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
$$ \chi : G\ \rightarrow \ \mathbb{C}$$ $$ g \mapsto \sqrt{|G|}\sum_{j=1}^k \frac{v_{sj}}{\sqrt{|C_j|}}\delta_{C_j}(g).$$
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)
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
$$ T^{(k,l)} := \Big(T^{(k,l)}{}_{i,j} \Big) = \Big(\frac{1}{|G|} \sum_{g\in G}\rho_{ki}(g)^*\rho_{lj}(g)\Big) = \Big((\rho_{ki},\rho_{lj})_G\Big)$$
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
$$ S^{(k,l)} := \Big(S^{(k,l)}{}_{ij} \Big) = \Big(\frac{1}{|G|} \sum_{g\in G}\rho_{ki}(g)^*\rho'_{lj}(g)\Big) $$
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 character. First prove claims 6.
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 on \( V_s\) 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 \(ch(L^W)(x) = \lambda \chi_s(x)\) 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)
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} (\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:
$$ ch(\rho) = \chi_s$$
\(\square\)
Note: this means that every irreducible representation of G is isomorphic to an irreducible representation that is found in the regular representation L.
