Library / Lie Groups and Lie Algebras / Chapter III / § 3
Lemma 2
Tag 0ABT, permanent. Cite it as [Bourbaki, Tag 0ABT].
Lie Groups and Lie Algebras, chapter III, § 3. Passage from a Lie group to its Lie algebra
Let G be a Lie group, U a symmetric open neighbourhood of e in G, E a complete normable space and ϕ : U 2 → E \phi : U^2 \to E ϕ : U 2 → E g ∈ U g \in U g ∈ U ω g \omega_g ω g h ↦ ϕ ( g − 1 h ) h \mapsto \phi(g^{-1}h) h ↦ ϕ ( g − 1 h ) ω \omega ω d e ϕ d_e \phi d e ϕ
Clearly ω e = d e ϕ \omega_e = d_e \phi ω e = d e ϕ g ∈ U g \in U g ∈ U t ∈ T e ( G ) t \in T_e(G) t ∈ T e ( G )
⟨ ω g , T e ( γ ( g ) ) t ⟩ = ⟨ d g ( ϕ ∘ γ ( g ) − 1 ) , T e ( γ ( g ) ) t ⟩ = ⟨ d e ϕ ∘ T g ( γ ( g ) − 1 ) , T e ( γ ( g ) ) t ⟩ = ⟨ d e ϕ , t ⟩
\langle \omega_g, T_e(\gamma(g))t \rangle = \langle d_g(\phi \circ \gamma(g)^{-1}), T_e(\gamma(g))t \rangle \\
= \langle d_e \phi \circ T_g(\gamma(g)^{-1}), T_e(\gamma(g))t \rangle = \langle d_e \phi, t \rangle
⟨ ω g , T e ( γ ( g )) t ⟩ = ⟨ d g ( ϕ ∘ γ ( g ) − 1 ) , T e ( γ ( g )) t ⟩ = ⟨ d e ϕ ∘ T g ( γ ( g ) − 1 ) , T e ( γ ( g )) t ⟩ = ⟨ d e ϕ , t ⟩ and hence ω g \omega_g ω g d e ϕ d_e \phi d e ϕ T e ( γ ( g ) ) T_e(\gamma(g)) T e ( γ ( g ))
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.