U.U.D.M. Project Report 2008:7

Examensarbete i matematik, 30 hpHandledare och examinator: Volodymyr Mazorchuk

Maj 2008

Decomposition of Certain [Sn]-modules into Specht Modules

Valentina Chapovalova

Department of MathematicsUppsala University

Abstract

Permutation modules are essential in the study of representationsof the symmetric group Sn. A natural action of a direct sum of sym-metric groups on a permutation module gives rise to its decompositioninto certain submodules Mµ

η (which correspond to the irreducible sub-modules of the semigroup FP+(Sn)). Since the permutation modulesplits into the direct sum of Specht modules in a known way, an in-teresting problem is finding multiplicities of the Specht modules inMµ

η . The background material is reviewed and the motivation for theproblem is presented. The results consist of several solved cases andsuggestions for further solution techniques. No general solution forthe problem is currently known.

Contents

1 Representations of Sn 2

2 The Origin of the Problem 7

3 Multiplicity Questions for Some Submodules of PermutationModules 83.1 Formulation of the Problem . . . . . . . . . . . . . . . . . . . 83.2 Case of Two-rowed Partitions . . . . . . . . . . . . . . . . . . 113.3 Hom(Sλ, Mµ) . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.4 Hook Tableaux . . . . . . . . . . . . . . . . . . . . . . . . . . 163.5 µ = (n, 1, . . . , 1) ` n + k, cases k = 2, 3 . . . . . . . . . . . . . 173.6 The Robinson-Schensted Algorithm . . . . . . . . . . . . . . . 223.7 µ = (2, 2, 1, 1) . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1 Representations of Sn

The object of our study is the symmetric group Sn, consisting of all bijectionsfrom the set {1, 2, . . . , n} to itself, with usual function composition as groupmultiplication. Call an element π ∈ Sn a permutation.

The main objects we will consider are modules.

Definition 1. Let V be a vector space over C and G be a group. Then Vis called a (left) G-module if there exists a group homomorphism ρ : G →GLC(V ), or equivalently, if there exists a a map from G× V to V such that:

(i) gv ∈ V

(ii) g(cv + dw) = c(gv) + d(gw)

(iii) (gh)v = g(hv)

(iv) ev = v

for all v, w ∈ V, g, h ∈ G, c, d ∈ C.A right G-module is defined in the similar way, but G acts from the rightand condition (iii) is changed into v(gh) = (vg)h.

In the following examples we consider the group G = Sn:

Example 2. Consider a one-dimensional vector space V , then πv = v for allv ∈ V, π ∈ Sn defines an Sn-module structure om V . We call V the trivialmodule.

Example 3. Consider a one-dimensional vector space V . Let

sgn(π) :=∏

1≤i<j≤n

π(i)− π(j)

i− j

for π ∈ Sn, sgn: Sn → C2 is then a group homomorphism. Then πv =sgn(π)v for all v ∈ V defines an Sn-module structure om V . We call V thesign module.

Example 4. Since Sn acts on S = {1, 2, . . . , n}, let

CS := {c11 + c22 + · · ·+ cnn : ci ∈ C}.

Then the action

π(c11 + c22 + · · ·+ cnn) = c1π(1) + c2π(2) + · · ·+ cnπ(n)

defines an Sn-module structure on CS and we call CS the natural module.

2

Example 5. If Sn = {π1, . . . , πm}, where m = |Sn| = n!, then

C[Sn] = {c1π1 + · · ·+ cmπm : ci ∈ C}

together with the action

π(c1π1 + · · ·+ cmπm) = c1(ππ1) + · · ·+ cm(ππm)

is an Sn-module and we call it the regular module. Also C[Sn] is called thegroup algebra of Sn.

For any group G its group algebra C[G] is defined similarly.

A submodule of a G-module V is a subspace W of V such that W is closedunder the action of G. Then a G-module V is called simple if it only hassubmodules V and {0}. A module is called indecomposable if it cannot bewritten as a direct sum of two non-zero submodules.

A G-homomorphism between G-modules is a linear transformation θ :V → W such that

θ(gv) = g(θv)

for all g ∈ G, v ∈ V . An isomorphism between G-modules is a bijectiveG-homomorphism.

Theorem 6 (Krull-Schmidt Theorem). Any finite-dimensional nonzero mod-ule can be expressed uniquely as a direct sum of finitely many indecomposablemodules, up to isomorphism and permutation of summands.

Theorem 7 (Maschke’s Theorem). If G is a finite group, then every inde-composable G-module is simple.

Krull-Schmidt Theorem together with Maschke’s Theorem imply that ifG is a finite group and V is a nonzero G-module, then V splits uniquely intoa direct sum of its simple submodules.

Therefore C[G] ∼=⊕

i miVi, where the Vi form a complete list of pairwiseinequivalent simple G-modules. Then proposition 1.10.1 in [Sa] states:

(i) mi = dim Vi

(ii)∑

i(dim Vi)2 = |G|

(iii) The number of Vi equals the number of conjugacy classes of G.

Thus, in particular, every simple G-module occurs in the decomposition ofC[G] at least once, which means the number of simple G-modules is finite.

3

In order to find all simple Sn-modules, we introduce the notion of apartition. If λ = (λ1, . . . , λl) is an array of natural numbers, such thatλ1 + · · · + λl = n and λ1 > . . . > λl, we call λ a partition of n an denote itλ ` n. Then a Ferrer’s diagram of shape λ is an array of n dots, with left-justified rows, row i having λi dots. A Young tableau of shape λ (or tableauof shape λ) is obtained from the corresponding Ferrer’s diagram by replacingthe dots by numbers 1, . . . , n bijectively. A tableau is called standard if itsrows and columns increase.

Example 8. The Ferrer’s diagram of shape λ = (4, 2, 1) looks like:

• • • •• ••

An example of a (non-standard) Young tableau of shape λ = (4, 2, 1):

1 4 3 67 25

We get an equivalence relation by saying that two Young tableaux arerow-equivalent if their corresponding rows contain the same numbers. Thenan equivalence class is represented by a tabloid of shape λ, which we shalldenote by {t}.

Example 9. An example of a tabloid of shape λ = (4, 2, 1):

1 3 4 62 75

We can define the action of Sn on a tabloid, namely (π(T ))ij = π(Tij),where Tij denotes the entry in the i:th row and j:th column of T.

Definition 10. Let λ ` n. Then

Mλ := C{{t1}, . . . , {tk}}.

where {t1}, . . . , {tk} is the complete list of λ-tabloids, is called the permuta-tion module corresponding to λ.

Definition 11. Let t be a tableau with columns C1, . . . , Ck and rows R1, . . . , Rk.Then the group Ct = SC1 × · · · × SC1 is called the column-stabilizer of t andRt = SR1 × · · · × SR1 the row-stabilizer of t.

4

Definition 12. The associated polytabloid to a tableau t is

et = κt{t} :=∑π∈Ct

sgn(π)π{t}

Example 13. If

t =1 2 34

then

et =1 2 34

− 4 2 31

.

Definition 14. If λ ` n, its corresponding Specht module, Sλ, is the sub-module of Mλ, which is spanned by polytabloids et, t running through alltableaux of shape λ.

Example 15. If λ = (2, 1) ` 3, then

Sλ = C{

1 23

− 3 21

,1 32

− 2 31

,2 13

− 3 12

}=

= C{

1 23

− 3 21

,1 32

− 2 31

},

since the rest of the tableaux of shape λ give no new polytabloids.

Theorem 16. The Specht modules Sλ for λ ` n form the complete list ofsimple Sn-modules.

Proposition 17. The set {et: t is a standard λ-tableau} is a basis of Sλ.This follows from Theorem 2.5.9 and Theorem 2.6.4 in [Sa]. Thereforedim(Sλ) is equal to the number of standard tableaux of shape λ.

A generalized Young tableau is obtained from a Ferrer’s diagram by re-placing the nodes with positive integers, but repetitions are allowed. Thecontent of a generalized tableau is a finite sequence (µ1, . . . , µm) where µi isequal to the number of times i occurs in the tableau. The set of all generalizedYoung tableaux of shape λ and content µ is denoted Tλµ.

A semistandard Young tableau is a generalized Young tableau, which rowsare increasing and columns are strictly increasing. The set of all semistandardYoung tableaux of shape λ and content µ is denoted T 0

λµ.

5

Theorem 18 (Young’s Rule).

Mµ ∼=⊕λ`n

KλµSλ,

where Kλµ = |T 0λµ|.

Definition 19. If λ ` n and v = (i, j) is a node in the corresponding Ferrer’sdiagram, then its hook is H(i,j) := {(i, j′) : j′ > j} ∪ {(i′, j) : i′ > i} with thecorresponding hooklength h(i,j) = |H(i,j)|.

Example 20. If λ = (4, 3, 1), then h(1,2) = 4:

· • • •· • ··

Theorem 21 (Hook Formula). If λ ` n, then:

dim(Sλ) =n!∏

(i,j)∈λ h(i,j)

.

Instead of modules one often considers matrix representations. If there ex-ists a group homomorphism ρ : G→ GLd, where GLd is the complex generallinear group of degree d, then ρ assigns to each g ∈ G a matrix ρ(g) ∈Matdand is called a matrix representation of G. The trivial module corresponds tothe trivial representation, the sign module to the sign representation, isomor-phic modules correspond to isomorphic representations etc. When a moduleis simple, we call the corresponding representation irreducible.

Since the notions of a representation and a module are equivalent, many ofthe previous theorems could be reformulated in the “matrix representation”form. Actually, proofs of several of the theorems are based on the notion ofthe group character.

Definition 22. Let ρ : G → Matd be a matrix representation. Then thecharacter of ρ is a function χ : G→ C such that:

χ(g) := tr(ρ(g))

Here are some properties of characters that will be used:

Proposition 23. Let ρ be a matrix representation of a group G of degree dand with character χ. Then(i) χ(id) = d(ii) If g and h are in the same conjugacy class of G, then χ(g) = χ(h)(iii) Suppose σ is a representation of G. Then σ is isomorphic to ρ iff χ(g) =σ(g) for all g ∈ G.

6

Theorem 24. Let ρ be a matrix representation of a group G with characterχ. Suppose

ρ ∼= m1ρ1 ⊕m2ρ2 ⊕ · · · ⊕mkρk,

where the ρi are pairwise inequivalent irreducible representations with char-acters χi. Then

χ = m1χ1 + m2χ2 + · · ·+ mkχk.

2 The Origin of the Problem

The problem that will be analyzed in chapter 3 comes from the theory ofsemigroups. This is a short introduction to its origin.

Let P(Sn) denote the power set of Sn. The subsets of Sn form a semigroupwith the operation

AB := {ab : a ∈ A, b ∈ B}for A, B ∈ P(Sn). If N = {1, . . . , n}, then P(Sn) naturally acts on P(N).We define an equivalence relation on the semigroup P(Sn):

A ∼ B ⇔ A(X) = B(X) for all X ∈ P(N).

Since ∼ is a congruence relation we can take the factor semigroup

FP(Sn) := P(Sn)/ ∼ .

Since ∅ is a completely isolated congruence class, we can remove it and stillget a semigroup, which we denote

FP+(Sn) := FP(Sn)\{∅}.

It is convenient to denote an element A ∈ FP(N) by tables(

kAk

), k ∈ N ,

where Ak = A(k). Then the product of two tables(

kAk

)and

(k

Bk

)corre-

sponding to classes A and B is the table(

kSi∈Ak

Bi

)corresponding to the class

AB.

Lemma 25. Idempotents of a semigroup S correspond bijectively to the max-imal subgroups of S. Namely if e ∈ S is an idempotent, then Ge = (eSe)∗,the group of units of eSe, is a maximal subgroup of S having e as the identityelement.

From the proof of Theorem 3 in [GM2] follows that the idempotents ofFP+(Sn) are exactly elements with a table

(i

(i)ρ

), where ρ is an equivalence

relation on N and (i)ρ is the block containing i. The following importantresult is Theorem 1 in [GM3]:

7

Theorem 26. Assume ρ is an equivalence relation on N , and ki is thenumber of parts equal to i, i.e. k1 + · · ·+ kn = m and k1 + 2k2 + · · ·+ nkn =n. Then the maximal subgroup Gρ of the semigroup FP+(Sn) with identity(

1 . . . n(1)ρ . . . (n)ρ

)is isomorphic to Sk1 × Sk2 × · · · × Skn.

Note that the notion of a representation of a semigroup and the corre-sponding module is very similar to that of a group. See Chapter 11 in [GM1]for details.

Let e ∈ FP+(Sn) be an idempotent as in Theorem 26 and Ge be thecorresponding maximal subgroup of FP+(Sn). Then e is associated with anequivalence relation ρ, which corresponds to a partition µ ` n (the parts arethe cardinalities of the equivalence classes). As in [GM1, Section 11.2] we canconstruct a so-called L-induced module V (Ge), a submodule of FP+(Sn).Then V (Ge) ↓Sn can be identified canonically with Mµ, the permutationmodule corresponding to µ. Following [GMS, Theorem 7], all the simpleFP+(Sn)-modules, associated with Ge, are submodules of V (Ge) and allare of the form V (M), where M is some simple Ge-module. Moreover, themultiplicity of V (M) in V (Ge) is equal to the dimension of M .

Sn can be seen as a subgroup of FP+(Sn) and each V (M) can be consid-ered an Sn-module by restriction. We want to find out about the structure ofthese V (M) ↓Sn . From Theorem 16 we get that all the simple submodules ofV (M) ↓Sn are Specht modules, therefore it remains to determine their mul-tiplicities. Though V (M) is not a canonical submodule of V (Ge), since weknow its multiplicity in V (Ge), we can instead look at its isotopic componentV (M) of V (Ge), that is the sum of all submodules of V (Ge), isomorphic toV (M), considered as Sn-modules. Then we determine the Specht modulemultiplicities in V (M). The maximal group Ge acts with right group multi-plication on the module V (Ge) by automorphisms. Therefore, the V (M) ↓Sn

will be isomorphic to the Mµη defined in Section 3.

3 Multiplicity Questions for Some Submod-

ules of Permutation Modules

3.1 Formulation of the Problem

Let µ ` n, such that µ contains k1 rows of length l1, k2 rows of length l2, . . . ,kt rows of length lt, i.e.

µ = (µ1, µ2, . . . , µr1 , l1, . . . , l1, µr1+1, . . . , µrt , lt, . . . , lt, µrt+1, . . . , µrt+1),

8

where

µ1 > µ2 > · · · > µr1 > l1 > µr1+1 > · · · > µrt > lt > µrt+1 > · · · > µrt+1 .

Then we can define the (right) action of G := Sk1 × Sk2 × · · · × Skt onMµ by permuting the rows of tabloids of shape µ in the following way: if Tconsists of the rows

a1, a2, . . . , ar1 , a11, . . . , a1k1 , ar1+1, . . . , art , at1, . . . , atkt , art+1, . . . , art+1

(in the given order), then T(π1, . . . , πt) consists of the rows

a1, a2, . . . , ar1 , a1π1(1), . . . , a1π1(k1), ar1+1, . . . ,

. . . , art , atπt(1), . . . , atπt(kt), art+1, . . . , art+1 .

In other words, the rows of unique length are unchanged and the rows ofequal length li are interchanged according to πi.

This action makes Mµ into a right C[G]-module: the row aij of a tabloidT(ππ) is equal to the row aiππ(j) of T. On the other hand, the row aij of atabloid (Tπ)(π) is the row aiπ(j) of the tabloid Tπ, and the row aiπ(j) of thetabloid Tπ is the row aiπ(π(j)) of T. Thus, the corresponding rows of T(ππ)and (Tπ)(π) are equal, and so the properties for Mµ being a right C[G]-module are satisfied. For reasons of convenience we write the action of Gfrom the right.

Lemma 27. The actions of Sn and G on Mµ commute.

Proof. G permutes the rows of a tabloid T , while Sn permutes the entries ofT independent of their placement in the tabloid.

Lemma 28. Mµ ∼= mC[G] as G-modules for some m ∈ N.

Proof. We can split Mµ into isomorphic C[G]-submodules in the followingway: namely for a tabloid T define

S(T ) := TG = {T (π1, ..., πt) : (π1, ..., πt) ∈ G}

Then T ′ ∈ S(T ) ⇒ S(T ) = S(T ′), therefore Mµ splits into orbits under theaction of G. Tπ 6=T for all π ∈ G\{(id)}, therefore the order of each orbit is|G| and each orbit is isomorphic to the regular representation of G, i.e. thegroup algebra C[G].

Thus Mµ ∼= mC[G] as G-modules, where m = dim(Mµ)|G| .

9

Let η := (η1, . . . , ηt), k := (k1, . . . , kt), and η ` k means ηi ` ki for all i,i = 1, . . . , t.

We know that

C[G] = C[Sk1 × Sk2 × · · · × Skt ] = C[Sk1 ]⊗ C[Sk2 ]⊗ · · · ⊗ C[Skt ] =

=⊕η`k

(dim Sη1) . . . (dim Sηt)(Sη1 ⊗ · · · ⊗ Sηt).

Then Lemma 28 gives us

Mµ ∼=⊕η`k

m(dim Sη1) . . . (dim Sηt)(Sη1 ⊗ · · · ⊗ Sηt) (1)

as G-modules. This suggests a splitting of Mµ into Sn-modules which corre-spond to simple G-submodules.

Therefore, define

Mµη :=

∑V

V ;

where V runs through all subspaces of Mµ such that V ∼= Sη1⊗· · ·⊗Sηt as G-modules. Then Mµ

η are closed under the usual action of Sn on tabloids, sinceif π ∈ Sn then left multiplication with π is a right G-module homomorphism,because of Lemma 27. Therefore π(Mµ

η ) ⊂Mµη by the definition of Mµ

η .

Proposition 29. Mµ ∼=⊕

η`k Mµη as Sn-modules.

Proof. Equality (1) shows there are no simple G-submodules of Mµ otherthan tensor products of Specht modules in Mµ

η , which are mutually non-isomorphic. Together with definition of Mµ

η we get that Mµ ∼=⊕

η`k Mµη as

vector spaces. Since each Mµη is stable under the action of Sn, the proposition

follows.

According to Young’s Rule

Mµ ∼=⊕λ`n

KλµSλ,

where Kλµ are the Kostka numbers. Proposition 29 implies

Mµη∼=

⊕λ

α(µ, η, λ)Sλ.

Problem: Determine α(µ, η, λ), i.e. decompose each module Mµη into

Specht C[Sn]-modules.

10

3.2 Case of Two-rowed Partitions

Consider µ = (n, n) ` 2n. Then C[S2] = S(2) ⊕ S(1,1), therefore we are

interested in the decompositions of modules M(n,n)(2) and M

(n,n)(1,1) over S2n.

Theorem 30. If n is odd,

M(n,n)(2)

∼=(n−1)/2⊕

l=0

S(2n−2l,2l)

and

M(n,n)(1,1)

∼=(n−1)/2⊕

l=0

S(2n−2l−1,2l+1)

If n is even,

M(n,n)(2)

∼=n/2⊕l=0

S(2n−2l,2l)

and

M(n,n)(1,1)

∼=(n−2)/2⊕

l=0

S(2n−2l−1,2l+1)

Proof. S2 = {id, (12)} acts on the base elements of M (n,n) by:

id · a1 . . . an

an+1 . . . a2n=

a1 . . . an

an+1 . . . a2n

(12) · a1 . . . an

an+1 . . . a2n=

an+1 . . . a2n

a1 . . . an

Since the regular representation of S2 corresponds to

id 7→(

1 00 1

), (12) 7→

(0 11 0

),

the matrices can be diagonalized by conjugating with

(1 11 −1

):

(1 11 −1

) (0 11 0

) (1 11 −1

)−1

=

(1 00 −1

).

11

Since S(2) corresponds to the trivial representation π 7→ (1), and S(1,1) corre-sponds to the sign representation π 7→ (sgn(π)), the matrix yields the bases

for M(n,n)(2) and M

(n,n)(1,1) :

M(n,n)(2) = C

{a1 . . . an

an+1 . . . a2n+

an+1 . . . a2n

a1 . . . an: {a1, . . . , an} ⊂ {1, . . . , n}

}

M(n,n)(1,1) = C

{a1 . . . an

an+1 . . . a2n− an+1 . . . a2n

a1 . . . an: {a1, . . . , an} ⊂ {1, . . . , n}

}Therefore M (n,n) = M

(n,n)(2) ⊕M

(n,n)(1,1) .

By Young’s Rule we have Mµ ∼=⊕

λ`2n KλµSλ, where Kλµ is the number

of semistandard tableaux of shape λ and content µ, i.e. containing n 1’s andn 2’s. Since the columns of a semistandard tableau are strictly increasing,Kλµ > 1 only if λ has at most two rows. Since the rows are non-strictlyincreasing, there is only one way of placing 1’s and 2’s in a tableau of suchshape:

1 1 . . . 1 1 . . . 1 2 . . . 22 2 . . . 2

Hence, we have a decomposition into Specht modules, each with mul-tiplicity one: M (n,n) ∼= S(n,n) ⊕ S(n+1,n−1) + · · · + S(2n). We now have todistribute these Specht modules between M

(n,n)(2) and M

(n,n)(1,1) .

Consider the characters of the the representations corresponding to themodules M

(n,n)(2) and M

(n,n)(1,1) , specifically the character of the conjugacy class

(2)n, taking the element (12)(34) . . . (2n− 1 2n) as its representative. Since

(12)(34) . . . (2n− 1 2n)

(a1 . . . an

an+1 . . . a2n+

an+1 . . . a2n

a1 . . . an

)=

=b1 . . . bn

bn+1 . . . b2n+

c1 . . . cn

cn+1 . . . c2n,

where (b1, . . . , b2n) and (c1, . . . , c2n) are permutations of (1, . . . , 2n), the char-

acter of M(n,n)(2) corresponding to (2)n, χ(2)n(M

(n,n)(2) ), is equal to the number of

fixed points of the action of (12)(34) . . . (2n− 1 2n) onto the basis of M(n,n)(2) .

Therefore, one of the following cases must occur:(case 1):

(12)(34) . . . (2n− 1 2n)a1 . . . an

an+1 . . . a2n=

a1 . . . an

an+1 . . . a2n

12

or(case 2):

(12)(34) . . . (2n− 1 2n)a1 . . . an

an+1 . . . a2n=

an+1 . . . a2n

a1 . . . an

ifa1 . . . an

an+1 . . . a2n+

an+1 . . . a2n

a1 . . . anis a fixed point.

Suppose n is odd. Then case 1 is impossible, since the elements of eachtransposition must be in the same row. Then only possibility is case 2, wherethe elements of each transposition must be in different rows. There are 2n

tabloids with this property, which gives 2n

2fixed points. Thus, χ(2)n(M

(n,n)(2) ) =

2n−1. The result is similar for M(n,n)(1,1) , though we can get both fixed points and

points with eigenvalue -1, the latter happening when the rows of a tableauare interchanged, i.e. in case 2. Thus, χ(2)n(M

(n,n)(1,1) ) = −2n−1.

Suppose n is even. Then case 2 gives the same number as for the odd n,but case 1 is possible as well. Since the transpositions must stay within thesame row, the number of tabloids with that property is

(n

n/2

)(we choose n

2

transpositions for the first row). Therefore case 1 yields( n

n/2)2

fixed points.

We get χ(2)n(M(n,n)(2) ) = 2n−1 +

( nn/2)2

. For M(n,n)(1,1) case 1 still gives points with

eigenvalue 1, thus χ(2)n(M(n,n)(1,1) ) = −2n−1 +

( nn/2)2

.

Now we shall find the characters χ(2)n(S(2n−k,k)) of the Specht modulesincluded in M (n,n). By Theorem 29.13 in [JL]

χ(2)n(S(2n−k,k)) = |fixIk((12) . . . (2n−1 2n))|−|fixIk−1

((12) . . . (2n−1 2n))|,

where Ik is the set of subsets of {(1, . . . , 2n)} of size k.Therefore |fixIk

((12) . . . (2n− 1 2n)| is equal to the number of choices for k2

transpositions.Hence χ(2)n(S(2n−k,k)) =

(n

k/2

)if k is even and χ(2)n(S(2n−k,k)) = −

(n

(k−1)/2

)if k is odd.

Thus if n is odd we have

χ(2)n(S(2n)) =

(n

0

),

χ(2)n(S(2n−1,1)) = −(

n

0

),

. . .

13

χ(2)n(S(n+1,n−1)) =

(n

n−12

),

χ(2)n(S(n,n)) = −(

nn−1

2

).

The maximal partial sum of these numbers is(

n0

)+

(n1

)+ · · · +

(n

n−12

)=

χ(2)n(M(n,n)(2) ), therefore we have the definite decomposition of M

(n,n)(2) , namely

M(n,n)(2)

∼=⊕(n−1)/2

l=0 S(2n−2l,2l). It follows that the rest of the terms are included

in M(n,n)(1,1) , i.e. M

(n,n)(1,1)

∼=⊕(n−1)/2

l=0 S(2n−2l−1,2l+1).And if n is even we have

χ(2)n(S(2n)) =

(n

0

),

χ(2)n(S(2n−1,1)) = −(

n

0

),

. . .

χ(2)n(S(n+1,n−1)) = −(

nn−2

2

),

χ(2)n(S(n,n)) =

(nn2

).

Again, the maximal partial sum is equal to χ(2)n(M(n,n)(2) ). Thus M

(n,n)(2)

∼=⊕n/2l=0 S(2n−2l,2l) and M

(n,n)(1,1)

∼=⊕(n−2)/2

l=0 S(2n−2l−1,2l+1).

3.3 Hom(Sλ, Mµ)

We have shown Mµ ∼=⊕

η`k Mµη as Sn-modules. Therefore Hom(Sλ, Mµ) ∼=

Hom(Sλ,⊕

η`k Mµη ) =

⊕η`k Hom(Sλ, Mµ

η ). We have dim(Hom(Sλ, Mµη )) =

α(µ, η, λ), the multiplicity of Sλ in Mµη , which is the number we want to

determine.If

Tλµ = {T : T is a generalized Young tableau of shape λ and content µ},

we can define an Sn-module structure on Tλµ if a tableau t of shape λ isfixed, namely: (πT )(i) := T (π−1(i)), where T (i) denotes the entry of T inthe same place as entry i of t. Then Proposition 2.9.2 in [Sa] states that Mµ

and C[Tλµ] are isomorphic.

14

Definition 31. For each T ∈ Tλµ, the homomorphism corresponding to Tis the map θT ∈ Hom(Mλ, C[Tλµ]) ∼= Hom(Mλ, Mµ) given by θT ({t}) =∑

S∈{T} S, with extension by the cyclicity of Mλ. Then θT := the restriction

of θT to Sλ, is an element of Hom(Sλ, Mµ).

By 2.10.1 in [Sa] we have Hom(Sλ, Mµ) ∼= C{θT : T ∈ T 0λµ}.

Take θ ∈ Hom(Sλ, Mµ). Then θ(gt) = g(θt) for all g ∈ Sn, t ∈ Sλ, sinceθ is an Sn-homomorphism. Then Hom(Sλ, Mµ) has a structure of a rightG-module, by (θπ)t := (θ(t))π, i.e permuting the rows of equal length of thecomponents of t ∈Mµ according to π. We check that θπ ∈ Hom(Sλ, Mµ):

(θπ)(gt) = θ(gt)π = gθ(t)π = g(θ(t)π) = g(θπ)t, ∀g ∈ Sn.

Now, since Hom(Sλ, Mµ) is a G-module, we can decompose it into a directsum of Specht modules Sη, η ` k (recall that η = (η1, . . . , ηt) and η ` kmeans ηi ` ki for all i = 1, . . . , t).

Lemma 32.

dim(Sη) · [Hom(Sλ, Mµ) : Sη] = dim(Hom(Sλ, Mµη )).

Proof. Proposition 29 gives us:

Hom(Sλ, Mµ) ∼= Hom(Sλ,⊕η`k

Mµη ) ∼=

⊕η`k

Hom(Sλ, Mµη ).

Hom(Sλ, Mµη ) is a G-module for each η if we define (τθ)(s) := τ(θ(s)), θ ∈

Hom(Sλ, Mµη ), s ∈ Sλ, π ∈ Sn, τ ∈ G:

(τθ)(πs) = τθ(πs) = τπθ(s) = πτθ(s) = π((τθ)(s)),

the next last step holding because of Lemma 27.To see that Sη is a submodule of Hom(Sλ, Mµ

η ), it is enough to show thatSη is a subspace of Hom(Sλ, Mµ

η ), which is closed under the action of G.From Proposition 29 and the definition of Mµ

η we get Mµη∼= m(dim Sη)Sη

as G-modules. Call this isomorphism ϕ. Then for any base vector et ∈ Sλ

let τ(et) := ϕ−1(τ, . . . , τ) and extend by cyclicity, which we can do accordingto Proposition 2.3.5 in [Sa]. Therefore τ is an Sn-homomorphism, which wewanted to prove.

Now take Sη′ where η′ 6= η. Suppose Sη′ is a submodule of Hom(Sλ, Mµη ),

i.e. we can consider elements of Sη′ as homomorphisms. But then Sη′(s) forsome s ∈ Sλ is a G-module: τθ(s) := (τθ)(s) for τ ∈ G, θ ∈ Sη′ . But since

15

Sη′(s) ⊂ Mµη , Mµ

η∼= m(dim Sη)Sη as G-modules and Sη′(s) ∼= Sη′ we have

Sη′ ∼= Sη, since both are simple modules. Contradiction.Hence Hom(Sλ, Mµ

η ) as a G-module consists of all the Sη in Hom(Sλ, Mµ)and nothing else. Thus dim(Hom(Sλ, Mµ

η )) = dim(Sη)[Hom(Sλ, Mµ) : Sη].

Therefore dim(Sη) · [Hom(Sλ, Mµ) : Sη] = [Mµη : Sλ], i.e. the decompo-

sition of Hom(Sλ, Mµ) into simple G-modules gives us the answer. We willapply this to the special cases in Chapters 3.5 and 3.7.

3.4 Hook Tableaux

A hook tableau is a Young tableau of shape µ = (n− k, 1, . . . , 1), µ ` n, .i.e.a Young tableau where the hook of the node h(1,1) is the whole tableau.

Fix µ such that n > 2k. Denote by µ� := (n−k+1, 1, . . . , 1), µ� ` n+1.In general, if λ = (λ1, λ2, . . . , λl) ` n, denote λ� := (λ1+1, λ2, . . . , λl) ` n+1.

Lemma 33.Hom(Sλ, Mµ) ∼= Hom(Sλ�, Mµ�)

as vector spaces.

Proof. Since Hom(Sλ, Mµ) ∼= C{θT : T ∈ T 0λµ} it is enough to determine a

one-to-one correspondence between T 0λµ and T 0

λ�µ�. Since the columns of asemistandard tableau are strictly increasing and rows are increasing, all 1’sare in the beginning of the first row of the tableau. Since the number of 1’sis n−k > k, all the boxes from row 2 and down are only in the columns withthe 1’s:

T ∈ T 0λµ ⇒ T =

1 1 1 . . . 1 1 1 ∗ . . . ∗∗ ∗ ∗ ∗ ∗∗ ∗ ∗∗ ,

which corresponds to

T ′ ∈ T 0λ�µ�, T ′ =

1 1 1 . . . 1 1 1 1 ∗ . . . ∗∗ ∗ ∗ ∗ ∗∗ ∗ ∗∗ ,

the tableau where the ∗-parts are unchanged, but en extra 1 in the first row isinserted. The correspondence is surjective, since one 1 can always be removeddue to the fact that the number of 1’s in any T ′ ∈ T 0

λ�µ� is n − k + 1 > k,

where k is the number of the rest of the boxes. Then θT 7→ θ′T , where T

16

is mapped to T’ in the way described, is a bijective correspondence of basevectors.

Conjecture 34. The bijection in the proof of Lemma 33 extends to an iso-morphism

Hom(Sλ, Mµ) ∼= Hom(Sλ�, Mµ�)

of Sk-modules.

Proposition 35. Assume that Conjecture 34 is true. Then for a hook tableauof shape µ = (n − k, 1, . . . , 1), µ ` n, such that n > 2k, and η ` k, thedecompositions of Mµ

η and Mµ�η are analogous, i.e.

[Mµη : Sλ] = [Mµ�

η : Sλ�].

Proof. Because of Conjecture 34

[Hom(Sλ, Mµ) : Sη] = [Hom(Sλ�, Mµ�) : Sη],

which by Lemma 32 implies the result.

3.5 µ = (n, 1, . . . , 1) ` n + k, cases k = 2, 3

For small k though, we can determine the decomposition merely by decidingwhere the sign module belongs and then distributing the rest of the Spechtmodules for dimensional reasons.

Case k = 2. Let µ = (n, 1, 1), n > 2, then S2 acts on Mµ and yieldsmodules Mµ

η1and Mµ

η2, where η1 = (2) and η2 = (1, 1). Possible partitions λ

such that there exist semistandard tableaux with shape λ and content µ (n1’s, one 2, one 3) are: λ1 = (n, 1, 1), λ2 = (n + 1, 1), λ3 = (n, 2), λ4 = (n +2) with multiplicities 1, 2, 1, 1 respectively since the possible semistandardtableaux with shape λi and content µ are:

T1 =

1 1 . . . 123 T2 =

1 1 . . . 1 23 T ′

2 =

1 1 . . . 1 32

T3 =

1 1 . . . 12 3 T4 = 1 1 . . . 1 2 3 .

Proposition 36.Mµ

η1∼= Sλ2 ⊕ Sλ3 ⊕ Sλ4 ,

Mµη2∼= Sλ1 ⊕ Sλ2 .

17

Proof. Consider T1 and the corresponding module Sλ1 . The only base ele-ment of Hom(Sλ1 , Mµ) in our standard presentation is θT1 . Therefore Hom(Sλ1 , Mµ)is a one-dimensional S2-module. For some standard tableau t of shape λ1,we take θT1(et), but as en element of C{T 0

λ1µ}:

θT1(et) =

1 1 . . .123 +

2 1 . . .131 +

3 1 . . .112 −

2 1 . . .113 −

3 1 . . .121 −

1 1 . . .132 .

S2 acts on C{T 0λ1µ} in the following way: (id) is the identity operation and

(12) permutes the 2 and the 3 in each tableau. We get that

Hom(Sλ1 , Mµ) ∼= S(1,1),

the sign module, since (id)θT1(et) = θT1(et) and (12)θT1(et) = −θT1(et).Hence, by Lemma 32, [Mµ

η2: Sλ1 ] = 1. Now use the Hook Formula to com-

pute the dimension of each module:

dim(Sλ1) =(n + 2)!

(n + 2) · (n− 1)! · 2 · 1=

(n + 1)n

2

dim(Sλ2) =(n + 2)!

(n + 2) · n! · 1= n + 1

dim(Sλ3) =(n + 2)!

(n + 1) · n · (n− 2)! · 2 · 1= (n + 2)(n− 1)

dim(Sλ4) =(n + 2)!

(n + 2)!= 1

and since

m =dim(Mµ)

|S2|=

(n+2

2

)· 2!

2!=

(n + 2)(n + 1)

2

and Mµηi

∼= mSηi as S2-modules and dim(Sηi) = 1, we get:

dim(Mµηi

) =(n + 2)(n + 1)

2,

for i = 1, 2. Since [Mµη2

: Sλ1 ] = 1, we get (n+2)(n+1)2

− (n+1)n2

= n + 1“dimension” left to distribute to Mµ

η2, which can only be filled by Sλ2 with

multiplicity one, since for n > 2 we have dim(Sλ3) = (n + 2)(n− 1) > n + 1and if we take Sλ4 , every dimension left would be larger than n. ThusMµ

η2∼= Sλ1 ⊕ Sλ2 and the rest of the modules must constitute Mµ

η1, therefore

the proposition is proved.

18

Case k = 3. Let µ = (n, 1, 1, 1), n > 4. By the Appendix in [Li]

C[S3] ∼= S(1,1,1) ⊕ 2S(2,1) ⊕ S(3).

ThusMµ ∼= Mµ

(1,1,1) ⊕Mµ(2,1) ⊕Mµ

(3).

The possible shapes for the Specht modules included in Mµ are λ1 =(n, 1, 1, 1), λ2 = (n, 2, 1), λ3 = (n + 1, 1, 1), λ4 = (n, 3), λ5 = (n + 1, 2), λ6 =(n + 2, 1), λ7 = (n + 3) with the multiplicities 1, 2, 3, 1, 3, 3, 1 respectively,since the possible semistandard tableaux are:

T1 =

1 1 . . . 1234 T2 =

1 1 . . . 12 34 T ′

2 =

1 1 . . . 12 43

T3 =

1 1 . . . 1 234 T ′

3 =

1 1 . . . 1 324 T ′′

3 =

1 1 . . . 1 423

T4 =

1 1 1 . . . 12 3 4 T5 =

1 1 . . . 1 23 4 T ′

5 =

1 1 . . . 1 32 4 T ′′

5 =

1 1 . . . 1 42 3

T6 =

1 1 . . . 1 2 34 T ′

6 =

1 1 . . . 1 2 43 T ′′

6 =

1 1 . . . 1 3 42

T7 = 1 1 . . . 1 2 3 4 .

HenceMµ ∼= Sλ1 ⊕ 2Sλ2 ⊕ 3Sλ3 ⊕ Sλ4 ⊕ 3Sλ5 ⊕ 3Sλ6 ⊕ Sλ7 .

Proposition 37.Mµ

(1,1,1)∼= Sλ1 ⊕ Sλ3

Mµ(2,1)∼= 2(Sλ2 ⊕ Sλ3 ⊕ Sλ5 ⊕ Sλ6)

Mµ(3)∼= Sλ4 ⊕ Sλ5 ⊕ Sλ6 ⊕ Sλ7 .

Proof. The dimensions are (by the Hook Formula):

dim(Sλ1) =(n + 3)!

(n + 3) · (n− 1)! · 3 · 2 · 1=

(n + 2)(n + 1)n

6

dim(Sλ2) =(n + 3)!

(n + 2) · n · (n− 2)! · 3 · 1 · 1=

(n + 3)(n + 1)(n− 1)

3

19

dim(Sλ3) =(n + 3)!

(n + 3) · n! · 2 · 1=

(n + 2)(n + 1)

2

dim(Sλ4) =(n + 3)!

(n + 1) · n · (n− 1) · (n− 3)! · 3 · 2 · 1=

(n + 3)(n + 2)(n− 2)

6

dim(Sλ5) =(n + 3)!

(n + 2) · (n + 1) · (n− 1)! · 2 · 1=

(n + 3)n

2

dim(Sλ6) =(n + 3)!

(n + 3) · (n + 1)! · 1= n + 2

dim(Sλ7) =(n + 3)!

(n + 3)!= 1

and since

m =dim(Mµ)

|S3|=

(n+3

3

)· 3!

3!=

(n + 3)(n + 2)(n + 1)

6

we get

Mµ(1,1,1)

∼= mS(1,1,1) ⇒ dim(Mµ(1,1,1)) =

(n + 3)(n + 2)(n + 1)

6,

Mµ(3)∼= mS(3) ⇒ dim(Mµ

(3)) =(n + 3)(n + 2)(n + 1)

6,

and

Mµ(2,1)∼= 2mS(2,1) ⇒ dim(Mµ

(3)) =4(n + 3)(n + 2)(n + 1)

6.

Note that since Mµ(2,1) can be subdivided into two non-isomorphic Sn+3-

modules, each Specht module contained in it must be present with evenmultiplicity.

As in the case k = 2, we look at the module Sλ1 . Take

θT1(et) =∑

σ∈CT1τ∈RT1

sgn(σ)στ(T1).

Then π ∈ S3 ⇒ ϕ(π) ∈ CT1 , if we consider the inclusion ϕ : S3 ↪→ S4 ×S1 × · · · × S1

∼= CT1 , which we get by ignoring the 1’s in the tableaux andconsidering the permutations of the set {2, 3, 4}. Then

π(θT1(et)) =∑

σ∈CT1τ∈RT1

ϕ(π)sgn(σ)στ(T1) =

=∑

σ∈CT1τ∈RT1

sgn(π)sgn(σ)στ(T1) = sgn(π)(θT1(et)),

20

for any π ∈ S3, therefore, since the module Hom(Sλ1 , Mµ) is one-dimensionalit is isomorphic to the sign module over S3. This holds of course for any k.The “dimension left” for Mµ

(1,1,1) is then

(n + 3)(n + 2)(n + 1)

6− (n + 2)(n + 1)n

6=

(n + 2)(n + 1)

2.

On the other hand, take T4 and T7. Since T7 only consists of one row, thecolumn stabilizer is trivial and the permutation group of {2, 3, 4} is includedin the row stabilizer, thus

π(θT7(et)) = π∑

τ∈RT7

τ(T7) = (θT7(et))

for all π ∈ S3, therefore Hom(Sλ7 , Mµ) is isomorphic to the trivial module.The same thing holds for Hom(Sλ4 , Mµ), since in case of T4 the row stabilizeris isomorphic to S3.

Thus “modules left” for Mµ(1,1,1) are Sλ2 , Sλ3 , Sλ5 and Sλ6 . But

dim(Sλ2) =(n + 3)(n + 1)(n− 1)

3>

(n + 2)(n + 1)

2

and

dim(2Sλ5) =2(n + 3)n

2>

(n + 2)(n + 1)

2for n > 4. If one Sλ5 is included, then there is only 1 “dimension left” andall remaining modules have larger dimension. Now if Sλ3 would be included,it would with multiplicity one fit the remaining dimension. Suppose that isnot the case. Then we must only have several Sλ6 ’s left. But

dim(2Sλ6) = 2n + 4 <(n + 2)(n + 1)

2

for n > 4 and

dim(3Sλ6) = 3n + 6 =(n + 2)(n + 1)

2⇒ n /∈ Z.

So we haveMµ

(1,1,1)∼= Sλ1 ⊕ Sλ3 .

We have 2Sλ2 , 2Sλ3 , 3Sλ5 and 3Sλ6 left to distribute. Because of an earlierremark about multiplicities in Mµ

(2,1) being even we get that one Sλ5 and one

Sλ6 are included in Mµ(3). Since

dim(Mµ(3)) = dim(Sλ4) + dim(Sλ5) + dim(Sλ6) + dim(Sλ7)

for all n, the distribution is determined. All modules that are left are includedin Mµ

(2,1), hence the proof is finished.

21

3.6 The Robinson-Schensted Algorithm

A partial tableau is a tableau which entries form a subset of {1, . . . , n} andwhich rows and columns increase. By the Robinson-Schensted algorithm onecan construct a standard tableau from any permutation π ∈ Sn, getting alarger partial tableau in each step. Every step is a row insertion, which isdone in the following recursive way:

Let P be a partial tableau and x an integer not in it. Then if x is largerthan any element in the first row of P, place x at the end of the first row,finish. If y is the smallest element in the first row of P which is larger thanx, replace y with x and now row-insert y into the partial tableau P\{the firstrow of P}.

Example 38. Row-insertion of 3 into1 2 46

:

(1 2 46

← 3 into the first row

)∼

(1 2 36

← 4 into the second row

)∼

∼(

1 2 34

← 6 into the third row

)∼

1 2 346

Suppose now that π ∈ Sn such that π(i) = xi. Then row-insertingx1, . . . , xn in that order into the empty partial tableau yields the insertiontableau of π, P (π). This is one part of the Robinson-Schensted algorithm.The following theorem is one part of the Robinson-Schensted-Knuth Theo-rem, proved for example in [Fu], chapter 4.1.

Theorem 39. P (π) is a standard tableau.

The algorithm can be defined for generalized tableaux as well, we onlydecide that for equal entries, the leftmost counts as the “least”.

Definition 40. The row word of a tableau is a sequence of the tableau’sentries, taken from the bottom row upwards, from the left to the right.

Example 41. The row word of1 2 3 446

is 641234.

Definition 42. The product of generalized tableaux T and U, T·U, is theresult of row-insertion of the row word of U into T.

22

Example 43. Our example is a little tricky. First take any semistandardtableau T of shape λ ` n and content µ = (n − 2, 1, 1) for n > 2. Takeaway the 1’s, then we get two generalized semistandard tableaux left, Ddownwards and R to the right. Let RS(T):=D·R. We compute this for everysemistandard tableau from the case k=2 from the previous chapter:

T1 =

1 1 . . . 123 T2 =

1 1 . . . 1 23 T ′

2 =

1 1 . . . 1 32

T3 =

1 1 . . . 12 3 T4 = 1 1 . . . 1 2 3 .

Then

RS(T1) =23·∅ =

23

RS(T2) = 3 · 2 = 3 ← 2 =23

RS(T ′2) = 2 · 3 = 2 ← 3 = 2 3

RS(T3) = 2 3 ·∅ = 2 3

RS(T4) = ∅ · 2 3 = (∅← 2)← 3 = 2 ← 3 = 2 3 .

We see that exactly one module of shape λ1 and one of shape λ2 correspondto the partition (1,1) of 2 and one module of shape λ2, one of shape λ3 andone of shape λ4 correspond to the partition (2). Compare with Proposition36.

Since similar computations for k=3 also give the correct decompositionof our modules, the following conjecture is reasonable:

Conjecture 44. For µ = (n, 1, . . . , 1) ` n + k

[Mµη : Sλ] = |{T : RS(T ) has shape η}|

3.7 µ = (2, 2, 1, 1)

So far, we have only looked at permutation modules with correspondingtableau having k rows of equal length and the rest of the rows being pair-wise non-equal. In the general case we have the right action of the groupSk1 × Sk2 × · · · × Skt , i.e. when the tableau has several blocks of rows ofequal length. The question is now if the multiplicities of the Specht mod-ules inside Mµ

η (where η = (ηk1 , . . . , ηkt), ηki` ki) can be determined, if they

23

are known inside Mµηk1

, . . . ,Mµηkt

. We study the following easiest non-trivial

example.Consider µ = (2, 2, 1, 1) ` 6, then S2 × S2 acts on Mµ. Since C[S2 ×

S2] ∼= C[S2] ⊗ C[S2], this action induces modules Mµ1 , Mµ

2 , Mµ3 and Mµ

4

corresponding to representations tr ⊗ tr, tr ⊗ sgn, sgn ⊗ tr and sgn ⊗ sgnrespectively, where tr is the trivial representation of S2 and sgn is the signrepresentation of S2. Since dim M =

(62

)(42

)(21

)= 180 and |S2 × S2| = 4 we

getMµ ∼= 45(Mµ

1 ⊕Mµ2 ⊕Mµ

3 ⊕Mµ4 ).

We will use the method described in Chapter 3.3 and in particular Lemma32, which means determining multiplicities of S2×S2-modules in the modulesHom(Sλ, Mµ), where λ ` 6. By Young’s Rule

Mµ ∼= S(2,2,1,1) ⊕ S(2,2,2) ⊕ 4S(3,2,1) ⊕ 2S(3,3)⊕

⊕4S(4,2) ⊕ 3S(5,1) ⊕ S(6) ⊕ S(3,1,1,1) ⊕ 3S(4,1,1).

Consider Hom(S(2,2,1,1), Mµ). It is one-dimensional and we get a representa-tion of one base vector as in Chapter 3.3:

v =

1 12 234

−

1 22 134

−

2 11 234

+

2 21 134

−

1 12 243

+

1 22 143

+

2 11 243

−

2 21 143

Then S2 × S2 acts on v by exchanging 1’s and 2’s respectively exchanging 3and 4. So

(id, id)v = v

(id, (12))v = −v

((12), id)v = v

((12), (12))v = −v,

which is the action of the module tr ⊗ sgn. Thus [Mµ2 : S(2,2,1,1)] = 1 and

[Mµi : S(2,2,1,1)] = 0 for i = 1, 3, 4.Continue with other modules of the form Hom(Sλ, Mµ) in a similar way,

i.e. by determining the corresponding representations and splitting them intoirreducible representations of S2 × S2. The result will be:

Mµ1∼= S(2,2,2) ⊕ S(3,2,1) ⊕ 2S(4,2) ⊕ S(5,1) ⊕ S(6)

Mµ2∼= S(2,2,1,1) ⊕ S(3,2,1) ⊕ S(5,1) ⊕ S(3,3) ⊕ S(4,1,1)

24

Mµ3∼= S(3,2,1) ⊕ S(3,3) ⊕ S(4,2) ⊕ S(5,1) ⊕ S(4,1,1)

Mµ4∼= S(3,2,1) ⊕ S(4,2) ⊕ S(3,1,1,1) ⊕ S(4,1,1)

Now, compare with the results for the parts (2,2) and (1,1) of µ, which areknown from Chapter 3.2:

M(2,2)(2)∼= S(2,2) ⊕ S(4)

M(2,2)(1,1)∼= S(3,1)

M(1,1)(2)∼= S(2)

M(1,1)(1,1)∼= S(1,1)

Partitions (2) and (1,1) correspond to the trivial and the sign representation

of S2 respectively. Look e.g. at M(2,2)(1,1) and M

(1,1)(1,1) , their simple submodules

correspond to the partitions (3,1) and (1,1) respectively. “Adding” them inall possible ways yield some of the partitions corresponding to the modulesgn ⊗ sgn: (3, 2, 1) = (3, 1, 0) + (0, 1, 1), (4, 2) = (3, 1) + (1, 1), (3, 1, 1, 1) =(3, 1, 0, 0) + (0, 0, 1, 1), ((4, 1, 1) = (3, 1) + (1, 0, 1) is a stranger “sum”). Thesame thing holds for other module combinations, namely we get some of thecorrect modules by “adding” original modules. In particular “adding” thepartitions of M

(2,2)(2) and M

(1,1)(2) gives the partition (4,2) in two ways ((4, 2) =

(2, 2) + (2, 0) and (4, 2) = (4, 0) + (0, 2)), agreeing with the multiplicity ofS(4,2) in Mµ

1 .This gives some motivation to the indicated method of solving the prob-

lem. That is, first determining the multiplicities of Specht modules in M(m,...,m)ηi

and then, if our partition µ consists of several blocks of rows of equal length,computing the answer for Mµ

η (η = (η1, . . . , ηt)) from what is known.

25

Acknowledgments

I am most grateful to my advisor Volodymyr Mazorchuk for suggesting theproblem for this work and guiding me through it, he has been great at ex-plaining new material in this interesting field of mathematics. I would alsolike to thank Johan Bjorklund for supporting and encouraging me in my lasttwo years of mathematical studies. Special thanks goes to Robert Rosen forhelp with editing.

26

References

[Fu] W. Fulton, Young Tableaux, London Mathematical Society StudentTexts 35, Cambridge University Press, 1997

[GM1] O.Ganyushkin, V.Mazorchuk, Introduction to Finite TransformationSemigroups, Springer-Verlag, to appear in 2008

[GM2] O.Ganyushkin, V.Mazorchuk, Factor Powers of Finite SymmetricGroups, Mat. Zametki 58 no.2, 176-188, 1995

[GM3] O.Ganyushkin, V.Mazorchuk, The Structure of Subsemigroups ofFactor Powers of Finite Symmetric Groups, Mat. Zametki 58 no.3,341-354, 1995

[GMS] O.Ganyushkin, V.Mazorchuk, B. Steinberg, On the Irreducible Rep-resentations of a Finite Semigroup, U.U.D.M. Report, 2007:63

[JL] G. James & M. Liebeck, Representations and Characters of Groups,Second Edition, Cambridge University Press, 2001

[Li] D. E. Littlewood, The Theory of Group Characters and Matrix Rep-resentations of Groups, Oxford at the Clarendon Press, 1940

[Sa] B. E. Sagan, The Symmetric Group: Representations, Combinato-rial Algorithms, and Symmetric Functions, Second Edition, Springer-Verlag, 2001

27