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Home , Group representation, Matrix representation, Unitary representation

REPRESENTATIONS OF FINITE GROUPS

SANG HOON KIM

Abstract. This paper provides the definition of a representation of a finite and ways to study it with several concepts and remarkable theorems such as an irreducible representation, the , and Maschke’s Theorem.

Contents 1. Introduction 1 2. Group representations 1 3. Important theorems regarding G-invariant Hermitian form and unitary representations 3 4. Irreducible representations and Maschke’s Theorem 7 5. Characters 8 6. representations 12 7. Regular representations 13 8. Schur’s Lemma and proof of Orthogonality relations 14 Acknowledgments 18 References 18

1. Introduction studies linear operators that are given by group elements acting on a . Thus, representation theory is very useful in that it makes it possible to delve into geometric symmetries effectively with using coordinates, which is not possible if abstract group elements are only considered. This paper assumes that readers prior knowledge of , including familiarity with Hermitian form and Spectral Theorem. Moreover, we deal only with finite groups.

2. Group representations We will start with a simple example of a representation of the group T of rotations of a tetrahedron. Suppose that the group T acts on a three-dimensional vector space V . Choose a (v1, v2, v3) so that each element of the basis passes through the midpoints A, B, C of three edges as follows. 1 2 SANG HOON KIM

v3

C D v B 2 A v1 Figure 1. Tetrahedron on a 3-dimensional space

Let x ∈ T be the counter clockwise around the vertex D by 2π/3 and y1, y2, y3 ∈ T be the counter clockwise rotations by π around the vertices A, B, C, respectively. The matrices that correspond to the group elements are

 1 1  −1 

(2.1) Mx = 1  ,My1 =  −1  ,My2 =  1  , 1 −1 −1 −1 

My3 =  −1  . 1

The group T is generated by {x, y1, y2, y3}. Therefore, the group of matrices isomorphic to the group T is generated by {Rx,Ry1 ,Ry2 ,Ry3 }. As in the example above, we define a representation of a group G as follows. Definition 2.2. A matrix representation of a group G is a

M : G → GLn(F) where GLn(F) is the general of degree n over a field F, the of n × n invertible matrices.

We will denote each image of g, an element of G, by Mg; the matrix repre- sentation M carries g to the Mg. Since M is a homomorphism, −1 −1 Mg−1 ,the image of g , is the inverse of Mg, which is thereby Mg . Also, the identity 1 ∈ G is carried to the n-dimensional identity matrix In = M1. Since a vector space does not have a unique basis, it is possible to deal with a representation without fixing a basis. Therefore, we introduce the definition of a group representation on a finite-dimensional vector space V . Definition 2.3. A representation of G on V is a homomorphism ρ : G → GL(V ) where GL(V ) is the group of invertible linear operators on V . Definition 2.4. The dimension of ρ is defined to be the dimension of the vector space V .

Choosing a basis, we can construct an Φ : GL(V ) → GLn(F) as follows: let ρg be the image of g via ρ. ρg defines a matrix Mg for a given basis B1 = (v1, v2, ..., vn). REPRESENTATIONS OF FINITE GROUPS 3

M : G / GLn(F) O isomorphic  ρ : G / GL(V )

If we treat the representation ρg with a different basis B2 = (w1, w2, . . . , wn), we get a different matrix with respect to the basis B2 via a change of basis matrix P in V as follows: 0 −1 (2.5) Mg = PMgP for every g ∈ G. A group representation can also be thought of as a on a vector space V . As long as a group element acts on a vector space, its action should be consistent with the vector space, which means it should act as a linear operator on V . Therefore, it should satisfy the two axioms of a group action, which are 1v = v and (gh)v = g(hv) for all g, h ∈ G and all v ∈ V, and the two conditions of linear operator, 0 0 g(v + v ) = gv + gv and g(cv) = cgv for c ∈ F. No matter what is given in the first place between a group representation and a group action, with the rule gv = ρg(v), we can define one by another. From now on, we will only deal with a that operates on a complex vector space.

3. Important theorems regarding G-invariant Hermitian form and unitary representations After defining a unitary representation, we will delve into several representations. Definition 3.1. A unitary representation is a homomorphism

M : G → Un from the group G to the unitary group Un. Let V be a Hermitian vector space. Then, a linear operator T is unitary if hv, wi = hT (v),T (w)i.

In the same way, we can say a representation ρ : G → GL(V ) is unitary if ρg is a for all g ∈ G, i.e. if

(3.2) hv, wi = hρgv, ρgwi, which is also (3.3) hv, wi = hgv, gwi for all v, w ∈ V . Given an orthonormal basis, a matrix representation, which is isomorphic to the unitary representation ρ, is unitary. Definition 3.4. Given a representation ρ of G on a vector space V , a form h·, ·i on V is called G-invariant if (3.2), or (3.3), is satisfied. Theorem 3.5. Let ρ be a representation of a finite group G on a complex vector space V . There exists a G-invariant, positive definite Hermitian form h·, ·i on V . 4 SANG HOON KIM

Proof. Choose an arbitrary positive definite Hermitian form on V , which is denoted by {·, ·}. We can then define a new form as follows:

1 X (3.6) hv, wi = {gv, gw} N g∈G where N is the of G. We will show by the next lemma that the new Hermitian form (3.6) is G-invariant and positive definite. 

Lemma 3.7. The form (3.6) is a G-invariant, positive definite Hermitian form.

Proof. For the form to be Hermitian, it should be linear in the second variable, conjugate-linear in the first variable, and Hermitian symmetric. (1) (Linearity in the second variable) Since the form {·, ·} is Hermitian and g acts as a linear operator on V ,

{gv, g(w + w0)} = {gv, gw + gw0} = {gv, gw} + {gv, gw0}.

Thus,

1 X hv, w + w0i = {gv, g(w + w0)} N g∈G 1 X 1 X = {gv, gw} + {gv, gw0} N N g∈G g∈G = hv, wi + hv, w0i.

Also,

{gv, g(cv0)} = {gv, c(gv0)} = c{gv, gv0}.

Thus,

1 X hv, cv0i = {gv, g(cv0)} N g∈G 1 X = c {gv, gv0} N g∈G = chv, v0i.

(2) (Conjugate linearity in the first variable) Similarly, we have that

{g(v + v0), gw} = {gv + gv0, gw} = {gv, gw} + {gv0, gw}. REPRESENTATIONS OF FINITE GROUPS 5

Thus,

1 X hv + v0, wi = {g(v + v0), gw} N g∈G 1 X 1 X = {gv, gw} + {gv0, gw} N N g∈G g∈G = hv, wi + hv0, wi.

Also,

{g(cv), gv0} = {c(gv), gv0} =c ¯{gv, gv0}.

Therefore,

1 X hcv, v0i = {c(gv), gv0} N g∈G 1 X =c ¯ {gv, gv0} N g∈G =c ¯hv, v0i.

(3) (Hermitian symmetry) Note that {gv, gv0} = {gv0, gv}.

1 X hv, v0i = {gv, gv0} N g∈G 1 X = {gv0, gv} N g∈G = hv0, vi.

1 Since h·, ·i is the sum of positive definite forms {·, ·} divided by the scalar N , h·, ·i is positive definite. To prove that the form h·, ·i is G-invariant, we should show that hv, wi = hg0v, g0wi for all v, w ∈ V when g0 is an element of G. Indeed,

1 X 1 X hg v, g wi = {g(g v), g(g w)} = {gg v, gg w} 0 0 N 0 0 N 0 0 g∈G g∈G

Since right multiplication by g0 is a on G, gg0 runs over every element in G. Therefore, we can write the equation above as follows:

1 X 1 X hg v, g wi = {gg v, gg w} = {g0v, g0w}. 0 0 N 0 0 N g∈G g0∈G 6 SANG HOON KIM

The only difference between the equation we get and the original equation of the form is just the change in the order of the sum in the original equation. 1 X hg v, g wi = {g0v, g0w} 0 0 N g0∈G 1 X = {gv, gw} N g∈G = hv, wi. The form h·, ·i is a G-invariant, positive definite Hermitian form.  Now we prove the next theorem, which leads to remarkable corollaries.

Theorem 3.8. Every matrix representation M : G → GLn of a finite group G is conjugate to a unitary representation. In other words, given M, there is a matrix −1 P ∈ GLn such that PMgP ∈ Un for every g ∈ G. Proof. Let V = Cn and a basis be the standard basis. Any homomorphism M : G → GLn is the matrix representation associated to a representation ρ. By Theorem 3.5, there exists a G-invariant form h·, ·i on V ; we choose an orthonormal basis for V with respect to the form. The matrix representation of ρ with respect to the orthonormal basis is conjugate to M by (2.5) and unitary. Therefore, the matrix representation M is conjugate to the unitary representation. 

Corollary 3.9. Every finite subgroup of GLn is conjugate to a subgroup of Un.

Proof. A matrix representation sends a finite group G to a finite subgroup of GLn. We think of the inclusion of the finite subgroup of GLn into GLn as a matrix representation of the group G. Then, every finite subgroup of GLn is conjugate to a subgroup of Un  Corollary 3.9 is so important that it will be used to prove wonderful theorems about group representations later.

r Corollary 3.10. Let A be an invertible matrix of finite order in GLn, i.e. A = I −1 for some r. Then, A is diagonalizable; there is a P ∈ GLn so that P AP is diagonal.

Proof. Matrix A generates a finite of GLn whose order is r. By Corol- lary 3.9, this finite group is conjugate to a subgroup of Un, which means A is conjugate to a . Since the Spectral Theorem for normal operators shows that every unitary matrix is diagonalizable, A is diagonalizable. 

Corollary 3.11. Let M : G → GLn be a representation of a finite group G. Then, for every g ∈ G, Mg is a diagonalizable matrix. Proof. Every element g of a finite group has finite order. Because the matrix representation M is a homomorphism, Mg also has finite order. The rest of the proof is same with the proof of Corollary 3.10.  By Corollary 3.11, given a representation of a finite group G on a vector space V , there is a basis of V for each element g in G so that the matrix of the operator ρg is diagonal. However, it is worth noting that the basis need not be same for every ρg to be diagonal. REPRESENTATIONS OF FINITE GROUPS 7

4. Irreducible representations and Maschke’s Theorem Definition 4.1. Let ρ be a representation of a group G on a vector space V .A subspace of V is called G-invariant if (4.2) gw ∈ W, for all w ∈ W and g ∈ G. If W is G-invariant, the operation by every group element g carries W to itself, i.e. gW ⊂ W . Moreover, the operation of G on V will restrict to its operation on W . Definition 4.3. If a representation ρ of a group G on a nonzero vector space V has no proper G-, it is called an irreducible representation. If there is a proper invariant subspace, ρ is called a reducible representation.

Definition 4.4. When V is the of G-invariant subspaces W1,W2,...,Wr, a representation ρ on V is called the direct sum of its restrictions ρi to Wi. Namely, r M (4.5) ρ = ρi = ρ1 ⊕ ρ2 ⊕ · · · ⊕ ρr. i=1

Given a representation ρ, suppose ρ = ρ1 ⊕ ρ2. Choose bases B1, B2 of G- invariant subspaces W1,W2. We choose a basis B = (B1, B2) by listing these two bases in order. The matrix Mg of ρg has a block form   Ag 0 (4.6) Mg = 0 Bg where Ag is the matrix of ρ1g with respect to B1 and Bg is the matrix of ρ2g with respect to B2. Conversely, if the matrix Mg has such a block form, the representation is the direct sum. 3 Take for an example the Dn operating on R by symmetries of an n-gon. Choose an orthonormal basis B so that v1 is perpendicular to the plane of n-gon and v2 passes through a vertex.

v3

B

A v2

v1 C

Figure 2. 3-gon on the v2v3-plane

Let xn = 1, y2 = 1 be the standard generators of the dihedral group. Then, the rotations corresponding to x, y are represented by the matrices 1  1  (4.7) Mx =  cn −sn ,My =  1  sn cn −1 8 SANG HOON KIM where cn = cos(2π/n) and sn = sin(2π/n). The matrix representation M is the direct sum of an one-dimensional represen- tation A,     (4.8) Ax = 1 ,Ay = −1 , and a two-dimensional representation B,     cn −sn 1 (4.9) Bx = ,By = . sn cn −1 2 The representation B is a two-dimensional representation of Dn operating on R as symmetries of n-gon on the plane. However, a matrix of a reducible representation might not have a block form un- less a given basis for V is consistent with the direct sum decomposition. Therefore, it is of significance to work with an appropriate basis to have a block form. Theorem 4.10. Let ρ be a unitary representation of G on a Hermitian vector space V so that the given form is G-invariant and a given basis is orthonormal, and let W be a G-invariant subspace. The orthogonal complement W ⊥ is also G-invariant, and ρ is the direct sum of its restrictions to W and W ⊥.

⊥ Proof. Suppose v ∈ W , i.e. v ⊥ W . The linear operators ρg for every g in G are unitary: for w ∈ W , we have

v ⊥ w ⇐⇒ hv, wi = 0 ⇐⇒ hρgv, ρgwi = hv, wi = 0. Therefore, gv ⊥ gW . Because W is G-invariant, W = gW , which means gv ⊥ W . Thus, gv ∈ W ⊥, and W ⊥ is G-invariant. Since V = W ⊕ W ⊥, by Definition 4.4, ρ ⊥ is the direct sum of its restrictions to W and W .  Theorem 4.10 tells us that if there is a proper G-invariant subspace W , a unitary representation ρ can be decomposed as the direct sum of its restrictions. Moreover, if either W or W ⊥ has a proper G-invariant subspace, ρ can be decomposed further, which is a process of using induction. In this way, we get the following corollary. Corollary 4.11. Every unitary representation ρ : G → GL(V ) on a Hermitian vector space V is the direct sum of irreducible representations. By Theorem 3.8 and Corollary 3.9, we found the important fact that every representation of a finite group G is conjugate to a unitary representation, which, by Corollary 4.11, is the direct sum of irreducible representations. This fact paves the way to one of the most significant and beautiful theorems in representations of finite groups. Theorem 4.12 (Maschke’s Theorem). Every finite-dimensional representation of a finite group G is the direct sum of irreducible representations.

5. Characters To study different properties of irreducible representations, it is crucial to work with the of a representation, which is called the character χ. But before we delve into the character, we first define isomorphic representations because as we will see later, representations in same isomorphic class share many properties together. REPRESENTATIONS OF FINITE GROUPS 9

Definition 5.1. Two representations ρ : G → GL(V ) and ρ0 : G → GL(V 0) of group G are ismorphic, or equivalent, if there exists an isomorphism of vector spaces T : V → V 0 which is consistent with the operation of G: 0 0 (5.2) gT (v) = T (gv) or ρgT (v) = T (ρg(v)) for all v ∈ V and g ∈ G.

Definition 5.3. The character χ of a representation ρ is a χ : G → C defined by

(5.4) χ(g) = trace(ρg). If M is a matrix representation isomorphic to the representation ρ with respect to a choice of basis for V ,

(5.5) χ(g) = trace(Mg) = λ1 + λ2 + ··· + λn where λi for 1 ≤ i ≤ n are eigenvalues of Mg. They are also eigenvalues of ρg since eigenvalues do not depend on a basis. The dimension of the character χ is defined to be the dimension of a represen- tation ρ. The character of an irreducible representation is called the irreducible character.

Now we will look at basic properties of the character. Proposition 5.6. Let χ be the character of a representation ρ of a finite group G on a vector space V . (a) χ(1) = dim V ,which is the the dimension of ρ. (b) χ(g) = χ(hgh−1) for all g, h ∈ G, i.e. the character is constant on each conjugacy class. (c) χ(g−1) = χ(g) for all g ∈ G. (d) If χ0 is the character of another representation ρ0, then the character of the direct sum ρ ⊕ ρ0 is χ + χ0. Proof.

(a) Since 1 is the identity element of group G, ρ1 is the n-dimensional identity In. Thus, χ(1) = trace In = dim V. (b) Since the matrix representation M isomorphic to ρ is a homomorphism,

Mhgh−1 = MhMgMh−1 .

We can think of Mh as a change of basis matrix, and since the trace does not depend on the change of basis,

trace (Mhgh−1 ) = trace Mg which means that χ(g) = χ(hgh−1). (c) Suppose that the eigenvalues of Mg are λ1, λ2, . . . , λn. From the equation −1 −1 Mgv = λv where v is an eigenvector, we can get Mg v = λ v so that the −1 −1 −1 r eigenvalues of Mg−1 are λ1 , λ2 , ··· , λn . G is finite. Thus, Mg = I for some r. By Corollary 3.11, Mg is a diagonalizable matrix, i.e. Mg can be −1 written down as PRgP where the matrix P consists of the eigenvectors of Mg and Rg is a diagonal matrix with the eigenvalues of Mg. Then, 10 SANG HOON KIM

r r −1 r Mg = PRgP = I, which means Rg = I. This implies that λ1, λ2, . . . , λn are roots of unity. Therefore, 2 −1 |λi| = 1 ⇐⇒ λiλi = |λi| = 1 ⇐⇒ λi = λi . (d) The representation ρ ⊕ ρ0 can be written as a block matrix with an appro- priate choice of basis; the trace of the block matrix is the sum of the traces 0 of Ag and Bg, which are the matrices forms of ρ and ρ .  From Proposition 5.6, we can discover two important facts that make compu- tation of the character significantly easier than simply calculating every character. The first one is that the character χ depends only on the conjugacy class of an ele- ment g ∈ G; we only need to choose one representative element for each conjugacy class to get the character of other elements in the same conjugacy class. Secondly, because the trace of a linear operator does not depend on a choice of basis, we can freely choose any basis that is convenient for us to calculate; we do not need to use a same basis for every element. Take (2.1) for an example of the character of the matrix representations of the rotations of the tetrahedral group T . There exist four conjugacy classes in T which are represented by 1, x, x2, y where x is the counter clockwise rotation around the front vertex and y is the counter clockwise rotation around the midpoint of an edge. The characters of T are (5.7) χ(1) = 3, χ(x) = 0, χ(x2) = 0, χ(y) = −1. Another way to think of the character χ is to treat it as a vector. Suppose g1, , ··· , gr are the representative elements for r conjugacy classes of a finite group G. Since value of the character is same for each conjugacy class, the character χ of a representation of G can be written as a vector as follows: T (5.8) χ = (χ(g1), χ(g2), ··· , χ(gr)) . The order of the conjugacy classes can be arbitrary but has to be consistent for all characters written as vectors. We write the character of T in (5.7) as a vector. Let the order of the conjugacy classes be the classes of 1, x, x2, y. The vector of the character is (5.9) χ = (3, 0, 0, −1)T .

Definition 5.10. A is a complex-valued function φ : G → C which is constant on each conjugacy class. The reason why the concept of class function is introduced is that it is deeply related to the character as the character is also a class function; the character is a complex valued function and constant on each conjugacy class. Given a class function, we can think of each conjugacy class having a different space. Thus, if there are r conjugacy classes, there are r distinct spaces. The spaces of the class function form a r-dimensional vector space VG. Since the class function can be written as a vector, the vector of the class function belongs to the vector space. Thus, for each representation ρ, the character χρ ∈ VG. We define a Hermitian form on VG 1 X (5.11) hχ, χ0i = χ(g)χ0(g) N g∈G REPRESENTATIONS OF FINITE GROUPS 11 where N = |G|. VG is now a Hermitian space. We will study important theorems that follow as we write the character in a vector form.

Theorem 5.12. Let G be a group of order N, let ρ1, ρ2,... represent the distinct isomorphism classes of irreducible representations of G, and let χi be the character of ρi.

(a) Orthogonality relations: The characters χi’s are orthonormal; hχi, χji = 0 if i 6= j, and hχi, χii = 1 for each i. (b) There are finitely many isomorphism classes of irreducible representations, the same number as the number of conjugacy classes in the group. (c) Let di be the dimension of the irreducible representation ρi, and let r be the number of irreducible representations. Then, di divides N, and 2 2 2 (5.13) N = d1 + d2 + ··· + dr. The complete proof of Theorem 5.12 will be given in the section 8 because it requires more than what we have studied so far.

Corollary 5.14. The irreducible characters of G form an orthonormal basis of VG. Proof. By (a) and (b) of Theorem 5.12, the characters are linearly independent as they are orthogonal to each other, and they span VG because the dimension of VG, which is the number of conjugacy classes, is the same as the number of irreducible characters. 

Now we have VG that contains the characters; the irreducible characters form a basis of VG. Therefore, by using orthogonal projection, we can decompose any character as a linear combination of the irreducible characters. By Theorem 4.12, given a representation ρ and the irreducible representations ρ1, ρ2, ··· , ρr, ρ = n1ρ1 ⊕ n2ρ2 ⊕ · · · ⊕ nrρr where ni are nonnegative and niρi is the direct sum of n copies of the representations ρi’s; thus, by Proposition 5.6 (d), the next corollary follows.

Corollary 5.15. Let χ1, . . . , χr be the irreducible characters of a finite group G, and let χ be any character. Then, χ = n1χ1 + ··· + nrχr where ni = hχ, χii. Corollary 5.16. If two representations ρ, ρ0 have the same character, they are isomorphic. 0 0 0 0 Proof. Suppose that ρ = n1ρ1 ⊕n2ρ2 ⊕· · ·⊕nrρr and ρ = n1ρ1 ⊕n2ρ2 ⊕· · ·⊕nrρr. 0 Then, the character of ρ is χ = n1χ1 + ··· + nrχr and the character of ρ is 0 0 0 χ = n1χ1 + ··· + nrχr. Since the irreducible characters χ1, . . . , χr are linearly 0 0 independent, ni = ni for each i as χ = χ .  Corollary 5.17. A character χ has the property hχ, χi = 1 if and only if it is irreducible. 2 2 Proof. hχ, χi = hn1χ1 + ··· + nrχr, n1χ1 + ··· + nrχri = n1 + ... + nr. If χ is the irreducible character, χ = χi for some i; thus, for hχ, χi, only one ni is 1 and the rest are zero.  By using Corollary 5.17, we can check whether the character (5.9) is irreducible. 1 · 32 + 4 · 02 + 4 · 02 + 3 · (−1)2 hχ, χi = = 1. 12 12 SANG HOON KIM

Thus, the character (5.9) is irreducible. Now we will look at the characters of the dihedral group D3. D3 has three conju- gacy classes, {1}, {y, xy, x2y}, {x, x2}, which are represented by 1, y, x, respectively. Along with the fact that it has three irreducible representations by Theorem 5.12 (b), the only possible solution to (5.13) in this case is 6 = 12 + 12 + 22. Thus, there exist two irreducible one-dimensional representations ρ1, ρ2 and one irreducible two- dimensional representation ρ3 of the dihedral group D3. Every finite group G has the trivial one-dimensional representation (Mg = 1 for every g ∈ G) as every group multiplication gh = g0 for g, h, g0 ∈ G can be represented as 1 · 1 = 1. So does the group D3. We will call the ρ1. The another one-dimensional representation is the sign representation, which is (4.8). We will call the sign repre- sentation ρ2. We define the two-dimensional representation, which will be called ρ3, by (4.9). We use the where the values of the characters are shown according to the representative elements of conjugacy classes. The numbers (1), (3), (2) above the representative elements are the orders of the conjugacy classes.

(1) (3) (2) 1 y x χ1 1 1 1 χ2 1 -1 1 χ3 2 0 -1

Table 1. Character table of D3

Since the top row is about the trivial representations, every element in the top row is 1. The first column is about χi(1), which is the dimension of the represen- tation ρi. Since 1 hχ , χ i = (1 · (χ (1)χ (1)) + 3 · (χ (y)χ (y)) + 2 · (χ (x)χ (x))) = 1, 3 3 6 3 3 3 3 3 3 ρ3 is irreducible.

6. Permutation representations

Let S be a set {s1, s2, . . . , sn}. Take (s1, s2, . . . , sn) for a basis for a vector space V (S). As a group G operates on S, X X (6.1) v = aisi where ai ∈ C =⇒ gv = aigsi. i i A representation ρ of the group G is equivalent to the group operation on the vector space V (S). A group operation that permutes vectors in a basis is called a permutation representation. For example, let G be the tetrahedral group T and S be a set of the faces of a tetrahedron, S = {A, B, C, D}. We define a four-dimensional permutation representation by an operation of G on S. As before, let x be the rotation by 2π/3 about a face A and y be the rotation by π about an edge. We have 1 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 (6.2) Mx =   ,My1 =   . 0 1 0 0 0 0 0 1 0 0 1 0 0 0 1 0 REPRESENTATIONS OF FINITE GROUPS 13

The interesting fact about a permutation representation is that unless S has only one element, it is always reducible because one-dimensional vector space spanned by w = s1 + ··· + sn is G-invariant. No matter how many times we permute, w stays unchanged. Since the vector space V (S) spanned by (s1, s2, . . . , sn) has a G-invariant proper subspace W , a permutation representation ρ of G is reducible. Therefore, if we find an orthonormal basis that includes w = s1 +···+sn, we will be able to write the permutation matrix ρ as a block matrix form. This block matrix has a 1 × 1 matrix as an entry having 1 because ρ has the one-dimensional trivial representation due to W . As (6.2) shows, every entry 1 of vectors that are not permuted stands still on the diagonal of a permutation matrix, but the others do not. Therefore, the character of a permutation representation is number of elements of S fixed, or not permuted, by g. (6.3) χ(g) = number of elements of S fixed by g.

7. Regular representations Definition 7.1. The ρreg of G is the representation asso- ciated to the operation of G on itself by left multiplication. We can also define the regular representation by right multiplication, but we will only consider left multiplication. The character χreg of the regular representation is as follows: (7.2) χreg(1) = N and χreg(g) = 0 if g 6= 1 where N = |G|. χ(1) = dim ρ for any representation ρ, and dimension of ρreg is N; thus, χreg(1) = N. The operation of G on itself as left multiplication is a permutation, which means no element is fixed as it is multiplied by an element other than 1. Along with (6.3), χreg(g) = 0 if g 6= 1. if we know how to compute hχreg, χi for the character χ of any representation ρ, we can express χreg as a linear combination of the irreducible characters by using Corollary 5.15. Lemma 7.3. hχreg, χi = dim ρ. Proof. 1 X hχreg, χi = χreg(g)χ(g) N g 1 = χreg(1)χ(1) N 1 = (dim ρreg dim ρ) dim ρreg = dim ρ.  reg reg Corollary 7.4. χ = d1χ1 + ··· + drχr and ρ ≈ d1ρ1 ⊕ · · · ⊕ drρr where di is the dimension of ρi. reg It is worth noting that χ (1) = d1χ1(1) + d2χ2(1) + ··· + drχr(1) is equivalent 2 2 2 with N = d1 + d2 ··· + dr, which is (5.13). Therefore, we can see that (5.13) is also based on Orthogonality relations. 14 SANG HOON KIM

Let’s look at how this beautiful formula in Corollary 7.4 is used. Using (7.2), we get the character table of the regular representation of D3 as follows.

1 x y χreg 6 0 0 Table 2. Character table of the regular representation of D3

reg With Table 1 and Table 2, we see that χ = χ1 +χ2 +2χ3, the equation derived from Corollary 7.4, is correct.

8. Schur’s Lemma and proof of Orthogonality relations We have studied how any representation can be decomposed as the direct sum of irreducible representations and many applications of it with the concept of the character. In that process, Theorem 5.12 including Orthogonality relations has been of most importance, but we have yet to give a sufficient proof of it. In this section, Schur’s Lemma will be introduced to prove Theorem 5.12. Definition 8.1. Let ρ, ρ0 be representations of a group G on two vector spaces V and V 0. A linear transformation T : V → V 0 is G-invariant if 0 −1 (8.2) gT (v) = T (gv) or ρg(T (v)) = T (ρg(v)) ⇐⇒ T (v) = g (T (gv)) for all g ∈ G and v ∈ V . The isomorphism T in Definition 5.1 is a bijective G-invariant transformation, and the linear transformation T in Definition 8.1 need not be an isomorphism. 0 0 0 Given bases B and B for V and V , respectively, we write Mg,Mg,L as the 0 matrices of ρg, ρg, and T with respect to the bases. Then, (8.2) is also described as 0 (8.3) MgL(v) = L(Mg(v)) for all g ∈ G. 0 If ρ = ρ , T and ρg commute for every g ∈ G as follows.

(8.4) ρg(T (v)) = T (ρg(v)). Proposition 8.5. The kernel and image of G-invariant linear transformation T : V → V 0 are G-invariant subspaces of V and V 0, respectively. Proof. To prove that ker T is G-invariant, we need to show gv ∈ ker T if v ∈ ker T or T (gv) = 0 if T (v) = 0. Since T is a G-invariant transformation, T (gv) = gT (v) = g0 = 0. Therefore, T (gv) = 0 as T (v) = 0. To prove that im T is G-invariant, we need to show gT (v) ∈ im T as T (v) ∈ im T . Since T is G-invariant transformation, gT (v) = T (gv). Since T (gv) ∈ im T by definition, gT (v) ∈ im T . Thus, im T is G-invariant.  Theorem 8.6 (Schur’s Lemma). Let ρ, ρ0 be two irreducible representations of G on vector spaces V,V 0, and let T : V → V 0 be a G-invariant transformation. (a) Either T is an isomorphism or else T = 0. (b) If V = V 0 and ρ = ρ0, then T is multiplication by a scalar. REPRESENTATIONS OF FINITE GROUPS 15

Proof.

(a) By definition, ρ has no proper G-invariant subspace of V . Since ker T is a G-invariant subspace by Proposition 8.5, ker T is an improper G-invariant subspace of V . Thus, either ker T = V or else ker T = 0. If ker T = V , then T v = 0 for every v ∈ V . Thus, T = 0. If ker T = 0, then T is injective, which means there exists an isomorphism between its domain and image. Thus, im T is not 0. Since ρ0 has no proper G-invariant subspace of V 0 and im T is a G-invariant subspace, im T = V 0. T is surjective and therefore an isomorphism. (b) Let λ be an eigenvalue of T . That T1 = (T − λI) is G-invariant is shown as follows: (T − λI)(ρgv) = T (ρgv) − λI(ρgv). Since ρ = ρ0 and T is a G-invariant transformation, the equation above becomes

ρgT (v) − ρg(λIv) = ρg(T (v) − λIv) =⇒ ρg(T − λI)(v).

Therefore, T1 is G-invariant. Because ρ is irreducible, either ker T = V or else ker T = 0. But given an eigenvector v of λ,(T − λI)(v) = 0; ker T1 is not 0 but contains the eigenvector v. Therefore, ker T = V . Then, T1 = 0, and T = λI.  The important point is that the lemma shows that if T is not an isomorphism so that the two irreducible representations ρ, ρ0 are not isomorphic, the G-invariant transformation T is 0. This fact will be used later to prove Orthogonality relations. As we can make a G-invariant, positive definite Hermitian form by choosing any positive definite Hermitian form by Theorem 3.5, we can construct a G-invariant linear transformation by using any linear transformation T : V → V 0 as follows. Lemma 8.7. Given an arbitrary linear transformation T , there exists a G-invariant linear transformation Te : V → V 0 which is 1 X (8.8) Te(v) = g−1(T (gv)) where N = |G|. N g We rewrite (8.8) in matrix form

1 X 0−1 (8.9) Le = M LMg. N g g

Proof. Because Te is a composition and sum of linear transformations, it is also a linear transformation. Given elements h, g ∈ G and g0 = gh,

1 X 1 X 0 (8.10) h−1Te(hv) = h−1g−1(T (ghv)) = g −1(T (g0v)) = Te(v). N N g g0

Thus, by (8.2), Te is a G-invariant linear transformation.  Proposition 8.11. Let ρ be a representation of a finite group G on a vector space V , and let T : V → V be a linear operator. Define Te by (8.8). Then, trace Te = trace T . 16 SANG HOON KIM

Proof. (8.9) becomes

1 X −1 (8.12) Le = M LMg. N g g Since the value of the trace of L does not depend on a choice of basis, i.e. trace L −1 = trace Mg LMg, 1 (8.13) trace Le = (Ntrace L) = trace L. N  Suppose that χ, χ0 are two irreducible representations of ρ, ρ0 of G which are not isomorphic. To prove Orthogonality relations, we have to show that hχ, χ0i = 0, and hχ, χi = 1. If hχ, χ0i = 0, since χ0(g−1) = χ0(g), 1 X (8.14) hχ, χ0i = χ(g−1)χ0(g) = 0. N g∈G Thus, it is enough to prove (8.14) to show hχ, χ0i = 0. The formula (8.14) can be written as

0 1 X X 0 (8.15) hχ, χ i = (M −1 ) (M ) = 0. N g ii g jj g∈G i,j Thus, if X 0 (8.16) (Mg−1 )ii(Mg)jj = 0, g∈G then hχ, χ0i = 0. To prove (8.16), we need Schur’s Lemma. According to Schur’s Lemma, for two nonisomorphic irreducible representations, every G-invariant linear transformation is 0, i.e. Te is 0. Then, (8.9) becomes X 0 (8.17) Mg−1 LMg = 0 g 0−1 for every matrix L that is compatible with right of Mg and left matrix multiplication Mg.

Lemma 8.18. Let M,N be matrices and let P = MeαβN where eαβ is a n × n matrix that has 0 in every entry except one entry having 1 on the ith row and jth column. Then, the entries of P are (P )ij = (M)iα(N)βj.

Proof. Let M be a a×α matrix and N be a β ×b matrix. Let K = Meαβ.(K)ij = Pα k=1(M)ik(eαβ)kj =⇒ Kiβ = (M)iα(eαβ)αβ for 1 ≤ i ≤ a. Then, (P )ij = Pβ d=1(K)id(N)dj = (K)iβ(N)βj = (M)iα(eαβ)αβ(N)βj = (M)iα(N)βj. 

We replace L in (8.17) by eij and get X 0 X 0 0 = (0)ij = (Mg−1 eijMg)ij = (Mg−1 )ii(Mg)jj g g which proves (8.16) and thereby hχ, χ0i = 0 for two nonisomorphic irreducible characters χ, χ0. REPRESENTATIONS OF FINITE GROUPS 17

We will prove now hχ, χi = 1. According to Schur’s Lemma, if V = V 0 and ρ = ρ0, then T is multiplication by a scalar. Therefore, (8.12) becomes 1 X (8.19) Le = Mg−1 LMg = kI for some scalar k. N g Since trace I = dim ρ, trace Le = k dim ρ. Proposition 8.11 shows that trace Le = trace L. Thus, k = trace L/dim ρ. By using Lemma 8.18 again, (8.19) becomes 1 X 1 X (8.20) (kI) = (M −1 e M ) = (M −1 ) (M ) ij N g ij g ij N g ii g jj g g where k = trace eij/dim ρ = 1/dim ρ.(kI)ij = 0 if i 6= j, and (kI)ij = 1/dim ρ if i = j. Then, 1 X X hχ, χi = (M −1 ) (M ) N g ii g ii g∈G i X  1 X  = (M −1 ) (M ) N g ii g ii i g∈G X 1 = = 1. dim ρ i Therefore, the irreducible characters are orthonormal. The only theorem that is still left unproven is Theorem 5.12 (b). It is equivalent with the statement that VG is spanned by the irreducible characters; by Orthog- onality relations, the irreducible characters are orthonormal, and dim VG is the number of conjugacy classes. Since the irreducible characters belong to VG, they ⊥ span a subspace of VG. Let the subspace be W . Then, VG = W ⊕ W . We should prove W ⊥ = 0 or that a class function φ that is orthogonal to every character is 0 so that a subspace spanned by the class function φ, which is orthogonal to W , is 0. Given such a class function and the character χ of a representation ρ, consider the linear operator T : V → V defined as 1 X (8.21) T = φ(g)ρ(g). N g Then, 1 X (8.22) trace T = φ(g)χ(g). N g The equation (8.22) is same with (5.11) and thereby can be written as follows. (8.23) trace T = hφ, χi = 0 because we assumed φ is orthogonal to every character. Lemma 8.24. T in (8.21) is G-invariant. Proof. Let g0 = h−1gh, a conjugate of g by h, for g, g0 ∈ G. Since the representation −1 0 ρ is a homomorphism, ρg0 = ρh ρgρh. Since φ is a class function, φ(g ) = φ(g).

−1 1 X −1 1 X 0 ρ T ρ = φ(g)ρ ρ ρ = φ(g )ρ 0 = T. h h N h g h N g g g0 Therefore, T is G-invariant.  18 SANG HOON KIM

By Schur’s Lemma, if ρ is irreducible, T = kI for some scalar k. Moreover, because trace T = 0 by (8.23), T = 0. By Maschke’s Theorem, every finite- dimensional representation is the direct sum of irreducible representations. Because the linear operator T is defined by using the finite-dimensional representation ρ, T is compatible with direct sum as well. Therefore, no matter what case we are dealing with, T = 0 because every summand of its direct sum is 0. Consider that ρ = ρreg is the regular representation; not only is an underlying set of a finite group G a basis for a vector space V (G), but the group G operates on itself, which is also its underlying set. Consider T (1) where 1 is a vector in the basis as the identity element of G. We already know T (1) = 0 as T = 0. Since ρreg is the regular representation, ρg(1) = g. 1 X 1 X (8.25) T (1) = 0 = φ(g)ρ (1) = φ(g)g. N g N g g The equation (8.25) is a linear combination. Because every element of G as a vector forms the basis, φ(g) = 0 for every g ∈ G. Therefore, φ that is orthogonal to every character is 0, and thereby we prove that the number of isomorphism classes of irreducible representations is same with the number of conjugacy classes in a finite group G.

Acknowledgments It is a pleasure to thank my mentor, Adan Medrano Martin del Campo, for all his great help in everything including writing this paper. He has always given me profound advice, and nothing could be possible without the help of my amazing mentor. I am particularly grateful for a great help provided by my older sister, April Kim, who proofread this paper and always encourages me throughout my study and whole life. I would also like to express my very great appreciation to Professor Jon Peter May for providing such a wonderful opportunity to participate in the UChicago Mathematics Research Experience for Undergraduates program (REU).

References [1] Michael Artin. Algebra. Prentice Hall. 1991. [2] Victor E. Hill. Groups, Representations, and Characters. Hafner Press. 1975.