On the Semisimplicity of the Brauer Centralizer Algebras

Journal of Algebra 211, 647]685 Ž1999. Article ID jabr.1998.7592, available online at http:rrwww.idealibrary.com on

On the Semisimplicity of the Brauer Centralizer Algebras William F. Doran IV and David B. Wales Department of Mathematics, California Institute of Technology, Pasadena, California 91125

and Philip J. Hanlon Department of Mathematics, Uni¨ ersity of Michigan, Ann Arbor, Michigan 48109 Communicated by Georgia Benkart Received June 24, 1997

1. INTRODUCTION The Brauer centralizer algebras, BnŽ Q ., are finite-dimensional algebras indexed by a positive integer n and a complex number Q. For integral values of Q, BnŽ Q . is the centralizer algebra for the orthogonal group or the symplectic group on the nth tensor powers of the natural representation. They were introduced by Richard Brauer in wBrx, where many of their properties are given. Earlier, Schur had used the group algebra of the symmetric groups to study the corresponding centralizer algebras for the general linear groups. The algebras BnŽ Q . are defined for any value of the parameter Q. But unlike the symmetric group algebras, these algebras are not semisimple for certain values of Q. Brauer wBrx, Brown wBrnx, and Weyl wWeyx proved results about semisimplicity. In wHW1, 2, 3, 4x Hanlon and Wales studies the algebras in the cases in which the algebras are not semisimple. They found many surprising combinatorial conditions that helped to describe the radical for values of Q when the radical was not zero. Together, these conditions led them to conjecture that the algebras are semisimple when Q is not an integer. This was proved by Hans Wenzl wWenx. In their work, Hanlon and Wales constructed certain matrices with polynomial entries that were the Gram matrices for certain representa647 0021-8693r99 $30.00 Copyright Q 1999 by Academic Press All rights of reproduction in any form reserved.

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tions. Semisimplicity meant that these matrices were nonsingular. The fact that BnŽ Q . is semisimple for Q not an integer means the determinant of these matrices has only integral roots. Some of these roots have combinatorial descriptions, which led to open problems in finding descriptions for all of them. The main problem when the radical is nonzero is to find an effective description of the radical or to find an algorithm that finds the radical, or even its dimension. A related question is to find the dimensions of the irreducible representations. Recent work of Paul Martin wMx on a related algebra, called the partition algebra, has suggested new techniques for studying the Brauer algebras. The partition algebras have a similar definition and are defined in terms of a parameter Q. The multiplication can be defined in a similar way, and in both cases the irreducible representations can be defined and parameterized by certain partitions. Indeed, the Brauer algebra is a subalgebra of the partition algebra with the same parameter Q. The purpose of this paper is to extend these techniques to the study of the Brauer algebras. In so doing, we are able to reprove Wenzl’s result about the semisimplicity for noninteger values of Q. We are also able to give conditions that must be satisfied for the existence of certain embeddings of the generic modules into others. In particular, we show that a necessary condition for the embedding of one generic irreducible into another is that Q be an integer, which implies Wenzl’s result. These embeddings were enough in the partition algebra case to determine an algorithm to find the dimensions of all of the irreducible representations. This does not yet seem to be the case for the Brauer algebras. The main new tools introduced by Martin are Frobenius reciprocity and the use of two functors, F and G. In the Brauer algebra case, these functors connect the modules for BnŽ Q . to those of Bnq2 Ž Q .. Martin used related functors extensively in his work. 1.1. Notation An integer partition l s Ž l1 G l2 G ??? G l l . is a weakly decreasing sequence of positive integers. If the sum of the l i ’s is n, then this is denoted by l & n and < l < s n. The integers l i are called the parts of l. The Ferrers diagram associated with a partition l is the collection of boxes w lx s Ž i, j .: 1 F j F l i 4 in Z 2 using matrix-style coordinates. A partition is e¨ en if all of its parts are even. For a box p s Ž i, j . in w lx, the content of p, denoted cŽ p ., is the value j y i. If l and m are two partitions such that l i G m i for all i, then we say l contains m , written m : l. If m : l Žso as sets w m x : w lx., then the skew-partition lrm is the set w lx _ w m x. A special case is when lrm contains one box. In this case, we say l co¨ ers m , denoted l d m Žor m is co¨ ered by l, denoted m e l..

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1.2. Brauer Centralizer Algebra The Brauer centralizer algebra, BnŽ Q ., is defined for every integer n and any complex number Q. A basis for BnŽ Q . is the set of all 1-factors on 2 n points. A 1-factor on 2 n points is a graph with 2 n vertices in which every vertex has degree 1. The set of 1-factors on 2 n points is denoted by Fn . We view elements of Fn by arranging the 2 n points in two rows, each containing n points, with the rows arranged one on top of the other. For example, a typical element of F7 is

When the 1-factor, d , is arranged this way, the top row of d is denoted topŽ d .. Similarly, we denote the bottom row by bot Ž d .. For reference purposes, the points of the 1-factor are numbered 1 to n from left to right in both the top and bottom. Lines joining two points, both of which are in the top or both of which are in the bottom, are called horizontal lines. Lines joining a point in the top to one in the bottom are called ¨ ertical lines. The multiplication of two 1-factors, d 1 and d 2 , is defined by placing d 1 above d 2 . Now draw an edge from point i in bot Ž d 1 . to point i in topŽ d 2 . for all i. The resulting graph consists of paths that start and finish in topŽ d 1 . and bot Ž d 2 ., as well as some cycles that use only points in the middle two rows. Let g Ž d 1 , d 2 . denote the number of these internal cycles. Form the 1-factor d in Fn by using the paths in the resulting diagram that start and stop on the bottom or the top. The product d 1 ? d 2 in BnŽ Q . is defined to be Qg Ž d 1 , d 2 .d . For example, if

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and

then g Ž d 1 , d 2 . s 1 and d 1 ? d 2 s Q1d , where The algebra BnŽ Q . is associative with identity and has dimension < Fn < s Ž2 n y 1.!!s Ž2 n y 1. ? Ž2 n y 3. ??? 3 ? 1. The identity is the 1-factor with each point i on the top joined to point i on the bottom. As shown by Brauer wBrx and used extensively in wHW1,2,3,4x, there is a series of ideals in BnŽ Q .. In particular, let Bnm Ž Q . be the span of the elements in Fn that have m or fewer vertical lines. As shown by Brauer, these elements form an ideal in BnŽ Q .. The notation here is different from that in earlier work, where the index was the number of horizontal lines on the top or the bottom. The difference in notation is made to be in conformity with the partition algebra, where the number of vertical lines is the relevant parameter. Notice these are an increasing series of ideals as Bnmy 2 Ž Q . ; Bnm Ž Q .. These ideals give a filtration of BnŽ Q .. In particular, let Inm Ž Q . s mŽ . Bn Q rBnmy 2 Ž Q . be the quotients of this filtration. A basis for Inm Ž Q . is the set of 1-factors that have exactly m vertical lines. The multiplication of d on this module is exactly as in BnŽ Q ., unless the resulting product has fewer than m lines, in which case the result is 0. So the multiplication by any element that is in Bnmy 2 Ž Q . gives 0. There are certain elements in Bnny 2 Ž Q ., denoted X i, j , which are needed later. The element X i, j has one horizontal line on the top and one on the bottom from point i to point j. The remaining lines join point k on the top to point k on the bottom for k / i or j.

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1.3. Representation Theory of the Symmetric Group The symmetric group on n objects is denoted by SymŽ n.. Notice that is isomorphic to the group algebra Cw SymŽ n.x because it is spanned by all of the 1-factors that have n vertical lines, which are themselves permutations. In fact these elements span a subalgebra isomorphic to SymŽ n., which we refer to as SymŽ n.. Any BnŽ Q .-module can be viewed as a SymŽ n.-module by restricting to the action of SymŽ n.. Notice that SymŽ n., along with  X i, j : 1 F i - j F n4 , generates BnŽ Q .. Thus, to show that two BnŽ Q .-modules are isomorphic, it suffices to show that they are isomorphic as SymŽ n.-modules and the action of each X i, j is the same in both modules. Several results on the representation theory of the symmetric group are needed. Here, we review some definitions and notation. For more details, see wJx, wJKx, and wSx. For l & n, let S l denote the irreducible SymŽ n.-module corresponding to l. The dimension of S l is denoted f l and its character by x l. In the group algebra Cw SymŽ n.x, let  el: l & n4 be a set of irreducible orthogonal idempotents where the indexing is such that S l s Cw SymŽ n.x el. A combinatorial description of one choice for el is as follows. Let Rl be the row stabilizer of the Ferrers diagram w lx. So, Rl ( SymŽ l1 . = ??? = SymŽ l l .. Let Cl be the column stabilizer of w lx; then Inn Ž Q .

el s

fl n!

Ý

Ý

sgn Ž s . st ,

sgCl tgR l

where sgnŽ s . is the sign of the permutation s . If V is a SymŽ n.-module, then for any ¨ g V, el ? ¨ is in the l-isotopic component of V. Let m & n and h & m. The Littlewood]Richardson coefficients cml, h are defined by nqm. S m m Sh ­ SS yy mŽ mŽ n.=S y mŽ m. (

[

l&nqm

cml, h S l ,

where the embedding of SymŽ n. = SymŽ m. in SymŽ n q m. is the obvious one. There is a combinatorial rule for computing these coefficients.

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Expositions of this ‘‘Littlewood]Richardson Rule’’ are given in Section 2.8 of wJKx and Section 4.9 of wSx. A basic result of this rule is that cml, h / 0 implies m : l. One special case that we need is h s Ž2.. In this case, cml, Ž2. s 1 if m ; l, where lrm contains two boxes that are in different columns, and is zero otherwise.

2. GENERIC IRREDUCIBLES OF THE BRAUER ALGEBRA This section gives a brief description of certain modules SmŽ n. of the Brauer centralizer algebras. The modules depend on the parameter Q. They are irreducible modules of BnŽ Q ., except for finitely many values of Q, as seen, for example, in wHW1x. Moreover, they are all irreducible if and only if BnŽ Q . is semisimple. In the nonsemisimple case, the irreducibles are quotients of these generic irreducibles by maximal submodules. Let n s m q 2 k. Define InmX Ž Q . to be the submodule of Inm Ž Q . generated by elements with the fixed bottom

If d 1 is any 1-factor on 2 n points, and d 2 is a 1-factor with bottom being Ž1., then either the product d 1 ? d 2 has bottom Ž1. or possibly two of the first m points in the bottom are now joined. In the second case, since this multiplication is taking place in Inm Ž Q ., the result is 0. So, InmX Ž Q . is indeed a submodule of Inm Ž Q .. Let SymŽ m. be the subgroup of SymŽ n., fixing all but the first m points 1, 2, . . . , m. Notice SymŽ m. acting on the right takes InmX Ž Q . to itself. Now given m & m, define SmŽ n. to be the BnŽ Q .-module InmX Ž Q . mS y mŽ m. S m , with the multiplication being d 1Ž d 2 m ¨ . s Ž d 1 ?I nm ŽQ. d 2 . m ¨ . The term d 1 ?I nm ŽQ. d 2 is 0 if d 1 ? d 2 is 0 in Inm Ž Q .. If it is not 0 in Inm Ž Q ., then d 1 ? d 2 has m vertical lines. These must be the ones connecting the first m points on the bottom to the top and in BnŽ Q ., d 1 ? d 2 s d 9, where d 9 g InmX Ž Q .. The BnŽ Q .-module SmŽ n. is generically irreducible. This means that for all but finitely many values of Q,  SmŽ n.: < m < F n, 2 <Ž n y < m <.4 is the set of distinct irreducible BnŽ Q .-modules. See wHW1x.

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Next we give an explicit basis for SmŽ n.. Again set n s m q 2 k. An Ž m, k . partial 1-factor on n points is a labeled graph with n vertices and k edges such that each vertex has degree 0 or 1. The m vertices of degree 0 are called free. Let Rm, k denote the set of Ž m, k . partial 1-factors. Associate with each x in Rm, k an element f Ž x . in InmX Ž Q . by letting the top of f Ž x . be x itself and the bottom be Ž1., and connect the m free vertices in the top and bottom from left to right in order. So, the vertical edges in f Ž x . do not cross. The action of d on f Ž x . m ¨ i is d Ž f Ž x . m ¨ i . s d ?I nm ŽQ. f Ž x . m ¨ i , and is 0 if d ?I nm ŽQ. f Ž x . s 0. Otherwise it is d Ž f Ž x . m ¨ i . s Qg Ž d , f Ž x .. f Ž y . m s ¨ i , where s g SymŽ m. is the unique permutation that straightens the vertical edges of d ? f Ž x . and y g Rm, k satisfies f Ž y . s Ž d ? f Ž x .. s . PROPOSITION 2.1. Take m & m F n. Let k s 12 Ž n y m. and Ž ¨ 1 , . . . , ¨ f m . be a basis for the SymŽ m.-module S m. Then  f Ž x . m ¨ i : x g Rm, k , 1 F i F f m4 is a basis for the BnŽ Q .-module SmŽ n.. In particular, dimŽ SmŽ n.. s Ž mn .Ž2 k y 1.!! f m. When l & n, the structure of SlŽ n. is essentially that of the SymŽ n.module S l. The space InnX Ž Q . is 1-dimensional and is generated by the identity 1. When s g SymŽ n. ; BnŽ Q ., s Ž1 m ¨ . s 1 m Ž s ¨ .. Thus as a SymŽ n.-module it is isomorphic to S l. If d g BnŽ Q . has a horizontal edge, then d ? 1 s 0 in Inn Ž Q .. Thus d Ž1 m ¨ . s d ?I nm ŽQ. 1 m ¨ s 0. Because the structure of SlŽ n. does not depend on Q in this case and is generically irreducible, it is always irreducible. Thus every nontrivial BnŽ Q .-homomorphism from SlŽ n. into another SmŽ n. is an embedding. Furthermore, given SmŽ n., an embedding of SlŽ n. in SmŽ n. is a subspace W ; SmŽ n., such that W ( S l as a SymŽ n.-module and X i, j w s 0 for all w g W and 1 F i j F n. The description here gives a way to identify BnŽ Q . as a cellular algebra in the same of wGLx. A basis for Inm Ž Q . can be given by all Ž x 1 , x 2 , s ., where the s is in SymŽ m. and x i are Ž m, k . partial 1-factors, where m q 2 k s n. Here a diagram, d 1 , with exactly m vertical lines is represented by topŽ d 1 . s x 1 , bot Ž d 1 . s x 2 , and s is the permutation that maps the ith free point from the left in bot Ž d 1 . to the s Ž i .th free point in topŽ d 1 .. Let Jnm be the complex vector space, with the basis being diagrams in Inm Ž Q . in which s is the identity. Now Inm Ž Q . can be identified with the vector space Jnm mS y mŽ m. V, where V is a space representing the regular representation of SymŽ n.. In particular, identify Ž x 1 , x 2 , s . with d x 1 , x 2 m s , where d x 1 , x 2 is the diagram in Jnm with top x 1 , bottom x 2 , and s is taken in V. Now pick a Kazhdan]Lusztig basis wt 1 , t 2 for V as described in wGLx, which is labeled by pairs of standard Young tableaux, t 1 , t 2 , of shape m for all partitions m of m. The cellular basis described in wGLx is  d x , x m wt , t 4 1

2

1

2

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for m ranging over all m, with 0 F m F n, for which 2 <Ž n y m.. The important condition C3 in wGL, Definition 1.1x follows from the action of d . In particular, we have seen above that d Ž f Ž x . m wi . s d ?I nm ŽQ. f Ž x . m wi is 0 if d ?I nm ŽQ. f Ž x . s 0, and is d Ž f Ž x . m wi . s Qg Ž d , f Ž x .. f Ž y . m s wi where s g SymŽ m. is as above. The same result is true here if we replace wi with the basis vectors wt 1 , t 2 and if we change the bottom from Ž1. to any fixed bottom, say x 2 , and replace f Ž x . with Ž x, x 2 , id .. The crucial fact that makes this a cellular basis is that the action of d does not depend on which bottom, x 2 , has been chosen. This leads to the defining relations C3 in wGLx, as shown in Section 5. When the action gives 0, it is modulo the span of diagrams with fewer vertical lines that are lower in the order L, as given in wGL, Section 5x. This set L is the set of m where m & m for all nonnegative m for which 2 N Ž n y m.. When the action does not give 0, the relation C3 in wGLx follows because the wi are a cellular basis for SymŽ m.. Some of the results referred to in wHW1,2,3,4x are properties common to cellular algebras. Furthermore, the representations SmŽ n. given here are the same as the modules W Ž m . for the elements of L.

3. NECESSARY CONDITIONS In this section, we give necessary conditions for an embedding of SlŽ n. into SmŽ n. when l & n. In the case of the partition algebra, Martin studied these embeddings and was able to characterize in wMx exactly when they occur. This is the first step that allowed him to describe the irreducible modules in that case. Let ² SlŽ n., SmŽ n.: denote the value of dimŽhom B n ŽQ.Ž SlŽ n., SmŽ n... for any l and m with 2 <Ž n y < l <. and 2 <Ž n y < m <.. In Section 5, following Martin’s work on the partition algebra, we show that for Q / 0, ² SlŽ n q n 0 ., SmŽ n q n 0 .: / 0 if and only if there is an embedding of SlŽ n. into SmŽ n. where l & n. Thus, it suffices to study only the case l & n. Recall that the subalgebra of BnŽ Q . generated by 1-factors with n vertical lines is isomorphic to Cw SymŽ n.x. We refer to this subalgebra as SymŽ n.. When l & n, SlŽ n. is essentially the SymŽ n.-module S l. Thus, if ² SlŽ n., SmŽ n.: / 0, then SmŽ n., when viewed as a SymŽ n.-module by restriction, must contain S l as a constituent. A decomposition of SmŽ n. as an SymŽ n.-module is given in wHW3x ŽTheorem 4.1.. THEOREM 3.1. Gi¨ en m : l. The multiplicity of S l in the decomposition of SmŽ n. as an SymŽ n.-module is

Ý

h&2 k h even

cml, h ,

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where 2 k s < lrm < and cml, h is a Littlewood]Richardson coefficient. If m ­ l, the multiplicity is 0. Thus a necessary condition for the embedding SlŽ n. ¨ SmŽ n. is that m : l. In the particular case < lrm < s 2, this multiplicity is one if the two boxes in lrm are in different columns and zero otherwise. We now give a stronger necessary condition for there to exist a BnŽ Q . embedding of SlŽ n. into SmŽ n. for l & n due to the annihilation by all X i, j ’s. Let m & m where n s m q 2 k. Suppose SlŽ n. ¨ SmŽ n.. Then there is a subspace W of SmŽ n., which as a BnŽ Q .-module is SlŽ n.. We know that by just considering SymŽ n. : BnŽ Q ., S l must be the action, and so m : l by Theorem 3.1. A further condition is that W is annihilated by Bnny 2 Ž Q .. To this end we form the element Ts

Ý

Xi , j .

1Fi-jFn

Recall X i, j has points i and j joined on the top and bottom, and the point l in the top is joined to the point l on the bottom for l / i or j. Thus the element is in Bnny 2 Ž Q . and acts as the identity, except at i and j. We first discuss how T acts on f Ž x . for x an Ž m, k . partial 1-factor in Rm, k . LEMMA 3.2. For a fixed Ž m, k . partial 1-factor x, let y s f Ž x . be the corresponding 1-factor in InmX Ž Q .. Then in Inm Ž Q ., Ty s k Ž Q y 1 . q

ž

Ý Ž i, j. y 1Fi-jFn

Ý 1Fa-bFm

/

Ž a, b . y,

where here the Ž i, j . are transpositions in SymŽ n. acting on the left of y and Ž a, b . are transpositions in SymŽ m. acting on the right of y and on the first m positions only. Proof. We consider how X i, j acts on y. There are four cases. Ža. Points i and j are joined in x. In this case, X i, j y s Qy. Žb. Points i and j are both free points in x. Here X i, j y s 0, as the resulting 1-factor has at least one more horizontal line and thus at most m y 1 vertical lines. Such terms are 0 in Inm Ž Q .. Žc. In x, one of the points i and j is free, say point i, and the other, point j, is joined to a third point l:

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Then X i, j y is the same as y, except that point i in the top is joined to point j in the top and point l in the top is joined to the point in the bottom that had been joined to point i. In other words, the line Ž jl . becomes the line Ž ji ., and l is joined to what i had been joined to. That is, X i, j y s Ž i, l . y. Žd. Neither point i nor j is free in x. Suppose points i and j are joined to points l and m, respectively:

Here points i and j become joined, and points l, m become joined. This could have been effected by interchanging j and l or i and m, i.e., X i, j y s Ž j, l . y s Ž i, m. y. Notice X l, m y gives the same term. We now consider the action of T on y. First partition Ž i, j .: 1 F i j F n4 into four sets A, B, C, and D, based on which of the four cases apply for this particular x. Now Ty s

Ý

Xi , j y q

Ž i , j .gA

Ý

Xi , j y q

Ž i , j .gB

Ý

Xi , j y q

Ž i , j .gC

Ý

X i , j y.

Ž i , j .gD

Since there are k edges in x, ÝŽ i, j.g A X i, j y s kQy. For each Ž i, j . g B, X i, j y s 0, and thus the second term makes no contribution. For any Ž i, j . g C or D, the effect of X i, j on y is to multiply by a transposition t i, j on the left. For Ž i, j . in D, recall that X i, j y s X l, m y s Ž j, l . y s Ž i, m. y, where points i and j are joined to points l and m, respectively. In this case, we leave the exact association between the Ž i, j . in D and transpositions t i, j vague. All that matters is that setwise, both are accounted for. Thus as a first approximation to the sum over C and D, we consider

Ý

Ž i , j .gC

Xi , j y q

Ý

Ž i , j .gD

Xi , j y f

Ý Ž i , j . y. 1Fi-jFn

However, the right-hand side ŽRHS. contains terms that do not arise from an Ž i, j . g C j D. There are two sources of overcounting. First, if points i

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and j in the top of y are joined, then Ž i, j . y s y is counted in the RHS but not in the LHS. To counteract this, we subtract ky from the RHS, since there are k edges in the top of y. The second source of overcounting arises if both points i and j in the top of y are joined to points in the bottom. Say point i in the top is joined to point a and j to b. Notice that Ž i, j . y s y Ž a, b .. To compensate for this overcounting, we subtract y Ž a, b .

Ý 1Fa-bFm

from the RHS. This gives

Ý

Ž i , j .gC

Xi , j y q

Ý

Xi , j y s

Ž i , j .gD

Ý Ž i , j . y y ky y 1Fi-jFn

Ý

y Ž a, b . .

1Fa-bFm

Adding everything together, we get the desired result. THEOREM 3.3. Take l & n and m & m - n. If ² SlŽ n., SmŽ n.: / 0, then k Ž Q y 1. q

Ý

c Ž p . s 0,

Ž w.

pg w l r m x

where n s 2 k q m. Proof. Now suppose ² SlŽ n., SmŽ n.: / 0. Here, we view SmŽ n. as a SymŽ n. = SymŽ m.-module by

Ž p , p 9 . ? Ž d m ¨ . s p ? Ž d m Ž p 9¨ . . , where p g SymŽ n., p 9 g SymŽ m., d g InmX Ž Q ., and ¨ g S m. The image of SlŽ n. is in the S l m S m component of SmŽ n.. The sums ÝŽ i, j.Ž i, j . and ÝŽ a, b.Ž a, b . each act as a scalar, as each is in the center of the respective group algebras. The values are the corresponding vlŽ1, 2. and vmŽ1, 2.. As is well known ŽSee Chapter 1 of wDx, for example.,

vl Ž 1, 2 . s

Ý cŽ p.

pg w l x

and

vm Ž 1, 2 . s

Ý cŽ p. ,

pg w m x

where here cŽ p . is the content of a box in l or m.

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If ¨ / 0 is in the S l m S m constituent of SmŽ n., T¨ s k Ž Q y 1 . q

ž

Ý cŽ p. y Ý cŽ p.

pg w l x

pg w m x

/

¨.

Since ¨ / 0, for this to be 0 we need k Ž Q y 1. q

Ý

c Ž p . s 0.

pg w l r m x

If n s m q 2 k with k G 0, l & n, m & m, and m : l, we say the pair Ž l, m . satisfies Žw. if k Ž Q y 1. q Ý p g w l r m x cŽ p . s 0, where as usual here cŽ p . is the content of p. In the case k s 1, for Ž l, m . to satisfy Žw., we also require that lrm has its two boxes in different columns. In general, satisfying Žw. does not imply that there is an embedding, but in the case k s 1 it does. Remark. Theorem 3.3 also follows from wN, p. 671, formula before Ž2.13.x when Q is a positive integer. They show that T y Ý1 F i - jF nŽ i, j . is central and acts as the appropriate scalar on each generic irreducible, and the result follows similarly. Our proof is direct and does not use the isomorphism with BnŽ Q . and the centralizer algebras, or any facts about their representations. It also works when Q is not a positive integer. THEOREM 3.4. Let l & n. If < lrm < s 2 and Ž l, m . satisfies Žw., then there exists an embedding of SlŽ n. into SmŽ n.. Proof. This follows for the case in which lrm is not in a row from results in wHW1x and wHW3x Theorem 4.8. However, an entirely new proof for < lrm < s 2 comes by using Theorem 3.3. The main arguments in wHW3x applied to this case involve computing the terms of a degree 1 polynomial. The fact it is degree 1 means there is a root, and this means there is an embedding. Now Theorem 3.4 specifies the value of Q by Q y 1 q cŽ p . q cŽ q . s 0, which gives the result. As described in wHW1x, the module SmŽ n. has a unique maximal submodule that is annihilated by all elements in Bnm Ž Q ., called the radical of SmŽ n.. When m s n y 2, the radical is annihilated by all X i, j , and so any SymŽ n. irreducible subspace of the radical is one of the irreducible SlŽ n. for some l & n. It is for this reason that when there is a radical for SmŽ n. with m & n y 2 that nontrivial maps from SlŽ n. occur. This is not the case when < m < - n y 2, except in special circumstances. The computations for the case of m & n y 2 in wHW3x use the matrix ZŽ x . s Zny 2, 1Ž x ., which is a square matrix with basis the labeled Ž n y 2, 1.

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partial 1-factors. Here if m q 2 k s n, a labeled Ž m, k . partial 1-factor is an Ž m, k . partial 1-factor for which the free points have been labeled with the integers 1, 2, . . . , m. Here the n y 1 free points are labeled with 1, 2, . . . , n y 2. The matrix ZŽ x . has x on the diagonals and constants off. Let the vector space with these as basis be V s Vny 2, 1. As described in wHW1,3x, there is a natural action of SymŽ n. = SymŽ n y 2. on V, where SymŽ n. permutes the vertices and SymŽ n y 2. permutes the labels. The matrix ZŽ x . commutes with this action. To identify the action of X i, j , let d be an Ž n y 2, 1. partial 1-factor and let d˜ be a labeled Ž n y 2, 1. partial 1-factor for which the free points of d have been labeled with the integers 1, 2, . . . , n y 2. Suppose the order when read from left to right is s Ž1., s Ž2., . . . , s Ž n y 2.. Here s is a permutation in SymŽ n y 2.. Identify d˜ with f Ž d . s , where here s acts from the right and on the first n y 2 positions. Recall f Ž d . has a line from n y 1 to n on the bottom and the vertical lines from the top to bottom do not cross. Then consider X i, j ? f Ž d . s . Recall this is 0 if i and j are both free; otherwise it is the usual BnŽ Q . multiplication. In this case it is of the form f Ž d#. s # or Q times it for an appropriate d# and s #. This, in turn, can be identified with d˜# for a unique Ž n y 2, 1. labeled partial 1-factor. This gives the action of X i, j by identifying X i, j d˜ with either d˜# or Q d˜# being the second case. As explained in wHW3x, if d˜1 corresponding to f Ž d 1 . s 1 is another Ž n y 2, 1. partial 1-factor, the Ž d˜, d˜1 . entry of ZŽ x . is obtained by evaluating w s Ž f Ž d 1 . s 1 .9 f Ž d . s .. Here Ž d 1 s 1 .9 is d 1 s 1 turned upside down. The map that turns the basis elements upside down generates an antisiomorphism of BnŽ Q .. The entry is 0 unless w s X ny1, n or QX ny 1, 1. It is 1 in the first case and x in the second. The second case occurs only on the diagonal. With the definitions as above, ZŽ Q . for any value of Q has the important property that X i, j d˜ s ZŽ Q . d˜. Now for a given value Q of x, elements in the null space of ZŽ Q . give rise under this correspondence to elements in Inny 2 Ž Q ., which are annihilated by all X i, j . Let U be a vector space on which SymŽ n y 2. acts as the regular representation. Now replace s by a vector space transformation representing the action of s on U. The space V can be identified with the space spanned by all d mS y mŽ ny2. U, where the d are Ž n y 2, 1. partial 1-factors. The matrix ZŽ x . commutes with the SymŽ n y 2. action for any value of x. Breaking V into irreducible SymŽ n y 2. invariant subspaces gives SmŽ n. with multiplicity f m. Each is isomorphic to the action on the span of all d˜ m S m, where d˜ are the Ž n y 2, 1. partial 1-factors and S m is a representation space for S m . Within each SmŽ n. is a unique copy of S l. As ZŽ x . commutes with this action, ZŽ x . must be a scalar matrix when acting on this subspace, with the scalar being a polynomial. As ZŽ x . has x on the diagonals, the determi-

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nant of the action on this subspace is the characteristic polynomial, and so the polynomial is linear of degree 1. The intersection of the null space of Z with this submodule is in the radical of SmŽ n.. But then when Q is the root of degree 1 polynomial, the subspace is in the radical. This means the S l subspace is in the radical for some value of Q. But as above, this is an embedding, and the value of Q is given by Theorem 3.4.

4. RESTRICTION IN THE BRAUER ALGEBRA In this section, we give a complete description of the restriction of the generic irreducibles of BnŽ Q .. The standard embedding of Bny1Ž Q . in BnŽ Q . is given as follows. For any 1-factor d 9 g Bny1Ž Q . append one point in both the top and bottom rows of d 9 and draw a vertical line between these points. Call this d . Extend this map linearly to get an embedding of Bny 1Ž Q . in BnŽ Q .. Given a BnŽ Q .-module M, the restriction of M to Bny 1Ž Q . via this embedding is denoted M x. In the semisimple case, it is well known that Sl Ž n . x (

[ Sm Ž n y 1. [ [ Sn Ž n y 1. .

me l

ndl

Ž 2.

However, when Bny 1Ž Q . is not semisimple, SlŽ n.x need not decompose into a direct sum of irreducibles. However, we show there is a two-step filtration of SlŽ n.x in which the first terms in the sum Ž2. appear as constituents of a subspace, and the terms in the second sum appear in the quotient. THEOREM 4.1. Fix n s m q 2 k and take l & m and assume m - n. Let V s SlŽ n.x. As a Bny1Ž Q .-module, V contains a submodule W ( [m e l SmŽ n y 1.. Furthermore, VrW ( [n d l Sn Ž n y 1.. Proof. Let f s f l and  ¨ 1 , . . . , ¨ f 4 be a basis for S l. Partition Rm, k into two sets A and B, where A is the set of Ž m, k . partial 1-factors in which vertex n is free Ži.e., has degree 0., and B is those in which vertex n is not free Ži.e., has degree 1.. Define W to be the space generated by  f Ž a. m ¨ i : a g A, 1 F i F f 4 . We show first that W is invariant under any permutation in SymŽ n y 1. and all X i, j for 1 F i - j F n y 1. As these generate Bny1Ž Q ., this shows W is invariant. Suppose s is a permutation in SymŽ n y 1.. When embedded in BnŽ Q ., these permutations have a vertical line from point n in the top to point n in the bottom. Thus, for any a g A, s Ž f Ž a.. s f Ž a9. s 9, where a9 is another Ž m, k . partial 1 factor and s 9 acts on the right and rearranges the vertical lines so they do not cross. Notice that in f Ž a. the

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line from point n on the top to point m in the bottom has not been affected by the action of s , so s 9 fixed m. Furthermore, a9 is in A. This means that s Ž f Ž a. m ¨ i . s f Ž a9. s 9 m ¨ i , which is f Ž a9. m s 9¨ i . Therefore, W is preserved by s . As in the proof of Lemma 3.2, there are four cases for the value of X i, j ? f Ž a., depending on whether or not the vertices i and j are free or are on lines in a. The result is either 0 if i and j are free, Qf Ž a. when i, j is a line in a, or t f Ž a. for a transposition t , which is Ž i, m. or Ž j, m.. In each case, W is preserved. This shows that W is a Bny 1Ž Q .-module. We now identify W as a Bny 1Ž Q .-module by finding an isomorphism l my 1 Ž . between W and Iny 1 Q mS y mŽ my1. S x. In particular, for any a g A, let w 1Ž a. be a with point n removed. Notice this is a bijection from A to the set of Ž m y 1, k . partial 1-factors. Now let w U1 Ž a. s f Ž w 1Ž a... For example,

my 1 Ž . l Now let w be the map from W to Iny given by 1 Q mS y mŽ my1. S U Ž Ž . . Ž . w f a m ¨ i s w1 a m ¨ i. This map is clearly an isomorphism of vector spaces as w 1 is a bijection. To show that it is a Bny 1Ž Q .-module map, we show that it commutes with permutations in SymŽ n y 1. and with all X i, j for 1 F i - j F n y 1. We saw above how X i, j acts on f Ž a.. The same description applies to the action of X i, j on w U1 Ž a.. The four cases are identical, and the actions are identical. In particular, if one is 0 or is multiplied by Q, so is the other. In the other cases, they are multiplied by the same transposition. Now we must check that X i, j Ž f Ž w 1Ž a. m ¨ i . s w Ž X i, j Ž f Ž a. m ¨ i ... The actions of X i, j are determined by the actions on f Ž a. and on f Ž w 1Ž a... This is immediate for the two cases being multiplied by 0 or Q, and once we show permutations commute with w , we will have shown that X i, j commutes. Suppose then that s g SymŽ n y 1.. We have seen that s f Ž a. s f Ž a9. s 9. Now w Ž s f Ž a. m ¨ i . s w Ž f Ž a9. m s 9¨ i . s w *Ž a9. m s 9¨ i . On the other hand, sw 1Ž a. s w 1Ž a9., as s does not move n. This means that s f Ž w 1Ž a.. s f Ž w 1Ž a9.. s 9, which is just sw U1 Ž a. s w U1 Ž a9. s 9. Now sw Ž f Ž a. m ¨ i . s sw U1 Ž a. m ¨ i s w U1 Ž a9. s 9 m ¨ i . But this is w *Ž a9. m s 9¨ i . This shows that the actions commute.

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To treat the action on VrW, we introduce a map similar to w 1. Let w 2 be the map from B to Rmq 1, ky1 , where w 2 Ž b . is b with vertex n and the edge incident to it removed. Recall that b is an Ž m, k . partial 1 factor with vertex n joined to another vertex. One of the vertices has been freed up, and a line taken away, so the resulting 1 factor is in Rmq 1, ky1. Given b9 g Rmq 1, ky1 , the elements of  b g B: w 2 Ž b . s b94 are obtained by appending an additional vertex to b9, called vertex n, and adding an edge from vertex n to any of the m q 1 free vertices of b9. Hence, w 2 is an Ž m q 1.-to-1 map. mq19Ž . Using w 2 , we define a map w U2 from B to Iny1 Q by letting the top of U Ž . Ž . w 2 b be w 2 b , draw an edge from the point in the top that was connected to vertex n in b to point m q 1 in the bottom, and draw in the remaining vertical lines so that they do not cross. For example,

More formally, let sj s Ž j, m q 1, m, . . . , j q 1. be the permutation in SymŽ m q 1. that maps j to m q 1 and then shifts the integers between j q 1 and m q 1 down by one. Suppose in b, vertex n is joined to a vertex l, and suppose there are j y 1 free points to the left of l. Then w U2 Ž b . s f Ž w 2 Ž b .. sj . Also from the description above, for any b9 g Rmq1, ky1 and 1 F j F m q 1, there is a b g B such that w U2 Ž b . s f Ž b9. sj . To prove VrW has the desired structure, we need some facts about induced representations in the symmetric group. For 1 F i F m, let t i s Ž i, m q 1. and let tmq 1 s 1. These form a set of coset representatives for SymŽ m q 1.rSymŽ m.. A basis for S l ­ is indexed by  ¨ 1 , . . . , ¨ f 4 =  1, . . . , m q 14 . Take s g SymŽ m q 1.. The action of s on S l ­ is given by

s ? Ž ¨i , j . s

y1 k st j

ž Žt

. ¨i , k / ,

Ž . where k is the unique value such that ty1 k st j g Sym m . We claim that mq 19 l w : VrW ª Iny 1 Ž Q . mS y mŽ mq1. S ­

f Ž b . m ¨ ¬ w U2 Ž b . m Ž ¨ , m q 1 . is a Bny 1Ž Q .-isomorphism. We first show that it is a bijection. Since w 2 is an Ž m q 1.-to-1 map and mq 19Ž . l dimŽ S l ­. s Ž m q 1.dimŽ S l ., VrW and Iny 1 Q m S ­ have the same dimension. Hence, it suffices to show that w is onto.

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mq 19Ž . l  Ž . Ž . A generating set for Iny 1 Q m S ­ is f b9 m ¨ , j : b9 g Rmq1, ky1 , l Ž . ¨ g S , 1 F j F m q 14 . Choose b9, ¨ , and j. Notice that sy1 j tj m q 1 s y1 Ž Ž . m q 1 or, equivalently, that tmq 1 sy1 ? ¨ , j . s Ž ¨ 9, j t j g Sym m . So, sj m q 1. for some ¨ 9 g S l. Then f Ž b9. m Ž ¨ , j . s f Ž b9. m sj ? Ž ¨ 9, m q 1. s f Ž b9. sj m Ž ¨ 9, m q 1.. Let b g B be such that w U2 Ž b . s f Ž b9. sj . Thus w Ž b m ¨ 9. s f Ž b9. m Ž ¨ , j ., which shows that w is onto. Now we show that w is a BnŽ Q .-homomorphism. Pick b g B and ¨ g S l. Arguing as above for W, notice that for all s g SymŽ n y 1., sw Ž f Ž b . m ¨ . s ws Ž f Ž b . m ¨ .. So, w is a SymŽ n y 1.-isomorphism. Thus, we only need to show that the action of X i, j is the same on both sides for all 1 F i - j F n y 1, as these generate BnŽ Q .. Let r be the vertex that is connected to vertex n in b. In w U2 Ž b ., point r in the top is connected to point m q 1 in the bottom. If neither i nor j is equal to r, then it is clear that X i, j w Ž f Ž b . m ¨ . s w X i, j Ž f Ž b . m ¨ .. Suppose i s r. Then there are two cases, based on whether vertex j in b is free or connected to some other vertex. Suppose first that vertex j is free in b. Then in X i, j f Ž b . point n in the top is connected to a point in the bottom. Thus, X i, j Ž f Ž b . m ¨ . g W and is equal to zero in VrW. On the other side, both points i and j in the top of w U2 Ž b . are connected to points in the bottom. So, X i, j w U2 Ž b . s 0, because mq 19Ž . the resulting multiplication in Iny 1 Q has fewer vertical lines. Thus, X i, j w Ž f Ž b . m ¨ . s 0, as desired. It may help to consult the picture above, form f Ž b ., and multiply by X 1, 3 . This shows also how the quotient plays a role. In the case in which vertex j is connected to, say, vertex k in b, X i, j f Ž b . s Ž r, k . f Ž b .. So, w X i, j Ž f Ž b . m ¨ . s Ž r, k . w Ž f Ž b . m ¨ . s X i, j w Ž f Ž b . m ¨ ..

EXAMPLE. We work an example that illustrates Theorem 4.1 and illustrates the situation for the next theorem. Let n s 4 and m s 2. Let V be S2 Ž4.x. The Ž2, 1.-partial 1 factors are the six elements in  t i, j ; 1 F i j F 44 , where t i, j has a line from vertex i to j and the remaining two vertices are free. A basis for SmŽ4. is the set f Ž t i, j . m ¨ , where ¨ spans the SymŽ2. module S 2 , which is the trivial module. Let w 1 s f Ž t 2, 3 . m ¨ , w 2 s f Ž t 1, 3 . m ¨ , and w 3 s f Ž t 1, 2 . m ¨ . Also let ¨ 1 s f Ž t 1, 4 . m ¨ , ¨ 2 s f Ž t 2, 4 . m ¨ , and ¨ 3 s f Ž t 3, 4 . m ¨ . Now W is the span of  w 1 , w 2 , w 34 . Notice that SymŽ3. acts naturally on w 1 , w 2 , and w 3 by permuting subscripts. It does the same for ¨ 1 , ¨ 2 , and ¨ 3 . Also note X 1, 2 w 3 s Qw 3 and X 1, 2 w 2 s X 1, 2 w 1 s w 3 . Similar relations hold for X 1, 3 and X 2, 3 . The identification with the action of B3 Ž Q . on S1Ž3. is now clear, as Theorem 4.1 predicts. The quotient by W is the direct sum of S 3 and S 2, 1. The vector y s ¨ 1 q ¨ 2 q ¨ 3 is fixed by SymŽ3.. Notice X 1, 2¨ 1 s Qw 3 , X 1, 2¨ 2 s w 3 , and X 1, 2¨ 3 s 0. Similar relations hold

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for X 1, 3 and X 2, 3 . These show that mod W, the span of  y4 , is S 3 Ž3.. The span of  ¨ 1 y ¨ 3 , ¨ 2 y ¨ 34 mod W gives S 2, 1Ž3.. Notice that both partitions that cover m occur, as Theorem 4.1 says they should. This also shows that y q W mod W does not split directly. However, it is possible that for some different choice of coset, representative y q w for some w g W splits. As SymŽ3. would fix y q w, SymŽ3. would have to fix w, and so w s g Ž w 1 q w 2 q w 3 .. Now notice X 1, 2 Ž y q g Ž w 1 q w 2 q w 3 .. s Ž2 q g Ž Q q 2.. w 3 . For this to be 0, we need 2 q g Ž Q q 2. s 0. Exactly the same equations must be satisfied if X 1, 3 or X 2, 3 is used where w 1 or w 2 occurs instead of w 3 . This can be solved if and only if Q / y2. But if Q y 2 s 0, the subspace spanned by  w 1 q w 2 q w 34 is an invariant module isomorphic to S 3 Ž3., which is an embedding of S 3 Ž3. into S1Ž3.. Notice the pair Ž3., Ž1. satisfies Žw. exactly when Q s y2. This shows that S 3 Ž3. is a submodule of S 2 Ž4.x whether Q s y2 or not. It occurs in different ways in the sense that if Q / y2, it splits from W. If Q s y2, it does not split from W, but occurs as a submodule of S1Ž3., which is itself a submodule. This phenomenon is quite general, which we demonstrate with the following theorem. We now consider the case in which l & n y 2 and examine SlŽ n.x more carefully. Using the notation of Theorem 4.1, we know VrW ( [n d l Sn Ž n y 1.. We may pick coset representatives for the cosets of VrW in the span of f Ž a. m ¨ i , where here a g A and 1 F i F f l by Theorem 4.1. Choose a specific n d l and find the subspace of VrW isomorphic to Sn Ž n y 1.. Set f s f l. Now pick coset representatives ¨ 1 , ¨ 2 , . . . , ¨ f in the span of f Ž a. m ¨ i for which ¨ 1 q W, ¨ 2 q W . . . , ¨ f q W is a basis for this subspace. This is how we chose them in the example above. There are now two possibilities: Ž1. For the given  ¨ 1 , . . . , ¨ f 4 there exist element  w 1 , . . . , wf 4 in W such that  ¨ 1 q w 1 , . . . , ¨ f q wf 4 is a basis for Sn Ž n y 1. in V. This is a splitting of Sn Ž n y 1.. Ž2. The module Sn Ž n y 1. appears as a submodule for some SmŽ n y 1. with m e l. Since < nrm < s 2, this occurs if and only if Ž n , m . satisfy condition Žw.. In particular, there is at most one value of Q for which this happens. We show that exactly one of these conditions holds for any given n d l, Q, and choice of ¨ i as indicated. Notice, incidentally, that for l & n, SlŽ n. is just the usual restriction for the symmetric group, and in the Brauer algebra, n y < l < must be divisible by 2, so this is the highest nontrivial case. THEOREM 4.2. Gi¨ en l & n y 2 and n d l, then exactly one of the conditions Ž1. or Ž2. abo¨ e holds for ¨ i as described abo¨ e.

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665

Proof. Let t i, j be the Ž n y 2, 1. partial 1-factor with an edge from vertex i to vertex j and Ti, j s f Ž t i, j . g Inny 2 9Ž Q .. Again set f s f l. A basis for SlŽ n. is Ti, j m sk 4 , where  s1 , . . . , s f 4 is a basis for S l. Denote SlŽ n.x by V. From Theorem 4.1 the Bny 1Ž Q .-module W is generated by Ti, j m sk : 1 F i, j F n y 14 and is isomorphic to [m e l SmŽ n y 1.. The quotient space VrW has coset representatives Ti, n m sk 4 and is isomorphic to [n d l Sn Ž n y 1. as a Bny1Ž Q.-module. Choose coset representative  ¨ 1 , . . . , ¨ j 4 in the span of Ti, n m sk 4 such that  ¨ 1 q W, ¨ 2 q W, . . . , ¨ f q W 4 rW ( Sn Ž n y 1.. To satisfy Ž1., we want to find  w 1 , . . . , wf 4 in W such that  ¨ 1 q w 1 , . . . , ¨ f q wf 4 is a basis for Sn Ž n y 1. in V. First, for s g SymŽ n y 1., s ? ¨ i s Ý a i,s j ¨ j , where w a i,s j x is the matrix representation for s in S n . Thus, we want s ? Ž ¨ i q wi . s Ý a si, j Ž ¨ j q wj ., which implies s ? wi s Ý a i,s j wj . This means that as a SymŽ n y 1.-module, ² w 1 , . . . , wf : ( S n . From comments after Theorem 3.1, there is such a unique subspace in SmŽ n y 1. for each m e l such that the two boxes in nrm are in different columns. When the two boxes of nrm are in the same column, Theorem 3.1 says that even as SymŽ n y 1.modules, Sn Ž n y 1. cannot be embedded in SmŽ n y 1.. We discuss the consequences of this later. Write wi s Ým e lŽ wi .m , where Ž wi .m is the m-isotropic component of wi in W. In general, for any w g W, Ž w .m stands for its m-isotropic component. By Schur’s Lemma, the Ž wi .m are unique up to a scalar multiple. Thus for any m with nrm in different columns, there is a unique choice of scalar am so that am w 1 , am w 2 , . . . , am wf gives the splitting. If nrm are in the same column, there is no choice. Up to this point, nothing in the proof has depended on Q. We have shown that Sn Ž n y 1. splits as a SymŽ n y 1.-module for any choice of am . To show that it splits as a Bny1Ž Q .-module, we need to show that there is some choice of scalars am such that Xr , s ¨ i q

ž

Ý

me l

am Ž wi . m s 0

/

for 1 F r, s F n y 1, and 1 F i F f. We examine this component by component. That is, we show that

Ž X r , s Ž ¨ i . . m q am X r , s Ž wi . m s 0 for all m. The choice of am’s depends on Q. In certain cases there are no solutions for the am’s Ži.e., condition Ž1. does not hold.. However, this happens exactly when Sn Ž n y 1. is a submodule of one of the SmŽ n y 1. Ži.e., condition Ž2. holds.. Since  ¨ 1 , . . . , ¨ f 4 is a basis for Sn Ž n y 1. in VrW and n & n y 1, we have X r, s ? ¨ i g W. Write this as X r, s ? ¨ i s Ým e lŽ X r, s ? ¨ i .m . Because of

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our choice of ¨ j spanned by linear combinations of Ti, n m sk and the fact that r, s - n, notice X r, s ŽTs, n m sk . s Tr, s m s) and X r, s ŽTi, n g sk . s 0 if i is not r or s. This means Ž X r, s ? ¨ i .m s Tr, s m s# for some s# in the S m component of S l x. Write Ž wi .m s Ý b k, l, mŽTk, l m sm .. Since W as a Bny1Ž Q .-module breaks up as a direct sum of SmŽ n., Ž X r, s ? wi .m s X r, s ? Ž wi .m . Break the sum for Ž wi .m into two parts: Ž k, l . s Ž r, s . and Ž k, l . / Ž r, s .. Let

Ž wi . m s Ý br , s, m Ž Tr , s m sm . q Ý m

m

s Tr , s m s## q

Ý m

Ý

Ý

b k , l , m Ž Tk , l m sm .

Ž k , l ./ Ž r , s .

b k , l , m Ž Tk , l m sm . ,

Ž k , l ./ Ž r , s .

where s## s Ý m br, s, m sm . The action of X r, s is different on these two parts. In particular, the terms with Tr, s are multiplied by Q. The action of multiplication on the left by X r, s is

X r , s Ž Tk , l

¡Q Ž T m s . , ¢0, m s###,

m s . s~T

r, s

r, s

if  k , l 4 s  r , s 4 , if <  k, l 4 l  r , s 4 < s 1, otherwise.

So, X r, s Ž wi .m s QŽTr, s m s##. q ŽTr, s m s###.. It is important to notice that each of these terms is independent of Q. To find a solution to X r, s Ž ¨ i .m q am X r, s Ž wi .m s 0, we need s# q Q am s## q am s### s 0.

Ž 3.

Remember, s#, s##, and s### are just vectors in the S m component of S l x. Because Bny 1Ž Q . is semisimple for almost all Q, a solution must exist for almost all Q. Suppose s## and s### were linearly independent. Then, by Ž3., s# is in their linear span and s# s cs## q ds### for unique c and d. But then the only time Ž3. has a solution is for am s yd and Q s crd. This contradicts the fact that there are an infinite number of solutions. Thus s## and s### are linearly dependent. By Ž3., it follows that s#, s##, and s### are linear multiples of one another. Suppose now the boxes of nrm are in the same column. There are no choices for Ž wi .m except 0. This means s## s s### s 0, as the Sn Ž n. splits for infinitely many Q. It follows that Ž X r, s¨ i . s 0 and the ¨ i are already split from this component. We now assume the boxes of nrm are in different columns. Recall that the value am does not depend on i. We need a technical fact about Ž X r, s ? ¨ i .m , which we prove below.

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Claim. For each m e l, with nrm not in a column, there exists an i such that Ž X r, s ? ¨ i .m / 0. We assume that this holds and continue with the proof. From the claim, we pick an i for which Ž X r, s Ž ¨ i ..m / 0. This implies s# / 0. So Ž3. becomes s# Ž 1 q Q amg 1 q am g 2 . s 0, where s## s g 1 s# and s### s g 2 s#. This applies for all i for which Ž X r, s Ž ¨ i ..m / 0. Because it can be solved for almost all Q, and because am is determined by it, g 1 and g 2 are the same for these i. This is a remarkable fact. In working examples, there seems to be no reason that these values should be the same. It is true, as we indicate, because of the solutions for infinitely many Q. Given an i for which Ž X r, s Ž ¨ i ..m s 0, s## and s### must also be 0, or Eq. Ž3. could not be solved for more than one Q. So

am s

y1 Qg 1 q g 2

gives the desired solution to Ž3. unless Qg 1 q g 2 happens to be 0. If this were the case, then X r, s Ž wi .m s 0 for all i. In other words, Ž w 1 .m , . . . , Ž wf .m 4 is an embedding of Sn Ž n y 1. in Sn Ž n y 1. ŽCondition Ž2. holds.. Furthermore, if there is such an embedding, then Qg 1 q g 2 must be 0. So, Condition Ž1. holds if and only if Qg 1 q g 2 / 0, and Condition Ž2. holds if and only if Qg 1 q g 2 s 0. Therefore, exactly one of these conditions holds. Proof of Claim. We now prove the technical claim. For each m e l, with nrm not in a column, there exists an i such that Ž X r, s¨ i .m / 0. Since  ¨ 1 , . . . , ¨ f 4 spans the S n component of VrW, all that is needed is to exhibit an element ¨ in the S n component such that Ž X r, s¨ .m / 0. Since X 1, 2 s s X r, s sy1 for some s g SymŽ n y 1., Ž X r, s¨ .m /0 m Ž s Ž X r, s¨ ..m / 0 m ŽŽ s X r, s sy1 .Ž s ¨ ..m / 0 m X 1, 2 Ž s ¨ .m . Thus, it suffices to show Ž X r, s¨ .m / 0 for just one pair  r, s4 . Since n & n y 1, Sn Ž n y 1. is essentially the SymŽ n y 1.-module S n . Thus given an element Ti, n m ¨ for VrW where ¨ g S l, en ? Ž Ti , n m ¨ . is in the Sn Ž n y 1. component of VrW. There is a vector ¨ in the S m-component, for which ¨ s em ? ¨ . By moving em through the tensor product, we get en ? ŽTi, n m ¨ . s Ž en ? Ti, n ? em . m ¨ . Thus, it suffices to

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show that X r , s Ž en ? Ti , n ? em . / 0 for some choice of r, s, and i. We are assuming that the boxes of nrm are in different columns. We visualize the terms in en ? Ti, n ? em as fillings of the Ferrers diagram of n . Start by associating the boxes in w n x with the first n y 1 points in any fashion, subject to the constraint that the boxes in the skew diagram nrm are associated with point i Žthe point connected to point n. and the point in the top that is connected to point n y 2 in the bottom. Now given a term in en ? Ti, n ? em , its associated diagram is obtained by looking at points 1 to n y 1 in the top and in its associated box, writing the number i if it is connected to point i in the bottom for 1 F i F n y 3, writing a if it is connected to point n in the top, and writing b if it is connected to point n y 2 in the bottom. For example, with n s 10, n s Ž4, 3, 1, 1., and m s Ž4, 2, 1., the diagram might be the following: 4

7

6

5

1

a

2

3 b

The idempotent en g Cw SymŽ n y 1.x acts on these diagrams by permuting positions, and em g Cw SymŽ n y 3.x acts by permuting values. In this framework, the row and column stabilizers used to define en and em are clear. They are just the row and column stabilizers of the diagram. Now we take a particular filling of the boxes of m with the top of Ti, n so that the diagram has 1 to n y 3 entered in order from left to right, then top to bottom. Then place a and b in lrm so that a is in a column to the right of b. The diagram looks like the following: 1

2

m1 q 1 .. . .. .

.. . b

my3

???

m1 m2

??? a

SEMISIMPLICITY OF BRAUER ALGEBRAS

669

Choose r, s to be the values of points associated with the two boxes nrm. The first thing to notice is that the only terms in en ? Ti, n ? em that are not zeroed by X r, s have a in nrm, since otherwise the resulting multiplication would have two horizontal lines and thus would be zero. When a term has a in one of the boxes of nrm, the effect of X r, s is to replace the edge in the top from point n to one of the points corresponding to nrm with an edge between the points r and s, and replace the vertical edge that uses the other point in nrm with one between point n in the top and the same point in the bottom:

To show that X r, s Ž en ? Ti, n ? em . is not zero, we examine just the coefficient on X r, s Ti, n . In the picture above, X r, s Ti, n has a line from a to b, and all other entries are as displayed. Suppose X r , s Ž Ž sgn Ž s . st . Ti , n Ž sgn Ž s * . s *t * . . s X r , s Ti , n ,

Ž 4.

where s g Cn , t g Rn , s * g Cm , and t * g Rm . There does exist such a choice of Ž s , t , s *, t *.: all four being the identity. For Ž4. to hold, the effect of Ž s , t , s *, t *. on Ti, n must be to either flip the positions of a and b or leave them fixed, and leave all of the other boxes fixed. Only s and t affect the boxes that initially contain a and b. Recall that a and b are in different columns. If they are also in different rows, then there is no choice of s and t that flips a and b. If a and b are in the same row, then it is possible for t to flip the two. However, s cannot move either of these positions. There are many ways for s < mt < m s *t * s id. In each case, sgnŽ s N m . s sgnŽ s *.; in fact, they must have the same cycle type. Since s cannot move either of the boxes in nrm, sgnŽ s . s sgnŽ s N m . s sgnŽ s *.. Thus, all of the terms that contribute to the coefficient on X r, s Ti, n have positive sign, and there is at least one. So, X r, s Ž en ? Ti, n ? m . / 0, which shows the claim. We can now use Theorem 4.2 to determine the top and bottom constituents of SlŽ n.x for l & n y 2. A bottom constituent G of a module U is a submodule U1 of U for which restriction to U1 is G. Theorem 4.1 proves that SmŽ n y 1. is a bottom constituent of SlŽ n.x if m e l. If n d l, we have shown that Sn Ž n y 1. is also a bottom constituent if there are no

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m e l for which nrm satisfies Žw., as it splits off as a direct summand of SlŽ n.x. However, if there is such a m , then Sn Ž n y 1. is also a bottom constituent, as it is embedded in SmŽ n y 1.. In particular, this proves that m e l or m d l, then ² SmŽ n y 1., SlŽ n.x: s 1. This analysis also enables us to determine the top constituents of SlŽ n.x. A top constituent G of a module U is a quotient of U that is isomorphic to G. That is, there exists a submodule U1 of U for which UrU1 affords G. The top constituents of SlŽ n.x are, of course, all Sn Ž n y 1. where n d l. Furthermore, if there are no n d l for which nrm satisfies Žw., then SmŽ n y 1. is a top constituent, as all Sn split from it. On the other hand, if there is some n d l for which nrm satisfies Žw., then SmŽ n. is not a top constituent of any indecomposable constituent by Theorem 4.2 and so cannot be a top constituent of V. This means all SmŽ n y 1. for m e l or m d l are bottom constituents. However, not all SmŽ n y 1. are top constituents. We summarize: COROLLARY 4.3. Ž1. Ž2.

Suppose l & n y 2.

For all m d l or m e l, ² SmŽ n y 1., SlŽ n.x: s 1. For all n d l, ² SlŽ n.x, Sn Ž n y 1.: s 1.

Ž3. If m e l and for all n d l, Ž n , m . does not satisfy Žw.; then SlŽ n.x has SmŽ n y 1. as a top constituent. If there is such a n , then it does not. Analyzing the above proves the following corollary. COROLLARY 4.4. For l & n y 2 and n d l, Sn Ž n y 1. is a direct summand of V s SlŽ n.x if and only if there is no m e l for which nrm satisfies Žw.. Remark. If l & n, then SlŽ n.x is the restriction for S l as a SymŽ n.module and thus is isomorphic to [m e l SmŽ n y 1..

5. THE FUNCTORS F AND G In this section we introduce the functors F and G, which relate Bny 2 Ž Q . modules to those of BnŽ Q .. The ideas make use of the work of Green wG, Section 6.2x and were fully utilized by Martin wMx. These are general arguments that apply to algebras with an associated idempotent. Here we use the idempotent e s Ž1rQ. X ny 1, n . When Q / 0, we cannot divide by 0, and so we define the functors differently and provide different arguments when necessary.

SEMISIMPLICITY OF BRAUER ALGEBRAS

671

Let BnŽ Q .-mod be the category of BnŽ Q .-modules and assume n ) 2. If Q / 0 we define F and G by F : Bn Ž Q . -mod ª Bny2 Ž Q . -mod M ¬ eM and G: Bny 2 Ž Q . -mod ª Bn Ž Q . -mod M ¬ Bn Ž Q . e mB ny 2 ŽQ. M. Properties of these functors can be found in Green wGx. The composition FG is the identity. The functor F also has the property of being exact. That is, if 0 ª V1 ª V2 ª V3 ª 0 is an exact sequence of BnŽ Q .-modules, then 0 ª F Ž V1 . ª F Ž V2 . ª F Ž V3 . ª 0 is an exact sequence of Bny2 Ž Q .modules. When Q s 0, the functor F is defined as above, except that we replace e by X ny 1, n . As Bny2 Ž Q . commutes with X ny1, n , X ny1, n M is a submodule of M. To define G, let HnŽ Q . be the left ideal of BnŽ Q . spanned by diagrams for which n y 1 in the bottom is joined to n in the bottom. Notice that if Q / 0, this is just BnŽ Q . e, as above. In any case, it is BnŽ Q . X ny1, n . Again as X ny1, n commutes with Bny2 Ž Q ., there is a natural right action of Bny 2 Ž Q . on HnŽ Q .. For Q s 0 and M a left Bny2 Ž Q . module defines GŽ M . s Hn mB ŽQ. M. ny 2 It is shown in wG, 6.2 d x that when Q / 0 and m / 0 in M, then e m m / 0. In any case, notice there is a map from HnŽ Q . m M to M by taking d 1 m m to Ž X ny1, n d 1 .o for any diagram d 1 in HnŽ Q .. Here Ž X ny 1, n d 1 .o is Ž X ny1, n d 1 ., with the Ž n y 1.st and nth vertices removed in both the top and the bottom. Now take d * s Ž n y 2, n y 1. X ny 1, n . The image of d * m m is m, and so d * m m is not 0 in GŽ M .. We now investigate some further properties of G. LEMMA 5.1.

Let w : M1 ª M2 be a Bny2 Ž Q .-homomorphism. Then

w#: G Ž M1 . ª G Ž M2 . d m m ¬ d m w Ž m. is a BnŽ Q .-homomorphism for elements d in HnŽ Q .. Moreo¨ er, if w is nonzero, then so is w#. Furthermore, there are elements d * in HnŽ Q . for which d * m m / 0 for any m / 0 in M1 or in M2 . Proof. Take d * as defined immediately above the statement of the theorem. If Q s 0 we can also take d * s e. Let d 1 be 1-factors in BnŽ Q .,

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let d be a diagram in HnŽ Q ., and let m g M1. Then the computation

w# Ž d 1 Ž d m m . . s w# Ž Ž d 1 d . m m . s Ž d1 d . m w Ž m. s d 1Ž d m w Ž m . . s d 1 w# Ž d m m . shows that w# is a BnŽ Q .-homomorphism. If w is not zero, then w Ž m1 . s m 2 / 0 for some m1 g M1 and m 2 g M2 . Now w#Ž d * m m1 . s d * m w Ž m1 . s d * m m 2 / 0, which shows that w# is also not zero. We show that F and G connect SlŽ n y 2. and SlŽ n. when < l < F n y 2. PROPOSITION 5.2. GŽ SlŽ n y 2.. s SlŽ n. and GŽ SlŽ n y 2.x. s SlŽ n.x. Proof. We do the case Q / 0 first. mX Ž . Claim. InmX Ž Q . ( BnŽ Q . e mB ŽQ. Iny2 Q . First, we examine the eleny 2X m ments in U ( BnŽ Q . e mB ŽQ. Iny2 Ž Q .. Suppose d 1 e m d 2 g U. Let « be mX Ž . ny 2 the element in Iny 2 Q for 1 F i F m, where point i on the top is joined to point i on the bottom. The remaining horizontal lines are also directly above one another in positions Ž m q 1, m q 2., Ž m q 3, m q 4., . . . , Ž n y 3, n y 2.. In particular, the top and the bottom are Ž1. with one line removed. For n s 9 and m s 3,

Choose a s g Sny 2 , for which sd 2 s « . This gives d 1 e m d 2 s d 1 sy1 e m « . Since X i, iq1 « s Q « for i s m q 1, m q 3, . . . ,

d 1 e m d 2 s Ž 1rQ k . d 1 sy1 e m Ž X mq1, mq2 X mq3, mq4 ??? X ny3, ny2 . « s d 1 sy1 Ž X mq1, mq2 X mq3, mq4 ??? X ny3, ny2 . e m « . Thus, any element of U can be written as a linear combination of terms of the form d 3 e m « , where d 3 has horizontal lines Ž m q 1, m q 2., . . . , Ž n y 3, n y 2., Ž n y 1, n. in the bottom. Furthermore, if there other horizontal lines, say Ž i, j . 1 F i, j F m, in the bottom, then d 3 e m « s Ž1rQ. d 3 X i, j e m « s Ž1rQ. d 3 e m X i, j « s Ž1rQ. d 3 e m 0 s 0. Thus, we need only consider d 3’s with the bottom being Ž1.. It is routine to show that the map d 3 e m « ¬ d 3 gives the desired isomorphism.

SEMISIMPLICITY OF BRAUER ALGEBRAS

673

Now a computation gives the result: mX G Ž Sl Ž n y 2 . . s Bn Ž Q . e mB ny 2 ŽQ. Ž Iny2 Ž Q . mS y mŽ m. S l . mX s Ž Bn Ž Q . e mB ny 2 ŽQ. Iny2 Ž Q . . mS y mŽ m. S l

s InmX Ž Q . mS y mŽ m. S l s Sl Ž n . . A similar computation gives GŽ SlŽ n y 2.x. s SlŽ n.x. We do the case Q s 0 in a similar way. Just as above, the essence is to mX Ž Q . Iny2 investigate HnŽ Q . mB . We want to show that this has the ny 2 same dimension as Inm Ž Q .9. In particular, suppose d 1 is a diagram in HnŽ Q .. This means it has a line joining n y 1 and n in the bottom. m Consider d 1 mB ŽQ. d 2 . Here d 2 is in Iny2 . We will show many of these ny 2 terms are zero because of the tensor product action. We will start by showing we need only take d 1 as a diagram in Inny 2 . To be in HnŽ Q . means it has a line from n y 1 in the bottom to n in the bottom, so the bottom and top each have at least one horizontal line. Suppose there are more with i, j as a line in the bottom other than the one from n y 1 to n. Let i9, j9 be a line on the top. Let d 1X be a diagram that is the same as d 1 , except that the lines from i9 to j9 and from i to j are replaced by vertical lines joining i to i9 and j to j9. Notice d 1 s d 1X X i, j . Now d 1 m d 2 s d 1X X i, j m ¨ i s d 1X m X i, j ¨ i . Continue until the resulting d 1U has n y 2 vertical lines. We can now act by an element of SymŽ n y 2. acting on topŽ d 2 ., so the lines are all lined up directly above the lines in Ž1., which are in positions Ž m q 1, m q 2., Ž m q 3, m q 4., . . . , Ž n y 3, n y 2. and act on the other side of the tensor with the inverse. Act further so that all of the vertical lines join j on the bottom to j on the top for 1 F j F m. We want to identify this d 1 m d 2 with an element of f Ž d . s , where d is an Ž m, k . partial 1-factor where m q 2 k s n. Notice d 1 already has one horizontal line in its top, which we take to be one of the lines in d. For the others identify topŽ d 2 . with bot Ž d 1 . in positions mq1, mq2, . . . ,n y 2. Let the k y 1 further lines for d be the images under d 1 of the lines in topŽ d 2 .. This gives k y 1 further lines and so a d with the right parameters. Now identify the elements i in topŽ d 2 . with the elements i in bot Ž d 1 . for 1 F i F m. This gives an element in InmX . The dimension is correct. It is only left to show that the left actions are the same. This is certainly true for elements of SymŽ n., as the points and lines are permuted in the same way. Now examine the action of X i, j . Once these agree the action is the same, as these generate BnŽ Q .. We first show that when i, j is a line in d, the action gives 0. This is certainly true for X i, j f Ž d ., as it is multiplied

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DORAN, WALES, AND HANLON

by Q, which is 0. We must consider X i, j Ž d 1 m d 2 . s Ž X i, j ? d 1 . m d 2 . If i, j is the horizontal line in topŽ d 1 ., the product X i, j ? d 1 s Q d 1 s 0. If i, j is one of the other horizontal lines in d, it means that i, j are joined by vertical lines in d 1 to, say, l, l q 1, which is one of the lines in topŽ d 2 .. Now X i, j d 1 s d 1 X l, lq1. Now passing this through the tensor gives d 1 m X l, lq1 d 2 . As Q s 0, this last term is 0. The other situation giving 0 in X i, j f Ž d . is when i, j are both isolated in d. In this case X i, j ? d 1 s 0. In this case neither i nor j is part of a line in d 1 , and so both are joined to, say, i9, j9 in bot Ž d 1 ., where i9 and j9 are at most m. Again X i, j d 1 s d 1 X i9, j9 and passing through the tensor gives 0 as X i9, j9 ? d 2 s 0. The remaining cases involve i, j, for which at least one of i, j Žsay i . is part of a line in d. In this case X i, j f Ž d . s Ž i, s . f Ž d .. It is only necessary to show that the same is true for X i, j Ž d 1 m d 2 .. Again one must distinguish between the different cases for i. Either i is part of the one horizontal line in topŽ d 1 . or it is an i in topŽ d 1 . that is joined to a point i9 that is part of a line in topŽ d 2 .. A short computation relating the various transpositions involved shows that this action is the same and we are done. PROPOSITION 5.3. Suppose < l < F n y 2. Then F Ž SlŽ n.. s SlŽ n y 2. and F Ž SlŽ n.x. s SlŽ n y 2.x. Proof. In Q / 0 we know from Proposition 5.2 that GŽ SlŽ n y 2.. s Ž Sl n.. Applying F to both sides and using the fact that FG is the identity gives SlŽ n y 2. s F Ž SlŽ n... The second statement is proved similarly. We give a different argument for Q s 0, which applies equally well for Q / 0 and is more direct than the above. Recall that SlŽ n. is spanned by InmX mS y mŽ m. ¨ i , where ¨ i is a basis for S l. If d is a diagram in Inm , consider X ny1, n ? d . This is zero if d has a line between n y 1 and n or if n y 1 and n are both free. If not it is Ž i, j . d , where j is n y 1 or n and i - n y 1. This element can be considered in mX Iny 2 by just taking away the four right-hand nodes that join n y 1 to n in top and bottom. In particular, X ny 1, nŽ d = ¨ i . is naturally an element of mX SlŽ n y 2.. It is onto, as any element d 1 in Iny2 can be obtained by mX extending it to In by adjoining four nodes with n y 1 and n joined in both the top and bottom, and then acting by a transposition that moves n y 1 in the top to j in the top for any j - n y 2. THEOREM 5.4. Let m & n and n 0 be a nonnegati¨ e e¨ en integer, and if Q s 0, suppose m is not the empty partition Ž i.e., assume m / 0.. Then ² SmŽ n q n 0 ., Sm 9Ž n q n 0 .: / 0 if and only if SmŽ n. can be embedded in Sm9Ž n.. Proof. Suppose first Q / 0 and suppose ² SmŽ n q n 0 ., Sm 9Ž n q n 0 .: / 0. Let w map SmŽ n q n 0 . to Sm 9Ž n q n 0 . with W the image of w . Apply the functor F with e s Ž1rQ. X nq n 0y1, nqn 0 . If eW / 0, this gives a non-

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SEMISIMPLICITY OF BRAUER ALGEBRAS

trivial map from SmŽ n q n 0 y 2. to Sm 9Ž n q n 0 y 2., and we can apply induction to get a nontrivial map from SmŽ n. into Sm 9 Ž n.. It is an embedding as SmŽ n. is irreducible. We suppose then that eW s 0. Notice W is a Bnq n 0Ž Q . submodule of Sm 9Ž n q n 0 ., which is annihilated by X nq n 0y1, nqn 0 . By applying elements of SymŽ n q n 0 . we see X i, jW s 0 also for any i, j. In particular, X s X 1, 2 X 3, 4 ??? X n 0y1, n 0W s 0. Because W is the image of w , W ( SmŽ n q n 0 .rU for some submodule U of Snq n 0 . By the results of wHW1x, SmŽ n q n 0 . contains a unique n Ž . maximal submodule annihilated by all elements in Bnq n 0 Q , particularly Ž . by X. Suppose first Q / 0. Let y be the n, n 0r2 partial 1-factor with lines Ž1, 2., Ž3, 4., . . . , Ž n 0 y1, n 0 . and with free points n 0 q1, n 0 q2, . . . , n 0 q n. Notice that ¨ s f Ž y . m sk is a nonzero element of SmŽ n q n 0 .. Notice also that X¨ s Q n 0 r2 ¨ / 0. This means ¨ is not in U. Also X Ž ¨ q U . s Q n 0 r2 Ž ¨ q U ., which is not 0 modulo U. This is a contradiction and shows eW / 0, which does this part. In the case Q s 0, recall that m is not empty. Now let y have free point 1, lines between Ž2, 3., Ž4, 5., . . . , Ž n 0 , n 0 q 1., and let the remaining points be free. As above, let ¨ s f Ž y . m sk . Now X¨ / 0 and argue as above. For the other direction, given an embedding of SmŽ n. in Sm 9Ž n., apply the functor G n 0r2 times. By Proposition 5.2, this gives a map from SmŽ n q n 0 . to Sm 9Ž n q n 0 .. By Lemma 5.1, this map is not zero, which shows the other direction. It should be noted that for n 0 ) 0, these maps obtained by applying G are not necessarily embeddings. THEOREM 5.5. SlŽ n.x: s 0.

Let l & m F n. If m e l or m d l, then ² SmŽ n y1.,

Proof. Proof by induction on n y m. We have already considered the cases n y m s 0 or 2 in Corollary 4.3 and the remark following it, so suppose n y m G 4. Recall n y m is even. Suppose m e l or m d l. By the induction hypothesis, ² SmŽ n y 3., SlŽ n y 2.x: / 0. Thus there is a nonzero homomorphism f : SmŽ n y 3. ª SlŽ n y 2.x. By Lemma 5.1, the homomorphism f#: GŽ SmŽ n y 3.. ª GŽ SlŽ n y 2.x. is nonzero and, by Proposition 5.2, f#: SmŽ n y 1. ª SlŽ n.x. So, ² SmŽ n y 1., SlŽ n.x: / 0. Notice that for < l < s n or n y 2, this occurs because SmŽ n y 1. is embedded in SlŽ n.x. For smaller < l < this need not be true. However, a homomorphic image of SmŽ n y 1. is embedded in SlŽ n.x. There is a stronger statement here that holds for certain values of m. It makes use of Corollary 4.4 and some properties of the tensor product. In general, if W1 is a Bny1Ž Q . submodule of W, the universality of the tensor product gives a map from BnŽ Q . mB ŽQ. ŽW1 . into BnŽ Q . mB ŽQ. ŽW .. ny 2

ny 2

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DORAN, WALES, AND HANLON

However, it is not, in general, an embedding. However, if W1 is a direct summand, this is indeed the case. In particular, if W ( W1 [ W2 , then BnŽ Q . mB ŽQ. ŽW . ( BnŽ Q . mB ŽQ. ŽW1 . [ BnŽ Q . mB ŽQ. ŽW2 .. ny 2

ny 2

ny 2

This gives rise to a strengthening of the previous theorem for certain m. THEOREM 5.6. Suppose l & m and m d l. Suppose further that there are no m9e l for which Ž m , m9. satisfies Žw.. Then SmŽ n y 1. is a direct summand of SlŽ n.x. If, on the other hand, m9e l and there are no m d l for which Ž m , m9. satisfies Žw., then Sm9Ž n. is a direct summand of SlŽ n.x. Proof. This follows from Corollary 4.4 and the property stated above.

6. INDUCTION IN THE BRAUER ALGEBRA Given a BnŽ Q .-module M, define the induced Bnq1Ž Q .-module M ­ to be Bnq 1Ž Q . mB ŽQ. M. In the semisimple case, n

Sl Ž n . ­ (

[ Sm Ž n q 1. [ [ Sn Ž n q 1. .

me l

ndl

As with restriction, when Bnq 1Ž Q . is not semisimple, such a direct sum decomposition need not exist. Here is a very nice method for connecting induction and restriction using the functor G. THEOREM 6.1. Let M be any BnŽ Q .-module. Then M ­ ( GŽ M .x as Bnq 1Ž Q .-modules. Proof. First, we show that Hnq 2 Ž Q . ( Bnq1Ž Q . as Bnq1Ž Q . = BnŽ Q .modules, where Bnq 1Ž Q . acts on the left and BnŽ Q . acts on the right. The isomorphism is given taking d g Bnq 1Ž Q ., moving point n q 1 in the bottom of d along with its adjacent edge to the point n q 2 in the top row and adding an edge in the bottom between points n q 1 and n q 2. The rest of the diagram remains the same. For example, gets mapped to It is straightforward to check that this is a Bnq 1Ž Q . = BnŽ Q .-isomorphism. The key is to notice that the ‘‘moved’’ point is effected by neither the right

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677

BnŽ Q . action when in the bottom of d nor by the left Bnq1Ž Q . action when in the top when mapped into Hnq 2 Ž Q .. Thus its physical location does not matter. Now recall that M ­ ( Bnq 1 mB ŽQ. M. Furthermore, GŽ M . ( n Hnq 2 mB ŽQ. M when Q s 0 and Bnq2 Ž Q . e mB ŽQ. M when Q / 0. In n n the second case, Bnq 2 Ž q . e is Hnq2 Ž Q .. Since both of the tensor products are right BnŽ Q .-actions, under this Bnq1Ž Q . = BnŽ Q .-isomorphism, we get Bnq 1 mB n ŽQ. M ( Hnq2 Ž Q . mB n ŽQ. M or Bnq 1 mB n ŽQ. M ( Bnq2 Ž Q . e mB n ŽQ. M as left Bnq 1Ž Q .-modules. This means M ­ ( GŽ M .x. This argument works both when Q s 0 and when it is not. COROLLARY 6.2.

Let < l < F n. Then Sl Ž n . ­ ( Sl Ž n q 2 . x

as Bnq 1Ž Q .-modules. Proof. Apply the previous theorem in the case M s S lŽ n. along with Proposition 5.2. Applying Theorem 5.5 to this gives the following result. THEOREM 6.3. Let m & m F n. The ¨ alue of ² SlŽ n q 1., SmŽ n.­: is not zero if m e l or m d l. To investigate induction in the next section, we examine the case l & n with a little more case. We find results analogous to Theorems 4.1 and 4.2. The analogue of Theorem 4.1 is quite easy to prove. COROLLARY 6.4. Let l & n and V s SlŽ n.­. As a Bnq1Ž Q .-module, V contains a submodule W ( [m e l SmŽ n q 1.. Furthermore, VrW ( [n d l Sn Ž n q 1..

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Proof. Combining Corollary 6.2 and Theorem 4.1 gives a proof. As in Theorem 4.2 there are two cases. For a given n d l, either Sn Ž n q 1. splits off from Sl ­ or the module Sn Ž n q 1. appears as a submodule of SmŽ n q 1. for some m e l. We know by Theorem 4.2 that exactly one of these occurs. Furthermore, from the previous section, the second occurs if and only if Ž m , n . satisfy condition Žw. and the two boxes in nrm are in different columns. Theorem 6.1 provides a method for connecting induction and restriction. Information about restrictions from Bnq 2 Ž Q .-modules is related to induction from BnŽ Q .-modules. In particular, when l & n and m e l, ² Sl Ž n . ­, Sm Ž n q 1 . : s ² Sl Ž n . , Sm Ž n q 1 . x: s 1. While this follows from Frobenius reciprocity, the knowledge about the structure of SlŽ n.­ uses Theorem 6.1. If one of the n d l has nrm satisfying Žw., this comes because Sn Ž n. is a top constituent of Sl ­ and a bottom constituent of SmŽ n q 1.. However, if none of the n d l has nrm satisfying Žw., SmŽ n q 1. is itself a top constituent of SlŽ n.­. We have proved the following. THEOREM 6.5.

Let l & n. Then

Ž1. For all n d l, ² SlŽ n.­, Sn Ž n q 1.: / 0. Ž2. If m e l and there does not exist a n d l such that Ž n , m . satisfy Žw., then ² SlŽ n.­, SmŽ n q 1.: / 0 7. FROBENIUS RECIPROCITY AND ² SlŽ n., SmŽ n.: / 0 WHERE l & n Besides Žw., there is another condition derived from Frobenius reciprocity that must be satisfied for ² SlŽ n., SmŽ n.: / 0 for l & n. THEOREM 7.1. Suppose l & n and ² SlŽ n., SmŽ n.: / 0. Then for e¨ ery l8e l, ² Sl8Ž n y 1., SmŽ n.x: / 0. Furthermore, there is a m8e m or m8d m such that ² Sl8Ž n y 1., Sm8Ž n y 1.: / 0. Proof. Suppose ² SlŽ n., SmŽ n.: / 0. Take l8e l. From Theorem 6.5, ² Sl8Ž n y 1.­, SmŽ n.: / 0. By Frobenius reciprocity, this implies ² Sl8Ž n y 1., SmŽ n.x: / 0, giving the first statement. As in Theorem 4.1, let V be SmŽ n.x and let W be the submodule of V isomorphic to [m8e m Sm8Ž n y 1.. Let S be the image of Sl8Ž n y 1. in V. Suppose first S : W. Since l8 & n y 1, S is irreducible. Thus it must be a submodule of some Sm8Ž n y 1.. This implies ² Sl8Ž n y 1., Sm8Ž n y 1.: / 0.

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Suppose otherwise that S ­ W. Then Ž S [ W .rW ( S because S is irreducible and is not in W. This gives a map in homŽ Sl8Ž n y 1., VrW ., which is nontrivial, as S ­ W. Indeed, Ž S [ W .rW is isomorphic to Sl8Ž n y 1.. Now by Theorem 4.1, VrW (

[

m8d m

Sm8 Ž n y 1 . .

Now just as before, S [ WrW : Sm8Ž n y 1. for some m8d m. Now we have ² Sl8Ž n y 1., Sm8Ž n y 1.: / 0. Remark. It is important to notice that the value of Q does not change. If l & n and ² SlŽ n., SmŽ n.: / 0 for a given value of Q, then ² Sl8Ž n y 1., Sm8Ž n y 1.: / 0 for the same value of Q. 8. SEMISIMPLICITY OF BnŽ Q . FOR NONINTEGER Q We now have enough information to show that if Q is not an integer, BnŽ Q . is semisimple. This was first conjectured by Hanlon and Wales in wHW1x and proved by Wenzl in wWenx. This has consequences for the integrality of roots of certain determinants, as described in wHW3x. This proof is quite different from Wenzl’s proof in a number of ways. In particular, it does not depend on the integrality of roots of certain polynomials associated with the orthogonal groups. THEOREM 8.1.

If Q is not an integer, BnŽ Q . is semisimple

Proof. Suppose BnŽ Q . is not semisimple. We show that Q must be an integer. In wHW1, Sect. 4Bx it is shown that BnŽ Q . not being semisimple implies ² SlŽ n., SmŽ n.: / 0 for some l and m with < m < - < l < F n. This is also inherent in wGL, 3.8iix. By Theorem 5.4, we can assume l & n when Q / 0. If Q s 0, we are done. Set k s < lrm
1 k

Ý

cŽ p. ,

pg w l r m x

which is an integer. If k ) 1, repeatedly apply Theorem 7.1 to obtain a ˜rm pair l ˜ with < l˜rm ˜ < s 2. Since each application of Theorem 7.1 reduces the size of the larger partition by one, this process will terminate in such a pair. This reduces the problem to the case k s 1, which we have just handled.

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9. THE CASE m s B The necessary conditions given in previous sections are not sufficient for showing the existence of an embedding. In general, we do not have a method for determining whether SlŽ n. embeds in SmŽ n.. However, we can treat the special case when m s B. Let a b denote the partition with b parts, all of which are equal to a. COROLLARY 9.1. If l & n, ² SlŽ n., SBŽ n.: / 0, then l s a b , where a is e¨ en and Q s b y a q 1. This means l has only one corner. Proof. Suppose l has two or more outer corners. Let p1 and p 2 be two of the outer corners of l. Applying Theorem 7.1 to l _ pi s l8e l, the only choice for m8 is the unique partition of 1, Ž1.. Thus, both Ž l _ p1 , Ž1.. and Ž l _ p 2 , Ž1.. satisfy condition Žw.. Hence, by subtraction, cŽ p1 . s cŽ p 2 ., which is a contradiction. Therefore, l must be a rectangle. We know from Theorem 3.1 that l must be an even partition on n if S l occurs as a constituent of SBŽ n. as a SymŽ n.-module. This means a is even. Now suppose n s ab and l s a b with a even. A basis for SBŽ n. is indexed by 1-factors on n vertices. The 1-factor corresponds to the top, the bottom is given by Ž1., and there are no vertical lines. When decomposed as a SymŽ n.-module, SBŽ n. contains S l with multiplicity 1. Let W be the unique copy of S l contained in SBŽ n.. Following the proof of the technical claim in Theorem 4.2, we visualize the basis of SBŽ n. as 1-factors of the boxes of the Ferrers diagram of a b. The points of the top row are associated with the boxes of the Ferrers diagram so that points 1 to a correspond to the boxes in the first row from left to right, points a q 1 to 2 a correspond to the boxes in the second row also from left to right, and so on. Note that the points n y 1 and n correspond to the two rightmost boxes in the bottom row. For example, with a s 4 and b s 3, the element

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681

is visualized as

In the following proof, the idempotent el is used. Following these diagrams, the obvious choices are made for the row and column stabilizers. THEOREM 9.2. Suppose Q s b y a q 1. Then X i, j w s 0 for all w g W and 1 F i - j F n. This shows that SlŽ n. embeds in SBŽ n.. Proof. We begin with a reduction that is for notational convenience. By definition of X i, j , it is easy to see that

s X i , j sy1 s Xs Ž i., s Ž j. for all s g SymŽ n.. Choose s so that s  i, j4 s  n y 1, n4 . Then X i, jW s 0 m s X i, j sy1 W s 0 m X ny1, nW s 0. So, it suffices to show that X ny1, nW s 0. Now consider W as a SymŽ n y 2. = SymŽ2.-module by restriction. We view SymŽ n y 2. as acting on the first n y 2 points and SymŽ2. on the last 2. By the Littlewood]Richardson rule and Frobenius reciprocity, we have W ( Y [ Z, m

where Y ( S m S with m s a by2 , Ž a y 1. 2 and Z ( S n m S Ž2. with by 1 n s a , a y 2. Here we are using the notation a by2 , Ž a y 1. 2 to denote the partition with b y 2 parts a, and a y 1 twice. Similarly a by 1 , a y 2 is the partition with b y 1 parts a and one part a y 2. Since S Ž1, 1. is the sign representation Ž1, 1.

Ys

1 2

Ž id y Ž n y 1, n . . Y .

Note that X ny 1, n s X ny1, nŽ n y 1, n.. So, X ny 1, n Y s  X ny1, n Ž id y Ž n y 1, n . . 4 Y s 0. So, it suffices to show that X ny 1, n Z s 0.

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Let U be the subspace of SBŽ n. spanned by all 1-factors in which point n y 1 and n are adjacent. Notice that U is a SymŽ n y 2. = SymŽ2.-module, where again SymŽ n y 2. acts on the first n y 2 points and SymŽ2. on the last 2. The decomposition of U as a SymŽ n y 2. = SymŽ2.-module is

[

h&ny2 h is even

Sh m S Ž2. .

In particular, U contains exactly one copy of the irreducible S n m S Ž2., ˆ which is denoted Z. The map X ny 1, n takes W to U. Moreover, it is a Ž SymŽ n y 2. = ˆ Notice the imporSymŽ2..-equivariant map. Thus it must map Z into Z. ˆ tance of the uniqueness of Z in W and Z in U. If this map is zero on Z, we are done. Otherwise, by Schur’s lemma, it is a scalar. Up to this point, nothing in this proof has depended on the value of Q. However, the value of this scalar is a function of Q and is denoted a Ž Q .. Pictorially, this is W X ny1, n , a Ž Q . I

Zˆ

6

Z

U

Let d be the special element of SBŽ n. whose top is Ž1.. For a s 4 and b s 3, the diagram of d is

We show below that the coefficient on d in X ny 1, n el d is zero when Q s b y a q y 1.

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683

We use this computation to show that a Ž Q . s 0 for Q s b y a y 1. We know that el d g W because W is isomorphic to S l as a SymŽ n.-module. Let z1 , . . . , z d be a basis for Z and ˆ z1 , . . . , ˆ z d be the corresponding basis for Zˆ such that X ny 1, n z d s a Ž Q . ˆ z d . Let el d s y q Ý c i z i , where y g Y and c i g C. Then X ny 1, n el d s 0 q a Ž Q . Ý c i ˆ zi . Let m i denote the coefficient of d in ˆ z i . The coefficient of d in X ny1, n el d is

a Ž Q . Ý ci mi . We show below that this is a nonzero multiple of Q y Ž b y a q 1.. Hence so is a Ž Q .. We now examine the coefficient of d in X ny 1, n el d . Recall el d s X ny 1, n el d s

fl n!

Ý

Ý

sgn Ž s . std

Ý

Ý

sgn Ž s . X ny1, n std .

sgCl tgR l

fl n!

sgCl tgR l

We need only consider s , t for which X ny 1, n std s Q jd . The power j is at most 1, as X ny 1, n has only one horizontal line. We ignore the coefficient f lrn!. If std has a line joining n y 1 and n, X ny 1, n std s Q std . This is Q d if and only if std s d . We count the number of pairs Ž s , t . for which std s d . Let R be the subgroup of Rl that fixes d . Suppose its size is r. If std s d , t must be in R. For if not, some line of td would not be in positions 2 i q 1, 2 i q 2 and no column permutations could put them in such positions. But all lines in d are in such positions. So for std s d , we must have t g R. This means td s d . Now we must count the number of s for which sd s d . For this to be so, the column permutations acting on d must map horizontal lines to horizontal lines. In particular, if s is the permutation in column 2 t q 1, the permutation in column 2 t q 2 must be the same. Suppose a s 2 a9. Then we may have any permutation in each odd column, as long as the same permutation occurs for the column to the right. There are b! such permutations for each odd column, and so Ž b!. a9 such terms. The total number is r Ž b!. a9. The sign is always q, as the permutations come in pairs. The total contribution is Q ? r ? Ž b!. a9. Suppose now that std does not have a line joining n y 1 and n. Then std has lines from n to i and from n y 1 to j, say. Here j and i are

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smaller than n y 1. Now X ny 1, n std has lines from i to j and from n to n y 1. Apart from these nodes, X ny 1, n std is the same as std . If this is to be d , i and j must be in the same row in positions 2 t q 1 and 2 t q 2. All of the remaining terms of X ny 1, n std are the same as in d . We must count the number of pairs s , t that give this. The situation is different if the images of n and n y 1 are not in the two rightmost columns. Suppose this for now. In particular, suppose 2 i q 1 and 2 i q 2 are adjacent in the columns s and s q 1, where s is odd and s F a y 3. Suppose s and t are such that std has lines from the pair n y 1, n to the pair 2 i q 1, 2 i q 2, and all other lines are as in d . Looking at td we see t that has interchanged the pair in the last two columns of some row, say row k, with a pair 2 t q 1, 2 t q 2 in the same columns as 2 i q 1, 2 i q 2. Applying s to td puts these lines into the proper rows. In particular, 2 t q 1 is mapped to 2 i q 1 and 2 t q 2 is mapped to 2 i q 2. Moreover, the points above n y 1, n in the kth row are mapped to n y 1, n. There are a9 y 1 such pairs of columns. There are two different ways to do each one: interchanging the term in the n y 1st column with the term in the column s or with the term in the column s q 1. This could have been from any of the b columns. All other column permutations must act on the lines together. The number of these not in the column containing 2 t q 1, 2 t q 2 and the last two is Ž b!. a9y2 . The total number of possibilities is r Ž b!. a9y2 ? ŽŽ b y 1.!. 2 ? Ž a9 y 1. ? 2 ? b 2 s 2Ž b!. a9Ž a9 y 1.. We now must account for the lines that go from n y 1, n to a pair i, i q 1, where i and i q 1 are in the last two columns. We are assuming X ny 1, n std s d , and std is the same as d , except that there is a line from n, n y 1 to a pair of points in the last two columns. As std has no vertical lines, n must be joined to a point in row s, but columns a y 1 and n y 1 must be joined to a point in row s and column a. As above, t must be in R. Notice that if g is the transposition interchanging n with the point in column s above it, gst s d . Let S be the subgroup of the Cl that fixes d . It follows that s must be of the form gs 9, where s 9 is in S. This is in the same coset of S as the transposition interchanging n y 1 with the point above it in the sth row. There are b y 1 choices for the row. The size of S is Ž b!. a9. The contribution here is r ? Ž b!. a9 ? Ž b y 1.. Here sgnŽ s . s y1, because the sign of elements in S is even and sgnŽg . s y1. Adding all contributions gives r Ž Ž b! .

a9

Ž Q q 2 Ž a9 y 1. y Ž b y 1. . . s r Ž Ž b! . a9 Ž Q q a y b y 1. . .

If Q q a y b y 1 s 0 we must have X ny 1, nW s 0. If Q q a y b y 1 / 0, X ny 1, nW / 0.

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ACKNOWLEDGMENT We thank the anonymous referee for the many helpful suggestions and critiques provided. The many hours of work went far beyond the norm. The paper is much better because of these efforts.

REFERENCES wBrx wBrnx wDx wGLx wGx wHW1x wHW2x wHW3x wHW4x wJx wJKx wMx wNx wSx wWenx wWeyx

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