Difference between revisions of "Aufgaben:Problem 2"

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(→‎(c): I meant the empty set)
 
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==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.
 +
 
==Task==
 
==Task==
  
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==Solution==
 
==Solution==
  
−
(a) let \(h \in G\)
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===(a)===
 +
let \(h \in G\)
  
 
$$ \pi \circ \rho(h) = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1})  \circ \rho(h)$$
 
$$ \pi \circ \rho(h) = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1})  \circ \rho(h)$$
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$$ = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1}h)$$
 
$$ = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1}h)$$
  
−
\(z = g^{-1} h \rightarrow g = h z^{-1}\), does not change the sum because inversion and group operation with h are bijections on the group
+
\(z = h^{-1} g \rightarrow g = hz\)
  
−
$$ = \frac{1}{|G|}\sum_{z \in G} \rho(h z^{-1})\circ \pi_{0} \circ \rho(z) = \frac{1}{|G|}\sum_{z \in G} \rho(h) \circ \rho(z^{-1})\circ \pi_{0} \circ \rho(z)$$
+
$$ = \frac{1}{|G|}\sum_{z \in G} \rho(hz)\circ \pi_{0} \circ \rho(z^{-1}) = \frac{1}{|G|}\sum_{z \in G} \rho(h) \circ \rho(z)\circ \pi_{0} \circ \rho(z^{-1}) = \rho(h) \circ \pi$$
  
−
\( y = z^{-1}\)
+
===(b)===
 +
$$ \pi^2 = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1}) \circ \frac{1}{|G|}\sum_{h \in G} \rho(h)\circ \pi_{0} \circ \rho(h^{-1})$$
  
−
$$ = \rho(h) \circ \frac{1}{|G|}\sum_{y \in G} \rho(y)\circ \pi_{0} \circ \rho(y^{-1}) = \rho(h) \circ \pi$$
+
$$ = \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1}) \circ \rho(h)\circ \pi_{0} \circ \rho(h^{-1})$$
  
−
(b)
+
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:
  
−
$$ \pi^2 = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1}) \circ \frac{1}{|G|}\sum_{h \in G} \rho(h)\circ \pi_{0} \circ \rho(h^{-1})$$
+
$$= \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(g)\circ \rho(g^{-1})\circ \rho(h)\circ \pi_{0} \circ \rho(h^{-1})$$  
  
−
$$ = \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1}) \circ \rho(h)\circ \pi_{0} \circ \rho(h^{-1})$$
+
$$= \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(e)\circ \rho(h)\circ \pi_{0} \circ \rho(h^{-1}) = \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} Id \circ \rho(h)\circ \pi_{0} \circ \rho(h^{-1})$$
  
−
$$ = \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1}h)\circ \pi_{0} \circ \rho(h^{-1})$$
+
$$= \frac{1}{|G|} \sum_{h \in G} \rho(h)\circ \pi_{0} \circ \rho(h^{-1}) = \pi$$
  
−
now we can see what happens if we apply this to \(v \in V\). We can think about it element wise because \(\rho(G) \in GL(V)\) and \(\pi_0\) as a projection are both linear. \(\rho(h^{-1}) v\) will just give back a diffrent \(v_h \in V\). \(\pi_0 v_h := u_h\) with \(u_h \in U\), applying \(\rho(g^{-1}h) u_h := u_{g,h}\) with \(u_{g,h} \in U\) because U is invariant. Now applying \(\pi_0 u_{g,h} = u_{g,h}\) because \(u_{g,h}\) is already in U. Thus we can cancel the second projection as it does nothing.
+
Where \(Id\) denote the identity map on \(V\)
  
−
$$ = \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(g)\circ \rho(g^{-1}h)\circ \pi_{0} \circ \rho(h^{-1}) = \frac{1}{|G|^2}\sum_{g \in G} \sum_{h \in G} \rho(h)\circ \pi_{0} \circ \rho(h^{-1})$$
+
===(c)===
  
−
$$ = \frac{1}{|G|} \sum_{h \in G} \rho(h)\circ \pi_{0} \circ \rho(h^{-1}) = \pi$$
+
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 \(V\)
  
−
(c)
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Let \(v \in V\) such that,
 +
$$ \pi (v) = 0 \Leftrightarrow v \in \mathrm{Ker}\pi$$
  
−
let \(v \in V\) such that ,
+
for any \(g\in G\):
−
$$ \pi v = 0 \Leftrightarrow v \in Ker\pi$$
+
−
using (a):
+
−
$$  \pi \circ \rho(g) v = \rho(g) \circ \pi v = \rho(g) 0 = 0$$
+
−
the last step is true because \(\rho\) is a homomorphism.
+
−
$$\Rightarrow \forall g \in G \ \ \rho(g) \in Ker\pi$$
+
  
−
(d) to show: \(Im\pi = U\)
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$$\pi ( \rho(g) (v) ) =  (\pi \circ \rho(g))(v) \overset{(a)}{=}    (\rho(g) \circ \pi)(v)= \rho(g) (\pi (v)) = \rho(g) (0) = 0$$
  
−
forward dircetion: let \(\psi \in Im \pi \Rightarrow \exists \phi \in V, \pi \phi = \psi\)
+
the last step is true because \(\rho(g) \in GL(V)\).
−
$$ \psi= \pi \phi =  (\frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1}))(\phi) = \frac{1}{|G|}\sum_{g \in G} (\rho(g)\circ \pi_{0} \circ \rho(g^{-1}))(\phi)$$
+
$$\Rightarrow \rho(g)(v) \in \mathrm{Ker}\pi$$
  
−
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.
+
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.
  
 
$$ \psi= \frac{1}{|G|}\sum_{g \in G} \psi_g^* $$
 
$$ \psi= \frac{1}{|G|}\sum_{g \in G} \psi_g^* $$
  
−
as a linear combination of elements in U, \(\psi \in U\) because U is a subspace.
+
::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 \(\psi \in U\)
 +
 
 +
$$ \pi (\psi) = \frac{1}{|G|}\sum_{g \in G} (\rho(g)\circ \pi_{0} \circ \rho(g^{-1}))(\psi) $$
  
−
backwards dircetion: let \(\psi \in U\)
+
::We always stay in the subspace \(U\) therefore the projection has no effect:
−
$$ \pi \psi = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \rho(g^{-1})\psi = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \pi_{0} \circ \psi_g$$
+
  
−
where \(\psi_g := \rho(g^{-1})\psi\), now because \(\psi_g \in U\) as U is invariant, the projection does nothing:
+
$$=  \frac{1}{|G|}\sum_{g \in G} (\rho(g) \circ \rho(g^{-1})) (\psi)= \frac{1}{|G|}\sum_{g \in G} \rho(e) \psi = \frac{1}{|G|}\sum_{g \in G} Id( \psi) = \psi$$
  
−
$$ = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \psi_g = \frac{1}{|G|}\sum_{g \in G} \rho(g)\circ \rho(g^{-1})\psi = \frac{1}{|G|}\sum_{g \in G} \rho(e) \psi = \frac{1}{|G|}\sum_{g \in G} e \psi = \psi$$
+
::\(\Rightarrow \psi \in \mathrm{Im} \pi\) \(\Rightarrow U \subseteq \mathrm{Im} \pi\)
  
−
\(\Rightarrow \psi \in Im \pi\)
+
\(\Rightarrow U = \mathrm{Im} \pi\)
  
−
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.

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.

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