Difference between revisions of "Aufgaben:Problem 2"
(Created page with "You might want to check out page 33 here: [http://math.mit.edu/~etingof/replect.pdf]. This identity is used to prove ''Maschke's Theorem'', so a Google search concerning this...") |
(→(c): I meant the empty set) |
||
| (24 intermediate revisions by 3 users not shown) | |||
| Line 1: | Line 1: | ||
| + | ==Note== | ||
| + | |||
You might want to check out page 33 here: [http://math.mit.edu/~etingof/replect.pdf]. This identity is used to prove ''Maschke's Theorem'', so a Google search concerning this might help. - Cheers, A. | You might want to check out page 33 here: [http://math.mit.edu/~etingof/replect.pdf]. 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: | ||
| + | |||
| + | $$= \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(g)\circ \rho(g^{-1})\circ \rho(h)\circ \pi_{0} \circ \rho(h^{-1})$$ | ||
| + | |||
| + | $$= \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(e)\circ \rho(h)\circ \pi_{0} \circ \rho(h^{-1}) = \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} Id \circ \rho(h)\circ \pi_{0} \circ \rho(h^{-1})$$ | ||
| + | |||
| + | $$= \frac{1}{|G|} \sum_{h \in G} \rho(h)\circ \pi_{0} \circ \rho(h^{-1}) = \pi$$ | ||
| + | |||
| + | 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. | ||
Latest revision as of 13:34, 30 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:
$$= \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(g)\circ \rho(g^{-1})\circ \rho(h)\circ \pi_{0} \circ \rho(h^{-1})$$
$$= \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(e)\circ \rho(h)\circ \pi_{0} \circ \rho(h^{-1}) = \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} Id \circ \rho(h)\circ \pi_{0} \circ \rho(h^{-1})$$
$$= \frac{1}{|G|} \sum_{h \in G} \rho(h)\circ \pi_{0} \circ \rho(h^{-1}) = \pi$$
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.
