Difference between revisions of "Aufgaben:Problem 13"

Latest revision as of 11:11, 6 August 2015

Task

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

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

$\square$

Difference between revisions of "Aufgaben:Problem 13"

Latest revision as of 11:11, 6 August 2015

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@@ Line 1: / Line 1: @@
 ==Task==
-Let G be a finite group. In the lectures we encountered two types of functions on G which were both called characters:
+Let G be a finite group. In the lectures we encountered two types of functions on \(G\) which were both called characters:
-. 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
+. 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
-$$ \chi : G\ \rightarrow \ \mathbb{C}$$
+$$ \chi_s : G\ \rightarrow \ \mathbb{C}$$
 $$ g \mapsto \sqrt{|G|}\sum_{j=1}^k \frac{v_{sj}}{\sqrt{|C_j|}}\delta_{C_j}(g).$$
-. For any represntation \(\rho\) we defines the map \(ch(\rho):x\rightarrow Tr(\rho(x))\).
+. 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
+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
 $$ V_s := Span\{x\mapsto \chi_s(xy^{-1})| y\in G\}$$
@@ Line 16: / Line 16: @@
 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
+(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}$$
@@ Line 22: / Line 22: @@
 i.e. the characters in the sense of 2. of irreducible representations are orthonormal.
-(b) Show that for every \( s \in {1,...,k}\) there is a nontrivial invariant subspace \( W \subset V_s\) w.r.t. \(L^{V_s}\) such that the restriction \( L_W = L^{V_s}|W \in GL(W)\) if the linear operators \(L^{V_s}\) from \(V_s\) to W defines an irreducible, unitary representation on G on the subspace W, and \(ch(L^W) = \chi_s\).
+(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.
-(c) Conclude that the characters in the sense of 2. of irreducible representations are
-exactly the characters in the sense of 1.
 ==Solution==
@@ Line 35: / Line 32: @@
 $$[\rho'](g) := \Big(\rho'_{ij}(g)\Big)_{1\leq i,j \leq m}$$
-be the matrices of \(\rho\) and \(\rho'\) relative to these bases. Set
+be the matrices of \(\rho\) and \(\rho'\) relative to these bases. Set (Shortcut for (a) see discussion)
-$$ T^{(k,l)} := \Big(T^{(k,l)}{}_{i,j} \Big) = \Big(\frac{1}{|G|} \sum_{g\in G}\rho_{ki}(g)^*\rho_{lj}(g)\Big) = \Big((\rho_{ki},\rho_{lj})_G\Big)\  \text{  for} \ 1\leq k,l\leq n$$
+$$ T^{(k,l)} := \Big(T^{(k,l)}{}_{i,j} \Big) = \Big(\frac{1}{|G|} \sum_{g\in G}\rho_{ki}(g)^*\rho_{lj}(g)\Big) \text{  for} \ 1\leq k,l\leq n$$
-Claim 1 : \([T^{(k,l)}, [\rho](g)] = 0\)
+'''Claim 1 :''' \([T^{(k,l)}, [\rho](g)] = 0\)
 Proof:
@@ Line 45: / Line 42: @@
 $$ \big(T^{(k,l)} [\rho](g)\big)_{ij} = \sum_{a=1}^n T^{(k,l)}{}_{ia} \rho_{aj}(g) = \sum_{a=1}^n \frac{1}{|G|} \sum_{h\in G}\rho_{ki}(h)^*\rho_{la}(h) \rho_{aj}(g)$$
-$$ \frac{1}{|G|} \sum_{h\in G}\rho_{ki}(h)^* \sum_{a=1}^n \rho_{la}(h) \rho_{aj}(g) = \frac{1}{|G|} \sum_{h\in G}\rho_{ki}(h)^* \rho_{lj}(hg) $$
+$$ = \frac{1}{|G|} \sum_{h\in G}\rho_{ki}(h)^* \sum_{a=1}^n \rho_{la}(h) \rho_{aj}(g) = \frac{1}{|G|} \sum_{h\in G}\rho_{ki}(h)^* \rho_{lj}(hg) $$
 \( z = hg \rightarrow h = zg^{-1}\)
@@ Line 53: / Line 50: @@
 $$ = \sum_{a=1}^n T^{(k,l)}{}_{aj} \rho_{ai}(g^{-1})^* $$
-since \([\rho](g)\) is a unitary matrix: \( \rho_{ai}(g^{-1})^* = \rho_{ia}(g^{-1})^{T^*} = \rho_{ia}(g^{-1})^{-1} = \rho_{ia}(g)\)
+since \([\rho](g)\) is a unitary matrix: \( \rho_{ai}(g^{-1})^* = \rho_{ia}(g^{-1})^{T*} = \rho_{ia}(g^{-1})^{-1} = \rho_{ia}(g)\)
-$$ = \sum_{a=1}^n T^{(k,l)}{}_{aj} \rho_{ia}(g) =  \big(T^{(k,l)}[\rho](g)\big)_{ij}$$
+$$ = \sum_{a=1}^n T^{(k,l)}{}_{aj} \rho_{ia}(g) =  \big([\rho](g)T^{(k,l)}\big)_{ij}$$
 <p style="text-align:right;">\(\square\)</p>
 Observe that if \( [A,B] = 0\) KerA is an invariant subspace of B: let \(v\in KerA\) then
-$$ABv = BAv = B0 = 0 \Rightarrow Bv \in KernA$$
+$$ABv = BAv = B0 = 0 \Rightarrow Bv \in KerA$$
-Claim 2: \((\rho_{kl},\rho_{lj})_G = \frac{1}{n}\delta_{kl}\delta_{ij}\)
+'''Claim 2:''' \((\rho_{kl},\rho_{lj})_G = \frac{1}{n}\delta_{kl}\delta_{lj}\)
-Proof: Let \(\lambda\) be an eigenvalue of \(T^{(k,l)}\)
+'''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.
@@ Line 75: / Line 72: @@
 $$ \lambda n = tr T^{(k,l)} = \sum_{j=1}^n \frac{1}{|G|}\sum_{g\in G} \rho_{kj}(g)^*\rho_{lj}(g) = \sum_{j=1}^n \frac{1}{|G|}\sum_{g\in G} \rho_{jk}(g^{-1})\rho_{lj}(g) = \frac{1}{|G|}\sum_{g\in G} \rho_{lk}(e) =\delta_{lk}$$
-because \( \rho(e) = e = \mathbb{I}\). For \( k\neq l \Rightarrow \lambda = 0 \Rightarrow T^{(k,l)} = 0\) and for \( k = k \Rightarrow \lambda = \frac{1}{n} \Rightarrow T^{(k,k)} = \frac{1}{n} \mathbb{I} \)
+because \( \rho(e) = e = \mathbb{I}\). For \( k\neq l \Rightarrow \lambda = 0 \Rightarrow T^{(k,l)} = 0\) and for \( k = l \Rightarrow \lambda = \frac{1}{n} \Rightarrow T^{(k,k)} = \frac{1}{n} \mathbb{I} \)
-$$\Rightarrow \Big(T^{(k,l)}{}_{ij} \Big) = \frac{1}{n} \delta_{lk} \delta_{ij}$$
+$$\Rightarrow \Big(T^{(k,l)}{}_{ij} \Big) = \frac{1}{n} \delta_{kl} \delta_{ij}$$
 <p style="text-align:right;">\(\square\)</p>
-Claim 3: \( ||ch(\rho)||_2 = 1\)
+'''Claim 3:''' \( ||ch(\rho)||_2^2 = 1\)
-Proof:
+'''Proof:'''
-$$ ||ch(\rho)||_2^2 = \frac{1}{|G|}\sum_{g\in G} |ch(\rho)|^2 = \frac{1}{|G|}\sum_{g\in G} tr[\rho](g)^*tr[\rho](g) = \frac{1}{|G|}\sum_{g\in G} \sum_{i=1}^n \rho_{ii}(g)^* \sum_{j=1}^n \rho_{jj}(g)$$
+$$ ||ch(\rho)||_2^2 = \frac{1}{|G|}\sum_{g\in G} |ch(\rho)(g)|^2 = \frac{1}{|G|}\sum_{g\in G} tr[\rho](g)tr[\rho](g)^* = \frac{1}{|G|}\sum_{g\in G} \sum_{i=1}^n \rho_{ii}(g) \sum_{j=1}^n \rho_{jj}(g)^*$$
-$$ = \sum_{i,j=1}^n (\rho_{ii}, \rho_{jj})_G = \sum_{i,j=1}^n \frac{1}{n}  \delta_{ij} = 1$$
+$$ = \sum_{i,j=1}^n \Big(T^{(j,i)}{}_{j,i} \Big) = \sum_{i,j=1}^n \frac{1}{n}  \delta_{ij} = 1$$
 <p style="text-align:right;">\(\square\)</p>
-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 \(\rho(g) = \phi^{-1} \circ \rho'(g) \circ \phi \).
+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.
-$$\Rightarrow ch(\rho)(g) = tr( [\rho](g)) = tr( T^{-1} [\rho'](g) T)$$
-using \( tr(AB) = tr(BA) \)
+Let \(T\) be the invertible matrix of \(\phi\) w.r.t our bases.
-$$ = tr( [\rho'](g) T T^{-1}) = tr( [\rho'](g)) = ch(\rho')(g)$$
+$$\Rightarrow ch(\rho)(g) = tr( [\rho](g)) = tr( T^{-1} [\rho'](g) T)= tr( [\rho'](g)) = ch(\rho')(g)$$
 $$ \Rightarrow (ch(\rho), ch(\rho'))_G = (ch(\rho), ch(\rho))_G = ||ch(\rho)||_2^2 = 1$$
@@ Line 108: / Line 101: @@
 which is the matrix of a linear transform from \(V'\) to \(V\).
-Claim 4: \( S^{(k,l)}[\rho'](g)= [\rho](g)S^{(k,l)}\)
+'''Claim 4:''' \( S^{(k,l)}[\rho'](g)= [\rho](g)S^{(k,l)}\)
-Proof:
+'''Proof:'''
-$$\big(S^{(k,l)}[\rho'](g)\big)_{ij} = \sum_{a=1}^m S^{(k,l)}{}_{ia} \rho'_{aj}(g) = \sum_{a=1}^n \frac{1}{|G|} \sum_{h\in G}\rho_{ki}(h)^*\rho'_{la}(h) \rho'_{aj}(g)$$
+$$\big(S^{(k,l)}[\rho'](g)\big)_{ij} = \sum_{a=1}^m S^{(k,l)}{}_{ia} \rho'_{aj}(g) = \sum_{a=1}^m \frac{1}{|G|} \sum_{h\in G}\rho_{ki}(h)^*\rho'_{la}(h) \rho'_{aj}(g)$$
 $$ = \frac{1}{|G|} \sum_{h\in G}\rho_{ki}(h)^* \rho'_{lj}(hg) $$
@@ Line 118: / Line 111: @@
 \( z = hg \rightarrow h = zg^{-1}\)
-$$ = \frac{1}{|G|} \sum_{z\in G}\rho_{ki}(zg^{-1})^* \rho'_{lj}(z) = \frac{1}{|G|} \sum_{z\in G} \sum_{a=1}^m \rho_{ka}(z)^* \rho_{ai}(g^{-1})^* \rho'_{lj}(z) $$
+$$ = \frac{1}{|G|} \sum_{z\in G}\rho_{ki}(zg^{-1})^* \rho'_{lj}(z) = \frac{1}{|G|} \sum_{z\in G} \sum_{a=1}^n \rho_{ka}(z)^* \rho_{ai}(g^{-1})^* \rho'_{lj}(z) $$
-$$ = \sum_{a=1}^m \rho_{ai}(g) S^{(k,l)}{}_{aj} = \big([\rho](g)S^{(k,l)}\big)_{ij} $$
+$$ = \sum_{a=1}^n \rho_{ia}(g) S^{(k,l)}{}_{aj} = \big([\rho](g)S^{(k,l)}\big)_{ij} $$
 <p style="text-align:right;">\(\square\)</p>
-Claim 5: \( S^{(k,l)} = 0\)
+'''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:
+'''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 \( S^{(k,l)}v' = v\)
@@ Line 135: / Line 128: @@
 $$ S^{(k,l)}\rho'(g)v' =  \rho(g)S^{(k,l)}v' = \rho(g)0 = 0 \Rightarrow \rho'(g)v' \in Ker S^{(k,l)}$$
-Since \(\rho\) is irreducible eighter
+Since \(\rho\) is irreducible either
-(i)\(Im S^{(k,l)} = 0 \Rightarrow  S^{(k,l)} = 0 \Rightarrow \) the claim is proven
+(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
+(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 eighter
+: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\(\Rightarrow  S^{(k,l)} = 0 \)
+:(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 \(\Rightarrow S^{(k,l)} = 0\) because of Claim 4 and because \(\rho\) and \(\rho'\) are not isomorphic.
+:(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.
 <p style="text-align:right;">\(\square\)</p>
-$$ \Rightarrow \big(S^{(k,l)}\big)_{ij} = \frac{1}{|G|} \sum_{g\in G}\rho_{ki}(g)^*\rho'_{lj}(g) = \big( \rho_{ki}, \rho'_{lj} \big)_G = 0$$
+$$ \Rightarrow \big(S^{(k,l)}\big)_{ij} = \frac{1}{|G|} \sum_{g\in G}\rho_{ki}(g)^*\rho'_{lj}(g) = 0$$
 Now we can easly prove the second part of (a):
-$$(ch(\rho), ch(\rho'))_G = \frac{1}{|G|} \sum_{g\in G}  \sum_{i = 1}^n \rho_{ii}(g)^* \sum_{j = 1}^m \rho'_{ii}(g) =  \sum_{i = 1}^n \sum_{j = 1}^m (\rho_{ii}, \rho'_{jj})_G = 0$$
+$$(ch(\rho), ch(\rho'))_G  = \frac{1}{|G|} \sum_{g\in G}  \sum_{i = 1}^n \rho_{ii}(g) \sum_{j = 1}^m \rho'_{ii}(g)^* =  \sum_{i = 1}^n \sum_{j = 1}^m  \big(S^{(i,j)}\big)_{ij}^* = 0$$
 ===(b)===
-We can use Claim 3 to determine wether \(L^{V_s}\) is irreducible, in order to do this we have to determine the character.
-Claim 6: \( \chi_s\) \(s \in {1,...,k}\) is an othonormal basis of the class funcitons on G.
+'''Claim 6:''' \( \chi_s\) \(s \in \{1,...,k\}\) is an othonormal basis of the class funcitons on \(G\).
-Proof: We know that \( \frac{\sqrt{|G|}}{\sqrt{|C_j|}} \delta_{C_j}(g)\) \(j \in {1,...,k}\) 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.
+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.
-$$ ( \chi_s, \chi_t )_G = \Big( \sqrt{|G|}\sum_{j=1}^k \frac{v_{sj}}{\sqrt{|C_j|}}\delta_{C_j}(g), \sqrt{|G|}\sum_{j=1}^k \frac{v_{tj}}{\sqrt{|C_j|}}\delta_{C_j}(g) \Big)_G$$
+$$ ( \chi_s, \chi_t )_G = \Big( \sqrt{|G|}\sum_{i=1}^k \frac{v_{si}}{\sqrt{|C_i|}}\delta_{C_i}(g), \sqrt{|G|}\sum_{j=1}^k \frac{v_{tj}}{\sqrt{|C_j|}}\delta_{C_j}(g) \Big)_G$$
-$$ \sum_{i,j=1}^k v_{si}^* v_{tj} \Big( \frac{\sqrt{|G|}}{\sqrt{|C_i|}}\delta_{C_i}(g), \frac{\sqrt{|G|}}{\sqrt{|C_j|}}\delta_{C_j}(g) \Big)_G = \sum_{i,j=1}^k v_{si}^* v_{tj} \delta_{ij} = \sum_{j=1}^k v_{sj}^* v_{tj} = \delta_{st}$$
+$$ = \sum_{i,j=1}^k v_{si} v_{tj}^* \Big( \frac{\sqrt{|G|}}{\sqrt{|C_i|}}\delta_{C_i}(g), \frac{\sqrt{|G|}}{\sqrt{|C_j|}}\delta_{C_j}(g) \Big)_G = \sum_{i,j=1}^k v_{si} v_{tj}^* \delta_{ij} = \sum_{j=1}^k v_{sj} v_{tj}^* = \delta_{st}$$
 <p style="text-align:right;">\(\square\)</p>
-Claim 7: Every classfunction \(\in V_s\) is a multiple of \( \chi_s\)
+'''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 \( \Pi_s = V_s\).
+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.
+\( \Rightarrow V_s\) are orthogonal subspaces of \(V\). Observe that \( \chi_s = \chi_{se} \in V_s\) for every \(s\).
 Now consider a class function \(\psi \in V_s\):
@@ Line 179: / Line 171: @@
 \(\forall r \in \{ 1,...,k \}, r\neq s\):
-$$ (\psi, \chi_r)_G = \sum_{j = 1}^k a_j (\chi_j, \chi_r)_G = a_r = 0$$
+$$ 0=(\psi, \chi_r)_G = \sum_{j = 1}^k a_j (\chi_j, \chi_r)_G = a_r$$
 $$ \Rightarrow \psi =  a_s \chi_s$$
 <p style="text-align:right;">\(\square\)</p>
-Claim 8: \( ch(L^{V_s})(x) = \sqrt{|G|} v_{s1} \chi_s(x)\)
+'''Claim 8:''' \((\chi_{sy}, \chi_{sx})_G = \frac{1}{v_{s1}\sqrt{|G|}} \chi_{sy}(x)\)
-The character is a class function \( \Rightarrow ch(L^{V_s})(x) =\lambda \chi_s(x)\). Consider:
+Proof:
-$$ch(L^{V_s})(e) = dim V_s = tr \Pi_s $$
+$$(\chi_{sy}, \chi_{sx})_G = \frac{1}{|G|} \sum_{z\in G} \chi_{sy}(z) \chi_s(zx^{-1})^*$$
-because the trace of a projection is the dimension of it's target space.
+$$= \frac{1}{v_{s1}\sqrt{|G|}} \frac{1}{|G|} \sum_{z\in G} v_{s1}\sqrt{|G|} \chi_s(zx^{-1})^* \chi_{sy}(z) = \frac{1}{v_{s1}\sqrt{|G|}} \frac{1}{|G|} \sum_{z\in G} \Pi_s(z,x)^* \chi_{sy}(z)$$
-$$ = \frac{1}{|G|} \sum_{x\in G} \Pi_s(x,x) = \frac{1}{|G|} \sum_{x\in G} \sqrt{|G|} v_{s1}\chi_s(xx^{-1}) = \frac{1}{\sqrt{|G|}} v_{s1} \sum_{x\in G}  \chi_s(e) = \sqrt{|G|} v_{s1} \chi_s(e)$$
+Fourier Script last page: \(\Pi_s(z,x)^* = \Pi_s(x,z)\)
+$$= \frac{1}{v_{s1}\sqrt{|G|}} \frac{1}{|G|} \sum_{z\in G} \Pi_s(x,z) \chi_{sy}(z) = \frac{1}{v_{s1}\sqrt{|G|}} (\Pi_s \chi_{sy} )(x) = \frac{1}{v_{s1}\sqrt{|G|}} \chi_{sy}(x)$$
-$$ \Rightarrow ch(L^{V_s})(x) = \sqrt{|G|} v_{s1} \chi_s(x)$$
 <p style="text-align:right;">\(\square\)</p>
-Claim 9: \( ||ch(L^{V_s})||_2^2 = dim V_s\)
+'''Claim 9:''' \( ch(L^W) \in V_s\) (Konstantin)
-$$||ch(L^{V_s})||_2^2 = \frac{1}{|G|} \sum_{x\in G} |ch(L^{V_s})(x)|^2 = \frac{1}{|G|} \sum_{x\in G} |\sqrt{|G|} v_{s1} \chi_s(x)|^2 = {|G|} v_{s1}^2 \frac{1}{|G|} \sum_{x\in G} \chi_s(x)^*\chi_s(x)$$
+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
-$$ = {|G|} v_{s1}^2 (\chi_s,\chi_s)_G  = {|G|} v_{s1}^2 = dim V_s$$
+\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}
-Now we get to the actual proof:
+<p style="text-align:right;">\(\square\)</p>
-$$||ch(L^{V_s})||_2 = \sqrt{dim V_s}$$
+Now for the actual proof:
-If \( dim V_s = 1 \Rightarrow  ch(L^{V_s})(x) = \chi_s(x)\) and from Claim 3 we know that \(V_s\) is irreducible in this case W = V and we are done.
+The character is a class function and with Claim 7 and 9 \( \Rightarrow ch(L^W) =\lambda \chi_s\). Since \(W\) is irreducible:
-If \( dim V_s > 1 \) we know that \(V_s\) is not irreducible, and there are at least two irreducible subspaces of \(W, W', ...\)
+$$||ch(L^W)||_2^2 = \frac{1}{|G|}\sum_{g\in G} |\lambda \chi_s(x)|^2 = |\lambda|^2 \overset{!}{=} 1 \Rightarrow |\lambda| = 1 $$
-W as a subspace of \(V_s\) is still orthogonal to all the other subspaces \(V_r,\ r\in \{1,...,k\}, r\neq s\) so by the same argument as before \(ch(L^W)(x) = \lambda \chi_s(x)\). Because W is irreducible:
-$$||ch(L^{V_s})||_2^2 = ... = |\lambda|^2 \overset{!}{=} 1 \Rightarrow |\lambda| = 1 $$
 also
-$$ch(L^{V_s})(e) = ... = \lambda \sqrt{|G|} v_{s1} \overset{!}{=} dim W$$
+$$ dim W = ch(L^W)(e) = \lambda \chi_s(e) = \lambda \sqrt{|G|} v_{s1}$$
-since \(\sqrt{|G|} v_{s1} > 0\) (Fourier Script page 5) and \(dim W\) is a positive integer \( \Rightarrow \lambda = 1\)
+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)
-which concludes the proof.
+Now for the second part we wanto show that the characters are equivalent:
-===(c)===
+From part (a) follows that  \(L^{W_s}, L^{W_r}\) are non isomorphic for \(r\neq s\) since:
+$$  (ch(L^{W_s}), ch(L^{W_r}))_G = (\chi_s, \chi_r)_G = \delta_{rs} = 0$$
-From part (a) follows that  \(L^{W_s}, L^{W_r}\) are non isomorphic for \(r\neq s\):
-$$  (ch(L^{W_s}), ch(L^{W_r}))_G = (\chi_s, \chi_r)_G = \delta_{rs}$$
 \(ch(\rho)\) is a classfunction
@@ Line 232: / Line 224: @@
 $$\Rightarrow ch(\rho) = \sum_{j=1}^k a_j \chi_j$$
-$$ (ch(L^{W_s}), ch(\rho))_G = \sum_{j=1} a_j (\chi_s, \chi_j)_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}$$
+$$ (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 zero for more then one \(s\) because that would implie that \(L^{W_s}, L^{W_r}\) are isomporhic, it follows that for one particular s:
+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
 $$ ch(\rho) = \chi_s$$
 <p style="text-align:right;">\(\square\)</p>
-Note: this means that every irreducible representation of G is isomorphic to an irreducible representation that is found in the regular representation L.