Aufgaben:Problem 2

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You might want to check out page 33 here: [1]. This identity is used to prove Maschke's Theorem, so a Google search concerning this might help. - Cheers, A.


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.



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

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

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

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


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

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

We apply \(\pi^2\) to an element of \(V\): after the first projection we stay in the invarinat subspace \(U\) and thefore second projction has no effect:

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

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

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

Where \(Id\) denote the identity map on \(V\)


As the Kernel of any endomorphism is a subspace of \(V\) ( \( \neq \emptyset\), closed under scalar multiplication and addition) Ker\(\pi\) is a subspace of \(V\)

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

for any \(g\in G\):

$$\pi ( \rho(g) (v) ) = (\pi \circ \rho(g))(v) \overset{(a)}{=} (\rho(g) \circ \pi)(v)= \rho(g) (\pi (v)) = \rho(g) (0) = 0$$

the last step is true because \(\rho(g) \in GL(V)\). $$\Rightarrow \rho(g)(v) \in \mathrm{Ker}\pi$$

So Ker\(\pi\) is an invariant subspace.


to show: \(\mathrm{Im}\pi = U\)

\(\mathrm{Im}\pi \subseteq U\): Let \(\psi \in \mathrm{Im} \pi \Rightarrow \exists \phi \in V, \pi (\phi) = \psi\)

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

similar discussion as before: \(\rho(g^{-1})(\phi) := \phi_g \in V\) and \(\pi_{0}(\phi_g) := \psi_g \in U\) and now \(\rho(g)(\psi_g) := \psi_g^* \in U\) 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. \(\Rightarrow \mathrm{Im}\pi \subseteq U\)
\(\mathrm{Im}\pi \supseteq U\): Let \(\psi \in U\)

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

We always stay in the subspace \(U\) therefore the projection has no effect:

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

\(\Rightarrow \psi \in \mathrm{Im} \pi\) \(\Rightarrow U \subseteq \mathrm{Im} \pi\)

\(\Rightarrow U = \mathrm{Im} \pi\)

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