Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 10 / c

Solution

⟨ w, v ⟩ = w^{†} v

. Since

W

is invariant under every

D_{2} (g)

, it is in particular invariant under

D_{2} (g^{- 1}) = [D_{1} (g)]^{†} .

Let

v \in W^{⊥}

and

w \in W

. Then

⟨ w, D_{1} (g) v ⟩ = ⟨ D_{1} (g)^{†} w, v ⟩ = 0,

because

D_{1} (g)^{†} w \in W

and

v

is orthogonal to

W

. Thus

D_{1} (g) v

is orthogonal to every

w \in W

, proving

D_{1} (g) W^{⊥} \subseteq W^{⊥} (g \in G) .

Therefore

W^{⊥}

D_{1}

. If

D_{1}

is irreducible,

W^{⊥}

is either

{0}

C^{n}

(W^{⊥})^{⊥} = W

, so these alternatives force

W = C^{n}

W = {0}

. Hence

D_{1} irreducible ⟹ D_{2} irreducible .

No unitarity of

D_{1}

was assumed; the inverse-adjoint representation is exactly what makes this orthogonal complement argument work.

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