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Home , Character (mathematics), Character theory

A NOTE ON THEORY

FAN HUANG

In this note, we explore two examples. We try to glimpse both the analogy between of finite groups and of the infinite compact , and the power of character theory for nonabelian groups.

1. Basic Facts of Character Theory of Finite Groups Let G be a finite group and ρ : G → GL(V ) be a linear representation of G in the complex finite dimensional V . For each g ∈ G, put χρ(g) = T r(ρ(g)). The complex valued χρ on G is called the character of the representation ρ and it is a – those are any functions from G into F which are constant on the conjugacy classes of G, i.e., f : G → F such that f(g−1xg) = f(x) for all g, x ∈ G. It characterizes the representation ρ in the sense that representations are determined by their characters, up to isomorphisms. Let ρ : G → GL(V ) τ : G → GL(W ) be two representations of a group G on vector spaces V and W . An isomorphism from V and W is a G-linear map f : V → W such that τ(g) ◦ f(v) = f ◦ ρ(g)(v), ∀g ∈ G, ∀v ∈ V i.e., for all g ∈ G, the below square commutes. ρ(g) V V

f f

W W τ(g)

In that case, we say that two representations ρ and τ are isomorphic. In terms of matrices, two matrix representations are isomorphic if and only if they describe the same representation in different bases. Specifically, choose a basis A = {a1, . . . , an} for V and a basis B = {b1, . . . , bn} for W , and let ρA : G → GLn(C) τB : G → GLn(C) be the two matrix representations obtained by writing ρ and τ in these bases. Then ρ and τ are isomorphic if and only if ρA and τB are isomorphic. Proof. It can be illustrated via the commutative graphs:

Date: August 2013. 1 A NOTE ON CHARACTER THEORY 2

τB(g) Cn Cn

B B

τ(g) W W

S f f S−1

ρ(g) V V

A A

Cn Cn ρA(g)

 There is an inner product on the vector space of class functions given by 1 X hf, hi = f(g)h(g), |G| g∈G where |G| is the of the group G. The characters of irreducible representations form an orthonormal basis for the space of class functions on G, denoted as Cl(G). Every represen- tation is a of irreducible representations. Let ρ be any representation of G and ρi be an isomorphism class of irreducible representations of G. Let χ and χi be the characters of ρ and ρi respectively. Then the multiplicity of ρi in ρ is the inner product hχi, χi.

2. The representation theory of the S1 (the simplest infinite ) Consider the unit circle S1 = {z ∈ C× | |z|= 1} = {(x, y) ∈ R2 | x2 + y2 = 1} in the complex plane. If we view the circle as the set of points {eiθ : θ ∈ R}, the natural group operation is multiplication of complex numbers. We treat S1 as an abelian under multiplication with its topology as a subset of C. (A topological group is a group G equipped with a topology such that the product map G×G → G and the inverse map G → G are both continuous. More generally, a representation of a topological group on a vector space V is a continuous ρ : G → GL(V ) with the topology of GL(V ) inherited from the space End(V ) of continuous linear operators). However, if we identify points on S1 with their angle θ, then S1 becomes R modulo 2π, denote it R/2πZ, where the operation is addition.The R/2πZ → S1 is given by t 7→ eit = cos t+i sin t. Math 597 Fan Huang FAN HUANG 3 From [JM74], since it is a continuous bijection between compact topological spaces, it is a homeomorphism. In this note we will focus on unitary representations of S1, i.e., representations on a Hilbert space which preserve the inner product. Unitary representations are those for which the operators ρg are unitary, i.e. preserve the inner product. On the complex plane, this t means hρgv, ρgwi = hv, wi = wv, ∀v, w ∈ V, g ∈ G. which means equivalently that ρg ρg = I. 1 1 We denote the unitary representation of S as ρn : S → U(V ). n It is not hard to see that for each integer n, the formula ρn(z) = z gives a 1-dimensional irreducible representation of S1. We shall show that (1) Every irreducible unitary representation of S1 has 1. (2) Every continuous irreducible representation of S1 is of this form. (3) For n 6= k, ρn and ρk are not isomorphic. 1 That is, Z is isomorphic to irreducible representations of S with bijection given by n 7→ ρn. First, we note that if W is an irreducible unitary representation of S1, then dim W = 1. To see this, we take as a fact that ρ is determined by ρ(g) where g ∈ S1 is a topological generator. Then by density, we know where every other element goes. The spectral theorem from linear algebra asserts that any unitary transformation is diagonalizable. Put another way, ρ0(g) = Sρ(g)S−1, where S ∈ U(V ) and ρ0(g) is a diagonal matrix. The eigenvalues of unitary matrices all have absolute value 1. It is obvious that for each i the diagonal term 0 Ln ρi(g) is a 1-dimensional unitary irreducible representation. This follows from ρ = i=1 ρi, Ln −1 ρ = i=1 SρiS and thus dim W = 1 as desired. Proposition 2.1. Every irreducible unitary representations of S1 are the continuous homo- morphisms of the form z 7→ zn for some n ∈ Z. To prove this proposition, we need two Lemmas. φ R R

ψ ψ

1 1 S ρ S

Lemma 2.1. Consider (R, +). If φ : R → R is a continuous homomorphism, then φ is a multiplication by a scalar. 1 1 c Proof. Set c = φ(1). Then φ(n) = nc, n ∈ Z. Also mφ( m ) = c and then φ( m ) = m , m ∈ Z. n n Thus φ( m ) = c m and so φ(x) = cx, for x ∈ Q. Since Q is dense in R and φ is continuous, we have φ(x) = cx for all x ∈ R.  Lemma 2.2. If ψ : R → S1 is a continuous homomorphism, then there exists c ∈ R such that ψ(x) = eicx for all x ∈ R. Proof. We claim that there is a unique continuous homomorphism l : R → R such that ψ(x) = ei·l(x). The exponential map ε : R → S1 given by ε(x) = eix maps the real line around the unit circle with period 2π and it is a universal cover. For any continuous ψ : R → S1 such that ψ(0) = 1, there exists a unique continuous lift l of this function to the real line such that l(0) = 0, In other words, there exists a unique continuous function l : R → R such that l(0) = 0 and ψ(x) = ε(l(x)) for all x, so the following diagram commutes: Math 597 Fan Huang A NOTE ON CHARACTER THEORY 4

R l ε

S1 R ψ

We also claim that if ψ is a homomorphism, then its lift l is also a homomorphism and by Lemma 2.1 l(x) = cx for some c. Note that ψ(s + t) = ψ(s)ψ(t), thus ε(l(s + t) − l(s) − l(t)) = 1 = e0. It follows that l(s + t) − l(s) − l(t) = 2πn for some n ∈ Z which depends only on s, t. Since s and t varies continuously, we find n is a constant. We set s = t = 0 to conclude that n = 0. Thus l is a homomorphism as promised, and so l(x) = cx for some c ∈ R by Lemma 2.2.  Proof of proposition 2.1. Going back to the irreducible unitary representations of S1, given a representation ρ : S1 → C∗, it has a compact and thus bounded image. It follows that the image lies on S1. Thus ρ : S1 → S1 is a continuous homomorphism. Precompose ρ with the exponential map ε, then we have the following diagram commutes: ρ S1 S1

ε ρε

R

By Lemma 2.2 there exists c ∈ R with ρε(x) = eicx for all x. Thus we have ρ(eix) = eicx. 2πi 2πic Note that 1 = ρ(1) = ρ(e ) = ρε(2π) = e and thus c ∈ Z. So we have ρ(z) = ρn(z) = n z . 

Finally, to show for n 6= k, that ρn and ρk are not isomorphic, because let f : C → C be the G-linear map. Note that znf(v) = f(znv) = zkf(v) for all z ∈ S1, then we have zn−k = 1, i.e., n = k. Because of the isomorphism from S1 to R/2πZ, another way to view the characters on S1 inx is as a map R/2πZ → C. Then the characters are given by en(x) = e with n ∈ Z. We shall show all characters have this form by following an exercise from [ST03]. First we denote by Gˆ the class of all characters of G, which is called the dual group of G. We observe that the trivial character e(g) = 1 plays the role of the unit, so Gˆ inherits the structure of an under the multiplication

χ1χ2(g) = χ1(g)χ2(g) for all g ∈ G and characters χ1, χ2. For x ∈ G the mapping χ 7→ χ(x) is an irreducible character of Gˆ and so an element of ˆ ˆ ˆ ˆ the dual G of the dual G of G. Let φx be the element be the map that maps x ∈ G to ˆ χx. Then consider the map Φ : G → G given by x 7→ φx. We see it is a homomorphism 0 0 since Φ(xy) = φxy(χ) = φx(χ)φy(χ) = Φ(x)Φ(y). Also if for x, x ∈ G, Φ(x) = Φ(x ), then Math 597 Fan Huang FAN HUANG 5 χ(x) = χ(x0) and thus χ(xx0−1) = 1 for χ ∈ Gˆ, which implies xx0−1 = 1. So Φ is injective. ˆ Since G and Gˆ have same orders, we conclude that it is an isomophism.

Lemma 2.3. If F : R → R is continuous and F (x + y) = F (x)F (y), then F is differentiable and F is of the form eAx for some constant A.

Proof. If F (x) = 0 is trivial, then we have F (0) = F (0)2. It follows that F (0) = 1 and it is clearly differentiable and is of the form e0x for constant 0. R δ R δ Consider F (0) 6= 0, then for an appropriate δ, c = 0 F (y) dy 6= 0. cF (x) = 0 F (y)F (x) dy = R δ R δ+x 0 F (x + y) = x F (y)dy. Then R δ+x+t R δ+x F (x + t) − F (x) F (y) dy − F (y) dy c lim = lim x+t x t→0 t t→0 t δ+x+t R F (y) dy R x+t F (y) dy = lim δ+x − lim x t→0 t t→0 t = F (δ + x) − F (x)

Thus F is differentiable and we have cF 0(x) = F (δ + x) − F (x) = F (δ)F (x) − F (x) F (δ) − 1 F 0(x) = F (x) c

Ax So F is of the form e for some constant A. 

By applying Lemma 2.3 and noting that the characters on S1 satisfy the assumptions, we see that all the characters are of the form eAx for constant A . The dual group of the circle 1 ˆ1 inx S is S = {en}n∈Z, where en(x) = e . Moreover, en 7→ n gives an isomorphism between ˆ1 S and the integers Z. Now we have a complete list of irreducible unitary representations of S1. From [Serre77], we already know the orthogonality relations of irreducible characters of finite groups. (a) If χ is the character of an irreducible representation, hχ, χi = 1 (i.e., χ is “of norm 1”). (b) If χ and χ0 are the characters of two nonisomorphic irreducible representations, hχ, χ0i = 0 (i.e., χ and χ0 are orthogonal). (c) Let V be a linear representation of G, with character φ, and suppose V decomposes into Lk a direct sum of irreducible representations: V = i=1 Wi. Then, if W is an irreducible representation with character χ, the number of Wi isomorphic to W is equal to the scalar P product hφ, χi. Moreover, for f ∈ Cl(G), f = khf, χiχ. As we shall see, the basic theorems from give the analogy of representation theory listed above for G = S1 and V = L2(S1). The space L2(S1) is the Hilbert space of complex measurable functions f on S1 with

1 Z 1  2 2 kfk= kfk2= |f| ≤ ∞. 0 Math 597 Fan Huang A NOTE ON CHARACTER THEORY 6

2 1 1 R 2 1 Also L (S ) is endowed with an inner product hf, gi = 2π t∈G f(t)g(t) dt for f, g ∈ L (S ). P inx Note that the linear combinations of these characters f(x) = n cne (x ∈ R) are trigono- metric polynomials. With complex coefficients an trigonometric polynomial is actually a Fourier series. Then naturally, we can find analogy with fourier analysis. Attention is now focused on the space L2(S1). By Stone-Weierstrass approximation theo- rem, the space spanned by the positive and negative powers of e2πix is dense in the space of continuous functions and the space C∞(S1) of infinitely differentiable functions is dense in L2(S1). Note S1 acts by left multiplication on f ∈ L2(S1), i.e., (ρ(t) · f)(θ) = f(t + θ) for t, θ ∈ S1. This is well-defined on L2 since Lebesgue measure is invariant under translation. Put another way, this action is indeed the and also the representation G induced from the identity element, denoted as Ind{1}. We will talk more of the in the next example. From [DM72],the basic theorems of Fourier series on S1 are:

Theorem 2.1. The orthogonal family of unit length (kenk= 1) 2πinx en(x) = e , n ∈ Z is a basis for L2(S1), that is, any function f ∈ L2(S1) can be expanded into a Fourier Series ∞ X ˆ f = f(n)en −∞ with coefficients Z Z ˆ −2πinx f(n) = (f, en) = fen dx = f(x)e dx, the sum being understood in the sense of distance in L2(S1). The map f → fˆ is therefore an isomorphism of L2(S1) onto L2(Z) and there is a Plancherel identity: Z 1 ∞ 2 2 2 X ˆ 2 kfk2= |f| = kfk = |f(n)| . 0 −∞ Theorem 2.2. For any 1 ≤ p < ∞ and any f ∈ Cp(S1), the partial sums X ˆ Sn = Sn(f) = f(k)ek |k|≤n

converge to f, uniformly as n ↑ ∞; in fact, kSn − fk∞ is bounded by a constant multiple of −p+ 1 n 2 .

1 −1 Theorem 2.3. For functions f of class C(S ), the arithmetic means n (S0 + ··· + Sn−1) P ˆ of the partial sums Sn = |k|≤n f(k)ek converge uniformly to f.

1 P 1 R 2π In our example, we replace the form “ |G| t∈G f(t)” with “ 2π 0 f(t) dt ”. So the ana- logues for the above in S1 are:

1 R 2π int −int 1 R 2π (a) The characters is of unit lenth hen, eni = 2π 0 e e dt = 2π 0 1 dt = 1.

1 R 2π int −ikt 1 R 2π i(n−k)t (b) For m 6= k, hen, eki = 2π 0 e e dt = 2π 0 e dt = 0.

Math 597 Fan Huang FAN HUANG 7

2 1 P∞ 1 R 2π −int (c) For f ∈ L (S ), f = −∞hf, enien, converges for fixed n, where hf, eni = 2π 0 e f(t) dt, leading to the Fourier inversion formula.

inx Among others, we see that the characters ρn = e form a complete orthonomal basis of L2(S1).

3. An example: the symmetric group Σ3 (the simplest nonabelian group)

The symmetric group Σ3 is isomorphic to the , D3, and has 6 elements (the identity 1, three transpositions of order two: (12), (13) and (23) and two permutations of order three: (123) and (132). It can be presented as hx, y | x3, y2, xyxyi, where x = (123) and y = (12). Its regular representation has dimension 3! = 6 and since it is a nonabelian finite group, its subrepresentation cannot be all 1-dimensional. In Σn, conjugacy classes are determined by cycle decomposition sizes: two permutations are conjugate if and only if they have the same number of cycles of each length. For Σ3, there are 3 conjugacy classes, so there are 3 different irreducible representations over C. In the decomposition into irreducible representations with respective dimensions d1, d2, d3, the order of the group 2 2 2 |Σ3|= 6 = d1 + d2 + d3. Thus there must be two 1-dimensional and one 2-dimensional representation . Explicitly, they are:

(a) The with character χT , that take the value 1 identically. (b) The sign representation with character χΣ given by πsgn(g)v = sgn(g)v, where sgn(g) = ±1 is the sign of the permutation. (c) The standard representation with character χA, that leaves invariant the orthogonal 3 subspace V = {(z1, z2, z3) ∈ C | z1 + z2 + z3 = 0}. This representation is 2-dimensional. The irreducible characters of a group can be assembled into a , where the first row in a character table lists representatives of conjugacy classes, the second row the numbers of elements in the conjugacy classes, and the other rows list the values of the characters on the conjugacy classes–which are obtained by explicitly computing traces in the irreducible representations. The characters are constant on conjugacy classes. The character table of Σ3 is:

Σ3 (1) (123) (12) # 1 2 3 χT 1 1 1 χΣ 1 1 -1 χA 2 -1 0

Let H be a subgroup of G. Let ρH be the restriction of ρ, the representation of G, to H. Let W be a sub representation of ρH , that is, a vector space of V -stable under the ρt, t ∈ H. Denote by θ : H → GL(W ) the representation of H in W thus defined. We call ρ of G in V is induced by the representation θ of H in W and write ∼ M V = σWσ σ∈G/H G where σW = gσW and gσ is any representative of the coset σ. Denote V =IndH (Wσ). Sup- pose (V, ρ) is induced by (W, θ) and let χρ and χθ be the corresponding characters of G and Math 597 Fan Huang A NOTE ON CHARACTER THEORY 8 of H.(W, θ) determines (V, ρ) up to isomorphism.

Example 1. Let H be the trivial subgroup {1}. Let W = C be the trivial representation. ∼ L Then G/H = G and thus V = g∈G gC. G acts by permuting these copies of W and we can see that V is the regular representation of G. As is shown in [Serre77]: Theorem 3.1. Let h be the order of H and let R be a system of representatives of G/H. For each u ∈ G, we have X 1 X χ (u) = χ (r−1ur) = χ (s−1us). ρ θ h θ r∈R s∈G r−1ur∈H s−1us∈H

In particular, χρ(u) is a linear combination of the values of χθ on the intersection of H with the of u in G.

Σ3 We shall apply above theorem to calculate the IndH for the subgroups of Σ3. We note r that characters do play an important role in the algorithm. Let {ρi}i=1 be the irreducible r representations of G and {χi}i=1 be the corresponding characters. To find how the induced representation decomposes in terms of irreducible representations of the group Σ3, that G =r is, IndH χρ(u) = i=1 aiχi, Since the characters follow the same relation and thus ai = 1 P hχρ, χii = |G| t∈G χρ(t)χi(t).

Example 2. Recall that the Cn, a finite abelian group, has n irreducible rep- resentations of degree 1 whoses characters are primitive roots of unity. Let H be the cyclic subgroup C3 = {1, (123), (132)}, which is a normal subgroup of Σ3. Thus there is short exact sequence of Σ3: ϕ ψ 1 H = C3 Σ3 G/H = Z2 1

The character table of C3 with 3 1-dimensional irreducible characters is

C3 (1) (123) (132) # 1 1 1 χ1 1 1 1 2 χ2 1 ω ω 2 χ3 1 ω ω where ω = e2πi/3. In this example we choose a system representatives of G/H: R = {(1), (12)}. Since −1 H E G, we have r ur ∈ H for all r ∈ R and u ∈ G. Let W1 be the 1-dimensional representation of H given by (123) 7→ 1, i.e, the trivial representation of C3. Then χρ(1) = χθ(1) + χθ(1) = 2

χρ(12) = χρ(13) = χρ(23) = 0

χρ(123) = χθ(123) + χθ(132) = 2

χρ(132) = χθ(132) + χθ(123) = 2 Math 597 Fan Huang FAN HUANG 9

Since IndΣ3 χ is of the form aχ + bχ + cχ and we can calculate C3 1 T Σ A 1 a = hχ , χ i = · [2 · 1 + 2 · 1 + 2 · 1 + 0 + 0 + 0] = 1 ρ1 T 6 1 b = hχ , χ i = · [2 · 1 + 2 · 1 + 2 · 1 + 0 + 0 + 0] = 1 ρ1 Σ 6 1 c = hχ , χ i = · [2 · 2 + 2 · (−1) + 2 · (−1) + 0 + 0 + 0] = 0 ρ1 A 6 So IndΣ3 ρ ∼ ρ ⊕ ρ . C3 1 = T Σ

Let W2 be the 1-dimensional representation of H given by (123) 7→ ω. Then

χρ(1) = χθ(1) + χθ(1) = 2

χρ(12) = χρ(13) = χρ(23) = 0 2 χρ(123) = χθ(123) + χθ(132) = ω + ω = −1 2 χρ(132) = χθ(132) + χθ(123) = ω + ω = −1 Since IndΣ3 χ is of the form aχ + bχ + cχ and we can calculate C3 2 T Σ A 1 a = hχ , χ i = · [2 · 1 + (−1) · 1 + (−1) · 1 + 0 + 0 + 0] = 0 ρ2 T 6 1 b = hχ , χ i = · [2 · 1 + (−1) · 1 + (−1) · 1 + 0 + 0 + 0] = 0 ρ2 Σ 6 1 c = hχ , χ i = · [2 · 2 + (−1) · (−1) + (−1) · (−1) + 0 + 0 + 0] = 1 ρ2 A 6 So IndΣ3 ρ ∼ ρ , and same for IndΣ3 ρ by symmetry. C3 2 = A C3 3

Example 3. Let H be the cyclic subgroup C2 = {1, (12)} and its character table is:

C2 (1) (12) # 1 1 χ1 1 1 χ2 1 -1 The three cosets are {(1), (12)}, {(13), (123)}, {(23), (132)}. We choose a system represen- tatives of G/H: R = {(1), (123), (132)}. A bit different from previous example,H is not a −1 normal subgroup in Σ3. We need to check if r ur ∈ H for all r ∈ R and u ∈ G. For example, if u = (12), then r ∈ R r−1ur in H? 1 (12) Yes (123) (13) No (132) (23) No if u = (123), then r ∈ R r−1ur in H? 1 (123) No (123) (123) No (132) (123) No Math 597 Fan Huang A NOTE ON CHARACTER THEORY 10

Let W1 be the 1-dimensional representation of H given by (12) 7→ 1, i.e, the trivial repre- sentation of C2. Then χρ(1) = χθ(1) + χθ(1) + χθ(1) = 3

χρ(12) = χρ(13) = χρ(23) = χθ(12) = 1

χρ(123) = χρ(132) = 0 Since IndΣ3 χ is of the form aχ + bχ + cχ and we can calculate C2 1 T Σ A 1 a = hχ , χ i = · [3 · 1 + 0 + 0 + 1 · 1 + 1 · 1 + 1 · 1] = 1 ρ1 T 6 1 b = hχ , χ i = · [3 · 1 + 0 + 0 + 1 · (−1) + 1 · (−1) + 1 · (−1)] = 0 ρ1 Σ 6 1 c = hχ , χ i = · [3 · 2 + 0 + 0 + 0 + 0 + 0] = 1 ρ1 A 6 So IndΣ3 ρ ∼ ρ ⊕ ρ . C2 1 = T A

Let W2 be the 1-dimensional representation of H given by (12) 7→ −1. Then

χρ(1) = χθ(1) + χθ(1) + χθ(1) = 3

χρ(12) = χρ(13) = χρ(23) = χθ(12) = −1

χρ(123) = χρ(132) = 0 Since IndΣ3 χ is of the form aχ + bχ + cχ and we can calculate C2 1 T Σ A 1 a = hχ , χ i = · [3 · 1 + 0 + 0 + (−1) · 1 + (−1) · 1 + (−1) · 1] = 0 ρ2 T 6 1 b = hχ , χ i = · [3 · 1 + 0 + 0 + (−1) · (−1) + (−1) · (−1) + (−1) · (−1)] = 1 ρ2 Σ 6 1 c = hχ , χ i = · [3 · 2 + 0 + 0 + 0 + 0 + 0] = 1 ρ2 A 6 So IndΣ3 ρ ∼ ρ ⊕ ρ . C2 2 = Σ A

Recall that Σ3 acts on the cosets of G/H by ggσH = gτ H and we can actually compute the explicit induced representation. For each g ∈ G, first note if it flips cosets to decide whether the matrix is diagonal or flipped. Then note which element of H appears and decides the actual entries in the matrix. Specifically, the matrix representation induced from W2 in Example 2 are as follows:

1 0 0 1  0 w 1 7→ , (12) 7→ , (13) 7→ , 0 1 1 0 w2 0 0 w2 w 0  w2 0 (23) 7→ , (123) 7→ , (132) 7→ w 0 0 w2 0 w

Again, it is clear that this induced representation is ρA by directly calculating the for each matrix . We see from the above examples that normal subgroups give an easier way to compute induced characters. Another way to look at induced representation is by seeing that W is a left C[H]-module and C[G] is a right C[H]-module. We define V := C[G] ⊗C[H] W . In fact C[G] is a free Math 597 Fan Huang FAN HUANG 11

C[H]-module via multiplication on the right with basis of gσ, representatives of left cosets of H. Here, for example, (123) · ((1) ⊗ e) = (123) ⊗ e = (1)((123) ⊗ e) = (1) ⊗ (123) · e. (123) · ((12) ⊗ e) = (123)(12) ⊗ e = (13) ⊗ e = (12)(132) ⊗ e These results lead to further questions. For example, why do characters decide the rep- resentation up to isomorphism? Another question is how to recover the symmetric function from the trace?

Acknowledgement The author would like to thank Professor Matthew Ando for precious and generous guid- ance, also appreciate undergraduate student Yan Zhou for helpful suggestions and graduate student Sarah Loeb for careful peer review.

References [Serre77] Jean-Pierre Serre, Linear Representations of Finite Groups, Springer-Verlag, NY,1977. [DM72] H.Dym and H.P.Mckean, Fourier Series and Integrals, Academic Press, NY, 1972. [ST03] Elias M. Stein and Rami Shakarchi, Princeton Lectures in Analysis Fourier Analysis: An Introduc- tion, Princeton University Press, NJ, 2003, 231–237. [MA10] Michael Artin, Algebra, Pearson,2 edition, 2010. [JM74] James R.Munkres, Topology; A First Course, Prentice Hall College Div, 1974. [CT05] Constantin Teleman, Lecture Notes of Representation Theory, CT, Lent 2005, available at math. berkeley.edu/~teleman/math/RepThry.pdf.

Department of Mathematics, University of Illinois, Urbana-Champaign, IL 61801, USA E-mail address: [email protected]

Math 597 Fan Huang