Bourbaki

Proposition 1

Tag 0AL9, permanent. Cite it as [Bourbaki, Tag 0AL9].

Lie Groups and Lie Algebras, chapter III, § 10. The automorphism group of a Lie group

Let K' be a non-discrete closed subfield of K, G a Lie group over K, V a manifold over K' and (v,g)↦vg(v, g) \mapsto vg a K'-*analytic mapping of* V ×\times G into G. Suppose that, for all v∈Vv \in V, the mapping g↦vgg \mapsto vg of G into G is an automorphism of G. Let ε\varepsilon be an element of V such that εg=g\varepsilon g = g for all g∈Gg \in G and a∈Tε(V)a \in T_\varepsilon(V). Then the vector field g↦agg \mapsto ag on G is an infinitesimal automorphism of G.

For v∈V, g1∈G, g2∈G, v(g1g2)=(vg1)(vg2)v \in V,\ g_1 \in G,\ g_2 \in G,\ v(g_1g_2) = (vg_1)(vg_2). Hence, for u1∈TG, u2∈TG, a(u1u2)=(au1)(au2)u_1 \in TG,\ u_2 \in TG,\ a(u_1u_2) = (au_1)(au_2) (§ 2, no. 1, Proposition 3). In particular, the mapping g↦agg \mapsto ag of G into TG is a group homomorphism. On the other hand, this mapping is analytic by Lemma 2.

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

Cited by

Nothing.

Cites

Transcribed from the printed volume. Not translated.