Partition algebras

Thomas Czempik

Born May 14th 1991 in Euskirchen, Germany

Tuesday 4th October, 2016

Master’s Thesis Mathematics

Advisor: Prof. Dr. Catharina Stroppel

Second Advisor: Prof. Dr. X Y

Mathematical Institute

Mathematisch-Naturwissenschaftliche Fakultät der

Rheinischen Friedrich-Wilhelms-Universität Bonn

Contents

1 Setup 51.1 Basic definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Restricted and induced representations . . . . . . . . . . . . . . . . . . . . . 91.3 Young tableau . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.3.1 Partitions and conjugacy classes in the symmetric group . . . . . . . 161.3.2 Young frames and Young tableau . . . . . . . . . . . . . . . . . . . . 181.3.3 The Young poset . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.3.4 Branching rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

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1 Setup

This chapter will build a basis for representation theory of finite groups in particular thesymmetric group Sn. We will introduce the concept of Young tableau to make calculationscombinatoric and easy. In the end we will give an idea of the classical Schur-Weyl duality.

1.1 Basic definitions

We will introduce some basic definitions and results from the general (finite dimensional)representation theory.

Definition 1.1.1. Let G be a group (resp. an algebra) with a vector space V , we call

ϕ : G→ GL(V )(resp. End(V ))

a representation of V , if ϕ is a group homomorphism (respectively a homomorphism ofalgebras).Notation: We write (ϕ, V ) or just V .

Note that, since a group consists of invertible elements, also the homomorphisms have tobe invertible, and therefore the image is a subset of GL(V ) and not just End(V )!

Definition 1.1.2. Let G be a group, we define the group algebra of G over the complexnumbers

C[G] :={∑g∈G

agg∣∣∣ag ∈ C, finitely many ag 6= 0

}with multiplication ∑

g∈Gagg

∑g′∈G

bg′g′

:=∑g,g′∈G

(agbg′

) (gg′).

The definition above works also for every ring R instead of C.

Definition 1.1.3. Let (ϕ, V ), (ρ,W ) be representations over an C-algebra A. A linear mapf : V →W is called A-equivariant if the following diagram commutes

Vf //W

V

ϕ(a)

OO

f//W

ρ(a)

OO

or in formulas: f(ϕ(a)v) = ρ(a)f(v) ∀a ∈ A,∀v ∈ V .Abusing notation we write: f(av) = af(v).

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A-equivariant maps are also called homomorphisms of representations. Lets define

HomA(V,W ) := {f ∈ HomC(V,W ) | f is A-equivariant}.

Notation: EndA(V ) = HomA(V, V )

Definition 1.1.4. • Let A be a C-algebra and (ϕ, V ) a representation of A. A vectorsubspace W ⊂ V is called a subrepresentation of A, if

∀a ∈ A, ∀w ∈W : ϕ(a)w ∈W.

Hence it makes sense to restrict ϕ onto W

ϕ�W : A→ EndC(W )

and we write (ϕ�W ,W ).

• A representation (ϕ, V ) of A is called irreducible, if there are only the trivial subrep-resentations (ϕ′, 0) and (ϕ, V ).

• Two representations (ϕ, V ) and (ρ,W ) of A are isomorphic, if there exists an A-equivariant vector space isomorphism Θ: V → W . In this case Θ is called an iso-morphism of representations.

• Let A be an algebra. We denote by

Irr(A) := {V irreducible finite dimensional representations of A}A := Irr(A)/ ∼

where ∼ means up to isomorphism.

The Definitions above are the same for a group G instead of an algebra A. The nextpart of this section will repeat definitions and basic facts about irreducible modules, somewithout proof. See also in [lang ].

Lemma 1.1.5 (Schur’s Lemma). Let (ϕ, V ) and (ρ,W ) be irreducible representations of agroup G. If f ∈ HomG(V,W ) then f = 0 or f is an isomorphism. In particular EndG(V ) =C idV .

Remark 1.1.6. a) Since kernel and image of a representation are subrepresentations,the proof of Schur’s Lemma is well known and written in every algebra book.

b) If (ϕ, V ) and (ρ,W ) are irreducible representations of G then

dim HomG(V,W ) =

{1 if ϕ ∼ ρ,0 if ϕ � ρ.

c) Since a representation (ϕ, V ) is equivalent to the action of a group (resp. a modulestructure) on V , Schur’s Lemma holds for simple modules. Here we not specify themodule structure because it holds for G-, R-, A-modules where G is a group, R a ringand A an algebra.

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Definition 1.1.7. A module V is called semisimple if one of the equivalent conditionsholds:

• V is a sum of a family of simple modules.

• V is a direct sum of a family of simple modules.

• Every submodule U of V is a direct summand.

Lemma 1.1.8. Every submodule and every factor module of a semisimple module V issemisimple.

Proof. Let U be a submodule. Let W be the sum of all simple submodules of U . Since Vis semisimple write V = W ⊕W ′. Every element u ∈ U can be written as u = w + w′ forw ∈W,w′ ∈W ′, but w′ = u− w ∈ U . Hence U is the direct sum U = W ⊕

(U ∩W ′

).

Assume U ∩W ′ 6= {0}. Let u ∈ U ∩W ′ be non zero. Then the simple submodule generatedby u lies in W , but W ∩W ′ = {0}, so u has to be zero, a contradiction. So U ∩W ′ = {0}and U = W is semisimple.For the factor module, we write V = U ⊕ W . As before W is the sum of its simplesubmodules, hence V → V/U induces an isomorphism onto W . Hence V/U is semisimple.

Definition 1.1.9. A ring R is called semisimple if 1 6= 0 and R is semisimple as a leftmodule over itself.

Lemma 1.1.10. A ring R is semisimple if and only if every R-module is semisimple.

Theorem 1.1.11 (Maschke). Let G be a finite group of order n and k a field with charac-teristic char(k) - n. Then the group algebra k[G] is semisimple, i.e. k[G] is a direct sum ofirreducible modules.

Proof. Let (ϕ, V ) be a finite dimensional representation of G and (ρ, U) a subrepresentationof G. Since k is a field, consider a basis of U (as k-vector space) and extend it to a basis ofV , then the matrix representation of ϕ can be viewed as

Mϕ :=

(Mρ 0

0 ∗

).

Then there exists as a k-vector subspace W such that

V = U ⊕W.

Let the homomorphism π : V → U be the projection onto U , i.e. π(u) = u ∀u ∈ U . Definethe trace TrG : V → V by TrG(v) =

∑g∈G[ϕ(g)](v). The trace is G-equivariant

ϕ(h) TrG(v) =∑g∈G

ϕ(h)ϕ(g)︸ ︷︷ ︸(∗)

(v),

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where (∗) is just a permutation of elements in G. If f : V → U is a homomorphism ofG-representations, then TrG ◦f is a G-equivariant map, where G acts by conjugation on f .Define ψ := 1

n TrG ◦π, so we get

0 // Uj // Vψoo

where j is just the inclusion and

ψ ◦ j(v)︸︷︷︸=v

=1

n

∑g∈G

[ϕ(g)](π(v))

(∗∗)=

1

n

∑g∈G

ϕ(g)(π(ϕ(g−1)(v)))

=1

nnv

= v.

For step (∗∗) we used that π is G-equivariant, and the G-action on the Hom-space is givenby conjugation. Since U is a G-subrepresentation, we have ϕ(g−1)(v) ∈ U for all v ∈ U , so(ψ ◦ j) is just the identity.We obtain that ψ splits, V is a G-direct sum of U and kerϕ. Hence by using lemma 1.1.8and lemma 1.1.10 k[G] is semisimple.

Notation: For the symmetric group Sn we denote a simple transposition by si

si =

i 7→ i+ 1,

i+ 1 7→ i,

k 7→ k else

for n ≥ 2 and i ≤ n− 1. Note, that we can see a simple transpositon si ∈ Sk as an elementof Sn for n ≥ k by sending all elements l ≥ k to itself.

Example 1.1.12. a) C[S2] is semisimple as a C[S2]-representation with a decompositioninto 1-dimensional irreducible representations:

C[S2] = {a(id +s1) | a ∈ C} ⊕ {a(id +bs1) | a, b ∈ C and b 6= 1}.

b) The following example shows, that we need the condition char(k) - n. Let k = F2 andG = S2, then F2[S2] is not semisimple.Assume A = F2[S2] is semisimple, then we can find for every subrepresentation Uof A a direct complement. Take U = span{id +s1}, then U has two elements, sinceid +s1 is nilpotent of order 2. So the complement have to contain {id, s1}, but then italready contains the sum of both elements, so it contains U , a contradiction. HenceF2[S2] is not semisimple.

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1.2 Restricted and induced representations

In this section we introduce the construction of the restricted and induced representationof a given representation (ϕ, V ) of G. The restricted representation forms a representa-tion from a subgroup H ≤ G of the representation of the whole group G. The restrictedrepresentation is easier to understand. On the other hand the induced representation goesthe other way round. There one construct a representation of a group G by extendinga representation of a subgroup H. This will be the basis for some theory like Frobeniusreciprocity, which describes the relation between the restricted and induced representation.Or the branching rule, which decribes the decomposition of an irreducible representationof G into irreducible representations of a subgroup H and by Frobenius reciprocity also forthe induced representation.

Definition 1.2.1. Let G be a finite group and H ⊂ G a subgroup. Let (ϕ, V ) be arepresentation of G. For h ∈ H the restriction of V on H is given by:

ResGH ϕ : H −→ Gϕ−→ Aut(V )

(ResGH ϕ)(h) = ϕ(h).

Then (ResGH ϕ,ResGH V ) is a representation of H, where ResGH V = V as a vector space.

For (ϕ, Vϕ) ∈ G and (ρ,Wρ) ∈ H irreducible representations let mϕ,ρ ∈ N be such that

ResGH Vϕ∼=⊕ρ∈H

mϕ,ρ⊕i=1

Wρ

and mϕ,ρ is the multiplicity of Wρ in ResGH Vϕ. Vϕ is an irreducible representation over G,but considered as a representation over H, ResGH Vϕ can be reducible.

Example 1.2.2. a) Let G = S3 the symmetric group, H = {e} the trivial subgroup,λ : S3 → End(C3) the left regular representation and Vλ = span{e1 − e2, e2 − e3}. Hhas only the trivial representation with Wρ = C

ResGH Vλ =⊕ρ∈H

mλ,ρ⊕i=1

Wρ =

mλ,ρ⊕i=1

C.

Since Wρ is one dimensional and ResGH Vλ = Vλ is two dimensional, we have mλ,ρ = 2,hence Vλ is reducible over H.

b) Let G = S3, H = S2, λ : S3 → End(C3) as above and again Vλ = span{e1−e2, e2−e3}.S2 acts by a sign on Wλ = span{e1 − e2}, so we get a one dimensional subspace andone can see that the direct complement is Uλ = span{e1 + e2 − 2e3}. Hence we getthat Vλ is irreducible as S3-representation, but as a S2-representation it is reducibleinto Vλ = Wλ ⊕ Uλ.

Definition 1.2.3. Let G be a finite group, H ⊂ G a subgroup and (ρ,W ) a representationof H. The induced representation (ϕ,Z) from (ρ,W ) of G is

Z = {f : G→W | f(gh) = ρ(h−1)f(g) ∀g ∈ G, h ∈ H}, (1.1)

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as a vector space with the following G-action

∀g1, g2 ∈ G, f ∈ Z : [ϕ(g1)(f)](g2) := f(g−11 g2). (1.2)

We write Z = IndGHW , ϕ = IndGH ρ.

Remark 1.2.4. First we have to check that ϕ as above is a representation ϕ : G→ GL(Z).Let g, g′ ∈ G, h ∈ H, f ∈ Z, then

[ϕ(g)(f)](g′h) = f(g−1(g′h))

= f((g−1g′)h)

= ρ(h−1)f(g−1g′)

= ρ(h−1)[ϕ(g)(f)](g′)

Hence ϕ(g)(f) ∈ Z.The definition of the vector space Z above is not very practical, so we give a constructiveway to describe Z. For w ∈ W (W is our representation of H) we define the map Fw asfollows:

Fw : G→W

g 7→

{ρ(g−1)(w) if g ∈ H,0 else.

Let g ∈ G, h ∈ H, then we obtain that Fw ∈ Z in (1.1)

Fw(gh) =

{ρ(h−1g−1)(w) if gh ∈ H,0 else,

=

{ρ(h−1)ρ(g−1)(w) if g, h ∈ H,0 else,

= ρ(h−1)Fw(g).

We claim that the map Fw and the vector space W ′ = {Fw : w ∈W} are invariant underthe action in (1.2) with elements in H. Let x, g ∈ G :

ϕ(x)Fw(g) := Fw(x−1g)

=

{ρ(g−1x)(w) if g ∈ xH,0 else.

With the additional assumption that x ∈ H we get

=

{ρ(g−1)ρ(x)(w) if g ∈ H,0 else,

= Fρ(x)w(g).

We even have (ρ,W ) ∼= (ResGH ϕ,ResGHW′) as representations of H.

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Let G/H be the left cosets of H in G, S a system of representatives of G/H. We get

G =∐s∈S

sH. (1.3)

Theorem 1.2.5. With the assumptions above, we get a direct sum decomposition of Z:

Z ∼=⊕s∈S

ϕ(s)W ′. (1.4)

Proof. Let f ∈ Z. We claim that f =∑

s∈S ϕ(s)Ff(s). Let g ∈ G arbitrary, x ∈ Srepresentant of the left coset of g. We have∑

s∈Sϕ(s)Ff(s)(g) =

∑s∈S

Ff(s)(s−1g)

(∗)= ρ(g−1x)f(x)

f∈Z= f(x(x−1g))

= f(g).

In (∗) we used s−1g ∈ H, if s and g are in the same left coset. Therefore only one summandis not necessary zero.

Hence we can view (ρ,W ) as a subrepresentation of (ResGH ϕ,ResGH Z) and the elementsof Z of the form ϕ(s)Ff(s) can be seen as the underlying representation ρ on the left cosetsH of G in (1.3) and zero else. So every element has at most one image not equal to zero.We can calculate the dimension of Z = IndGH V by dim IndGHW = dim

⊕s∈S ϕ(s)W ′ =

|G/H|dimW .

Example 1.2.6. Let S2 ⊂ S3 and (ρ,C[S2]) the left regular S2-representation given byleftmultiplication. Then the right cosets are S3/S2 = {id = s1, s2 = s2s1, s1s2s1 = s1s2}.We choose for our representatives S = {id, s2, s1s2}. Then by the construction above wehave W ′ = {Fw : w ∈ C[S2]} ∼= C[S2], where Fw 7→ (w, σw) 7→ w.Let ϕ : S3 → GL(Z) and Z =

⊕s∈S ϕ(s)C[S2]. We claim: Z ∼= C[S3] To prove the claim

note first of all that a basis of C[S2] is given by {id, s1} = S2. Similar S3 is a basis for C[S3].A basis of Z is given by {ϕ(s)Fσ | s ∈ S, σ ∈ S2} by using (1.4). Then the isomorphism ofvector spaces is given by

ψ :⊕s∈S

ϕ(s)W ′ → C[S3]

ϕ(s)Fσ 7→ s · σ.

This isomorphism depends on the chosen representative system S.

Let H ⊂ G be a subgroup, (ρ,W ) a representation of H and (ϕ,Z) the induced repres-entation of G and G =

∐s∈S sH as in (1.3). In the following, we want to consider how the

endomorphisms between EndH(W ) and EndG(Z) are related. For T ∈ EndH(W ) we defineT : Z → Z for w ∈W, s ∈ S arbitrary by

T (ϕ(s)w) = ϕ(s)(T (w))

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where we consider (ρ,W ) as a subrepresentation of (Resϕ,ResZ) and extend T linearly toZ by using the decomposition in (1.3).

Theorem 1.2.7. The map ζ : EndH(W )→ EndC(Z) with T 7→ T has the following prop-erties.

(i) The definition of ζ does not depend of the chosen representative system S.

(ii) ζ is linear and injectiv.

(iii) ζ is an embedding from EndH(W ) to EndG(Z), i.e. im(ζ) ⊂ EndG(Z).

Therefore EndH(W ) is a subalgebra of EndG(Z).

Proof. We show the properties for the generating set {ϕ(s)w : s ∈ S,w ∈W}.

(i) Let s ∈ S, h ∈ H, w ∈W arbitrary. Let sh be a representant of the left coset of s.

Tϕ(sh)w = Tϕ(s)(ϕ(h)w)

= ϕ(s)T (ϕ(h)w)

= ϕ(s)ϕ(h)T (w)

= ϕ(sh)T (w)

Therefore the definition does not depend of the chosen representative system S.

(ii) Linearity follows directly by definition and for injectivity take T,U ∈ EndH(W ) suchthat T = U :

∀w ∈W : T (w) = Tϕ(1G)w = Uϕ(1G)w = U(w).

We get T = U , so ζ is injective.

(iii) We want to show T ∈ EndG(Z). Let g ∈ G, s ∈ S, w ∈W , T ∈ EndH(W ) and t ∈ S,h ∈ H such that gs = th, so t is the representant of the left coset of gs in S.

T (ϕ(g)(ϕ(s)w)) = T (ϕ(gs)w)

= T (ϕ(th)w)

= Tϕ(t)(ϕ(h)w)

def. of T= ϕ(t)T (ϕ(h)w)

(∗)= ϕ(th)T (w)

= ϕ(gs)T (w)

def. of T= ϕ(g)T (ϕ(s)w)

In (∗) we used, that T is H-invariant. Hence T is G-equivariant and T ∈ EndG(Z)

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Example 1.2.8. In example 1.2.6, that (ρ,C[S2]) is the left regular representation of S2given by leftmultiplication. Every endomorphism T ∈ EndS2(C[S2]) is determined by theimage of T (id) = a · id +b · s1, since the image of T (s1) is given by

ρ(s1)T (id) = T (ρ(s1) id) = T (s1).

Similar for T ∈ EndS3(C[S3]) with the induced representation. Here we use the isomorphismfrom example 1.2.6 to easily calculate the dimension of EndS3(C[S3]).

dim EndS3(C[S3]) =dim EndC(C[S3])

| S3 |=

6 · 66

= 6,

where the factor | S3 | denotes the dimension of the image of the representation. For thetrivial representation we will get the factor 1.

Theorem 1.2.9. Let (ϕ, V ) be a finite dimensional representation of a finite group G and(ψ,W ) a finite dimensional representation of a subgroup H ⊂ G. The following represent-ations are isomorphic as representations of G:

V ⊗ IndGHW∼= IndGH((ResGH V )⊗W ). (1.5)

Proof. Let S be a system of representatives of G/H, {v1, . . . , vn} a basis of V , {w1, . . . , wk}a basis of W . On the left hand side of (1.5) we have the basis

{vi ⊗ ϕ(s)Fwj | i ∈ {1, . . . , n}, s ∈ S, j ∈ {1, . . . , k}}. (1.6)

The basis of the right hand side of (1.5) is given by

{ϕ(s)Fvi⊗wj : G→ V ⊗W | i ∈ {1, . . . , n}, s ∈ S, j ∈ {1, . . . , k}},

but also by

{ϕ(s)F(ϕ(s−1)vi)⊗wj : G→ V ⊗W | i ∈ {1, . . . , n}, s ∈ S, j ∈ {1, . . . , k}} (1.7)

since all s ∈ S are invertible and therefore ϕ(s) is an isomorphism.We define the map Φ: V ⊗IndGHW → IndGH((ResGH V )⊗W ) on the basis elements as follows:For i ∈ {1, . . . , n}, s ∈ S, j ∈ {1, . . . , k} and g ∈ G we set

Φ(vi ⊗ ϕ(s)Fwj )(g) := (ϕ(g−1)vi)⊗ (ϕ(s)Fwj )(g)

and extends Φ linearly onto V ⊗IndGHW . Now we need to show (1) that Φ is an isomorphismisomorphism of vector spaces and (2) that Φ is G-equivariant. For (1) show that Φ sendsthe basis in (1.6) to the basis in (1.7).Let g ∈ G arbitrary.

Φ(vi ⊗ ϕ(s)Fwj )(g) = (ϕ(g−1)vi)⊗ (ϕ(s)Fwj )(g)

= ϕ(g−1s)(ϕ(s−1)vi)⊗ Fwj (s−1g)

????? = F(ϕ(s−1)vi⊗wj)(s−1g)

= ϕ(s)F(ϕ(s−1)vi⊗wj)(g).

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For (2) let x, g ∈ G and vi ⊗ ϕ(s)Fwj a basis element of (1.6), we shorten the notation tov ⊗ F .

Φ(ϕ(x)(v ⊗ F ))(g) = Φ((ϕ(x)v)⊗ ϕ(x)F )(g)

= ϕ(g−1)ϕ(x)v ⊗ ϕ(x)F (g)

= ϕ((x−1g)−1)v ⊗ F (x−1g)

= Φ(v ⊗ F )(x−1g)

= ϕ(x)Φ(v ⊗ F )(g).

Hence Φ is an isomorphism of representations of G.

Alternative proof of Theorem 1.2.9. An alternative proof is obtained directly by the defini-tion of the induced representation in (1.2.3). The left-hand side is spanned by all productsv ⊗ f , where v ∈ V and f : G→W satisfies

f(gh) = ψ(h−1)f(g)

for all h ∈ H and g ∈ G.The right-hand side is made up of all functions F : G→ V ⊗W such that

F (gh) = [ϕ(h−1)⊗ ψ(h−1)]F (g).

Then the same map Φ: V ⊗ IndGHW → IndGH((ResGH V ) ⊗ W ) with [Φ(v ⊗ f)](g) :=(ϕ(g−1)v)⊗ f(g) defines a G-equivariant linear isomorphism.

Corollary 1.2.10. Let (ϕ, V ) be a representation of a group G, H a subgroup of G andX = G/H then

IndGH(ResGH V ) ∼= V ⊗ L(X)

as representations of G. Here L(X) := {f : X → C} the vector space of all complex-valuedfunctions defined on X.

Proof. We use Theorem 1.2.9 with ψ the trivial representation of H. Then

V ′ = {Fv : G→ C | v ∈ C, Fv(g) = v for g ∈ H} ∼= C

Z =⊕s∈S

ϕ(s)V ′ ∼= L(X).

Therefore we get

V ⊗ L(X) = V ⊗ IndGH C= IndGH(ResGH(V )⊗ C)

= IndGH(ResGH V ).

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Example 1.2.11. From the corollary above we obtain that the induced trivial represent-ation of a subgroup H on a group G coincides with L(X), where X = G/H. By usingDefinition 1.2.3 we get the space {f : G → C | f(gh) = f(g),∀g ∈ G, h ∈ H} of all H-invariant functions. Hence all functions only depends on the values of X which leads ofcourse to the same result.

The Frobenius reciprocity decribes the relation between the restriced and induced rep-resentation of a group G and a subgroup H.

Theorem 1.2.12 (Frobenius reciprocity). Let G be a finite group, H ≤ G a subgroup,(ϕ, V ) a representation of G and (ρ,W ) a representation of H. For every T ∈ HomG(V, IndGKW )define T : V →W by setting for every v ∈ V

T v = [Tv](1G). (1.8)

Then T ∈ HomH(ResGH V,W ) and the map

HomG(V, IndGKW )→ HomH(ResGH V,W )

T 7→ T

is a linear isomorphism.

Proof. Let T ∈ HomG(V, IndGKW ), h ∈ H and v ∈ V . We first check, that T ∈ HomH(ResGH V,W ).

Tϕ(h)v = [T (ϕ(h)v](1G)

(∗)= [(IndGH ρ)(Tv)](1G)

(∗∗)= [Tv](h−1)

(∗∗∗)= ρ(h)[Tv(1G)]

= ρ(h)T v.

Note that we use in step (∗) that T is G-equivariant, in step (∗∗) the action of G on IndGHWand finally in step (∗ ∗ ∗) definition 1.2.3 of IndGHW . To show that this is an isomorphism,we give the inverse map. Let U ∈ HomH(ResGH V,W ), v ∈ V and g ∈ G. We defineU : ResGH V →W by [Uv](g) = Uϕ(g−1)v. Then

[Uv](g) = Uϕ(g−1)v

= (IndGH ρ)(g−1)Uv

= (IndGH ρ)(g−1)[Uv](1G),

hence U ∈ HomG(V, IndGKW ).Now we check, that the maps are inverse to each other

[ˇT ]v = [T v](1G) = Tϕ(1−1G )v = Tv,

[ ˆUv](g) = [Uϕ(g−1)v](1G) = Uv(g)

and therefore we have the required isomorphism.

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Example 1.2.13. Let G be a group, which acts transitively on a finite set X. Fix a pointx0 ∈ X and let H = {g ∈ G | gx0 = x0} be the stabilizer of x0. Then X = G/H is the setof left cosets of H.Take the trivial representation (ι,C) on H and the permutation representation (ϕ, V )on G. Then we get an isomorphism HomG(V,L(X)) ∼= HomH(ResGH V,C). By Riesz-representation theorem there exists for every map T ∈ Hom(V,C) an H-invariant vector vTsuch that T can be expressed as T (v) = 〈v, vT 〉V . Therefore if T ∈ HomG(V,L(X)), vT isthe corresponding H-invariant vector of T and the action of ϕ looks like:

Tϕ(g−1)v = [Tv](gx0) = 〈v, ϕ(g)x0〉V ∀g ∈ G, v ∈ V.

1.3 Young tableau

1.3.1 Partitions and conjugacy classes in the symmetric group

In this section we will introduce the concept of a Young tableau, which are directly relatedto the partition of an integer n ∈ N and by an 1 : 1 bijection also to the finite dimensionalrepresentation of the symmetric group. The advantage of a Young tableau is the intuitiveway to calculate decompositions of representations just by deleting or adding boxes in theYoung tableau.Let Sn be the symmetric group of degree n, that is, the group of all permutations of the

set {1, . . . , n}.

Definition 1.3.1. • A permutation π ∈ Sn is called a cycle of length t, and we denoteit by π = (a1, a2, . . . , at), with 1 ≤ ai 6= aj ≤ n for 1 ≤ i 6= j ≤ t[≤ n], if

π(a1) = a2, π(a2) = a3, . . . , π(at−1) = at, π(at) = a1

and

π(b) = b if b ∈ {1, 2, . . . , n} \ {a1, a2, . . . , at}.

• Two cycles π = (a1, a2, . . . , at) and τ = (b1, b2, . . . , bs) are called disjoint if {a1, a2, . . . , at}∩{b1, b2, . . . , bs} = ∅. Two disjoint cycles clearly commute.

• A transposition is a cycle of length 2.

Remark 1.3.2. • Every permutation π ∈ Sn can be written as a product of disjointcycles.

π = (a1, a2, . . . , aµ1)(b1, b2, . . . , bµ2) . . . (c1, c2, . . . , cµh)

where {a1, a2, . . . , aµ1 , b1, b2, . . . , bµ2 , . . . , c1, c2, . . . , cµh} = {1, 2, . . . , n} and we sup-pose, that µ1 ≥ µ2 ≥ · · · ≥ µh and µ1 + µ2 + · · ·+ µh = n.

• If τ ∈ Sn is another permutation then

τπτ−1 = (τ(a1), τ(a2), . . . , τ(aµ1))(τ(b1), τ(b2), . . . , τ(bµ2)) . . . (τ(c1), τ(c2), . . . , τ(cµh))(1.9)

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To see the second point, take a permutation π ∈ Sn in cycle notation. Now conjugatewith an other permutation τ ∈ Sn. By multipling with τ−1τ between every two cycle, onecan reduce to the case, that π has only one cycle. Now apply from the right τ on both sidesof (1.9). We obtain

τ(a1, a2, . . . , aµ)!

= (τ(a1), τ(a2), . . . , τ(aµ1)) ◦ τ

Let l ∈ {1, 2, . . . , n}, then we have two cases.

Case 1 There exists an i such that l = ai, which also means τ(l) = τ(ai). Then we have on theleft side τ [π(l)]

l=ai= τ(ai+1) and on the right side (τ(a1), τ(a2), . . . , τ(aµ1))[τ(ai)] =τ(ai+1).

Case 2 There exists not such an i such that l = ai. In other words, l is a Fixpoint of π. Thenwe have on the left side τ [π(l)] = τ(l) and on the right side (τ(a1), τ(a2), . . . , τ(aµ1))[τ(l)] =τ(l).

From (1.9) we see that two permutations π and π′ are conjugate if and only if they havethe same cycle structure, that is

π′ = (a′1, a′2, . . . , a

′ν1)(b′1, b

′2, . . . , b

′ν2) . . . (c′1, c

′2, . . . , c

′νk

)

with h = k and µi = νi for all i = 1, 2, . . . , h.

Definition 1.3.3. Let n be an positive integer. A partition of n is a sequence λ =(λ1, λ2, . . . , λh) of positive integers such that λ1 ≥ λ2 ≥ · · · ≥ λh and λ1+λ2+ · · ·+λh = n.We write λ ` n.

That is already enough to show the following lemma, which we will use in the next section.

Lemma 1.3.4. Let n ∈ N. There exists an 1 : 1 bijection between

{conjugacy classes of Sn}1:1←→ {partitions of n}

πλ ←[ λπ = (a1, . . . , aµ1) . . . (b1, . . . , bµh) 7→ µ

Proof. Let n ∈ N and λ = (λ1, λ2, . . . , λh) a partition of n. Since it only depends on thecycle structure because of (1.9) we take the following constructive bijection: There are λ1cycles

πλ = (1, . . . )(2, . . . ) . . . (λ1, . . . )

λ2 cycles of length ≥ 2 and λ1 ≥ λ2, (here we assume λ1 > λ2). Hence we can write

πλ = (1, λ1 + 1, . . . )(2, λ1 + 2, . . . ) . . . (λ2, λ1 + λ2, . . . )(λ2 + 1) . . . (λ1),

where we have (λ1 − λ2) cycles of length 1.We repeat this process until we have λh cycles of length ≥ h. Since λh+1 = 0 we have λhcycles of length h. This results in:

πλ = (1, . . . , n− λh + 1) . . . (λh, . . . , n)︸ ︷︷ ︸λh cycles of length h

(λh + 1, . . . ) . . . (2λh − λh−1, . . . )︸ ︷︷ ︸λh−1−λh cycles of length h−1

. . . (λ1 − λ2) . . . (λ1)︸ ︷︷ ︸λ1−λ2 Fixpoints

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For the inverse map we take a permutation π = (a1, a2, . . . , aµ1)(b1, b2, . . . , bµ2) . . . (c1, c2, . . . , cµh)and once more, only the cycle structure matters. We get µ1 summands, µ2 summands ≥ 2,that means, we have µ1 − µ2 summands equal to 1. We repeat this process until we haveµk summands ≥ k. Since there is no µk+1 and µh ≥ µh−1 for h = 2, . . . k, this results in

µ = (k, . . . , k︸ ︷︷ ︸µk times

, k − 1, . . . , k − 1︸ ︷︷ ︸(µk−1−µk) times

, . . . , 1, . . . , 1︸ ︷︷ ︸µ2−µ1 times

).

There is also a bijection to the finite dimensional representation of Sn up to isomorphism,so we have in total the following:

Sk1:1←→ {conjugacy classes of Sk}

1:1←→ {partitions of k} (1.10)

but we will not prove this.

1.3.2 Young frames and Young tableau

In the last section we saw the correspondence between partitions of a number k and theconjugacy classes of the symmetric group Sk. To visualize this and make combinatoricseasier, we will introduce the Young frames and later the Young tableau.

Definition 1.3.5. Let λ = (λ1, λ2, . . . , λh) be a partition of n. A Young frame of shapeλ is an array formed by n boxes with h rows, where the ith row contains λi boxes for alli = 1, 2, . . . , h. In particular we have λ1 columns.

Example 1.3.6. Let λ = (4, 2, 2, 1) ` 9 be a partition, then the associated Young framelooks like:

(1.11)

The rows and columns are numbered from top to bottom and from left to right, like therows and columns of a matrix, respectively. Our example looks like:

(1,1) (1,2) (1,3) (1,4)

(2,1) (2,2)

(3,1) (3,2)

(4,1)

Definition 1.3.7. • Let λ ` n be a partition, then a box in position (i, j) is said tobe removable if i < h and j = λi > λi+1 or i = h and j = λh. That means thatthe remaining Young frame is still a young frame for a partition λ′ ` n − 1. In ourexample, the boxes (1, 4), (3, 2) and (4, 1) are removable.Similarly a box in position (i, j) is addable if λi = j < λi−1 or i = h+ 1 and j = 1.

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• Let λ ` n be a partition of n. A Young tableau of shape λ is a bijection between theboxes of the Young frame of shape λ and the numbers {1, 2, . . . , n}. The bijection isillustrated by filling the boxes of the Young frame with exactly one of the numbers1, 2, . . . , n.Our example 1.3.6 could be filled like this:

5 3 9 12 46 78

• A Young tableau is called standard if the numbers in the boxes are filled such thatthe numbers from left to right and from top to bottom are increasing.Our example above is not standard, but the following is:

1 3 5 92 46 78

(1.12)

We see, that the number 1 has to be in the box (1, 1), while 2 is either in (2, 1) or(1, 2) and n has to be in a removable box.

For any partition λ ` n, we denote by Tab(λ) the set of all standard tableau of shape λ.Finally we set

Tab(n) =⋃λ`n

Tab(λ).

1.3.3 The Young poset

Denote by Y = {λ | λ ` n, n ∈ N} the set of all partitions. Since we get for everypartition a different Young frame, we can also see Y as the set of all Young frames. We canequip Y with a partial order by setting: Given two partitions λ = (λ1, λ2, . . . , λh) ` n andµ = (µ1, µ2, . . . , µk) ` m, then

λ � µ

if h ≤ k, n ≤ m and λi ≤ µi for all i = 1, 2, . . . , h. Or to describe this definition morevisual, the Young frame of λ is contained in the Young frame of µ (If the Young frame ofλ has a box in position (i, j), then also the Young frame of µ).For example, if µ = (5, 4, 3, 1) and λ = (4, 2, 2, 1), then λ � µ.If µ � λ we denote by µ/λ the Young frame of µ where we delete the boxes, which obtain

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to λ.

µ λ µ/λ

In the special case, where µ/λ contains exactly one box, we say that µ covers λ. Equivalentywe can say:

For λ, µ, ν ∈ Y with λ ≤ ν ≤ µ⇒ ν = µ or ν = λ

We write µ→ λ.The Young graph of Y is the oriented graph with vertex set Y and an arrow from µ→ λ ifand only if µ covers λ.

A path in the Young graph is a sequence λ = (λ(n) → λ(n−1) → · · · → λ(1)) of partitionsλ(k) ` k such that λ(k) covers λ(k−1) for k = 2, 3, . . . , n. A path always ends in the trivialpartition λ(1) ` 1. The length of a path is the number l(λ) = n. We denote by Πn(Y) theset of all paths of length n in the Young graph and we set

Π(Y) =∞⋃n=1

Πn(Y)

Given a partition λ ` n and a path λ = (λ(n) → λ(n−1) → · · · → λ(1)) we associate thestandard tableau T of shape λ obtained by placing the integer k ∈ {1, 2, . . . , n} in the boxλ(k)/λ(k−1). Of course this process can be reversed and we receive a bijection

ΠnY↔ Tab(n)

between the paths of length n and Tab(n) which can extend to a bijection

ΠY↔∞⋃n=1

Tab(n)

In our example (1.12) we associate the standard tableau with the path

(4, 2, 2, 1)→ (3, 2, 2, 1)→ (3, 2, 2)→ (3, 2, 1)→ (3, 2)→ (2, 2)→ (2, 1)→ (1, 1)→ (1).

1.3.4 Branching rule

Definition 1.3.8. Let λ ` n be a partition. We denote by Sλ the irreducible representationof Sn corresponding to the standard tableau of shape λ.

The next theorem is very important, but we will only use the result and refer to [symmgrp].

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Theorem 1.3.9. The dimension of the irrducible representation Sλ is equal to the numberof standard young tableau of shape λ

dimSλ =| Tab(λ) |

Corollary 1.3.10. Let 0 ≤ k < n, λ ` n, and µ ` k. Then the multiplicity mµ,λ of Sµ

in ResSn

SkSλ is equal to zero if µ � λ and it equals the number of paths in Y from λ to µ,

otherwise. In any case, mµ,λ ≤ (n− k)! and this estimate is sharp.

Proof. The restriction can be devided into restrictions of the form Sl → Sl−1 which aremultiplicity-free in each step. Hence we have

ResSn

SkSλ = ResS

k+1

SkResS

k+2

Sk+1. . .ResS

n

Sn−1Sλ

According to the Young graph, the multiplicity of Sµ in ResSnSk Sλ is equal to the number

of paths in Y that start in λ and end in µ. It is also equal to the possibilities by addingsuccessively n − k boxes to the diagram of µ. (Via this way we have always a diagram ofa partition). In particular, this multiplicity is bounded above by (n− k)!. This estimate issharp when the boxes can be added to different rows and columns.

Example 1.3.11. The following example will always give (n−k)! paths in the Young graphrespectively the multiplicity mµ,λ = (n − k)!. Define m = n − k, then take the partitionsλ = (m,m− 1, . . . , 1) ` n and µ = (m− 1,m− 2, . . . , 1) ` k. In this case, we will have inthe l-th step (m− l) removable boxes to obtain a partition of µ with l = 1, 2, . . . ,m. Hencewe have (n−k)! = m! paths from λ to µ. Below you see the example for m = 4 respectively(10− 6)! possible paths.

XX

XX

−−−−−−−−−→4! = 24 paths

The following theorem is very useful for the combinatorics of the irreducible represent-ations of the symmetric group respectively multiplicity-free group extensions because wecan draw a young tableau and get an arbitrary (multiplicity-free) inductive or restrictiverepresentation just by adding or removing boxes and calculating the possible ways to dothat.

Theorem 1.3.12 (Branching rule). For every partition λ ` n we have

ResSnSn−1Sλ =

⊕µ`n−1λ→µ

Sµ, (1.13)

that is, the sum runs over all partitions µ ` n − 1 such that λ covers µ. Furthermore forevery µ ` (n− 1) we have

IndSnSn−1Sµ =

⊕λ`nλ→µ

Sλ. (1.14)

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Proof. Equation (1.13) is a special case from corollary 1.3.10. For equation (1.14) we usethe isomorphism

HomG(⊕λ`nλ→µ

Sλ, IndSnSn−1Sµ) ∼= HomH(ResSnSn−1

⊕λ`nλ→µ

Sλ, Sµ)

by Frobenius reciprocity. Since Sµ is an irreducible representation, we get by Schur’s lemmaan isomorphism for every possibility to add a box in the corresponding Young tableau. Thatis enough because the path to add a box and then delete the same one to get back to thegiven partition µ ` (n− 1) is unique.

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