rings Af n (F p [L/]). So, we do not get an analogue of Theorem 8.9 in the modular case. However, a very strong link, of another type, with the representation theory of G is a consequence of Theorem 8.16 below. In what follows F = F p . We will need some background on modular representations of groups of Lie type. First recall that irreducible modular representations of G are in one-to-one correspondence with ordered pairs (/, x ) , where I C. S and x '• Li —> F* is any homomorphism. If (j> : G —> GLn(F) is an irreducible representation, then there exists a unique 1-dimensional subspace Y of the column space V = Fn which is stabilized by B. Then I C S is such that P 7 = {x e G\{x)Y = Y}. Moreover Y is a 1-dimensional F[L/]-module, yielding x '• Lj —> F*, [16]. Let Vui ={veV\ {JJj)v = v}. Then V = VUl ® (1 - (U]r))V and both summands are F[L/]-submodules of V. Here 1 — 4>(Uf) stands for the F-subspace spanned by all matrices 1 — <j){u),u G Uj. In par ticular, eV = VUl = Y and (1 - e)V = (1 - <j>{Uj))V for a unique idempotent e £ Mn(F) of rank one, [12]. It is also known that every irreducible representation of G is the restriction of a rational repre sentation (f) of the corresponding reductive group H C GLn(F) [131].
8.3.
CONNECTIONS
WITH GROUP REPRESENTATIONS
247
So G — Ha for a Frobenius map a. Moreover, there exists a i?TV-pair NH, BH for H such that a(BH) = BH and cr(NH) = NH. In particular, we may assume that B C BH and AT C NH. A similar decomposition of V is also known in this case, [12]. By [40], Theorem 31.3, the unique J5-invariant subspace determined by <j> is eV — lin(v), where v is a 'heighest weight vector'. Conjugating in M n ( F ) , we may assume that the image (T#) of a maximal torus T# of H containing T is diagonal and e = I
. We have V = © A VA, where the 'weight spaces' V\
are defined by V\ = {v G V \ <j>(t)v — \(t)v}, with A running over the set of homomorphisms T —> F*. Then (1 — e)V is the direct sum of the weight spaces VA other than eV, and every (n) either satisfies e(n)e — 0 or ecj){n) — ecj)(n)e — (j)(n)e. P r o p o s i t i o n 8.15 Let G be a finite group of Lie type. Assume that (f) : G —> GLn(F) is an irreducible modular representation. Then there is a nonzero idempotent e G Mn(F) and a subset I C S such that Pj — {x G G\(x)e = e(x)e},Pf = {x G G\e<j>{x) = ecj)(x)e},UI C {x G G | (j)(x)e = e}, and Uf C {x G G \ e(x) = e}. For g G G, a G M n ( F ) , /e£ ga = cf)(g)a,ag — a
M^GuGeGU{«} is a monoid of Lie type on G and there is a representation Mn(F) such that <j>(g) = (j)(g) for g G G and <^(e) = e.
: M^ —>
Proof. We know that there exists e = e2 G Mn(F) of rank one and a subset J C S such that eV = Vu* and (1 - e)V = (1 - {Uf))V. Moreover Pi = {x G G | (:r)eV = eV} = {x G G | <£(z)e = e(x)e} and ([//)e = e. Then e(l - ^(C/f))!^ = 0 implies that e(j)(Uj) = e. Since (1 — e)V is L/-invariant, we also have (L/)(l —e) = (1 — e)(/>(L/)(l — e). Hence e(L7) = e(L7)e. Since P f = L/C/f and e<j>(Uf) = e, we get e ^ P f ) = e(Pf)e. So, in M 0 we have P 7 = P{e),Pf C P " ( e ) and t/jCt/(e),t/fC[f-(e). Now, 0(G) C Mn(F) acts irreducibly by right multiplication on the row vector space W. Let C = {x G G | e<£(x) = e0(z)e}. Then C is a subgroup and B~ C P f C G, so G = P j for some J C S. Clearly
248
CHAPTER 8. MONOIDS OF LIE
TYPE
I C J. But We is a 1-dimensional B -invariant subspace. Dually to [16],[12] we see that W = We 0 W(l - {Uj)) and the summands are ^(Lj)-invariant'. Since Uj C t//, we have <j)(Uj)e = e. Therefore W(l-(/)(Uj))e = 0, so that W{l-(j){Uj)) C W{l-e). Hence, comparing the dimensions, we get W(\ — (£//)) — W(\ — e). Now, an argument symmetric to that used in the first paragraph of the proof shows that cj)(Pj)e — e(/)(Pj)e. In view of the definition of Pi this implies that J — I. We have shown that Pf = P~(e) in M^. Next, we prove that M^ is closed under multiplication. Consider the representation <j> : H —> GLn(F) of the reductive group H correspond ing to G, which is an extension of (j>. Let NJJ be the normalizer in H of a maximal torus T# of H containing T. Note that N = NH^G. Let n G N. We have seen above that either e(n)e = 0 or e<j>(n) = e{n)e. But the latter is equivalent to n G I / . Let x G G\Pf Pi. By the Bruhat decomposition G — B~NB, so that x = anb for some a G B~,b G B,n G N. Then n $ Lj. So e{x)e — ecj){anb)e — {e^{a)e)(j){n){e<j){b)e) = 0. Thus, it is easy to see that eGe = e(Pf P/)e U {6>} = eLj U {<9} C GeG U {(9} in M^, so that M^ is a monoid. Now, as in the proof of Corollary 8.5, it follows that the rules x H> <j)(x)e(j)(y), for x,y £ G, determine a homomorphism M / = G U Ge/G U {<9} —> M n ( F ) , which factors through Afy. Hence, all idempotents of GeG are conjugate, because this is so in GeiG by Corollary 8.6. Therefore, the conditions defining a monoid of Lie type, checked above for e only, are also satisfied for the remaining idempotents of Mff,. It follows that M^ is a monoid of Lie type. This completes the proof of the theorem. □ We are now ready for the main result on modular representations, first obtained in [111]. Note that from Theorem 8.3 and the remark following Theorem 4.28 it follows that the irreducible modular repre sentations of A4 are in one-to-one correspondence with ordered pairs (/,(/>), where I Q S and <j> is an irreducible modular representation of Li. T h e o r e m 8.16 Every irreducible modular representation of the monoid M. restricts to an irreducible representation of the unit group G of Ai. An irreducible representation of G of type (/, x) extends to 2' 5 ^ 7 ' nonequivalent irreducible representations of M.. If aj denotes the number
8.3.
CONNECTIONS
WITH GROUP REPRESENTATIONS
of homomorphisms Li —> F*, then the number of nonequivalent ducible representations of M is J2ics^S^ai-
249 irre
Proof. Let <j> : G —> GLn(F) be an irreducible representation of type (7,X)- From Corollary 8.5 and Proposition 8.15 it follows that there exists e = e2 G Mn(F) of rank one and K C S such that <j> : MK —> Mn(F) given by (xeKy) = {x)e(/)(y)^(x) = c/>(x), for x,y G G, is well defined and it is a homomorphism. Since also P # = {x G G | (x)e = ec/>(x)e}, we must have K = I by the definition of the pair ( / , x ) Since J ° is an ideal of Mi, the column space V = Fn is also an irreducible F 0 [Jj]-module. But J ° is a principal factor of M. Hence, from Section 4.3 we know that V has a structure of F 0 [.M]-module such that eKV = 0 for all K C S with / % K. Now, let I C K. Then £K > e J implies that JKV ^ 0. Since MK is a submonoid of M con taining G, V is also an irreducible F 0 [-Mx]-module. So, as above, V is also an irreducible F 0 [.M]-module in a new way such that e^V ^ 0 and ex'V — 0 for K <2 K' C S. Thus, we have produced 2' 5 \ 7 ' nonequivalent irreducible F 0 [.M]-modules, which are equal as F[G]-modules. Hence, choosing all possible x '■ Lj —> F*, we have produced J2ics^S^^ai nonequivalent irreducible representations of M, each restricting to an irreducible representation of G. On the other hand, we know that the number of nonequivalent irreducible representations of Lj is Y^KCiaKBy the comment preceding the theorem, the number of irreducible rep resentations of M is therefore J2ICSJ2KCI aK- Since this is equal to J2ics 2' 5 ^ 7 'a/, the assertion follows. □ The following result is now a direct consequence of the remark after Theorem 4.28. Corollary 8.17 Let (j) be an irreducible modular representation of G of type (/,x)? where I C S and x '• Li —> F* is a homomorphism. Then the degree of (j) is equal to the rank of the matrix x{Qi)i for ^e sandwich matrix Qj of GejG. Corollary 8.18 Let M be a monoid of Lie type on G. Then any ir reducible modular representation of M restricts to an irreducible repre sentation of G. Proof. Let V be an irreducible F 0 [M]-module. Then there exists a unique J-class J of M such that JV ^ 0 and M0V = 0, where M 0 =
250
CHAPTER
8. MONOIDS
OF LIE
TYPE
{a e M\J <£ MaM}. So V is an irreducible F0[M(J)]-module, where M(J) = G U J U {#}. By Corollary 8.5 it is also an irreducible F0[Mi]module. Therefore, Theorem 8.16 implies that V is an irreducible F[G]module. □ Recall that every irreducible modular representation of G is the restriction of the representation of the corresponding reductive group H C GLn(F). The Zariski closure H of H in Mn(F) is a monoid, called a reductive monoid, satisfying all axioms of a monoid of Lie type except for being finite (see [108] and the example after Lemma 8.2). The same applies to (H) for any rational modular representation <j) of H. This provides a strong link between H and monoids M of Lie type on G, in particular providing another approach to modular representations [111]. The cross section lattice of M, see Theorem 8.7, can be also recovered via the modular representations of G by considering the image of G in the semisimple algebra F[M}/J(F[M]) and applying Theorem 8.16 and its proof. For other results, and bibliography, on monoids of Lie type we refer
to [110].
Chapter 9 Applications One of the motivations for developing the theory of linear semigroups comes from problems on associative rings that have 'many' finite dimen sional linear representations. Rings satisfying polynomial identities and Noetherian rings are the main classes of this type. However, they often lead to representations over a division algebra, which is not necessarily a field. So, an extension of our main ideas and results to the case of 'skew linear semigroups' is needed. Throughout this chapter D denotes a division algebra. We shall be concerned with subsemigroups of Mn(D). Our aim in Section 1 is to present an extension of Theorem 3.5 to this case and to discuss the dif ferences. In particular, this leads to certain invariants for the division algebra D. Our second aim is to present in Section 3 examples of appli cations of this structural approach to various problems on graded rings. As an intermediate step, Section 2 provides a motivation, supported by some technical results, focusing on the case of gradations by cancellative semigroups. The main idea is to reduce problems to the case of group - graded rings, or preferably to the special case of group crossed products, for which the general theory is very well developed, [103].
9.1
Skew linear semigroups and division algebras
Because of the role of matrix algebras Mn(D) over a division algebra D, a natural question that arises is whether the statement of our main 251
CHAPTER
252
9.
APPLICATIONS
structural result, Theorem 3.5, is valid for skew linear semigroups, see [87], Problem 35. The numerical bounds established in this theorem for S C Mn(K) were proved through an exterior power argument. One might expect that another proof can be found, that would use basic linear algebra only, and so it would cover the case of skew linear semi groups as well. We will show that in general this is not true, but all non-numerical ingredients of Theorem 3.5 remain valid in this more general setting. We will also find some necessary conditions on D for the above structural description to be true for semigroups S C Mn(D). Most of the material of this section comes from [119]. All spaces will be right spaces over D. So the set of column vectors Dn is a right Z)-space and Mn(D) acts on Dn be left multiplication. The structure of Mn(D) is exactly as described in Lemma 2.1 and The orem 2.3. Keeping the notation of Chapter 2, we denote by Mj the subsemigroup of Mn(D) consisting of all matrices of rank at most j , and we identify the nonzero elements of the Rees factor Mj/Mj-i with the corresponding elements of Mj \ Mj_i. Recall from Section 2.2 that a linearly ordered set T of nonzero idempotents of a semigroup S is called a triangular set of idempotents if e -< f implies that ef = 0 for e, / £ T, e -< / . The main structural theorem for S C Mn(K) may be now extended in the following way. T h e o r e m 9.1 Let S C Mn(D)
be a semigroup.
Put
Sj = {a £ S\ rank(a) < j} Tj = {a e Sj\SaS
does not intersect non-null H-classes of
Mj/Mj_i}.
Then S0 C 7\ C S, C T2 C • • • C Sn = S are ideals of S (if nonbelianempty). Moreover, 1. Nj — Tj/Sj-i is a union of nilpotent ideals of S/S3_x of nilpotency index two, and S3/T3 - (Sj \ Tj) U {9} C Sj/Sj^ is a 0-disjoint union of uniform subsemigroups Ua ofMj/Mj-i that intersect dif ferent H-classes and different C-classes ofMj/Mj-i. In particular UaUpCN3 for a ^ p. 2. Define Sj
— {a e Mj/Mj_1 | there exist images b,c of elements of S in Mj/Mj-i such that aCb and aTZc}.
9.1. SKEW
LINEAR
SEMIGROUPS
253
/ / every triangular set of idempotents in Sj has at most n2 ele ments, where rij < oo, then Nj is nilpotent of index < rij + 2 and Sj/Tj is a 0-disjoint union of at most nj uniform ideals. Proof. The proof is similar to that of Theorem 3.5. We will only say what changes should be made there. Let T C Mn(D) be a set of idempotents of rank j . Assume that ( e x / ) 2 = 9 for every x G (T) and every e , / G T with e / / (where x denotes the image of x in Mj/Mj_i). As in Corollary 2.13 we see that there exists a linear order -< on T such that ef = 9 whenever e -< / . (i) Tj/Sj_i C Sj is a nil subsemigroup and Sj is a semigroup of matrix type. Moreover every triangular set of idempotents in Sj has at most nj elements. Then Lemma 2.10 may be applied. Thus, we obtain that Tj/Sj-i is nilpotent of index < rij + 2 if rij < oo. (ii) By the proof of Theorem 3.5 we know that: if W / V are two equivalence classes of p and x G W, y G V, moreover e, / are idempotents in Mj/Mj-i such that eHx^fHy and g G Mj/Mj-i is 'H-related to an 2 element of Sj/Sj-i, then (egf) = 0. So, let T be a set of idempotents in Mj/Mj-i chosen from different p-classes. If e , / G T, e ^ / and x G (T 1 ), then (exf) = #. As explained above, T is a triangular set of idempotents. Since T C 5J, |T| < rij. Thus the number of uniform components in Sj/Tj does not exceed rij. □ In other words, if 5 C Mn{D) for a division algebra D, then the assertion of Theorem 3.5 holds with two qualitative differences: - there might be infinitely many uniform components consisting of matrices of a given rank j , - the nilpotent component Nj — Tj/Sj-i is only proved to be a union of nilpotent ideals of index two, (but we do not know whether it is nilpotent). Assume that all n j , defined as in Theorem 9.1, are finite. By this theorem there exists a chain of ideals J\ C J2 C • • • C Jt = 5, where t < Z)j = i rij + n such that J x and all factors J t -/J t -_i, i = 2,. . . ,£, are uniform or nilpotent of index not exceeding max(rij + 2 ) . If we assume that S is 7r-regular, then non-nilpotent factors are completely 0-simple by Corollary 3.2. In this case the set of completely 0-simple factors of the chain is equal to the set of all ^-classes containing idempotents. Thus, the number of such classes is at most 5Dj rij. The following fact, analogous to Proposition 2.14, is true: if / is an ideal of S and S/I is a
254
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nil semigroup, then S/I is nilpotent of index < Tij(nj + 2 ) . In this case / contains all Ji/Ji-i which are completely 0-simple. The remaining factors are nilpotent. A simple proof of the above fact will be omitted. We note also that the closure cl(5) can be defined for any semigroup S C Mn(D) and its construction is the same as in the case of linear semigroups, see Lemma 3.11 and the remark following it. E x a m p l e Assume that there exists an infinite triangular set of idem potents T in some Mj/Mj_i,j < n, (such an Mn(D) will be con structed in the example before Theorem 9.16). Define U = {s G Mj/Mj-i | there exist e , / G T such that e •< f,se — e a n d / s = / } . From the remark after Lemma 2.10 we know that S — U U Mj_i is a semigroup. Moreover, it has infinitely many uniform components con sisting of matrices of rank j and the remaining matrices of rank j yield a nil ideal of S/Mj-i = Sj/Sj-i which is not nilpotent. Therefore, in general, one cannot strengthen the statement of Theorem 9.1. Let D be a division algebra. To D we can associate a series of nu merical invariants dnj = dUij(D), where j < n, n > 1, defined as follows: dnj = the cardinality of a maximal triangular set of idempotents in the principal factor of M2\M2_X of Mn(D). The notion of a triangular set of idempotents in the case where D = K is a field was studied in Lemma 2.11. Namely, if T C Mn(K) is a set of elements of Mj whose images in Mj/Mj_i form a triangular set of idempotents. Then \T\ < (n). Moreover, considering diagonal idempotents we come to dnj(K) = (n). The next step is to consider finite dimensional division algebras. P r o p o s i t i o n 9.2 Assume that a division algebra D is of dimension k < oo over its centre K — Z(D). Fix a basis of D over K. Let 7r : Mn(D) —> Mnk(K) be the representation arising from left multipli cation Dn —y Dn by elements of Mn(D), written in the corresponding basis of Dn. Assume that S C Mn(D) is a semigroup. Then 1. IT maps uniform (nilpotent, respectively) uniform (nilpotent) components ofir(S),
components
of S onto
2. the cardinality of maximal triangular sets of idempotents M2jM2_x does not exceed ft£), so that dn%j(D) < ( ^ ) .
in each
9.1. SKEW
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Proof. First note that dimD(aDn)k = d'imK(aDn) for a G M n ( D ) , so that rank(7r(a)) = rank(a)/c. Therefore each S3 \ Sj-i is mapped onto (n(S))jk \ {n{S))jk-i. From Theorem 9.1 it follows that the structural components of S C Mn(D) are mapped onto those of ir(S) C Mnk(K). It is clear that any triangular set E of idempotents of Sj/Sj-i is mapped onto such a subset in the corresponding principal factor of Mnk(K). The result follows. □ Consequently, to get an example of a skew linear semigroup with in finitely many uniform components, one has to consider division algebras which are infinite dimensional over their centres. P r o p o s i t i o n 9.3 Assume that T is a triangular set of idempotents Mi or M n _ i / M n _ 2 for Mn(D). Then we have \T\ < n.
in
Proof. We will assume that T is a triangular set of idempotents of M\. The second case may be reduced to the former by applying the mapping a H-> / — a. Suppose that \T\ > n. Then there exist ei, e 2 ,. . . , en_|_x G T with e^-ej = 0 for i < j . Let 0 ^ V{ G Im(e 7 ). Since i>i,t>2} • • • ? *Vn a r e linearly dependent, there exist k < n + 1 and a*. +1 ,... , a n + 1 G D such that vk - vk+iak+1 + • • • + vn+1an+1. Thus ekvk = ekvk+iak+i + • • • + e£i>n+1an_|_i = 0 because ekek+i = 0 for i > 0. This contradicts the fact that ekek — ek. O It follows that the first nontrivial case to consider is rank two ma trices in M±(D). In this case, existence of a bound on the cardinalities of triangular sets of idempotents may be expressed in terms of chains of centralizers in D. To prove this result we will need a sequence of auxiliary lemmas. Noncommutativity of D interferes only in the proof of Lemma 9.13. The notation used in the lemmas is justified by the structure of proof of Theorem 9.14. Lower case Greek letters will de note elements of JD, if not stated otherwise. If t>i,... , vk G Dn, by Lin(t;i,... , vk) we mean the right 7}-subspace of Dn spanned by these vectors. L e m m a 9.4 Let V^V^Va be 2-dimensional subspaces of D4. If Vi,V2 and V3 are all different and Vx D Vj ^ 0 for all z',j, then one of the following conditions holds
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1. there exist linearly independent vectors Vi,v2,v3 £ D4 with V\ — Lin(v 2 ,v 3 ), V2 = Lin(ui,v 3 ) and V3 = Lin(ui,u 2 ),
2. Vi n v2 n v3 ^ o. Proof. Choose nonzero elements v\ E V2 P\ V3 v2 E Vi Pi V3,v3 £ Vi fl V2. If Vx fl V2 n V3 = 0, then v 2 , ^3 are linearly independent. Since dim(Vi) = 2 and v2,v3 E Vu it follows that Vi — Lin(i>2, t>3). Similarly V2 = Lin(vi,f3) and V3 = L'm{vi^v2). If v i, v 2 , v3 are linearly dependent, then we must have V\ = V2 — V3, a contradiction. The assertion follows.
□
L e m m a 9.5 Assume that Vi,v2^v3 E D4 are linearly independent. De fine Vi = L i n ( v 2 , ^ 3 ) , ^ — Lin(^i5^3) and MB = Lm(vi,v2)If W 0 Vt■ ^ 0,i = 1,2,3, for a 2-dimensional subspace W of D4, then W C Lm(vuv2,v3). Proof. Let V\ai-\-v2a2 be a nonzero element of WDV^. Similarly, choose nonzero v2(32 + v3/33 E W H Vi. If a x 7^ 0, then t^ax + ^ 2 a 2 , v2(32 + ^3^3 are linearly independent. Since dim(VK) = 2, we get W — L i n ^ a i + t>2a2, V2A2+^3/^3)5 a n d hence W C Lin(u 1 , ^2,^3), as desired. So, suppose that a i = 0. Then 0 ^ v2a2 E W. Let 0 ^ v1^1 + v3^3 E W 0 V2. Then v2 and v^i + V373 are linearly independent elements of W. Hence W — Lin(v2, ^i7i + ^373), and also W C \j'm(vi,v2,v3). D L e m m a 9.6 Assume that V ^ V s , . . . , K 5 W4, W 5 ) . . . , W n , n > 4, are 2-dimensional subspaces of D4 such that Vi 0 Wj 7^ 0 for i < j and Vi n W{ = 0 for 4 < z, j < n. Assume further that Wj Q W,j = 4 , 5 , . . . , n, for a 3-dimensional subspace W of D4. Then n < 6. Proof. Since dim(K-)+dim(W r ) > 4 and 1/,, W C D4 for i < n, it follows that K-HW ^ 0. Suppose that d\m(VtnW) = 2. Then V, C W. But W, C W by the hypothesis. Since dim(W) = 3, V- Pi Wz- ^ 0, a contradiction. Hence dim(V5 ClW) = 1. Let 0 ^ ut- G V5 H W for i < n. If n > 6, then there exists A;, 4 < fc < rz, such that v^ £ L i n ^ , t> 5 ,... ,Vk-i) because t? 4 ,... ,u n E W and dim(VF) = 3. Let i < j , z, j E { 4 , . . . , n } . We know that 0 ■£ V{ n Wj C V{ Cl W. Since dim(K- n W) = 1, we have VJ fl Wj = Vi- fl W. Therefore ut- G VJ H VK implies that i;,- E W3. Hence v 4 , . . . , Ujfc-i E Wk, so that t^ E Wk because vj, E Lin(u 4 ,... ,^/c-i). Since Vk E Vk, it follows that 14 fl ^ / 0, a contradiction. This completes the proof. □
9.1. SKEW
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L e m m a 9.7 Let W,Vi,V2,V3 be 2-dimensional subspaces of D4. As sume that Vi H V2 0 V3 + 0 and Vx ^ V2. If dim(Vi + V2 + V3) < 3,W CiVi ^ 0 andWCiV2^ 0, then also ^ f l V 3 / 0 . Proof. Let ?;, Vi,v2 be nonzero elements of Vifl V2H V3: ViPlVF, V2C\W re spectively. If u,i>! are linearly dependent, then u G VK because Vi G VK, so that V3C\W^0. So, we may assume that v,v\ are linearly inde pendent. Similarly, we may assume that v,v2 are independent. Now, if v\,v2 are linearly dependent, then V\ G V\ C\ V2. Since v E ViC\V2 and v, Vi are linearly independent, we have dim(Vr1 C\V2) > 2. Then Vi = y 2 , a contradiction. So, we may assume that Vi,v2 are linearly independent. Since v\,v2 G W by our choice, W = L i n ^ i , ^ ) . Hence W C VL+V^ + V^ because ui G V\ and v2 £ V2. Clearly V3 C Vi + V2 + V3. But by the hypothesis dim(W) = dim(V 2 ) = 2 and dim(V1 + V2 + V3) < 3. Therefore l V n V 3 / 0 , a s desired. □ L e m m a 9.8 Let W^Vi^V2:Vs be 2-dimensional sume that dim(Vi + V2 + V3) = 4 and W H V-^
subspaces of DA. As 0 /or i = 1, 2, 3. 7% en
Vi n y2 n v3 c w. Proof. We may assume that Vi D V2 PI V3 7^ 0. Let 0 ^ 1; G Vi D V2 n V3. Choose ut- G .D4 such that VJ = Lin(v,v ? ) for i — 1,2,3. Since Vi + V2 + V3 = Lin(u, ux) + Lin(v, u2) + Lin(v, u3) C Lin(v, vuv2, v3) is of dimension 4 by the hypothesis, u, t>i, t>2, U3 must be linearly independent. Let a{,fa G D be such that 0 ^ 1;^ + t;t-/?t- G W Pi VJ for z = 1,2,3. If there exists i such that fa = 0, then v G VK, which yields the assertion. So, suppose that fa ^ 0 for i = 1,2,3. Then vat + Vifa^i = 1,2,3, are linearly independent elements of W. But dim(W) = 2, a contradiction. This completes the proof. □ L e m m a 9.9 Assume that V4, V5,... , K , W4, W 5 , . . . , Wn C D 4 , n > 4, /iaue dimension 2. Moreover, let V{ DWj ^ 0 for i < j and V{ f! W{ = 0, 4 < z, j < n. / / rU<j 1 and 0 ^ u G ri4<j3. Clearly dim(Wj) = 1 because v G W} and dim(V7) = 2 because V; D Lin(u) = 0 (otherwise u G V; H VKt-, a contradiction). We have VI H Wj ^ 0 for i < j because 1; g !<■ n W,- ?£ 0. So W\ C VJ'
258
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for i < j because dim(W-) = 1. Suppose that V- n l ^ ' / O for some %. Then there exist 0 ^ v{ G K , 0 ^ iut- G Wi and # G D such that wt + u # . But w;,t> G W;, so that v% G W2. Since v{ G VJ, we Vi get Vi H Wi; 7^ 0, a contradiction. Hence V/ H W/ = 0 for 4 < z < n. Since n > 7 and dim(Z') = 3, there exists k G {4, 5 , . . . , n } such that W£ C W£ +1 + W£ +2 + • • • + W'n. But W^ C V£ whenever 1 < i < n - k. Therefore W'k C 14'. This contradicts the fact that V'k H W£ = 0. The result follows. □ L e m m a 9.10 LetVil^Vi2JWjn_l,Wjn be 2-dimensional subspaces of D4. Assume that Vla fl W3b ^ 0 for a = 1,2 and o = n - l , n . 7/ V21 0 K2 = 0 and W^n_! fl Wj n = 0, £/ien £nere ezz's* linearly independent vectors ^1,^2,^3,^4 € 7) 4 suc/i tfia* Vn - Um(vl,v2),Vl2 = Lin(u 3 ,u 4 ), W in _ 1 = L i n ^ i , ^ ) , ana7 Wj n = L i n ^ , ^ ) Proof. Let T;1?T;2 G D 4 be such that 0 ^ vx G V%1 D Wjn_x and 0 / ^2 £ K'i n Wjn. Then u l5 t>2 are linearly independent, because otherwise v\ G Wjn_l fl Wj n , contradicting the hypothesis. Since vi,v2 G V^, w e get V^ = L i n O i , ^ ) - Let 0 ^ v3 e Vl2 H WJn and 0 ^ v4 e Vl2 H Wjn_x. Similarly as above we check that V{2 = Lin(t>3, v 4 ). Since V{x C\Vi2 = 0 by the hypothesis, it follows that ^1,^2,^3,^4 are linearly independent. Then Wjn = Lin(i>2,^3) because ^2,^3 £ Wjn, and similarly Wjn_l = Lin(t;i,U4). D L e m m a 9.11 Let ^1,^2,^3,^4 be linearly independent vectors of D4. Define spaces Vil = Lin(v 1 ,u 2 ), Vi2 = Lin(v 3 , i>4), W Jn -i — L i n ^ , ^ ) , Wj n = Lin(u 2 ,t> 3 ). Le£ V, VK 6e 2-dimensional subspaces of D4 such that the following condition holds: V 7^ Via, V{a fl W 7^ 0 /or a = 1,2 ana7 W 7^ Wjfa, V fl Wjb 7^ 0 /or b — n — 1, n. 7/ moreover there exists k G {1,2,3,4} suc/i tfia* ufc € W for u* G VJ then V nW ^ 0 if and only if Vk G V or Vk+2 G V (vk £ W or vk+2 G W, respectively), with indices taken modulo 4. Proof. By symmetry we may assume that vx G W. Then we have to show that V D W 7^ 0 is equivalent to ^ G V or u3 G V. We know that W H Vi2 7^ 0. So there exist e, <£ G 7) such that O / ^ + ^ ^ H K 2 • By the hypothesis t>i G W, so VF = Lin(i;i,i;4C + v3(/>). Similarly, there exist a,/3 € D such that 0 7^ vxa + v4(3 e V P\ WJn_1 and 5,7 G D such that O ^ ^ T + ^ ^ V n W} n . Therefore V = Lin(uia + u4jS, v 2 7 + ^ ) -
9.1. SKEW LINEAR
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259
Suppose first that VC\W 7^ 0. Then there exist p, q, r, s G £>, (p, g, r, 5) ^ (0, 0, 0, 0), such that vxp + (v4e + v3)q = (uxa + u4/?)r + (v 2 7 + ^3^)5. Comparing the coefficients corresponding to the vectors V{ we come to (1) p = ar (3) q = 8s
(2) 0 = 75 (4) eg = /?r.
By (2) 7 = 0 or s — 0. If 7 = 0, then v3 G V, so that the right side of the desired equivalence holds. So, assume that 5 = 0. Then (3) yields 4>q = 0, hence q = 0 or <j> = 0. If — 0, then VK = Lin(u l5 v4e) = W j n _ n a contradiction. So we have q — 0. Then (4) implies that fir = 0, so that r = 0 or (i = 0. If r = 0, then from (1) it follows that p = 0, so that (p,q,r,s) = (0,0,0,0), which is not possible. Therefore (i = 0. This means that v\ £ V, so again the right side of the equivalence holds. It remains to prove the converse implication. If v\ G V, then V C\ W 7^ 0 because v1 £ W. If ^3 G V, then 7 = 0 and therefore V = Lin(u 1 a + v4(i, v3S). Hence V C L'm(vi,v3,v4). Since VK = L i n ^ ^ e + v3<j)), we also get W C Lin(ui, ^3,^4). In view of the fact that dim(V) = 2,dim(iy) = 2 and dim(Lin(ui,t;3, v4)) = 3 this implies that VdW ^ 0, as desired. This completes the proof of the lemma. □ L e m m a 9.12 LetVi^V^^V^^V^W^W^^^Wj^^W^ be 2-dimensional subspaces of D4. Assume that Vix f| V£2 = 0 and Wjn_1 D WJn = 0. Let Vil, K2 ,Vi3jV be all different and similarly let all W, Wjn_2, Wj n _j, Wj n 6e different. Assume also that Vlaf\W]h ^ 0, K a H W / 0 andVnWJb ^ 0, /or a — 1,2,3 anc! 6 = n — 2,rz — l , n . If one of the spaces V£3, V, VK, Wjn_2 contains one of V{a C\ WJb (a = 1,2,6 = n — l , n j ; £/ien 1/ Pi W 7^ 0. Proof. Spaces V^, K 2 , Wjn-l, W}n satisfy the hypotheses of Lemma 9.10. Thus there exist linearly independent vectors Vi,v2,v3,v4 G D4 such that Vn = Lin(ui,t;2),K- 2 = Lin(v 3 , ^)> Wj'n-i ^ L i n ^ i , ^ ) , and WJn = Lin(v2,f 3 ). Consider the sequence of spaces V, Wjn_2, VJ3, W. Neigh bouring spaces of this sequence have nonzero intersections. By the hypothesis we know that one of these spaces contains one of V{1 fl Wjn_, = L i n ( ^ ) , Vn H W in = L h u S ) , K-2 H W ^ = Lin(i; 4 ), VJ2 n W}B = Lin(u 3 ). This space will be denoted by Z. Thus, Vk G Z for some /c G { 1 , 2 , 3 , 4 } . Let Z' be a space following or preceding Z in the sequence V, WJri_2, Vi3, W. If we rename these spaces by putting V = Z, W = Z' if Z = K , Z ' = W* and V = Z',W = Z if Z' = K , Z = VK* (where
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each * stands for an index or a dash), then we meet the notation and the hypotheses of Lemma 9.11. Since we know that V C\ W ^ 0, the "only i f part of this lemma yields Vk G Zf or Vk+2 G Zf. If Z" is a space neighbouring Z' in the sequence V, Wjn_2, VJ3, W, then a similar reasoning leads to vk G Z" or Vk+2 G Z " (note that the indices again are taken modulo 4). Repeating this reasoning we eventually get Vk G V or Vk+2 G V and ^ G VK or Ufc+2 G W. By the " i f part of Lemma 9.11 this implies that V Pi W ^ 0, as desired. □ We denote by N(Z)) the supremum of the set of integers k such that there exist a i , a 2 , . . . ,flfc,&i, °2, • • • , ^ G -D with a;6j = bjdi for z < j and afii ^ 6tat- for i = 1, 2 , . . . , k. L e m m a 9.13 Let n > 6 and let a collection of 2-dimensional subspaces V V- V■ VA V o WA W* W -x W W W of D4 be given. Assume that V{x 0 V{2 = 0 and VK7n_1 D Wjn = 0. Le£ moreover Via D Wk jt 0 for a = 1,2,3 and 4 < k < n - 3, V-a fl WJb ^ 0 /or a = 1,2,3 and b = n - 2,n - l , n , 14 H Wjb ^ 0 /or 4 < fc < n — 3 and b = n — 2,n — l , n . Assume a/so £/m£ VJ Pi VFj 7^ 0 for 4 < z < j < n — 3 and Vk H W^ = 0 for A < k < n — 3. If none of the spaces VJ3, V4,... , K - 3 , W 4 , VK 5 ,... , W n _ 3 , Wj n _ 2 contains one of Via H H^-6, w/iere a = 1,2,6 = n — 1,72, £/ien n < N(D) + 6. Proof. By the hypothesis V^ PiVt2 = 0, Wrin_1 n ^ J n = 0, and VianWjb ^ 0 for a = 1,2 and 6 = n — l , n . So Lemma 9.10 may be applied to Vi1,Vi2^Wjri_l,WjTi. Then there exist linearly independent vectors vi,v2,v3,v4 G -D4 such that V{l = Lin(?;i, t>2), V;-2 = L i n ^ , ^ ) , W/Jn_1 = L i n ^ i , ^ ) and W}n = Lin(t>2,^3)- This implies that V^ H VKjn_1 = Lin(ui), Vn O W jn - Lin(*;2), VZ2 0 Wrjn_1 = L i n ^ ) and Vl2 0 W i n = Lin(v 3 ). We rename the spaces by putting V3 — Vi3,Wn-2 = WJri_2. By the hypothesis none of V3, V 4 , . . . , K - 3 , W4, VK5,... , Wn-2 contains one of the vectors Vi,v2,v3,V4. Define a2-, /?t-, 7,-, o"-t G D, 3 < z < n — 3, such that 0 ^ vxat + v^H G VJ 0 W ^ and 0 7^ u2/3t- + u3<Jt- G K" H VKjn. Then Vi — Lin(t;iat- + v4rfi,v2f3i + v35t). Let a^fJ^^^Sj G D, 4 < j < n - 2, be such that 0 7^ ^ i ^ - + v2pj G Kx H W3 and 0 ^ ^ 7 ^ + v35j G VJ2 n Wj. Hence Wj = L'm(viaj + v2j3^v^)- + v35j). By the assumptions a
are a iiPiiliifiii®jiPjiljifij ll nonzero. We will find a condition for the inequality V{ 0 Wj ^ 0 in terms of these elements of D.
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First note that the condition V, C)Wj ^0 is equivalent to the exis tence of (p, q, r, s) € D4 with (p, q, r, s) ^ (0,0, 0, 0) and (via, + v2'PJ)p + (vffj + v36j)q = (Vlai + v4-ji)r + ( u 2 # + v3Si)s. Then we get (1) ajP = atr (3) Sjq = SiS
(2) 0jP = (3%s (4) JJq = -flr
It is easy to see that p ^ 0 , g ^ 0 , r ^ 0 and s ^ 0. (If for instance p = 0 then r = 0 by (1). This implies that q = 0 by (4), so that s = 0 by (3). We come to (0,0,0,0) = (p, g,r, 5), a contradiction). From (1) we get p = a 7 -" 1 a t T and by (2) we have p = /?." /3zs. Then a j _ 1 a 2 r = /^ l r
=
/?t-s. Also, eliminating g from (3) and (4) we get
s
lj~ li $j $i - Therefore rs'1 may by written in the form
= a~la.j(3~
(3{ = J^lj^j"
( 7l a- 1 )(a^- 1 )(AA- 1 )(^77 1 ) = l-
£•■
Tnis
(9-1)
Conversely, assume that (9.1) holds. It is easy to check that 5 = l , r = a~x7x$- (3i,q = Sj 5i,p = f3- ^ satisfy (1) to (4). Therefore V- D Wj 7^ 0 is equivalent to (9.1). Assume that 3 < i < n — 3 and 4 < j < n — 2. Define at- = 7 ; a ~ \ 6 j = o.jf3- , Cj = f3i8~l,dj = Sj^J1. We rewrite condition (9.1) as follows ciibjCidj — 1.
(9.2)
From the hypotheses of the lemma we know that (9.2) holds for 3 < i < j < n — 2 and does not hold for i = j and A < i < n — 3. In particular a3b3c3dj — 1 for 4 < j < n — 2, so that dj = c3lb~la3l. By substituting this to (9.2) we get aibjCi^b^a^1 a%b3cxcixb-xa3x
= 1 for 4 < z < j < n - 2 ± 1 for z = j , 4 < % < n - 3.
,
, l
'
;
In particular az-6n_2QC3 1b~?_2a3~x = 1 for 4 < i < n - 3. Thus, at- = a 3 6 n _2C3q _1 6^2- By substituting this to (9.3) we see that the element ^K-2CzC~lb~l_2b3CiC~lb~la3l
CHAPTER
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is equal to 1 for 4 < i < j < n — 3 and it is ^ 1 for i = j , 4 < i < n — 3. Transforming this equation we come to
b^icscr^b^b^c-^bj1
= 1,
and next to (b-l2b])(ctc^)
= (cic^)(b-i2b]).
(9.4)
Here (9.4) holds for 4 < i < j < n — 3 and does not hold for 4 < i = j < n — 3. Since (ciC^1) and (b~^_2bj) are as in the definition of N(D), we must have n — 6 < N(D). This establishes the assertion of the lemma.
□ We are now ready for the second main result of this section. Note that 6 = (t) = d4i2(K): and N(D) can be considered as a noncommutativity measure of D. T h e o r e m 9.14 Let T be a set of idempotents of rank 2 in M4(D) such that the image of T in M2/Mi is a triangular set of idempotents. Then \T\ < N(D) + 6, so that d4)2(D) < N(D) + 6. Proof. Suppose that \T\ > N ( D ) + 6 . Then there exist ei, e 2 , . . . , en £ T with n > N(Z)) + 6, such that rank(e z ej) < 2 for 1 < i < j < n. Let Vk = ker(e^) and Wk = Im(e^). So we have (*)
Vi H ^ / 0
for i < j and V{ 0 W{ = 0 for ij
= 1,2,... ,n.
It is clear that dim(1l4) = dim(Wk) = 2. Assume that Vi H Wj ^ 0 for all 1 < i" ^ j < 3. Lemma 9.4 may be applied to ^ , 1 4 , ^ 3 (we know they are all different since, if V{ — Vj for 1 < i < j < n, then Vj fl Wj = Vi fl Wj ^ 0, a contradiction). We consider two cases (see Lemma 9.4.) (a) There exist independent vectors t>i, u 2 , v3 with Vi = Lin(i>2, ^3), V2 = Lin(^3,v 4 ) and V3 = L i n ^ , v2). By condition (*) we have Vi fl W ^ 0 and V2 fl W 7^ 0 whenever W = W^ for arbitrary 4 < A; < n. Hence Lemma 9.5 yields Wk Q Lin(^ 1? v2l v3) for k = 4 , 5 , . . . , n. Define W = L'm(vuv2,v3). Spaces V 4 ,V 5 ,... , K , W 4 , W 5 ,-• - , ^ n and W satisfy the hypotheses of Lemma 9.6, so that n < 6. This con tradicts the fact that n > N(£>) + 6. Thus (a) cannot hold.
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(b) Vx H V2 n V3 ^ 0. Assume first that dim(Vi + V2 + V3) < 3. By (*) Vi fl W + 0 and V2 n W ^ 0 for W = W3. By Lemma 9.7 we have y 3 n l f / 0 , thus V3DW3^ 0. This contradicts (*). Therefore we may assume that dim(Vi + V2 + V3) = 4. By condition (*) \<- Pi WP ^ 0 for i = 1,2,3, whenever W = Wk for 4 < A; < n. Applying Lemma 9.8 to Vi, y 2 , V3 and this W we get 0 7^ Vx H V2 n V3 C W. Then U4<j-» Wm+i-i,Wi i->» 14+1-,- for z = 1,2,... , rc. Hence, by a similar reasoning we see that there exist j n _ 2 , j n - i , jn with { j n _ 2 , j n - i , jn} = {n - 2,ra- 1, n } and W ^ fl Wjn = 0. First assume that there exists k G { 4 , 5 , . . . , n — 3} such that one of the spaces V£3, T4, W^, Vjn_2 contains one of VJa fl W^6, where a = 1,2,6 = n - l , n - 2 . Put F = Vk,W = Wk. Now Lemma 9.12 may by applied to the sequence Vl, V{2, V{3, V, W, Wjn_2, W} n _ 1 , W/jn. This yields F n F / 0 , s o that VkPiWk^ 0. This contradicts (*). Therefore none of the spaces V{3, V4,... , V^-3, W 4 , VK 5 ,... , Wn-$, Wjn_2 contains one of Via fl Wj b , where a — 1, 2, 6 = n — 1, n. So Lemma 9.13 may be applied to the sequence V
V
V
VA
V
*
WA W*
W
* VK
w
w
It follows that n < [Vj + N(D), a contradiction. This completes the proof of the theorem. D It is an open problem to find necessary and sufficient conditions for dnj(D) to be finite, and to determine bounds in terms of the division algebra Z), in case n > 4 and j' =fi 1, n — 1, n. R e m a r k It is easy to see that N(D)
=
sup{n | there exists a chain Co D C\ D • • • D Cn of centralizers of subsets of D}.
In fact, assume that a\, a 2 , . . . dk^i-, b2,... bk of N(D). Then C(a0) D C ( a 0 , a i ) D ••• D a 0 G Z(D), because b{ G C ( a 0 , a i , . . . ,a t --i) \ 1,2,... ,fc. On the other hand, assume that
are as in the definition C ( a 0 , a i , . . . ,«;), where C ( a 0 , a i , . . . ,a 2 ) for i = C 0 D C\ D • • • D Ck,
CHAPTER
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where C — C(Ai), Ai C D. By induction we construct elements a, G Atsuch that C(ao) 3 C(a 0 ,cti) 3 • • • I> C ( a 0 , . . . ,ak). Choose any a 0 G AQ. Assume that the elements ax G Au ... , a z - i G A _ i , z < A;, have been defined. Suppose that C(a0,... , at-_i) = C(ao> • • • , at-_i, A,-). Then C ( A ) 2 C(a 0 )nC f (a 1 )n---nC(a,_ 1 ) 2 C ( A _ i ) , since C, 2 C i - i f o r j = 0 , 1 , . . . , z - 1 . This contradicts the fact that C(A t _i) D C ( A ) . Therefore there exists a{ G A- with C ( a 0 , . . . ,flt-i) D C ( a 0 , . . . ,a t -). This proves the inductive claim. Let b{ G C(a0,au... , a 2 _i) \ C(aQ,au... ,a 2 ) for i = 1,2 . . . fc. Then a i , . . . , a ^ , 6 i , . . . ,6* are as in the definition of N(D).
Thus A; < N(£>). Corollary 9.15 If centralizers of noncentral elements of D are com mutative, then N(Z)) < 2. Proof. Define a relation ~ o n D \ ^(£>) as follows: a ~ b if and only if a6 = 6a. By the assumption on £), ~ is an equivalence relation. Note that C(a) = Z ( £ ) U {6 G £> \ Z(D) | 6 ~ a} whenever a 0 Z ( D ) . Thus every centralizer of a subset of D is equal to Z(D), D or Z(D)U X for a ~ - class X. Therefore the length of every chain of centralizers is less than 3. By the above remark N(D) < 2. □ E x a m p l e Let D be the quotient ring of the skew polynomial ring K[x,5], where S is a derivation of the field K. It is known that if t is a noncentral element of Z), then its centralizer C(t) is commutative, (see [65], Lemma 1). By Corollary 9.15 N(L>) < 2. Our next example shows that there exist division algebras D with infinite triangular sets of idempotents in M2/M1 for M4(D). Its role is explained in the example after Theorem 9.1. E x a m p l e Define D = Dx ®K D2 ®K ''' , where Dt-,i > 1, are finite dimensional division algebras over their common centre K. If their ranks are pairwise relatively prime, then D is a division algebra (see [36], Lemma 4.4.8). Let a t -,# be any noncommuting elements of D{. Put ai = 1(8) ••• ® l ® q , 1 • • • , bi = 1 ® • • • ® 1®/?, ® 1 ® • • • . t-i times
z-i times
Thus at-6j = fya; for i < j and a{bi ^ 6,-a,-. Let u 1 ? . . . , v4 be a basis of D 4 . Consider V{ = Lin(vi + u4at-, v2 + u3at-) and Wj = L i n ^ + ^ & n ^ + ^ j ) .
9.1. SKEW LINEAR
SEMIGROUPS
265
As in the proof of Lemma 9.13 it follows that V{ D Wj ^ 0 if and only if ctibj = bjdi (this is a direct consequence of the equivalence of condition (9.1) with Vi fl Wj ^ 0). By our choice of a{ and b{ we have VJ f l ^ / 0 for i < j and VJ D W{ — 0. Let e be the idempotents in M4(D) such that ker(et-) = VJ,Im(et-) = W2-. The images of et- in M2/Mi form an infinite triangular set of idempotents (e,- -< e^ if and only if z < j ) . We conclude with an application of Theorem 9.1 to the Burnside problem for certain skew linear semigroups. It extends the assertion of Theorem 7.3. T h e o r e m 9.16 Let S C Mn(D) be a n-regular semigroup. subgroup of S is locally finite, then S is locally finite.
If every
Proof. We may assume that S is finitely generated. Define Sj^S^Tj as in Theorem 9.1. Let j be the least integer such that S C Mj. Since 5 = Sj is finitely generated, Sj intersects finitely many £-classes of M = Mj/Mj-i. There exists rij < oo such that every triangular set of idempotents in Sj has at most Uj elements because all elements of a triangular set of idempotents belong to different ^-classes, (if ef = 6 and Me = Mf: then there exists x G M such that e = xf, thus 9 — ef = (xf)f = xf = e, a contradiction). By Theorem 9.1 Sj/Tj is a 0-disjoint union of finitely many completely 0-simple ideals. Let C — A4(G, / , M, P) be any of them. By assumptions on 5, the group G is locally finite. So .M(G, / , M, P) is also locally finite by Lemma 2.3 in [87]. Since C is finitely generated, C must be finite. Therefore Sj/Tj is also finite. By Lemma 2.1 of [87], Tj is finitely generated. Since Tj/Sj-i is nilpotent, Tj/Sj-i is finite. Thus Sj-i is finitely generated. The above reasoning may by repeated for Sj-i and so forth. This shows that Sj/Tj and Tj/Sj-i are all finite. Thus S is finite. □ Corollary 9.17 Let G be a torsion-free almost polycyclic group and D the field of fractions of the group algebra K[G}. If S C Mn(D) is a periodic semigroup, then S is locally finite. Proof. We know that every periodic subgroup of Mn(D) is locally finite (see [72]). Since S is a 7r-regular semigroup, Theorem 9.16 may by applied. Thus S is locally finite. □
266
9.2
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APPLICATIONS
Graded rings: motivation and technical results
The aim of this section is to present a general technique that can be applied, in particular, to several problems in ring theory. It is based on the structure theory of skew linear semigroups. The significance of the theory of linear groups is well known. Their structure and proper ties are exploited repeatedly in many important applications. On the other hand, while semigroups of endomorphisms show up often in var ious problems, and are a natural object to study, until recently their structure had not been exploited in applications. Our aim is to present the key ingredients of such an approach and to illustrate it with some applications to graded rings. We start with preliminary motivating re marks on graded rings, leading to semigroups of matrices. We will consider rings R graded by a semigroup S. That is, R = ®ses Rs as additive groups, and RsRt C Rst for every s,£ G S. H(i2) = Uses Rs is the set of homogeneous elements of R. Clearly, H(i?) is a multiplicative subsemigroup. If X C H(fl), then by Z { X } we denote the (homogeneous) subring of R generated by X. For r G R,r ^ 0, we put supp(r) = {s G S\rs ^ 0}. If I C R is a subset, then supp(Z) = {supp(r) | 0 / r G / } , while the graded part of I is defined by Igr = © s G 5 ( 7 fl Rs). The degree map deg : H(i?) \ {0} —> S is defined by deg(/i) = s for h G RsGroup crossed products R = A * G, for a ring A and a group G, form the simplest class of graded rings, [103]. Here Rg = Ag for g G G, so each component Rg contains a unit of R provided that A is a unital ring. On the other hand, if for a graded ring R the set H = H(fl) \ {0} consists of regular elements (that is, non - zero divisors in /?), then under many classical finiteness conditions on 7?, H is an Ore subset of R and one can form the homogeneous localization RH~l. If additionally S is cancellative, then the image deg(H) of H under the degree map is an Ore semigroup, so RH~l has a natural structure of a group - graded ring with components {Js,tes,s=gt RS{H H Rt)~\ for g G G = d e g ( # ) deg(ff)" 1 . Since each component contains a unit, the following result shows that RH'1 = A * G, a group crossed product of the group G. L e m m a 9.18 Assume that every component of R contains an invertible element. Then S = supp(i£) is a group and R ~ Re * 5, a group crossed
9.2. GRADED
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267
product. Proof. Let us G Rs,s G 5, be an invertible element. Let G be the group generated by all us. It is clear that deg(G) = S is a group. Next Rs-ius C i? e , so that Rs-i C ReuJx. Hence i? = Z)SG5 ^ e ^ J 1 and this is a direct sum. Write gs = u~x. Clearly gsgt = ustgsi for some ust G G D Re. Therefore, for r,r' G Re and s , s ' G 5 we get (rgs)(r'gsi) = T gsr'g~lgsgs' = {rgsr'gjlussl)gss,. Also (r')(s> = ^r'flfj 1 G Re. Hence u s t yield the desired cocycle and conjugations by gs are the desired automorphisms of Re. D Thus, a natural idea is to study R via RH~l, or more generally to reduce problems on an arbitrary graded ring R to certain crossed products derived from R. Such an approach is quite natural and it turns out to be very fruitful even in the simplest case where R is a domain and S = Z, see [3],[137]. What are the most general circumstances this can be applied? For example, if H(R) C Mn(D) for a division algebra D, and F is the group of units of the monoid eMn(D)e for some idempotent e G M n ( D ) , then H ( # ) n F C F - GLj(D), where j = rank(e), (if nonempty) is a natural candidate for a 'nice' set of homogeneous elements of the graded subring Z{R(R) Pi F}. Our main motivating ideas are built along these lines, aiming at applications to rings with many homomorphic images embeddable into simple Artinian rings. L e m m a 9.19 Let P be a prime ideal of a graded ring R such that is a right Goldie ring. Then
R/P
1. the image H' of H(Z2) in R/P has a unique ideal uniform com ponent / , consisting of all matrices of the least nonzero rank in the classical quotient ring Qci(R/P) and the zero matrix. Further more, if I is the completely 0-simple closure of I, then H' = H'\JI is a semigroup in which I is an ideal, 2. if a G H(7?) is such that its image a' in R/P belongs to I and is nonnilpotent, then a'R'a' C R/P is the subring generated by V — a'H'a' \ {0}, it is a prime right Goldie ring and V is a cancellative semigroup.
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Proof. 1) Let Qci(Rf) = Mn(D) be the classical ring of quotients of R' = R/P. Consider the set / of elements q in H' such that the matrix q is equal to zero or has a minimal positive rank in Mn(D). Clearly 7 ^ 0 , because R! ^ 0. Suppose that J is a nonzero nil ideal of H'. Then J is nilpotent by Theorem 9.1 and Proposition 9.3, (see also [24, 17.20, 17.22]). Hence J generates a nilpotent subring TV of R'. Therefore N is a nonzero nilpotent ideal of R'. This contradicts the primeness of R' and shows that H' has no nonzero nil ideals. Furthermore, a similar argument shows that for any nonzero ideals I\ and I2 of H' we have I1I2 7^ {0}. Hence applying the structure theorem for skew linear semigroups, Theorem 9.1, to the subsemigroup H of Qd{R) it follows that I is the unique ideal uniform component of H'. The first part of the lemma now follows as in Lemma 3.11. 2) The % - class of a' in / C QC\{R!) contains an idempotent e, so that a'Qci(R')a' = eQci(R')e is again a simple Artinian ring and a' is invertible in this ring. Similarly, each nonzero b £ a'H'a' C a'la' is a regular element of a'R'a', so T' is a cancellative semigroup. Suppose 0 ^ x £ (r1a'R' + • • • + rka'R')r\ra'R', for some r,ri,... , r^ £ a'R'a'. Then, because R' is prime, xR'a' / {0}. Therefore (ria'R'a! + • • • + rna'R'a')n ra'R'a' ^ {0}. Since by assumption R' is right Goldie, this implies that a'R'a' has finite right Goldie dimension. Suppose that J, L are ideals of a'R'a' such that JL = {0}. Clearly Ja'R'a'L = {0} implies that Ja' = {0} or a'L = {0}. Thus J = {0} or L = {0} because a' is a unit in a'Qd(R)a'. Therefore a'R'a' is prime. Since a'R'a' is a subring of Qci(R), it also satisfies the ascending chain condition on right annihilators. It follows that a'R'a' is a prime right Goldie ring. □ The above lemma applies in particular to prime homomorphic im ages of a graded ring R which satisfies a polynomial identity or is right Noetherian. For the rest of this section we fix a prime ideal P of R such that R/P is a right Goldie ring. Let Qcl(R/P) = Mn(D) for a division algebra D. Our aim is to study R/P via a homogeneous subring of R with a nicer graded structure. For any x £ R and I C i ? , the images of x and X in the ring R/P will be denoted by x and T, respectively. Choose a minimal nonzero idempotent e £ Mn(D) such that F =
9.2. GRADED
RINGS
269
{a £ eMn(D)e\ rank(a) =_rank(e)} intersects E(R). Let T be the inverse image in E(R) of T = F D H(i2). Note that F is a (max imal) subgroup of the multiplicative semigroup Mn(D). The degree map deg : E(R) \ {0} —> S restricted to T is a homomorphism T —> deg(T) C S. Consider the subring RT = Z{T} generated by T. It is an deg(T) - graded ring. In many cases one can show that T is a right Ore subset of RT and the homogeneous localization R' = RTT~l is graded by the group G of quotients of deg(T). Hence R' = A * G, a group crossed product. Moreover, as we will see in Section 9.3, the im age RT of RT in R/P to a large extent is responsible for the properties of R/P. However, in general, one also needs to consider the intersection of H(i?) with other maximal subgroups of Mn(D), and not only those coming from the matrices of minimal nonzero rank. Then, in order to implement the above motivating ideas, one has to use the full strength of the structural approach to subsemigroups of Mn(D). For the remainder of this section we focus on the most important case where S is a cancellative semigroup. If a £ H(J?), then clearly aE(R)a C E(R) and aH(iJ)a is a graded subring of R. Moreover, if S is cancellative, then sts = stf's, £,£,£' £ S implies that t — £', so that aE(R)a = H(ai?a). We will fix an element a in T C H(fl). Let Ha = aE(R)a \ P. Then a belongs to a subgroup F of Mn(D). Moreover, atl(R)a C F U {0}. Hence Ha is the set of all elements q in aR(R)a such that q £ F. It follows that Ha C T is a multiplicative subsemigroup of R. Since Ha does not contain zero, evidently Ha has no zero divisors, that is, xy ^ 0 for any x, y £ i/ a . The additive subgroup R^ generated in R by Ha is a subring of Z{T}. Obviously, the image i?(a) C R/P is equal to aRa. Since i7 a consists of homogeneous elements, i?(a) is a homogeneous subring of aRa. Note that RlPgr is graded and U(R/Pgr) = l)ses Rs/(RsnP). Hence, the semigroup of homogeneous elements of R/Pgr trivially intersects P/Pgr. Clearly P/Pgr is a prime ideal of R/Pgr. So, studying prime homomorphic images of R we may replace R by R/Pgr and hence con sider the special case where P has no nonzero homogeneous elements, that is P H E(R) = 0. Then we get Ha = aE(R)a \ {0}, R[a) = aRa and E(R{a)) = aE(R)a. Hence rs £ (Ha)s = Ha n i? s , for any element r £ i?(a) and any 5 £ supp(r).
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L e m m a 9.20 If P has no nonzero homogeneous elements, then all el ements of Ha are regular in i2(a). In particular, Ha is a cancellative semigroup. Proof. Suppose that rh = 0 for some elements h G Ha, r G i?( a ). The cancellativity of 5 implies that rsh — 0 for each s G S. We know that rs G Ha. Since Ha has no zero divisors, this yields rs — 0 for all s; whence r — 0. We have shown that all elements of Ha are right regular in i? (a ). If we take any elements x,y,z in Ha such that yx — zx, then (y — z)x — 0 and y - z G R(a) imply y — z. Thus j / 0 is right can cellative. Similarly, Ha consists of left regular elements of i?(a) and it is left cancellative. □ A multiplicative subsemigroup T of of regular elements of R is said to be a right (left) Ore subset if, for each r G R and t G T, there exist r' G R and *' G T such that rt' = tr' (respectively, t'r = r't). By B(R) we denote the prime radical of R. L e m m a 9.21 Assume that P has no nonzero homogeneous and one of the following conditions holds
elements,
1. R is an algebra with no free subalgebras of rank two, 2. R has finite right Goldie
dimension.
Then Ha is a right Ore semigroup and it is a right Ore subset of the ring R(a). If 1) holds, then it is also a left Ore subset of i?(a). Proof. Take any elements x,y G Ha. 1) We may assume that £,y are not invertible in Ha since otherwise xHa fl yHa 7^ 0. By the hypothesis there exists a nonzero polynomial / with zero constant term and with coefficients in the base field K such that f(x,y) = 0. Assume that / is of minimal degree. Clearly, f(x,y) = xfi(x^y) + yf2(x->y) for some polynomials fx and / 2 . By symmetry we may assume that fx is nonzero. Here c = fi(x,y) and d = f2(x,y) are elements of the algebra R1 obtained by adjoining an identity to R in case R has no identity element, and R1 = R otherwise. Assume first that xc 7^ 0. In view of Lemma 9.20 we may replace xc.yd by xcx.ydx respectively. Then c G R. Choose s G S such that (xc)s ^ 0. The
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271
cancellativity of S implies that (xc)s — x(ct) for some t £ S. Similarly, (yd)s = y(du) for some u £ S. We get x(ct) + y(du) = f(x,y)s = 0, whence 0 7^ # ( Q ) = y{—du). On the other hand, suppose that xc = 0. If the constant term of /1 is zero, then c £ R and from Lemma 9.20 it follows that c = / i ( x , y) = 0, contradicting the choice of / . So suppose that the constant term of fi is nonzero, so / x — g -\- \ for some A £ K and a polynomial g with zero constant term. Then xc = 0 can be rewritten as x = xe, where c = g(x,y) + A and e — —g(x,y)\~1 £ R(a). This leads to x = xev for some ev £ Ha. By Lemma 9.20 e — ev is the identity of i7 a . But then g(x, y) = — A 7^ 0 and we have xgi(x, y) + yg2{x, y) — — A for some polynomials #i,#2- An argument as in the first paragraph of the proof shows that one of the elements £,y is invertible in # a , again leading to a contradiction. Since x,y £ Ha, the definition of c and c/ implies that ct £ Ha D Rt and — du £ Ha P\ Ru. Thus xHa H yHa 7^ 0. This means that # a is a right Ore semigroup. Similarly, Ha is a left Ore semigroup. 2) By the hypothesis the right ideals x^yaR^j = 1,2,... , are de pendent. So xkyark + • • • + xnyarn = 0 for some rz- £ R and n > k > 1 with xnyarn 7^ 0. Choosing maximal such k we have xkyar^ 7^ 0. So we have b' + b" — 0, where V = xkyc1b" — xk+1d for some c, d £ a P . Also 6' 7^ 0 by the choice of k. As in 1) we see that xkycs — xk+1dt 7^ 0 for some s £ supp(c) and t £ supp(d). So u = a^yCs ^ P because P fl H(P) = 0. Hence there exists z £ H(P) such that uza 0 P. So xky(csza) = xk+1(dtza) £ ffa and both elements in the parentheses lie in aRa. Hence, Lemma 9.20 allows us to cancel xk on the left. Therefore xHan yHa 7^ 0, as desired. So, we have shown that Ha is a right Ore semigroup. Therefore it has the group of right quotients HaH~l. Since R^ is the subring generated by Ha, it follows easily that Ha is a right Ore subset in R(a)- Namely, if /i, /&!,... , hn £ i7 a , then choosing a common denominator g £ Ha we can write /i_1/iz- = g ^ - 1 for some gi £ i7 a . So, for r = hi + • • • + hn we get rg = h(gi + • • • + gn), as desired. Similarly H^ is a left Ore subset of R{a) if 1) holds. D Denote by Sa the image deg(Ha) C 5. Since deg is a homomorphism when restricted to Ha-> under the hypotheses of Lemma 9.21 it follows that Sa is a (cancellative) right Ore semigroup. It has a group
CHAPTER
272
9.
APPLICATIONS
of quotients Ga = SaS'1. Clearly, R(a) is Ga - graded with compo nents (R(a))x = 0 for x G Ga \ Sa and (R(a))x = aRta for x G Sa with x = deg(a)£deg(a). Also Sa Q deg(T) and the latter is a subsemigroup of S. Since deg(a) deg(T) deg(a) C 5 t t , it follows that T is also right Ore and the groups of quotients of T, Sa are equal. Since Ha is a right Ore subset, we can consider localization R^H'1. 1 is x l Obviously, Ra = R(a)H~ Ga = SaS~ - graded, because H~ con sists of homogeneous elements. Clearly, the identity component Da of Ra is equal to J2sesa(R{a))s{Ha)71 ■ Consider ruht G (Ha)t,rs,hs G (77 a ) s . Then rth^1 = rthu(hthu)~l and rsh~l = rshv{hshv)~l, where /i u G (Ha)u,hv G (i/ a )v are such that /i t /i u = hshv. Since tu = su, we 1 see that rshj ,rth^~1 G (Ha)tu{Ha)Tu • This implies easily that Da = \JseSa(Ha)s(Ha)j1 U {0}. Therefore Da is a division algebra. Besides, every homogeneous component of Ra contains an invertible element from HaH~l. This means that Ra — Da * Ga a crossed product of Da and G a , as Lemma 9.18 shows. Note that the natural map from R^a) onto aRa C R/ P extends to Ra = Da* Ga —> aQci(R/P)a. In order to use the above observations, one needs to establish some relations between the properties of i?, i?( a ), and Ra. L e m m a 9.22 Assume that P has no nonzero homogeneous and Sa is a right Ore semigroup. Then
elements
1. If M C M' are right ideals of R^ then also (M H R(a))aR C ( M ' D R(a))aR- In particular, if R has right Krull dimension, or is right Noetherian, then R^ has right Krull dimension, not exceeding that of i?, respectively R(a) is right Noetherian. 2. Assume that prime homomorphic images of R(a) are right Goldie. U Q C Q' are prime ideals of R(a), then there exist prime ideals W C W ofR such that WnR{a) = QnR{a),WfnR{a) = Q'nR{a). Moreover, the classical Krull dimension of R^ does not exceed
that of R. Proof. Assume that M C M' are right ideals of R{a). Then ( M n R(a))aR C (M'nR(a))aR. Suppose that (MnR{a))aR = (MfnR{a))aR. Then ( M n R{a))R(a) = (M n R{a))aRa = (Mf n R{a))aRa = (AT n
9.2. GRADED
RINGS
273
R(a))R(a)- Since R(a) = R{a)Ha\ we come to M = ( M D R(a))R(a) n i2(o) = ( M ' D R(a))Ra n fl(o) = M', a contradict ion. This shows that a strictly decreasing (increasing) chain of right ideals of R(a) yields a strictly decreasing (increasing) chain of right ideals of R. So the first assertion follows. Q^R(a) ^ Q'^R(a) are prime ideals of R^ by the hypothesis on R(ay Since (Q 0 R(a))R(a) = Q, the inclusion above is proper. Let J be the ideal generated by Q H i2 (a) in i2. Since /2(Q D R(a))RnR(a) CQn /2( a) , we must have J H i2(a) = Q C\ R(ay So, let VK be an ideal of R which is maximal with respect to the condition W 0 i2(a) = Q 0 i2(a). It is clear that VF is a prime ideal because Q Pi i2(a) is a prime ideal of /2(a). Similarly, let J ' be the ideal of i2 generated by Qf D i2(a). Write I=(W + J')n R{a). Then a / a C aWa + aJ'a C M ^ n R(a) + J ' H R{a) = Q n i?(a) + <2' fl R{a) = Q' n i? ( a ) . Then / c g ' n i2 (a) since the latter is prime. So we get / = Q' fl i?(a) because J ' D i2(a) = Q' fl i2(a). Therefore there exists an ideal W of i2 which is maximal with respect to W C W and W fl /2(a) = Q' 0 i2(a). Again it is clear that W is a prime ideal of 72. Since Q fl i2(a) C ^ ' f l R(a)i w e must have VF ^ W'. The above can also be applied to finite ascending chains of prime ideals of i?(a), so the remaining assertion follows. □ We note that the additional hypothesis in assertion 2) of Lemma 9.22 is satisfied whenever R is a PI - algebra or, by 1) of this lemma, if R has right Krull dimension, [81]. The theory of group - graded rings is especially well developed for gradations by groups of some special types. Our next aim is to show that certain important ring theoretic properties lead to gradations by such groups. We say that R is a graded algebra if R is an algebra over a field K and all components are Ji-subspaces. L e m m a 9.23 Assume that R is an algebra of finite Gelfand - Kirillov dimension. Then Ga is locally almost nilpotent. Proof. Choose a subalgebra B generated by finitely many nonzero ho mogeneous elements r\- £ 72St, i = 1 , . . . , /c, of i?( a ). Let w\,... , wn £ Ha be some words in rt- such that d e g ( w i ) , . . . ,deg(iu n ) £ Sa are differ ent. Since the components of R are subspaces, tu 1? . .. , wn are linearly independent over K. Therefore, the growth function of the semigroup
CHAPTER
274
9.
APPLICATIONS
S7 = ( s i , . . . ,Sjfc) C Sa is bounded above by the growth function of B. So S' has polynomial growth. Since it is cancellative, it generates an almost nilpotent subgroup of the group Ga = SaS'1 by Theorem 7.1 and Theorem 7.2. □ L e m m a 9.24 Let R be an S - graded algebra satisfying a polynomial identity and let A be a multiplicative subsemigroup ofH(R). If A does not contain zero, then the subsemigroup deg(A) of S has the permuta tion property. In particular, Ga has a subgroup of finite index whose commutator subgroup is finite. Proof. Let H = deg(A). Every PI - algebra satisfies a multilinear identity, that is, an identity of the form (*)
X1'--Xn+
Yl
KXa(l)'"X^n)
= 0,
where Sn is the symmetric group, ka are elements of the base field, [117]. Take any elements a^c^?--- , o>n in A. Suppose that a% G i? St , for i = 1 , . . . , n. Applying (*) to the elements a i 7 . . . , an we get ax • • • an e RSlSn
n
Yl
^(i)'"^(n)'
Given that A does not contain 0 it follows that ax • • • an ^ 0. Therefore
o ^ ai • • • an e R s l S n n
Rs^ys^
for some a ^ 1. Hence sx • • • sn = sa^ • • • s a ( n ). This means that H has the permutation property, as claimed. Finally, the assertion on Ga follows from Lemma 5.25 since we can take A = Ha. □
9.3
Applications to graded rings
This section is devoted to examples of applications of the approach presented before. We shall discuss four different topics: prime Goldie semigroup algebras, homogeneity of the Jacobson radical of a graded ring, graded rings with finiteness conditions and semigroup algebras that are principal ideal rings. Here the semigroup algebra R — K[S], and the contracted semigroup algebra A'olS'], of a semigroup S over a
9.3.1. PRIME
GOLDIE SEMIGROUP
ALGEBRAS
275
field K are viewed as graded rings with components Rs = Ks for all s G S, respectively for all nonzero s £ S. So, it is natural to consider the subsemigroup S of H(i?) rather than entire H(R). A number of other important and easy-to-state problems have been recently treated via representations in simple Artinian rings. In par ticular, several problems concerning presentability of idempotents as a sum of nilpotents in an algebra over a field of characteristic zero have been solved within the algebras M4(D) [121]. Clearly, the methods go beyond the multiplicative structure, but they carry a lot of the flavour of this chapter. This, together with the material of Section 9.1, magnifies the feeling of an abundance of algebraic structures within the algebras Mn(D) (compared to the case where D is a field), providing more moti vation for a study of skew linear representations of associative algebras, [122]. Since a complete bibliography would have to be very long, only the key references are provided, from which the main results come and from which an extensive bibliography can be traced. Our general references to ring theory and to graded rings are [81] and [103]. For basic facts on group rings and semigroup rings we refer to [102],[87]. In many cases we give a sketch of the proof only, the emphasis being on the way semigroups of matrices come to the picture.
9.3.1
P r i m e Goldie semigroup algebras
In [136] Zelmanov described prime semigroup algebras that satisfy a polynomial identity. Our first aim is to extend his result to a descrip tion of prime Goldie semigroup algebras. We do this in the more general context of contracted semigroup algebras. The results come from [48]. The formalism of Munn algebras will be used, see Section 4.2. In par ticular, in this context o will stand for the usual matrix multiplication. We begin with a technical result, which is an extension of Lemma 3.27. L e m m a 9.25 Let J be a uniform subsemigroup of a completely 0-simple semigroup with closure J = M(G,X,Y,P) such that \Y\ - r < oo. Assume that for any maximal subgroup H of J, the semigroup J H H satisfies the right Ore condition and K[G] is semiprime right Goldie. Then
276
CHAPTER
9.
APPLICATIONS
1. G is the group of right quotients of a subsemigroup T such that J has a subsemigroup I isomorphic to M{T,X,Y,P') for a sandwich matrix 2. if Ko[J] is semiprime,
CM{G,X,Y,P')~J,
P\ then \X\ < | y | ,
3. if \X\ = r and P o a / 0 for every nonzero a G A' 0 [J], then the rings K0[I] C K0[J] C K0[J] have the same classical ring of right quotients which is isomorphic to the classical ring of quotients of Mr(K[G)). Proof. (1) To construct T we introduce some notation. For x G X,y G y , denote by Jxy the set of (g,£,y) G J , where g G G. Choose u G X,v G y such that Juv is contained in a maximal subgroup H of J. Let T —
\j
JuyJxv
C Juv U {9}.
xeX,y£Y
For each x £ X,y £Y,
choose zxv G Jxv and r*uy G J wy . Define
/= (
(J
2^TrttyJ U { J } C J .
As in [136] (or see the proof of Theorem 22.5 in [87]), / is a semigroup which is isomorphic to the semigroup of matrix type A4(T, X, y, P ' ) , where P ' is the r x r matrix with (y,x)-entry r ^ z ^ . Since U \o:EX,yey
zxvJuvruy
U {19} = J, /
we get that J can be identified with M(H,X,Y,P'). By assumption, 77 is the group of right quotients oi J f] H. So, it is easily verified that H is as well the group of right quotients of T. (2) Suppose \X\ > r. Let Q be the (semisimple Artinian) ring of quotients of K[G]. There exists a nonzero (r + 1) x r matrix b over Q
9.3.1. PRIME
GOLDIE SEMIGROUP
ALGEBRAS
277
such that Pi ob — 0, where P x is a fixed r x (r + 1) submatrix of P. Since Q is the ring of right quotients of K[T], there exists an r x r matrix c over A"[T] such that b o c ^ 0 is a matrix over K[T]. Then 6 o c can be viewed as an element of K0[I] which annihilates K0[J] on the right. This contradicts the semiprimeness of A ^ J ] . (3) Since, by assumption, K[H] is semiprime right Goldie, the matrix ring Mr(K[H]) has a right classical ring of quotients, say Q'. Hence, for any a £ Mr(K[H}) there exists a regular diagonal matrix 5 £ Mr(K[T]) such that a o 5 £ Mr(K[T]). The assertion on the sandwich matrix P is equivalent with the right annihilator of A'0[
/
A' 0 [/]oP = M r (A^[r])oP C K0[J]oP'
Since
C K0[J\oP'
C Mr(AT[ff]) C Q\
we obtain that Q' is the ring of right quotients of all these rings. As P' is invertible in Q\ the natural homomorphism Ao[ KQ[J] O P' is an isomorphism. Hence, K0[T\ C A^0[J] C A^0[J] have a common ring of right quotients isomorphic to Q'. D T h e o r e m 9.26 Let K be a field and S a semigroup. following conditions: 1. the contracted semigroup Goldie,
Consider
the
algebra KQ[S] is prime left and right
2. S does not contain noncommutative is prime left Goldie,
free subsemigroups and KQ[S]
3. S contains a right ideal A and a left ideal B such that AB and BA are both uniform subsemigroups in a common completely 0simple semigroup with equal closures, AB = M.(T,r,r,P), where T is a cancellative semigroup such that K[T] is prime left and right Goldie, and the right and left annihilators of the contracted semigroup ring KQ[AB] in KQ[S] are trivial.
CHAPTER
278
9.
APPLICATIONS
Then, (1) and (3) are equivalent, and (2) implies (3). Proof. Assume R = K0[S] is prime left Goldie. It is easy to see that the assertions of Lemma 9.19 remain valid for the subsemigroup S of H(P) = KS = {As | A G K,s G 5 } . This is because sUXs, if A ^ 0, in the left quotient ring Q of R. Hence S has an ideal I which is uniform in a completely 0-simple semigroup / = M(G, X, Y, P ) contained in Q. It follows also that, for any maximal subgroup H of / , H is a group of left quotients of T = H n / and K[H(M] is prime left Goldie. Note that then also the ring K[H] is prime left Goldie. Because the 'columns' of / define, in a natural way, left ideals in the left Goldie ring A"o[5], it is clear that Y is finite. For simplicity we write Y = { 1 , . . . , r } . As an ideal of a prime left Goldie ring, KQ[I] is prime left Goldie as well. Assume now that Ko[S] is also right Goldie or that S does not contain noncommutative free subsemigroups, that is, we assume condition (1) or (2) of the statement of the theorem is satisfied. We claim that then Ko[I] is also right Goldie. The claim clearly holds if A'o[5] is right Goldie. So, we deal with the case that S does not contain noncommutative free subsemigroups. From Lemma 9.25 and its right-left dual we know that \X\ = \Y\ = r and Ko[I) is right Goldie. Therefore K[H C\ I] also is right Goldie. Moreover, we may assume that
X = Y =
{l,...,r}.
Using the notation of the proof of Lemma 9.25, let A=
z U xvluy U {#} , B = (J Ixvruy xex,yeY xex,yeY
U {#}.
Then the right (left) annihilator of R = AB in K0[S] is trivial because it coincides with the right (left) annihilator of K0[I]. Thus, (3) follows. We now prove that (3) implies (1). Let A be a right ideal of S and B a left ideal of S satisfying the assumptions stated in (3). Since K[T] is prime left and right Goldie, the cancellative semigroup T has a left and right group of quotients, say G. It follows that AB and BA are uniform in the completely 0-simple semigroup J — M(G, r, r, P). Now K0[AB] is isomorphic to the Munn algebra M(K[T],r,r, P), and also KQ[J] ~ M(K[G],r,r, P). Since, by assumption, K0[AB] has zero right and left annihilators in KQ[S], and thus also in K0[AB], it is easily seen that P o c / 0 and a o P ^ 0 for any nonzero a G KQ[J]. Hence, by Lemma 9.25, K0[AB] C K0[J] have the same prime (left and
9.3.2. SEMILOCAL
AND PERFECT
RINGS
279
right) classical ring of quotients. So both rings are prime left and right Goldie. Clearly we then also obtain that K0[(AB)2] is prime (left and right) Goldie. Let L be a nonzero ideal of /lot*?]. Then ALB is an ideal of AB and clearly ALB C L. If ALB = 0, then BALBA = 0, and there fore Ko[(AB)2]LK0[(AB)2] = 0. The hypothesis on the annihilator of Jio[Ai?] implies that L = 0, a contradiction. Therefore, every nonzero ideal of K0[S] intersects /\0[^4-i?] nontrivially. The primeness of / ^ [ A B ] implies that /^ 0 [5] is prime. For each 1 < i < r, choose ez- = (c z -,z,l) £ BA, that is we pick an element from each row of the first column of BA C J C J. Then the free right K[G]-module M = ®ri=1 K[G] with basis ez, 1 < i < r, can be identified with K0[J]ei. Let
EndK[G](M)
be the ring morphism defined via left multiplication, that is, for a G A'of'S'], ipip1) 1S ^ n e rna^rix of the left multiplication by a in the above basis of M. Since /\o[S] is prime, EndK[G](M). Abusing notation we will denote this map also by tp. Hence
Mr(K[G\).
By the previous we know that (p(K0[AB]) and tp(K0[J]) are prime (left and right) Goldie, and they both have the same classical ring of quo tients. It follows that ip(K0[S]), a n d thus also Ar0[S'], is prime (left and right) Goldie. D
9.3.2
Semilocal and perfect group - graded rings
Following [46], we consider rings satisfying certain classical finiteness conditions. Such conditions have been studied by many authors, but in most cases not from a structural point of view. However, most of the known results follow from the structure theorem presented below. The first result shows that, roughly speaking, the case where H(i?) is a completely 0-simple semigroup is crucial. Recall that a semigroup N with zero 6 is left T-nilpotent if for every x l 5 x 2 , . . . in TV there exists n > 1 such that 9. A ring R is semilocal if R/ J(R) is
280
CHAPTER
9.
APPLICATIONS
semisimple Artinian. R is called left perfect if it is semilocal and is left T-nilpotent.
J{R)
P r o p o s i t i o n 9.27 Let R be a left perfect (semilocal, respectively) ring graded by a semigroup S. Then H(i?) has a finite ideal chain whose factors are either left T-nilpotent (nil) or completely 0-simple. The idea of the proof: i) first show that H(i?) is 7r-regular, that is, a power of every a G R lies in a subgroup of the semigroup H(i?). (This is an extension of the well-known fact that a simple Artinian ring has this property, extending Proposition 1.3). The proof uses the following useful lemma: (*) If a ring R graded by a semigroup S is semilocal and a G Rs is not nilpotent, then s is periodic in S. ii) next, use the assumption on R together with the fact that a 0-simple semigroup which is 7r-regular must be completely 0-simple. □ We use the above proposition to reduce the study of R to that of 5-graded rings whose homogeneous elements form a semigroup with a nicer structure. The largest homogeneous ideal contained in the Jacobson radical J(R) of R = ® s 6 5 Rs will be denoted by Jgr{R). Let N be the maximal nil ideal of H(i?) for a left perfect 5-graded ring R. Then N is left T-nilpotent. So the subring Z{N} generated by N in R is a left T-nilpotent homogeneous ideal of R. We denote this ring by Jx. It fol lows that Jx — Jgr{R). By the proposition, the semigroup R(R)/N has a minimal completely 0-simple ideal / . Let Ix = ( Z { / } + Jgr{R)) /Jgr{R) be the image of Z{I} in the S-graded ring Rl = R/Jgr(R). Then the im age of I in I\jJ\ is a completely 0-simple ideal in H ( / i / J \ ) . Continuing this process we obtain a chain /o = { 0 } C J 1 C / 1 C J 2 C / 2 C . . . C J
m
C/
r o
C Jm+l = R
of homogeneous ideals of it!, with m less than or equal to the length of the i?-module R/J(R), and such that each Jt-//t- is left T-nilpotent and each R(Ii/Ji) has a completely 0-simple ideal L that generates U/ J{ as an additive group, and in particular supp(/ 2 /J 2 ) = supp(L). Next, we need the following technical lemma.
9.3.2. SEMILOCAL
AND PERFECT
RINGS
281
L e m m a 9.28 Let G be a group with identity e and let R be a semilocal G-graded ring with J(Re) nil. If L is a homogeneous left ideal of R withLDRe C J(Re), then L C J(R). In particular, if Rg-irg C J(Re) for some rg G Rg, then rg G J{R). Hence J(Re) C J(R) and J(Re) = J(R) n Re. Proof: Using (*), first show that H(i?) is nil modulo Re. Then the assumptions imply that \Jg^o Lg must be nil, and consequently L C
J(R). If Rg-irg C J(Re), then the above applied to L = it!?^ imphes that rg e J(R). This yields J ( i ? e ) C J(R)PiRe (the converse is well known).
□
The main result in the group - graded case shows that the rings in question are built of group crossed products. T h e o r e m 9.29 ([46]) Let G be a group and let R be a G-graded ring with J(Re) C Jgr[R) ^ jR. Then the G-graded ring R/' Jgr[R) is semilocal (respectively left perfect, semiprimary, left Artinian) if and only if the following conditions are satisfied 1. Re/J(Re) algebra,
~ Mni(Di)
x • • • x M n r ( D r ) , where each D{ is a division
2. for any complete set of orthogonal idempotents eu of Re/J(Re), 1 < u < q = ni + • • • + n r , the ring R' — R/Jgr(R) is the direct product of matrix rings over crossed products over some periodic subgroups Hi of G Mmi(D(i)*ffi) x ... x Mmi(0(i)*#i)> with q = mi + • • ■ + mj, and for each 1 < i < /, M m ,-(D W * Hi) = R!e3R!,
D{t) =
e3(ReU(Re))e3
for some 1 < j < q. In particular these matrix rings are homoge neous subrings. 3. each crossed product D^) * Hi is semilocal (respectively left perfect, semiprimary, left Artinian).
282
CHAPTER
9.
APPLICATIONS
The idea of the proof: we can assume that Jgr(R) — 0. Then Re has a unity and one verifies that this is the unity of R as well. Write 1 = J2 ei for primitive orthogonal idempotents e2- of Re. Each ezi?ee2 is a division algebra. But H(e 2J Re z )\{0} is cancellative and Lemma 9.28 implies that the grading is non-degenerate (Rg-\rg ^ 0 ^ rgRg-i for 0 ^ rg G Rg,g G G). Via Lemma 9.18 this allows us to show that ezi?ez ~ e{Reei * Hi for a group i/ z . R semilocal implies that Hi is periodic. Glue the pieces together, writing first R as a generalised matrix ring J2ij eiRej-> to get the structure of R. □ The hypothesis J{Re) ^ Jgr(R) is always satisfied for a left perfect ring R, because of Lemma 9.28 and Lemma 9.30 below. Clearly, the condition Jgr(R) ^ R is satisfied for a group - graded ring with unity. For an arbitrary group - graded ring R one obtains from the theorem that R is left perfect if and only if Jgr[K) is left T-nilpotent and R/ Jgr(R) satisfies conditions (1) - (3) if R ^ Jgr(R)An extension of Theorem 9.29 to rings graded by a semigroup S is also known, [46]. It turns out that R modulo J{R) can be covered by finitely many subrings of the form aRob, where a, 6 G H(i?) and RG — ®geG Rg i s a G-graded (semilocal) ring, for a subgroup G of S. The following lemma is easy in case of rings with unity, but creates unexpected problems in the general case, which in particular is needed to handle semigroup gradations. L e m m a 9.30 Let H be a subgroup of an S-graded ring R. If R is semilocal (left perfect), then so is the ring RJJ = 0 / i e / / Rh-
9.3.3
T h e radical of a graded ring
The homogeneity problem for the radical of R asks whether Jgr(R) — J{R) under some conditions on R and on S. The theorems of Bergman on Z-graded rings and of Cohen and Montgomery on algebras over a field of characteristic zero graded by a finite group G are the most basic results in this area. For these and various generalizations we refer to [59]. _ _ S is said to be a u.p. (unique product) semigroup if for every non empty finite subsets A, B C S there exists an element of AB which has a unique presentation in the form a&, where a G A, 6 G 5 , [87]. Such an
9.3.3. THE RADICAL
OF A GRADED
RING
283
S is cancellative. If it has a group of quotients, then this group must be torsion - free. It is an open problem whether the radical of any 5-graded ring is homogeneous. Our main result reads as follows. T h e o r e m 9.31 ([61]) The Jacobson radical of an S-graded ring R is homogeneous if either of the following conditions holds 1. S is a cancellative semigroup and R is a Pi-algebra over a field of characteristic zero. 2. S is a u.p.-semigroup is satisfied:
and at least one of the following
(a) all nil subsemigroups
ofH(RlJgr(R))
conditions
are locally nilpotent,
(b) every nil subsemigroup of every right primitive image of R is locally nilpotent,
homomorphic
(c) for every minimal prime ideal P of R, the ring RjP is a domain or embeds into a matrix ring over a division algebra. Conditions (a),(b),(c) cover in particular the case where R is PI, semilocal, or it is right Goldie modulo the prime radical. The same proof shows also that the prime radical of R is homogeneous in any of the above cases. We will outline some ideas of the proof. For an element x £ R by H(x) we denote the subsemigroup of H(i?) generated by the set of homogeneous components of x. The first lemma, roughly speaking, allows us to consider two cases only: the case of a group gradation and the case where the study of quasi inverses of elements x £ J(R) requires only the ring generated by H(x). L e m m a 9.32 Let S be a cancellative semigroup, R an S-graded ring, I the ideal of nonunits of S and G = S \ I (if S has no identity, then S = I). Assume that x + y = yx for some x £ /?/, y £ R. Then y £ A, where A = Z{R(x)} is the subring generated by H(x). An element a £ R is called rigid if xay = 0 implies xaty = 0 for every t £ supp(a) and x, y £ R(R). This is a useful notion, derived from the often exploited properties of nonzero elements of shortest length in a given ideal of R.
CHAPTER
284
9.
APPLICATIONS
L e m m a 9.33 Let S be a cancellative semigroup, R an S-graded ring, r a rigid element of R. IfH(r) contains 0, then H(r) is nilpotent. So, when studying the 'local' properties of elements r G J{R) (those expressible in terms of the ring Z{H(r)}), the only other case to consider is when H(r) has no zero divisors, and hence deg is a homomorphism on H(r). This, together with a reduction to semigroups of matrices (pos sible because of the assumptions on R) allows us to apply the ideas of Sections 9.1,9.2. The case of a PI - algebra R is the easiest to explain. Namely, Lemma 9.24 leads to the case of a gradation by a group which is locally almost abelian. So known results on gradations by such groups (extending the theorems of Bergman and of Cohen and Montgomery, [59]) can be applied. If J(R/Jgr(R)) ^ 0, this yields a nonzero nil ideal of the semigroup R(R/J(R)) constructed from certain 0 / r G J(R), which leads to a contradiction. On the other hand, the proof in case (c) requires an induction that involves all 'layers' of the skew linear semi groups which are the images of H(i2) in the quotient rings of the prime rings i ? / P , for minimal primes P of R.
9.3.4
Principal ideal semigroup algebras
Our last application is again concerned with semigroup algebras K[S] viewed as 5-graded rings with components Ks,s G S. We assume that K[S] has a unity and that every left ideal of K[S] is principal. The main problem is to find a structural description of such algebras and of the underlying semigroups. The group ring case is our point of departure. T h e o r e m 9.34 ([101]) Let G be a group and K a field. The following conditions are equivalent: 1. K[G] is a principal left ideal ring, 2. if char K = 0 then G is finite or finite-by-infinite cyclic, if char K = p > 0 then G is (finite p')-by- (cyclic p) or G is (finite p')-by-infinite cyclic. Note that such rings are finitely generated A"-algebras, they satisfy a polynomial identity and have Gelfand-Kirillov dimension 1. The fol lowing extension to cancellative semigroups has been obtained in [49].
9.3.4. PRINCIPAL
IDEAL SEMIGROUP
ALGEBRAS
285
T h e o r e m 9.35 Let T be a cancellative monoid and K a field of charac teristic p (not necessarily nonzero). The following conditions are equiv alent: 1. K[T] is a principal left ideal ring, 2. T is a semigroup satisfying one of the following
conditions:
(a) T is a group satisfying the conditions of Theorem 9.34, (b) T contains a finite p'-subgroup H and a nonperiodic element x such that xH = Hx, T = U ; E N Hxl and the central idempotents of K[H] are central in K[T]. The first major step of our approach to the general case is the fol lowing result. T h e o r e m 9.36 ([47]) Let K[S] be a principal left ideal ring. K[S] satisfies a polynomial identity.
Then
The idea of the proof: for simplicity we discuss only the case where S C Mn(D) and the 7\-subalgebra A generated by S is an order in Mn(D). Consider a maximal subgroup G ~ GLj(D) of Mn(D). General techniques allow to lift (right) ideals of K[gp(S 0 G)] to (right) ideals of A"[5], [87], Lemma 7.21 (compare Lemma 9.22). The assumption on K[S] implies then that K[gp(S 0 G)] is right Noetherian. The main difficulty is in showing that this group ring actually is a principal ideal ring. Then it is PI by Theorem 9.34. Finally, the structure of S C Mn(D) together with the Goldie condition for A allow us to show that K[S] is PI as well. □ Corollary 9.37 If K[S] is a principal left ideal ring, then S is finitely generated and K[S] embeds into a matrix ring Mn(F) over a field ex tension F of K. In particular, S is a linear semigroup. The groups associated to S are either finite or finite-by-infinite cyclic. Proof. K[S] is finitely generated because it is a left Noetherian PIalgebra, [87], Theorem 19.14. A theorem of Ananin implies that a fi nitely generated left Noetherian PI - algebra embeds into a matrix ring over a (finitely generated) commutative algebra A, [2]. A classical result
286
CHAPTER
9.
APPLICATIONS
of Malcev then implies that K[S] embeds into matrices over a field, [76]. Theorem 9.34 is used to get the remaining assertion. □ The following is an easy consequence. Corollary 9.38 If K[S] is a principal left ideal ring, then the GelfandKirillov dimension of K[S] is equal to its classical Krull dimension and it is 0 or 1. In the former case S is finite. Moreover, every prime Artinian homomorphic image of K[S] is finite dimensional over K. The following construction plays a crucial role in our final result. E x a m p l e 1 Assume that H is a finite group whose order is not divis ible by the characteristic of K. Let T be a monoid with group of units H such that T — (Jz>o ^x% for sorne x £ T, and either this union is disjoint or xn = 6 for some n > 1. Assume also that Hx — xH and the central idempotents of K[H] commute with x. Then K[T] is a principal ideal ring. In fact, one can check that K[T] ~ A © B , where A is a semisimple Artinian ring and B is a finite direct sum of algebras of the type C\x,a\l(xk) for a semisimple Artinian ring C, an automorphism a of C and some k>l. Note that this extends the construction of 2 b) in Theorem 9.35. Let C be a prime homomorphic image of a principal left ideal ring K[S). We use the fact that C is a finitely generated prime PI - algebra of classical Krull dimension 0 or 1. First, we claim that, if an ideal J of C is idempotent, then either J — 0 or J — C. Let J = Ca. Then CaCa — Ca2, so that J = Ca2 = Ja. Therefore ba — a for some b £ J. Thus bax = ax for all x £ C and consequently b is a left identity of J. Therefore it is an identity of J because C is prime, so J = 0 or J — C. Secondly, if J is a nonzero ideal of C, then C/J is of finite dimension over K. Indeed, the classical Krull dimension of C/J is 0. Hence GK(C/J) = clKdim (C/J) = 0. The former observation will allow us to decompose certain algebras into a direct product of its simpler blocks. The latter leads to the strat egy of considering the finite dimensional case first, and then applying this to the general case. It is well known that finite dimensional algebras which are princi pal left ideal rings also are principal right ideal rings. Moreover, they
9.3.4. PRINCIPAL
IDEAL SEMIGROUP
ALGEBRAS
287
are exactly finite direct sums of matrix rings over local algebras whose radical is a principal ideal, [21], Theorem 9.4.1. This decomposition of K[S] turns out to be a refinement of a decomposition on the semigroup level, and leads to a complete description of K[S] as follows. P r o p o s i t i o n 9.39 Let S be a finite semigroup and N a nilpotent ideal of S, or N — 0, such that S/N is completely 0-simple. Then the con tracted semigroup algebra K0[S] is a principal ideal ring if and only if one of the following conditions is satisfied 1. S ~ M.{G, n,n,Q) for a finite group G satisfying the conditions of Theorem 9.34 and for a sandwich matrix Q that is invertible over K[G], 2. S ~ M.(T, n, rc, Q) for a finite monoid T satisfying the conditions of Example 1 and for a sandwich matrix Q that is invertible over K0[T\. P r o p o s i t i o n 9.40 Let S be a finite semigroup. Then Ko[S] is a princi pal ideal ring if and only if S has a chain of ideals I\ C I2 C • • • C Ir = S such that Ii and every factor Ij/Ij_i is of the type described in Propo sition 9.39. The remaining step is to show that principal ideal rings /\o[5] can be approached as inverse limits of finite dimensional semigroup algebras. This leads to our main result. T h e o r e m 9.41 ([47]) The following conditions are equivalent: 1. KQ[S] is a principal (left an right) ideal ring, 2. there exists an ideal chain I1C--Cli
= S
such that 1\ and every factor I3/1)_x is of the form M(T,n,n, P) for an invertible over KQ\T\ sandwich matrix P, and one of the following conditions holds: (a) T is a group of the type described in Theorem 9.34,
CHAPTER
288
9.
APPLICATIONS
(b) T is a monoid of the type described in Example 1. In case the equivalent conditions are satisfied it follows that K0[S] ~ K0[h] © Ko[h/h]
© • • • ffi K0[It/It-i]
and each component is isomorphic to some Mn(K[T]). Moreover, is a finite module over its centre, which is finitely generated.
K0[S]
It may be checked that each of the semigroups / i , 7j//j_i, yielding the structural blocks in the above theorem, is of the form M.(T', n, rc, Q), where T" is a homomorphic image of a cancellative monoid T satisfying the conditions of Theorem 9.35. In case T is not a group, one of the conditions for K[T] to be of such a type is that the central idempotents of the semisimple ring K[H] are central in /^[T], where H is a finite subgroup of T = (Ji>o Hxx. In case K[H] is split (that is, a direct product of matrix rings over K) this means that the centre of K[H] is contained in the centre of i^[G], where G is the group of quotients of S. Therefore, this condition can be given an intrinsic formulation in terms of S only: the S - conjugacy class of every element h of H coincides with the conjugacy class of h in H. The methods used in the proof of Theorem 9.41 allow us to estab lish the left - right symmetry of the principal ideal condition for any semiprime algebra Ar[S']. This leads to the following result. T h e o r e m 9.42 Let S be a semigroup and K a field. Then KQ[S] is a semiprime principal left ideal ring if and only if there exists an ideal chain
/iC-C/
t
= S
such that Ii and every factor Ij/Ij_l is of the form M(T,n,n,P) for an invertible over KQ[T] sandwich matrix P and a monoid T such that 1. either T is an infinite group as in Theorem 9.34, 2. orT — Ui Hxl is of the type described in Example 1 and such that for every primitive central idempotent e £ /^[-ff], either K[H]ex — 0 or K[H]exi ^ 0 for all t > 1. Moreover, if the equivalent conditions are satisfied, then Ko[S] is a prin cipal right ideal ring and the direct sum decomposition of Theorem 9.41 occurs.
9.3.4. PRINCIPAL
IDEAL SEMIGROUP
ALGEBRAS
289
If G is as in Theorem 9.34 and K[G] is prime, then G is trivial or infinite cyclic. If T is as in Example 1 and Ao[T] is prime, then H is trivial. Therefore, the following is an easy consequence of Theorem 9.42. Corollary 9.43 A ^ S ] is a prime principal left ideal ring if and only if S~M({l},n,ra,Q),
S~M((x)\n,n,Q)
or S ~
M((x,x~l),n,n,Q)
where the matrix Q is invertible in Mn(K), Mn(K[x]) or Mn(K[x,x~1}) respectively. Hence, K0[S] ~ Mn(K), Mn(K[x\), or Mn(K[x, x'1]).
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Index algebra Artinian, 267, 281, 286 division, 110, 129, 251, 264 Hecke, 55 locally finite, 184 Munn, 5, 50, 98, 111, 164, 275, 278 Noetherian, 166, 272, 285 PI, 157, 163, 166, 184, 187, 274, 283, 285 principal ideal, 284, 287 semigroup, 4, 50, 98, 157, 163, 237, 284, 287 contracted, 4, 111,150,277 semisimple, 50, 105, 237 separable, 110 almost nilpotent semigroup, 172 almost semisimple semigroup, 209, 219 almost unipotent semigroup, 138, 141, 188 associated group, 59, 67, 71, 82, 83, 93, 95, 133, 137, 142, 143, 151, 173, 179, 190, 192, 199, 219, 285
bottom layer, 123, 125 Bruhat decomposition, 25, 226, 235 closed set, 5 closure completely 0-simple, 59, 77 7r-regular, 8, 78, 80, 81, 105, 145, 147, 148, 180, 192, 254 Zariski, 5, 8, 80, 95, 131,137, 158, 161, 180 commutator subgroup, 158, 168 completely 0-simple closure, 59, 77 completely 0-simple semigroup, 2, 30, 62, 68, 86, 120, 162, 237, 280 completely reducible semigroup, 54, 102 connected component, 8, 168, 174 constructible set, 169 Coxeter generators, 225, 230 cross section lattice, 235, 250 crossed product, 266, 281 degree map, 266 dimension Gelfand-KiriUov, 184, 190, 203, 211, 273, 286
basic ideal, 116 £ TV-pair, 225 Borel subgroup, 25, 34, 55, 226 301
302 Goldie, 268, 270 Krull, 191, 272 distance function, 171 admissible, 171 egg-box pattern, 3, 23, 67, 69 exterior power, 9, 26, 175 Frobenius map, 226, 246 Gelfand-Kirillov dimension, 184, 190, 203, 211, 273, 286 generating function, 16 Goldie dimension, 268, 270 Goldie ring, 267, 275,277 graded ring, 266, 279, 282 homogeneous semigroup of, 280, 283 Green's relations, 1, 19, 23, 160, 228 group almost abelian, 163 almost cyclic, 197, 199, 200, 203, 208 almost diagonalizable, 222 almost N-, 131 almost nilpotent, 134, 137, 145, 157, 167, 172, 179, 180, 184, 190, 192, 198, 217, 219 almost solvable, 9, 133, 143, 173 almost unipotent, 222 associated, 59, 67, 71, 82, 83, 93,95,133,143,151,179 connected, 136 finite-by-infinite cyclic, 284 free, 107, 133 linear projective, 171
INDEX locally almost nilpotent, 273 locally N-, 131 nilpotent, 136, 144, 169 of Lie type, 226 of quotients, 60, 71, 132, 158, 172, 185, 198, 217, 220, 269,271,276,283 reductive, 226, 229, 246, 250 semisimple, 168, 174 solvable, 8, 169 Weyl, 25 growth, 183 linear, 203 polynomial, 184, 187, 190, 211, 218 growth function, 184 for the number of ^-classes, 210 'H-class, 2 Hecke algebra, 55 homogeneous element, 266 ideal uniform component, 67, 77, 92, 95, 103, 267 idealizer, 125 idempotent, 3, 15, 17, 20, 21, 27, 84, 86, 87, 114, 117, 126, 145, 147, 160, 212, 214, 228, 230, 235, 237, 247, 253-255, 262, 268, 281, 285, 286 identity polynomial, 157, 163, 166, 184, 274, 285 semigroup, 132, 133,136, 137, 162, 173, 192 inversion, 151 irreducible component, 159
INDEX irreducible module, 54, 116, 245, 248 irreducible representation, 54, 120, 121, 236, 238, 245, 248 irreducible semigroup, 33, 101, 109, 147 irreducible set, 8 Jacobson radical, 104, 163, 280, 282 ^-class, 2 Krull dimension, 191, 272 layer, 67 bottom, 123 £-class, 2 length, 35, 42, 48, 226 Levi decomposition, 226 Levi factor, 226 linear recursion, 15 characteristic polynomial of, 16 linear semigroup, 5 associated group of, 67, 71, 82, 83, 93, 95, 133, 137, 142, 143, 151, 173, 179, 190, 192, 199, 219, 285 ideal uniform component of, 67, 77, 92, 95, 103, 267 layer of, 67 nilpotent component of, 67, 69, 80, 253, 254 structural chain of, 72, 210 uniform component of, 67, 68, 74, 77, 91, 93, 104, 105, 121, 122, 127, 129, 151, 154, 208, 210, 253, 254 locally compact field, 170, 175
303 matrix almost unipotent, 138 block monomial, 147 monomial, 147 semisimple, 95, 168 unipotent, 95, 129, 138 maximal subgroup, 1, 7, 12, 32, 58, 143, 147, 233 monoid connected, 154, 229 full linear, 19 local, 74, 234 of Lie type, 228 universal, 230, 233, 243, 246 Renner, 25 symmetric inverse, 25 Munn algebra, 5, 50, 98, 111, 164, 275, 278 basic ideal of, 116 identity of, 113 irreducible module of, 116, 120 nilpotent component, 67, 69, 80, 253, 254 nilpotent semigroup Malcev, 132, 144, 191 power, 27, 32, 144, 191 norm, 171 Ore condition, 60, 98, 275 Ore semigroup, 124, 266, 270, 272 Ore set, 270 parabolic subgroup, 226 opposite, 226 partial map, 50 permutation property, 157, 163, 166, 187, 274
INDEX
304 7r-regular closure, 8, 78, 80, 81, 105, 145, 147, 148, 180, 192, 254 7r-regular semigroup, 6, 73, 162, 163, 280 polynomial identity, 157, 163, 166, 184, 274, 285 power factorization, 193 prime radical, 184, 270 principal factor, 2, 23, 86, 233, 234 principal ideal, 73 rank of a matrix, 5, 9 of a sandwich matrix, 118 of a semigroup, 218 rank sequence, 49 7£-class, 2 rectangular band, 161, 162 reduced row elementary form, 20 Rees factor, 2 Renner monoid, 25, 34, 235 representation cuspidal, 239 induced, 236, 239 irreducible, 54, 116, 120, 121, 236, 238, 245, 248 modular, 56, 246 rigid element, 283 ring Goldie, 267, 275, 277 graded, 266, 279, 282 of quotients, 267, 268, 276 perfect, 280-282 semilocal, 280-282 root of unity, 10, 136, 170, 216 root subgroup, 43, 226
row space, 102 sandwich matrix, 2, 50, 92, 98, 113, 123, 233, 236 rank of, 118 row module of, 115 similar rows of, 123, 162 semigroup absolutely irreducible, 101 almost nilpotent, 172, 173 almost semisimple, 209, 219 almost unipotent, 138, 141, 188 bicyclic, 84 block monomial, 147 block upper triangular, 104 Brandt, 145, 150, 156 cancellative, 70, 158, 172, 185, 2 1 1 , 2 1 8 , 2 6 9 , 2 8 3 , 285 completely 0-simple, 2, 30, 62, 68, 86, 120, 162, 237, 280 completely reducible, 54, 102, 150 connected, 8, 160 divided by, 198 free, 71, 84, 86, 134, 136,173, 179, 190, 277 inverse, 114, 146, 235 irreducible, 33, 101, 109,147 linear, 5 locally finite, 185, 265 monomial, 147 nil, 27, 30, 32, 253, 283 nilpotent (Malcev), 132, 144, 191 nilpotent (power), 27, 32, 144, 191
INDEX of finite index, 172 of matrix type, 2 column of, 2 egg-box pattern of, 3 idempotents of, 3 row of, 2 sandwich matrix of, 2 Ore, 124, 266, 270, 272 periodic, 141, 185 7r-regular, 6, 73, 162,163,280 Rees factor, 2 regular, 6 repetitive, 193 simple, 107 skew linear, 252 T-nilpotent, 279 triangularizable, 126, 127 u.p., 282 unipotent, 127 0-simple, 2, 84 semigroup algebra, 4, 50, 98, 157, 163, 237, 284, 287 contracted, 4, 111, 113, 150, 277 semigroup identity, 132, 133, 136, 137, 162, 173, 192 semiidempotent, 49 similar rows, 123 skew linear semigroup, 252 solvable radical, 137, 144, 168, 174 structural chain, 72, 210 Tits alternative, 9, 133, 167 Tits system, 34, 225 torus, 168, 176 maximal, 137 trace, 109
305 translational hull, 124 triangular set, 30, 252, 254, 255, 262, 265 uniform component, 67, 68, 74, 77, 80, 91, 93, 104, 105, 121, 122, 127, 129, 151, 154, 208, 210, 253, 254 uniform subsemigroup, 59, 87, 124, 145, 188, 194, 275, 277 unipotent radical, 8, 23, 169, 192, 217, 219, 226 unipotent semigroup, 127 van der Waerden's theorem, 15, 196 weight space, 247 Weyl group, 25, 225 Zariski closure, 5, 8, 80, 81, 95, 131, 137, 158, 161 Zariski topology, 5, 174
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Main notation Chapter 1 (A)
gpM S1
S°
e Mn{K) GLn(K) \A\
n,c,j,u S/I K{A) K[S]
Ko[S] supp(a) AoB S = M(G,X,Y,P) {g,x,y) S(x) S(y) q(y)
M(K[G\,X,Y,P) K rank(a) Rn
ker(a),Im(a) 5
subsemigroup generated by A subgroup generated by A semigroup with identity adjoined semigroup with zero adjoined the zero element of a semigroup n x n full linear monoid over a field K full linear group the cardinality of A Green's relations Rees factor semigroup i^-linear span of A semigroup algebra contracted semigroup algebra the support of a £ K[S] ordinary product of (generalized) matrices semigroup of matrix type over a group G typical element of M(G,X, Y, P) row determined by x 6 X column determined by y £ Y intersection of S{x), S™ in M{G, X, Y, P) Munn algebra algebraic closure of K rank of a matrix a the space of column vectors kernel, image of a linear map Zariski closure of S
307
308 cl(S) Z,Q,R,C F ch(/\) Gc A> EndR(V) AutH(V) Mn
NOTATION 7r-regular closure of S integers, rationals, reals, complexes field of q elements characteristic of K the connected component of G j - t h exterior power endomorphisms of an ii-module V automorphisms of an R-module V matrices of rank < j in Mn(K)
Chapter 2 M{H3,X3,Y3,Q3) L'mK(vu... ,vk)
u3,u~ GLn{K)/P3 GLn(K)/pR W B U T W3 Eij
a* 8(a,b) cr(c?(a,6))
a Rees matrix presentation of Mj/Mj-i A'-span of vectors u 1 ? . . . , vk maximal parabolic subgroups of GLn(K) unipotent radicals of P j , P~ Levi factor of Pj, P~ left coset representatives of Pj right coset representatives of P~ Renner monoid for Mn(K) Weyl group for GLn(K) upper triangular matrices in GLn(K) unipotent matrices in B diagonal matrices in GLn(K) matrices of rank < j in W (i,j)-th matrix unit transpose of a distance on R rank sequence associated to a, b
Chapter 3 U Si Tj
Ua K*
completely 0-simple closure of U matrices of rank < j in S a distinguished ideal of Sj uniform components of S multiplicative group of K
NOTATION
309 the semigroup generated by 5, G 7r-regular closure of 5
Sa c\(S) Chapter 4 C(A),R(A) tr( 5 ) [D : K] R = p>(y)
M{R,X,Y,P)
p>Y>
-"-(x)' "-X'
Mrx°w(R) lR(Z),rR(Z) Row(P) Col(P) B{J) V(P),V0(P) X.r P rank(P) J{K[S\) I(S) 0(5)
column space, row space of a subset A C M n (tf) trace of 5 dimension of D over K Munn algebra over R subalgebras of R determined by xeX,yeY and by X'CX,Y'CY X x X matrices over R with finitely many nonzero rows left, right annihilators of Z in R left i?-module generated by rows of P right i2-module generated by columns of P basic ideal determined by J R-modules associated to R-module V r copies of the set X P viewed as Y.r x X.r matrix rank of the sandwich matrix P Jacobson radical of K[S] idealizer of S translational hull of S
Chapter 5 the coordinate ring of
K[xij]
Mn(K)
%m \% •> V-> ^ 1 ? • • • ? ^ m / ?
ym(x,y,uu... Y
—V
Gn £(A)
V ' 7 7 1 5
V '
[H,H]
,um)
Malcev words Malcev identity unipotent elements of G set of idempotents in A
permutation properties commutator subgroup of H
NOTATION
310 Chapter 6
n(G) Gs
IMI A(g),A'(g) PGLn(K) X
solvable radical of G semisimple elements of G norm of g characteristic polynomial attracting space and repulsing space projective linear group topological closure of X
Chapter 7 GK(5) ds(m) Z(H) L(S,X,k) \w\ fA{m) G(L/E)
MS)
Gelfand - Kirillov dimension of S growth function centre of H repetitivity constant length of a word w growth function with respect to a sub set A of S Galois group of L over E rank of a semigroup S
Chapter 8 B,N W
s
1(a) B~ Pi,PT Ui,UT Li Ru(P)
Y
Y ■
P(e),P~(e)
BN-paiv for a group G Weyl subgroup system of simple reflections length of a parabolic subgroup of W opposite Borel subgroup standard opposite parabolics unipotent radicals of P j , Pf Levi factor unipotent radical of P root subgroups index sets for root subgroups a map G —> Lj U {9} parabolics associated to an idempotent
NOTATION U(e),U- "(e), L(e) CG(A),. NG(A) Ma M{G),M ej
J? Mi M(J) U{M) A(M) GL{V)
rt
311 distinguished subgroups of P(e), P~(e) centralizer, normalizer of A in G finite reductive monoid universal monoid of Lie type on G standard idempotent of M. principal factor of M. GUJ,° three ^7-class monoid of Lie type set of ^-classes of M cross section of idempotents automorphisms of the space V induced representation
Chapter 9 Mn(D) < , • ( £ ) , N(D) (BsesRs H(R) supp(r) Z{X} J-gr
deg(fe) A*G Qd(R) Ha R(a) Sa Ga R(a) J(R) Jgr(R) clKdim(i?)
full skew linear monoid over D numerical invariants for D 5-graded ring semigroup of homogeneous elements of a graded ring R support of r in R additive subgroup generated by X graded part of / degree of a homogeneous element h crossed product classical quotient ring of R a subsemigroup of H(i?) related to a additive group generated by Ha image of Ha under the degree map group of quotients of Sa localization of i2(a) with respect to Ha Jacobson radical of R graded part of J{R) classical Krull dimension of R
