Proposition 3

Tag 0A6W, permanent. Cite it as [Bourbaki, Tag 0A6W].

Lie Groups and Lie Algebras, chapter II, § 8. Convergence of the Hausdorff series (ultrametric case)

Let $G$ be the ball $\{ x \in g | \| x \| < a^{\theta} \}$ . The analytic mapping

h : G \times G \to G

makes $G$ into a group in which $0$ is the identity element and $-x$ is the inverse of $x$ for all $x \in G$ . Moreover, if $R$ is a real number such that $0 < R < a^{\theta}$ , the ball

\{ x \in g | \| x \| < R \}

(resp. $\{ x \in g | \| x \| \leq R \}$ ) is an open subgroup of $G$ .

As $H(U, -U) = 0$ and $H(0, U) = H(U, 0) = U$ , $h(x, -x) = 0$ and

h(0, x) = h(x, 0) = x

for all

x \in G

. It therefore remains to prove the associativity formula

h(h(x, y), z) = h(x, h(y, z)) \quad \text{for } x, y, z \text{ in } G.

H(H(U, V), W) = H(U, H(V, W))

\hat{L}(\{ U, V, W \})

(§ 6, no. 5, Proposition 4), we have

\tilde{H} \circ (\tilde{H} \times \mathrm{Id}_g) = \tilde{H} \circ (\mathrm{Id}_g \times \tilde{H})

\hat{P}(g \times g \times g; g)

(no. 2) and (19) implies (18) by (16) and Differentiable and Analytic Manifolds, R, 4.1.5.

In other words (Chapter III, § 1), $G$ with the Hausdorff function is a Lie group.

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

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Lemma 3, § 4

Proposition 3

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Cites