Proposition 13

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

Lie Groups and Lie Algebras, chapter III, § 6. Real and complex Lie groups

Let $G$ be a connected Lie group and $\varphi$ an analytic linear representation of $G$ on a complete normable space $E$ . Let $E_1, E_2$ be two closed vector subspaces of $E$ such that $E_2 \subset E_1$ . The following conditions are equivalent:

(i) $\varphi(g)x \equiv x \pmod{E_2}$ for all $g \in G$ and all $x \in E_1$ ; (ii) $L(\varphi)(L(G))$ maps $E_1$ into $E_2$ .

\begin{align*} &\varphi(g)x \equiv x \quad (\text{mod } E_2) \quad \text{for all } g \in G \text{ and all } x \in E_1 \\ \Leftrightarrow &\varphi(\exp a)x \equiv x \quad (\text{mod } E_2) \quad \text{for all } a \in L(G) \text{ and all } x \in E_1 \\ \Leftrightarrow &(\exp L(\varphi)a)x \equiv x \quad (\text{mod } E_2) \quad \text{for all } a \in L(G) \text{ and all } x \in E_1. \end{align*}

On the other hand, if $u \in \mathscr{L}(E)$ , then

\begin{align*} &\exp(\lambda u)x \equiv x \quad (\text{mod } E_2) \quad \text{for all } \lambda \in K \text{ and all } x \in E_1 \\ \Leftrightarrow &u(E_1) \subset E_2 \end{align*}

whence the proposition.

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

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