Bourbaki

Lemma 1

Tag 0AL6, permanent. Cite it as [Bourbaki, Tag 0AL6].

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

Let G be a Lie group and α\alpha a vector field on G. For all g∈Gg \in G, let

β(g)=α(g)g−1∈L(G). \beta(g) = \alpha(g)g^{-1} \in L(G).
The following conditions are equivalent: (i) α\alpha is a homomorphism of the group G into the group T(G); (ii) for all g,g′g, g' in G, α(gg′)=α(g)g′+gα(g′)\alpha(gg') = \alpha(g)g' + g\alpha(g'); (iii) for all g,g′g, g' in G, β(gg1)=β(g)+(Ad g)β(g′)\beta(gg_1) = \beta(g) + (\mathrm{Ad}\,g)\beta(g').

Condition (i) means that, for all g,g′g, g' in G, we have in the group T(G):

β(g)gβ(g′)g′=β(gg′)gg′ \beta(g)g\beta(g')g' = \beta(gg')gg'
or
β(g)((Ad g)β(g′))gg′=β(gg′)gg′. \beta(g)((\mathrm{Ad}\,g)\beta(g'))gg' = \beta(gg')gg'.
But the product of β(g)\beta(g) and (Ad g)β(g′)(\mathrm{Ad}\,g)\beta(g') in T(G) is just the sum of β(g)\beta(g) and (Ad g)β(g′)(\mathrm{Ad}\,g)\beta(g') in L(G)L(G) (§ 2, no. 1, Proposition 2). Hence (i) ⇔\Leftrightarrow (iii). On the other hand, condition (ii) may be written β(gg′)gg′=β(g)gg′+gβ(g′)g′\beta(gg')gg' = \beta(g)gg' + g\beta(g')g', or
β(gg′)=β(g)+(Ad g)β(g′) \beta(gg') = \beta(g) + (\mathrm{Ad}\,g)\beta(g')
and hence (ii) ⇔\Leftrightarrow (iii).

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.