Let G be a Lie group germ, X a manifold of class Cr (r≥2) and x0 a point of X. Let a↦Da be a law of left infinitesimal operation of class Cr−1 of L(G) on X.
(i) There exists an open neighbourhood X′ of x0 in X and a law chunk of left operation of class Cr−1 of G on X′ such that the associated law of infinitesimal operation is a↦Da∣X′.
(ii) Let there be two law chunks of left operation of class Cr−1 of G on an open neighbourhood X′′ of x0; if they admit a↦Da∣X′′ as associated law of infinitesimal operation, they coincide on a neighbourhood of (e,x0).
Assertion (ii) follows from the Corollary to Proposition 11. We prove (i). For all (g,x)∈G×X and all a∈L(G), we write
Qa(g,x)=(ag,Da(x))∈Tg(G)×Tx(X). Let
S(g,x) be the set of
Qa(g,x) for
a∈L(G). By Lemma 5 of no. 6, the
S(g,x) are the fibres of a vector subbundle S of
T(G)×T(X). Let
a,b be in
L(G); then
[Qa,Qb](g,x)=([Ra,Rb](g),[Da,Db](x))=(−R[a,b](g),−D[a,b](x))(§3,18.6)=Q−[a,b](g,x) and hence S is integrable (
Differentiable and Analytic Manifolds, R, 9.3.3, (iv)).
By Differentiable and Analytic Manifolds, R, 9.3.7, there exist an open neighbourhood G1 of e in G, an open neighbourhood X1 of x0 in X and a mapping (g,x)↦gx of class Cr−1 of G1×X1 into X such that ex=x for all x∈X1 and
(ag)x=Da(gx)for a∈L(G),g∈G1,x∈X1. In particular
ax=Da(x). Let
G2 be an open neighbourhood of e in
G1 and
X2 an open neighbourhood of
x0 in
X1 such that
gg′ is defined and belongs to
G1 for
g,g′ in
G2 and
gx is defined and belongs to
X1 for
(g,x)∈G2×X2. Consider the mappings
α1,α2 of
G2×(G2×X2) into
X defined by
α1(g,(h,x))=g(hx),α2(g,(h,x))=(gh)x. They are of class
Cr−1. Then
α1(e,(h,x))=hx=α2(e,(h,x)). On the other hand
T(α1)(ag,0(h,x))T(α2)(ag,0(h,x))=(ag)(hx)=Da(g(hx))by (6)=Da(α1(g,(h,x))),=(agh)x=Da((gh)x)by (6)=Da(α2(g,(h,x))). By
Differentiable and Analytic Manifolds, R, 9.3.7,
α1 and
α2 coincide on a neighbourhood of
(e,(e,x0)). Then (i) follows from (7) and Proposition 23 of § 1, no. 11.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.