Difference between revisions of "Aufgaben:Problem 13"

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In the first part of the solution for problem one, the derivative w.r.t time of \(\xi\) is never used... it either be changed to the actual usecase, or left away as it is shown in the next steps.
 
  
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Also one may should show that \(\frac{d}{dt}\left(\int_0^t\phi\left( s\right)ds\right) = \phi\left( t\right)\)
+
==Task==
  
−
Regards Patrick
+
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
  
−
Thank you for your hint. My previous attempt used this derivative, so you're absolutely right, I can certainly delete that.
+
$$ \chi : G\ \rightarrow \ \mathbb{C}$$
−
I think the second statement is actually a formulation of the Fundamental Theorem of Calculus, so I will put that as a remark behind the step where I used it.
+
$$ g \mapsto \sqrt{|G|}\sum_{j=1}^k \frac{v_{sj}}{\sqrt{|C_j|}}\delta_{C_j}(g).$$
  
−
Cheers Valentin
+
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
  
−
Ah and I'm sorry I put an extra in \( x \) to clarify that the function is \( 2\pi-\)periodic in \( x \). Did forget to mention it.
+
$$ V_s := Span\{x\mapsto \chi_s(xy^{-1})| y\in G\}$$
  
−
Regards Patrick
+
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
  
−
In the Solution of a) after we know that the derivative of xi is less than or equal to zero, i would just say that since xi is continues it is monotonically decreasing and thus \( \xi(t) \leq \xi(0) \), the claim follows. "[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 22:06, 17 January 2015 (CET)"
+
$$(Tr(\rho), Tr(\rho'))_G = \begin{cases}  1, & \text{if \(\rho\) and \(\rho'\) are isomorphic} \\ 1, & \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)===
 +
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
 +
 
 +
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
 +
 
 +
$$ T^{(k,l)} := \Big(T^{(k,l)}{}_{i,j} \Big) = \Big(\frac{1}{|G|} \sum_{g\in G}\rho_{ki}(g)^*\rho_{lj}(g)\Big) \ \ , 1\leq k,l\leq n$$
 +
 
 +
 
 +
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: \( \rho_{ai}(g^{-1})^* = \rho_{ia}(g^{-1})^{T^*} = \rho_{ia}(g^{-1})^{-1} = \rho_{ia}(g)\)
 +
 
 +
$$ = \sum_{a=1}^n T^{(k,l)}{}_{aj} \rho_{ia}(g) =  \big(T^{(k,l)}[\rho](g)\big)_{ij}$$
 +
<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
 +
$$ABv = BAv = B0 = 0 \rightarrow Bv \in KernA$$
 +
 
 +
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 \( k = k \Rightarrow \lambda = \frac{1}{n} \Rightarrow T^{(k,k)} = \frac{1}{n} \mathbb{I} \)
 +
 
 +
$$\Rightarrow \Big(T^{(k,l)}{}_{ij} \Big) = \frac{1}{n} \delta_{lk} \delta_{ij}$$
 +
 
 +
$$ (\rho_{kl},\rho_{lj})_G = \frac{1}{n}\delta_{kl}\delta_{ij} $$
 +
<p style="text-align:right;">\(\square\)</p>
 +
 
 +
Claim 3: \( ||ch(\rho)||_2 = 1\)
 +
 
 +
Proof:
 +
 
 +
$$ ||ch(\rho)||_2^2 = \frac{1}{|G|}\sum_{g\in G} |ch(\rho)|^2 = \frac{1}{|G|}\sum_{g\in G} tr[\rho](g)^*tr[\rho](g) = \frac{1}{|G|}\sum_{g\in G} \sum_{i=1}^n \rho_{ii}(g)^* \sum_{j=1}^n \rho_{jj}(g)$$
 +
 
 +
$$ = \sum_{i,j=1}^n (\rho_{ii}, \rho_{jj})_G = \sum_{i,j=1}^n \frac{1}{n}  \delta_{ij} = 1$$
 +
<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 \(\rho(g) = \phi^{-1} \circ \rho'(g) \circ \phi \).
 +
 
 +
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
 +
 
 +
$$ S^{(k,l)} := \Big(S^{(k,l)}{}_{ij} \Big) = \Big(\frac{1}{|G|} \sum_{g\in G}\rho_{ki}(g)^*\rho'_{lj}(g)\Big) $$
 +
 
 +
for \(1\leq k\leq n\) and \( 1\leq l\leq m\) 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:
 +
 
 +
$$\big(S^{(k,l)}[\rho'](g)\big)_{ij} = \sum_{a=1}^m S^{(k,l)}{}_{ia} \rho'_{aj}(g) = \sum_{a=1}^n \frac{1}{|G|} \sum_{h\in G}\rho_{ki}(h)^*\rho'_{la}(h) \rho'_{aj}(g)$$
 +
 
 +
$$ = \frac{1}{|G|} \sum_{h\in G}\rho_{ki}(h)^* \rho'_{lj}(hg) $$
 +
 
 +
\( z = hg \rightarrow h = zg^{-1}\)
 +
 
 +
$$ = \frac{1}{|G|} \sum_{z\in G}\rho_{ki}(zg^{-1})^* \rho'_{lj}(z) = \frac{1}{|G|} \sum_{z\in G} \sum_{a=1}^m \rho_{ka}(z)^* \rho_{ai}(g^{-1})^* \rho'_{lj}(z) $$
 +
 
 +
$$ = \sum_{a=1}^m \rho_{ai}(g) S^{(k,l)}{}_{aj} = \big([\rho](g)S^{(k,l)}\big)_{ij} $$
 +
<p style="text-align:right;">\(\square\)</p>
 +
 
 +
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\(\Rightarrow  S^{(k,l)} = 0 \)
 +
 
 +
(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.
 +
<p style="text-align:right;">\(\square\)</p>
 +
 
 +
 
 +
$$ \Rightarrow \big(S^{(k,l)}\big)_{ij} = \frac{1}{|G|} \sum_{g\in G}\rho_{ki}(g)^*\rho'_{lj}(g) = \big( \rho_{ki}, \rho'_{lj} \big)_G = 0$$
 +
 
 +
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 charackter. First prove claims 6 and 7.
 +
 
 +
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 classfunctions. And \( v_{s} \) \(s \in {1,...,k}\) are orhonormal w.r.t the standard complex inner product.
 +
 
 +
$$ ( \chi_s, \chi_t )_G = \Big( \sqrt{|G|}\sum_{j=1}^k \frac{v_{sj}}{\sqrt{|C_j|}}\delta_{C_j}(g), \sqrt{|G|}\sum_{j=1}^k \frac{v_{tj}}{\sqrt{|C_j|}}\delta_{C_j}(g) \Big)_G$$
 +
 
 +
$$ \sum_{i,j=1}^k v_{si}^* v_{tj} \Big( \frac{\sqrt{|G|}}{\sqrt{|C_i|}}\delta_{C_i}(g), \frac{\sqrt{|G|}}{\sqrt{|C_j|}}\delta_{C_j}(g) \Big)_G = \sum_{i,j=1}^k v_{si}^* v_{tj} \delta_{ij} = \sum_{j=1}^k v_{sj}^* v_{tj} = \delta_{st}$$
 +
<p style="text-align:right;">\(\square\)</p>
 +
 
 +
Remember the definition
 +
 
 +
$$ V_s := Span\{x\mapsto \chi_s(xy^{-1})| y\in G\}$$
 +
 
 +
Claim 7: Every classfunction on \( V_s\) is a multiplol 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 on \(V_s\):
 +
 
 +
$$ \psi = \sum_{j = 1}^k a_j \chi_j, \quad a_j \in \mathbb{C} $$
 +
 
 +
\(\forall r \in \{ 1,...,k \}, r\neq s\):
 +
 
 +
$$ (\psi, \chi_r)_G = \sum_{j = 1}^k a_j (\chi_j, \chi_r)_G = a_r = 0$$
 +
$$ \Rightarrow \psi =  a_s \chi_s$$
 +
 
 +
<p style="text-align:right;">\(\square\)</p>
 +
 
 +
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.
 +
 
 +
$$ = \frac{1}{|G|} \sum_{x\in G} \Pi_s(x,x) = \frac{1}{|G|} \sum_{x\in G} \sqrt{|G|} v_{s1}\chi_s(xx^{-1}) = \frac{1}{\sqrt{|G|}} v_{s1} \sum_{x\in G}  \chi_s(e) = \sqrt{|G|} v_{s1} \chi_s(e)$$
 +
 
 +
$$ \Rightarrow ch(L^{V_s})(x) = \sqrt{|G|} v_{s1} \chi_s(x)$$
 +
<p style="text-align:right;">\(\square\)</p>
 +
 
 +
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 \( dim V_s = 1 \Rightarrow  ch(L^{V_s})(x) = \sqrt{|G|} v_{s1} \chi_s(x) =  \chi_s(x)\) 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 invariant, 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:
 +
 
 +
$$|ch(L^{V_s})||_2^2 = ... = \lambda^2 \overset{!}{=} 1$$
 +
 
 +
because W is now irreducible. Also
 +
 
 +
$$ch(L^{V_s})(e) = ... = \lambda \sqrt{|G|} v_{s1} \overset{!}{=} dim W$$
 +
 
 +
since \(\sqrt{|G|} v_{s1} \) is positive (Fourier Script at the very end) and \(dim W\) is a positive integer \( \Rightarrow \lambda = 1\)
 +
 
 +
which concludes the proof.
 +
 
 +
===(c)===
 +
\(ch(\rho)\) is a classfunction on G
 +
 
 +
$$\Rightarrow ch(\rho) = \sum_{j=1}^k a_j \chi_j$$
 +
 
 +
$$ (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} \\  1, & \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
 +
 
 +
$$ ch(\rho) = \chi_s$$
 +
 
 +
for one partivular 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.

Revision as of 19:24, 13 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} \\ 1, & \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)

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

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

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


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

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

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

\(\square\)

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

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

for \(1\leq k\leq n\) and \( 1\leq l\leq m\) 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\(\Rightarrow S^{(k,l)} = 0 \)

(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.

\(\square\)


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

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 charackter. First prove claims 6 and 7.

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 classfunctions. And \( v_{s} \) \(s \in {1,...,k}\) are orhonormal w.r.t the standard complex inner product.

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

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

\(\square\)

Remember the definition

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

Claim 7: Every classfunction on \( V_s\) is a multiplol 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 on \(V_s\):

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

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

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

\(\square\)

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 \( dim V_s = 1 \Rightarrow ch(L^{V_s})(x) = \sqrt{|G|} v_{s1} \chi_s(x) = \chi_s(x)\) 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 invariant, 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:

$$|ch(L^{V_s})||_2^2 = ... = \lambda^2 \overset{!}{=} 1$$

because W is now irreducible. Also

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

since \(\sqrt{|G|} v_{s1} \) is positive (Fourier Script at the very end) and \(dim W\) is a positive integer \( \Rightarrow \lambda = 1\)

which concludes the proof.

(c)

\(ch(\rho)\) is a classfunction on G

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

$$ (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} \\ 1, & \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

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

for one partivular 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.

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