Difference between revisions of "Aufgaben:Problem 2"
(→(b): shortening) |
m (→Solution: Took the liberty to set "Ker" and "Im" upright) |
||
| Line 47: | Line 47: | ||
===(c)=== | ===(c)=== | ||
Let \(v \in V\) such that, | Let \(v \in V\) such that, | ||
| − | $$ \pi v = 0 \Leftrightarrow v \in Ker\pi$$ | + | $$ \pi v = 0 \Leftrightarrow v \in \mathrm{Ker}\pi$$ |
for any \(g\in G\): | for any \(g\in G\): | ||
$$ \pi \circ \rho(g) v = \rho(g) \circ \pi v = \rho(g) 0 = 0$$ | $$ \pi \circ \rho(g) v = \rho(g) \circ \pi v = \rho(g) 0 = 0$$ | ||
the last step is true because \(\rho(g) \in GL(V)\). | the last step is true because \(\rho(g) \in GL(V)\). | ||
| − | $$\Rightarrow \rho(g)v \in Ker\pi$$ | + | $$\Rightarrow \rho(g)v \in \mathrm{Ker}\pi$$ |
===(d)=== | ===(d)=== | ||
| − | to show: \(Im\pi = U\) | + | to show: \(\mathrm{Im}\pi = U\) |
| − | \(Im\pi \subseteq U\): let \(\psi \in Im \pi \Rightarrow \exists \phi \in V, \pi \phi = \psi\) | + | \(\mathrm{Im}\pi \subseteq U\): let \(\psi \in \mathrm{Im} \pi \Rightarrow \exists \phi \in V, \pi \phi = \psi\) |
$$ \psi= \pi \phi = (\frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1}))(\phi) = \frac{1}{|G|}\sum_{g \in G} (\rho(g)\circ \pi_{0} \circ \rho(g^{-1}))(\phi)$$ | $$ \psi= \pi \phi = (\frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1}))(\phi) = \frac{1}{|G|}\sum_{g \in G} (\rho(g)\circ \pi_{0} \circ \rho(g^{-1}))(\phi)$$ | ||
| Line 64: | Line 64: | ||
$$ \psi= \frac{1}{|G|}\sum_{g \in G} \psi_g^* $$ | $$ \psi= \frac{1}{|G|}\sum_{g \in G} \psi_g^* $$ | ||
| − | as a linear combination of elements in U, \(\psi \in U\) because U is a subspace. \(\Rightarrow Im\pi \subseteq U\) | + | as a linear combination of elements in U, \(\psi \in U\) because U is a subspace. \(\Rightarrow \mathrm{Im}\pi \subseteq U\) |
| − | \(Im\pi \supseteq U\): let \(\psi \in U\) | + | \(\mathrm{Im}\pi \supseteq U\): let \(\psi \in U\) |
$$ \pi \psi = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1})\psi = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \psi_g$$ | $$ \pi \psi = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1})\psi = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \psi_g$$ | ||
| Line 73: | Line 73: | ||
$$ = \frac{1}{|G|}\sum_{g \in G} \rho(g) \psi_g = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \rho(g^{-1})\psi = \frac{1}{|G|}\sum_{g \in G} \rho(e) \psi = \frac{1}{|G|}\sum_{g \in G} e \psi = \psi$$ | $$ = \frac{1}{|G|}\sum_{g \in G} \rho(g) \psi_g = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \rho(g^{-1})\psi = \frac{1}{|G|}\sum_{g \in G} \rho(e) \psi = \frac{1}{|G|}\sum_{g \in G} e \psi = \psi$$ | ||
| − | \(\Rightarrow \psi \in Im \pi\) \(\Rightarrow U \subseteq Im \pi\) | + | \(\Rightarrow \psi \in \mathrm{Im} \pi\) \(\Rightarrow U \subseteq \mathrm{Im} \pi\) |
| − | \(\Rightarrow U = Im \pi\) | + | \(\Rightarrow U = \mathrm{Im} \pi\) |
| − | 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. | + | 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. |
Revision as of 10:05, 26 July 2015
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.
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)
$$ \pi^2 = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1}) \circ \frac{1}{|G|}\sum_{h \in G} \rho(h)\circ \pi_{0} \circ \rho(h^{-1})$$
$$ = \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1}) \circ \rho(h)\circ \pi_{0} \circ \rho(h^{-1})$$
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:
$$= \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(g)\circ \rho(g^{-1})\circ \rho(h)\circ \pi_{0} \circ \rho(h^{-1})$$
$$= \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(\mathbb{I})\circ \rho(h)\circ \pi_{0} \circ \rho(h^{-1}) = \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(h)\circ \pi_{0} \circ \rho(h^{-1})$$
$$ \stackrel{\sum_{g \in G} \mathbb{I} = |G|}{=} \frac{1}{|G|} \sum_{h \in G} \rho(h)\circ \pi_{0} \circ \rho(h^{-1}) = \pi$$
(c)
Let \(v \in V\) such that, $$ \pi v = 0 \Leftrightarrow v \in \mathrm{Ker}\pi$$
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 \(\rho(g) \in GL(V)\). $$\Rightarrow \rho(g)v \in \mathrm{Ker}\pi$$
(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 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. \(\Rightarrow \mathrm{Im}\pi \subseteq U\)
\(\mathrm{Im}\pi \supseteq U\): let \(\psi \in U\) $$ \pi \psi = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1})\psi = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \psi_g$$
where \(\psi_g := \rho(g^{-1})\psi \in U\), because U is invariant.
$$ = \frac{1}{|G|}\sum_{g \in G} \rho(g) \psi_g = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \rho(g^{-1})\psi = \frac{1}{|G|}\sum_{g \in G} \rho(e) \psi = \frac{1}{|G|}\sum_{g \in G} e \psi = \psi$$
\(\Rightarrow \psi \in \mathrm{Im} \pi\) \(\Rightarrow U \subseteq \mathrm{Im} \pi\)
\(\Rightarrow U = \mathrm{Im} \pi\)
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.
