Bourbaki

Lemma 6

Tag 038V, permanent. Cite it as [Bourbaki, Tag 038V].

Théories spectrales, chapter V, § 1. Représentations unitaires

Let π\pi and ϱ\varrho be unitary representations of a topological group G in Hilbert spaces E and F respectively. Let D be a dense subspace of E which is stable under π\pi. Let uu be a closed partial operator from E into F with domain D such that u∘π(g)=ϱ(g)∘uu\circ \pi (g) =\varrho (g)\circ u for every g∈Gg\in G. Then the domain of u∗u^* is stable under ϱ\varrho and one has the relation u∗∘ϱ(g)=π(g)∘u∗u^*\circ \varrho (g) =\pi (g)\circ u^* for every g∈Gg\in G.

Let g∈Gg\in G and x∈x\in dom(u∗u^*). For every y∈y\in dom(uu), one has

⟨ϱ(g)x∣u(y)⟩=⟨x∣ϱ(g−1)u(y)⟩=⟨x∣u(π(g−1)y)⟩ \langle \varrho (g)x|u(y)\rangle =\langle x|\varrho (g^{-1})u(y)\rangle =\langle x|u(\pi (g^{-1})y)\rangle
=⟨u∗(x)∣π(g−1)y⟩=⟨π(g)(u∗(x))∣y⟩ =\langle u^*(x)|\pi (g^{-1})y\rangle =\langle \pi (g)(u^*(x))|y\rangle

since ϱ(g)∗=ϱ(g)−1=ϱ(g−1)\varrho (g)^*=\varrho (g)^{-1}=\varrho (g^{-1}). This proves that ϱ(g)x∈\varrho (g)x\in dom(u∗u^*) and that u∗(ϱ(g)x)=π(g)(u∗(x))u^*(\varrho (g)x) =\pi (g)(u^*(x)). In particular, the domain of u∗∘ϱ(g)u^*\circ \varrho (g) contains the domain of u∗u^*, and π(g)∘u∗⊂u∗∘ϱ(g)\pi (g)\circ u^*\subset u^*\circ \varrho (g).

But moreover, if x∈x\in dom(u∗∘ϱ(g)u^*\circ \varrho (g)), then x=ϱ(g−1)(ϱ(g)x)x=\varrho (g^{-1})(\varrho (g)x) belongs to dom(u∗u^*) by what precedes applied to g−1g^{-1}. It follows that u∗∘ϱ(g)=π(g)∘u∗u^*\circ \varrho (g) =\pi (g)\circ u^*.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Cites

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Machine translation of the English, by gpt-5.4, gpt-5-6-mini, against glossary version 34. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.