Bourbaki

Proposition 11

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

Lie Groups and Lie Algebras, chapter III, § 7. Lie groups over an ultrametric field

Suppose that p≠0p \neq 0. Let x,yx, y be two permutable elements of GfG_f. Then xy∈Gfxy \in G_f and log⁡(xy)=log⁡x+log⁡y\log(xy) = \log x + \log y.

The fact that xy∈Gfxy \in G_f follows from Lemma 1. Let UU be an open subgroup of the additive group L(G)L(G) and ϕ:U→ϕ(U)\phi : U \to \phi(U) an exponential mapping of GG with the properties of Proposition 3 of no. 2; UU can be assumed to be small enough for log⁡∣ψ(U)∣\log|\psi(U)| to be the inverse mapping of ϕ\phi. For n∈Z−{0}n \in \mathbf{Z} - \{0\} suitably chosen, xn∈ϕ(U)x^n \in \phi(U), yn∈ϕ(U)y^n \in \phi(U). Let u=log⁡xnu = \log x^n, v=log⁡ynv = \log y^n, whence xn=ϕ(u)x^n = \phi(u), yn=ϕ(v)y^n = \phi(v). By formula (2), [u,v]=0[u, v] = 0. The Hausdorff formula proves then that ϕ(λ(u+v))=ϕ(λu)ϕ(λv)\phi(\lambda(u + v)) = \phi(\lambda u) \phi(\lambda v) for ∣λ∣|\lambda| sufficiently small; hence, for every sufficiently large integer ii,

ϕ(pi(u+v))=ϕ(piu)ϕ(piv) \phi(p^i(u + v)) = \phi(p^i u) \phi(p^i v)
that is
pi(log⁡xn+log⁡yn)=log⁡(xnpiynpi) p^i (\log x^n + \log y^n) = \log(x^{np^i} y^{np^i})
or
npi(log⁡x+log⁡y)=npilog⁡(xy). np^i (\log x + \log y) = np^i \log(xy).

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

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Cites

Transcribed from the printed volume. Not translated.