Let M be an analytic manifold of finite dimension n. Let ω1,…,ωn be analytic differential forms of degree 1 on M, with scalar values, which are linearly independent at each point of M and such that, for all k=1,…,n, dωk is a linear combination with constant coefficients of the ωi∧ωj. Then, for all m0∈M, there exists an open neighbourhood M′ of m0 in M and a Lie group germ structure on M′ compatible with the manifold structure on M′, with identity element m0 and such that ω1∣M′,…,ωn∣M′ form a basis of the space of left invariant differential forms on M′ of degree 1 with scalar values.
If M1′ and M2′ are two such group germs, M1′ and M2′ have a common open subgroup germ.
Let X1,…,Xn be the vector fields on M such that, at each point m of M, the (Xt)m constitute the basis of Tm(M) dual to ((ω1)m,…,(ωn)m). These fields are analytic. By hypothesis, there exist cijk∈K (1≤i,j,k≤n) such that cijk=−cjik and dωk=∑i<jcijkωi∧ωj. By Differentiable and Analytic Manifolds, R, 8.5.7, formula (11),
⟨[Xi,Xj],ωk⟩=−(dωk)(Xi,Xj)=−(r<s∑crskωr∧ωs)(Xi,Xj)=−cijk and hence
[Xi,Xj]=−∑kcijkXk. It then follows that the
−cijk are the constants of structure of a Lie algebra
g relative to a basis
(e1,…,en). For all
m∈M, let
αm be the linear mapping of
Tm(M) into
g which maps
(X1)m to
e1,…,(Xn)m to
en. Then
α is an analytic differential form on
M of degree 1 with values in
g and
αm is an isomorphism of
Tm(M) onto
g. On the other hand,
α=∑k=1nωkek and hence
dα=k=1∑n(dωk)ek=k=1∑n(i<j∑cijkωi∧ωj)ek and
[α]2=k=1∑n[ωkek]2+i<j∑(ωiei)∧(ωjej)(§3,formula (30))=i<j∑(ωi∧ωj)[ei,ej]=−k=1∑ni<j∑(cijkωi∧ωj)ek=−dα. It then suffices to apply Proposition 10.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.