Bourbaki

Lemma 2

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

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

Let L be a complete normed Lie algebra over R\mathbf{R} or C\mathbf{C}. Let G be the set of x∈Lx \in L such that ∥x∥<13log⁡32\|x\| < \frac{1}{3} \log \frac{3}{2}. Let θ\theta be the mapping x↦−xx \mapsto -x of G into G. Let H be the restriction to G×GG \times G of the Hausdorff function of L (Chapter II, § 7, no. 2). (i) (G,0,θ,H)(G, 0, \theta, H) is a Lie group germ. (ii) Let ϕ\phi be the identity mapping of G into L. The differential of ϕ\phi at 0 is an isomorphism of the normable Lie algebra L(G)L(G) onto L. (i) follows from Chapter II, § 7, no. 2. As ϕ\phi is a chart on G, the differential ψ\psi of ϕ\phi at 0 is an isomorphism of normable spaces. On the other hand, the expansion as an integral series H=∑i,j≥0HijH = \sum_{i,j \geq 0} H_{ij} of the mapping H is such that H11(x,y)=12[x,y]H_{11}(x, y) = \frac{1}{2}[x, y]. By § 3, Proposition 24, for all a,ba, b in L(G)L(G),

ψ([a,b])=H11(ψ(a),ψ(b))−H11(ψ(b),ψ(a))=[ψ(a),ψ(b)] \psi([a, b]) = H_{11}(\psi(a), \psi(b)) - H_{11}(\psi(b), \psi(a)) = [\psi(a), \psi(b)]
which proves (ii).

G is called the Lie group germ defined by L.

Suppose that K is ultrametric. Let pp be the characteristic of the residue field of K. If p≠0p \neq 0, let λ=∣p∣1/(p−1)\lambda = |p|^{1/(p-1)}; if p=0p = 0, let λ=1\lambda = 1.

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

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