Branching Rules and Little Higgs models

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Branching Rules and Little Higgs models

Cathelijne ter Burg

Bachelor Thesis year 3

Supervisors: prof. dr. J. Stokman & prof. dr. E. Laenen

July 16, 2015

Image is from [28]

KdVI & IoP-ITFA

Faculteit der Natuurwetenschappen, Wiskunde en Informatica Universiteit van Amsterdam

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Abstract

This thesis will discuss branching rules for GL(n, C) and Little Higgs models and will revolve around the concept of symmetry breaking. An introduction to representation theory is given after which the results are applied to Sn and GL(n, C). The irreducible characters of GL(n, C) will be related to the

Schur polynomials which enables us to derive the branching rules for GL(n, C). The results are then extended to SU (N ) and some examples are given. Then I discuss spontaneous symmetry breaking in physics, introduce Nambu-Goldstone bosons (NGB) and discuss the Higgs mechanism applied to the Standard Model gauge group. The hierarchy problem and the need to search for physics beyond the Standard model are discussed. We focus on Little Higgs models that are a partial solution to the hierarchy problem by postulating new physics at the TeV scale, and yield a naturally light Higgs through a mechanism called collective symmetry breaking. Collective symmetry breaking will be introduced via an SU (3) based toy model after which the ”Littlest Higgs”, based on a global SU (5) symmetry is discussed. Branching rules for SU (5) will be also be discussed in the framework of Grand Unified Theories.

Title: Branching rules and Little Higgs models. Author: Cathelijne ter Burg, 10422722

Supervisors: Prof. dr. J. Stokman & Prof. dr. E. Laenen Final date: 17-07-2015

IoP-ITFA & KdVI University of Amsterdam

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1 Introduction

Symmetries and symmetry groups play an important role in modern science, as well in mathemat-ics as in physmathemat-ics. Mathematically we say an object obeys a symmetry when it is invariant under a transformation. A 3D sphere for one, obeys rotational symmetry i.e. is invariant under the contin-uous symmetry group SO(3). In most cases though, contincontin-uous symmetry groups are not easy to work with. Therefore, mathematicians ”represent” their elements as linear transformations between vectorspaces. Representation theory goes back to the late eighteen hundreds and finds applications in many fields, ranging from mathematics and statistics, both pure and applied, to physics. In the latter it has among others applications in particle physics. There, it proved to be convenient to associate the transformations of different particles under a symmetry group with its different representations. Each particle is assigned to a certain representation and is said to transform as the representation or ”lie in the representation”. As important as symmetries is the notion of symmetry breaking, meaning that the symmetry group is reduced to a smaller group. This will be an integral part throughout this thesis. Mathematically, symmetry breaking is described by branching rules. These describe how the restriction of an irreducible representation decomposes into irreducible representations of the sub-group. Put in terms of particles, they tell us how these will transform under the reduced symmetry group. A symmetry group that is of special importance in particle physics is the group SU (N ) as the Standard Model symmetry group is SU (3)C× SU (2)W × U (1)Y.

The Standard Model describes the universe in terms of fermions (matter) and the forces between them by interchanging bosons1. With the discovery of the Higgs boson in 2012 all particles predicted by the standard model have been observed. It is highly consistent with experimental data, stands as one of the biggest successes of modern science. However, it still has some unanswered questions, one of them being the nature of dark matter, and physicists nowadays believe that it is only an effective theory, meaning that at some high energy scale it must be replaced by a more fundamental theory. One reason for expecting physics beyond the standard model comes from measurements of the coupling constants. They are very different at low energies but at higher energies seem to converge to a single point at around 1015_{GeV indicating that the three forces were once united. Here we discuss a different}

motivation for expecting physics not far beyond the Standard Model, namely the hierarchy problem. Put briefly, the hierarchy problem refers to the vast difference between the energy scales of different physical theories and causes the Higgs mass to acquire quadratically divergent quantum corrections. If the Standard Model is assumed to remain valid up to energies many orders of magnitude above the electroweak symmetry breaking scale, it becomes an extremely fine-tuned theory to keep the Higgs mass at its measured value ∼ 100 GeV, which is highly unnatural. In this thesis I will discuss a class of models that form a partial solution to the hierarchy problem. These are called ’Little Higgs models’ and they postulate new physics at the 1 TeV scale by introducing new particles. Little Higgs models will be the subject of the second part.

The full content of this thesis will be organized as follows: The first part will have a focus on math-ematics whilst the second will be focusing on physics. I start by providing an overview of results from representation theory that will be needed. In section 3 I discuss the symmetric group Sn. I

introduce the Young diagram corresponding to the partitions of n and the Young tableaux and show

1

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that by constructing a particular element in its group algebra, called the Young symmetrizer, we can construct all the irreducible representations of Sn, which will be parametrized by the partitions. Then

we turn our attention to the group GL(V ) ∼_{= GL(n, C) in section 4. It turns out that the same Young} symmetrizer can be used to also construct many of the irreducible representations of GL(n, C). In particular I show that the irreducible characters are given by certain symmetric polynomials, called Schur polynomials. Once we have made this identification, determining branching rules in section 5 will come down to using known identities between these Schur polynomials. Needed results about these Schur polynomials can be found in Appendix A. Once we have these branching rules, I will discuss some results from Lie theory to argue that we can extend all results to SU (N ) ⊂ U (N ) ⊂ GL(n, C). Then, in section 6 I will introduce the field theoretic Lagrangian, discuss symmetry breaking and in-troduce Nambu-Goldstone bosons. Section 7 will inin-troduce local symmetries, the covariant derivative, gauge fields and discuss the Higgs mechanism in the Standard Model responsible for the masses of the elementary fermions and the W±, Z0 bosons. Section 8 will focus on the hierarchy problem. In section 9 I start by discussing the first Little Higgs model that will be based on SU (3). It will act as a toy model to comprehend the physics. Later in section 11 I will discuss the ’Littlest Higgs’, based on SU (5) that is the minimal model that could act as a viable extension of the Standard Model at the TeV scale. In section 10 I will give a general introduction to the group SU (5) by discussing how the fundamental particles can be distributed over the irreducible SU (5) representations. For this, we will use the results of one of the branching rules as derived in section 5.

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2 Necessities from representation theory

This section will serve as an overview of all results from representation theory that will be needed in the following sections. The results are mainly based on [1], [4], [5] and [6]2_{. We begin by defining what}

we mean by a representation.

2.1 Representations

Definition 2.1. Let G a group. A representation of a group G on a C-vectorspace V is a homo-morphism ρ : G → GL(V ) of G to the group of automorfisms of V , such that ρ(gh) = ρ(g)ρ(h) and ρ(1) = 1 for all g, h ∈ G.

The dimension of the representation is the dimension of the vectorspace V . Here GL(V ) is the group of all invertible linear transformations φ : V → V . We will often call V the representation of G and omit the symbol ρ, that is, write gv for ρ(g)v. A few representations that will be important are the following.

Definition 2.2. The trivial representation of G is the representation ρ : G → GL(C) such that ρ(g) = 1 for all g ∈ G.

All groups have this one-dimensional representation. In the case where G = Sn, the symmetric group

on n letters, there is a second one-dimensional representation.

Definition 2.3. The sign representation of Sn is the representation sgn : G → GL(C) such that:

sgn(g) = (

1 if g is an even permutation

-1 if g is an odd permutation (2.1)

Definition 2.4. If X is a finite set and G acts on X on the left, then there is an associated permutation representation. If V is the vectorspace with basis {ex: x ∈ X}, then G acts on V by

g ·X x∈X axex= X x∈X axegx. (2.2)

Definition 2.5. The permutation representation corresponding to the left action of G on itself is called the regular representation.

This representation is of dimension |G| and has a set basis vectors given by {eg: g ∈ G}.

Definition 2.6. A sub-representation of a representation V is a sub-vectorspace W of V which is invariant under the action of G, i.e. g · w ∈ W for all w ∈ W .

Definition 2.7. A representation V is called irreducible (or simple) if the only sub-representations are {0} or V itself. It is called indecomposable if it cannot be written as a direct sum of two nonzero sub-representations. It is called reducible if it has a proper sub-representation.

2_{Proofs that have been omitted are in accordance with Prof. Stokman, mainly due to length and actual}

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Example 2.1. Consider the symmetric group on n letters and let {e1, e2, . . . , en} be the standard basis

of Cn_{. S}

n is a permutation group and thus has a natural permutation representation, where it acts on

Cn by permuting the indices. Note that the one-dimensional subspace spanned by e1+e2+. . .+en is left

invariant under the action of Sn. It has a complementary subspace {(x1, . . . , xn) : x1+ . . . + xn = 0}.

This subspace is n − 1 dimensional and is also invariant and hence a sub-representation. It is called the standard representation of Sn.

Making new representations

Once we have two representations V and W it is possible to construct new representations and this is most easily done by taking their direct sum, i.e. V ⊕ W . Also the tensor product V ⊗ W is a representation via g(v ⊗ w) = gv ⊗ gw. It is then easily deduced that the nth tensor power V⊗n is also a representation. This nth _{tensor power has the exterior power and symmetric powers, denoted}

Λ⊗nV and Sym⊗nV respectively, as sub-representations. They are defined as: Definition 2.8. The nth _{symmetric power is the subspace of V}⊗n _{spanned by}

{X

σ∈Sn

vσ(1)⊗ . . . ⊗ vσ(n)| vi∈ V }

Definition 2.9. The nth _{exterior power is the subspace of V}⊗n_{spanned by}

{X

σ∈Sn

sgn(σ)vσ(1)⊗ . . . ⊗ vσ(n) | vi∈ V }

A completely different way of constructing a new representation is through the dual representation V∗ of V . This is the space of all linear maps φ : V → C.

Definition 2.10. If ρV : G → GL(V ) is a representation of G on a vectorspace V , then the dual

representation ρV∗: G → GL(V∗) is defined by

ρV∗(g)(φ) = φ ◦ ρ_V(g−1)

The vectorspace Hom(V, W ) can also be made into a representation through the map ρHom(V,W )(g)(φ) = ρW(g) ◦ φ ◦ ρV(g−1)

Now, if we compare this equation with the defining map for the dual representation, then we can deduce that in the case of W = C, the trivial representation, we can make the identification of: V∗∼_{= Hom(V, C). And this is in fact a special case of the general case where we have}

Hom(V, W ) ∼= V∗⊗ W

Complete reducibility and Schur’s lemma

Given a representation we would now like to know how it is build up in terms of its irreducible sub-representations, or put differently, how it decomposes in terms of its irreducible sub-representations. We start with the following result.

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Proposition 2.1. If W is a sub-representation of a representation V of a group G, then there is a complementary invariant subspace W⊥ of V such that we have V = W ⊗ W⊥.

Proof To prove this, we define a Hermitian inproduct H(·, ·) on V such that H(gv, gw) = (v, w) for all g ∈ G, v ∈ V, w ∈ W . We can get such an Hermitian inner product by taking any Hermitian inner product H0 and averaging it over G. That is, we define it as

H(v, w) = 1 |G|

X

g∈G

H0(gv, gw)

Then, if W is a sub representation of V , then W⊥ is also a sub representation of V since: gv ∈ W⊥⇔ H(gv, w) = 0, for all w ∈ W ⇔ H(v, g−1w) = 0, for all w ∈ W

since g−1w ∈ W , and we can thus describe W⊥ using W . This proposition tells us that we can in fact consider any representation as a direct sum of sub representations, and by induction on the dimension it can be concluded that it can in fact be written as a direct sum of irreducible sub representations.

Corollary 2.1. Any representation of a finite group G can be written as a direct sum of irreducibles. This is a property that all finite groups have and it is called complete reducibility. It does not tell us however, how it decomposes as a direct sum of its irreducibles and whether this decomposition will in fact be unique. This we are told by Schur’s lemma. To formulate this we need to introduce the G-homomorphism/intertwining operator.

Definition 2.11. Given two representations (ρ, V ) and (π, W ) of the same group G, a intertwining operator or G-homomorphism is a linear map ψ : V → W that intertwines with the action of G, i.e. for which the following holds:

ψ(ρ(g)v) = π(g)(ψ(v)), for all g ∈ G, v ∈ V

Remark Both the kernel and image of φ denoted Ker(φ) and Im(φ) are sub representations of V and W respectively. The vectorspace of all intertwining operators is denoted by HomG(V, W ) which is a

subspace of Hom(V, W ). We can now formulate Schur’s lemma, telling us under what conditions two irreducible representations V and W will be equivalent.

Lemma 2.1. Schur’s lemma If V and W are two irreducible representations of G and φ : V → W is a G-homomorfism, then

1. φ is an isomorphism or φ = 0.

2. If V = W , then φ = λI for some complex scalar λ, where I is the identity map.

Proof The first part of the theorem follows from the fact that both the kernel and image of an intertwiner are invariant subspaces of V and W respectively. Since V and W are both irreducible they can not be proper subspaces. Therefore, Ker(φ) is either {0} or V . If Ker(φ) = V then φ = 0, if φ 6= 0 then Ker(φ) = {0} meaning φ is injective. Similarly, Im(φ) is either {0} or W . If φ = 0 Im(φ) = 0, else Im(φ) = W and φ is surjective. Thus, if both V and W are irreducible, φ is an isomorphism or the zero map.

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For (2) we let λ an eigenvalue3

of φ. It exists since C is algebraically closed. Then the operator φ−λ·Id has a non-zero kernel. But then (1) implies that φ − λ·Id= 0. Thus φ is scalar multiplication.

Note that with this lemma it follows that the intertwiners between irreducible representations satisfy: • HomG(V, W ) = {0} if V is not isomorphic to W .

• HomG(V, W ) ∼= C if V ∼= W .

Proposition 2.2. For any representation V of a finite group G there is a decomposition V = V⊕a1

1 ⊕ . . . ⊕ V ⊕ak

k

where the Vi are non-isomorphic irreducible representations. This decomposition of V into a direct

sum of k factors is unique up to isomorphism, and so are the Vi that occur in the decomposition and

their multiplicities ai.

proof Suppose W is another representation of G with decomposition W =L W⊕bj

j . Suppose further

that φ : V → W is a G-homomorphism. Then by Schur φ must map the factor V⊕ai

i into that factor

W⊕bj

j for which Vi ∼= Wj since if it was mapped to more that one, Vi would not be irreducible. By

applying this to the Identity map φ : V → V the uniqueness follows.

2.2 Character theory

This section will discuss character theory, a convenient way to characterize representations.

Definition 2.12. If (ρ, V ) is a representation of G, then its character χV is the function χV : G → C

defined by

χV(g) = Tr(ρ(g)|V)

i.e. the trace of ρ(g) on V .

The characters represent class functions on the group G. The set class functions on G, written Cclass(G)

is the set of functions that are constant on the conjugacy classes of G. It can be seen as follows: χV(hgh−1) = Tr(hgh−1) = Tr(h−1hg) = Tr(g) = χV(g).

A few more results about characters that we will need are given by the following proposition. Proposition 2.3. Let V and W two representations of G. Then

χV ⊕W = χV + χW, χV ⊗W = χV · χW χV∗(g) = χ_V(g), χ_∧2_V(g) = 1 2[χV(g) 2_{− χ} V(g2)] 3

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proof To prove this we consider a fixed element g ∈ G and compute the values of these characters on g. For the action of g we let {λi} and {µi} be the eigenvalues of V and W respectively. Then the

first two formulas follow from the observation that {λi+ µi} are the eigenvalues of V ⊕ W and {λiµi}

those of V ⊗ W . Similarly λ−1_i = λi are the eigenvalues of g on V∗, since all eigenvalues are nthroots

of unity, with n the order of g. Regarding the last formula we observe that {λiλj : i < j} are the

eigenvalues for g onV2_{V and}

X

i<j

λiλj =

(P λi)2−P λ2i

2 Then the formula follows since g2 _{has eigenvalues {λ}2

i}.

Characters have many applications. For one, they can be used to explicitly find the decomposition of a representation into a direct sum of its irreducible sub-representations. In order to achieve this we want to find a way to project V onto the irreducible representations to find out if those irreducibles are in V , and if so, to determine the multiplicity that they appear with. For this we introduce the first projection formula by setting:

φ = 1 |G|

X

g∈G

ρ(g) ∈ End(V ). (2.3)

Here |G| represents the number of elements of the group G. φ thus represents an average of all endomorphisms ρ(g) : V → V . We further set for any representation V of G

VG= {v ∈ V : gv = v, for all g ∈ G} (2.4) This is again a representation of G and it is a direct sum of trivial sub-representations of V .

Proposition 2.4. The map φ defined in (2.3) is a projection of V onto VG_.

proof Suppose first that v = φ(w) = _|G|1 P gw. Then:

hv = 1 |G| X hgw = 1 |G| X gw for any h ∈ G

so Im(φ) ⊂ VG_{. To prove ” ⊃ ”, we let v ∈ V}G_{, then φ(v) =} 1

|G|P v = v, so V

G _{⊂ Im(φ). Further,}

φ ◦ φ = φ.

With formula (2.3) we can explicitly find the direct sum of the trivial sub representations in a given representations. In particular, the multiplicity of the trivial representation appearing in the decompo-sition of V is the dimension of VG_{. Since φ is a projection onto V}G_{, this dimension is the trace of φ.}

Therefore, writing m for the multiplicity, we have:

m = Trace(φ) = 1 |G| X g∈G Trace(g) = 1 |G| X g∈G χV(g) (2.5)

We can do more with this idea. We let

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Then it follows by Schur that if V is irreducible, dim(Hom(V, W )G_{) is the multiplicity of V in W .}

Also, is V and W are both irreducible we have

dim(Hom(V, W )G) = (

1 if V ∼= W

0 if V 6= W (2.6)

Using proposition 2.3 we have

χ_{Hom(V,W )}G = χV(g) · χW(g)

where we used the fact Hom(V, W ) = V∗⊗ W . By now applying (2.5) we deduce: 1 |G| X g∈G χV(g)χW(g) = ( 1 if V ∼= W 0 if V 6= W (2.7)

where V and W are irreducible. This relation looks a lot like some kind of inner product < χV(g), χW(g) >

and that is precisely what it is. It represents an hermitian inner product on Cclass:

< α, β >= 1 |G|

X

g∈G

α(g)β(g) (2.8)

Therefore we can reformulate (2.7) as:

Theorem 2.1. In terms of the inner product (2.8), the characters of the irreducible representations are orthonormal.

We now let V ∼= V⊕a1

1 ⊕. . .⊕V ⊕ak

k with the Vidistinct irreducible representations, then χV =P aiχVi.

Since the {χVi} is linearly independent, we can conclude the following.

Corollary 2.2. Any representation is determined by its character. By using (2.7) we can further deduce the following results.

Corollary 2.3. A representation V is irreducible iff (χV, χV) = 1.

proof The implication from left to right follows immediately with (2.7). For the other implication we let V ∼= V⊕a1

1 ⊕, . . . , ⊕V ⊕ak

k with the Vi distinct irreducible representations. Then, (χV, χV) =P a2i

which is 1 only if ai= 1 for all i and n = 1.

Corollary 2.4. Let V have a decomposition as above. Then the multiplicity aiof Vi in V is the inner

product ai= (χV, χVi).

proof We have that

(χV, χVi) =

X

j

aj(χVj, χVi) = ai

since (χVj, χVi) = 0 for i 6= j and 1 when i = j.

Another important result follows from the fixed point formula applied to the regular representation. Theorem 2.2. Fixed point formula Let G be a finite group, and X a finite set. Let V be the permutation representation as in definition 2.4. Then for all g ∈ G, χV(g) is the number of elements

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proof Observe that the matrix M that is associated with the action of g is a permutation matrix. Suppose first that X = {x1, x2, x3} and ρ(g) permutes the basis vectors of V by sending ex1 → ex3,

ex2 to itself and ex3 → ex1. Then

M =    0 0 1 0 1 0 1 0 0   

In the general case: if gexi = egxi = exj then the matrix M will have a 1 in the i−th column and

j−th row, and zeros in all other entries of that column. In particular, when xi is held fixed by g, then

gexi = egxi = exi then M has a 1 in the i-th row and i-th column. Therefore, the trace is the number

of 1’s on the diagonal, i.e. the number of elements left fixed by g.

Then we deduce that the character of the regular representation, χR, is given by:

χR(g) =

(

0 if g 6= e

|G| if g = e (2.9)

Thus only when G = e is R irreducible and if we let R = L Vai

i a decomposition into distinct

irreducibles Vi we find

ai= (χVi, χR) =

1

|G|χVi(e)|G| = dim(Vi),

which gives us the following corollary:

Corollary 2.5. Any irreducible representation V of G appears in the regular representation dim (V ) times. In particular this means that the regular representation contains all irreducibles.

Another consequence is the following:

|G| = dim(R) =X i aidim(Vi) = X i (dim(Vi))2

To conclude this section on character theory there is one more result that we will need.

Proposition 2.5. The number of irreducible representations of G is equal to the number of conjugacy classes of G. Equivalently, the characters form an orthonormal basis for the set of class functions CClass(G).

proof Take α : G → C a class function and (α, χV) = 0 for all irreducible representations V . Then it

is to show that α = 0. For this consider the endomorphism φα,V =

X

g∈G

α(g)g V → V

We now want to apply Schur’s lemma. For this we first have to show that φα,V is a G-homomorphism/intertwining

operator. φα,V(hv) = X α(g)g(hv) =Xα(hgh−1)hgh−1(hv) =hXα(hgh−1)g(v) =hXα(g)g(v) =h(φα,V(v))

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Then it now follows by Schur’s lemma (2.1) that φα,V = λ· Id, whereby: λ = 1 dimVTrace(φα,V) = 1 dimV X α(g)χV(g) = |G| dimV(α, χV∗) =0

Therefore φα,V = 0 and hence Pg∈Gα(g)g = 0. This also holds for the regular representation R and

in this representation the elements of G are linearly independent, implying α(g) = 0 for all g ∈ G as

was to be shown.

2.3 Modules and the group algebra

There is one particular choice for the vectorspace V that turns out to be very convenient. This is when V is taken to the the group algebra C[G]. Before getting into the necessary details about the group algebra, I will first review the concepts of algebra’s and modules.

Definition 2.13. An associative algebra over C is a vectorspace A over C together with a bilinear map A × A → A, (a, b) → ab such that (ab)c = a(bc).

Definition 2.14. A left A-module with unit 1A, is a finite dimensional vector space V over C together

with a function φ : A × V → V , (a, v) → av which is bilinear and satisfies a(bv) = (ab)v for all a, b ∈ A, v ∈ V .

Just as is we can define a representation of a group G, we can define a representation of an algebra. Definition 2.15. A representation of an algebra A (or equivalently a left A-module) is a vectorspace V together with an algebra homomorphism

φ : A → End(V ) such that φ(ab) = φ(a)φ(b) and φ(1A) = 1.

Definition 2.16. The regular representation of A (also called the left regular A-module) is the vec-torspace A itself, made into an A-module through the map A × A → A, given by (a, b) → ab for all a, b ∈ A.

As said, we now consider the particular case where the vectorspace A is taken to be the group algebra C[G]. This is the vectorspace with {eg|g ∈ G} as set of basis vectors and multiplication defined by

eg· eh= egh. It consists of all element of the form

C[G] =    X g∈G egg|eg∈ C    .

In this case it holds that C[G] modules correspond directly to representations of G over C since any representation ρ : G → GL(V ) can be linearly extended to a map ˆ_{ρ : C[G] → End(V ) via the map:}

ˆ ρ : C[G] → End(V ), ρˆ   X g∈G egg  = X g∈G egρ(g) ∈ End(V ). (2.10)

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Therefore the correspondence ρ 7→ ˆρ gives us an equivalence between the representations of G and C[G] modules. Further, sub-representations correspond to submodules, irreducible representations to simple submodules etc. All statements about representations of G have an equivalent statement in terms of its group algebra.

Proposition 2.6. If Wi are the irreducible representations of G then we have an isomorphism of

algebra’s:

C[G]∼= M

End(Wi)

proof As mentioned above, a map G → GL(V ) of groups extends linearly to a map C[G] → End(V ) of algebra’s. By applying this to each of the Wi we find the canonical map:

φ : C[G] →MEnd(Wi)

This map is an isomorphism. It is injective since the representation on the regular representation is faithful. Surjectivity follows from the observation that both have dimensionP(dim(Wi))2.

Remark. We can alternatively formulate this in terms of n × n matrix algebra’s over a division ring C because endomorphisms between vectorspaces can be seen as matrices. If we denote ni as the

dimension of Wi, then:

C[G]∼= M

Matni(C).

This relation holds in fact for the more general case where A is a semisimple algebra. It is a result due to Wedderburn. A proof can be found in [4] pp 26. Since it is rather long I will not include it here. It introduces the opposite algebra to show that A ∼=L End(Vi) and uses the observation that if we can

decompose Vi =L_iniUi into irreducibles Ui with a certain multiplicity, then

End(Vi) = End k M i=1 niUi ! ∼ = M 1≤i,j≤k Hom(niUi, njUj) ∼= k M i=1 End(niUi) ∼= k M i=1 Matni(C)

where the third equality follows from Schur’s lemma (2.1), since intertwiners between non-isomorphic irreducibles are zero. Any semisimple algebra A can in this way be written as a direct sum of matrix algebra’s over C.

Primitive idempotents in the center of the group algebra

I will now introduce an important type of elements, called idempotent elements, in an algebra A. These we will need in the irreducible representations of the symmetric group in the next section. In the following A will always be a unital, finite dimensional, associative, commutative algebra over C. Definition 2.17. An idempotent element p ∈ A is an element that satisfies p2_{= p. Two idempotents}

p1, p2 are called mutually orthogonal if p1p2 = 0 = p2p1. An idempotent p is called primitive or

minimal if p = p1+ p2 implies p1= 0 or p2 = 0, where p1 and p2 are mutually orthogonal. A set of

mutual orthogonal idempotents p1, . . . pn is complete if p1+ . . . + pn= 1

These idempotent elements generate left ideals in a commutative algebra A and these left ideals are precisely its submodules. The irreducible submodules of A correspond to the minimal left ideals generated by primitive idempotents. In fact we have the following lemma:

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Lemma 2.2. Let A be an algebra. If V = U ⊕ W is a decomposition of V as direct sum of A -submodules, then the projection of V onto U and W , denoted pu and pw respectively, satisfy

• pu and pw and mutually orthogonal idempotents in A.

• 1 = pu+ pw

• if p ∈ A is an idempotent then 1 − p is an idempotent as well, 1 = p + (1 − p) is a decomposition of 1 ∈ A as sum of orthogonal idempotents and V = pV ⊕ (1 − p)V is a decomposition of V as a direct sum of A-submodules.

Proof Since (1) and (2) are obvious, we only prove (3). Suppose that p ∈ A is an idempotent, then also 1 − p ∈ A is an idempotent, since we have (1 − p)2 _{= 1 − 2p + p}2 _{= 1 − p. They are clearly}

orthogonal since p(1 − p) = (1 − p)p = p − p2_{= 0. Therefore, Im(p)∩Im(1 − p) = pV ∩ (1 − p)V = 0}

and V =Im(p)+Im(1 − p) = pV + (1 − p)V . Thus V = pV ⊕ (1 − p)V is a direct sum decomposition

of V .

In the case where {pi}i=1,...,nrepresents a complete set of orthogonal idempotents, then by the previous

lemma we have that V = ⊕n

i=1piV is a decomposition of V as direct sum of A-submodules. There a a

few more results about these idempotents that we will need.

Lemma 2.3. Suppose p1, p2∈ A are primitive idempotents. Then p1p2= 0 iff p16= p2.

proof We will prove p1p26= 0 iff p1= p2. For ” ⇒ ” this let p1, p2∈ A be primitive idempotents such

that p1p26= 0. Then

p1= p1p2+ p1(1 − p2)

is a decomposition of p1in mutually orthogonal idempotents. Note that A is commutative. Now, since

p1 is primitive and p1p2 6= 0 we must have p1(1 − p2) = 0 and thus p1 = p1p2. Also p2 = p2p1 by

interchanging p1 and p2 and thus p1= p2. Conversely, if p1= p2, then p1p2= p21= p16= 0.

Corollary 2.6. The primitive idempotents of an algebra A form a finite, linear independent set. Proof SupposeP

iλipi = 0, then 0 = pjPiλipi= λjpj by lemma 2.3. Thus, λj = 0 for all j and the

set is this linear independent. Since A is finite dimensional the set pi is a finite set.

Proposition 2.7. Let ai be the set of primitive idempotents in A. Then

1 =X

i

ai

Proof We prove this with induction to the dimension of A. If dim(A) = 1 there is nothing to prove since then 1 is the only primitive idempotent. Suppose now dim(A) > 1. If 1 ∈ A is primitive, then it is the only primitive idempotent. For this, suppose a ∈ A where another primitive idempotent. Then 0 6= a 6= a · 1 and thus a = 1 by lemma 2.3. Therefore, it remains to prove the induction step in the case that 1 ∈ A is not primitive. Then there exist nonzero, pairwise orthogonal idempotents b, c ∈ A such that 1 = b + c. Now set A(b) = Ab = {ab : a ∈ A} and A(c) = Ac = {ac : a ∈ A}. Then A(b), A(c) ⊂ A are subalgebras of A with unit elements b and c respectively. Further, we have that

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A = A(b) + A(c), since 1 = b + c and A(b) ∩ A(c) = {0} since bc = 0. Thus, viewed as vectorspaces we have

A = A(b) ⊕ A(c)

We further conclude that A is isomorphic to the direct sum of the two subalgebras A(b) and A(c) since A(b)A(c) = 0. Now, since b ∈ A(b) and c ∈ A(c) we have A(b) 6= 0 6= A(c) and thus dimA(b), dimA(c) <dim(A). By the induction hypothesis, bP

ibi and c =

P

jcj with {bi} and {cj} the sets of

primitive idempotents of A(b) and A(c) respectively. Then the observation that {bi} ∪ {cj} is the set

primitive idempotents of A(b) ⊕ A(c) completes the proof. Proposition 2.8. Let p ∈ A an idempotent element and A an algebra. Then the left ideal Ap is indecomposable if p is a primitive idempotent.

Proof Suppose Ap where decomposable. Then Ap = I1⊕ I2 for two nonzero A−submodules of Ap.

Since, p ∈ Ap, we can find unique p1 ∈ I1 and p2 ∈ I2 that satisfy p = p1+ p2. If p1 = 0 then

p = p2 ∈ I2 implying that Ap = Ap2 ⊆ I2, since I2 is a left ideal, whereby I1⊕ I2 ⊂ I2, which

contradicts the assumption I1 6= {0}. Similar reasoning for p2 = 0. Further, since p ∈ Ap we have

p1= p1p, and thus

(1 − p1)p1= p1− p21= p1p − p12= p1(p1+ p2) − p21= p1p2

However, p1p2∈ I2and (1−p1)p1∈ I1and we thus conclude that p1−p21= p1p2= 0 since I1∩I2= {0}.

Thus p2

1= p1and p1p2= 0. Similarly, p22= p2and p2p1= 0 by interchanging p1 and p2. Thus p1 and

p2 are orthogonal idempotents with p1+ p2= p. Therefore p is not primitive.

Definition 2.18. A finite dimensional, associative, unital algebra A over C is semi simple if A is the sum of its simple left ideals.

The group algebra C[G], for one, is semi-simple. We can thus conclude that finding a complete set of primitive orthogonal idempotent elements in Z(C[G]), the center of the group algebra, and determining the left ideals they generate will give a decomposition in terms of its simple left ideals, i.e. simple submodules. That the idempotents be in the center is important, since a the group algebra is in general not commutative. These idempotents are defined as:

pπ= dim(Vπ) |G| X g∈G χπ(g)eg∈ Z(C[G]) (2.11)

where π is an irreducible representation of G. Then4: Corollary 2.7. 1. {pπ} is a linear basis of Z(C[G])

2. P

πpπ = ee where ee∈ C[G] is the unit element

3. {pπ} is the set of primitive orthogonal idempotent elements of Z(C[G])

4

I will not prove this here since its proof is rather long. It can be found in [5] The proof relies on the observation that f 7→ ψf ≡P_g∈Gf (f )eg defines an isomorphism between the set of class functions F (G) and

Z(C[G]) and the observations that the characters span the set of class functions and form a set of idempotent elements of F (G)

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Now, we have also seen that the left regular C[G] module contains all simple submodules in its decompositions. Thus, the problem of finding all irreducible representations amounts to constructing such a complete set of idempotents of Z(C[G]). However, constructing such a set is not as easy as it might seem. What we know is that there are as many irreducibles as conjugacy classes, but these are not always easy to determine for a general group. In the case of the symmetric group though, we will see in the next section that these conjugacy classes are in bijection with the partitions of n, and these we can determine easily. First though, we will need to take a look at restricted and induced representations.

2.4 Restricted and induced representations

Given a representation (ρ, V ) of a group G and a subgroup H ⊂ G we can consider the restricted representation, denoted ResG_H(ρ). This is the representation of H defined by

ResG_H(ρ) : H → GL(V ) ResG_H(ρ) := ρ|H (2.12)

In the same way that the above operation of restricted representations provides us to construct repre-sentations of subgroups, we can consider induced reprerepre-sentations. This operation produces represen-tations of G from represenrepresen-tations of H. Here I will briefly discuss this particular construction. For this we let V a representation of G and W ⊆ V an H−invariant subspace. We write G/H for the left cosets of H in G, i.e. it is the set of equivalence classes of G w.r.t. the equivalence relation g ∼ g0

iff g−1g0 ∈ H. Its elements are thus the left cosets gH = {gh : h ∈ H, g ∈ G}. Now, for any g ∈ G, the subspace g · W depends only on the left coset gH of g modulo H, since ghW = g(hW ) = gW . Let now σ ∈ G/H, a coset and write σW for the subspace of V . Then the induced representation is defined as follows:

Definition 2.19. Let V and W be two representations of G and H respectively with H ⊆ G a subgroup and W ⊆ V . Then we say that V is induced by W if:

V = M

σ∈G/H

σW

I this case we write V = IndG_HW , of simply Ind W if there is no ambiguity.

From the previous section we now that representations of a group G have an exact equivalence in terms of its group algebra C[G]. In this sense then, the induced representation IndGHW is defined as

the left C[G] module5_.

C[G] ⊗C[H]W with action ρ(a)(a 0_⊗

C[H]w) = (aa 0_{) ⊗}

C[H]w

Two important induced representation that we will see later are the following.

Example 2.2. The permutation representation of G is induced from the trivial one dimensional representation W of H.

Example 2.3. The regular representation of G is induced from the trivial representation on the trivial subgroup.

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3 The irreducible representations of S

n

Equipped with all the results of the previous section we now turn to the symmetric group and in particular to the construction of its irreducible representations. This construction will also allow us to construct the irreducible representations of GL(V ) in section 4. First thought, we need some results concerning the symmetric group and in particular we discuss the Young diagrams. Results are from [1], [4], [7] and [8].

3.1 The symmetric group

Recall that the symmetric group was defined as:

Definition 3.1. Let n ≥ 1 and write Sn for the symmetric group on n letters. Sn is the group that

consists of all bijections of Ωn = {1, 2, . . . , n} into itself under composition. We have #Sn = n!. The

elements of Sn are the permutations. If σ and π represent two permutations, then π ◦ σ means that

we first apply σ and then π.

We will write a permutation σ ∈ Sn using cycle notation. Take for example, the permutation σ =

(142)(53)(6). This notation means that σ maps 1 → 4 → 2 , 2 → 1 5 → 3 and 3 → 5 and maps 6 to itself. The content of a cycle (i1, . . . , ir) we denote by I and this is an ordered subset of {i1, . . . , ir} ⊆

Ωn of cardinality r. The permutation σ consists of 3 disjoint cycles, where by disjoint we mean that

their contents have trivial intersection. Note that disjoint cycles commute. The length of a cycle is the number of elements it contains and the identity element is the cycle of length 1. The permutation σ thus consists of one 3-cycle, one 2-cycle and one 1-cycle.

Lemma 3.1. Any permutation σ ∈ Sn can be written as a product of disjoint cycles.

Definition 3.2. A partition λ of n is a sequence λ = (λ1, λ2, . . . ) of nonnegative integers such that

P

iλi = n and λ1≥ λ2≥ λ3≥ . . . . We write λ ` n. We refer to the length of λ by writing l(λ) where

l(λ) is the largest i such that λi6= 0.

Definition 3.3. Let σ ∈ Snand write σ as a product of disjoint cycles such that each i ∈ {1, 2, . . . , n}

is in one of the cycles. Collect the lengths of the disjoint cycles and put them in nondecreasing order. This again defines a partition c(σ) of n, which is called the cycle type of σ.

Lemma 3.2. Two permutations σ and τ are conjugate iff they have the same cycle type.

Recall that by proposition 2.5 we have for any group G that the number of irreducible representations is equal to its number of conjugacy classes. In the case of G = Snwe further have the following result.

Proposition 3.1. The conjugacy classes of the symmetric group are in bijection with the partitions. proof Write Sn/ ∼ for the set conjugacy classes and Pn for the set of partitions of n. Now consider

the map

Sn/ ∼→ Pn; Ad(Sn)(τ ) 7→ c(τ )

with c(τ ) the cycle type of τ as in the above definition and Ad(Sn)(τ ) the orbit of τ under conjugation

with σ ∈ Sn, i.e. Ad(σ)(τ ) = στ σ−1. There orbits represent the conjugacy classes of Sn. It is now to

show that the map Ad is well defined and bijective. Consider now a cycle (i1, i2,· · · , ir), then

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for all σ ∈ Sn and all subsets I = {i1, . . . , ir} ⊆ Ωn of cardinality r. It then follows that:

Ad(Sn)σ = {τ ∈ Sn: c(τ ) = c(σ)}

and thus Ad is a well defined and injective map. To show surjectivity, let λ = (λ1, . . . , λm) be a

partition of n, where m = l(λ). Choose further subsets Ij = {ij1, i

j 2, . . . , i

j

λj} ⊂ Ωn

of cardinality λj such that Ij∩ Ij0 = ∅ if 1 ≤ j 6= j0 ≤ m. For the corresponding cycles of length λ_j

let σj be the cycle of content Ij, i.e.

σj= (ij1, i j 2,· · · , i j λj) Then σ ≡ σ1σ2· · · σm∈ Sn

is a product of disjoint cycles such that c(σ) = λ. Thus Ad(Sn)σ → c(σ) maps onto Pnand surjectivity

is also shown.

Corollary 3.1. The number of irreducible representations of the symmetric group is equal to the number of partitions.

This tells us that we can parametrize the irreducible representations with the partitions λ of n. In the next subsection we will discuss an alternative way to look at these partitions by associating them Young diagrams. These are a combinatorial tool named after the British mathematician Alfred Young who first introduced them, in particular to study the representations of the symmetric group.

3.2 Young diagrams and Young tableau’s

The Young diagram is an array of boxes that is associated to a given partition λ = (λ1, λ2, ...) of n. A

Young diagram has λi boxes in the ith row which are lined up on the left. The conjugate partition λ0

of the partition λ is defined by interchanging the rows and columns of the Young diagram associated to λ. Note that (λ0)0 = λ.

Example 3.1. The Young diagrams of the partition λ = (5, 4, 2) and that of its conjugate partition λ0 = (5, 4, 2)0= (3, 3, 2, 2, 1) are respectively given by

and

Ordering on partitions

Given two partitions of n, λ = (λ1, λ2, ..., λl) and µ = (µ1, µ2, ..., µk) we distinguish two different

orderings. The dominance ordering and the lexicographic ordering.

Definition 3.4. We say that λ dominates µ in dominance ordering, written as λ D µ if

m X i=1 λi≥ m X j=1

µj for all 1 ≤ m ≤ max{k, l}.

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In terms of Young diagrams we say that the Young diagram for λ dominates that for µ if there are more boxes in the first m rows for λ than in the first m rows for µ for all 1 ≤ m ≤ max{k, l}. Example 3.2. This dominance ordering is a partial ordering for all n. However, for n ≤ 5, we can consider it also as a total ordering on the partitions. This is easily verified by drawing all the possible young diagrams for the first n = 1, . . . 5. For example if n = 2 we see that the only possible Young diagrams satisfy

D For n = 3 we have:

D D

Up to n = 5 we can construct such a tree of Young diagrams to compare the Young diagrams in total ordering. However, when arriving at n = 6, the the dominance ordering becomes a partial ordering, which can be seen by considering the Young diagrams for λ = (2, 2, 2) and µ = (3, 1, 1, 1).

Definition 3.5. We say that λ dominates µ in lexicographic ordering, written as λ µ if the first non vanishing λi− µi is positive.

Note, that these two orderings are almost the same. Further, the dominance ordering implies the lexicographic ordering, i.e. if λ D µ then also λ µ. The other implication does not hold.

We obtain the Young tableau of a given Young diagram by numbering the boxes with the numbers {1, . . . , n} where each number may be assigned once. We refer to a Young tableau λ by writing Tλ

and write Tλ(i, j) for the numbers in the i-th row and j-th column (1 ≤ j ≤ λi).

Definition 3.6. Let Tλ a Young tableau and λ ` n. Then Tλ is called

1. row standard if its filling is increasing along each row, 2. column standard if its filling is increasing along each column, 3. standard if it row standard as well as column standard.

A Young tableau is further called semi-standard if its filling is nondecreasing along each row and strictly increasing along each column. Note that here it is allowed to place the same number in multiple boxes. Example 3.3. The λ-tableau defined by

tλ(i, j) =    Pi−1 k=1λk+ j if i 6= 1 j if i = 1

for (1 ≤ j ≤ λk)is a standard λ-tableau. For the partition λ = (4, 3, 2) it corresponds to:

1 2 3 4 5 6 7 8 9

Having defined the tableau we write T (λ) for the set of all λ- tableaux. Given a permutation σ ∈ Sn

we can obtain a new tableau σT by defining this to be the tableau with the number σ(T (i, j)) in the (i, j)-th box of its tableaux. This defines a left action of Sn on T (λ) i.e. Sn× T (λ) → T (λ). In the

next section we will use these Young tableau for the construction of the irreducible representations of the symmetric group, following [1].

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3.3 Constructing the irreducible representations of S

n

The first step is to define the row and column stabilizer of the Young tableau Tλ. That is, we define

the following subgroups:

P = Pλ= {σ ∈ Sn: σ preserves each row of Tλ}, (3.1)

and

Q = Qλ= {σ ∈ Sn: σ preserves each column of Tλ}. (3.2)

Corresponding to these subgroups we introduce two elements in the group algebra C[Sn] by setting:

aλ= X σ∈Pλ eσ and bλ= X σ∈Qλ sgn(σ) · eσ (3.3)

and we further define the Young symmetrizer cλ∈ CSn to be

cλ= aλ· bλ=

X

σ∈Pλ,τ ∈Qλ

sgn(τ ) · eστ. (3.4)

This cλ generates a left ideal in the group algebra C[Sn]. This left ideal is a sub representation of the

regular Sn representation and we define it as follows:

Definition 3.7. We call Vλ= C[Sn]cλ the Specht module.

Example 3.4. this example demonstrates how Vλ is computed. For λ = (n) we have cλ = aλ =

P σ∈Sneσ, so V(n)= C[Sn] X σ∈Sn eσ= C · X σ∈Sn eσ

which is the 1-dimensional trivial representation. For n ≥ 2 we have a second 1-dimensional repre-sentation which we find by taking λ = (1, 1, . . . , 1) Then cλ= bλ and we have

V(1,1,...,1)= C[Sn] X σ∈Sn sgn(σ)eσ= C · X σ∈Sn sgn(σ)eσ

which is the sign representation. Taking λ = (2, 1), we find for c(2,1)∈ CS3,

c(2,1)= (e(1)+ e(12)) · (e(1)− e(13)) = e(1)+ e(12)− e(13)− e(132).

To find out which subspace this is we multiply c(2,1) by the basis elements of C[S3]. Then we find:

e(1)(e(1)+ e(12)− e(13)− e(132)) = e(1)+ e(12)− e(13)− e(132) e(12)(e(1)+ e(12)− e(13)− e(132)) = e(12)+ e(1)− e(132)− e(13) e(13)(e(1)+ e(12)− e(13)− e(132)) = e(13)+ e(123)− e(1)− e(23) e(23)(e(1)+ e(12)− e(13)− e(132)) = e(23)+ e(132)− e(123)− e(12) e(123)(e(1)+ e(12)− e(13)− e(132)) = e(123)+ e(13)− e(23)− e(1) e(132)(e(1)+ e(12)− e(13)− e(132)) = e(132)+ e(23)− e(12)− e(123)

Thus C[S3] · c(12) is the 2-dimensional subspace spanned by the first and third vector, and we conclude

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We will see that, after a normalization, these cλ form the complete set of primitive orthogonal

idem-potents we set out for and that thus the Vλ are the irreducible representations.

Theorem 3.1. cλ is an idempotent up to scalar multiplication, i.e. c2λ= nλcλ and the Specht module

Vλ is an irreducible representation of Sn. Every irreducible representation can be obtained in this way.

We need some more results to prove this theorem. In the following, the subscript λ is omitted when it is clear it should be there, i.e. write a for aλ, etc. The idea behind the proof is to show that the

cλ are primitive idempotents up to a scalar multiple which we will call nλ. Further, we show that for

different partitions, λ, µ the product of the corresponding Young symmetrizers yields zero, meaning they are mutually orthogonal. Lastly we show that the left ideals they generate are irreducible. We start by observing that P and Q satisfy the following property which is clear from the way they are defined.

1. For p ∈ P we have p · a = a · p = a

2. For q ∈ Q we have (sgn(q)q) · b = b · (sgn(q)q) = b

Lemma 3.3. For all p ∈ P and q ∈ Q we have p · c · (sgn(q)q) = c, and c is the only such element in C[Sn] up to scalar multiplication.

We first note that since P and Q have trivial intersection, an element of Sn can be written as the

product p · q, p ∈ P, q ∈ Q in at most one way. Therefore, c =P ±eg where the sum is over al cycles

g ∈ Snthat can be written as the product p·q with the coefficient being sgn(q). For one, the coefficient

of e1in c is 1. In the proof we consider the tableau Tλ = tλ defined in example 3.3. Proof IfP ngeg

satisfies this condition, then we have that npgq = sgn(q)ng for all g, p, q. In particular we have npq=

sgn(q)ng. We therefore have to verify that ng = 0 if g /∈ P Q. For such a g it is sufficient to find a

transposition t such that p = t ∈ P and q = g−1tg ∈ Q, since then g = pgq, so ng = −ng. We now

define T0 = gT , i.e. the tableau obtained by replacing each entry i of T by g(i). The claim is that there are two distinct integers appearing in the same row of T and in the same column of T0 and that the element t is the transposition of these two integers. It is now to verify that if such a pair of integers did not exist, that one could then write g = pq for some p ∈ P, q ∈ Q. To show this, we take p1 ∈ P

and q0₁∈ Q0 _{= gQg}−1 _{such that the tableaux p}

1T and q01T0 have the same first row. This we repeat

on the rest of the tableau. Then one gets p ∈ P and q0 ∈ Q0_{so that pT = q}0_T0_{. Then pT = q}0_{gT from}

which it follows that p = q0g and thus g = pq where q = g−1q0−1g ∈ Q.

In the following we will use the lexicographic ordering on the partitions λ and µ.

Lemma 3.4. _{1. If λ > µ, then for all x ∈ C[S}n] we have that aλ· x · bµ= 0. In particular we have

cλ· cµ= 0.

2. For all x ∈ C[Sn], cλ· x · cλ is a scalar multiple of cλ. In particular we have cλ· cλ= nλ· cλ,

for some nλ∈ C.

Proof

1. Take x = g ∈ Sn. Then, since gbµg−1 is the element constructed from gT0, with T0 the tableau

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two integers in the same row of T and in the same column of T0. Let now t is as in the previous lemma be the transposition of these two integers. Then aλ = aλ· t and t · bµ = −bµ, hence

aλ· bµ = aλ· t · t · bµ= −aλ· bµ as was required to show.

2. This follows from Lemma 3.3

Corollary 3.2. If λ < µ, then cλ· C[Sn] · cµ = 0; in particular cλ· cµ= 0.

Proof We use the anti-involution6 _{map ˆ of C[S}n] that is induced by the map g 7→ g−1, g ∈ Sn.

Noting that its fixed points are aλ, bλ, aµ, bµ , i.e. ˆcλ = (aλbλ)ˆ = ˆbλˆaλ = bλaλ, we have (cλxcµ)ˆ =

(aλbλxaµbµ)ˆ= ˆbµaˆµxˆˆbλˆaµ= bµaµxbˆ λaµ= 0 since aµxbˆ λ= 0.

Having showed that the cλare idempotent elements that are mutually orthogonal, we now show that

Vλ they define are the irreducible. That is we proof:

Lemma 3.5. 1. Each Vλ is an irreducible representation of Sn.

2. If λ 6= µ, then Vλ and Vµ are not isomorphic.

Proposition 3.2. Let R be a ring and I 6= 0 a left ideal of R. If I is a direct summand of R, then I2_{6= 0.}

Proof Suppose that I is a direct summand of R, then there exists a left ideal J such that I ⊕ J = R. In particular, we can find i ∈ I and j ∈ J such that i + j = 1. Then i = i2_{+ ij by multiplying both}

sides on the left with i. Then I2 6= 0, for else we had i = ij ∈ I ∩ J = 0, 1 = j ∈ J , and hence

J = R, I = 0.

Proof lemma 3.5.

1. We begin by noting that cλVλ ⊂ Ccλ by Lemma 3.4. If W ⊂ Vλ is a sub representation, then

either cλW is Ccλ or 0. If the first is true, then cλ ∈ cλW ⊆ W so Vλ = C[Sn]cλ ⊂ W .

Otherwise W · W ⊂ C[Sn] · cλW = 0, but then W = 0 with proposition 3.2. This shows in

particular that cλVλ6= 0, i.e. that the number nλ6= 0.

2. We may assume λ > µ. Then cλVλ= Ccλ 6= 0, but cλVµ= cλC[Sn]cµ = 0. So they can not be

isomorphic as C[Sn] modules.

As a final step, we determine the factor nλ in c2λ= nλcλ.

Lemma 3.6. For any partition λ, cλcλ= nλcλ with nλ=_Dim(Vn! _λ₎.

Proof Let F be right multiplication by cλ on C[Sn]. Then, since F is multiplication by nλon Vλ, and

zero on Ker(cλ), the trace of F is nλ times the dimension of Vλ. But the coefficient of eg in egcλis 1,

so trace(F) = |Sn| = n!

We have thus shown that the elements cλ = dim(V_n!λ)cλ7 form a mutually orthogonal set of

primi-tive idempotents of Z(C[Sn]). This therefore proves the theorem since they give us all the irreducible

representations by letting λ vary over the partitions. In the remaining of the chapter I will discuss some more properties about the Specht modules. In particular, I will introduce the Young-subgroup and discuss Young’s Rule [1].

6_{An (anti-)involution map is a function f that is it’s own inverse, i.e. f (f (x)) = x for all x in the domain}

of f .

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3.4 Young subgroups, induced representations and Young’s Rule

We now introduce the Young subgroup. This is a subgroup of Sn that is isomorphic to

Sλ= Sλ1× . . . × Sλk

for some partition λ = (λ1, . . . , λk). There is one specific Young subgroup that is called the standard

Young subgroup. It is defined as

Sλ= Sλ1× Sλ2× . . . × Sλk

where Sλ1 acts on the set {1, 2, ..., λ1},

Sλi acts on the set

   i−1 X j=1 λj+ 1, i−1 X j=1 λj+ 2, . . . , i−1 X j=1 λj+ λi    and Sλk acts on    k−1 X j=1 λj+ 1, k−1 X j=1 λj+ 2, . . . , n    .

Since this a subgroup, we can induce representations on it to representations of Sn. In particular,

inducing the trivial representation on each of the Sλi to Sn gives us the permutation representation.

Definition 3.8. We write Mλ for the permutation representation obtained by inducing the trivial

representation on Sλ to Sn, i.e. Mλ= Ind ↑SSnλ (1).

This Mλ we can equivalently define as Mλ = C[Sn]aλ, with aλ as before. Further, since we have a

surjection

Mλ= C[Sn]aλ Vλ= C[Sn]aλbλ, x 7→ x · bλ

and an isomorphism

Vλ= C[Sn]aλbλ∼= C[Sn]bλaλ⊂ C[Sn]aλ= Mλ

we note that Vλappears in the decomposition of Mλ for every partition λ. To see the second equality

note that right multiplication by aλ gives a map C[Sn]aλbλ→ C[Sn]bλaλ and right multiplication by

bλ gives a map back. These compositions are multiplications by non-zero scalars.

There is in fact an explicit formula, known as Young’s Rule, that tells us how the permutation module decomposes in terms of the irreducible Specht modules.

Theorem 3.2. (Young’s Rule) The permutation module Mλ _{decomposes as}

Mλ=

M

µDλ

KµλVµ

Definition 3.9. The numbers Kµλ are called the Kostka numbers.

These Kostka numbers are defined combinatorially as the number of semi-standard tableaux of shape µ and content λ. That is, it is the number of ways we can fill the boxes of the Young diagram for µ with λ11’s, λ22’s, up to λk, k’s, in such a way that the entries in each row are nondecreasing and the

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Proposition 3.3. Suppose λ, µ ` n. Then the Kostka number Kµλ is non vanishing if and only if

µ D λ. Further, Kλλ= 1

The property of the Kostka numbers as stated in proposition 3.3 is important. Suppose we consider an ordering on the partitions λ1, λ2, . . . from (n) up to (1n)8. Then this will also gives us an ordering

on Mλ_{. Young’s rule now says that the first module M}λ1 _{will be equal to one copy of V}λ1_{. The next}

module, Mλ2 _{will contain this same V}λ1 _{with a certain multiplicity, plus one copy of a new irreducible}

Vλ2 _etc.

Example 3.5. Consider the partition λ = (1, . . . , 1) Then Mλ _{is easily seen to be the regular}

rep-resentation since we induce the trivial reprep-resentation from the trivial subgroup. It therefore follows that in this case Kµ(1,...,1) = dim(Vµ) since for the regular representation the irreducibles occur with

multiplicity being their dimension. This thus provides a way to determine the dimension of Vλ. It is

the number of ways to fill the Young diagram of λ with the numbers 1 to n, in such a way that all rows and columns are increasing.

Example 3.6. Observe that K(n)λ= 1 since there is only one semi-standard tableau. Then by Young’s

Rule we conclude that each permutation module Mλ _{contains exactly one copy of the trivial}

represen-tation S(n)_{. See example 3.4.}

The Kostka numbers are usually notated in a table. In the next example I will apply Young’s rule to compute this table for S5.

Example 3.7. I will begin by giving the table of Kµλ and then work out how its entries are obtained.

Kµλ λ → (5) (4,1) (3,2) (3,1,1) (2,2,1) (2,1,1,1) (1,1,1,1,1) µ ↓ (5) 1 1 1 1 1 1 1 (4,1) 0 1 1 2 2 3 4 (3,2) 0 0 1 1 2 3 5 (3,1,1) 0 0 0 1 1 3 6 (2,2,1) 0 0 0 0 1 2 5 (2,1,1,1) 0 0 0 0 0 1 4 (1,1,1,1,1) 0 0 0 0 0 0 1

The values of the Kostka numbers are determined by counting the number of Young tableaux of shape µ and content λ. When λ = (5) there is only one tableau that is semi standard and has 5 times the number 1 as numbering of its boxes. Namely:

1 1 1 1 1

For the next partition λ = (4, 1) we can draw the following semi standard tableau with 4 times a 1 and one 2. These are

1 1 1 1 2

8

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Thus K(5)(4,1) = 1 and K(4,1)(4,1)= 1, all other Kostka numbers being zero, so we indeed get a new

Specht module V(4,1)_{. I will not write out the diagrams for all partitions, since the result of doing}

so can be read of from the table but I will write out the semi standard tableaux for µ for the filling λ = (2, 1, 1, 1). 1 1 2 3 4 1 1 2 3 4 1 1 2 4 3 1 1 3 4 2 1 1 2 3 4 1 1 3 2 4 1 1 4 2 3 1 1 2 3 4 1 1 3 2 4 1 1 4 2 3 1 1 2 3 4 1 1 2 4 3 1 1 2 3 4

We thus have M(2,1,1,1)∼= V(5)⊕3V(4,1)_⊕3V(3,2)_⊕3V(3,1,1)_⊕2V(2,2,1)_⊕V(2,1,1,1)_{, and as decomposition}

for the regular representation M(15) _{we find}

M(15)∼= V(5)⊕ 4V(4,1)⊕ 5V(3,2)⊕ 6V(3,1,1)⊕ 5V(2,2,1)⊕ 4V(2,1,1,1)⊕ V(15). Recall we defined the permutation module as Ind(↑Sn

Sλ, 1). We can also consider

M_λ0 = Ind ↑Sn

Sλ0( sgn), i.e. the representation induced from the sign representation on the young

sub-group Sλ0. This induced representation we can also realize as: Mλ0 = C[Sn]bλ and this representation

also includes Vλ. In [2] it is shown that the Specht module is the only irreducible constituent that these

two induced modules have in common and that this common constituent occurs with multiplicity one (which is reflected by the diagonal Kostka number being one) . That is, it is shown that:

Ind ↑Sn

Sλ (1) ∩ Ind ↑

Sn

Sλ0(sgn) = Vλ

That this multiplicity is 1 is crucial in the construction. It ensures us that when we consider this intersection we get exactly one copy of Vλ. One final result about the Specht modules concerns a

formula for its dimension.

Theorem 3.3. (The Hook Length formula)

dimVλ= n! l1!· · · lk! Y 1<i<j<≤k (li− lj) = n! Q i≤λjhi,j

where hi,j is the hook length of the box with label (i, j) and is the number of boxes directly below of to

the right including the box once, and li= λi+ k − i.

The first equality is a result that follows from the Frobenius character formula. We did not discuss this here and a proof can be found in [1] or [4]. The second equality follows from the observation that

l1! Q 1<j≤k(l1− lj) = Y 1≤m≤l1,m6=l1−lj m

and noting that the factors m in this product are precisely the hook lengths hi,1. Deleting the first

row of the diagram and proceeding by induction proves the statement.

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6 4 3 1 4 2 1 1

Then for the dimension of the corresponding S8 representation we find:

dim V(4,3,1)=

8!

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4 The irreducible representations of GL(V )

In this section we will focus on the irreducible representations of the group GL(V ) ∼= GL(n, C). It turns out that there is a connection between these and the Specht modules we considered in the pre-vious section. In particular, we will determine the irreducible characters of these irreducible GL(V ) representations. We will follow [1]

Given a group G we have a representation of G on a vectorspace V which we denote by g(v) 7→ gv. We can now consider the nth tensor power V⊗n which is also a representation and both G and Sn

act on this space. We have a left-action of G given by g(v1⊗ . . . ⊗ v2) 7→ (gv1) ⊗ . . . ⊗ (gv2). We

also have a right-action of Sn on V⊗n given by (v1⊗ . . . ⊗ vn)σ = vσ(1)⊗ . . . ⊗ vσ(n) and it is easily

seen that their actions commute. This V⊗n is not irreducible though and we would therefore like to break it up into irreducible representations of G. We will see that in the case where G = GL(V ) this can actually be accomplished9_{. Due to the commutativity of the actions of S}

n and GL(V ) on V⊗n

we expect there to be some kind of relation between the decomposition into irreducibles of V⊗nwhen viewed as Sn representation and its decomposition when viewed as GL(V ) representation. This, we

will see is indeed the case, and to construct these irreducible GL(V ) representations we will use the Young symmetrizer cλ. Recall that it was defined as cλ= aλ· bλ=Pσ∈Pλ,τ ∈Qλsgn(τ ) · eστ. We can

now define a new representation of GL(V ), which we will denote as SλV , by computing the image of

cλ on V⊗n. Thus:

SλV = Im(cλ|V⊗n).

This SλV is also a subrepresentation of V⊗n.

Definition 4.1. We call the functor V SλV , that sends a representation V to SλV the

Schur-functor. The representation SλV is called the Weyl-Module.

By a functor we mean that a linear map φ : V → W between two vectorspaces determines a map Sλ(φ) : SλV → SλW with Sλ(φ ◦ ψ) = Sλ(φ) ◦ Sλ(ψ) and Sλ(IdV) = IdV.

Example 4.1. In this example I will demonstrate how cλ acts on V⊗n, by decomposing V⊗2. We

start by recalling that the nth_{symmetric power and exterior powers denoted Sym}n_{V and}Vn_{V are} sub-representations of V⊗n. We can realize them as GL(V ) representations, by considering the partitions (n) and (1n_{). In those cases we have for the Young symmetrizer c}

λ that c(n)= a(n) and c(1n₎= b₍₁n₎.

Then, if v1⊗· · · ⊗ vn∈ V⊗n, cλ acts on this tensor by permuting the indices and we have for any n:

c(n)V⊗n= a(n)(v1⊗· · · ⊗ vn) = X σ eσ(v1⊗· · · ⊗ vn) = X σ (vσ(1)⊗· · · ⊗ vσ(n)) = SymnV and c(1n₎V⊗n= b₍₁n₎(v₁⊗· · · ⊗ v_n) = X σ sign(σ)eσ(v1⊗· · · ⊗ vn) = X σ sign(σ)(vσ(1)⊗· · · ⊗ vσ(n)) = ^n V

by definitions 2.8 and 2.9. Therefore these two partitions (n) and (1n) correspond respectively for any n to the functors

V Sym⊗nV and _V^⊗nV

9

In the case of a general group G though, the best we can then hope for is to break it up into some sub-representations.

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This also immediately gives us the decomposition for n = 2: VN V = Sym⊗2_V L V⊗2_{V . For n > 2} there will be a additional spaces in the decomposition that appear with a certain multiplicity m. For example, when n = 3, we have the additional symmetrizer

c(2,1)= 1 + e(12)− 1(13)− e(132).

Its image is on V⊗3 is the vectorspace spanned by the vectors

v1⊗ v2⊗ v3+ v2⊗ v1⊗ v3− v3⊗ v2⊗ v1− v3⊗ v1⊗ v2.

In the next subsection we will formulate a theorem that will enable us to determine the multiplicity m and we will see that this multiplicity is related to the Specht module corresponding to the same partition.

4.1 _{The irreducible characters of S}

λ

V .

We will now take a closer look at the representations SλV of GL(V ) we have constructed. As we

will see, they are indeed irreducible representations and their characters will be identified with certain symmetric polynomials, called Schur polynomials. The needed results about these Schur polynomials, some other symmetric polynomials and needed relations between them can be found in appendix A. Theorem 4.1. 10

1. Let mλ be the dimension of the irreducible representation Vλ of Sn corresponding to λ. Then

V⊗n∼=M

λ

(SλV )⊕mλ

2. Let k = dim V . For any semisimple g ∈ GL(V ), the trace of g on SλV is the value of the Schur

polynomial on the eigenvalues x1, . . . , xk of g on V , i.e.

χ_S_λV(g) = sλ(x1, . . . , xk)

3. Each SλV is an irreducible representation of GL(V ).

4. Let k = dim V . Then SλV is zero if λk+16= 0. If λ = (λ1≥ . . . ≥ λk), then

dim SλV = sλ(1, . . . , 1) =

Y

1≤i<j≤k

λi− λj+ j − i

j − i

Before turning to the proof of this theorem, we have a closer look at what (1) and (2) say. We already stated that we expected that the representations of GL(V ) and Sn will be connected in some way due

to their commuting actions on V⊗n. This is indeed reflected by (1). It says that as GL(V ) module V⊗n decomposes in irreducible sub-GL(V ) modules that occur with the multiplicity of the corresponding Specht module, the irreducible of Sn11.

10_{The theorem and proof of the theorem is from [1]} 11

In fact, this duality also holds the other way around, (a fact that will not be proven here). As Sn

representation V⊗n∼=L

λ(Vλ) ⊕n_λ

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Regarding (2) we can already say something about a special cases that is easy to see, namely for the case λ = (n). We let ρ(g) a semisimple endomorphism on V , then we know this leads to an endomorphism of SλV and we want to compute the trace of this endomorphism. For this we let

x1, x2, . . . , xk be the eigenvalues of ρ(g) on V , where k = dim (V ). Now in the case where ρ(g) is the

diagonal matrix, we have χV(g) = x1+ x2+ . . . + xk. Then in the case λ = (n), SλV = SymnV and

χSymn_V is the complete symmetric polynomial of degree n obtained by multiplying the k factors in all

possible orders, which is clearly symmetric due to commutativity. Thus we have the special case of: χSymn_V = h_(n)(x₁, . . . , x_k) (4.1)

We now turn to the proof. For this we translate the fact that the actions of GL(V ) and Sn on V⊗n

commute, to the language of algebras. This we will do by introducing the commutator algebra. We will formulate the results for the general case and then apply it to the situation we have in the theorem. We consider a finite group G, later to be taken Snand let U be a right module over an algebra A = C[G].

We define the commutator algebra B as:

B = HomG(U, U ) = {φ : U → U : φ(v · g) = φ(v)g, for all v ∈ U, g ∈ G} (4.2)

It is the algebra of of all the endomorphisms φ of U that commute with the action of G. B acts on U from the left, and this action commutes with the right action of A on U . If now U =L

iU ⊕ni

i is

decomposition into non-isomorphic irreducible right A-modules, then we can apply to find: B =M i HomG(Ui⊕ni, U ⊕ni i ) ∼= M i Matni(C)

which follows from Schur’s lemma (2.1) in the same way as before. If we now consider an additional left A module W , we can construct a left B module through the tensor product:

U ⊗AW = U ⊗CW/subspace generated by {va ⊗ w − v ⊗ aw}

This defines a left B module by acting on the first factor: b(v ⊗ w) = (bv) ⊗ w. Having defined these we can now formulate the first lemma:

Lemma 4.1. Let U a finite dimensional right A- module

1. For any c ∈ A, the canonical map U ⊗AAc → U c is an isomorphism of left B-modules.

2. If W = Ac is an irreducible left A-module, then U ⊗AW = U c is an irreducible left B-module

3. If Wi= Aci are all distinct irreducible left A-modules, with mi being the dimension of Wi, then

U ∼=M i (U ⊗AWi)⊕mi ∼= M i (U ci)⊕mi

is the decomposition of U into irreducible left B-modules.

A first observation to make is that Ac is a direct summand of A due to semi-simplicity. proof

Branching Rules and Little Higgs models