Some new decomposable Specht modules
aa r X i v : . [ m a t h . R T ] M a r This is the second author’s version of a work that was accepted forpublication in the Journal of Algebra. Changes resulting from the publish-ing process, such as peer review, editing, corrections, structural formatting,and other quality control mechanisms may not be reflected in this docu-ment. Changes may have been made to this work since it was submittedfor publication. A definitive version was subsequently published in
J. Algebra (2012) 235–62.http: // dx.doi.org / / j.jalgebra.2012.01.035 ome new decomposable Specht modules Craig J. Dodge
Department of Mathematics, University at Bu ff alo, SUNY,244 Mathematics Building, Bu ff alo, NY 14260, U.S.A. Matthew Fayers
Queen Mary, University of London, Mile End Road, London E1 4NS, U.K.2010 Mathematics subject classification: 20C30, 05E10
Abstract
We present (with proof) a new family of decomposable Specht modules for the sym-metric group in characteristic 2. These Specht modules are labelled by partitions of theform ( a , , b ), and are the first new examples found for thirty years. Our method of proofis to exhibit summands isomorphic to irreducible Specht modules, by constructing explicithomomorphisms between Specht modules. Let n be a positive integer, and let S n denote the symmetric group on n letters. For any field F , the Specht modules form an important family of modules for F S n . If F has characteristic zero,then the Specht modules are precisely the irreducible modules for F S n . If F has positive char-acteristic, the simple F S n -modules arise as quotients of certain Specht modules. In addition,the Specht modules arise as the ‘cell modules’ for Murphy’s cellular basis of F S n .A great deal of e ff ort is devoted to determining the structure of Specht modules; in partic-ular, finding the composition factors of Specht modules and the dimensions of the spaces ofhomomorphisms between Specht modules. In this paper, we consider the question of whichSpecht modules are decomposable. It is known that in odd characteristic the Specht modulesare all indecomposable, so we can concentrate on the case where char( F ) =
2. In fact, since anyfield is a splitting field for S n , we can assume that F = F . In this case, there are decompos-able Specht modules, but remarkably few examples are known. Murphy [M1] analysed theSpecht modules labelled by ‘hook partitions’, i.e. partitions of the form ( a , b ), computing theendomorphism ring of every such Specht module (and thereby determining which ones aredecomposable). However, in the last thirty years no more progress seems to have been made.Our main result is the discovery of a new family of decomposable Specht modules, the firstexamples of which were discovered by the two authors independently using computationswith GAP and MAGMA. These new decomposable Specht modules are labelled by partitionsof the form ( a , , b ), where a , b are even. So in this paper we make a case study of partitions ofthis form; we are unable to apply Murphy’s method to determine exactly which of these Specht1 Craig J.DodgeandMatthewFayersmodules are decomposable, but by considering homomorphisms between Specht modules,we are able to show which irreducible Specht modules arise as summands of these Spechtmodules. We then apply this result to determine which of our Specht modules have a summandisomorphic to an irreducible Specht module.We now briefly indicate the layout of this paper. In the next section, we recall some basicdefinitions and results in the representation theory of the symmetric group, which enable usto state our main results in Section 3. In Section 4 we go into more detail on homomorphismsbetween Specht modules. In Sections 5 and 6 we consider the two classes of irreducible Spechtmodules which can occur as summands of our decomposable Specht modules. We then applythese results in Section 7 to complete the proof of our main results. Finally, we make someconcluding remarks in Section 8. Acknowledgements.
The authors are indebted to David Hemmer, who first made us aware ofeach other’s work and initiated this collaboration, and also invited the second author to SUNYBu ff alo in September 2011, where some of this work was carried out. This work continuedduring the ‘New York workshop on the symmetric group’; we are grateful to Rishi Nath ofCUNY for inviting us to this conference.The research of the first author was supported in part by NSA grant H98230-10-1-0192. In this section, we summarise some basic results on the representation theory of the sym-metric group. For brevity, we specialise some results to characteristic 2, referring the reader tothe literature for general results.We begin by fixing a field F ; all our modules will be modules for the group algebra F S n .We assume familiarity with James’s book [J2]; in particular, we refer the reader there forthe definitions of partitions, the dominance order, the permutation modules M λ , the Spechtmodules S λ and the simple modules D λ . We shall also briefly use the Nakayama Conjecture[J2, Theorem 21.11] which describes the block structure of the symmetric group.We also need the following two results; recall that if λ is a partition then λ ′ denotes theconjugate partition. Lemma 2.1.
Suppose char( F ) = and λ is a partition such that S λ is irreducible. Then S λ (cid:27) S λ ′ . Proof.
By [J2, Theorem 8.15] we have S λ (cid:27) ( S λ ′ ) ∗ , since the sign representation is trivial incharacteristic 2. But by [J2, Theorem 11.5], all simple modules for the symmetric group areself-dual. (cid:3) Lemma 2.2. If λ, µ are partitions of n, then dim F Hom F S n ( S λ , S µ ) = dim F Hom F S n ( S µ ′ , S λ ′ ) . Proof.
This also follows from [J2, Theorem 8.15]. (cid:3) omenewdecomposableSpechtmodules 3
We recall here a useful lemma which we shall use later; this is due to James, although it doesnot appear in the book [J2]. We concentrate on the special case where F has characteristic 2,referring to [J1] for the full result.For any l >
1, the l th ladder in N is L l = n ( i , j ) (cid:12)(cid:12)(cid:12) i + j = l + o . If λ is a partition, the 2 -regularisation of λ is the partition λ reg whose Young diagram is obtainedby moving the nodes in [ λ ] as high as possible within their ladders. For example, (8 , , ) reg = (8 , , λ reg is a 2-regular partition, and we have the followingresult. Theorem 2.3. [J1, Theorem A]
Suppose λ and µ are partitions of n, with µ -regular. Then [ S λ : D λ reg ] = , while [ S λ : D µ ] = if µ S λ reg . In this paper we shall be concerned with the Specht modules labelled by partitions of theform ( a , , b ); so we compute the regularisations of these partitions. Lemma 2.4.
Suppose a > and b > . Then ( a , , b ) reg = ( a , b + ,
2) ( a > b )( b + , a − ,
2) ( a b ) . It will be very helpful to know the classification of irreducible Specht modules, which (incharacteristic 2) was discovered by James and Mathas [JM2]. If k is a non-negative integer welet l ( k ) denote the smallest positive integer such that 2 l ( k ) > k . Theorem 2.5. [JM2, Main Theorem]
Suppose µ is a partition of n and char( F ) = . Then S µ isirreducible if and only if one of the following occurs: • µ i − µ i + ≡ − l ( µ i + − µ i + ) ) for each i > ; • µ ′ i − µ ′ i + ≡ − l ( µ ′ i + − µ ′ i + ) ) for each i > ; Craig J.DodgeandMatthewFayers • µ = (2 ) . Note that µ satisfies the first condition in the theorem if and only if µ ′ satisfies the second.In view of Lemma 2.1 (and since we shall only be considering values of n greater than 4) wemay assume that any irreducible Specht module is of the form S µ where µ satisfies the firstcondition in the theorem. In this section, we describe the new family of decomposable Specht modules discussed inthis paper, and the method we use to prove decomposability.
For the rest of this paper, we assumethat F has characteristic . Computer calculations show that the first few decomposable Specht modules which are notlabelled by hook partitions have labelling partitions of the form ( a , , b ) (and their conjugates)with a , b even positive integers. So in this paper we make a case study of this family ofpartitions. Our technique is di ff erent from that of Murphy [M1], and is weaker in the sense thatwe cannot always when tell for certain whether one of our Specht modules is decomposable.However, in the cases where we can show decomposability, we have the advantage of beingable to describe one summand explicitly.More specifically, our main result is a determination of exactly which irreducible Spechtmodules occur as summands of the Specht modules S ( a , , b ) . The technique we use is toconsider homomorphisms between Specht modules, and the set-up for computing such homo-morphisms is described in Section 4.We use homomorphisms between Specht modules in the following way. Suppose λ, µ arepartitions of n . Then it is a straightforward result that S µ occurs as a summand of S λ if andonly if there are homomorphisms γ : S µ → S λ and δ : S λ → S µ such that δ ◦ γ is the identityon S µ . If we assume in addition that S µ is irreducible, then by Schur’s Lemma we just need toshow that δ ◦ γ is non-zero.Some e ff ort has been devoted to computing the space of homomorphisms between twoSpecht modules, beginning with the paper of the second author and Martin [FM]. In fact, thereis now an explicit algorithm which computes the homomorphism space Hom F S n ( S λ , S µ ) exceptwhen char( F ) = λ is 2-singular. Even in this exceptional case, this technique can be usedto construct some homomorphisms between S λ and S µ , though only if λ dominates µ .In our situation, the partitions µ we shall consider are always 2-regular, because (as longas n ,
4) every irreducible Specht module in characteristic 2 has the form S µ for µ F S n ( S µ , S λ ). Computing the space Hom F S n ( S λ , S µ ) isharder, because λ is 2-singular, so we fall foul of the exception above. To circumvent this, weuse Lemma 2.1, which allows us to take δ to be an element of Hom F S n ( S λ , S µ ′ ). By Lemma 2.2this has the same dimension as Hom F S n ( S µ , S λ ′ ), which we can compute because µ is 2-regular.Having established the dimension of this space, we can construct all possible homomorphisms δ , and then check the condition δ ◦ γ , S λ which are isomorphic to irreducible Spechtmodules S µ . In fact, we can restrict attention to a small set of candidate Specht modules S µ , asfollows. Assuming S µ is irreducible and µ is 2-regular, S µ is isomorphic to the simple module D µ ; therefore in order for S µ to appear as a summand of S λ , the decomposition number [ S λ : D µ ]must be non-zero. Therefore by Theorem 2.3, µ must dominate λ reg , and by Lemma 2.4 thisomenewdecomposableSpechtmodules 5has the form ( x , y ,
2) for some x , y . So we may assume that µ has the form ( u , v , w ) for some u > v > w
2. Furthermore, S µ and S λ must lie in the same block of F S n . Using the NakayamaConjecture, this means that u , w must be even, while v is odd. So we can restrict attention to µ of the form ( u , v ) or ( u , v ,
2) where u is even (and hence v is odd). We apply the techniquedescribed above to these two types of partition in Sections 5 and 6. Our main result is thefollowing. Theorem 3.1.
Suppose λ = ( a , , b ) is a partition of n, where a , b are positive even integers with a > ,and suppose µ is a partition of n such that S µ is irreducible. Then S λ has a direct summand isomorphicto S µ if and only if one of the following occurs.1. µ or µ ′ equals ( u , v ) , where v ≡ and (cid:16) u − va − v (cid:17) is odd.2. µ or µ ′ equals ( u , v , , where (cid:16) u − va − v (cid:17) is odd. Using this result, we can show that most of the Specht modules under consideration aredecomposable. Specifically, we have the following result.
Corollary 3.2.
Suppose a , b are positive even integers with a > , and let λ = ( a , , b ) . Then S λ has asummand isomorphic to an irreducible Specht module if and only if at least one of the following occurs: • a + b ≡ or , a > and b > ; • a + b ≡ and (cid:16) a + b − a − (cid:17) is odd; • a + b ≡ and (cid:16) a + b − a − (cid:17) is odd. In this section, we explain the set-up for computing the space of homomorphisms betweentwo Specht modules. We begin with a revision of some material from [J2], before citing someresults of the second author and Martin.
Suppose µ and λ are partitions of n . Since S λ M λ , any homomorphism from S µ to S λ canbe regarded as a homomorphism from S µ to M λ . This is very useful, because if µ is 2-regular (orif char( F ) , F S n ( S µ , M λ ) can be described explicitly. Furthermore, usingthe Kernel Intersection Theorem below, one can check whether the image of a homomorphism θ : S µ → M λ lies in S λ .We now make some more precise definitions. We take λ, µ as above, but we now allow λ tobe any composition of n , not necessarily a partition. A µ -tableau of type λ is a function T fromthe Young diagram [ µ ] to N with the property that for each i ∈ N there are exactly λ i nodes Craig J.DodgeandMatthewFayersof [ µ ] mapped to i . Such a tableau is usually represented by drawing [ λ ] with a box for eachnode n , filled with the integer T ( n ). T is row-standard if the entries in this diagram are weaklyincreasing along the rows, and is semistandard if the entries are weakly increasing along therows and strictly increasing down the columns.We write T r ( µ, λ ) for the set of row-standard µ -tableaux of type λ , and T ( µ, λ ) for the setof semistandard µ -tableaux of type λ . For each T ∈ T r ( µ, λ ), James defines a homomorphism Θ T : M µ → M λ (over any field), whose precise definition we do not need here. The restrictionof Θ T to S µ is denoted ˆ Θ T . Now we have the following. Theorem 4.1. [J2, Lemma 13.11 & Theorem 13.13]
The set n ˆ Θ T (cid:12)(cid:12)(cid:12) T ∈ T ( µ, λ ) o is linearly independent, and spans Hom F S n ( S µ , M λ ) if µ is -regular. Now return to the assumption that λ is a partition. As a consequence of Theorem 4.1, in orderto compute Hom F S n ( S µ , S λ ) when µ is 2-regular, we just need to find all linear combinations θ of the homomorphisms ˆ Θ T for T ∈ T ( µ, λ ) for which the image of θ lies in S λ . Even when µ isnot 2-regular, homomorphisms from S µ to S λ can very often be expressed in this way.In order to determine whether the image of such a homomorphism θ lies in S λ , we useanother theorem of James which provides an alternative definition of S λ . For any pair ( d , t ) with d > t λ d + , there is a homomorphism ψ d , t : M λ → M ν , where ν is a compositiondepending on λ, d , t . Again, we refer the reader to [J2, §
17] for the definition; we warn thereader that the homomorphism ψ d , t is called ψ d ,λ d + − t in [ loc. cit. ]. The importance of thehomomorphisms ψ d , t is in the following. Theorem 4.2. The Kernel Intersection Theorem [J2, Corollary 17.18]
Suppose λ is a partition ofn. Then S λ = \ d > t λ d + ker( ψ d , t ) . This provides a clear strategy for computing Hom F S n ( S µ , S λ ): find all linear combinations θ of the homomorphisms ˆ Θ T such that ψ d , t ◦ θ = d , t . Fortunately, it is known howto compute the composition ψ d , t ◦ ˆ Θ T when T ∈ T r ( µ, λ ). For our next few results, we needto introduce some more notation. For any multiset A of positive integers, let A i denote thenumber of i s in A . If A , B are multisets, we write A ⊔ B for the multiset with ( A ⊔ B ) i = A i + B i for all i . Given a row-standard tableau T , we write T j for the multiset of entries in row j of T . Lemma 4.3. [FM, Lemma 5]
Suppose λ, µ are partitions of n, T ∈ T r ( µ, λ ) , d ∈ N and t λ d + .Let S be the set of all row-standard tableaux which can be obtained from T by replacing t of the entriesequal to d + in T with ds. Then ψ d , t ◦ Θ T = X S ∈S Y j > S jd T jd ! Θ S . omenewdecomposableSpechtmodules 7The slight di ffi culty with using this lemma to compute homomorphism spaces is that thetableaux S in the lemma are not always semistandard; so it can be di ffi cult to tell whether aparticular linear combination is zero when restricted to S µ . To circumvent this, we recall anotherlemma from [FM] which gives certain linear relations between the homomorphisms ˆ Θ T , andoften enables us to write a homomorphism ˆ Θ T in terms of semistandard homomorphisms. Lemma 4.4. [FM, Lemma 7]
Suppose µ is a partition of n and λ a composition of n, and i , j , k arepositive integers with j , k and µ j > µ k . Suppose T ∈ T r ( µ, λ ) , and let S be the set of all S ∈ T r ( µ, λ ) such that: • S ji = T ji + T ki ; • S jl T jl for every l , i; • S l = T l for all l , j , k.Then ˆ Θ T = ( − T ki X S ∈S Y l > S kl T kl ! ˆ Θ S . Informally, a tableau in S is a tableau obtained from T by moving all the i s from row k torow j , and moving some multiset of entries di ff erent from i from row j to row k .One very simple case of Lemma 4.4 which we shall apply frequently is the following: if T ∈ T r ( µ, λ ) and for some i , j , k we have T ji + T ki > max { µ j , µ k } , then ˆ Θ T = Lemma 4.5.
Suppose λ and µ are partitions of n, and T is a row-standard λ -tableau of type µ . Supposei > , and A , B , C are multisets of positive integers such that | B | > λ i and A ⊔ B ⊔ C = T i + T i + . Let B be the set of all pairs ( D , E ) of multisets such that | D | = λ i − | A | and B = D ⊔ E. For each such pair ( D , E ) , define T D , E to be the row-standard tableau withT jD , E = A ⊔ D ( j = i ) C ⊔ E ( j = i + T j ( otherwise ) . Then X ( D , E ) ∈B Y i > A i + D i D i ! C i + E i E i ! ˆ Θ T D , E = . This lemma appears in the second author’s forthcoming paper [F] where it is proved inthe wider context of Iwahori–Hecke algebras; however, a considerably easier proof exists inthe symmetric group case. In [F], Lemma 4.5 is used to provide an explicit fast algorithm forwriting a tableau homomorphism as a linear combination of semistandard homomorphisms.We now give another result which will help us in showing that a linear combination ofrow-standard homomorphisms is non-zero without having to go through the full process of Craig J.DodgeandMatthewFayersexpressing it as a linear combination of semistandard homomorphisms. This concerns the dominance order on tableaux.Suppose µ is a partition, and S , T are row-standard µ -tableaux of the same type. We saythat S dominates T , and write S Q T , if it is possible to get from S to T by repeatedly swappingan entry of S with a larger entry in a lower row (and re-ordering with each row). We warnthe reader that this is not quite the same as the dominance order described in [J2, 13.8]. Forexample, the dominance order on T r (cid:16) (3 , , (2 , (cid:17) is given by the following Hasse diagram.2 2 31 11 2 31 2 1 2 21 31 1 32 2 1 1 22 3Now we have the following lemma. Lemma 4.6.
Suppose µ is a partition of n and λ a composition of n, and T ∈ T r ( µ, λ ) . If we write ˆ Θ T = X S ∈T ( µ,λ ) a S ˆ Θ S , then a S , only if S Q T. Proof.
For this proof, we adopt the set-up of [J2, § µ -tableau t (of type (1 n )),then t determines a natural bijection between the set of λ -tabloids and the set of µ -tableaux oftype λ . We identify tabloids with tableaux according to this bijection. We also let e t denote thepolytabloid indexed by t , which generates S µ . Given λ -tableaux U , V , we write U row ←→ V if V can be obtained from U by permuting entries within rows, and we define U column ←→ V similarly.Now given U ∈ T r ( µ, λ ), we have ˆ Θ U ( e t ) = X ( V , W ) W , where we sum over all pairs ( V , W ) of λ -tableaux such that U row ←→ V column ←→ W and V hasdistinct entries in each column. (In general there are also signs determined by the columnpermutations, but in characteristic 2 we can neglect these.)omenewdecomposableSpechtmodules 9Now if S is a semistandard tableau with U row ←→ V column ←→ S for some V , then (since theentries in each column of S are increasing) we must have S Q U . On the other hand, if U issemistandard, then the only tableau V such that U row ←→ V column ←→ U is U itself. Hence for anysemistandard U , the coe ffi cient of U in ˆ Θ U ( e t ) is 1.Now suppose S ∈ T ( µ, λ ) is such that a S , S is minimal (with respect to Q ) withthis property. Then the coe ffi cient of S inˆ Θ T ( e t ) = X S ∈T ( µ,λ ) a S ˆ Θ S ( e t )is a S . So the coe ffi cient of S in ˆ Θ T ( e t ) is non-zero, and hence S Q T . Any S ∈ T ( µ, λ ) forwhich a S , S , and so dominates T . (cid:3) Using the results in this section, it will be possible to compute Hom F S n ( S µ , S λ ) in the casesof interest to us. We remark that it is often easier to express such homomorphisms as linearcombinations of non-semistandard homomorphisms; in particular, the conditions ψ d , t ◦ θ = It will also be important in this paper to compute compositions of homomorphisms betweenSpecht modules. It is well understood how to compose two tableau homomorphisms; indeed,computing this composition is the same as computing the structure constants for the Schuralgebra. We give this result, of which Lemma 4.3 is a special case. This result is easy toprove and well known (indeed, ‘quantised’ versions appear in the literature) but it does notseem to appear explicitly. However, translating to the language of the Schur algebra (where Θ T corresponds to a basis element ξ i , j ) it amounts to the Multiplication rule (2.3b) in Green’smonograph [G].Recall that if S is a tableau, then S j denotes the multiset of entries in row j of S , and inparticular S ji denotes the number of entries equal to i in row j of S . If x , x , . . . are non-negativeintegers with finite sum x , we write (cid:16) xx , x , . . . (cid:17) for the corresponding multinomial coe ffi cient. Proposition 4.7.
Suppose λ, µ, ν are compositions of n, S is a λ -tableau of type µ and T is a µ -tableauof type ν . Let X be the set of all collections X = ( X ij ) i , j > of multisets such that | X ij | = S ji for each i, j, G j > X ij = T i for each i.For X ∈ X , let U X denote the row-standard λ -tableau with ( U X ) j = F i > X ij . Then Θ T ◦ Θ S = X X ∈X Y i , j > X ji + X ji + X ji + . . . X ji , X ji , X ji , . . . ! Θ U X . S ( u , v ) In this section, we find all cases where one of our Specht modules S ( a , , b ) has a summandisomorphic to an irreducible Specht module of the form S ( u , v ) , where u is even and v is odd.Throughout, we continue to assume that a , b are positive even integers with a >
4, and we let n = a + b +
3. By Theorem 2.3 and Lemma 2.4, D ( u , v ) cannot appear as a composition factor of S ( a , , b ) unless ( u , v ) Q ( a , , b ) reg , which is the partition (max { a , b + } , min { a − , b + } , v min { a + , b + } . For easyreference, we set out notation and assumptions for this section. Assumptions and notation in force throughout Section 5: λ = ( a , , b ) and µ = ( u , v ), where a , b , u , v are positive integers with a , b , u even, a > u > v , n = a + b + = u + v and v min { a + , b + } . S λ to S µ ′ In this subsection we consider F S n -homomorphisms from S λ to S µ ′ . We begin by construct-ing such a homomorphism in the case where 3 v a − U be the set of λ -tableaux having the form1 2 3 v ⋆ ⋆ ⋆ ⋆⋆⋆ in which the ⋆ s represent the numbers from 2 to u , and in which • the entries along each row are strictly increasing, • the entries down each column are weakly increasing.Now define σ = X T ∈U ˆ Θ T . Proposition 5.1.
With the assumptions and notation above, we have ψ d , t ◦ σ = for each d , t. Proof.
First take v < d u and t =
1. If T ∈ U , then T contains a single d and a single d + T , then ψ d , ◦ ˆ Θ T = T ′ ∈ U obtained by interchanging the d andthe d +
1. By Lemma 4.3 we have ψ d , ◦ ( ˆ Θ T + ˆ Θ T ′ ) =
0. Hence by summing ψ d , ◦ ˆ Θ T over all T ∈ U , we get zero. A similar argument applies in the case d = v .If 1 d < v and t =
2, then we have ψ d , t ◦ ˆ Θ T = T ∈ U just using Lemma 4.3. Nowtake 2 d < v and t =
1, and consider a tableau T ∈ U . There are a single d and a single d + ψ d , ◦ ˆ Θ T =
0. Otherwise, let T ′ bethe tableau obtained by interchanging the d and the d + ψ d , ◦ ( ˆ Θ T + ˆ Θ T ′ ),and we are done.omenewdecomposableSpechtmodules 11We are left with the case d = t =
1. Applying Lemma 4.3, we find that ψ , ◦ θ is the sum ofhomomorphisms labelled by tableaux1 2 3 v ⋆ ⋆ ⋆ ⋆ ⋆⋆ in which the ⋆ s now represent the numbers from 3 to u , and where the entries are strictlyincreasing along rows and weakly increasing down columns. Now we apply Lemma 4.4 toeach of these homomorphisms to move the 1 from row 3 to row 2, and then to reorder rows3 , . . . , b +
2. We obtain a sum of tableaux of the form1 2 3 v ⋆ ⋆ ⋆⋆⋆ , but each tableau occurs b times in this way. Since b is even, we have zero. (cid:3) Now we need to check that σ ,
0, which is not obvious because the tableaux involved arenot semistandard.
Proposition 5.2.
With the notation above, σ , . Remark.
The version of this paper published in the Journal of Algebra includes a fallaciousproof of Proposition 5.2; the proof below replaces it. The authors are grateful to Sin´ead Lylefor pointing out the error.
Proof.
We’ll use Lemma 4.6. Consider the semistandard tableau S = v b + b + u b + b + b + . We’ll show that when σ is expressed as a linear combination of semistandard homomorphisms,ˆ Θ S occurs with non-zero coe ffi cient, and hence σ ,
0. Given T ∈ U , consider expressing ˆ Θ T asa linear combination of semistandard homomorphisms. By Lemma 4.6, ˆ Θ S can only occur if S Q T ; so we can ignore all T ∈ U for which S S T . In particular, we need only consider thosetableaux in U which have b + , . . . , u in the first row and b + , b + v < b +
3, then the tableaux T ∈ U that we need to consider arethose of the following forms: T [ i ] = v i b + b + u b + b + vv + ı b + for v < i b + U [ i ] = v b + b + b + u i b + ı b + for 2 i b + V [ i ] = v b + b + b + u i b + ı b + for 2 i b + ı in the first column indicates that i does not appear in that column.First consider the tableau T [ i ], and apply Lemma 4.4 to move the 1 from row 2 to row 1. Ofthe tableaux appearing in the resulting expression, the only ones dominated by S are T ′ [ i ] = v b + b + ui b + b + vv + ı b + omenewdecomposableSpechtmodules 13and the tableaux T ′ [ i , j ] = v i b + b + uj b + b + vv + ı b + for 2 j v .Applying Lemma 4.4 to T ′ [ i ] to move the 2 from row 3 to row 2, we obtain three tableaux, buttwo of these are not dominated by S . The other one is T ′′ [ i ] = v b + b + u b + b + i vv + ı b + , and i − Θ T ′′ [ i ] = ˆ Θ S .Now consider applying Lemma 4.4 to T ′ [ i , j ], to move the 2 from row 3 to row 2. If j = S . If j >
2, then two of the threetableaux obtained are not dominated by S ; the third has rows 3 and j + j , sothe resulting homomorphism is zero by Lemma 4.4.So we conclude that ˆ Θ T [ i ] equals ˆ Θ S plus a linear combination of homomorphisms indexedby tableaux not dominated by S .Now we consider applying Lemma 4.4 to U [ i ], moving the 1 up from row 2. The tableauxobtained that are dominated by S are T ′ [ i ] and the tableaux U ′ [ i , j ] = v b + b + b + uj i b + ı b + for 2 j v with i , j .4 Craig J.DodgeandMatthewFayers(Note that if i < j then i and j in the second row should be written the other way round; thecase i = j does not occur because the accompanying coe ffi cient would be (cid:0) (cid:1) = i =
2, then U ′ [ i , j ] is a semistandard tableau di ff erent from S . If i > S and(in the case j >
2) neglecting the tableau with two rows equal to j , the only tableau we get is U ′′ [ i , j ] = v b + b + b + u j b + i ı b + ; i − Θ U ′′ [ i , j ] equals a semistandard homomorphismdi ff erent from ˆ Θ S .We conclude that ˆ Θ U [ i ] equals ˆ Θ S plus a linear combination of homomorphisms indexedby tableaux which are either not dominated by S or semistandard and di ff erent from S . Thehomomorphism ˆ Θ V [ i ] is analysed in exactly the same way, interchanging b + b + ffi cient of ˆ Θ S in σ is the total number oftableaux of the form T [ i ], U [ i ] or V [ i ], i.e. ( b + − v ) + b + v = b +
3. In this case only the tableaux V [ i ] appear, butthe analysis of these tableaux is exactly the same, so the coe ffi cient of ˆ Θ S in σ is the number oftableaux V [ i ], i.e. b +
1, which again is odd. (cid:3)
It turns out that up to scaling, σ is the only homomorphism from S λ to S µ . Proposition 5.3.
With λ, µ as above, dim F Hom F S n ( S λ , S µ ′ ) = v = or v = a + v a − . Proof.
The construction of the homomorphism σ above shows that the dimension of the ho-momorphism space is at least that claimed. So we just have to show the reverse inequality. ByLemma 2.2, we have dim F Hom F S n ( S λ , S µ ′ ) = dim F Hom F S n ( S µ , S λ ′ ) , and we can use the technique outlined in § µ is 2-regular.So suppose θ is a linear combination of semistandard homomorphisms ˆ Θ T : S µ → M λ ′ suchthat ψ d , t ◦ θ = d , t .To begin with, we consider ψ , ◦ ˆ Θ T for each T . Using Lemma 4.3, this equals zero if T has a 2 in each row, because the homomorphisms occurring in Lemma 4.3 each appearwith a coe ffi cient (cid:0) (cid:1) , which is zero in F . Otherwise, ψ , ◦ ˆ Θ T is either a single semistandardhomomorphism or a sum of two semistandard homomorphisms. Moreover, the semistandardomenewdecomposableSpechtmodules 15tableaux that occur for the various T are distinct. Hence in order to have ψ , ◦ θ = θ can onlyinvolve semistandard homomorphisms ˆ Θ T for those T having a 2 in each row. In particular, θ = v = a +
1, since in this case there is only one semistandard tableau, whose first rowconsists entirely of 1s.Now we consider ψ , ◦ ˆ Θ T for each of these T . If T has a 2 and a 3 in each row, we get ψ , ◦ ˆ Θ T =
0, while if T has a 2 in each row and two 3s in the same row, ψ , ◦ ˆ Θ T is a semi-standard homomorphism. Again, all the semistandards that occur in this way are di ff erent, so θ cannot involve any tableaux of the latter type. In particular, if v = θ = ψ d , ◦ ˆ Θ T where 4 d < a and T is a semistandard tableau having a 2 and a3 in each row. T contains a single d and a single d +
1. If these both lie in the same row of T ,then ψ d , ◦ ˆ Θ T =
0. Otherwise, ψ d , ◦ ˆ Θ T is a semistandard homomorphism ˆ Θ T ′ . If U is anothersemistandard tableau and ψ d , ◦ ˆ Θ U is a semistandard homomorphism ˆ Θ U ′ , then T ′ = U ′ if andonly if U is obtained from T by interchanging d and d +
1; hence any two such tableaux mustoccur in θ with equal coe ffi cients. Applying this for all d > T , we find that θ must bea scalar multiple of the sum of all semistandard homomorphisms ˆ Θ T for T having a 2 and a 3in each row. Hence dim F Hom F S n ( S µ , S λ )
1, and we are done. (cid:3)
Example.
We provide an example to illustrate the above proof for the benefit of the reader whomay not be familiar with this technique. We take ( a , b , u , v ) = (4 , , , S (8 , is reducible, so is ultimately irrelevant to our main theorem, but it serves well forthis example.)We suppose we have a linear combination θ of semistandard homomorphisms such that ψ d , t ◦ θ = d , t . For this example, we abuse notation by identifying a tableau with thecorresponding homomorphism. The first step of the proof is to eliminate most of the possiblesemistandard homomorphisms by taking d = t =
1. For example, Lemma 4.3 gives ψ , ◦ = , and no other semistandard tableau can give the semistandard tableau on the right in this waywith non-zero coe ffi cient; note that the tableau1 1 1 1 2 3 4 62 3 5 7 8does give this tableau, but with a coe ffi cient of (cid:0) (cid:1) =
0. So since ψ , ◦ θ =
0, our initial tableaucannot possibly occur in θ . Arguing in this way, one finds that the only semistandard tableauxwhich can occur in θ are those with a 2 in each row, i.e. those of the form1 1 1 1 2 3 3 ⋆ ⋆ ⋆ ⋆ ⋆ , ⋆ ⋆ ⋆ ⋆ ⋆ or 1 1 1 1 2 ⋆ ⋆ ⋆ ⋆ ⋆ . Now the first and last of these three types can be ruled out using the same argument with ψ , .So θ can only involve tableaux with a 2 and a 3 in each row; call these usable tableaux. Nowwe consider ψ d , ◦ θ for d >
4. Now for each usable tableau T , ψ d , ◦ ˆ Θ T is either zero (if d d + T ) or a semistandard homomorphism. Furthermore, thesesemistandard homomorphisms ‘pair up’; for example, with d = ψ , ◦ = ψ , ◦ = . Since the semistandard tableau on the right can only arise in this way from the two semi-standard tableaux on the left, these two semistandard homomorphisms must occur with equalcoe ffi cients in θ . Now we observe that we can get from any usable tableau to any other by asequence of steps in which we interchange the integers d , d + d >
4. So ifwe apply the above argument for all d >
4, we see that all usable tableaux occur with the samecoe ffi cient in θ . S µ to S λ Now we consider homomorphisms from S µ to S λ , where λ, µ are as above. In view ofProposition 5.3, we assume for the rest of this section that v a −
1. It turns out that all suchhomomorphisms can be expressed as linear combinations of ˆ Θ A and ˆ Θ B , where A , B are thefollowing µ -tableaux of type λ : A = b +
21 1 1 1 ; B = b +
21 1 2 2 . Note that our assumptions on the parameters a , b , u , v mean that these tableaux really do exist,i.e. there are enough 1s to fill the bottom row. Lemma 5.4. ˆ Θ A and ˆ Θ B are non-zero, and are linearly independent if v b + . Proof.
It is straightforward to express ˆ Θ A and ˆ Θ B as linear combinations of semistandardhomomorphisms using a single application of Lemma 4.4; in each case we get at least onesemistandard appearing, so the homomorphisms are non-zero. If in addition v b +
1, then inthe expression for ˆ Θ A there is at least one semistandard tableau with two 2s in the first row; thereis no such tableau appearing in the expression for ˆ Θ B , so ˆ Θ A , ˆ Θ B are linearly independent. (cid:3) Proposition 5.5. • If a − v ≡ or v = b + , then ψ d , t ◦ ˆ Θ A = for all admissible d , t. • If v ≡ , then ψ d , t ◦ ˆ Θ B = for all admissible d , t. • If a ≡ , then ψ d , t ◦ ( ˆ Θ A + ˆ Θ B ) = for all admissible d , t. Proof.
Lemma 4.3 immediately gives ψ d , ◦ ˆ Θ A = ψ d , ◦ ˆ Θ B = d >
2. Using the fact that A , B each have an odd number of 1s in each row, we also get ψ , t ◦ ˆ Θ A = ψ , t ◦ ˆ Θ B = t = ,
3. Finally, we have ψ , ◦ ˆ Θ B = v ≡ ψ , ◦ ˆ Θ A = a − v ≡ v = b + ψ , ◦ ˆ Θ A = ψ , ◦ ˆ Θ B if a ≡ (cid:3) Proposition 5.6. dim F Hom F S n ( S µ , S λ ) = if a ≡ , v ≡ and v b + otherwise ) . Proof.
By Lemma 5.4 and Proposition 5.5 the dimension of the homomorphism space is atleast that claimed. Now we show the reverse inequality, by considering linear combinations ofsemistandard homomorphisms.Throughout this proof, we’ll write T [ i ] for the set of semistandard µ -tableaux of type λ having exactly i
2s in the first row, for i = , , ,
3, and let τ i = P T ∈T [ i ] ˆ Θ T .Suppose we have a linear combination θ of semistandard homomorphisms ˆ Θ T : S µ → M λ such that ψ d , t ◦ θ = d , t .First consider ψ d , ◦ ˆ Θ T for T ∈ T ( µ, λ ) and d >
3. By Lemma 4.3, ψ d , ◦ ˆ Θ T is either zero or asemistandard homomorphism (according to whether the d and the d + T occur in the samerow). If it is non-zero, then there is exactly one other T ′ ∈ T ( µ, λ ) such that ψ d , ◦ ˆ Θ T = ψ d , ◦ ˆ Θ T ′ ,namely the tableau obtained by interchanging the d and the d + T . Hence ˆ Θ T and ˆ Θ T ′ mustoccur with the same coe ffi cient in θ . Applying this for all d >
3, we find that for a fixed i ∈ { , , , } , all the homomorphisms ˆ Θ T for T ∈ T [ i ] occur with the same coe ffi cient in θ . Inother words, θ is a linear combination of τ , τ , τ , τ .We can apply a similar argument in which we consider ψ , ◦ ˆ Θ T for T ∈ T ( µ, λ ). Again ψ , ◦ ˆ Θ T is either zero or a semistandard homomorphism; and if it is non-zero, then the onlyother T ′ having ψ , ◦ ˆ Θ T ′ = ψ , ◦ ˆ Θ T is the tableau obtained by exchanging the 3 in T with a 2in the other row. ˆ Θ T and ˆ Θ T ′ occur with the same coe ffi cient in θ , and we deduce that θ mustbe a linear combination of τ + τ and τ + τ .Finally, we consider ψ , ◦ θ . Each µ -tableau T of type ( a + , b + ) contains a single 2; let φ denote the sum of ˆ Θ T for all those T having the 2 in row 1, and χ the sum of all ˆ Θ T for T havingthe 2 in row 2. Using Lemma 4.3 and Lemma 4.4 (and recalling that a is even and v is odd), wehave ψ , ◦ τ = (cid:16) v − (cid:17) χ,ψ , ◦ τ = (cid:16) v (cid:17) φ,ψ , ◦ τ = (cid:18)(cid:16) a + (cid:17) + (cid:19) φ + χ,ψ , ◦ τ = (cid:16) a + (cid:17) φ. So if v ≡ ψ , ◦ ( τ + τ ) ,
0, so θ cannot equal τ + τ . If a ≡ ψ , ◦ ( τ + τ ) ,
0, so θ cannot be τ + τ . Hence dim F Hom F S n ( S µ , S λ ) F Hom F S n ( S µ , S λ ) b = d −
3, since in this case T [2] and T [3]are empty, so τ + τ = (cid:3) Now we complete the analysis of when S µ is a summand of S λ , by composing the homo-morphisms from the preceding subsections. This will be straightforward, using Proposition4.7.Recall that the space of homomorphisms from S λ to S µ ′ is one-dimensional, spanned bythe homomorphism σ = P T ∈U ˆ Θ T . On the other hand, the space of homomorphisms from S µ to S λ has dimension one or two, each homomorphism being a linear combination of thehomomorphisms ˆ Θ A and ˆ Θ B . So it su ffi ces to compute the compositions σ ◦ ˆ Θ A and σ ◦ ˆ Θ B .Let D be the µ -tableau D = u v of type µ ′ . Then we have the following. Lemma 5.7.
Suppose T ∈ U , and let x be the entry in the (2 , -position of T. Then ˆ Θ T ◦ ˆ Θ A = ˆ Θ D , ˆ Θ T ◦ ˆ Θ B = ˆ Θ D ( x v )0 ( x > v ) . Furthermore, ˆ Θ D , . Proof.
The fact that ˆ Θ D , T ∈ U and recall the notation of Proposition 4.7, with S equal to either A or B .Suppose X ∈ X . Since each T i is a proper set, each X ij must be as well. This means that ifsome integer i appears in two sets X kj , X lj , then the multinomial coe ffi cient (cid:16) X jj + X jj + X jj + . . . X jj , X jj , X jj , . . . (cid:17) from Proposition 4.7 will include a factor (cid:0) (cid:1) , which gives 0.So in order to get a non-zero coe ffi cient in Proposition 4.7, we must have X j , X j , X j , . . . pairwise disjoint for each j , which means that we will have X ⊔ X ⊔ · · · = { , . . . , u } , X ⊔ X = { , . . . , v } ; ( † )so U X will equal D .If S = A , the only way to achieve this is to have X = T \ { , . . . , v } , X = { , . . . , v } , X i = T i for i > . Thus we have ˆ Θ T ◦ ˆ Θ A = ˆ Θ D .In the case S = B , let y be the (2 , T . Then y > x . X must contain either x or y ,so if x > v then we cannot possibly achieve ( † ). So we get ˆ Θ T ◦ ˆ Θ B = x v < y ,then the only way to achieve ( † ) is to have X = { , x } and X = { , . . . , ˆ x , . . . , v } , and this yieldsˆ Θ T ◦ ˆ Θ B = ˆ Θ D . Finally, if y v , then there are three possible ways to achieve ( † ); each of thesegives a coe ffi cient of 1, and again we have ˆ Θ T ◦ ˆ Θ B = ˆ Θ D . (cid:3) This result is very helpful: it tells us that the composition of σ with a combination of ˆ Θ A and ˆ Θ B is a scalar multiple of ˆ Θ D ; hence this composition is non-zero if and only if this scalar isnon-zero. In order to use this result, we need to find the number of tableaux in U , and also thenumber of tableaux in U in which the (2 , v . This is a straightforward count.omenewdecomposableSpechtmodules 19 Lemma 5.8. • The number of tableaux in U is (cid:16) u − va − v (cid:17)(cid:16) u + v − a − (cid:17) . • The number of tableaux in U whose (2 , -entry is greater than v is (cid:16) u − va − v (cid:17)(cid:16) u − a (cid:17) . Proof of Theorem 3.1(1).
Suppose S µ = S ( u , v ) is irreducible, with u + v = a + b +
3. Throughoutthis proof, all congruences are modulo 4.Suppose first that u , v satisfy the given conditions, i.e. v ≡ (cid:16) u − va − v (cid:17) is odd. The secondcondition implies in particular that 0 a − v u − v , which gives v min { a − , b + } ; so theassumptions in force in this section are satisfied. In addition, Theorem 2.5 gives u ≡ S µ γ −→ S λ δ −→ S µ ′ such that δ ◦ γ ,
0. Since3 v a −
1, we can take δ = σ .If a ≡
0, take γ = ˆ Θ A + ˆ Θ B . By Proposition 5.5, γ is a homomorphism from S µ to S λ . ByLemma 5.7 and Lemma 5.8, δ ◦ γ = (cid:16) u − va − v (cid:17)(cid:16) u − a (cid:17) ˆ Θ D . The first term is odd by assumption; the second term is odd because u − a ≡
2, and ˆ Θ D , a ≡
2, take γ = ˆ Θ A . Then γ is a homomorphism from S µ to S λ , and δ ◦ γ = (cid:16) u − va − v (cid:17)(cid:16) u + v − a − (cid:17) ˆ Θ D . Again, the first term is odd by assumption, the second term is odd because now u + v − a − ≡ Θ D , γ, δ such that δ ◦ γ ,
0. By Proposition 5.3we can assume that 3 v a − δ = σ . From Proposition 5.6 we can take γ to beˆ Θ A , ˆ Θ B or ˆ Θ A + ˆ Θ B , according to the congruences in Proposition 5.5. Then δ ◦ γ will be a scalarmultiple of ˆ Θ D , and the scalar will include (cid:16) u − vu − a (cid:17) as a factor. So this binomial coe ffi cient mustbe odd, and all that remains is to show that v ≡ v >
1, Theorem 2.5 gives u − v ≡ a − v ≡ v = b + γ = ˆ Θ A In this case the coe ffi cient of ˆ Θ D in δ ◦ γ is (cid:16) u − va − v (cid:17)(cid:16) u + v − a − (cid:17) . The second binomial coe ffi cient must be odd, so u + v − a ≡
3. In the case a − v ≡
3, this isthe same as saying u ≡
2, so that v ≡
3. In the case v = b +
3, we have a = u , so that again v ≡ a ≡ γ = ˆ Θ A + ˆ Θ B Now the coe ffi cient of ˆ Θ D in δ ◦ γ is (cid:16) u − va − v (cid:17)(cid:16) u − a (cid:17) . The second binomial coe ffi cient is odd only if u ≡ a +
2, which is the same as saying v ≡ v ≡ γ = ˆ Θ B In this case the coe ffi cient of ˆ Θ D in δ ◦ γ is (cid:16) u − va − v (cid:17) (cid:18)(cid:16) u + v − a − (cid:17) + (cid:16) u − a (cid:17)(cid:19) . Since v ≡
1, we have u + v − a − ≡ u − a , so that (cid:16) u + v − a − (cid:17) and (cid:16) u − a (cid:17) have the sameparity. Hence δ ◦ γ =
0, a contradiction. (cid:3) S ( u , v , In this section, we find when one of our Specht modules S ( a , , b ) has a summand isomorphicto an irreducible Specht module of the form S ( u , v , , where u is even and v is odd. By Theorem2.3 and Lemma 2.4, D ( u , v , cannot appear as a composition factor of S ( a , , b ) unless ( u , v , Q ( a , , b ) reg . So we may assume that this is the case, which is the same as saying v min { a − , b + } . We set out notation and assumptions for this section. Assumptions and notation in force throughout Section 6: λ = ( a , , b ) and µ = ( u , v , a , b , u , v are positive integers with a , b , u even, a > u > v > n = a + b + = u + v + v min { a − , b + } . S λ to S µ ′ We begin by constructing a homomorphism from S λ to S µ ′ . As in § U be the set of λ -tableaux having the form1 2 3 v ⋆ ⋆ v ⋆⋆ in which the ⋆ s represent the numbers from v + u , and in which the entries are weaklyincreasing along the first row and down the first column. Let σ = P T ∈U ˆ Θ T . Proposition 6.1.
With the notation and assumptions above, we have ψ d , t ◦ σ = for all d , t. Proof.
For d > v and t =
1, we use the same argument as that used in several proofs above: for T ∈ U either ψ d , ◦ ˆ Θ T =
0, or there is a unique other T ′ ∈ U with ψ d , ◦ ˆ Θ T ′ = ψ d , ◦ ˆ Θ T .The cases where 2 d v are easier: in this case Lemma 4.3 and Lemma 4.4 imply that wehave ψ d , t ◦ ˆ Θ T = T ∈ U .omenewdecomposableSpechtmodules 21So we are left with the cases where d = t ∈ { , , } . For T ∈ U we have ψ , ◦ ˆ Θ T immediately from Lemma 4.3, while ψ , ◦ ˆ Θ T is a homomorphism labelled by a tableau of theform 1 2 3 v ⋆ ⋆ v ⋆⋆ . But this homomorphism is zero by Lemma 4.4. Finally, ψ , ◦ ˆ Θ T is the sum of the homomor-phisms labelled by two tableaux1 2 3 v ⋆ ⋆ v ⋆⋆ , v ⋆ ⋆ v ⋆⋆ . But these two homomorphisms are equal by Lemma 4.4, and we are done. (cid:3)
Now, as in Section 5.1 we have to show that σ ,
0. Again, we use a dominance argument.
Proposition 6.2.
With the notation above, σ , . Proof.
We’ll show that when σ is expressed as a linear combination of semistandard homo-morphisms, the homomorphism ˆ Θ S occurs with non-zero coe ffi cient, where S = v b + u b + . Recall that when ˆ Θ T is expressed as a linear combination of semistandard homomorphisms,the coe ffi cient of ˆ Θ S is zero unless S Q T . The only elements of U which are dominated by S T [ i ] = v i b + u vv + ı b + for v + i b +
2. Consider applying Lemma 4.4 to T [ i ], to move the two 1s from row 2to row 1. Of the tableaux appearing in that lemma with non-zero coe ffi cient, the only onesdominated by S are those having no more than four entries less than 4 in the first row; theseare the tableaux T ′ [ i ] and T ′ [ i , j ] for 4 j v , where T ′ [ i ] = v b + u i vv + ı b + , T ′ [ i , j ] = v i b + u j vv + ı b + . So, modulo homomorphisms labelled by tableaux not dominated by S , we have σ = P i ˆ Θ T ′ [ i ] + P i , j ˆ Θ T ′ [ i , j ] . However, two applications of Lemma 4.4 show that ˆ Θ T ′ [ i , j ] = i , j , andLemma 4.4 also gives ˆ Θ T ′ [ i ] = ˆ Θ S .So the coe ffi cient of ˆ Θ S in σ is b + − v , which is odd; so σ , (cid:3) As before, we find that σ is the only homomorphism from S λ to S µ ′ up to scaling.omenewdecomposableSpechtmodules 23 Proposition 6.3.
With λ, µ as above, dim F Hom F S n ( S λ , S µ ′ ) = . Proof.
The existence of the homomorphism σ shows that the space of homomorphisms isnon-zero. To show that it has dimension at most 1, we again use the fact thatdim F Hom F S n ( S λ , S µ ′ ) = dim F Hom F S n ( S µ , S λ ′ ) . So suppose θ is a linear combination of semistandard homomorphisms ˆ Θ T : S µ → S λ ′ suchthat ψ d , t ◦ θ = d , t .First of all, consider ψ , ◦ θ . Given a semistandard tableau T , we can use Lemma 4.3 tocompute ψ , ◦ ˆ Θ T , and then if necessary use Lemma 4.4 (to move a 2 from row 3 to row 2) toexpress this composition as a linear combination of semistandard homomorphisms. We findthat if T has two 2s in its first row, then ψ , ◦ ˆ Θ T involves a semistandard tableau which doesnot occur in any other ψ , ◦ ˆ Θ T ′ ; hence the coe ffi cient of ˆ Θ T in θ must be zero.Now we do the same thing with ψ , : in this case we find that if T is a semistandard tableauhaving two 3s in its first row, then ψ , ◦ ˆ Θ T involves a semistandard homomorphism whichdoes not occur in any other ψ , ◦ ˆ Θ T ′ (except possibly for a tableau T ′ already ruled out in theparagraph above). So we may restrict attention to those T having at most one 2 and one 3 inthe first row.Now return to ψ , ◦ ˆ Θ T , for T of the form1 1 2 3 x x s z z z t k , where x , . . . , x s , z , . . . , z t , k are the integers 4 , . . . , a in some order. When we express ψ , ◦ ˆ Θ T as a linear combination of semistandard homomorphisms, we find that the homomorphismlabelled by 1 1 2 3 x x s z z t z k occurs with non-zero coe ffi cient; but this homomorphism does not occur in any other ψ , ◦ ˆ Θ T ′ (except for T ′ having two 3s in its first row). So for any T of the above form, the coe ffi cient ofˆ Θ T in θ must be zero.Now any semistandard tableau which contributes to θ must be of one of the following eight4 Craig J.DodgeandMatthewFayersforms.1 . ⋆ ⋆ ⋆ ⋆ ⋆ . ⋆ ⋆ ⋆ ⋆ ⋆ . ⋆ ⋆ ⋆ ⋆ . ⋆ ⋆ ⋆ ⋆ ⋆ . ⋆ ⋆ ⋆ ⋆⋆ ⋆ . ⋆ ⋆ ⋆ ⋆⋆ ⋆ . ⋆ ⋆ ⋆ ⋆⋆ ⋆ . ⋆ ⋆ ⋆ ⋆ ⋆ In each case, the ⋆ s represent the numbers from 4 to a . Note that in each of these tableaux,the entries 4 , . . . , a must all occur in di ff erent columns (the assumption v b + ψ d , for d >
3, and repeat the argument used in the last paragraph of the proof of Proposition5.3, to show that if T , T ′ are two tableaux which have their 1s, 2s and 3s in the same positions,then ˆ Θ T and ˆ Θ T ′ occur with the same coe ffi cient in θ . Hence θ is a linear combination of thehomomorphisms τ , . . . , τ , where τ i is the sum of all homomorphisms ˆ Θ T for T of type i .Once more we can consider ψ , ◦ θ : when we compute ψ , ◦ τ , we obtain (in addition tosome other semistandard tableaux) the sum of the semistandard tableaux of the form1 1 1 1 3 ⋆ ⋆ ⋆ ⋆⋆ ⋆ , which do not occur in any other ψ , ◦ τ i (note these tableaux do occur when we compute ψ , ◦ ˆ Θ T for T of type 4, but each one occurs twice when we sum over tableaux of type 4, sothe contributions cancel). So τ does not appear in θ .Next we consider ψ , ◦ τ i for each i . For i = , ψ , ◦ τ i involvessemistandard tableaux which do not occur in any other ψ , ◦ τ i ; so τ , τ , τ cannot occur in θ . Moreover, we find that ψ , ◦ τ = ψ , ◦ τ and ψ , ◦ τ = ψ , ◦ τ , and that these twohomomorphisms are linearly independent. So θ must be a linear combination of τ + τ and τ + τ .Finally we return once more to ψ , ◦ θ . We find that ψ , ◦ τ = ψ , ◦ τ ,
0, while ψ , ◦ τ = ψ , ◦ τ =
0. So τ and τ must appear with the same coe ffi cient in θ ; so θ must bea scalar multiple of τ + τ + τ + τ , and so the homomorphism space has dimension at most1. (cid:3) omenewdecomposableSpechtmodules 25 S µ to S λ Now we consider homomorphisms from S µ to S λ . We begin by constructing a non-zerohomomorphism. Let C be the µ -tableau b +
21 1 12 2 of type λ . Proposition 6.4.
With C as above, we have ψ d , t ◦ ˆ Θ C = for all d , t, and ˆ Θ C , . Proof.
Showing the first statement is very easy, using Lemma 4.3. The only homomorphismsthat occur in that lemma with non-zero coe ffi cient are labelled by tableaux with more than v Θ C , v + b +
22 2 2 5 v +
13 4 (for example) labels a homomorphism occurring with non-zero coe ffi cient. (cid:3) Proposition 6.5.
With λ, µ as above, dim F Hom F S n ( S µ , S λ ) = . Proof.
The existence of the homomorphism ˆ Θ C above shows that the space of homomorphismsis non-zero. So we just need to show the upper bound on the dimension. So suppose θ is alinear combination of semistandard homomorphisms ˆ Θ T : S µ → S λ such that ψ d , t ◦ θ = d , t .For 3 d b +
1, say that a semistandard tableau T is d-bad if the entries d , d + T . Note that this must be column 1 or 2, because the assumption v a − T is d -bad, then ˆ Θ T cannot appear in θ . To show this, we consider ψ d , ◦ ˆ Θ T for every semistandard T . If T is not d -bad, then by Lemma 4.3 ψ d , ◦ ˆ Θ T is either zero or ahomomorphism labelled by a semistandard tableau with two d s in di ff erent rows. If T is d -bad,then we can express ψ d , ◦ ˆ Θ T as a linear combination of semistandard homomorphisms usingLemma 4.3 together with Lemma 4.5. For example, if T = ψ , ◦ ˆ Θ T = ˆ Θ T ′ , where T ′ = A = { } , B = { , , , , , } , C = ∅ to express ˆ Θ T ′ as a sum of fourteen semistandard homomorphisms.Doing this for each d -bad tableau T , we find that ψ d , ◦ ˆ Θ T is a sum of homomorphismslabelled by semistandard tableaux with the same first row as T ; furthermore, at least one ofthese tableaux will have two d s in the second row. Moreover, each d -bad tableau will yield atableau of this kind which does not occur for any other d -bad tableau T ′ . To see this, supposefirst of all that d , d + T . Then there is no other d -bad tableauwith the same first row as T , so any tableau occurring in ψ d , ◦ ˆ Θ T with two d s in the secondrow can only possibly occur in ψ d , ◦ ˆ Θ T . Alternatively, if d , d + T ,then T has the form x x s d z z z t d + k . There are v − d -bad tableaux with the same first row as T , and they all also have the same(2 , T . Hence when we apply Lemma 4.3 and Lemma 4.5 (or equivalently Lemma4.4), we find that the homomorphism labelled by1 1 2 2 2 x x s d d z z t z k occurs in ψ d , ◦ ˆ Θ T but not in ψ d , ◦ ˆ Θ T ′ for any other T ′ .So in θ the coe ffi cient of ˆ Θ T is zero for any d -bad tableau. In particular, this means thatfor any ˆ Θ T occurring in θ , ψ d , ◦ ˆ Θ T is either zero or a single semistandard homomorphism.Now we claim that the coe ffi cient of ˆ Θ T is zero whenever T has two or three 2s in its first row.Supposing this is false, take a T with at least two 2s in its first row such that ˆ Θ T appears withnon-zero coe ffi cient in θ , and suppose that T is minimal (with respect to the dominance order)subject to this property. Suppose the (3 , T is greater than 3; then this entry equals d + d >
3, and the entry equal to d cannot be the (2 , T is not d -bad). So the d and the d + T lie in di ff erent rows and di ff erent columns, and ψ d , ◦ ˆ Θ T is the semistandardhomomorphism obtained by replacing the d + d . The only other semistandard tableau T ′ such that ψ d , ◦ ˆ Θ T ′ = ψ d , ◦ ˆ Θ T is the tableau obtained by interchanging the d and the d + T , so ˆ Θ T ′ must also occur with non-zero coe ffi cient. But T Q T ′ , contradicting the choice of T .So the (3 , T must be 3 (and hence the (2 , , d +
1. Then d >
4, and the d in T cannot occur in the (2 , T isomenewdecomposableSpechtmodules 27not d -bad). So we can repeat the above argument and show that that there is a tableau T ′ P T such that ˆ Θ T ′ occurs in θ ; contradiction.We now know that every semistandard homomorphism occurring in θ has at least two 2sin the second row. This means in particular that the entries 3 , . . . , b + ff erent columns.So we can repeat the argument from Proposition 5.3 and Proposition 6.3 to show that θ mustbe a linear combination of τ and τ , where τ i is the sum of all homomorphisms labelled bysemistandard tableau with i
2s in the first row. If u = a , then τ =
0, and so the space ofhomomorphisms S µ → S λ has dimension at most 1. if u > a , then ψ , ◦ τ = ψ , ◦ τ ,
0, so θ must be a scalar multiple of τ + τ and again the homomorphism space has dimension atmost 1. (cid:3) We have constructed homomorphisms S µ ˆ Θ C −→ S λ σ −→ S µ ′ , and shown that these homo-morphisms are unique up to scaling. To complete this section, we just need to compute thecomposition of these homomorphisms.Let E be the µ -tableau v v + u v of type µ ′ . Then we have the following. Proposition 6.6.
For T ∈ U , we have ˆ Θ T ◦ ˆ Θ C = ˆ Θ E , , and therefore we have σ ◦ ˆ Θ C , if andonly if (cid:16) u − va − v (cid:17) is odd. Proof.
It is easy to express ˆ Θ E as a linear combination of semistandard homomorphisms usingthree applications of Lemma 4.4, from which it follows that ˆ Θ E , Θ T ◦ ˆ Θ C = ˆ Θ E , use the notation of Proposition 4.7, with S = C . Suppose X ∈ X is such that the coe ffi cient of ˆ Θ U X in Proposition 4.7 is non-zero. Since X must be { } , wecannot have X = { } (because this would give a factor (cid:0) (cid:1) ), so X = { } and hence X = { , } .Now if X contains any of the numbers 1 , . . . , v then again we get a factor (cid:0) (cid:1) . So we have X = { , . . . , v } , which determines X , and we find that ˆ Θ T ◦ ˆ Θ C = ˆ Θ E as required.So we have σ ◦ ˆ Θ C = |U| ˆ Θ E , which is non-zero if and only if |U| is odd. But it is easy to seethat |U| = (cid:16) u − va − v (cid:17) , and the proposition is proved. (cid:3) Proof of Theorem 3.1(2).
Suppose S µ = S ( u , v , is irreducible.Suppose first that (cid:16) u − va − v (cid:17) is odd. This implies in particular that 0 a − v u − v , so v min { a − , b + } . So the assumptions of this section are valid, and we have homomorphismsˆ Θ C : S µ → S λ and σ : S λ → S µ ′ . By Proposition 6.6, σ ◦ ˆ Θ C ,
0, so S µ is a summand of S λ .Conversely, suppose we have homomorphisms S µ γ −→ S λ δ −→ S µ ′ with δ ◦ γ ,
0. ByPropositions 6.3 and 6.5, δ must be a scalar multiple of σ , and γ must be a scalar multiple ofˆ Θ C . Hence by Proposition 6.6, (cid:16) u − va − v (cid:17) is odd. (cid:3) In this section, we prove Corollary 3.2, which answers the question of which Specht modulesare shown to be decomposable by Theorem 3.1. First we consider the case where a + b ≡ Proposition 7.1.
Suppose n ≡ , and a is even, with a n − . Let b = n − a − . ThenS ( a , , b ) has an irreducible summand of the form S ( u , v , . Proof.
Using Theorem 3.1, we need to show that there is a pair u , v with u + v + = n such that S ( u , v , is irreducible and (cid:16) u − vu − a (cid:17) is odd. By Theorem 2.5, ( u , v ,
2) is irreducible if and only if v ≡ , u − v ≡ − l ( v − ) , where l ( k ) = ⌈ log ( k + ⌉ for an integer m .We use induction on n , with our main tool being the following well-known relations modulo2 on binomial coe ffi cients: x y ! ≡ x + y ! ≡ x + y + ! ≡ xy ! , x y + ! ≡ . We consider three cases. a = , , n − or n − v = u = n − n ≡ u ≡ u − v ≡ S ( u , v , is irreducible. Furthermore,the binomial coe ffi cients u − ! , u − ! , u − ! , u − ! are all odd, which means that (cid:16) u − u − a (cid:17) must be odd. n ≡
11 (mod 16) In this case, let n ′ = n + , a ′ = a + ( a ≡ a + ( a ≡ . Then n ′ , a ′ satisfy the conditions of the proposition, and n ′ < n (note that the conditionson a mean that n > u ′ , v ′ such that v ′ ≡ , u − v ≡ − l ( v ′ − ) , (cid:16) u ′ − v ′ u ′ − a ′ (cid:17) ≡ . Note that because u ′ − v ′ is odd and u ′ − a ′ is even, this also gives (cid:16) u ′ − v ′ u ′ − a ′ + (cid:17) odd.We let u = u ′ − v = v ′ −
5. Then u + v + = n , and we have v ≡ u − v = u ′ − v ′ ) + ≡ − l ( v ′ − + ) , with l ( v − l ( v ′ − +
1. So S ( u , v , is irreducible. Furthermore u − vu − a ! = u ′ − v ′ + u ′ − a ′ ( + ! ≡ u ′ − v ′ u ′ − a ′ ( + ! ≡ , and we are done.omenewdecomposableSpechtmodules 29 n ≡ , a n −
11 In this case, let n ′ = n + , a ′ = a + ( a ≡ a ( a ≡ . Then n ′ , a ′ satisfy the conditions of the proposition, and n ′ < n . So by induction there isa pair u ′ , v ′ such that v ′ ≡ , u − v ≡ − l ( v ′ − ) , (cid:16) u ′ − v ′ u ′ − a ′ (cid:17) ≡ . Again, because u ′ − v ′ is odd and u ′ − a ′ is even, (cid:16) u ′ − v ′ u ′ − a ′ + (cid:17) is also odd.We let u = u ′ and v = v ′ −
1. Then u + v + = n , v ≡ u − v = u ′ − v ′ ) + ≡ − l ( v ′ − + ) , with l ( v − l ( v ′ − +
1. So S ( u , v , is irreducible. Furthermore u − vu − a ! = u ′ − v ′ + u ′ − a ′ ( + ! ≡ u ′ − v ′ u ′ − a ′ ( + ! ≡ . (cid:3) The next result addresses most of the cases where a + b ≡ Proposition 7.2.
Suppose n ≡ , and a is even, with a n − . Let b = n − a − . ThenS ( a , , b ) has an irreducible summand of the form S ( u , v ) with v > . Proof.
The proof is very similar to the proof of Proposition 7.1. We need to show that there isa pair u , v such that S ( u , v ) is irreducible, v > v ≡ (cid:16) u − vu − a (cid:17) is odd. The conditionfor S ( u , v ) to be irreducible is u − v ≡ − l ( v ) ).Again, we need three cases. a = , , n − or n − v = u = n − n ≡ u ≡ u − v ≡ S ( u , v ) is irreducible), and thebinomial coe ffi cients u − ! , u − ! , u − ! , u − ! are all odd, which means that (cid:16) u − u − a (cid:17) will be odd. n ≡ , a n −
11 In this case, let n ′ = n + , a ′ = a + ( a ≡ a + ( a ≡ . Then n ′ , a ′ satisfy the conditions of the proposition, and n ′ < n . So by induction there isa pair u ′ , v ′ such that v ′ ≡ , v ′ > , u − v ≡ − l ( v ′ ) ) , (cid:16) u ′ − v ′ u ′ − a ′ (cid:17) ≡ . u ′ − v ′ is odd and u ′ − a ′ is even, this also gives (cid:16) u ′ − v ′ u ′ − a ′ + (cid:17) odd.We let u = u ′ − v = v ′ −
3. Then u + v = n , and we have u − v = u ′ − v ′ ) + ≡ − l ( v ′ ) + )and l ( v ) l ( v ′ ) +
1. So S ( u , v ) is irreducible. Furthermore, v > v ≡ u − vu − a ! = u ′ − v ′ + u ′ − a ′ ( + ! ≡ u ′ − v ′ u ′ − a ′ ( + ! ≡ , as required. n ≡
13 (mod 16) In this case, let n ′ = n + , a ′ = a + ( a ≡ a + ( a ≡ . Then n ′ , a ′ satisfy the conditions of the proposition, and n ′ < n . So by induction there isa pair u ′ , v ′ such that v ′ ≡ , v ′ > , u − v ≡ − l ( v ′ ) ) , (cid:16) u ′ − v ′ u ′ − a ′ (cid:17) ≡ . Because u ′ − v ′ is odd and u ′ − a ′ is even, this also gives (cid:16) u ′ − v ′ u ′ − a ′ + (cid:17) odd.We let u = u ′ − v = v ′ −
7. Then u + v = n , and we have u − v = u ′ − v ′ ) + ≡ − l ( v ′ ) + ) , and l ( v ) l ( v ′ ) +
1. So S ( u , v ) is irreducible. Furthermore, we have v ≡ v > u − vu − a ! = u ′ − v ′ + u ′ − a ′ ( + ! ≡ u ′ − v ′ u ′ − a ′ ( + ! ≡ . (cid:3) Now we can prove our main result.
Proof of Corollary 3.2.
Suppose we have a pair a , b of positive even integers with a >
4. If a > b > a + b ≡ a > b > a + b ≡ a = a + b ≡ S ( a + b , is an irreducible summand of S λ .The second case of the corollary is precisely the condition for S ( a + b , to be an irreduciblesummand of S λ , while the third case is the condition for S ( a + b − , , to be a summand. So in anyof the given cases, S λ certainly has an irreducible Specht module as a summand. To completethe proof, it su ffi ces to show that if a = b = a + b ≡ S λ are S ( a + b , and S ( a + b − , , .Suppose S λ has an irreducible summand S ( u , v ) with v >
3. Then v ≡ u − v ≡ u + v ≡ a + b ≡ u , v ) Q λ reg , which implies that a > b > S λ has an irreducible summand S ( u , v , with v >
5, then v ≡ u − v ≡ u + v + ≡ a + b ≡ u , v , Q λ reg implies that a > b > (cid:3) omenewdecomposableSpechtmodules 31 The results in this paper do not give anything like a complete picture; this work is in-tended as a re-awakening of a long-dormant subject. Given how small the first example ofa decomposable Specht module in this paper is, it is surprising that it has taken thirty yearsfor this example to be found. We hope that this paper will be the start of a longer study ofdecomposable Specht modules.We conclude the paper by making some speculations about decomposable Specht modules;these are based on calculations and observations, but we do not have enough evidence to makeformal conjectures.
Our main results show that in certain cases summands of Specht modules are isomorphic toirreducible Specht modules. In fact, reducible Specht modules can also occur as summands; forexample, the first new decomposable Specht module S (4 , , ) found in this paper decomposesas S (6 , ⊕ S (4 , , , with the latter Specht module being reducible.However, it is certainly not the case that every summand of a decomposable Specht moduleis isomorphic to a Specht module. But in the cases we have been able to calculate, everysummand appears to have a filtration by Specht modules. If this is true in general, it meansthat our main results are stronger, in that we have found all irreducible summands of Spechtmodules in our family.In fact, we speculate that every Specht module has a filtration in which the factors areisomorphic to indecomposable Specht modules; this would imply in particular that everyindecomposable summand has a Specht filtration. This speculation is certainly true in the caseof Specht modules labelled by hook partitions; this follows from [M2, § -quotient separated partitions In [JM1, Definition 2.1], James and Mathas make the following definition: a partition λ is2 -quotient separated if it can be written in the form( c + x c , c − + x c − , . . . , d + x d , d y d , ( d − y d − + , . . . , y + ) , where c + > d > x c > · · · > x d > y , . . . , y d >
0. (Note that the definition includes thecase c =
0, where we have λ = (2 x ) if d =
0, or (1 y ) if d = λ can bedecomposed as in the following diagram, where horizontal ‘dominoes’ can appear in the first c − d + d columns.2 Craig J.DodgeandMatthewFayersThe definition of a 2-quotient separated partition was made as part of the study of decom-position numbers: for the Iwahori–Hecke algebra H C , − ( S n ), whose representation theory isvery similar to that of S n in characteristic 2, the composition factors of a Specht module labelledby a 2-quotient separated partition are known explicitly. The reason we recall the definitionhere is that every known example of a decomposable Specht module is labelled by a 2-quotientseparated partition. (Note that the partition ( a , , b ) considered in this paper is 2-quotientseparated precisely when a and b are even.) It is interesting to speculate whether the 2-quotientseparated condition is necessary for a Specht module to be decomposable. References [F] M. Fayers, ‘An algorithm for semistandardising homomorphisms’,
J. Algebra (2012),38–51.[FM] M. Fayers & S. Martin, ‘Homomorphisms between Specht modules’,
Math. Z. (2004),395–421.[G] J. Green,
Polynomial representations of GL n , Lecture notes in mathematics , Springer-Verlag, New York / Berlin, 1980.[J1] G. James, ‘On the decomposition matrices of the symmetric groups II’,
J. Algebra (1976),45–54.[J2] G. James, The representation theory of the symmetric groups , Lecture notes in mathematics , Springer-Verlag, New York / Berlin, 1978.[JM1] G. James & A. Mathas, ‘Hecke algebras of type A with q = − J. Algebra (1995),102–58.[JM2] G. James & A. Mathas, ‘The irreducible Specht modules in characteristic 2’,
Bull. LondonMath. Soc. (1999), 457–62.[M1] G. Murphy, ‘On decomposability of some Specht modules for symmetric groups’, J.Algebra (1980), 156–68.[M2] G. Murphy, ‘Submodule structure of some Specht modules’, J. Algebra74