Theorem 4

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

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

Let G be a finite-dimensional Lie group. There exists an open subgroup of G isomorphic to a standard group.

By replacing G by a group isomorphic to an open subgroup of G, the problem is reduced to the case where G is an open subset of K', with identity element 0 and where the coordinates of the product x.y and the inverse x^{[-1]} are given by formulae

(3)

(x.y)_i = x_i + y_i + \sum_{|\alpha| \geq 1, |\beta| \geq 1} c_{\alpha \beta i} x^\alpha y^\beta \quad (i = 1, 2, \ldots, r)

(4)

(x^{[-1]})_i = -x_i + \sum_{|\alpha| > 1} d_{\alpha i} x^\alpha \quad (i = 1, 2, \ldots, r)

where the series on the right hand side are convergent for x, y in G (§ 5, no. 1). Let λ ∈ K* and let the group law be transported from G to G' = λG by the homothety of ratio λ. For x', y' in G', the product x'.y' and the inverse x'^{[-1]} evaluated in G' have coordinates

(x'.y')_i = x'_i + y'_i + \sum_{|\alpha| \geq 1, |\beta| \geq 1} c'_{\alpha \beta i} {x'}^\alpha {y'}^\beta \quad (i = 1, 2, \ldots, r)

({x'}^{[-1]})_i = -x'_i + \sum_{|\alpha| > 1} d'_{\alpha i} {x'}^\alpha \quad (i = 1, 2, \ldots, r)

where

c'_{\alpha \beta i} = \lambda^{-|\alpha|-|\beta|+1} c_{\alpha \beta i}, \qquad d'_{\alpha i} = \lambda^{-|\alpha|+1} d_{\alpha i}.

As the series (3) and (4) are convergent, we see that, for |λ| sufficiently large, for all α, β, i,

|c'_{\alpha \beta i}| \leq 1, \qquad |d'_{\alpha i}| \leq 1

that is c'_{\alpha \beta i} ∈ A and d'_{\alpha i} ∈ A; and on the other hand G' ⊃ m × m × ... × m. Then m × m × ... × m is an open subgroup of G' and is a standard group.

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

Cited by

Cites

§ 5, no. 1