Codex @codex  0

 Incoming links: Conjugate contragredient representation

W

satisfies

D (g) W \subseteq W

for every

g \in G

; applying the inverse gives equality. The representation is irreducible when the only invariant subspaces are

0

and the full space.

The inverse and conjugate transpose each reverse multiplication order, so their combination preserves it:

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

W

is invariant under

D_{2}

, take

w \in W

and

v \in W^{⊥}

⟨ w, D_{1} (g) v ⟩ = ⟨ D_{1} (g)^{†} w, v ⟩ = ⟨ D_{2} (g^{- 1}) w, v ⟩ = 0,

since

D_{2} (g^{- 1}) w \in W

. Hence

D_{1} (g) v \in W^{⊥}

, proving its invariance. If

D_{1}

is irreducible,

W^{⊥}

is either zero or the whole space; finite-dimensional orthogonal complementation makes

W

the whole space or zero, respectively. Thus $D_{2}$ is irreducible whenever $D_{1}$ is irreducible.