Bourbaki

Proposition 44

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

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

For all a∈L(G)a \in L(G),

(L(Ad))(a)=adL(G)a. (L(\mathrm{Ad}))(a) = \mathrm{ad}_{L(G)}a.
Let b∈L(G)b \in L(G). By Proposition 42 (ii) of no. 11 and Corollary 3 to Proposition 43,
(L(Ad))(a).b=−[Ra−La1Lb](e). (L(\mathrm{Ad}))(a).b = -[R_a - L_{a_1}L_b](e).
Now Ra∘Lb=Lb∘RaR_a \circ L_b = L_b \circ R_a (no. 6, Proposition 23 (ii)), whence [Ra,Lb]=0[R_a, L_b] = 0; then, using Proposition 23 (ii),
(L(Ad))(a).b=[La,Lb](e)=L[a,b](e)=[a,b]=(adL(G)a)b. (L(\mathrm{Ad}))(a).b = [L_a, L_b](e) = L_{[a, b]}(e) = [a, b] = (\mathrm{ad}_{L(G)}a)b.

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.