# Aufgaben:Problem 13

## Contents

Let G be a finite group. In the lectures we encountered two types of functions on $$G$$ which were both called characters:

1. Let $$C_1 = \{e\}, C_2,...,C_k$$ be the conjugacy classes, and let $$v_1,..., v_k$$ be the normalized eigenvectors of the Burnside matrices of $$G$$, then for all $$s \in \{1,...,k\}$$ we defined the maps

$$\chi_s : G\ \rightarrow \ \mathbb{C}$$ $$g \mapsto \sqrt{|G|}\sum_{j=1}^k \frac{v_{sj}}{\sqrt{|C_j|}}\delta_{C_j}(g).$$

2. For any representation $$\rho$$ we defined the map $$ch(\rho):x \mapsto Tr(\rho(x))$$.

We want to show that the characters in the sense of 2. of irreducible representations are exacly the characters in the sense of 1. For all $$s \in \{1,...,k\}$$ let

$$V_s := Span\{x\mapsto \chi_s(xy^{-1})| y\in G\}$$

Recall that $$V_s$$ is an invariant subspace of all linear transformations $$L(g), g\in G$$

(a) Let $$\rho$$ and $$\rho'$$ be unitary irreducible representations of $$G$$ on finite dimensional complex inner product space $$V$$ and $$V'$$ respectivly. Show that

$$(Tr(\rho), Tr(\rho'))_G = \begin{cases} 1, & \text{if $$\rho$$ and $$\rho'$$ are isomorphic} \\ 0, & \text{if $$\rho$$ and $$\rho'$$ are not isomorphic,} \end{cases}$$

i.e. the characters in the sense of 2. of irreducible representations are orthonormal.

(b) Assume that for every $$s \in \{1,...,k\}$$ there is an invariant subspace $$W \subseteq V_s$$ w.r.t. $$L^{V_s}$$ such that the restriction $$L_W = L^{V_s}|_W \in GL(W)$$ of the linear operator $$L^{V_s}$$ from $$V_s$$ to $$W$$ defines an irreducible, unitary representation on $$G$$ on the subspace $$W$$. Show that then $$ch(L^W) = \chi_s$$, and conclude that the characters in the sense of 2. of irreducible representations are exactly the characters in the sense of 1.

## Solution

### (a)

Choose an orthonormal basis of $$V, V'$$ with dimension $$n, m$$ respectively. Let

$$[\rho](g) := \Big(\rho_{ij}(g)\Big)_{1\leq i,j \leq n}$$ $$[\rho'](g) := \Big(\rho'_{ij}(g)\Big)_{1\leq i,j \leq m}$$

be the matrices of $$\rho$$ and $$\rho'$$ relative to these bases. Set (Shortcut for (a) see discussion)

$$T^{(k,l)} := \Big(T^{(k,l)}{}_{i,j} \Big) = \Big(\frac{1}{|G|} \sum_{g\in G}\rho_{ki}(g)^*\rho_{lj}(g)\Big) \text{ for} \ 1\leq k,l\leq n$$

Claim 1 : $$[T^{(k,l)}, [\rho](g)] = 0$$

Proof:

$$\big(T^{(k,l)} [\rho](g)\big)_{ij} = \sum_{a=1}^n T^{(k,l)}{}_{ia} \rho_{aj}(g) = \sum_{a=1}^n \frac{1}{|G|} \sum_{h\in G}\rho_{ki}(h)^*\rho_{la}(h) \rho_{aj}(g)$$

$$= \frac{1}{|G|} \sum_{h\in G}\rho_{ki}(h)^* \sum_{a=1}^n \rho_{la}(h) \rho_{aj}(g) = \frac{1}{|G|} \sum_{h\in G}\rho_{ki}(h)^* \rho_{lj}(hg)$$

$$z = hg \rightarrow h = zg^{-1}$$

$$= \frac{1}{|G|} \sum_{z\in G}\rho_{ki}(zg^{-1})^* \rho_{lj}(z) = \frac{1}{|G|} \sum_{z\in G} \sum_{a=1}^n \rho_{ka}(z)^* \rho_{ai}(g^{-1})^* \rho_{lj}(z)$$

$$= \sum_{a=1}^n T^{(k,l)}{}_{aj} \rho_{ai}(g^{-1})^*$$

since $$[\rho](g)$$ is a unitary matrix: $$\rho_{ai}(g^{-1})^* = \rho_{ia}(g^{-1})^{T*} = \rho_{ia}(g^{-1})^{-1} = \rho_{ia}(g)$$

$$= \sum_{a=1}^n T^{(k,l)}{}_{aj} \rho_{ia}(g) = \big([\rho](g)T^{(k,l)}\big)_{ij}$$

$$\square$$

Observe that if $$[A,B] = 0$$ KerA is an invariant subspace of B: let $$v\in KerA$$ then $$ABv = BAv = B0 = 0 \Rightarrow Bv \in KerA$$

Claim 2: $$(\rho_{kl},\rho_{lj})_G = \frac{1}{n}\delta_{kl}\delta_{lj}$$

Proof: Let $$\lambda$$ be an eigenvalue of $$T^{(k,l)}$$ $$[T^{(k,l)} -\lambda \mathbb{I}, [\rho](g)] = 0$$ since the identity commutes with everything.

$$\Rightarrow Ker(T^{(k,l)} -\lambda \mathbb{I}) \neq \{0\}$$

is an invariant subspace of $$\rho$$ and because $$\rho$$ is irreducible is equal to $$V$$

$$\Rightarrow T^{(k,l)} = \lambda \mathbb{I}$$

$$\lambda n = tr T^{(k,l)} = \sum_{j=1}^n \frac{1}{|G|}\sum_{g\in G} \rho_{kj}(g)^*\rho_{lj}(g) = \sum_{j=1}^n \frac{1}{|G|}\sum_{g\in G} \rho_{jk}(g^{-1})\rho_{lj}(g) = \frac{1}{|G|}\sum_{g\in G} \rho_{lk}(e) =\delta_{lk}$$

because $$\rho(e) = e = \mathbb{I}$$. For $$k\neq l \Rightarrow \lambda = 0 \Rightarrow T^{(k,l)} = 0$$ and for $$k = l \Rightarrow \lambda = \frac{1}{n} \Rightarrow T^{(k,k)} = \frac{1}{n} \mathbb{I}$$

$$\Rightarrow \Big(T^{(k,l)}{}_{ij} \Big) = \frac{1}{n} \delta_{kl} \delta_{ij}$$

$$\square$$

Claim 3: $$||ch(\rho)||_2^2 = 1$$

Proof:

$$||ch(\rho)||_2^2 = \frac{1}{|G|}\sum_{g\in G} |ch(\rho)(g)|^2 = \frac{1}{|G|}\sum_{g\in G} tr[\rho](g)tr[\rho](g)^* = \frac{1}{|G|}\sum_{g\in G} \sum_{i=1}^n \rho_{ii}(g) \sum_{j=1}^n \rho_{jj}(g)^*$$

$$= \sum_{i,j=1}^n \Big(T^{(j,i)}{}_{j,i} \Big) = \sum_{i,j=1}^n \frac{1}{n} \delta_{ij} = 1$$

$$\square$$

now we are prepaired to prove the first part of (a): let $$\rho, \rho'$$ be isomorphic $$\Rightarrow \exists \phi$$ an isomorphism form $$V$$ onto $$V'$$ such that $$\rho(g) = \phi^{-1} \circ \rho'(g) \circ \phi$$.

Let $$T$$ be the invertible matrix of $$\phi$$ w.r.t our bases.

$$\Rightarrow ch(\rho)(g) = tr( [\rho](g)) = tr( T^{-1} [\rho'](g) T)= tr( [\rho'](g)) = ch(\rho')(g)$$

$$\Rightarrow (ch(\rho), ch(\rho'))_G = (ch(\rho), ch(\rho))_G = ||ch(\rho)||_2^2 = 1$$

Now for the second part: from now on let $$\rho, \rho'$$ be not isomorphic. Set

$$S^{(k,l)} := \Big(S^{(k,l)}{}_{ij} \Big) = \Big(\frac{1}{|G|} \sum_{g\in G}\rho_{ki}(g)^*\rho'_{lj}(g)\Big) \text{ } 1\leq k\leq n,\ 1\leq l\leq m$$

which is the matrix of a linear transform from $$V'$$ to $$V$$.

Claim 4: $$S^{(k,l)}[\rho'](g)= [\rho](g)S^{(k,l)}$$

Proof:

$$\big(S^{(k,l)}[\rho'](g)\big)_{ij} = \sum_{a=1}^m S^{(k,l)}{}_{ia} \rho'_{aj}(g) = \sum_{a=1}^m \frac{1}{|G|} \sum_{h\in G}\rho_{ki}(h)^*\rho'_{la}(h) \rho'_{aj}(g)$$

$$= \frac{1}{|G|} \sum_{h\in G}\rho_{ki}(h)^* \rho'_{lj}(hg)$$

$$z = hg \rightarrow h = zg^{-1}$$

$$= \frac{1}{|G|} \sum_{z\in G}\rho_{ki}(zg^{-1})^* \rho'_{lj}(z) = \frac{1}{|G|} \sum_{z\in G} \sum_{a=1}^n \rho_{ka}(z)^* \rho_{ai}(g^{-1})^* \rho'_{lj}(z)$$

$$= \sum_{a=1}^n \rho_{ia}(g) S^{(k,l)}{}_{aj} = \big([\rho](g)S^{(k,l)}\big)_{ij}$$

$$\square$$

Claim 5: $$S^{(k,l)} = 0$$

Proof: Observe that $$Ker S^{(k,l)}$$ and $$Im S^{(k,l)}$$ are invariant subspaces of $$V'$$, $$V$$ respectivly:

let $$v\in Im S^{(k,l)} \Rightarrow \exists v' \in V'$$ such that $$S^{(k,l)}v' = v$$

$$\rho(g)v = \rho(g) S^{(k,l)}v' = S^{(k,l)}\rho'(g)v' = S^{(k,l)} w' \Rightarrow \rho(g)v \in Im S^{(k,l)}$$

let $$v'\in Ker S^{(k,l)}$$

$$S^{(k,l)}\rho'(g)v' = \rho(g)S^{(k,l)}v' = \rho(g)0 = 0 \Rightarrow \rho'(g)v' \in Ker S^{(k,l)}$$

Since $$\rho$$ is irreducible either

(i)$$Im S^{(k,l)} = \{0\} \Rightarrow S^{(k,l)} = 0 \Rightarrow$$ the claim is proven

(ii) $$Im S^{(k,l)} = V \Rightarrow S^{(k,l)}$$ is surjective, because $$S^{(k,l)}$$ is a linear transfrom from $$V'$$ to $$V$$

Since $$\rho'$$ is irreducible either
(iii)$$Ker S^{(k,l)} = V'$$ but since again $$S^{(k,l)}$$ is a linear transfrom from $$V'$$ to $$V$$ $$\Rightarrow S^{(k,l)} = 0$$
(iv)$$Ker S^{(k,l)} = \{0\} \Rightarrow S^{(k,l)}$$ is injective, with (ii) $$\Rightarrow S^{(k,l)}$$ is an isomorphism, which is a contradiction, because of Claim 4 and because $$\rho$$ and $$\rho'$$ are not :isomorphic.

$$\square$$

$$\Rightarrow \big(S^{(k,l)}\big)_{ij} = \frac{1}{|G|} \sum_{g\in G}\rho_{ki}(g)^*\rho'_{lj}(g) = 0$$

Now we can easly prove the second part of (a):

$$(ch(\rho), ch(\rho'))_G = \frac{1}{|G|} \sum_{g\in G} \sum_{i = 1}^n \rho_{ii}(g) \sum_{j = 1}^m \rho'_{ii}(g)^* = \sum_{i = 1}^n \sum_{j = 1}^m \big(S^{(i,j)}\big)_{ij}^* = 0$$

### (b)

Claim 6: $$\chi_s$$ $$s \in \{1,...,k\}$$ is an othonormal basis of the class funcitons on $$G$$.

Proof: We know that $$\frac{\sqrt{|G|}}{\sqrt{|C_j|}} \delta_{C_j}$$ $$j \in \{1,...,k\}$$ is an orthonormal basis of the class functions. And $$v_{s}$$, $$s \in \{1,...,k\}$$ are orhonormal w.r.t the standard complex inner product.

$$( \chi_s, \chi_t )_G = \Big( \sqrt{|G|}\sum_{i=1}^k \frac{v_{si}}{\sqrt{|C_i|}}\delta_{C_i}(g), \sqrt{|G|}\sum_{j=1}^k \frac{v_{tj}}{\sqrt{|C_j|}}\delta_{C_j}(g) \Big)_G$$

$$= \sum_{i,j=1}^k v_{si} v_{tj}^* \Big( \frac{\sqrt{|G|}}{\sqrt{|C_i|}}\delta_{C_i}(g), \frac{\sqrt{|G|}}{\sqrt{|C_j|}}\delta_{C_j}(g) \Big)_G = \sum_{i,j=1}^k v_{si} v_{tj}^* \delta_{ij} = \sum_{j=1}^k v_{sj} v_{tj}^* = \delta_{st}$$

$$\square$$

Claim 7: Every classfunction $$\in V_s$$ is a multiple of $$\chi_s$$

Proof: We know that $$\Pi_s 1\leq s\leq k$$ are a family of orthogonal projections on $$V$$ and range $$\Pi_s = V_s$$.

$$\Rightarrow V_s$$ are orthogonal subspaces of $$V$$. Observe that $$\chi_s = \chi_{se} \in V_s$$ for every $$s$$.

Now consider a class function $$\psi \in V_s$$:

$$\psi = \sum_{j = 1}^k a_j \chi_j, \quad a_j \in \mathbb{C}$$

$$\forall r \in \{ 1,...,k \}, r\neq s$$:

$$0=(\psi, \chi_r)_G = \sum_{j = 1}^k a_j (\chi_j, \chi_r)_G = a_r$$ $$\Rightarrow \psi = a_s \chi_s$$

$$\square$$

Claim 8: $$(\chi_{sy}, \chi_{sx})_G = \frac{1}{v_{s1}\sqrt{|G|}} \chi_{sy}(x)$$

Proof:

$$(\chi_{sy}, \chi_{sx})_G = \frac{1}{|G|} \sum_{z\in G} \chi_{sy}(z) \chi_s(zx^{-1})^*$$

$$= \frac{1}{v_{s1}\sqrt{|G|}} \frac{1}{|G|} \sum_{z\in G} v_{s1}\sqrt{|G|} \chi_s(zx^{-1})^* \chi_{sy}(z) = \frac{1}{v_{s1}\sqrt{|G|}} \frac{1}{|G|} \sum_{z\in G} \Pi_s(z,x)^* \chi_{sy}(z)$$

Fourier Script last page: $$\Pi_s(z,x)^* = \Pi_s(x,z)$$

$$= \frac{1}{v_{s1}\sqrt{|G|}} \frac{1}{|G|} \sum_{z\in G} \Pi_s(x,z) \chi_{sy}(z) = \frac{1}{v_{s1}\sqrt{|G|}} (\Pi_s \chi_{sy} )(x) = \frac{1}{v_{s1}\sqrt{|G|}} \chi_{sy}(x)$$

$$\square$$

Claim 9: $$ch(L^W) \in V_s$$ (Konstantin)

Proof: Choose an orthomnormal basis $$e_1,\dots e_n$$ of $$W$$ w.r.t $$(\ ,\ )_G\Rightarrow e_i = \sum_{y\in G} \lambda_{iy} \chi_{sy}$$. Notice that we don't know if all the $$\chi_{sy}$$ are in $$W$$ but we can still write $$e_i$$ this way

\begin{align} ch(L^W)(g) &= tr(L^W(g)) \\ &= \sum_{i= 1}^n (e_i, L^W(g) e_i)_G =\sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} e_i(h) (L^W(g) e_i)(h)^* =\sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} e_i(h) e_i(g^{-1}h)^* \\ &= \sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} \sum_{z\in G} \lambda_{iz} \chi_{sz}(h) \sum_{y\in G} \lambda_{iy}^* \chi_{sy}(g^{-1}h)^* = \sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} \sum_{z\in G} \lambda_{iz} \chi_{sz}(h) \sum_{y\in G} \lambda_{iy}^* \chi_{sgy}(h)^*\\ &= \sum_{i= 1}^n \sum_{z\in G} \sum_{y\in G} \lambda_{iz} \lambda_{iy}^* (\chi_{sz}, \chi_{sgy} )_G = \sum_{i= 1}^n \sum_{z\in G} \sum_{y\in G} \lambda_{iz} \lambda_{iy}^* \frac{1}{v_{s1}\sqrt{|G|}} \chi_{sz}(gy) = \sum_{i= 1}^n \sum_{z\in G} \sum_{y\in G} \frac{ \lambda_{iz} \lambda_{iy}^*}{v_{s1}\sqrt{|G|}} \chi_{szy^{-1}}(g) \\ \end{align}

$$\square$$

Now for the actual proof:

The character is a class function and with Claim 7 and 9 $$\Rightarrow ch(L^W) =\lambda \chi_s$$. Since $$W$$ is irreducible:

$$||ch(L^W)||_2^2 = \frac{1}{|G|}\sum_{g\in G} |\lambda \chi_s(x)|^2 = |\lambda|^2 \overset{!}{=} 1 \Rightarrow |\lambda| = 1$$

also

$$dim W = ch(L^W)(e) = \lambda \chi_s(e) = \lambda \sqrt{|G|} v_{s1}$$

since $$\sqrt{|G|} v_{s1} > 0$$ (Fourier Script page 5) and $$dim W$$ is a positive integer $$\Rightarrow \lambda = 1$$. And we have proven the first part of (b)

Now for the second part we wanto show that the characters are equivalent:

From part (a) follows that $$L^{W_s}, L^{W_r}$$ are non isomorphic for $$r\neq s$$ since: $$(ch(L^{W_s}), ch(L^{W_r}))_G = (\chi_s, \chi_r)_G = \delta_{rs} = 0$$

$$ch(\rho)$$ is a classfunction

$$\Rightarrow ch(\rho) = \sum_{j=1}^k a_j \chi_j$$

$$(ch(\rho), ch(L^{W_s}))_G = \sum_{j=1} a_j (\chi_j, \chi_s)_G = a_s = \begin{cases} 1, & \text{if $$\rho$$ and $$L^{W_s}$$ are isomorphic} \\ 0, & \text{if $$\rho$$ and $$L^{W_s}$$ are not isomorphic,} \end{cases}$$

since $$a_s$$ can't be zero for all the $$s\in \{1,...,k\}$$ and can't be $$\neq 0$$ for more then one $$s$$ because that would imply that $$L^{W_s}, L^{W_r}$$ are isomporhic. Therefore $$\rho$$ is isomorphic to one particular $$L^{W_s}$$ and

$$ch(\rho) = \chi_s$$

$$\square$$