Difference between revisions of "Aufgaben:Problem 13"

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m (corrections)
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===(a)===
 
===(a)===
let \(v_1,...,v_n, n=dim_{\mathbb{C}}V ,\ v'_1,...,v'_m, m=dim_{\mathbb{C}}V'\) be the orthonormal bases of V, V' respectivly
+
Choose an orthonormal basis of  \(V, V'\) with dimension \(n, m\) respectively. Let
  
let
 
 
$$[\rho](g) := \Big(\rho_{ij}(g)\Big)_{1\leq i,j \leq n}$$
 
$$[\rho](g) := \Big(\rho_{ij}(g)\Big)_{1\leq i,j \leq n}$$
 
$$[\rho'](g) := \Big(\rho'_{ij}(g)\Big)_{1\leq i,j \leq m}$$
 
$$[\rho'](g) := \Big(\rho'_{ij}(g)\Big)_{1\leq i,j \leq m}$$
  
be the matrices of \(\rho\) and \(\rho'\) relative to these bases.
+
be the matrices of \(\rho\) and \(\rho'\) relative to these bases. Set
 
+
set
+
 
+
$$ T^{(k,l)} := \Big(T^{(k,l)}{}_{i,j} \Big) = \Big(\frac{1}{|G|} \sum_{g\in G}\rho_{ki}(g)^*\rho_{lj}(g)\Big) = \Big((\rho_{ki},\rho_{lj})_G\Big)$$
+
 
+
for \( 1\leq k,l\leq n\)
+
  
 +
$$ T^{(k,l)} := \Big(T^{(k,l)}{}_{i,j} \Big) = \Big(\frac{1}{|G|} \sum_{g\in G}\rho_{ki}(g)^*\rho_{lj}(g)\Big) = \Big((\rho_{ki},\rho_{lj})_G\Big)\  \text{  for} \ 1\leq k,l\leq n$$
  
 
Claim 1 : \([T^{(k,l)}, [\rho](g)] = 0\)
 
Claim 1 : \([T^{(k,l)}, [\rho](g)] = 0\)
Line 69: Line 63:
 
Claim 2: \((\rho_{kl},\rho_{lj})_G = \frac{1}{n}\delta_{kl}\delta_{ij}\)
 
Claim 2: \((\rho_{kl},\rho_{lj})_G = \frac{1}{n}\delta_{kl}\delta_{ij}\)
  
Proof: let \(\lambda\) be an eigenvalue of \(T^{(k,l)}\)
+
Proof: Let \(\lambda\) be an eigenvalue of \(T^{(k,l)}\)
 
$$ [T^{(k,l)} -\lambda \mathbb{I}, [\rho](g)] = 0$$
 
$$ [T^{(k,l)} -\lambda \mathbb{I}, [\rho](g)] = 0$$
 
since the identity commutes with everything.
 
since the identity commutes with everything.
Line 75: Line 69:
 
$$ \Rightarrow Ker(T^{(k,l)} -\lambda \mathbb{I}) \neq \{0\}$$
 
$$ \Rightarrow Ker(T^{(k,l)} -\lambda \mathbb{I}) \neq \{0\}$$
  
is an invariant subspace of \(\rho\) and because \(\rho\) is irreducible is equal to \(V\) \(\Rightarrow T^{(k,l)} = \lambda \mathbb{I}\)
+
is an invariant subspace of \(\rho\) and because \(\rho\) is irreducible is equal to \(V\)
  
$$ \lambda n = tr T^{(k,l)} = \sum_{j=1}^n \frac{1}{|G|}\sum_{g\in G} \rho_{kj}(g)^*\rho_{lj}(g) = \sum_{j=1}^n \frac{1}{|G|}\sum_{g\in G} \rho_{jk}(g^{-1})\rho_{lj}(g) = \frac{1}{|G|}\sum_{g\in G} \rho_{lk}(e) =\delta_{lk}$$
+
$$\Rightarrow T^{(k,l)} = \lambda \mathbb{I}$$
  
because \( \rho(e) = e = \mathbb{I}\).
+
$$ \lambda n = tr T^{(k,l)} = \sum_{j=1}^n \frac{1}{|G|}\sum_{g\in G} \rho_{kj}(g)^*\rho_{lj}(g) = \sum_{j=1}^n \frac{1}{|G|}\sum_{g\in G} \rho_{jk}(g^{-1})\rho_{lj}(g) = \frac{1}{|G|}\sum_{g\in G} \rho_{lk}(e) =\delta_{lk}$$
  
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 \( k = k \Rightarrow \lambda = \frac{1}{n} \Rightarrow T^{(k,k)} = \frac{1}{n} \mathbb{I} \)
  
 
$$\Rightarrow \Big(T^{(k,l)}{}_{ij} \Big) = \frac{1}{n} \delta_{lk} \delta_{ij}$$
 
$$\Rightarrow \Big(T^{(k,l)}{}_{ij} \Big) = \frac{1}{n} \delta_{lk} \delta_{ij}$$
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$$ \Rightarrow (ch(\rho)(g), ch(\rho')(g))_G = (ch(\rho)(g), ch(\rho)(g))_G = ||ch(\rho)||_2^2 = 1$$
 
$$ \Rightarrow (ch(\rho)(g), ch(\rho')(g))_G = (ch(\rho)(g), ch(\rho)(g))_G = ||ch(\rho)||_2^2 = 1$$
  
now for the second part: from now on let \(\rho, \rho'\) be not isomorphic
+
Now for the second part: from now on let \(\rho, \rho'\) be not isomorphic. Set
 
+
set
+
  
$$ S^{(k,l)} := \Big(S^{(k,l)}{}_{ij} \Big) = \Big(\frac{1}{|G|} \sum_{g\in G}\rho_{ki}(g)^*\rho'_{lj}(g)\Big) $$
+
$$ S^{(k,l)} := \Big(S^{(k,l)}{}_{ij} \Big) = \Big(\frac{1}{|G|} \sum_{g\in G}\rho_{ki}(g)^*\rho'_{lj}(g)\Big) \text{  } 1\leq k\leq n,\ 1\leq l\leq m$$
  
for \(1\leq k\leq n\) and \( 1\leq l\leq m\) which is the matrix of a linear transform from V' to V
+
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: \( S^{(k,l)}[\rho'](g)= [\rho](g)S^{(k,l)}\)
Line 153: Line 145:
 
(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\(\Rightarrow  S^{(k,l)} = 0 \)
  
(vi)\(Ker S^{(k,l)} = 0 \Rightarrow S^{(k,l)}\) 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 \(\Rightarrow S^{(k,l)} = 0\) because of Claim 4 and because \(\rho\) and \(\rho'\) are not isomorphic.
 
<p style="text-align:right;">\(\square\)</p>
 
<p style="text-align:right;">\(\square\)</p>
  
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<p style="text-align:right;">\(\square\)</p>
 
<p style="text-align:right;">\(\square\)</p>
  
Claim 7: Every classfunction on \( V_s\) 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 \( \Pi_s = V_s\).
Line 209: Line 201:
 
$$||ch(L^{V_s})||_2^2 = \frac{1}{|G|} \sum_{x\in G} |ch(L^{V_s})(x)|^2 = \frac{1}{|G|} \sum_{x\in G} |\sqrt{|G|} v_{s1} \chi_s(x)|^2 = {|G|} v_{s1}^2 \frac{1}{|G|} \sum_{x\in G} \chi_s(x)^*\chi_s(x)$$
 
$$||ch(L^{V_s})||_2^2 = \frac{1}{|G|} \sum_{x\in G} |ch(L^{V_s})(x)|^2 = \frac{1}{|G|} \sum_{x\in G} |\sqrt{|G|} v_{s1} \chi_s(x)|^2 = {|G|} v_{s1}^2 \frac{1}{|G|} \sum_{x\in G} \chi_s(x)^*\chi_s(x)$$
  
$$ = {|G|} v_{s1}^2 (\chi_s,\chi_s)_G  = {|G|} |v_{s1}|^2 = dim V_s$$
+
$$ = {|G|} v_{s1}^2 (\chi_s,\chi_s)_G  = {|G|} v_{s1}^2 = dim V_s$$
  
 
Now we get to the actual proof:  
 
Now we get to the actual proof:  
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$$||ch(L^{V_s})||_2 = \sqrt{dim V_s}$$
 
$$||ch(L^{V_s})||_2 = \sqrt{dim V_s}$$
  
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.
+
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.
  
if \( dim V_s > 1 \) we know that \(V_s\) is not irreducible, and there are at least two irreducible subspaces of W, W', ...
+
If \( dim V_s > 1 \) we know that \(V_s\) is not irreducible, and there are at least two irreducible subspaces of \(W, W', ...\)
  
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 \(ch(L^W)(x) = \lambda \chi_s(x)\) again we look at the norm of the characters:
+
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:
  
$$||ch(L^{V_s})||_2^2 = ... = \lambda^2 \overset{!}{=} 1$$
+
$$||ch(L^{V_s})||_2^2 = ... = |\lambda|^2 \overset{!}{=} 1 \Rightarrow |\lambda| = 1 $$
  
because W is now irreducible. Also
+
also
  
 
$$ch(L^{V_s})(e) = ... = \lambda \sqrt{|G|} v_{s1} \overset{!}{=} dim W$$
 
$$ch(L^{V_s})(e) = ... = \lambda \sqrt{|G|} v_{s1} \overset{!}{=} dim W$$
  
since \(\sqrt{|G|} v_{s1} \) is positive (Fourier Script at the very end) 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 \( \Rightarrow \lambda = 1\)
  
 
which concludes the proof.
 
which concludes the proof.
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$$\Rightarrow ch(\rho) = \sum_{j=1}^k a_j \chi_j$$
 
$$\Rightarrow ch(\rho) = \sum_{j=1}^k a_j \chi_j$$
  
$$ (ch(L^{W_s}), ch(\rho))_G = \sum_{j=1} (\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(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}$$
  
since \(a_s\) can't be zero for all the \(s\in \{1,...,k\}\) and can't be zero for more then one 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 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:
  
 
$$ ch(\rho) = \chi_s$$
 
$$ ch(\rho) = \chi_s$$

Revision as of 17:33, 14 June 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

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

2. For any represntation \(\rho\) we defines the map \(ch(\rho):x\rightarrow Tr(\rho(x))\).

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) 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 \(ch(L^W) = \chi_s\).

(c) 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

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

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: \( \rho_{ai}(g^{-1})^* = \rho_{ia}(g^{-1})^{T^*} = \rho_{ia}(g^{-1})^{-1} = \rho_{ia}(g)\)

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

\(\square\)

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

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

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 \( k = k \Rightarrow \lambda = \frac{1}{n} \Rightarrow T^{(k,k)} = \frac{1}{n} \mathbb{I} \)

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

\(\square\)

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

Proof:

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

$$ = \sum_{i,j=1}^n (\rho_{ii}, \rho_{jj})_G = \sum_{i,j=1}^n \frac{1}{n} \delta_{ij} = 1$$

\(\square\)

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.

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

using \( tr(AB) = tr(BA) \)

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

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

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:

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

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

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

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

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

\(\square\)

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 eighter

(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 eighter

(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.

\(\square\)


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

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

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

(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.

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.

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

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

\(\square\)

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 \(\psi \in V_s\):

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

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

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

\(\square\)

Claim 8: \( ch(L^{V_s})(x) = \sqrt{|G|} v_{s1} \chi_s(x)\)

The character is a class function \( \Rightarrow ch(L^{V_s})(x) =\lambda \chi_s(x)\). Consider:

$$ch(L^{V_s})(e) = dim V_s = tr \Pi_s $$

because the trace of a projection is the dimension of it's target space.

$$ = \frac{1}{|G|} \sum_{x\in G} \Pi_s(x,x) = \frac{1}{|G|} \sum_{x\in G} \sqrt{|G|} v_{s1}\chi_s(xx^{-1}) = \frac{1}{\sqrt{|G|}} v_{s1} \sum_{x\in G} \chi_s(e) = \sqrt{|G|} v_{s1} \chi_s(e)$$

$$ \Rightarrow ch(L^{V_s})(x) = \sqrt{|G|} v_{s1} \chi_s(x)$$

\(\square\)

Claim 9: \( ||ch(L^{V_s})||_2^2 = dim V_s\)

$$||ch(L^{V_s})||_2^2 = \frac{1}{|G|} \sum_{x\in G} |ch(L^{V_s})(x)|^2 = \frac{1}{|G|} \sum_{x\in G} |\sqrt{|G|} v_{s1} \chi_s(x)|^2 = {|G|} v_{s1}^2 \frac{1}{|G|} \sum_{x\in G} \chi_s(x)^*\chi_s(x)$$

$$ = {|G|} v_{s1}^2 (\chi_s,\chi_s)_G = {|G|} v_{s1}^2 = dim V_s$$

Now we get to the actual proof:

$$||ch(L^{V_s})||_2 = \sqrt{dim V_s}$$

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.

If \( dim V_s > 1 \) we know that \(V_s\) is not irreducible, and there are at least two irreducible subspaces of \(W, W', ...\)

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

$$ch(L^{V_s})(e) = ... = \lambda \sqrt{|G|} v_{s1} \overset{!}{=} dim W$$

since \(\sqrt{|G|} v_{s1} > 0\) (Fourier Script page 5) and \(dim W\) is a positive integer \( \Rightarrow \lambda = 1\)

which concludes the proof.

(c)

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

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

$$ (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}$$

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:

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

\(\square\)

Note: this means that every irreducible representation of G is isomorphic to an irreducible representation that is found in the regular representation L.