Weyl modules for the Schur algebra of the alternating group

Journal of Algebra 257 (2002) 168–196 www.academicpress.com

Weyl modules for the Schur algebra of the alternating group A. Henke ∗,1 and A. Regev 2 Department of Mathematics, The Weizmann Institute of Science, Rehovot 76100, Israel Received 8 November 2001 Communicated by Gordon James

Abstract Let F be the field of complex numbers and V = V0 ⊕ V1 a vector space over F with dim V0 = dim V1 . The symmetric group acts on V ⊗n by the sign-permutation ∗ (V , n) be the action [Berele, Regev, Adv. in Math. 46 (2) (1987)]. Let S ∗ (V , n) ⊂ SA corresponding Schur algebras of Sn and of An ⊂ Sn , respectively, where An is the alternating group [Regev, J. Algebra, in print]. Following the fundamental work of H. Weyl, the explicit decomposition of V ⊗n as an S ∗ (V , n)-module was given in [Berele, Regev, Adv. in Math. 46 (2) (1987)]. By applying the character theory and the representation ∗ (V , n)-module. theory of An we give here the explicit decomposition of V ⊗n as an SA  2002 Elsevier Science (USA). All rights reserved.

1. Introduction 1.1. Let F be the field of complex numbers and assume that all objects (modules, algebras, vector spaces) are finite dimensional. It should be noted that many of the arguments also work for a field of positive characteristic. Consider the * Corresponding author. Current address: University of Leicester, Leicester LEI 7RH, England,

UK. E-mail addresses: [email protected] (A. Henke), [email protected] (A. Regev). 1 Partially supported by the EC TMR network ‘Algebraic Lie Representations’ grant no. ERB FMRX-CT97-0100. 2 Partially supported by ISF Grant 6629 and by Minerva Grant No. 8441. 0021-8693/02/$ – see front matter  2002 Elsevier Science (USA). All rights reserved. PII: S 0 0 2 1 - 8 6 9 3 ( 0 2 ) 0 0 1 0 6 - 0

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vector space V = V0 ⊕ V1 where dim V0 = dim V1 = k and the n-fold tensor space W := V ⊗n with n
2. The symmetric group acts on W via place permutations with a sign and the corresponding representation is ϕ ∗ : F Sn → EndF (V ⊗n ). For example, let n = 2, σ = (1, 2) ∈ S2 and let y1 , y2 ∈ V be homogeneous (i.e. y1 , y2 ∈ V0 ∪ V1 ), then ∗ ϕ(1,2) (y1 ⊗ y2 ) = (−1)(deg y1 )(deg y2 ) (y2 ⊗ y1 ).

For a precise definition of this sign action see [1]. The Schur superalgebra is defined by
 S ∗ (V , n) = Endϕ ∗ (F Sn ) V ⊗n . More generally, given a subgroup G of the symmetric group Sn , the Schur superalgebra of G is given by
 SG (V , n) = Endϕ ∗ (F G) V ⊗n . Of course, SG (V , n) depends on the particular embedding of G in Sn . In the present paper we are interested in SA∗ (V , n), the Schur superalgebra of the alternating group A = An . 1.2. The aim of this paper is to understand the structure of W = V ⊗n as an For S ∗ (V , n) (i.e. for G = Sn ) this is well understood: it is a superalgebra generalization of the classical work of Schur and Weyl for V = V0 and V1 = 0. It involves the decomposition of W into a direct sum of its isotypic components and into the so-called Weyl modules. Moreover, these decompositions are given explicitly in terms of Young tableaux (see [1, Section 3]). The Schur– Weyl theory and its super-generalization in [1], both require a detailed description of the group algebra F Sn . In particular, an explicit decomposition of F Sn into a direct sum of minimal one-sided ideals is required, as well as a description in terms of tableaux for the generators of such left ideals. The analogue decomposition of W as an SA∗ (V , n)-module requires an analogue description for FAn . In Section 2 we review some generalities and some known facts about FAn . Then in Section 3 we develop the analogue structure theory for FAn . We describe explicit generators gTλ for its minimal left ideals, where the Tλ ’s are certain (half of the) standard tableaux. The elements gTλ are similar to the well known generators eTλ = RT+λ CT−λ of the minimal left ideals of F Sn ; here RTλ (CTλ ) denotes the row (column) stabilizer of Tλ . It follows from the well known-von Neumann Lemma (see Section 4.3) that the eTλ ’s are semi-idempotents, and this fact is of great importance in the structure theory of F Sn . Similar to the Sn -case, most of the above gTλ ’s are semi-idempotents. In particular, if λ = λ then gTλ is indeed semi-idempotent. However, there is a subset of the self-conjugate partitions λ = λ for which the gTλ ’s are not semiidempotents. We do show that this subset is very small: in every infinite (k × k)hook of partitions there are only finitely many such elements λ. The details SA∗ (V , n)-module.

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are given in Appendix A, which was written together with Tamar Seeman. We nevertheless show, in Section 4, that there are infinitely many partitions λ for which gTλ is not semi-idempotent. In the rest of the paper we describe the explicit structure and decompositions of W = V ⊗n as an SA∗ (V , n) module. We describe three types of (related) decompositions of W : (1) into its isotypic components, (2) into a direct sum of Weyl-modules, (3) into a direct sum of irreducible SA∗ (V , n)-modules. For S ∗ (V , n), the eTλ ’s define the Weyl-submodules of W both in the classical case and in the supercase. Similarly, for SA∗ (V , n) the gTλ ’s define (the analogues of) the Weyl-submodules of W . In Section 5 we describe the decomposition of W into a direct sum of such SA∗ (V , n)-Weyl-submodules—in terms of partitions and certain standard tableaux. For S ∗ (V , n), the Weyl modules are all irreducible. For SA∗ (V , n), some of these Weyl modules are reducible: those corresponding to non-self-conjugate partitions remain irreducible, but those corresponding to selfconjugate partitions split as a direct sum of two non-isomorphic submodules. The details are given in Theorem 5.3. Note that this is closely related to a similar phenomena between the characters of Sn and of An . In Section 6 we examine the various Weyl-modules through the eigenvalues of the actions of permutations on W . We show that these eigenvalues are closely related to—and can be calculated by—the values of Sn and of An characters. In particular, for a self-conjugate partition this yields the decomposition of the corresponding Weyl-module into a direct sum of two eigenspaces. In Section 7 we describe explicit bases for the irreducible SA∗ (V , n)-submodules of W in terms of partitions and of semi-standard tableaux. As for the irreducible S ∗ (V , n)-submodules, these bases allow us to attach to each irreducible SA∗ (V , n)-module a certain supersymmetric function (see [1, Section 6]). However, unlike for S ∗ (V , n), in the case of a self-conjugate partition these functions do not characterize the corresponding irreducible SA∗ (V , n)-modules. 2. Generalities 2.1. Generalities All modules in this paper are left modules. The following is well-known: Let M be a vector space over F and assume that R is a semisimple ring with q (nonzero) simple components Ri and such that R ⊆ EndF (M). The decomposition q q R = i=1 Ri implies a decomposition of the identity in R, say 1 = i=1 ui with ui ∈ Ri . Then ui uj = δi,j ui . The Ri ’s are the minimal two-sided ideals

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in R, each isomorphic to a certain matrix algebra over F , while  the ui ’s are the minimal central idempotents in R. Let Mi = ui M then M = Mi and Rj Mi = 0 for i = j . Moreover, the Mi ’s are the isotypic components of M also as an Rmodule. Consider q the ring S := EndR (M) and set Si = EndRi (Mi ) = EndR (Mi ). Then S = i=1 Si where each Si is simple with ui ∈ Si . Again, the Si ’s are the minimal two-sided ideals in S, each isomorphic to a certain matrix algebra over F , while the ui ’s are the minimal central idempotents in S, and the Mi ’s are the isotypic components of M as an S-module. So in particular this shows that M has the same isotypic decomposition as an R and as an S-module. Moreover, Ri = EndSi (M iq) and EndF (Mi )  Ri ⊗F Si . So we have R = EndS (M) and EndF (M) = i=1 (Ri ⊗F Si ). Let 1  i  q. As an Ri -module, Mi  Ni⊕a where Ni is Ri -irreducible and Ni is unique up to isomorphism. Similarly Mi  Ki⊕b where Ki is Si -irreducible, and unique up to isomorphism. We apply this in Section 2.2 to R = ϕ ∗ (FAn ) and M = W = V ⊗n . 2.2. Minimal two-sided ideal decompositions In order to apply the situation described in Section 2.1 to R = ϕ ∗ (FAn ) and M = W = V ⊗n we need to understand the isotypic decomposition of R. Recall the classical and unique decomposition of the group algebra F Sn into minimal two-sided ideals:  F Sn = Iλ . (1) λn

We say a partition λ is self-conjugate if it equals its conjugate partition λ , otherwise we call the partition non-self-conjugate. The group algebra FAn has an analogue decomposition into minimal two-sided ideals as F Sn . To the pair λ = λ corresponds in FAn a minimal ideal I¯λ = I¯λ
Iλ ⊕ Iλ (by (2.4.2) in [6]) and I¯λ  Iλ  Iλ as left An -modules; and to a self-conjugate partition λ correspond two minimal ideals I¯λ+ and I¯λ− with I¯λ+ ⊕ I¯λ−
Iλ and where dim I¯λ+ = dim I¯λ− = (1/4) dim Iλ (see [6, Lemma 2.5]). Let λ > λ denote the left lexicographic order. Then       FAn = (2) I¯λ ⊕ I¯λ+ ⊕ I¯λ− . λn,λ>λ

λn,λ=λ

Denote the set of partitions which lie inside the (k, l)-hook by H (k, l; n), that is H (k, l; n) = {(λ1 , λ2 , . . .)  n | λk+1  l}. By [1], Theorem 3.20 (since here l = k),  ϕ ∗ (F Sn ) = ϕ ∗ (Iλ ) λ∈H (k,k;n)

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and ϕ ∗ restricted to Iλ for a partition λ lying  in the (k, k)-hook is an isomorphism. Hence by Section 2.1 we have W = λ∈H(k,k;n) Wλ where Wλ = ϕ ∗ (Iλ )W are the isotypic components of W as ϕ ∗ (F Sn )-module. An analogue of this decomposition also holds for ϕ ∗ (FAn ). Let H1 (k, k; n) be the set of all nonself-conjugate partitions λ lying in the (k, k)-hook such that λ is bigger than λ in the left lexicographic order. Let H2 (k, k; n) denote the set of all self-conjugate partitions lying in the (k, k)-hook. Then      
 R = ϕ ∗ (FAn ) = ϕ ∗ (I¯λ ) ⊕ ϕ ∗ I¯λ+ ⊕ I¯λ− . λ∈H1 (k,k;n)

λ∈H2 (k,k;n)

Moreover, since a minimal ideal is a simple algebra, ϕ ∗ applied to a minimal ideal is either zero or isomorphic to that same minimal ideal. Since here ϕ ∗ (Iλ )  Iλ we have ϕ ∗ (I¯λ )  I¯λ and ϕ ∗ (I¯λ% )  I¯λ% for % = ± which says that       ϕ ∗ (FAn )  I¯λ ⊕ I¯λ+ ⊕ I¯λ− . λ∈H1 (k,k;n)

λ∈H2 (k,k;n)

2.3. Conjugacy classes and characters of An Let µ be a partition of n and Cµ be its corresponding conjugacy class in Sn consisting of permutations of cycle type µ. We denote by cµ the class sum of Cµ . The subgroup An of Sn is given by the even permutations of Sn . We call a partition µ of n even if any permutation in Cµ is even. Recall that the An -conjugacy classes are determined by the conjugacy classes Cµ of Sn as follows (see [5, Section 1.2]): (1) If µ = (µ1 , µ2 , . . .) is a partition of n where all µi are distinct and odd then the Sn -conjugacy class Cµ splits into two An -conjugacy classes which we denote as Cµ+ and Cµ− . Denote the corresponding An -class sums as cµ+ and cµ− . (2) If µ is an even partition of n but is not of the type described in (1) then Cµ is also the conjugacy class in An . The irreducible characters of Sn , denoted χ λ are parametrised by the partitions λ of n. Write χ λ (µ) for the character value of χ λ on an element of the conjugacy class Cµ . Suppose λ is non-self-conjugate then the Sn -character χ λ restricted to An is irreducible (and we denote it again by χ λ ) whereas, for λ self-conjugate it + − splits into two irreducible characters, χ λ = χ λ + χ λ (see [5, Section 2.5]). Let µ be an even partition and denote by h(λ) := (h11 (λ), . . . , hkk (λ)) the partition recording the hook lengths hi,i (λ) of λ. Then χ λ (µ) is a character value for An if λ is non-self-conjugate and Cµ non-split. If Cµ splits (but not χ λ ) or χ λ splits with µ = h(λ) then
 χ λ µ± = χ λ (µ)/2, χ λ± (µ) = χ λ (µ)/2, and

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 χ λ± µ± = χ λ (µ)/2, where the left-hand side denotes the An -character and the right-hand side the Sn character. For λ self-conjugate and µ = h(λ) the conjugacy class Cµ splits with (see [5, 2.5.13])  
 
 λ 1 λ
hii , χ h(λ) ± χ λ h(λ) χ λ± h(λ)+ = i 2  
 
 λ 1 λ
λ± − λ h(λ) = h . χ χ h(λ) ∓ χ h(λ) i ii 2

n

Note that since χ λ = χ λ ⊗ χ (1 ) we have χ λ (µ) = χ λ (µ) for µ even. Hence in

particular χ λ (1) = χ λ (1). 2.4. Minimal central idempotents We describe next the minimal central idempotents of F Sn and FAn . For Sn , the minimal central idempotent lying in Iλ is denoted by eλ or eλSn . An explicit expression for eλSn is given for example in [4, Lemma 3]. For An , the minimal central idempotent lying in I¯λ = I¯λ for λ non-self-conjugate is denoted by eλAn = eλA n ; the minimal central idempotent of FAn lying in I¯λ% for λ self-conjugate is n denoted by eλ% where % = + or e = −. Note that eλ+ + eλ− = eλS . In describing explicit expressions for these idempotents we follow [4, Theorems 4 and 5].3 For a non-self-conjugate partition λ we have eλAn = eλSn +eλS n ; for a self-conjugate partition λ the central minimal idempotents eλ± are given as follows: Let dλ be the number of standard tableaux of shape λ (see [2,7,8]) and define pλ = (−1)(n−k)/2

k

hii (λ).

i=1

Then the minimal central idempotents eλ± are given by   dλ 1 (ch(λ)+ − ch(λ)− ) , eλ+ = eλSn 1 + √ 2 2 pλ |Ch(λ)+ |   dλ 1 Sn − (ch(λ)+ − ch(λ)− ) . eλ = eλ 1 − √ 2 2 pλ |Ch(λ)+ | 3 The equation in the last line of p. 473 (respectively the first line of p. 474) in the proof of Theorem 5 in [4] is not correct. However, the authors of [4] pointed out that the weaker equation needed for the proof can be obtained by using Eq. (33.11) in [3].

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2.5. Further notation Consider the algebra automorphism f : F Sn → F Sn given by linearity and by f (σ ) = sgn(σ )σ for a permutation σ ∈ Sn . It is well known that f (Iλ ) = Iλ . Also, x + f (x) ∈ FAn for any x ∈ F Sn . Let G be a subgroup of Sn . We define elements G+ and G− in the group algebra F G by
 g and G− = f G+ = sgn(g)g. G+ = g∈G

g∈G

Given a tableau Tλ of shape λ where λ is a partition of n, we write RTλ for the group of row permutations and CTλ for the group of column permutations determined by the tableau Tλ . Moreover, corresponding to the tableau Tλ we have the elements eTλ ∈ Iλ and eT λ ∈ Iλ given by eTλ = RT+λ · CT−λ

and eT λ = f (eTλ ) = RT−λ · CT+λ .

(3)

For the rest of this paper we fix the following notation: V = V0 ⊕ V1 with dim V0 = dim V1 = k, and W = V ⊗n . We shall delete the ϕ ∗ from the action of F Sn on V ⊗n . For example, for the Schur superalgebras and the super-Weyl modules (see later sections) we write

 S ∗ (V , n) = Endϕ ∗ (F Sn ) V ⊗n = EndF Sn V ⊗n ,

 SA∗ (V , n) = Endϕ ∗ (FAn ) V ⊗n = EndFAn V ⊗n , eTλ W := ϕ ∗ (eTλ )W.

3. The structure of FAn 3.1. Description of results Let Tλ be a tableau of shape λ. We are interested in the elements eTλ introduced in Section 2.5. They are the fundamental objects for analyzing the structure of the group algebra F Sn : The elements eTλ (and eT λ ) are semi-idempotents, that is eT2λ = αeTλ for some non-zero α. Here α equals the product of the hook numbers of the diagram λ. A complete but in general not unique decomposition of F Sn into minimal left ideals is given by Eq. (1) and by the decomposition  Iλ = F Sn eTλ . Tλ standard

The set of semi-idempotents {eTλ | Tλ is standard} consists in general of mutually non-orthogonal elements. A replacement of this set by a set of orthogonal idempotents is given explicitly for example in [4, Lemma 3].

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In this section we introduce elements gTλ in FAn which share most of the above properties of the eTλ . Theorem 3.2 shows how the decomposition of the minimal two-sided ideals of FAn (given by Eq. (2)) into a direct sum of minimal left ideals is given explicitly by these elements gTλ . The gTλ in general are not semi-idempotents but, however, Theorem 4.1 and Proposition 4.2 indicate that for most λ these elements are indeed semi-idempotents. Definition 3.1. We define gTλ := eTλ + eT λ = eTλ + f (eTλ ) = RT+λ CT−λ + RT−λ CT+λ . Let λ = λ and let T1 , . . . , Th be the standard tableaux of shape λ with one and two in their first column. Define Ti to be the tableau obtained from Ti by conjugation. The conjugate tableau Tλ of a standard tableau of shape λ is a standard tableau of shape λ which means in this special situation of shape λ. Moreover Ti is a tableau whose first row starts with one and two. This implies that the set of all Ti and all Ti is the set of all standard tableaux of shape λ and hence dλ = 2h. The main result of this section is: Theorem 3.2. Let λ be a partition of n. With the notation from above we have: (1) Let λ be non-self-conjugate. Then I¯λ ⊂ Iλ ⊕ Iλ and I¯λ decomposes into a direct sum of minimal left ideals of FAn as follows:  I¯λ = FAn gTλ . Tλ standard

(2) Let λ be self-conjugate and let % = + or % = −. Then I¯λ+ ⊕ I¯λ− ⊂ Iλ and I¯λ% decomposes into a direct sum of left ideals of FAn as follows: I¯λh =

h 

FAn eλ% gTi .

i=1

3.2. Proof of main result In this section we prove Theorem 3.2. We assume the notation introduced in Section 3.1 and we start with the following simple lemma which is frequently applied throughout the paper. Lemma 3.3. Let T be a tableau on 1, . . . , n, let 1  a = b  n and assume a and b appear in the same column in T (a similar statement can be formulated for the rows). Then

 1 + (a, b) CT+ = CT+ 1 + (a, b) = 2CT+ ,

 1 + (a, b) CT− = CT− 1 + (a, b) = 0,

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 1 − (a, b) CT− = CT− 1 − (a, b) = 2CT− ,

 1 − (a, b) CT+ = CT+ 1 − (a, b) = 0. Proof. This follows since the transposition (a, b) is in the column stabilizer CTi , so we can write CT+ = (1 + (a, b))q + = q + (1 + (a, b)) and CT− = (1 − (a, b))q − = q − (1 − (a, b)) for some q + , q − ∈ F Sn . ✷ Lemma 3.4. Let λ be a self-conjugate partition of n and Tλ any tableau of shape λ. Then
 eλ± gTλ = eλ± eTλ + eT λ = 0. Proof. We prove the claim for eλ+ . Since eλ+ is central in FAn and eTλ + eT λ is an element of FAn we have eλ+ (eTλ + eT λ ) = (eTλ + eT λ )eλ+ . Assume for a contradiction that eλ+ (eTλ + eT λ ) = 0. Since n
2 and λ is self-conjugate, therefore λ = (1n ). Thus, without loss of generality we assume the tableau Tλ has a first row starting with entries one and two. Multiply the equation (eTλ + eT λ )eλ+ = 0 from the left by a := (1 + (1, 2)); since aRT+λ = 2RT+λ while aRT−λ = 0 (see Lemma 3.3), together with (eTλ +eT λ )eλ+ = 0 it implies that eTλ eλ+ = 0, which in turn implies (F Sn eTλ )eλ+ = 0. This provides the contradiction: By the character theory of An the simple F Sn -module F Sn eTλ splits over FAn into two (non-zero) irreducible modules, one of which is (F Sn eTλ )eλ+ (see [5, Section 2.5]). ✷ Lemma 3.5. Suppose λ is self-conjugate and let Tλ be a tableau of shape λ. Then FAn gTλ is the direct sum of two non-zero left ideals (we show later that these are minimal left ideals): FAn gTλ = FAn eλ+ gTλ ⊕ FAn eλ− gTλ .

(4)

Proof. In Lemma 3.4 we have seen that eλ± gTλ = 0. Hence the two ideals on the right-hand side of Eq. (4) are non-zero. Their sum is direct since the elements eλ± are central in FAn with eλ+ · eλ− = 0. The right-hand side of Eq. (4) is included in the left-hand side since FAn eλ% gTλ ⊆ FAn gTλ . Also, since eλ+ + eλ− = eλSn and hence gTλ = eλ+ gTλ + eλ− gTλ , we obtain that the left-hand side of Eq. (4) is contained in the right-hand side. ✷ Lemma 3.6. (1) Let λ be a self-conjugate partition. Then the following sum of left ideals of FAn is direct h i=1

FAn gTi =

h  i=1

FAn gTi .

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(2) Let λ be a non-self-conjugate partition. Then the following sum of left ideals of FAn is direct  FAn gTλ = FAn gTλ . Tλ standard

Tλ standard

Proof. Let f : F Sn → F Sn be as in Section 2.5.

 (1) Recall Lemma 3.3 and Definition 3.1. Assume that hi=1 ai gTi = 0 for some elements ai ∈ FAn . We multiply this equation by (1 − (1, 2)) from the right. Since CT−i (1 − (1, 2)) = 2CT−i and CT+i (1 − (1, 2)) = 0, it implies 0=

h

h
 ai gTi 1 − (1, 2) = 2 ai eTi .

i=1

i=1



But since the sum F Sn eTλ running over all standard tableaux Tλ is a direct  sum, it implies that all ai eTi = 0. Next, deduce that hi=1 ai eT i = 0 and apply the  h automorphism f : 0 = f ( i=1 ai eT i ) = hi=1 f (ai )eTi . By the same argument as above all f (ai )eTi = f (ai eT i ) = 0 hence all ai eT i = 0. Since gTi = eTi + eT i it follows that all ai gTi = 0. (2) Note that for λ non-self-conjugate eλSn gTλ = eTλ since eT λ ∈ Iλ and  aTλ gTλ = 0 where the sum runs over all standard eλSn Iλ = 0. Assume that tableaux Tλ and aTλ ∈ FAn . We multiply this equation with the central element  Sn   eλSn and obtain 0 = eλ aTλ gTλ = aTλ eTλ . But since the sum F Sn eTλ running over all standard tableaux Tλ is a direct sum, it implies that all aTλ eTλ = 0. Applying f and arguing as in the first part, deduce that all aTλ gTλ = 0. Hence  FAn gTλ = FAn gTλ . ✷ Tλ standard

Tλ standard

Proof of Theorem 3.2. (1) Note that for λ non-self-conjugate I¯λ is isomorphic to the matrix algebra Mdλ (F ), hence any direct sum of dλ non-zero left ideals in I¯λ must equal I¯λ , and these ideals necessarily are irreducible. The proof now follows from part 2 of the last lemma. (2) Since eλSn = eλ+ + eλ− we obtain using Lemmas 3.5 and 3.6 h


FAn eλ+ gTi

 + FAn eλ− gTi

⊇

i=1

h i=1

=

FAn gTi =

h 

FAn gTi

i=1

h 
 FAn eλ+ gTi ⊕ FAn eλ− gTi i=1

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=

h 
i=1

h   FAn eλ+ gTi ⊕ (FAn eλ− gTi ) i=1

and by dimensionality we obtain equality in this last equation. The equation also implies (multiplying by eλ+ ) that h

FAn eλ+ gTi =

i=1

h 

FAn eλ+ gTi .

i=1

I¯λ+

Since is isomorphic to the matrix algebra Mh (F ), any direct sum of h nonzero left ideals in I¯λ+ must equal I¯λ+ , and these ideals necessarily are irreducible. Similarly for I¯λ− . ✷ Corollary 3.7. Let Tλ be any tableau of shape λ then dim FAn gTλ = dλ , the cardinality of the standard tableaux of shape λ. Proof. (1). Let λ be non-self-conjugate. Since I¯λ ∼ = Mdλ (F ), all minimal left ideals in the decomposition in Theorem 3.2 (1) are dλ dimensional. (2) Let λ be self-conjugate and % = ±. If Tλ = Ti this follows since dim FAn eλ% gTi = h = (1/2)dλ and FAn gTi = FAn eλ+ gTi ⊕ FAn eλ− gTi . Let now Tλ be any tableau of shape λ, then there exist σ ∈ Sn such that Tλ = σ T1 . It follows that gTλ = σgT1 σ −1 , and since σ An σ −1 = An , FAn gTλ = σ FAn σ −1 σgT1 σ −1 = σ FAn gT1 σ −1 ∼ = FAn gT1 , a vector space isomorphism x → σ xσ −1 . ✷ We can now prove Proposition 3.8. Let λ be any partition of n with corresponding minimal twosided ideal Iλ . Let Tλ be a tableau of shape λ with corresponding eTλ and gTλ , and let Jλ ⊆ Iλ be a minimal left ideal in F Sn . Then (1) As FAn -left modules Jλ = FAn gTλ . (2) F Sn eTλ = FAn eTλ . (3) F Sn gTλ = F Sn eTλ ⊕F Sn eT λ and in particular dim F Sn gTλ = 2 dim F Sn eTλ = 2dλ . Proof. Let η = id + f where f was defined in Section 2.5, then η : F Sn → FAn is an FAn -module homomorphism, and η(eTλ ) = gTλ . As F Sn and therefore also as left FAn -modules, Jλ = F Sn eTλ . Apply η to show that F Sn eTλ ∼ = FAn gTλ : η(FAn eTλ ) = FAn gTλ therefore η is surjective, hence Corollary 3.7 implies that η : F Sn eTλ → FAn gTλ is an isomorphism, which proves the first part. This also shows that η(F Sn eTλ ) = η(FAn eTλ ) and since η is an isomorphism, we

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can deduce the second part. To prove the third part, notice that FAn gTλ is a left ideal in FAn , and for any non-zero left ideal L ⊆ FAn , dim F Sn L = 2 dim L (since F Sn L = L⊕(1, 2)L). Since F Sn gTλ = F Sn (FAn gTλ ), deduce that dim F Sn gTλ = 2 dim FAn gTλ = 2dλ . Now F Sn gTλ ⊆ F Sn eTλ + F Sn eT λ and since both sides have the same dimension, they are equal and the sum is direct. ✷ 3.3. Let T1 , . . . , Tdλ be the standard tableaux of shape λ, ordered as in [4, Section 2] (i.e. lexicographically, by the rows), and with eTi the corresponding semi-idempotents: eT2i = αeTi where α equals the product of the hook numbers in λ (with 1  i  dλ ), and eTj eTi = 0 if i < j . If λ is non-self-conjugate then also gT2i = αgTi . In this case denote ETi =

1 eT α i

and GTi =

1 gT , α i

λ ai ETi , then {a1 ET1 , . . . , then ETi and GTi are idempotents. Write eλSn = di=1 adλ ETdλ } is a set of mutually orthogonal idempotents. The following explicit expression for the ai ETi is given in [4, Lemma 3]:   i−1 ETj + ETj ETk − · · · + (−1) ET1 . . . ETi−1 ETi . ai ETi = 1 − j
j
In the non-self-conjugate case this extends to the elements GTi as follows. Proposition 3.9. Let λ be non-self-conjugate and T1 , . . . , Tdλ be the standard λ tableaux of shape λ, ordered as above, and let eλAn = di=1 ai GTi . Then {a1 GT1 , . . . , adλ GTdλ } forms a set of mutually orthogonal idempotents with   i−1 ai GTi = 1 − GTj + GTj GTk − · · · + (−1) GT1 . . . GTi−1 GTi . j
j
Proof. Recall that gTλ = eTλ + f (eTλ ), hence GTλ = ETλ + f (ETλ ), and eλAn = eλSn + f (eλSn ), where f (σ ) = sgn(σ )σ for σ ∈ Sn . Adding the above expressions for ai ETi and f (ai ETi ), we only need to verify that, for example, ETj ETk + f (ETj ETk ) = (ETj + f (ETj ))(ETk + f (ETk )) (and similarly for longer products). This follows from ETj f (ETk ) = f (ETj )ETk = 0, since ETj ∈ Iλ , f (ETk ) ∈ Iλ and λ non-self-conjugate. ✷

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4. Semi-idempotents gTλ 4.1. Semi-idempotents gTλ We assume the notation as established in Section 3. As eT2λ = αeTλ , hence also eT 2λ = αeT λ (see Definition 3.1). So if eTλ eT λ = 0 then also eT λ eTλ = 0 and it follows that gT2λ = αgTλ , that is, gTλ is semi-idempotent. However, if eTλ eT λ + eT λ eTλ = 0 then gTλ is not semi-idempotent. These properties are independent of the particular tableau Tλ but depend only on the shape λ itself since, if Tλ is another tableau of shape λ, then Tλ = ηTλ for some η ∈ Sn , and gηTλ = ηgTλ η−1 . We henceforth look for conditions on λ such that eTλ eT λ = 0 and hence gTλ is a semi-idempotent. If λ is non-self-conjugate then eTλ eT λ = 0 since eTλ ∈ Iλ and eT λ ∈ Iλ , hence gTλ is semi-idempotent. We therefore consider for the rest of this subsection selfconjugate partitions λ. In this case not every element gTλ is a semi-idempotent. For example, let λ = (2, 1) and let the tableau Tλ be given by T2,1 =

1 2 3

then gTλ = 2 − 2 · (1, 3, 2), and it is easy to see that this is not a semi-idempotent. Theorem 4.1 and Proposition 4.2 give different (and independent) conditions on λ, which guarantee that gTλ is a semi-idempotent. Since most λ’s satisfy these conditions, most gTλ ’s are semi-idempotents. Nevertheless Proposition 4.3 yields an infinite sequence of λ’s with gTλ not semi-idempotent. Theorem 4.1. Let λ be a partition of n lying in a (k, k)-hook and assume λ1
2k + 1 (this in particular holds if n > 3k 2 ). Let Tλ be a tableau of shape λ with corresponding e = eTλ and e = eT λ . Then ee = e e = 0, hence gTλ is semi-idempotent. We shall give two different proofs of Theorem 4.1; because of the length and complexity of these proofs we postpone it to Appendix A. The following proposition can be applied in many cases to show that eTλ eT λ = 0. Proposition 4.2. Let λ be self-conjugate and assume there exist at least two rows (hence also columns) in λ with equal and odd length (this holds in particular if λ1 − λ2
2), and let Tλ be a tableau of shape λ. Then eTλ eT λ = 0, hence gTλ is semi-idempotent. Proof. Recall that eTλ eT λ = RT+λ CT−λ RT−λ CT+λ . We show that CT−λ RT−λ CT+λ = 0. The two rows of equal and odd length (say length r) split Tλ into three parts T1 , T2 , and T3 as indicated in Fig. 1. The parts T1 and T3 might be empty. Part T2

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Fig. 1. The diagram Tλ .

consists of the two rows of equal and odd length r. Suppose these two rows have entries ai and bj , respectively. Let σ = (a1 , b1 ) · · · (ar , br ) be the product of the r transpositions (ai , bi ) defined by T2 . Then σ lies in CTλ with σ 2 = id and sgn(σ ) = −1. The row stabilizer of Tλ factors as RTλ = RT1 × RT2 × RT3 . Since σ is disjoint from the entries of T1 and T3 , therefore σ τ = τ σ for any τ ∈ RT1 ∪ RT3 . Also, RT2 = Sr (a1 , . . . , ar ) × Sr (b1 , . . . , br ),

  ηRT2 η−1 = Sr η(a1 ), . . . , η(ar ) × Sr η(b1 ), . . . , η(br ) where Sr (x1 , . . . , xr ) denotes the symmetric group on the r symbols xi and η ∈ Sn arbitrary. It follows that σ RT2 σ −1 = RT2 . In particular, this implies that σ RT−2 = RT−2 σ . Since RT−λ = RT−1 · RT−2 · RT−3 we conclude that (1 − σ )RT−λ = RT−λ (1 − σ ). We apply Lemma 3.3 and its proof to a σ ∈ CTλ ; hence we can write CT+λ = (1 + σ )q1+ and CT−λ = q2− (1 − σ ) for some q1+ , q2− ∈ F Sn . Since (1 − σ )(1 + σ ) = 0 this implies CT−λ RT−λ CT+λ = q2− (1 − σ )RT−λ (1 + σ )q1+ = q2− RT−λ (1 − σ )(1 + σ )q1+ = 0. ✷ Application. Note that for a large n, a random λ would have a tail of (many) 1’s, hence gTλ is semi-idempotent. 4.2. Non-semi-idempotents gTλ The next proposition shows that for infinitely many partitions λ we obtain eTλ eT λ + eT λ eTλ = 0 and hence gTλ is not a semi-idempotent. Let t1 , . . . , tk ∈ V0 and u1 , . . . , uk ∈ V1 be bases of V0 and V1 , respectively, with V = V0 ⊕ V1 . Call w = z1 ⊗ · · · ⊗ zn ∈ W monomial if z1 , . . . , zn ∈ {t1 , . . . , tk , u1 , . . . , uk }.

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Following [9] (see also Section 3 in [1]), Tλ identifies a monomial in V ⊗n with a (k, k)-tableau of shape λ on t1 , . . . , tk , u1 , . . . , uk , and a row or a column permutation of Tλ acts on such a tableau (i.e. monomial) accordingly. Proposition 4.3. Call λ a stair-case partition if λ = (m, m − 1, . . . , 1). Let λ be such a partition and let Tλ be a tableau of shape λ. Then eTλ eT λ + eT λ eTλ = 0. Proof. Let n = 1 + · · · + m with λ a stair-case partition of n, and let Tλ be a tableau on 1, . . . , n. In this proof we only use the ti ’s, hence we identify ti ≡ i, for 1  i  k. Denote by w0 = w0 (t) and w1 = w1 (t) the monomials identified by Tλ with the following tableaux: m ··· 2 1 .. · · w0 = . 2 · 1

1 2 .. .. and w1 = . . 1 2 1

··· m ·

.

Denote R = RTλ , C = CTλ , e = eTλ and e = eT λ . We will prove that e ew1 = 0 while ee w1 = 0, thus proving that ee + e e = 0. Note that C − w1 = 0 since there are repetitions in the columns of w1 . Hence we obtain e ew1 = R − C + R + C − w1 = 0. We show next that ee w1 = 0. For r := m!(m − 1)! · · · 1! we have C + w1 = r · w1 , and since ee = R + C − R − C + it suffices to show that R + C − R − w1 = 0. We need the following two properties of w0 and w1 . Property 1. Let p ∈ R, let M = pw1 and assume M has no repetitions in its columns. Then M = w0 . To see this, check the rows of M from the bottom to the top: The bottom row in w1 is 1, which is also the bottom row in M = pw1 . For the same reason, the next row in M contains one and two, and since M has no repetitions in its columns, this row has to be 2, 1. Continue this process to obtain M = w0 . Property 2. Let p ∈ R, q ∈ C such that pqw0 = w0 ; we claim p = q = 1. To prove this we show first that q must fix the first column of w0 ; otherwise, in the first column of qw0 there is some j below the (m − j + 1)th row, and therefore in pqw0 there is some j below the (m − j + 1)th row, contradicting pqw0 = w0 . By the same argument q must also fix the second column of w0 . Continue to conclude firstly that q = 1, then secondly that p = 1. We can now show that R + C − R − w1 = 0: There is a unique element p ∈ R such that pw1 = w0 . Therefore, R − w1 = ±w0 + sum of monomials M with column repetitions (by Property 1). For each such monomial M, C − M = 0, which implies C − R − w1 = ±C − w0 . Thus R + C − R − w1 = ±R + C − w0 . By the second property, R + C − w0 = w0 + other monomials, hence R + C − R − w1 = ±R + C − w0 = 0. ✷

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4.3. Von Neumann’s Lemma An important tool in analyzing the structure of F Sn is von Neumann’s Lemma. It implies that the eTλ ’s are semi-idempotents. In this section we discuss a vonNeumann-type lemma for FAn . However, it does not imply that the gTλ ’s are semi-idempotents. For convenience we restate the result for Sn before proving an analogue for An . Theorem 4.4 (Von Neumann’s Lemma for Sn ). Let λ be a partition of n, T be a tableau of shape λ and let a ∈ F Sn . Then paq = sgn(q)a for all p ∈ RT and q ∈ CT , if and only if a = αeT for some α ∈ F . Proposition 4.5 (Von Neumann’s Lemma for An ). Let λ be a partition of n, T = Tλ be a tableau of shape λ and let a ∈ FAn . Then paq = sgn(q)a for all p ∈ RT and q ∈ CT such that pq ∈ An , if and only if a = αgT for some α ∈ F . Proof. The proof is essentially identical to the proof in the Sn -case. Denote R = RT and C = CT . First, since sgn(p) = sgn(q), we have 
pgT q = p R + C − + R − C + q = sgn(q)R + C − + sgn(p)R − C + = sgn(q)gT .  Next, let a = σ ∈An ασ σ and assume a satisfies the above condition. Deduce that αpσ q = sgn(q)ασ for all σ ∈ An . If σ = 1 then αpq = sgn(q)a where α = α1 . If σ is not of the form σ = pq for any p ∈ R and q ∈ C then there exist two transpositions p ∈ R and q ∈ C—hence pq ∈ An —such that σ = pσ q (see [2, Theorem 2.8]). This implies that ασ = 0. ✷ ∗ (V , n) 5. The Weyl modules for SA

5.1. The (super-) Weyl modules for the symmetric group of degree n (denoted in [1] as W ∗ eTλ ) are given by ϕ ∗ (eTλ )W = eTλ W . Recall from Corollary 3.18 of [1] that these are irreducible over S ∗ (V , n) and that by [1, Section 3].   W= Wλ and Wλ = eTλ W. (5) λ∈H (k,k;n)

Tλ standard

We give an explicit construction of SA∗ (V , n)-submodules LTλ ⊆ V ⊗n which are natural generalisations of these Weyl modules. We also call these modules Weyl modules. When λ is non-self-conjugate then LTλ is irreducible over SA∗ (V , n), whereas when λ is self-conjugate, LTλ is a sum of two irreducible and nonisomorphic submodules SA∗ (V , n)-modules. We show that V ⊗n is a direct sum of such LTλ .

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We introduce the map T : W → W , which is the map T in [6, Lemma 1.8]. We recall its definition and basic properties. Let t1 , . . . , tk ∈ V0 and u1 , . . . uk ∈ V1 be bases of V0 and V1 , respectively, with V = V0 ⊕ V1 . Let τ ∈ EndF (V ) be given by τ ti = ui and τ ui = ti for i = 1, . . . , k. Note that the degree of τ in EndF (V ) is one. Now define T = τ ⊗n . For example, if n = 2 and y1 , y2 ∈ V are homogeneous, T (y1 ⊗ y2 ) = (τ ⊗ τ )(y1 ⊗ y2 ) = (−1)deg y1 · τy1 ⊗ τy2 . It follows that ϕσ∗ ◦ T = sgn(σ )T ◦ ϕσ∗ for σ ∈ Sn , which clearly implies that T ∈ SA∗ (V , n). It also implies that for w ∈ W , T eTλ w = eT λ T w hence T eTλ W = eT λ W , and that T : Wλ → Wλ , which is a vector space isomorphism. 5.2. We begin by describing the isotypic decomposition of V ⊗n as an Using Section 2.1 (and notations of Sections 2.2 and 2.4) we deduce: SA∗ (V , n)-module.

Proposition 5.1. Let λ be a partition lying in the (k, k)-hook. Then the following holds: (1) Suppose λ is non-self-conjugate. As FAn -modules (i.e. as R = ϕ ∗ (FAn )modules), Wλ and Wλ are isomorphic, and Wλ ⊕ Wλ is an isotypic component of W both as ϕ ∗ (FAn )-module and as SA∗ (V , n)-module. (2) Suppose λ is self-conjugate. As FAn -module and as SA∗ (V , n)-module, Wλ is the direct sum of two isotypic components Wλ = Wλ+ ⊕ Wλ− with Wλ+ = eλ+ W and Wλ− = eλ− W . So, the isotypic decomposition of W as ϕ ∗ (FAn )-module and as SA∗ (V , n)-module is      
+ − Wλ ⊕ Wλ . W= (Wλ ⊕ Wλ ) ⊕ λ∈H1 (k,k;n)

λ∈H2 (k,k;n)

5.3. The super-Weyl-modules eTλ W are irreducible over S ∗ (V , n), see [1]. Replacing eTλ by gTλ we obtain the analogous SA∗ (V , n)-modules. These may be reducible, and are given in the following definition. Definition 5.2. Let λ be a partition of n and Tλ a tableau of shape λ. Then define LTλ = gTλ W = (eTλ + eT λ )W . Since gTλ is an element of FAn hence LTλ is an SA∗ (V , n)-module; we call it the Weyl module for the Schur superalgebra SA∗ (V , n) of the alternating groups, corresponding to the tableau Tλ . We prove that LTλ satisfies the following properties.

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Theorem 5.3. Let λ be a partition of n
2 and let Tλ be a tableau of shape λ with LTλ the corresponding Weyl module for SA∗ (V , n). Then the following holds: (1) LTλ = eTλ W ⊕ eT λ W . (2) If λ is non-self-conjugate then LTλ is an irreducible SA∗ (V , n)-module, LTλ ⊆ Wλ ⊕ Wλ , and any irreducible SA∗ (V , n)-submodule of Wλ ⊕ Wλ is isomorphic to LTλ . − (3) If λ is self-conjugate then LTλ splits: LTλ = L+ Tλ ⊕ LTλ , a direct sum of + two irreducible and non-isomorphic SA∗ (V , n)-modules. Here L+ Tλ = eλ gTλ W − − + − and LTλ = eλ gTλ W with dim LTλ = dim LTλ . Moreover, any irreducible − SA∗ (V , n)-submodule of Wλ is isomorphic to either L+ Tλ or LTλ . Note that by Proposition 5.1, up to isomorphism, the irreducible SA∗ (V , n)modules from (2) and (3) above are the only irreducible SA∗ (V , n)-submodules of W . Proof. To prove part 1, we show first that the sum eTλ W + eT λ W is direct. If λ is non-self-conjugate this follows since eTλ W ⊆ Wλ and eT λ W ⊆ Wλ , and moreover Wλ ∩ Wλ = 0. If λ is self-conjugate then as n
2, it follows that λ = (1n ) and so RTλ is non-trivial. Let (a, b) ∈ RTλ then, by Lemma 3.3, (1 + (a, b))eTλ = 2eTλ while (1 + (a, b))eT λ = 0. This  implies that the sum is direct. Decompose the identity 1 = λ∈H(k,k;n) 1λ , where   Iλ , ϕ ∗ (F Sn ) = ϕ ∗ (Iλ ), F Sn = λn

λ∈H (k,l;n)

 and each 1λ ∈ Now, = λ∈H(k,k;n) S(λ) where each S(λ) is isomorphic to a matrix algebra. Since S ∗ (V , n) is the centralizer of ϕ ∗ (F Sn ) it also follows that each 1λ is the identity of S(λ). Since S ∗ (V , n) ⊆ SA∗ (V , n) and LTλ is an SA∗ (V , n)-module, 1λ LTλ ⊆ LTλ . Obviously LTλ ⊆ eTλ W + eT λ W . We show that equality between these modules holds, and this will complete the proof of part 1. Consider first a non-self-conjugate partition λ. Then 1λ LTλ ⊆ LTλ while 1λ gTλ = eTλ therefore eTλ W ⊆ LTλ , and similarly, eT λ W ⊆ LTλ . Hence LTλ = eTλ W + eT λ W for λ non-self-conjugate. We consider next a self-conjugate partition λ. Let % = ±. Since eλ% ∈ SA∗ (V , n), % eλ LTλ ⊆ LTλ . By Lemma 3.4, eλ% gTλ = 0 and since eλ+ eλ− = 0, we deduce that eλ+ gTλ W ⊕ eλ− gTλ W ⊆ LTλ ⊆ eTλ W + eT λ W . Since eλ% gTλ W is an SA∗ (V , n)submodule, it is also an S ∗ (V , n)-module. Since eTλ W and eT λ W are irreducible over S ∗ (V , n), any S ∗ (V , n)-submodule of eTλ W + eT λ W which is a direct sum of two non-zero modules must equal eTλ W + eT λ W . It follows that ϕ ∗ (Iλ ).

S ∗ (V , n)

eλ+ gTλ W ⊕ eλ− gTλ W = LTλ = eTλ W + eT λ W.

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Also, the above equality and the irreducibility of eTλ W and eT λ W as S ∗ (V , n)modules imply that the SA∗ (V , n)-modules eλ+ gTλ W and eλ− gTλ W are irreducible. In particular, dim eλ% gTλ W = dim eTλ W . This completes the proof of part 1, as well as of part 3. To complete the proof of the theorem we only need to show that the Weyl module LTλ is irreducible over SA∗ (V , n) for λ non-self-conjugate. Suppose z = 0 is an arbitrary element of LTλ , say z = x + T y with x, y ∈ eTλ W (see Section 5.1). We show that SA∗ (V , n)z = LTλ . Without loss of generality we assume that x = 0. Then for any λ = µ we have 1λ Wµ = 0. In particular 1λ Wλ = 0. It follows that 1λ z = x. Since 1λ ∈ S ∗ (V , n) ⊆ SA∗ (V , n), we have 0 = x ∈ SA∗ (V , n)z. Since x ∈ eTλ W and eTλ W is an irreducible S ∗ (V , n)-module we have eTλ W = S ∗ (V , n)x ⊆ SA∗ (V , n)z. Applying T we also have T eTλ W = eT λ W ⊆ SA∗ (V , n)z and hence LTλ ⊆ SA∗ (V , n)z. The opposite inclusion is obvious. ✷ 5.4. In Section 5.2 we defined Weyl modules for SA∗ (V , n). We now give an explicit decomposition of each isotypic component of W as a direct sum of Weyl modules. Proposition 5.4. Let λ be non-self-conjugate and let T1 , . . . , Tdλ be the standard tableaux of shape λ. Then Wλ ⊕ Wλ =

dλ 

LTj .

j =1

Proof. Recall the expression for Wλ given in Eq. (5). Applying the vector space isomorphism T : Wλ → Wλ (see Section 5.1) to this expression, we obtain a similar equation for Wλ . Using Theorem 5.3 this implies dλ dλ   Wλ ⊕ W = (eTj W ⊕ T eTj W ) = LTj , λ

j =1

j =1

which is a direct sum of simple modules. ✷ We consider henceforth a self-conjugate partition λ. Let T1 , . . . , Th be the standard tableaux of shape λ whose first row starts with the entries one and two. As remarked in Section 3.1, the number of standard tableaux of shape λ is dλ = 2h. Proposition 5.5. Let λ be self-conjugate and let T1 , . .  . , Th be the standard tableaux of shape λ with one and two in the first row. Then hi=1 LTi = hi=1 LTi and hence h  Wλ = LTi . i=1

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 Proof. By a standard argument, the sum hi=1 eTi W is direct. Since T is a vector  space isomorphism (see Section 5.1), the sum hi=1 T eTi W is also direct. The proof of this proposition follows once we show that h  i=1

eTi W ∩

h 

T eTi W = 0.

i=1

Note that for each 1  i  j we have (1 + (1, 2))RT+i = 2RT+i and (1 + (1, 2)) × RT−i = 0 (see Lemma 3.3). Suppose x is an element in the intersection (eT1 W ⊕ · · · ⊕ eTh W ) ∩ T (eT1 W ⊕ · · · ⊕ eTh W ). We calculate the element (1 + (1, 2))x in two different ways: The element x has the form x = hi=1 RT+i CT−i wi for some elements wi ∈ W and, by the above, we obtain (1 + (1, 2))x = 2x. On the other hand, since x also has the form  x = hi=1 RT−i CT+i wi for some wi ∈ W we obtain (1 + (1, 2))x = 0. Thus 2x = 0 implies that x = 0. ✷ 6. Splitting the Weyl modules via eigenvalues 6.1. In this section we discuss the splitting of the reducible Weyl modules via eigenvalues. Let Z(FAn ) denote the center of FAn . Recall from Section 2.5 that when acting on W , F Sn means ϕ ∗ (F Sn ). Also, recall the notation and facts given in Section 2.3 on the conjugacy classes Cµ of An and its characters χ λ and χ λ± . In case we need to distinguish the minimal central idempotents e of An and Sn we write instead of e either eAn or eSn . We begin with the following simple lemma on the left multiplication: Lemma 6.1. Let λ be a partition of n lying in the (k, k)-hook and let fa : W → W be the left multiplication fa (x) = ax. (1) If a ∈ Z(F Sn ) then the restriction of fa to the S ∗ (V , n)-Weyl module eTλ W is an endomorphism. (2) If a ∈ Z(FAn ) then the restriction to the SA∗ (V , n)-Weyl module LTλ satisfies aLTλ ⊆ LTλ . Let α be an eigenvalue of the restriction fa to LTλ and let Lα be the corresponding eigenspace. Then Lα is an SA∗ (V , n)-submodule of LTλ . Proof. We prove only the second part. It follows from the definition of SA∗ (V , n) that Z(FAn ) ⊆ SA∗ (V , n), hence, since LTλ is a left SA∗ (V , n)-module, the restriction of fa to LTλ is fa : LTλ → LTλ . Also sa = as for any s ∈ SA∗ (V , n), which implies for an element x ∈ LTλ that fa (sx) = asx = sax = α(sx). ✷ Since F denotes the field of complex numbers, such fa has precisely one eigenvalue on the S ∗ (V , n)-module eTλ W and at least one eigenvalue on the

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SA∗ (V , n)-module LTλ . The irreducibility of LTλ when λ = λ , and its being the direct sum of two irreducibles when λ = λ , imply that the linear map fa has precisely one eigenvalue for λ non-self-conjugate and at most two distinct eigenvalues in the other case. 6.2. It is well known that the class sums of a finite group form a basis for the center of the group algebra. We proceed with describing the map fa operating on the Weyl modules of S ∗ (V , n) and SA∗ (V , n) where a is a class sum for either Sn or An . We fix the following notation: For an eigenvalue α of fa we denote the eigenspace of fa with respect to α by Lα , and moreover, we define α(λ, µ) by α(λ, µ) =

|Cµ |χ λ (µ) . χ λ (1)

(6)

In this equation the involved characters are Sn -characters and the conjugacy class is an Sn -conjugacy class. Theorem 6.2 (The Sn -case). The map fcµ : eTλ W → eTλ W is given by fcµ (x) = αx and eTλ W = Lα for α = α(λ, µ). Proof. Denote by eν the minimal central idempotent of Sn which is the identity of Iν . By the orthogonality relations for the irreducible characters of Sn , we can express (see [3, p. 237]) the class sum cµ of the conjugacy class Cµ by cµ =

|Cµ |χ ν (µ) ν

χ ν (1)

eν .

(7)

As eTλ ∈ Iλ we obtain eν eTλ = 0 for ν = λ and eν eTλ = eTλ for ν = λ. This implies cµ eTλ = αeTλ with α = α(λ, µ) defined as in Eq. (6). Hence, using Lemma 6.1, we obtain fcµ (x) = α(λ, µ)x for any element x ∈ eTλ W . ✷ Theorem 6.3 (The An -case). Consider the map fcµ : LTλ → LTλ for an even partition µ of n. (1) Assume that the Sn -conjugacy class Cµ does not split in An . Then fcµ (x) = αx for x ∈ LTλ and LTλ = Lα for α = α(λ, µ). (2) Assume that the conjugacy class Cµ splits in An . Let either λ be non-selfconjugate or let λ be self-conjugate such that µ = h(λ) (see Section 2.3). Then for x ∈ LTλ we have fcµ± (x) = αx and LTλ = Lα for α = 12 α(λ, µ). Proof. (1) By Theorem 6.2, fcµ (x) = α(λ, µ)x for any x ∈ eTλ W and where α(λ, µ) is given as in Eq. (6). Next, let x ∈ T eTλ W = eT λ W (see Section 5.1) where eT λ ∈ Iλ . Similar as before, by applying Eq. (7), we obtain that fcµ (x) = α(λ , µ)x. Since µ is even, α(λ, µ) = α(λ , µ).

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(2) We give the proof only for the class sum cµ+ (a similar proof holds for cµ− ). (a) Suppose first that λ is non-self-conjugate. Rewriting Eq. (7) for An we obtain |Cµ+ |χ λ (µ+ ) An eλ + other summands cµ+ = χ λ (1) where the other summands involve eνAn with ν = λ, λ . Since λ is non-selfconjugate and by Section 2.4 we obtain eλAn eTλ = eλSn eTλ = eTλ . Thus for x = eTλ w ∈ eTλ W we have cµ+ x = cµ+ eTλ w =

|Cµ+ |χ λ (µ+ ) 1 eTλ w = α(λ, µ)eTλ w λ χ (1) 2

where the last equation follows since χ λ (µ+ ) = 12 χ λ (µ) (see Section 2.3). If x ∈ T eTλ W = eT λ W the argument is similar and the above statement for a nonself-conjugate partitions λ follows. (b) Suppose next that λ is self-conjugate with µ = h(λ). We have eλSn = An and the irreducible = χ λ+ + χ λ− . Then

eλ+ + eλ− where e± are minimal central idempotents in Sn -character splits into two irreducible An -characters χ λ +

cµ+ =

|Cµ+ |χ λ (µ+ ) +

χ λ (1)

−

eλ+

+

|Cµ+ |χ λ (µ+ ) −

χ λ (1)

eλ− + other summands.

±

Since µ = h(λ) and χ λ (µ± ) = 12 χ λ (µ) it follows that 1 cµ+ = α(λ, µ)eλSn + other summands 2 and the proof follows by continuing as above. ✷ 6.3. So far we described the conditions on λ and µ such that the map fcµ : LTλ → LTλ has one eigenvalue and is given by a scalar multiple of the identity matrix. In case λ is self-conjugate with µ = h(λ) we obtain two eigenvalues for fcµ± and this case is considered in this subsection. Define for % = ±
 |Ch(λ) | λ%
 α % = α % λ, h(λ) = λ · χ h(λ)+ . χ (1)

(8)

Theorem 6.4 (The An -case). The left multiplications fch(λ)± : LTλ → LTλ both have the two eigenvalues α % defined in Eq. (8). They are given by

α + x for x ∈ L+ Tλ , fch(λ)+ (x) = and α − x for x ∈ L− Tλ ,

α + x for x ∈ L− Tλ , fch(λ)− (x) = − α x for x ∈ L+ Tλ ,

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where in the first case L%Tλ = Lα % and in the second case L%Tλ = Lα −% . Proof. Note that eλ+ annihilates elements in L− Tλ and acts as identity on elements − + in L+ . Similarly, e annihilates L and acts as identity on L− λ Tλ Tλ Tλ . Moreover, +

ch(λ)+ =

|Ch(λ)+ |χ λ (h(λ)+ ) +

χ λ (1) + other summands.

−

eλ+

+

|Ch(λ)+ |χ λ (h(λ)+ ) −

χ λ (1)

eλ−

With this it follows that ch(λ)+ x = α % x for an element x ∈ L%Tλ . A similar argu+

−

ment holds for ch(λ)− using that χ λ (h(λ)+ ) = χ λ (h(λ)− ) (see Section 2.3). ✷

7. Bases for the Weyl modules In this section we discuss bases of Weyl modules for the Schur algebra of the alternating groups. Theorem 7.1 describes a basis for every LTλ , Theorem 7.3 describes another such basis when n is large enough. When λ is self-conjugate, − Theorem 7.2 describes bases for L+ Tλ and LTλ . These bases allow us to attach to each irreducible module a certain supersymmetric function. Unlike in the Sn case, in the case of a self-conjugate partition these functions do not characterize the corresponding modules. 7.1. Based on Corollary 3.18 of [1] we now describe an explicit basis for the Weyl module LTλ . Let t1 , . . . , tk ∈ V0 and u1 , . . . , uk ∈ V1 be a fixed basis of V0 ⊕ V1 . Recall the definition of monomial given in Section 4.2. Then a monomial w ∈ W is called semi-standard if its corresponding (k, k)-tableau is semi-standard. Call a monomial w ∈ W dual-semi-standard if Tλ identifies w with a tableau T which becomes semi-standard upon exchanging the t’s and the u’s (that is after applying T , see Section 5.1). Denote the dual of a semi-standard tableau w by w . Theorem 7.1. Let λ be a partition of n lying in a (k, k)-hook and let Tλ be a tableau of shape λ. Then LTλ has basis    eTλ w, eT λ w   w ∈ W is (k, k)-semi-standard of shape λ . Proof. By Corollary 3.18 of [1], {eTλ w | w is (k, k)-semi-standard} is a basis of the vector space eTλ W . Now, up to a ± sign, T interchanges between the t’s and the u’s in a monomial, hence the vector space isomorphism eTλ W ∼ = T eTλ W = eT λ T W = eT λ W implies that {eT λ w | w is (k, k)-dual-semi-standard} is a basis of the vector space T eTλ W = eT λ W . The proof follows from the first part of Theorem 5.3. ✷

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191

For a self-conjugate partition λ the Weyl module splits by Theorem 5.3 into two non-isomorphic irreducible components. Explicit bases for these irreducible components are given in the following theorem. Theorem 7.2. Let λ be a self-conjugate partition, Tλ a tableau of shape λ and eλ% ∈ FAn with % = ± be the corresponding minimal central idempotents. Then a basis of L%Tλ is given by    % eλ eTλ w  w ∈ W is (k, k)-semi-standard of shape λ . Proof. We show the theorem for % = +. Suppose the tableau has first row starting with the entries a and b. Denote by τ the transposition τ = (a, b). Then τ ∈ RTλ and hence τ RT+λ = RT+λ and τ eTλ = eTλ . It is well known that for any odd permutation η in Sn we have eλ− = ηeλ+ η−1 . See, for example, [5, 2.5.7]. In particular, we therefore have eλ− = τ eλ+ τ which also can be seen directly from the expressions for e% , see Section 2.4. Let X denote the set of the (k, k)-semistandard tableaux of shape λ. Assume that αw eλ+ eTλ w = 0. (9) w∈X

We calculate from this assumption the value for the same equation with eλ+ replaced eλ− : αw eλ− eTλ w = αw τ eλ+ (τ eTλ )w = τ αw eλ+ eTλ w = 0. w∈X

w∈X

Hence Eq. (9) implies that also αw eλ− eTλ w = 0.

w∈X

(10)

w∈X

Recall  that eλ+ + eλ− = eλ and eλ eTλ = eTλ . Thus adding Eqs. (9) and (10) we obtain w∈X αw eTλ w = 0. By Theorem 7.1 this implies that all αw = 0. ✷ 7.2. The elements gTλ , introduced and discussed in Sections 3 and 4 give the decomposition of the two-sided ideals of FAn into minimal left ideals (see Theorem 3.2) and therefore play a similar role in the An -theory as the elements eTλ in the Sn -theory. As a basis for the super-Weyl modules in the Sn -case is described by the elements eTλ it suggests to look for a basis of the super-Weyl modules LTλ in the An -case described by the gTλ . Theorem 7.3. Let n > 7k 2 and let λ be a partition of n lying in a (k, k)-hook and Tλ a tableau of shape λ. Then a basis of the super-Weyl module LTλ of SA∗ (V , n) is given by    gT w, gT w   w ∈ W is (k, k)-semi-standard of shape λ . λ

λ

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Note that in the case λ = (3, 1) and k = 1 the corresponding set is not a basis for LTλ . Proof. Let X denote the set of the (k, k)-semi-standard tableaux of shape λ. By Corollary 3.15 of [1] the elements {eTλ w | w ∈ X} are linearly independent. We show first that the elements {gTλ w | w ∈ X} are linearly independent. The case λ = (1n ) is easy: Let w ∈ X. Since n > 2k and each of t1 , . . . , tk appears at most once in w, there exist a uj which appears at least twice in w. Thus CT+λ w = 0 and hence gTλ w = eTλ w, which clearly implies the above linear independence. Assume next that λ = (1n ), that is λ1
2, and let (a, b) ∈ RTλ . Then (1 + (a, b))gTλ = 2eTλ (see Lemma 3.3) and so (1 + (a, b))gTλ w = 2eTλ w and we deduce that the elements {gTλ w | w ∈ X} are also linearly independent. Deduce  | w ∈ X}: Indeed, here now that this holds also for the elements in the set {gTλ w (see Section 5.1)  = eT λ w  = ±eT λ T w = ±T eTλ w, eλS n gTλ w and independance follows since T : eTλ W → T eTλ W is a vector space isomorphism. Recall that each gTλ is a signed sum of the permutations pq where p ∈ RTλ , q ∈ CTλ (and pq ∈ An ). Thus a linear combination of elements of the set {gTλ w | w ∈ X} is a linear combination of such monomials pqw. Similarly  | w ∈ X}. Let S denote the linear span of {gTλ w | w ∈ X} and  S that for {gTλ w  | w ∈ X}. The theorem will be proved once we show that S ∩  S = 0, and of {gTλ w this will follow once we show that no monomial of the form p1 q1 w1 can equal 2 where p1 , p2 ∈ RTλ and q1 , q2 ∈ CTλ and w1 , w2 ∈ X. For a monomial p2 q2 w a contradiction we assume the opposite, that is assume that p1 q1 w1 = ±p2 q2 w 2 . As usual, the fixed tableau Tλ identifies a monomial w with its (k, k)-tableau. Suppose µ denotes the first k rows of a tableau of shape λ. Then we call this the µ-part of this tableau. Consider the µ-part of p1 q1 w1 . First, the number of ui ’s in that µ-part of w1 is less than or equal to k 2 since w1 semi-standard. Now, the column permutation q1 might bring at most k 2 new ui ’s into that µ-part, while the row permutation p1 does not change the amount of ui ’s in that µ-part. It follows that the number of ui ’s in the µ-part of p1 q1 w1 is less than or equal to 2k 2 : degu [(p1 q1 w1 )µ ]  2k 2 . The total ti ’s degree of the µ-part therefore satisfies degt [(p1 q1 w1 )µ ]
|µ| − 2k 2 . Now, in a dual semi-standard tableau, the role of the ti ’s and the ui ’s is 2 )µ ]  2k 2 . It follows that reversed. Hence, by the same argument, degt [(p2 q2 w 2 if |µ| > 4k then degt [(p1 q1 w1 )µ ] > degt [(p2 q2 w 2 )µ ] and hence p1 q1 w1 = 2 . p2 q2 w Next, denote the first k columns of λ by ν. By the same argument, if |ν| > 4k 2 then p1 q1 w1 cannot be of the form p2 q2 w 2 . Finally, if |λ| > 7k 2 then either 2 2 |µ| > 4k or |ν| > 4k , and the proof of the theorem follows. ✷

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7.3. The supersymmetric functions We follow [1, Section 6] whose basic idea is due to I. Schur. Let P = diag(x1 , . . . , xk , y1 , . . . , yk ) be an endomorphism in End(V ). Given an irreducible SA∗ (V , n)-submodule M of W , one calculates the trace of P ⊗n on M to obtain a supersymmetric function SM (x; y) = SM (x1 , . . . , xk , y1 , . . . , yk ). (1) Let λ be non-self-conjugate and M = LTλ . By Theorem 7.1 and by Lemma 6.9 in [1] it follows that SLTλ (x; y) = H Sλ (x; y) + H Sλ(y; x). Since H Sλ (y; x) = H Sλ (x; y) we get SLTλ (x; y) = H Sλ (x; y) + H Sλ (x; y). − (2) Let λ be self-conjugate. For either M = L+ Tλ or M = LTλ Theorem 7.3 and again Lemma 6.9 in [1] imply that

SL+ (x; y) = SL− (x; y) = H Sλ (x; y). Tλ

Tλ

Remark. In the Sn -case the functions H Sλ (x; y) characterize the irreducible S ∗ (V , n)-submodules of W (see Section 6 in [1]). In the An -case this however is not true: it is still true for the non-self-conjugate partitions, but for a self-conjugate − partition λ, to the two irreducible and non-isomorphic submodules L+ Tλ and LTλ corresponds the same supersymmetric function H Sλ (x; y).

Appendix A. (Written together with T. Seeman) In this appendix we give two proofs of Theorem 4.1. We recall this theorem. Theorem A.1. Let λ be a self-conjugate partition lying in a (k, k)-hook and assume λ1
2k + 1 (this in particular holds if n > 3k 2 ). Let Tλ be a tableau of shape λ with corresponding elements e = eTλ and e = eT λ . Then ee = e e = 0 and hence gTλ is a semi-idempotent. First proof. We show that ee = 0. Here we employ the classical action of Sn on V ⊗n , i.e. V = V0 and V1 = 0. Assume dim V
n and let M ∈ V ⊗n denote the monomial M = x1 ⊗ · · · ⊗ xn ≡ x1 · · · xn , where x1 , . . . , xn ∈ V are linearly independent. For simplicity we denote ϕσ (M) = σ M. As vector spaces, F Sn ∼ = span{σ M | σ ∈ Sn }; hence for a ∈ F Sn , a = 0 if and only if aM = 0. As

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was remarked in Section 4.1 to prove that ee = 0, we can choose any Tλ , then prove that eTλ eT λ = 0. Here we choose h1 + 1 h1 + 2 .. .

1 2 . Tλ = .. · h1

h1 + h2

· · ·

·

·

where λ j = hj , j = 1, 2, . . . . Now, Tλ identifies monomials with tableaux. λ

λ

Consider the monomials r = x1 1 x2 2 · · · and the above M. Rename x1 , . . . , xn by xi,j ’s so that M and r are identified with the following tableaux: x1,1 x1,2 M= · · x1,h1

x2,1 x2,2 · x2,h2

· · · ·

·

x1 x1 and r = · · x1

x2 x2 · x2

· · · ·

· .

Linearization. Let m ∈ V ⊗n be a monomial in the variables x1 , . . . , xk of degree di = ∂xi (m) in xi , where i = 1, . . . , k. Write m = m1 xi m2 xi m3 · · · mdi xi mdi +1 where m1 , . . . , md+1 do not involve xi . Then the operator Lxi (linearization of xi ) is defined by linearity and by: Lxi (m) = m1 xi,σ (1)m2 xi,σ (2) · · · xi,σ (di ) mdi +1 . σ ∈Sdi

It is easy to verify that for i = j we have Lxi Lxj = Lxj Lxi . Now define the linearization operator L by linearity and by L(m) = Lx1 · · · Lxk (m). For example, using the identification of monomials m ∈ V ⊗n and tableaux, induced by the above Tλ , the above monomials r and M are related by L(r) = C + M where C = CTλ . Note the following basic property: Let a ∈ F Sn then, as operators on V ⊗n , aL = La. (Indeed we may assume that a = σ ∈ Sn then verify, for a given i, that σ Lxi = Lxi σ , which is straightforward, since σ permutes the places in a monomial.) In Lemma A.2 we deduce the basic tool for proving Theorem 4.1. The proof of Theorem 4.1 then follows from Lemma A.3. ✷ Lemma A.2. Recall that ee = R + C − R − C + and let the tableau r of shape λ be defined as above. If R + C − R − r = 0 then ee = 0. Proof. Firstly, ee = 0 if and only if ee M = 0. Since by above C + M = L(r) and (R + C − R − )L = L(R + C − R − ) we deduce that ee M = R + C − R − C + M = (R + C − R − )L(r) = L(R + C − R − r), and the proof follows. ✷

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Lemma A.3. Assume that λ1
2k + 1 and let r be as above. Then C − R − r = 0. Proof. Identify r with the tableau

r=

A1 A3

A2

1 2 · · 1 2 where A3 = · · · · 1

· · · · ·

· k · · · k ·

and thenumber of ones in the first column of A3 is at least k + 1. Then R − r = ±mi where mi =

Ai,1 Ai,3

Ai,2

.

Since each row in Ai,3 is a permutation of 1, . . . , k, each row contains a one; hence the number of ones in Ai,3 is greater than or equal to k + 1. Since the number of columns in Ai,3 is  k, it follows that there is at least one column in Ai,3 —and therefore also in m1 —with at least two ones. This implies that C − mi = 0. ✷ Second proof. Let R = RTλ , C = CTλ , and recall that ee = R + C − R − C + . We show that ee = 0 by proving (see the beginning of Section 4.1) that C − R − C + Tλ = sgn(qp)qpqT ¯ λ= sgn(p) sgn(q)qpqT ¯ λ = 0. q,q∈C ¯ p∈R

p∈R

q,q∈C ¯

To do so it suffices to prove the following: Given p ∈ R, q, q¯ ∈ C, let S = (qm , q¯m )} be the set of all pairs qi , q¯i ∈ C satisfying qi pq¯i Tλ = {(q1 , q¯1 ), . . . ,  qpqT ¯ λ . Then m i=1 sgn(qi ) = 0. Let T1 be the first k rows of Tλ , and T2 the remaining rows. It follows from the assumptions on λ that T2 consists of at most k columns, with the leftmost column of T2 being of length k + 1 or greater. A pair of elements is said to be co-column if they appear in the same column in a given tableau. We claim that for every i ∈ {1, . . . , m}, there exists a pair of elements which are co-column in both q¯i Tλ and pq¯i Tλ . Define T1 (q¯i ) to be the first k rows of q¯i Tλ , and T2 (q¯i ) the remaining rows, and similarly define T1 (pq¯i ) and T2 (pq¯i ). Consider the first k + 1 elements in the leftmost column of T2 (q¯i ). Since p ∈ R, each of these elements appears in T2 (pq¯i ). But T2 (pq¯i ) consists of at most k columns, hence it follows by the pigeonhole principle that at least two of these k + 1 elements must also be co-column in T2 (pq¯i ). We can therefore let a and b be a pair of elements which are co-column in both q¯1 Tλ and pq¯1 Tλ , with a being higher than b in q¯1 Tλ . Let x be the number of pairs (qi , q¯i ) such that a and b are co-column in both q¯i Tλ and pq¯i Tλ , with a higher than b in q¯i Tλ . Similarly, let y be the number of pairs (qi , q¯i ) such that a and b

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are co-column in both q¯i Tλ and pq¯i Tλ , but with b higher than a in q¯i Tλ . Without loss of generality, the above set S is ordered so that a and b are co-column in both q¯i Tλ and pq¯i Tλ for all i ∈ {1, . . . , x + y}. Furthermore, if i ∈ {1, . . . , x}, then a is ∈ {x + 1, . . . , x + y}, however, then b is higher than a. higher than b in q¯i Tλ ; if i  x+y We claim that y = x, and i=1 sgn(qi ) = 0. Given (qi , q¯i ), i  x, let τi ∈ C be the permutation that, when applied to q¯i Tλ , swaps a and b and leaves all other entries unchanged. Then q¯i Tλ and τi q¯i Tλ are identical except with a and b in opposite locations; hence pq¯i Tλ and pτi q¯i Tλ are also identical except with a and b in opposite locations. It follows that a and b are co-column in both τi q¯i Tλ and pτi q¯i Tλ , with b higher than a in τi q¯i Tλ . Now, given such a pair (qi , q¯i ) with τi as above, let σi be the column permutation that swaps a and b in pq¯i Tλ and leaves all other entries unchanged. Thus pq¯i Tλ = σi pτi q¯i Tλ , hence (qi σi )p(τi q¯i )Tλ = qi pq¯i Tλ = qpqT ¯ λ . Note that σi , τi ∈ C implies qi σ, τi q¯i ∈ C, hence (qi σi , τi q¯i ) ∈ S. Moreover, since a and b are co-column in both τi q¯i Tλ and pτi q¯i Tλ , with b higher than a in τi q¯i Tλ , (qi σi , τi q¯i ) = (qj , q¯j ) for all 1  j  x, hence we can reorder S such that (qx+i , q¯x+i ) = (qi σi , τi q¯i ). For all i  x, both τi and σi are column permutations which swap two elements in a tableau, leaving all other elements fixed. Thus τi2 = σi2 = 1, so (qi , q¯i ) = (qx+i σi , τi q¯x+i ). This defines a bijection between the sets {(q1 , q¯1 ), . . . , (qx , q¯x )} and {(qx+1, q¯x+1 ), . . . , (qx+y , q¯x+y )}, which implies that y = x. Moreover, it also x+y  implies that sgn(qx+i σi ) = sgn(qi ). Thus i=1 sgn(qi ) = 2x i=1 sgn(qi ) = 0. Recall that S = {(q1 , q¯1 ), . . . , (qm , q¯m )}. Thus if x + y = m, then we are done. If x + y < m, however, then repeat this procedure on {(qx+y+1, q¯x+y+1), . . . , (qm , q¯m )}. ✷

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