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 Incoming links: Inverse-adjoint representation

Every

D_{2} (g)

is invertible because both inversion and the conjugate transpose preserve invertibility. For

g, h \in G

, the group homomorphism property of

D_{1}

and reversal of multiplication under the adjoint give

D_{2} (g h) = [D_{1} ((g h)^{- 1})]^{†} = [D_{1} (h^{- 1}) D_{1} (g^{- 1})]^{†} = [D_{1} (g^{- 1})]^{†} [D_{1} (h^{- 1})]^{†} = D_{2} (g) D_{2} (h) .

Also

D_{2} (e) = I

. Hence $D_{2}$ is a group representation. It is the inverse-adjoint representation of

D_{1}

; the inverse and adjoint each reverse product order, so together they preserve it.

⟨ 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.