Cyclotomic q{Schur algebras 1 Introduction - CiteSeerX

Cyclotomic q{Schur algebras Richard Dipper Gordon James Mathematische Institut B Department of Mathematics Universitat Stuttgart Imperial College Postfach 80 11 40 Queen's Gate 70550 Stuttgart London SW7 2BZ Deutschland England Andrew Mathas Fakultat fur Mathematik Universitat Bielefeld P.O. Box 100 131 33501 Bielefeld Deutschland 15th August, 1997

1 Introduction

Recently, we [9] and, independently, Du and Scott [11], de ned an analogue of the q{Schur algebra [6, 7] for an Iwahori{Hecke algebra of type B. In this paper we study an analogue of the q{Schur algebra for an arbitrary Ariki{Koike algebra. The Ariki{Koike algebra H is a cyclotomic algebra of type G(r; 1; n) [2], and it becomes the Iwahori{Hecke algebra of type A or B when r = 1 or 2 respectively. By working over a ring R which contains a primitive rth root of unity, and by specializing the parameters appropriately, the Ariki{Koike algebra turns into the group algebra R(Cr o Sn) of the wreath product of the cyclic group Cr of order r with the symmetric group Sn of degree n. For each multicomposition of n, we construct a right ideal M of H? (see De nition 3.8). The cyclotomic q{Schur algebra is then de ned to be S (r ) = EndH M : (Under the A.M.S. subject classi cation (1991): 16G99, 20C20, 20G05

This paper is a contribution to the DFG project on \Algorithmic number theory and algebra". The authors acknowledge support from DFG and the Isaac Newton Institute; the third author was also supported in part by SERC grant GR/J37690 and by the Sonderforschungsbereich 343 in Bielefeld.

1

2

Richard Dipper, Gordon James and Andrew Mathas

specialization above where H = R(Cr o Sn), the module M becomes a module induced from a subgroup of the form (Cr Cr ) o S.) In this paper we construct a cellular basis for the cyclotomic q{Schur algebra. As a consequence we obtain a Weyl module W for each multipartition of n. We show that W has simple head F and that the set fF g, as ranges over the multipartitions of n, is a complete set of non{isomorphic irreducible S (r ){modules. Using the cellular structure of S (r ), it is now easy to see that the cyclotomic q {Schur algebra is quasi{hereditary. In order to prove these results about the cyclotomic q{Schur algebra, we need to examine the ideals M in some detail. Using the cellular structure of the Ariki{Koike algebra H (cf. [10, 13]), we obtain a basis of M and a special series of submodules of M , known as a Specht series. From the Specht series of M we construct the cellular basis of S (r ).

2 The Ariki{Koike algebra

Throughout this paper, r and n will be xed positive integers with r 1 and n 1. Let R be a commutative ring with 1 and let q; Q1 ; : : : ; Qr be elements of R with q invertible. The Ariki{Koike algebra H is the associative unital R{algebra with generators T0 ; T1; : : : ; Tn?1 subject to the following relations (T0 ? Q1 ) (T0 ? Qr ) T0 T1T0 T1 (Ti + 1)(Ti ? q) Ti+1 TiTi+1 Ti Tj

= = = = =

0 T1 T0 T1T0 0 Ti Ti+1Ti Tj Ti

for 1 i n ? 1 for 1 i n ? 2 for 0 i < j ? 1 n ? 2.

Suppose that 2 R is a primitive rth root of 1 and that q = 1 and Qk = k for k = 1; 2; : : : ; r. Then it follows from the de nition that H is isomorphic to R(Cr o Sn). For each nite set X , let S(X ) be the group of permutations of X ; thus, the symmetric group Sn = S(f1; 2; : : : ; ng). Let si = (i; i + 1) for 1 i < n; then s1 ; s2; : : : ; sn?1 are the standard Coxeter generators of Sn. If w 2 Sn then a word w = si : : : sik for w is a reduced expression for w if k is minimal; in this case we say that w has length k and write `(w) = k. Given a reduced expression si : : : sik for w 2 Sn, let Tw = Ti : : : Tik ; the relations in H ensure that Tw is independent of the choice of reduced expression. We denote by H (Sn) the subalgebra of H generated by T1 ; T2; : : : ; Tn?1. Then H (Sn) has basis f Tw j w 2 Sn g, and it is isomorphic to an Iwahori{Hecke algebra of type A. De ne elements Lm = q1?m Tm?1 : : : T1 T0T1 : : : Tm?1 for m = 1; 2; : : : ; n; these are analogues of the q{Murphy operators of the Iwahori{Hecke algebras of type A [5, 15]. Easy calculations using the relations in H (cf. [5, (2.1), (2.2)]) show that we have the following results. 1

1

(2.1) Suppose that 1 i n ? 1 and 1 m n. Then 1. Li and Lm commute.

1

Cyclotomic q {Schur algebras

3

2. Ti and Lm commute if i 6= m ? 1; m. 3. Ti commutes with Li Li+1 and Li + Li+1. 4. If a 2 R and i 6= m then Ti commutes with (L1 ? a)(L2 ? a) : : : (Lm ? a). Using the elements Tw and Lm de ned above, Ariki and Koike gave a basis for H as follows.

(2.2) Theorem (Ariki{Koike [1, (3.10)]) The algebra H is a free R{module with basis f Lc1 Lc2 : : : Lcnn Tw j w 2 Sn and 0 cm r ? 1 for m = 1; 2; : : : ; n g: 1

2

In particular, H is free of rank rnn!

(2.3) Let be the R{linear antiautomorphism of H determined by Ti = Ti for all i with 0 i n ? 1. Then Tw = Tw? and Lm = Lm for all w 2 Sn and m = 1; 2; : : : ; n. 1

We therefore have the following result.

(2.4) f Tw Lc1 Lc2 : : : Lcnn j w 2 Sn and 0 cm r ? 1 for m = 1; 2; : : : ; n g is a basis 1

2

of H .

3 A cellular basis of H

In their paper [12], which introduced the concept of cellular algebras, Graham and Lehrer gave a cellular basis of H , using the Kazhdan{Lusztig basis of H (Sn). We require a dierent cellular basis, namely one similar to the basis of H (Sn) introduced by Murphy [16]. Although the construction of the cellular basis of H in this section is similar to that in [10, 13], we are obliged to keep track of new information concerned with the cellular basis (see Corollary 3.24 below). Consider the R{submodule of H which is spanned by

f Lc1 Lc2 : : : Lcnn j 0 ci r ? 1 for 1 i n g: 1

2

When q = 1, this is a subalgebra of H , but one of the main diculties of working with the Ariki{Koike algebra is that it is not a subalgebra in general. (To see this, consider L22 when r = 2.) We shall need certain elements u+a of this R{submodule of H , and we introduce these now.

(3.1) De nition Suppose that a = (a1 ; : : : ; ar ) is an r{tuple of integers ai such that 0 ai n for all i. Let u+a = ua;1ua;2 : : : ua;r where ua;k =

ak Y m=1

(Lm ? Qk ) for 1 k r:

4


(3.2) Remarks ua;k; so

u+ a

1. Suppose that every ak is non{zero. Then (L1 ? Qk ) is a factor of each has a factor r Y

k=1

(L1 ? Qk ) =

r Y

(T0 ? Qk ) = 0:

k=1

Therefore, u+a is zero in this case. 2. Rearranging the order of a1 ; a2; : : : ; ar amounts just to reordering the parameters Q1 ; Q2; : : : ; Qr . For example, if we de ne u?a = u+(ar ;ar? ;:::;a ) then u?a is obtained from u+a by replacing Qk by Qr?k+1 for 1 k r. 3. In practice, we shall use u+a only for r{tuples a = (a1 ; a2; : : : ; ar ) such that 0 = a1 a2 : : : ar n. Our last two remarks show that there is no loss in doing this, and that we could equally well work with the elements u?a de ned in Remark (ii). (3.3) Example Suppose that r = 4, n 5 and a = (0; 2; 4; 5). Then u+a = (L1 ? Q2 )(L2 ? Q2) (L1 ? Q3 )(L2 ? Q3)(L3 ? Q3)(L4 ? Q3 ) (L1 ? Q4 )(L2 ? Q4)(L3 ? Q4)(L4 ? Q4 )(L5 ? Q4 ): Our rst lemma relates u+a and u+b when b is obtained from a by increasing a single part by one. (3.4) Lemma Let a = (a1; a2 ; : : : ; ar ) and assume that 1 k r and ak + 1 n. Let b = (a1 ; : : : ; ak?1 ; ak + 1; ak+1; : : : ; ar ). Then, for some h1 ; h2 2 H (Sn), we have u+a Tak Tak ?1 : : : T1 T0 = u+a h1 + u+b h2: Proof: The de nition of u+b gives u+b = u+a (Lak +1 ? Qk ). Hence, u+a Tak Tak ?1 : : : T1T0 T1 : : : Tak? Tak = qak u+a Lak +1 = qak Qk u+a + qak u+b : The desired result follows by postmultiplying by Ta?k1 : : : T1?1. 1

1

1

We next turn our attention to the subalgebra H (Sn) of H . Here we shall need the notation and combinatorics of multipartitions. A composition = (1; 2; : : :) is a nite sequence of non{negative integers; we denote by jj the sum of this sequence. A P multicomposition of n is an ordered r{tuple = ((1) ; : : : ; (r) ) ( k ) of compositions such that rk=1 j(k)j = n. We call (k) the kth component of . A partition is a composition whose parts are non{increasing; a multicomposition is a multipartition if all its components are partitions. For each composition = (1 ; 2; : : :) with jj = m we have a Young subgroup S = S S of Sm. Similarly, to each multicomposition = ((1) ; : : : ; (r)) of n we associate the Young subgroup S = S S S r of Sn. We are now in a position to de ne certain key elements m of H . The element m depends upon the Young subgroup S and also involves one of the elements u+a de ned above. 1

2

(1)

(2)

( )

5


(3.5) De nition of n and de ne a = (a1 ; a2 ; : : : ; ar ) P is a multicomposition + Pk?1 (i) Suppose that by ak = i=1 j j. Let x = w2S Tw and set m = ua x. ? (3.6) Example Suppose that = (2; 1); (12); (2) . Then S = S2 S1 S1 S1 S2, a = (0; 3; 5), and m = (L1 ? Q2 )(L2 ? Q2 )(L3 ? Q2 ) (L1 ? Q3 )(L2 ? Q3 )(L3 ? Q3 )(L4 ? Q3 )(L5 ? Q3) (1 + T1 )(1 + T6): (3.7) Remark If = (j(1) j; j(2)j; : : : ; j(r) j) then all of the elements in H (S) commute with u+a by (2.1)(iv). In particular, m = u+a x = x u+a . Hence, m = m, where is the antiautomorphism of (2.3).

The H {modules which will be our main concern in this paper are the right ideals generated by the m , as varies over the multicompositions of n. The cyclotomic q{Schur algebra will be built from endomorphisms between such right ideals.

(3.8) De nition Suppose that is a multicomposition of n. Let M = m H . We leave the proof of the following remarks to the reader.

(3.9) Remarks

1. If the multicomposition is obtained from by reordering the parts in each component then M = M . 2. Suppose that q = 1 and that Qk = k , for k = 1; 2; : : : ; r, where is a primitive rth root of unity in R. Then H = R(Cr o Sn). Let? k = j(k)j for 1 k r. Then M is induced from a module U for the subgroup Cr Cr Crr o S; where Cri = Cr : : : Cr (i factors). The restriction of U to the subgroup Cr Crr has the form U1 Ur r where Uk has rank k for 1 k r. The restriction of U to S is the trivial module. 1

2

1

1

We shall construct a basis of M , and study H {homomorphisms between the various modules M . To this end, we introduce {tableaux. The diagram of a composition = (1 ; 2; : : :) is f (i; j ) j 1 i and 1 j i g, which we regard as an array of nodes, or boxes, in the plane. The diagram of a multicomposition is the ordered r{tuple of the diagrams of its components. Let be a multicomposition of n. A {tableau t = (t(1) ; : : : ; t(r) ) is obtained from the diagram of by replacing each node by one of the integers 1; 2; : : : ; n, allowing no repeats. We call the tableaux t(k) the components of t.

(3.10) De nition

1. A {tableau is row standard if the entries in each row of each component increase from left to right. 2. A {tableau t is standard if is a multipartition of n, t is row standard and the entries in each column of each component of t increase from top to bottom.

6


3. If is a multipartition of n, then let Std() be the set of standard {tableaux. Note, particularly, that we de ne standard {tableaux only when is a multipartition whereas row standard tableaux are de ned for all multicompositions. We require partial orders on the set of multicompositions and on the set of row standard tableaux. If t is a row standard {tableau and 1 m n, then the entries 1; 2; : : : ; m in t occupy the diagram of a multicomposition; let t # m denote this multicomposition. For example, t # n = . We use this notation in our next de nition.

(3.11) De nition Suppose that = ((1) ; : : : ; (r) ) and = ((1) ; : : : ; (r) ) are multicompositions of n. 1. We say that dominates , and write D , if k?1 X i=1

j(i) j +

j X (k)

k?1 X

i=1

i=1

i

j(i) j +

j X (k) i=1

i

for all k and j with 1 k r and j 0. If D and 6= then we write B . 2. Suppose that s is a row standard {tableau and that t is a row standard {tableau. We say that s dominates t, and write s D t if s # m D t # m for all m with 1 m n. If s D t and s 6= t then we write s B t. ?

if n = ?r = 2, then of 2 are ordered by (2); (0) B ? 2For example, ? ? the multipartitions (1 ); (0) B (1); (1) B (0); (2) B (0); (12) . Note that if s is a row standard {tableau and t is a row standard tableau such that s D t then D . Our next de nition gives another relation between tableaux.

(3.12) De nition Suppose that s is a tableau and that 1 j n. We write comps(j ) = k

if j appears in the kth component s(k) of s. ? Suppose that t is another tableau. Then we write comps = compt if comps j ) = compt(j ) for all j with 1 j n. We also write comps compt if comps(j ) compt (j ) for all 1 j n; and comps > compt if comps compt and comps 6= compt.

(3.13) Example Let

s=

1 4 ; 5 6

3 8 9 ; 2 7

t=

2 3 ; 4 7

1 6 8 5 9

and

u=

1 2 3 ; 4 6 7

5 9 : 8

Then s is row standard, and t and u are standard. We have u B s and u B t but s and t are incomparable in the dominance order. Also, compt > compu, but there are no other equations or inequalities between comps, compt and compu.


7

Let be a multicomposition of n. The symmetric group Sn acts from the right on the set of {tableaux by permuting the entries in each tableau. Let t be the {tableau where 1; 2; : : : ; n appear in order along the rows of the rst component, and then along the rows of the second component,? and so on. The row stabilizer of t is the Young subgroup S of Sn. For example, if = (3; 2); (12); (3) then

t =

1 2 3 ; 4 5

6 ; 7

8 9 10 :

For a row standard {tableau s, let d(s) be the element of Sn such that s = td(s). Then d(s) is a distinguished right coset representative of S in Sn and we obtain, in this way, a correspondence between the set of row standard {tableaux and the set of right coset representatives of S in Sn. Recall the antiautomorphism of H from (2.3).

(3.14) De nition Suppose that is a multicomposition of n and that s and t are row

standard {tableaux. Let mst = Td(s) m Td(t) .

Note that m = mtt . Also, mst = mts (cf. Remark 3.7). One of the aims of this section is to show that the elements mst, as (s; t) varies over the ordered pairs of standard tableaux of the same shape, give a cellular basis of H . We shall also establish useful properties of the right ideals M . Initially, though, we concentrate upon the two{sided ideal generated by m .

(3.15) Lemma Suppose that is a multicomposition of n and that s and t are row standard {tableaux. Let h 2 H (Sn). Then msth is a linear combination of terms of the form msv where each v is a row standard {tableau.

2 Sn. Then there exist y 2 S and a distinguished right coset representative d of S in Sn such that w = yd and `(w) = `(y) + `(d). Hence, xTw = x Ty Td = q`(y) x Td . If m = u+a x then

Proof: Suppose that w

mst Tw = Td(s) u+a x Tw = q`(y) Td(s) u+a x Td = q`(y) msv where the tableau v = td is row standard. Now, msth = mst Td(t) h, and this is a linear combination of terms of the form mst Tw with w 2 Sn. Therefore, the required result follows. In order to apply a result of Murphy in Proposition 3.18 below, we require a combinatorial lemma which concerns the dominance order on row standard tableaux. A preliminary de nition sets the scene.

(3.16) De nition We say that a tableau s is of the initial kind (for ) if comps = compt .

8


Note that a {tableau s can be of the initial kind for even though 6= . (3.17) Lemma Suppose that = ((1) ; : : : ; (r)) is a multicomposition of n, and let = (j(1) j; : : : ; j(r)j). Suppose that w is a distinguished right coset representative of S in Sn and that s is a row standard {tableau of the initial kind. Then the following hold. 1. The tableau sw is row standard. 2. If s is standard then sw is standard. 3. If t is a row standard {tableau of the initial kind with s B t then sw B tw. Proof: The fact that w is a distinguished right coset representative of S in Sn implies that whenever x and y are integers such that x and y are in the same component of s and x < y, then xw < yw. Hence, (i) and (ii) are true. Now assume that t is a row standard {tableau of the initial kind, and that s B t. Then d(s) B d(t) in the Bruhat{Chevalley order B on Sn, by [4, (1.3)]. (Our notation for the order is such that 1 D v for all v 2 Sn.) Since w is a distinguished right coset representative of S in Sn, we also have `(d(t)w) = `(d(t)) + `(w). The well{known \cancellation" property of the Bruhat{Chevalley order now lets us conclude that d(s)w B d(t)w. But sw and tw are row standard, so d(s)w = d(sw) and d(t)w = d(tw). By applying [4, (1.3)] again, we deduce that sw B tw.

(3.18) Proposition Suppose that is a multicomposition of n and that s and t are row

standard {tableaux. Then mst is a linear combination of terms of the form muv where u and v are standard {tableaux for some multipartition of n, and 1. u D s and compu = comps; and, 2. v D t and compv = compt.

Proof: When r = 1, this is a Theorem of Murphy [16, (4.18)]. We deduce the general

case from this. Let = (j(1) j; : : : ; j(r)j). We may write s = s0w1 and t = t0 w2 where s0 and t0 are row standard {tableau of the initial kind, and w1 and w2 are distinguished right coset representatives for S in Sn. P ?1 (i) j j, as in the de nition of m . We have mst = Let a = (a1; : : : ; ar ) where ak = ki=1 Tw ms0t0 Tw and ms0t0 = Td(s0 )u+a x Td(t0 ) = u+a Td(s0 ) xTd(t0 ) since d(s0) 2 S. We may write Td(s0) x Td(t0 ) as a product of r commuting terms, one for each component of ; say, Td(s0 ) xTd(t0 ) = x1 x2 : : : xr , where the kth term xk involves only elements Tw with w 2 S(fak + 1; ak + 2 : : : ; ak+1g). For example, x1 = Td(s0 ) x Td(t0 ) where s01 is the rst component of s0 and t01 is the rst component of t0. By applying Murphy's result [16, (4.18)] to the Hecke algebra H (S ) we may write x1 as a linear combination of terms Td(u0 ) x Td(v0 ) where u01 and v01 are standard (1) {tableaux for some partition (1) of a1 , and u01 D s01 and v01 D t01 . We can apply the same technique for the other factors x2; : : : ; xr , to conclude that Td(s0 )x Td(t0 ) is a linear combination of terms of the form Td(u0 ) x Td(v0 ) where u0 and v0 are 1

2

1

1

(1)

1

1

(1)

1


9

standard {tableaux for some multipartition of n, and u0 D s0 and v0 D t0. Also, u0 and v0 are of the initial kind, and j(k)j = j(k)j for 1 k r. Therefore, ms0t0 is a linear combination of terms mu0 v0 where the sum runs over the same set of pairs (u0; v0). Consequently, mst is a linear combination of terms Tw mu0v0 Tw . Now, Td(v0 ) Tw = Td(v0 )w = Td(v0 w ), so Tw mu0 v0 Tw = Tw mu0v where v = v0w2. By Lemma 3.17, since v0 is of the initial kind, v = v0w2 is standard, and v0 w2 D t0w2; that is, v D t. Moreover, compv0 w = compt0w since v0 and t0 are of the initial kind. Similar remarks applied to Tw Td(u0 ) now complete the proof of the Proposition. 2

1

2

2

2

2

1

2

1

2

1

(3.19) Corollary Suppose that is a multicomposition of n and that s and t are row standard {tableaux. If h 2 H (Sn) then mst h is a linear combination of terms of the form muv where u and v are standard {tableaux for some multipartition of n, and D , u D s and

compu = comps. Proof: Combine Lemma 3.15 and Proposition 3.18, and recall that u D s implies that D .

Corollary 3.19 provides the kind of information we need when we multiply mst by elements of H (Sn). More complicated is our next proposition, which shows what happens when we multiply mst by T0. It is vital to the proposition that is a multipartition, not merely a multicomposition. (3.20) Proposition Suppose that is a multipartition of n, and that s and t are standard {tableaux. Then mstT0 = x1 + x2 where 1. x1 is a linear combination of terms of the form muv where u and v are standard { tableaux, with u D s and compu = comps, and 2. x2 is a linear combination of terms of the form muv where u and v are standard { tableaux for some multipartition of n, with B and comps > compu . P ?1 (i) Proof: Let = (j(1) j; : : : ; j(r) j) and let a = (a1 ; a2 ; : : : ; ar ) where ak = ki=1 j j for 1 k r. Write d(t) = yc with y 2 S and c a distinguished right coset representative for S in Sn. Then the {tableau tc is row standard. Assume that 1 is in row k of t c; thus, 1 k r. Now, c = (ak ; ak + 1)(ak ? 1; ak ) : : : (1; 2)c0, where `(c) = ak + `(c0) and c0 xes 1. Thus, Tc = Tak Tak? : : : T1 Tc0 and Tc0 T0 = T0 Tc0 . Let b = (a1; : : : ; ak + 1; : : : ; ar ). Then u+a TcT0 = u+a Tak Tak? : : : T1T0 Tc0 = u+a h1 + u+b h2 for some h1 ; h2 2 Sn by Lemma 3.4. Since y 2 S and t y is standard, y xes ak + 1; therefore, Ty commutes with u+b . Hence, using Remark 3.7, we have mstT0 = Td(s) u+a x Td(t) T0 = Td(s) x u+a Ty TcT0 = Td(s) x Ty u+a TcT0 = Td(s) x Ty (u+a h1 + u+b h2) = Td(s) x (u+a Ty h1 + u+b Ty h2 ): 1

1

10


The rst term, Td(s) x u+a Ty h1, is a linear combination of terms of the required form by Corollary 3.19. If k = 1 then u+b = 0 by Remark 3.2(i), and the proof is complete in this case. Assume therefore that k 2. We turn out attention to the term Td(s) x u+b Ty h2 . We now need to digress in order to prove that Td(s) x u+b Ty h2 has the required form. De ne the multicomposition = ( (1) ; : : : ; (r) ) of n by (i) = (i) for i 6= k ? 1; k, (k?1) = ((1k?1) ; (2k?1); : : : ; 1) and (k) = (1(k) ? 1; (2k); : : :). (Thus we reduce the rst part of the kth component of by 1, and adjoin a new part of size 1 to the end of the (k ? 1)th component of .) Note that B . Let l = (1k) be the rst part of (k). The entries in the rst row of the kth component of t are then ak + 1; ak + 2; : : : ; ak + l. Let G1 be the symmetric group on these numbers, and let G2 be the symmetric group on ak + 2; : : : ; ak + l. Let c1; c2; : : : ; cl be the distinguished right coset representatives for G2 in G1 , ordered in terms of increasing length. Then the tableaux t ci are row standard, and they agree with t except on the last row of the (k ? 1)th component and the rst row of the kth component. The last row of the (k ? 1)th component of t ci contains the single entry ak + i. We are given that s = td(s) is a row standard {tableau. Let ui = t ci d(s) for 1 i l. Each ui agrees with s except on the last row of the (k ? 1)th component of ui and on the rst row of the kth component of ui. Furthermore, since (ak + 1)d(s) < < (ak + l)d(s), it follows that each ui is row standard. We also have compui (j ) = comps(j ) if j 2= f(ak + 1)d(s); : : : ; (ak + l)d(s)g and, for j 2 f(ak + 1)d(s); : : : ; (ak + l)d(s)g, we have compui (j ) = k ? 1 or k and comps(j ) = k. Therefore, comps > compui . Now x has a factor X

Hence, x =

w2G1

Pl

i=1 Tci x .

T

d(s) x

l X i=1

Tci

X

w2G2

Tw :

Note that m = x u+b . Thus,

u+ b

Tw =

=

l l l X X X + + Td(s) Tci x ub = Tcid(s) x ub = muit : i=1

i=1

i=1

Hence, by Corollary 3.19, Td(s) x u+b Ty h2 is a linear combination of the required form. This completes the proof of the Proposition.

(3.21) De nition Suppose that is a multicomposition of n. 1. Let N be the R{module spanned by n

mst

s and t are standard {tableaux for some o: multipartition of n with D

11


2. Let N be the R{module spanned by n t are standard {tableaux for some o: mst s andmultipartition of n with B We now apply Propositions 3.18 and 3.20 to obtain a sequence of useful results.

(3.22) Proposition Suppose that is a multicomposition of n. Then N and N are two{

sided ideals of H .

Proof: Corollary 3.19 shows that N is closed under postmultiplication by T1 ; T2 ; : : : ; Tn?1

and Proposition 3.20 shows that N is also closed under postmultiplication by T0. Because H is generated by T0 ; T1 ; : : : ; Tn?1 , we deduce that N is a right ideal of H . Since mst = mts , by applying the antiautomorphism we see that N isPalso a left ideal of H . Finally, N is a two sided ideal of H since N = B N . A proof very similar to that of Proposition 3.22 gives our next result.

(3.23) Proposition Suppose that is a multicomposition of n and that s is a row standard {tableau. Let Is be the R{module spanned by n standard {tableaux for some multipartition muv u and v are of n with D and comps compu Then Is is a right ideal of H .

o

:

Recall that M = m H .

(3.24) Corollary Suppose that is a multicomposition of n. Then every element of M

is a linear combination of terms of the form muv where u and v are standard {tableaux for some multipartition of n with D and compt compu. Proof: Since m = mt t , Proposition 3.18 shows that m 2 It (note that may not be

a multipartition). Therefore, M It and the desired result follows by Proposition 3.23.

(3.25) Proposition Suppose that is a multipartition of n and that t is a standard { tableau. Let h 2 H . Then for every standard {tableau v there exists rv 2 R such that, for all standard {tableau s, we have

msth

X

v2Std()

rv msv mod N :

12


Proof: Let U be the R{module spanned by f mtt0 j t0 is a standard {tableau g. Note

that t is the unique {tableau u such that u D t . Therefore, U + N is closed under postmultiplication by T1 ; T2; : : : ; Tn?1 by Proposition 3.18, and it is closed under postmultiplication by T0 by Proposition 3.20. Hence U + N is a right ideal of H . Thus, for each v 2 Std() there exists rv 2 R such that

mtt h

X

v2Std()

rv mtv mod N :

Multiply this congruence on the left by Td(s) , and use the fact that N is an ideal, to obtain the congruence of the proposition.

(3.26) Theorem ( [13, (1.7)]) The algebra H is a free R{module with basis n

M = mst

s and t are standard {tableaux for some o: multipartition of n

Moreover, M is a cellular basis of H . ?

Proof: Since m = 1 when = (0); : : : ; (0); (1n) , Proposition 3.25 shows that M spans

H . By Theorem 2.2, H is free of rank rn n! Since this is also the cardinality of M , we deduce that M is a basis of H . Finally, the properties that M must satisfy in order to be a cellular basis of H (as given in [12, (1.1)]), are covered by Proposition 3.25 and the fact that mst = mts.

(3.27) De nition We call M the standard basis of H . We can now apply Graham and Lehrer's theory of cellular algebras [12] to describe the representation theory of H .

(3.28) De nition Suppose that is a multipartition of n. Let z = (N + m)=N . The

Specht module S is the submodule of H =N given by S = z H .

Directly from Proposition 3.25 and Theorem 3.26, S is a free R{module with basis f zTd(t) j t a standard {tableau g. De ne a bilinear form h ; i on the Specht module S by

m Td(s) Td(t) m hzTd(s) ; z Td(t) im mod N : By Proposition 3.25 (and the version of Proposition 3.25 obtained by applying the antiautomorphism ), the bilinear form is well de ned (cf. [12]). Moreover, the bilinear form is symmetric and huh; vi = hu; vhi for all u; v 2 S and all h 2 H [12, (2.4)]. Consequently, rad S , the radical of the bilinear form, is an H {module.


13

(3.29) De nition Suppose that is a multipartition of n. Let D = S = rad S . (3.30) Theorem Suppose that R is a eld. Then the non{zero H modules in f D j a multipartition of n g form a complete set of non{isomorphic irreducible H {modules. Moreover, each irreducible module D is absolutely irreducible. Proof: Since M is a cellular basis of H the Theorem follows from the general theory of cellular algebras [12, (3.4)].

The theory in [12, (3.6)] also shows that if D 6= 0 and D is a composition factor of S then D . When r = 1 the partitions for which D 6= 0 have been classi ed in [4, (7.6)]. When r > 1, if q = 1, or if the parameters Qk are powers of q, the multipartitions for which D 6= 0 are classi ed by the results of [12, (5.9)], [13, (1.17)] and [14, (3.7)]; in type B, see also [8, (5.3)] and [10, (8.19)].

4 A Specht series for M In the last section, we used various elements m of H to construct a cellular basis of H . We saw that the right ideal M generated by m is an analogue of an induced module; indeed, if r = 1 and q = 1 then M is the permutation module of Sn on the Young subgroup S. We shall use the right ideals M of H to construct the cyclotomic q{Schur algebra, but, before that, we study the individual modules M in more detail. In particular, we shall give a semistandard basis of M and construct a Specht series for M . We are going to de ne a new kind of {tableau. This will have entries which are ordered pairs (i; k) where i is a positive integer and 1 k r. We denote such tableaux by capital letters.

(4.1) De nition Suppose that is a multipartition of n and that is a multicomposition of n. A {tableau S has type if its entries are ordered pairs (i; k), as above, and for all m and k the number of times (i; k) is an entry in S is (ik) .

Next, we introduce a function which converts standard {tableaux (or, indeed any tableau whose entries are 1; 2; : : : ; n) into a {tableau of type .

(4.2) De nition Suppose that is a multipartition of n and that is a multicomposition of n. Let s be a standard {tableau. De ne (s) to be the {tableau obtained from s by replacing each entry m in s by (i; k) if m is in row i of the kth component of t .

In our examples, we shall always write ik for the ordered pair (i; k).

14

Richard Dipper, Gordon James and Andrew Mathas ?

(4.3) Example Suppose that = (3; 1); (2); (2; 12) . Then

t =

1 2 3 ; 4

5 6 ;

7 8 9 10

Suppose that

s1 = and Then and

1 2 3 5 ; 4

6 10 ;

s3 =

7 8 9

(s3) =

;

1 2 5 9 ; 7

(s1) = (s2) =

and (t

) =

s2 =

3 10 ;

11 11 11 ; 12 12 ; 13 13 : 21 23 33

1 2 3 6 ; 4 4 6 8

5 10 ;

7 8 9

:

11 11 11 12 ; 12 33 ; 13 13 21 23

11 11 12 23 ; 11 33 ; 21 12 : 13 13

Given two pairs (i1; k1) and (i2 ; k2), we write (i1; k1) < (i2 ; k2) if k1 < k2, or k1 = k2 and i1 < i2. Note that if s is a standard {tableau, then (s) is a {tableau of type whose entries are weakly increasing along rows and down columns. We next single out some of these tableaux (s).

(4.4) De nition Suppose that is a multipartition of n and that is a multicomposition of n. Let S = (S(1) ; : : : ; S(r) ) be a {tableau of type . Then S is semistandard if 1. the entries in each row of each component S(k) of S are non{decreasing; and 2. the entries in each column of each component S(k) of S are strictly increasing; and 3. for each k with 1 k r, no entry in S(k) has the form (i; l) with l < k. Let T0 (; ) denote the set of semistandard {tableaux of type .

For example, (s1) in Example 4.3 is semistandard, but (s3) is not, since the third part of De nition 4.4 is not satis ed. The point of the third condition in De nition 4.4 is the following result (cf. Corollary 3.24).

(4.5) Suppose that s is a standard {tableau. Then (s) satis es De nition 4.4(iii) if and only if compt comps If is a multipartition and S 2 T0 (; ), then it is clear that there exists a standard

{tableau s with S = (s). In order to say more about this, we introduce another de nition.

15


(4.6) De nition If s is a standard {tableau and 1 m n then let rows(m) = (i; k) if m belongs to row i of the kth component of s.

The de nition of rows allows us to recover s from the function rows Suppose that S 2 T0 (; ) and let s be a standard {tableau such that (s) = S. If there exists an integer i such that rows(i) 6= rows(i +1) and i and i +1 are in the same row of t then the tableau s1 = s(i; i + 1) is standard and (s1 ) = S. Also, s B s1 if rows(i) < rows(i + 1), and s1 B s if rows(i) > rows(i + 1). Hence we have shown the following.

(4.7) Suppose that S 2 T0 (; ). Then there exist standard {tableaux rst(S) and last(S) such that 1. ( rst(S)) = (last(S)) = S; and, 2. if s is any standard {tableau such that (s) = S then rst(S) D s D last(S). (4.8) De nition Suppose that is a multipartition of n and that is a multicomposition of n. Let S 2 T0 (; ) and t 2 Std(). Then mSt is the element of H given by mSt =

X

2 (s)=S

s Std()

mst:

Note that mSt is a sum of standard basis elements mst where rst(S) D s D last(S). ?

?

(4.9) Example Suppose that = (4; 3); (2; 1); (2; 1) , = (32; 1); (12); (2; 12) and S

=

11 11 11 21 ; 12 33 ; 13 13 : 21 21 31 22 23

Then S 2 T0(; ) and for any t 2 Std() we have mSt = ms t + ms t + ms t where 1

s1 =

1 2 3 4 ; 5 6 7

and

8 13 ; 10 11 ; 9 12

s3 = 14 25 37 6 ; Here s1 = rst(S) and s3 = last(S).

s2 =

2

1 2 3 5 ; 4 6 7

3

8 13 ; 10 11 9 12

8 13 ; 10 11 : 9 12

(4.10) Lemma Suppose that is a multipartition of n and that is a multicomposition of n. Let S 2 T0 (; ) and t 2 Std(). Then mSt 2 M . Proof: Let P k?1 j(i) j. i=1

a = (a1 ; : : : ; ar ) where ak =

Pk?1

i=1

j(i) j, and b = (b1 ; : : : ; br ) where bk =

16


Let s1 = rst(S) and let d = d(s1). Then, as in [4], X

2 (s)=S

s Std()

P

x Td(s) =

X

w2SdS

Tw = hTd x ;

where h = Tv , the sum running over certain elements v 2 S. Since h 2 H (S), we have h u+b = u+b h ; so we obtain

mSt =

X

2

s Std()

mst =

(s)=S

X

2

s Std()

Td(s) xu+b Td(t) = x Tdu+b h Td(t) :

(s)=S

Therefore, since m = x u+a , it is sucient to prove that TdP u+b 2 u+a H . ?1 j(i) j and b = Pk?1 j(i) j Assume that 1 k r. Recall that s1 = rst(S), ak = ki=1 k i=1 and note that ak bk since S is semistandard. De ne tk to be the standard {tableau such that 1; 2; : : : ; ak occupy the same positions in s1 and tk ; bk + 1; : : : ; n occupy the same positions in t and tk ; and that for ak + 1 i < j bk we have rowtk (i) rowtk (j ) | thus the numbers ak +1; : : : ; bk are inserted into tk in \row order". The fact that S is semistandard ensures that tk is well de ned. Note that t1 = t and tr = s1 . Now de ne wk 2 Sn inductively by requiring that tk = tw1w2 :::wk . Then w1 = 1 and d = w1 w2:::wr ; moreover, `(d) = `(w1) + ::: + `(wr ). By construction, the element wk xes each of the numbers 1; 2; : : : ; ak?1 and it also xes each of bk + 1; bk + 2; : : : ; Q n. For 1 k r, let ua;k and ub;k be as de ned in De nition 3.1, and let vk = bmk=ak +1(Lm ? Qk ). Then ub;k = ua;k vk . Now, Twk commutes with ua;l for l < k (since wk xes 1; 2; : : : ; ak?1), and Twk commutes with ub;l for l k (since wk xes bk + 1; bk + 2; : : : ; n); see (2.1)(iv). Therefore,

Td u+b = Twr : : : Tw ub;1ub;2 : : : ub;r = Twr : : : Tw ub;1ub;2 : : : ub;r Tw = ua;1Twr : : : Tw ub;2 : : : ub;r v1 Tw = ua;1ua;2Twr : : : Tw ub;3 : : : ub;r v2 Tw v1 Tw ... ... = ua;1 : : : ua;r vr Twr vr?1 Twr? : : : v2 Tw v1 Tw : 1

2

1

2

1

3

2

1

2

1

1

Thus, Tdu+b 2 u+a H , as required. Having shown that mSt 2 M , our next aim is to prove that the set of elements of the form mSt give a basis of M . P ?1 (i) Let a = (a1 ; : : : ; ar ) where ak = ki=1 j j as in the previous proof; then m = u+a x . + Since ua and x commute, we have M = m H u+a H \ x H . We shall show that we have equality here. We have seen that H has a standard basis which consists of elements mst with s and t standard {tableaux for some multipartition of n. Suppose that h 2 H and let h =


17

P

s;t rst mst , with rst 2 R, be the unique expression for h in terms of the standard basis. We say that mst is involved in h if rst 6= 0.

(4.11) Lemma Suppose that is a multipartition of n and that m = u+a x . Let h 2 x H \ u+a H and suppose that (s; t) is a pair of standard tableaux of the same shape such

that 1. mst is involved in h; and, 2. if (u; v) is a pair of standard tableaux of the same shape such that s D u, t D v and (s; t) 6= (u; v) then muv is not involved in h. Let S = (s). Then S is semistandard and s = last(S). Proof: Since s is standard, the entries in S are non{decreasing down rows and columns. Suppose that i and i + 1 belong to the same row of t. Then Tix = qx, so Ti h = qh. Therefore, by [16, (4.19)], i and i + 1 do not belong to the same column of s. Hence the entries in S are strictly increasing down columns. Since h 2? u+a H we have comp t comps by Corollary 3.24 (applied to the multicomposition (1 ); (1 ); : : : ; (1r ) , where = (j(1) j; : : : ; j(r)j)). Hence, by (4.5), S is semistandard. Finally, suppose that s 6= last(S). Then there exist integers i and i + 1 in the same row of t such that the tableau s0 = s(i; i + 1) is standard and s B s0 . As in [16, (4.19)], ms0t is involved in Tih = qh, contradicting part (ii) of our hypothesis. Therefore, s = last(S). 1

2

(4.12) Corollary Suppose that is a multicomposition of n and m = u+a x . Then u+a H \ x H is spanned by

n

mSt

S

2 T0 (; ) and t 2 Std() for some o: multipartition of n

Proof: Note that mSt 2 M by Lemma 4.10.

For each multipartition of n let Tl (; ) = f s 2 Std() j (s) = last(S) for some S 2 T0 (; ) g: By Lemma 4.11 every non{zero element of u+a H \ x H involves a standard basis element mst where s 2 Tl (; ) and t 2 Std() for some multipartition . + Now P suppose that h 2 ua H \ x H and write h in terms of the standard basis; say, h = ruvmuv with ruv 2 R. Let X X X h0 = h ? rstm(s)t : s2Tl (;) t2Std()

Then h0 2 u+a H \ x H , but h0 does not involve any term mst for any s 2 Tl (; ). Therefore, h0 = 0, and we have obtained an expression for h as a linear combination of the required form.

18


(4.13) Corollary Suppose that is a multicomposition of n and m = u+a x . Then M = u+a H \ x H . Proof: We have that M

u+a H \ x H . The inclusion u+a H \ x H M follows

from Corollary 4.12 and Lemma 4.10

The next theorem for Iwahori{Hecke algebras of type A (that is, the case r = 1), is due to Murphy [16]; for Iwahori{Hecke algebras of type B (that is, the case r = 2), the result is due to Du and Scott [11]

(4.14) Theorem Suppose that is a multicomposition of n. Then M is free as an R{ module with basis o n mSt S 2 T0 (; ) and t 2 Std() for : some multipartition of n

Proof: By Lemma 4.10 each element mSt belongs to M . Since distinct elements mSt

involve distinct standard basis elements mst, the elements mSt are linearly independent. Finally, Corollaries 4.12 and 4.13 show that the elements mSt span M .

(4.15) Corollary Suppose that is a multicomposition of n. Then there exists a ltration

of M ,

M = M1 > M2 > > Mk+1 = 0 such that for each i with 1 i k there exists a multipartition i of n with Mi =Mi+1 = S i . Moreover, if is a multipartition of n, then the number of i such that = i is equal to the number of semistandard {tableaux of type . > S2 > > Sk on the set of semistandard tableaux of type such that j > i if i B j where Si 2 T0(i; ) and Sj 2 T0 (j ; ). Let Mi be the R{submodule of M with basis f mSj t j j i and t 2 Std(j ) g. Then Proof: Choose an ordering

S1

M = M1 > M2 > > Mk+1 = 0; and, by Proposition 3.25 and Theorem 4.14, each Mi is a right ideal of H . Suppose that 1 i k. Then Mi \ N i Mi+1 . Hence, there is a well de ned H { homomorphism from S i onto Mi =Mi+1 such that (zi ) = mSi ti + Mi+1 . Since both S i and Mi =Mi+1 have rank equal to the number of standard i{tableaux, is an isomorphism. The Corollary now follows.


19

5 The double annihilator of m The purpose of this section is to compute the double annihilator of the element m of H for any multicomposition of n. This will enable us to calculate a basis for HomH (M ; M ) in the next section. (5.1) De nition Suppose S is a subset of H and de ne r(S ) = f h 2 H j Sh = 0 g and l(S ) = f h 2 H j hS = 0 g. The double annihilator of S is lr(S ) = l(r(S )). It is elementary that lr(S ) contains the left ideal of H generated by S . If H is a quasi{ Frobenius algebra, then lr(S ) is equal to the left ideal of H generated by S [3, (61.2)]. Although H is quasi{Frobenius for r 2 (see, for example, [4, section 2]), we do not know whether it is quasi{Frobenius for r > 2. If H is quasi{Frobenius for all r, then the main result of this section, namely Theorem 5.16, is immediate. A connection between double annihilators and homomorphisms is provided by the following easy lemma. (5.2) Lemma Suppose that m 2 H and that lr(m) = H m. Let I be a right ideal of H . 1. For all ' 2 HomH (mH ; I ) there exists h' 2 H such that '(m) = h'm. 2. HomH (mH ; I ) = H m \ I. Proof: For all x 2 r(m) we have '(m)x = '(mx) = 0. Therefore, '(m) 2 lr(m) = H m; so '(m) = h'm for some h' 2 H . The proof of part (ii) is now straightforward. The main result of this section is that lr(m ) = H m . Because the Iwahori{Hecke algebra H (Sn) is quasi{Frobenius, our rst step is easy. (5.3) Lemma Suppose that is a multicomposition of n. Then lr(x ) = H x . Proof: Since H (Sn) is a quasi{Frobenius algebra, we have lr(x ) \ H (Sn) = H (Sn )x . Now suppose that z 2 lr(x ). By Theorem 2.2, we may write X z = Lc1 ; : : : Lcnn hc 1

c2C

where C = f c = (c1; : : : ; cr ) j 0 cj < r for 1 j n g and hc 2 H (Sn). Then, for all y 2 r(x ) \ H (Sn), we have X 0 = zy = Lc1 ; : : : Lcnn hcy: 1

c2C

Therefore, by Theorem 2.2, we have hc y = 0 for all c 2 C . Thus, hc 2 lr(x ) \ H (Sn) = H (Sn )x , and so z 2 H x . For the of this section we x a multicomposition and de ne a = (a1; : : : ; ar ) Pkremainder ? 1 (i) by ak = i=1 j j; so 0 = a1 a2 : : : ar n. Our main task in this section is to prove that lr(u+a ) = H u+a . To achieve this goal, we use the next result.

20


(5.4) Lemma Assume that I is a left ideal of H and suppose that S is a subset of r(I ) such that l(S ) I . Then lr(I ) = I . Proof: By assumption S r(I ) and l(S ) I . Also, I lr(I ). Therefore, lr(I ) l(S )

I lr(I ); so lr(I ) = I as required.

Our de nition of u+a expresses u+a as a double product

u+ a

=

ak r Y Y k=1 m=1

(Lm ? Qk ):

We now introduce notation which allows us to reverse the order of the multiplication here.

(5.5) De nition Suppose that 1 i n.

1. Let i = k if k is maximal such that ak < i. Let = ( 1; 2; : : : ; n). r Y (Li ? Qk ). 2. Let vi = k= i +1

Then u+a = v1 v2 : : : vn and all of these factors commute.

i Y

(5.6) De nition Suppose that 1 i n. Let yi = Ti?1 : : : T2T1 (L1 ? Qk ). k=1

(5.7) Example Suppose that r = 4, n = 5 and a = (0; 2; 4; 5). Then u+a = (L1 ? Q2 )(L2 ? Q2) (L1 ? Q3 )(L2 ? Q3)(L3 ? Q3)(L4 ? Q3 ) (L1 ? Q4 )(L2 ? Q4)(L3 ? Q4)(L4 ? Q4 )(L5 ? Q4 ): Also = (1; 1; 2; 2; 3) and, for 1 i n, the product of the factors in the ith column of the above array is vi . We have

y1 y2 y3 y4 y5

= (L1 ? Q1 ) = T1(L1 ? Q1 ) = T2 T1(L1 ? Q1 )(L1 ? Q2 ) = T3T2 T1(L1 ? Q1 )(L1 ? Q2 ) = T4 T3T2 T1(L1 ? Q1 )(L1 ? Q2 )(L1 ? Q3 ):

(5.8) Lemma Suppose that 1 i n. Then v1 v2 : : : vi yi = 0.

Cyclotomic q {Schur algebras Proof: Assume that k satis es i + 1

21

k r. Then v1 v2 : : : vi has a factor (L1 ?

Qk )(L2 ? Qk ) : : : (Li ? Qk ). By (2.1)(iv), Q this factor commutes with Ti?1 : : : T2 T1 . Hence, v1 v2 : : : viTi?1 : : : T2 T1 has a right factor rk= i +1(L1 ? Qk ). However, r Y k=1

(L1 ? Qk ) =

r Y

(T0 ? Qk ) = 0;

k=1

so it follows that v1 v2 : : : viyi = 0 as claimed. Since u+a = v1v2 : : : vn we have the following Corollary.

(5.9) Corollary Suppose that 1 i n. Then u+a yi = 0. It turns out that y1; y2; : : : ; yn generate r(u+a ). More importantly, together with Lemma 5.4 the elements yi will allow us to prove that lr(u+a ) = H u+a . For our proof, we require the technical Lemma 5.13 below; for this result we need some preparation.

(5.10) Lemma Suppose that 1 i < n and that a; b 2 f0; 1; : : :; r ? 1g. b?a X +c . 1. If a b then Lai Lbi+1 Ti = TiLbi Lai+1 + (q ? 1) Lbi ?cLia+1 c=1 aX ?b?1

2. If a > b then Lai Lbi+1Ti = qTi?1Lbi Lai+1 ? (q ? 1)

c=1

?c . Lbi +cLia+1

Proof: (i) Assume that a b. If b = a then Ti commutes with Lai Lai+1 by (2.1)(iii), so

the result is correct in this case. Now suppose that b > a. Since Li+1 Ti = Ti Li + (q ? 1)Li+1 we have ?1 ?T L + (q ? 1)L ; Lai Lbi+1Ti = Lai Lbi+1 i i i+1 which gives the required result by induction on b ? a. (ii) Either argue similarly, or apply the antiautomorphism to the result of part (i) and rearrange, interchanging a and b. We next generalize Lemma 5.10 as follows.

(5.11) Lemma Suppose that 1 i n and bj 2 f0; 1; : : : ; r ? 1g for j = 1; 2; : : : ; n. Then there exists an integer b and 1 ; : : : ; i?1 2 f1g such that Lb1 Lb2 : : : Lbi i Ti?1 : : : T2T1 = x1 + x2 ; 1

2

where x1 = q b Ti?i?1 : : : T2 T1 Lb1i Lb2 Lb3 : : : Libi? and x2 is a linear combination of terms of the form Tw Lc1 Lc2 : : : Lci i where 1. w 2 Sn; and 1

1

2

2

1

1

2

1

22


2. c1; : : : ; ci 2 f0; 1; : : : ; r ? 1g with c1 + c2 + + ci = b1 + b2 + + bi ; and 3. either

i Y j =1

(cj + 1)