Aufgaben:Problem 13
Contents
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
$$ ch(\rho) = \chi_s$$
\(\square\)
