Proposition 3

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

Lie Groups and Lie Algebras, chapter III, § 9. Commutators, centralizers and normalizers in a Lie group

Let G be a finite-dimensional Lie group and H_1 and H_2 subgroups of G. Let h_1, h_2 and h be the Lie subalgebras tangent at e to H_1 H_2 and (H_1, H_2) respectively. Then [h_1, h_2] ⊂ h.

Let a ∈ h_1, b ∈ h_2. There exist an open neighbourhood I of 0 in K and analytic mappings f_1, f_2 of I into G such that

f_1(0) = f_2(0) = e, \quad f_1(I) \subset H_1, \quad f_2(I) \subset H_2,

(T_0 f_1)1 = a, \quad (T_0 f_2)1 = b.

We write

f(\lambda, \mu) = (f_1(\lambda), f_2(\mu)) \in (H_1, H_2) \quad \text{for } \lambda, \mu \text{ in I}.

We identify an open neighbourhood of e in G with an open subset of K' by means of a chart which maps e to 0. Then L(G) is identified with K'. By § 5, no. 2, Proposition 1, the expansion of f(\lambda, \mu) as an integral series about the origin is

f(\lambda, \mu) = \lambda \mu [a, b] + \sum_{i \geq 1, j \geq 1, i+j \geq 2} \lambda^i \mu^j a_{ij}

where a_{ij} ∈ K' (the terms in \lambda^i and \mu^j in the expansion of f(\lambda, \mu) are zero because f(\lambda, 0) = f(0, \mu) = 0). We fix \mu in I. Letting \lambda tend to 0, we see that

\mu [a, b] + \sum_{j \geq 2} \mu^j a_{1j} \in h.

Since this is true for all \mu ∈ I, it follows that [a, b] ∈ h.

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

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Cites

§ 5, no. 2, Proposition 1