Bourbaki

Corollary 2

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

Lie Groups and Lie Algebras, chapter III, § 8. Lie groups over R or Qp

Let GG be a finite-dimensional Lie group over KK, HH a subgroup of GG, VV an open neighbourhood of ee in GG and (Mi)i∈I(M_i)_{i \in I} a family of analytic manifolds over KK; for all i∈Ii \in I, let fif_i be a KK-analytic mapping of VV into MiM_i such that

H∩V={x∈V∣fi(x)=fi(e) for all i∈I}. H \cap V = \{ x \in V \mid f_i(x) = f_i(e) \text{ for all } i \in I \}.

(i) If K=CK = \mathbf{C}, HH is a Lie subgroup of GG. (ii) If KK is a finite extension of Qp\mathbf{Q}_p and II is finite, HH is a Lie subgroup of GG.

(i) Suppose that K=CK = \mathbf{C}. We consider GG as a real Lie group. Then HH is a real Lie subgroup of GG (Theorem 2). Let a∈L(H)a \in L(H). There exists a connected open neighbourhood WW of 00 in C\mathbf{C} such that exp⁡λa∈V\exp \lambda a \in V for all λ∈W\lambda \in W. Let i∈Ii \in I. Then fi(exp⁡λa)=fi(e)f_i(\exp \lambda a) = f_i(e) if λ∈R∩W\lambda \in \mathbf{R} \cap W. Hence fi(exp⁡λa)=fi(e)f_i(\exp \lambda a) = f_i(e) if λ∈W\lambda \in W by analytic continuation. Thus, exp⁡λa∈H\exp \lambda a \in H for λ∈W\lambda \in W and therefore μa∈L(H)\mu a \in L(H) for all μ∈C\mu \in \mathbf{C}. Therefore HH is a Lie subgroup of the complex Lie group GG (§ 4, no. 2, Proposition 2).

(ii) Suppose that KK is a finite extension of Qp\mathbf{Q}_p. We consider GG as a Lie group over Qp\mathbf{Q}_p. It is finite-dimensional and Theorem 2 implies that HH is a pp-adic Lie subgroup of GG. Since II is finite, ∏i∈IMi\prod_{i \in I} M_i is a manifold and it can be assumed that the family (fi)(f_i) reduces to a single mapping ff. Let a∈L(G)a \in L(G). Let ϕ\phi be an exponential mapping of GG. Then f(ϕ(λa))=f(e)f(\phi(\lambda a)) = f(e) for λ∈Qp\lambda \in \mathbf{Q}_p and ∣λ∣|\lambda| sufficiently small. Since ff is K-analytic, it follows that f(ϕ(λa))=f(e)f(\phi(\lambda a)) = f(e) for λ∈K\lambda \in K and ∣λ∣|\lambda| sufficiently small. Hence ϕ(λa)∈H\phi(\lambda a) \in H for λ∈K\lambda \in K and ∣λ∣|\lambda| sufficiently small and therefore μa∈L(H)\mu a \in L(H) for all μ∈K\mu \in K. The proof is completed as in (i).

Corollary 2 (ii) is no longer true if we omit the hypothesis that I is finite.

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

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