Bourbaki

Proposition 40

Tag 0AB3, permanent. Cite it as [Bourbaki, Tag 0AB3].

Lie Groups and Lie Algebras, chapter III, § 3. Passage from a Lie group to its Lie algebra

Let GG be a Lie group, EE a complete normable space, π\pi an analytic linear representation of GG on EE and FF a closed vector subspace of EE stable under π(G)\pi(G). Suppose that KK is of characteristic 0, or that FF is a direct factor of EE.

(i) The subrepresentation π1\pi_1 and the quotient representation π2\pi_2 of π\pi defined by FF are analytic representations.

(ii) FF is stable under L(π)(L(G))L(\pi)(L(G)).

(iii) Let ϱ1\varrho_1 and ϱ2\varrho_2 be the subrepresentation and quotient representation of L(π)L(\pi) defined by FF. Then L(π1)=ϱ1,L(π2)=ϱ2L(\pi_1) = \varrho_1, L(\pi_2) = \varrho_2.

Let AA be the set of u∈L(E)u \in \mathscr{L}(E) such that u(F)⊂Fu(F) \subset F. Then AA is a closed vector subspace of L(E)\mathscr{L}(E) and π\pi takes its values in AA. By virtue of the hypotheses on KK and FF, the mapping π′:G→A\pi': G \to A with the same graph as π\pi is analytic (Differentiable and Analytic Manifolds, R, 5.8.5). The canonical mappings θ1:A→L(F)\theta_1 : A \to \mathscr{L}(F) and θ2:A→L(E/F)\theta_2 : A \to \mathscr{L}(E/F) are continuous and linear and hence analytic. This proves (i). The mappings Te(π)T_e(\pi) and Te(π′)T_e(\pi') have the same graph and hence L(π)(L(G))⊂AL(\pi)(L(G)) \subset A, which proves (ii). We have

Te(π1)=Te(θ1∘π′)=θ1∘Te(π′)=ϱ1Te(π2)=Te(θ2∘π′)=θ2∘Te(π′)=ϱ2. T_e(\pi_1) = T_e(\theta_1 \circ \pi') = \theta_1 \circ T_e(\pi') = \varrho_1 \\ T_e(\pi_2) = T_e(\theta_2 \circ \pi') = \theta_2 \circ T_e(\pi') = \varrho_2.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed volume. Not translated.