Bourbaki

Lemma 1

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

Lie Groups and Lie Algebras, chapter III, § 4. Passage from Lie algebras to Lie groups

Let G be a Lie group germ and h\mathfrak{h} a Lie subalgebra of L(G)L(G) admitting a topological supplement. The union of the ghg\mathfrak{h} (resp. hg\mathfrak{h}g) for g∈Gg \in G is an integrable vector subbundle of T(G)T(G).

By considering the left trivialization of T(G)T(G) (§ 2, no. 3), it is seen immediately that the ghg\mathfrak{h}, for g∈Gg \in G, are the fibres of a vector subbundle E of T(G)T(G). Let g∈Gg \in G. The set of (La)g(L_a)_g, where a∈ha \in \mathfrak{h}, is equal to ghg\mathfrak{h}. Now, if aa and bb belong to h\mathfrak{h}, then [La,Lb]=L[a,b][L_a, L_b] = L_{[a, b]} and [a,b]∈h[a, b] \in \mathfrak{h}. Hence E is integrable (Differentiable and Analytic Manifolds, R, 9.3.3 (iv)). The argument is similar for the hg\mathfrak{h}g.

The integral foliation (Differentiable and Analytic Manifolds, R, 9.3.2) of the union of the ghg\mathfrak{h} (resp. hg\mathfrak{h}g) is called the left (resp. right) foliation of G associated with h\mathfrak{h}.

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.