Proposition 38

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

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

Consider $G$ as operating on $E$ on the left by the mapping $(g, x) \mapsto \pi(g)x$ . Let $b \in E$ and $\varphi(b)$ be its orbital mapping. Let $T_b(E)$ be canonically identified with $E$ . For all $t \in L(G)$ ,

(L(\pi)t)(b) = \langle t, \varphi(b) \rangle = \varphi(b)*t = t * \varepsilon_b.

In particular, the vector field defined by $t$ on $E$ is the field $b \mapsto (L(\pi)t)(b)$ .

$L(\pi)t = \langle t, \pi \rangle$ (no. 9, Proposition 34). As the mapping $A \mapsto Ab$ of $\mathscr{L}(E)$ into $E$ is continuous and linear, it follows that

\begin{align*} (L(\pi)t)(b) &= \langle t, g \mapsto \pi(g)b \rangle \\ &= \langle t, \mathrm{Id}_E \circ \varphi(b) \rangle \\ &= \langle \varphi(b)*t, \mathrm{Id}_E \rangle \quad (\text{Diff. \& Anal. Man., R, 13.2.3}) \\ &= \varphi(b)*t. \end{align*}

Finally, $\varphi(b)*t = t * \varepsilon_b$ (no. 3, Proposition 14 (ii)).

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

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Proposition 41, § 3

Proposition 38

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