Bourbaki

Proposition 8

Tag 0A9J, permanent. Cite it as [Bourbaki, Tag 0A9J].

Lie Groups and Lie Algebras, chapter III, § 3. Passage from a Lie group to its Lie algebra

Let G, H be Lie groups and ϕ\phi a morphism of G into H. If t∈T(∞)(G)t \in \mathscr{T}^{(\infty)}(G), then ϕ∗(t∨)=(ϕ∗(t))∨\phi_*(t^\vee) = (\phi_*(t))^\vee.

Let θ\theta (resp. θ′\theta') be the mapping g↦g−1g \mapsto g^{-1} of G into G (resp. of H into H). Then ϕ∘θ=θ′∘ϕ\phi \circ \theta = \theta' \circ \phi, whence ϕ∗(θ∗(t))=θ∗′(ϕ∗(t))\phi_*(\theta_*(t)) = \theta'_*(\phi_*(t)).

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

Cited by

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Cites

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Transcribed from the printed volume. Not translated.