Difference between revisions of "Aufgaben:Problem 2"

Revision as of 13:05, 15 June 2015

Task

Let G be a fintite group Let \( \rho\ :G \rightarrow GL(V)\) be a representation on a finite dimensinal complex vectorspace V. Assume that U is an invariant subspace of V with \(U \neq \{0\},\ V\). Let W be any vector space complement of U in V. Let \(\pi_{0}\ :\ V \rightarrow V\) denote the projection of V onto U along W. Consider the linear map \(\pi\) defined by

$$ \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1})$$

a) Prove that \(\pi \circ \rho(g) = \rho(g) \circ \pi\) for all \(g \in G\).

b) Prove that \(\pi\) is a projection, i.e. \(\pi^2 = \pi\).

c) Prove that the kernel of \(\pi\) is an invariant subspace of V.

d) As a projection, \(\pi\) induces a decomposition \(V = Ker \pi \oplus Im \pi\). Prove that \(Im \pi = U\) and conclude that we have decomposed V into a direct sum of two invariant subspaces.

Solution

(a)

let \(h \in G\)

$$ \pi \circ \rho(h) = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1}) \circ \rho(h)$$

because \(\rho\) is a homomorphism

$$ = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1}h)$$

\(z = h^{-1} g \rightarrow g = hz\)

$$ = \frac{1}{|G|}\sum_{z \in G} \rho(hz)\circ \pi_{0} \circ \rho(z^{-1}) = \frac{1}{|G|}\sum_{z \in G} \rho(h) \circ \rho(z)\circ \pi_{0} \circ \rho(z^{-1}) = \rho(h) \circ \pi$$

(b)

$$ \pi^2 = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1}) \circ \frac{1}{|G|}\sum_{h \in G} \rho(h)\circ \pi_{0} \circ \rho(h^{-1})$$

$$ = \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1}) \circ \rho(h)\circ \pi_{0} \circ \rho(h^{-1})$$

$$ = \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1}h)\circ \pi_{0} \circ \rho(h^{-1})$$

now we can see what happens if we apply this to \(v \in V\). \(\rho(h^{-1}) v =: v_h \in V\). \(\pi_0 v_h =: u_h \in U\). \(\rho(g^{-1}h) u_h := u_{g,h} \in U\) because U is invariant. Finally \(\pi_0 u_{g,h} = u_{g,h}\) because \(u_{g,h}\) is already in U. Thus we can cancel the second projection as it does nothing.

$$ = \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(g)\circ \rho(g^{-1}h)\circ \pi_{0} \circ \rho(h^{-1}) = \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(g)\circ \rho(g^{-1})\circ \rho(h)\circ \pi_{0} \circ \rho(h^{-1})$$

$$ = \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(\mathbb{I})\circ \rho(h)\circ \pi_{0} \circ \rho(h^{-1}) = \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(h)\circ \pi_{0} \circ \rho(h^{-1})$$

$$ \stackrel{\sum_{g \in G} \mathbb{I} = |G|}{=} \frac{1}{|G|} \sum_{h \in G} \rho(h)\circ \pi_{0} \circ \rho(h^{-1}) = \pi$$

(c)

Let \(v \in V\) such that, $$ \pi v = 0 \Leftrightarrow v \in Ker\pi$$

for any \(g\in G\): $$ \pi \circ \rho(g) v = \rho(g) \circ \pi v = \rho(g) 0 = 0$$ the last step is true because \(\rho(g) \in GL(V)\). $$\Rightarrow \rho(g)v \in Ker\pi$$

(d)

to show: \(Im\pi = U\)

\(Im\pi \subseteq U\): let \(\psi \in Im \pi \Rightarrow \exists \phi \in V, \pi \phi = \psi\) $$ \psi= \pi \phi = (\frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1}))(\phi) = \frac{1}{|G|}\sum_{g \in G} (\rho(g)\circ \pi_{0} \circ \rho(g^{-1}))(\phi)$$

similar discussion as before: \(\rho(g^{-1})\phi := \phi_g \in V\) and \(\pi_{0}\phi_g := \psi_g \in U\) and now applying \(\rho(g)\psi_g := \psi_g^* \in U\) because U is invariant.

$$ \psi= \frac{1}{|G|}\sum_{g \in G} \psi_g^* $$

as a linear combination of elements in U, \(\psi \in U\) because U is a subspace. \(\Rightarrow Im\pi \subseteq U\)

\(Im\pi \supseteq U\): let \(\psi \in U\) $$ \pi \psi = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1})\psi = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \psi_g$$

where \(\psi_g := \rho(g^{-1})\psi \in U\), because U is invariant.

$$ = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \psi_g = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \rho(g^{-1})\psi = \frac{1}{|G|}\sum_{g \in G} \rho(e) \psi = \frac{1}{|G|}\sum_{g \in G} e \psi = \psi$$

\(\Rightarrow \psi \in Im \pi\) \(\Rightarrow U \subseteq Im \pi\)

\(\Rightarrow U = Im \pi\)

We conclude that \(V = Ker \pi \oplus Im \pi = Ker \pi \oplus U\) is indeed a decompsition of V into a direct sum of invariant subspaces.