Bourbaki

Remark

Tag 0A7R, permanent. Cite it as [Bourbaki, Tag 0A7R].

Lie Groups and Lie Algebras, chapter III, § 1. Lie groups

With the above hypotheses, further let HH be a manifold of class CrC^r and (x,h)↦m(x,h)(x, h) \mapsto m(x, h) a mapping of class CrC^r of X×HX \times H into XX such that m(gx,h)=gm(x,h)m(gx, h) = gm(x, h) for x∈X, g∈G, h∈Hx \in X,\ g \in G,\ h \in H. Let nn be the mapping of (X/G)×H(X/G) \times H into X/GX/G derived from mm by taking quotients. We show that nn is of class CrC^r. Consider the diagram

X×H→mXπ×1↓↓π(X/G)×H→nX/G \begin{array}{ccc} X \times H & \xrightarrow{m} & X \\ \pi \times 1 \downarrow & & \downarrow \pi \\ (X/G) \times H & \xrightarrow{n} & X/G \end{array}

It is commutative, π∘m\pi \circ m is of class CrC^r and π×1\pi \times 1 is a surjective submersion; it then suffices to apply Differentiable and Analytic Manifolds, R, 5.9.5.

Let GG be a Lie group, XX a manifold of class CrC^r and (g,x)↦xg(g, x) \mapsto xg a law of right operation of class CrC^r of GG on XX. Let τ(g)x=ρ(x)g=xg\tau(g)x = \rho(x)g = xg for g∈G, x∈Xg \in G,\ x \in X. Then this time

(3′)(3')
ρ(x)=ρ(xg)∘γ(g−1),ρ(x)=τ(g)∘ρ(x)∘δ(g) \rho(x) = \rho(xg) \circ \gamma(g^{-1}), \qquad \rho(x) = \tau(g) \circ \rho(x) \circ \delta(g)
and hence
(4′)(4')
Tg(ρ(x))=Te(ρ(xg))∘Tg(γ(g−1)) T_g(\rho(x)) = T_e(\rho(xg)) \circ T_g(\gamma(g^{-1}))
(5′)(5')
Tg(ρ(x))=Tx(τ(g))∘Te(ρ(x))∘Tg(δ(g)). T_g(\rho(x)) = T_x(\tau(g)) \circ T_e(\rho(x)) \circ T_g(\delta(g)).

On the other hand, if η\eta denotes the mapping (g,x)↦xg(g, x) \mapsto xg of G×XG \times X into XX, formula (6) remains valid. Proposition 9, its Corollary and Proposition 10 remain equally true (with "principal left fibre bundle" replaced by "principal right fibre bundle" in the last).

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

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Cites

Transcribed from the printed volume. Not translated.