Difference between revisions of "Aufgaben:Problem 13"

Revision as of 11:19, 27 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 $$ABv = BAv = B0 = 0 \Rightarrow Bv \in KerA$$

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

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

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

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:

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

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

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

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

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

\(\square\)

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: \((\chi_{sy}, \chi_{sx})_G = \frac{1}{v_{s1}\sqrt{|G|}} \chi_{sy}(x)\)

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: \( ch(L^{V_s}) \in V_s\) (Konstantin)

Proof: Choose an orthomnormal basis \(e_1,\dots e_n\) of \(V_s\ w.r.t\ (\ ,\ )_G\Rightarrow e_i = \sum_{y\in G} \lambda_{iy} \chi_{sy}\)

\begin{align} ch(L^{V_s})(g) &= tr(L^{V_s}(g)) \\ &= \sum_{j= 1}^n (e_j, L^{V_s}(g) e_j)_G = \sum_{j= 1}^n (\sum_{z\in G} \lambda_{iz} \chi_{sz}, L^{V_s}(g) \sum_{y\in G} \lambda_{iy} \chi_{sy} )_G = \sum_{j= 1}^n \sum_{z\in G} \sum_{y\in G} \lambda_{iz} \lambda_{iy}^* (\chi_{sz}, L^{V_s}(g) \chi_{sy} )_G \\ &= \sum_{j= 1}^n \sum_{z\in G} \sum_{y\in G} \lambda_{iz} \lambda_{iy}^* (\chi_{sz}, \chi_{sgy} )_G = \sum_{j= 1}^n \sum_{z\in G} \sum_{y\in G} \lambda_{iz} \lambda_{iy}^* \chi_{sz}(gy) = \sum_{j= 1}^n \sum_{z\in G} \sum_{y\in G} \lambda_{iz} \lambda_{iy}^* \chi_{szy^{-1}}(g) \\ \end{align}

\(\square\)

Claim 10: \( ch(L^{V_s})(x) = \sqrt{|G|} v_{s1} \chi_s(x)\)

The character is a class function and with Claim 9 \( \Rightarrow ch(L^{V_s})(x) =\lambda \chi_s(x)\). 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 11: \( ||ch(L^{V_s})||_2^2 = dim V_s\)

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

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

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 \(W, W', ...\)

\(W\) is a subspace of \(V_s\), 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 \(L\).