# Aufgaben:Problem 2

## Note

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.

## 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)

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

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

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 $$V$$

### (c)

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.

### (d)

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.