Bourbaki

Proposition 48

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

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

Let G be a Lie group and H and E Lie subgroups of G. Suppose that G is, as a Lie group, the semi-direct product of H by E. Let φ\varphi be the linear representation g↦(Ad g)∣L(E)g \mapsto (\mathrm{Ad}\, g) \mid L(E) of the Lie group G on L(E)L(E) (cf. Proposition 47) and let σ\sigma be the restriction of φ\varphi to H. Then: (i) L(G)L(G) is the topological direct sum of L(H)L(H) and L(E)L(E); (ii) L(H)L(H) is a subalgebra of L(G)L(G) and L(E)L(E) is an ideal of L(G)L(G); (iii) L(σ)L(\sigma) is a linear representation of L(H)L(H) on the Lie algebra of derivations of L(E)L(E); (iv) L(G)L(G) is the semi-direct product of L(H)L(H) by L(E)L(E) defined by L(σ)L(\sigma) (Chapter I, § 1, no. 8).

(i) is obvious and (ii) follows from Proposition 47. L(σ)=L(φ)∣L(H)L(\sigma) = L(\varphi) \mid L(H). Now by Propositions 40 (no. 11) and 44 (no. 12), L(φ)(t)L(\varphi)(t) is, for all t∈L(G)t \in L(G), the restriction of adL(G)t\mathrm{ad}_{L(G)} t to L(E)L(E). This proves (iii). Using (i) and (ii), this also proves (iv).

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

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