Difference between revisions of "Aufgaben:Problem 6"
m (inserted task) |
(→a) |
||
| Line 19: | Line 19: | ||
== Problem 6 (Textbook) == | == Problem 6 (Textbook) == | ||
===a=== | ===a=== | ||
| − | From \([A, \rho(g)] = 0 ~ \forall g \) it follows that \(Ker(A)\) is an invariant subspace of \(V\) and since the representation is irreducible that \(A=0=0\cdot | + | From \([A, \rho(g)] = 0 ~ \forall g \) it follows that \(Ker(A)\) is an invariant subspace of \(V\) and since the representation is irreducible that \(A=0=0\cdot \mathbb{I}\) or \(A\) is an isomorphism. Suppose the second case: \(A\) has an eigenvalue \(\lambda \in \mathbb{C}\), therefor \(A-\lambda \cdot \mathbb{I}\) is not injective. But it holds too that \([A-\lambda \cdot \mathbb{I}, \rho(g)] = 0 \; \forall g\), and since \(Ker(A-\lambda \mathbb{I}) \neq \{0\}\) it follows that \(A - \lambda \cdot \mathbb{I} = 0 \) |
\(\blacksquare\) | \(\blacksquare\) | ||
| − | |||
===b=== | ===b=== | ||
Revision as of 19:54, 29 June 2015
Follow the link! [1]
An independently written, second solution is here: [2]
Enjoy!
Contents
Task
Let \(G\) be a finite group. Let \(V\) be a finite dimensional complex vector space.
a) Let \(\rho: G \rightarrow GL(V)\) be an irreducible representation. Let \(A \in End(V)\) with \([A, \rho(g)] = 0, \; \forall g \in G \). Show that then \(A = \lambda \mathbb{I}, \; \lambda \in \mathbb{C}\).
b) Let \(\rho: G \rightarrow GL(V)\) be unitary w.r.t. \(\langle \cdot, \cdot \rangle \) on \(V\). Assume that it holds $$A \in End(V), [A, \rho(g)] = 0 \; \forall g \in G \Rightarrow A = \lambda \mathbb{I}, \; \lambda \in \mathbb{C}.$$ Show that then \(\rho\) is irreducible.
Problem 6 (Textbook)
a
From \([A, \rho(g)] = 0 ~ \forall g \) it follows that \(Ker(A)\) is an invariant subspace of \(V\) and since the representation is irreducible that \(A=0=0\cdot \mathbb{I}\) or \(A\) is an isomorphism. Suppose the second case: \(A\) has an eigenvalue \(\lambda \in \mathbb{C}\), therefor \(A-\lambda \cdot \mathbb{I}\) is not injective. But it holds too that \([A-\lambda \cdot \mathbb{I}, \rho(g)] = 0 \; \forall g\), and since \(Ker(A-\lambda \mathbb{I}) \neq \{0\}\) it follows that \(A - \lambda \cdot \mathbb{I} = 0 \)
\(\blacksquare\)
b
Let \(\rho\) be unitary but reducible and let \(U\) be a not trivial subspace. If \(u \in U, u^{\perp} \in U^{\perp}\) it holds that \( 0 = (\rho(g)u,u^{\perp}) = (u, \rho(g^{-1}u^{\perp}) ~ \forall g\) and therefor \(U^{\perp}\) is invariant too. Let \(A\) be the orthogonal projection on \(U\). Then it holds that \(A \neq \lambda \cdot 1\) (since \(U\) is not trivial), but \([A,\rho(g)]v = [A,\rho(g)](u + u^{\perp}) = A(\rho(g)u + \rho(g)u^{\perp}) - \rho(g)u = 0 ~ \forall g \).
\(\blacksquare\)
