Bourbaki

Theorem 4

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

Lie Groups and Lie Algebras, chapter III, § 4. Passage from Lie algebras to Lie groups

Let G be a Lie group germ, L its Lie algebra, V an open neighbourhood of 0 in L, φ an analytic mapping of V into G such that φ(0) = 0 and T_0(φ) = Id_L. The following conditions are equivalent: (i) For all b ∈ L, φ((λ + λ')b) = φ(λb)φ(λ'b) for |λ| and |λ'| sufficiently small. (ii) For all b ∈ L and every integer n > 0, φ_*(b^n) is homogeneous of degree n in U(G) (T_0^{(∞)}(L) is identified with TS(L) and b^n is evaluated in TS(L)). (iii) The mapping φ_* of TS(L) into U(G) is compatible with the graduations of TS(L) and U(G). (iv) The mapping φ_* of TS(L) into U(G) is the canonical mapping of TS(L) into the enveloping algebra of L. (v) There exist a norm on L defining the topology of L and such that

∥[x,y]∥≤∥x∥∥y∥ \|[x, y]\| \leq \|x\|\|y\|
for all x, y in L and an open subgroup germ W ⊂ V of the Lie group germ defined by L (no. 2) such that φ|W is an isomorphism of W onto an open Lie subgroup germ of G.

(v) ⇒ (i): obvious, for (λb).(λ'b) = (λ + λ')b in W for |λ| and |λ'| sufficiently small. (i) ⇒ (ii): suppose that condition (i) is satisfied. Let b ∈ L. Let ψ be the restriction of φ to V ∩ Kb. By hypothesis, there exists a symmetric neighbourhood T of 0 in the additive Lie group Kb such that ψ|T is a morphism of the Lie group germ T into G. Hence

φ∗(bn)=(ψ∣T)∗(bn)=((ψ∣T)∗(b))n=(φ∗(b))n, \varphi_*(b^n) = (\psi|T)_*(b^n) = ((\psi|T)_*(b))^n = (\varphi_*(b))^n,
so that φ_*(b^n) is homogeneous of degree n in U(G). (ii) ⇒ (iii): this follows from the fact that TS^n(L) is the vector subspace of TS(L) generated by the n-th powers of the elements of L (Algebra, Chapter IV, § 5, Proposition 5). (iii) ⇒ (iv): the canonical mapping of TS(L) into the enveloping algebra of L is the unique morphism of graded cogebras mapping 1 to 1 and extending Id_L (Chapter II, § 1, no. 5, Remark 3). Now φ_* is a cogebra morphism and φ_*|L = Id_L by hypothesis. If condition (iii) holds, it is seen that condition (iv) also holds. (iv) ⇒ (v): suppose that condition (iv) is satisfied. Choose a norm on L defining the topology of L and such that \|[x, y]\| \leq \|x\|\|y\| for all x, y in L. Let H be the Lie group germ defined by the normed Lie algebra L. By Theorem 1, there exist an open subgroup germ S ⊂ V of H and an isomorphism φ' of S onto an open subgroup germ of G. As we know already that (v) ⇒ (iv), the mapping φ_* of TS(L) into U(G) is the canonical mapping of TS(L) into the enveloping algebra of L. Thus φ_*(t) = φ'_*(t) for all t ∈ T_0^{(∞)}(L). As φ and φ' are analytic, φ and φ' coincide on a neighbourhood of 0.

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

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