Symmetries

Suppose that for every orthonormal basis

b_{1}, \dots, b_{n}

for

E

and all

j, k

there is some

g \in G

such that

⟨ Ψ (g) b_{j}, b_{k} ⟩ \neq 0

. Prove that

Ψ : G \to U (E)

is irreducible.

Any self-adjoint operator $A \in H o m (E)$ can be written as $A x = \sum a_{j} ⟨ x, b_{j} ⟩ b_{j}$ and $A Ψ (g) = Ψ (g) A$ is equivalent to $a_{j} ⟨ Ψ (g) b_{j}, b_{k} ⟩ = ⟨ Ψ (g) A b_{j}, b_{k} ⟩ = ⟨ Ψ (g) b_{j}, A b_{k} ⟩ = a_{k} ⟨ Ψ (g) b_{j}, b_{k} ⟩ .$ By assumption we conclude that for all $j, k$ : $a_{j} = a_{k}$ , i.e. $A = 1$ .