Revision as of 23:24, 16 June 2015

Follow the link! [1]

An independently written, second solution is here: [2]

Enjoy!

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 \(A=0=0\cdot 1\) or \(A\) is an isomorphism. Suppose the second case: \(A\) has a eigen value \(\lambda \in \mathbb{C}\), therefor \(A-\lambda \cdot 1\) is not injective. But it holds too that \([A-\lambda \cdot 1, \rho(g)] = 0 ~ \forall g\), and since \(Ker(A) \neq 0\) it follows that \(A - \lambda \cdot 1 = 0 \blacksquare \)

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 \([A,\rho(g)]v = [A,\rho(g)](u + u^{\perp}) = A(\rho(g)u + \rho(g)u^{\perp}) - \rho(g)u = 0 ~ \forall g \blacksquare \).