Proposition 4

Tag 0AD1, permanent. Cite it as [Bourbaki, Tag 0AD1].

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

Let G be a Lie group germ and $\phi$ an injective exponential mapping of G. For all x, y in $L(G)$ ,

(1)

x + y = \lim_{\lambda \in K^*, \lambda \to 0} \lambda^{-1} \phi^{-1}(\phi(\lambda x) \phi(\lambda y))

(2)

[x, y] = \lim_{\lambda \in K^*, \lambda \to 0} \lambda^{-2} \phi^{-1}(\phi(\lambda x) \phi(\lambda y) \phi(-\lambda x) \phi(-\lambda y))

(note that

\phi^{-1}(\phi(\lambda x) \phi(\lambda y))

and

\phi^{-1}(\phi(\lambda x) \phi(\lambda y) \phi(-\lambda x) \phi(-\lambda y))

are defined for

|\lambda|

sufficiently small).

Let $L = L(G)$ be given a norm defining the topology of L and such that $\| [x, y] \| \leq \| x \| \| y \|$ for all $x, y$ in L. Using Theorems 2 and 4, it can be assumed that G is the Lie group germ defined by L and that $\phi = \mathrm{Id}_G$ . Let $(x, y) \mapsto x \cdot y$ denote the product in the group G. The formulae to be proved can then be written (3)

x + y = \lim_{\lambda \in K^*, \lambda \to 0} \lambda^{-1}((\lambda x) \cdot (\lambda y))

(4)

[x, y] = \lim_{\lambda \in K^*, \lambda \to 0} \lambda^{-2}((\lambda x) \cdot (\lambda y) \cdot (-\lambda x) \cdot (-\lambda y)).

There exists an open neighbourhood V of 0 in K such that the function

\lambda \mapsto f(\lambda) = (\lambda x) \cdot (\lambda y)

is defined and analytic on V. By Chapter II, § 6, no. 4, Remark 2, the expansion of

f

as an integral series about the origin is

\lambda(x + y) + \frac{1}{2} \lambda^2 [x, y] + \cdots

and this proves (3). On the other hand, for

u, v

in G and

\|u\|, \|v\|

sufficiently small,

u \cdot v

is an analytic function of

(u, v)

and the terms of degrees 1 and 2 in the expansion of this function as an integral series about the origin are

u + v + \frac{1}{2}[u, v]

. By Differentiable and Analytic Manifolds, R, 3.2.7 and 4.2.3, the terms of degrees 1 and 2 in the expansion of the function

f(\lambda) \cdot f(-\lambda)

as an integral series about the origin are the terms of degrees 1 and 2 in

f(\lambda) + f(-\lambda) + \frac{1}{2}[f(\lambda), f(-\lambda)]

or also in

\lambda(x + y) + \frac{1}{2} \lambda^2 [x, y] - \lambda(x + y) + \frac{1}{2} \lambda^2 [x, y] \\ + \frac{1}{2} [\lambda(x + y), -\lambda(x + y)] = \lambda^2 [x, y]

and this proves (4).

EXponential mappings

§ 4.3

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

Cited by

Cites

Chapter II, § 6, no. 4