Bourbaki

Proposition 6

Tag 0AD3, permanent. Cite it as [Bourbaki, Tag 0AD3].

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 and ϕ:V→G\phi : V \to G an exponential mapping of G. For all x∈Vx \in V, let Tx(L)T_x(L) be identified with L, so that the right differential ϖ(x)\varpi(x) of ϕ\phi at x is a linear mapping of L into L. For x sufficiently close to 0,

ϖ(x)=∑n≥01(n+1)!(ad⁡x)n. \varpi(x) = \sum_{n \geq 0} \frac{1}{(n+1)!} (\operatorname{ad} x)^n.

Let L be given a norm compatible with its topology and such that ∥[x,y]∥≤∥x∥∥y∥\| [x, y] \| \leq \| x \| \| y \| for all x,y∈Lx, y \in L. It suffices to consider the case where G is the Lie group germ defined by L and ϕ=Id⁡G\phi = \operatorname{Id}_G. By definition, ϖ(x)\varpi(x) is then the tangent mapping at x to the mapping y↦y.x−1y \mapsto y . x^{-1} of G into G. If H(X,Y)H(X, Y) denotes the Hausdorff series, ϖ(x)\varpi(x) is therefore, for ∥x∥\| x \| sufficiently small, the tangent mapping at 0 to the mapping y↦H(x,+y,−x)y \mapsto H(x, +y, -x) of G into G. In H(X+Y,−X)H(X + Y, -X), the sum of terms of the first degree in Y is

∑m≥01(m+1)!(ad⁡X)mY \sum_{m \geq 0} \frac{1}{(m+1)!} (\operatorname{ad} X)^m Y

(Chapter II, § 6, no. 5, Proposition 5). The proposition then follows from Differentiable and Analytic Manifolds, R, 3.2.4 and 4.2.3.

Let G be a Lie group germ and t∈Kt \in K. A t-th power mapping of G is any mapping, defined and analytic on an open neighbourhood of e, with values in G and coinciding on a neighbourhood of e with a mapping

g↦ϕ(tϕ−1(g)) g \mapsto \phi(t \phi^{-1}(g))

where ϕ\phi is an injective exponential mapping of G.

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

Cited by

Cites

Transcribed from the printed volume. Not translated.