Bourbaki

Lemma 1

Tag 0A65, permanent. Cite it as [Bourbaki, Tag 0A65].

Lie Groups and Lie Algebras, chapter II, § 6. The Hausdorff series

Let g\mathfrak{g} be a filtered Lie algebra (§ 4, no. 1), (gα)α∈R(\mathfrak{g}_\alpha)_{\alpha \in \mathbf{R}} its filtration and let α∈R\alpha \in \mathbf{R}. Let P be a homogeneous Lie polynomial of degree n in the indeterminates (Γi)i∈I(\Gamma_i)_{i \in I} (§ 2, no. 4). Then P((ai))∈gnαP((a_i)) \in \mathfrak{g}_{n\alpha} for every family (ai)i∈I(a_i)_{i \in I} of elements of gα\mathfrak{g}_\alpha.

Every Lie polynomial of degree n⩾2n \geqslant 2 is a finite sum of terms of the form [Q,R][Q, R] where QQ and RR are of degree <n< n and the sum their degrees is equal to nn (§ 2, no. 7, Proposition 7). The lemma follows by induction on nn.

A Lie formal power series† (with coefficients in KK) in the indeterminates (Ti)i∈I(T_i)_{i \in I} is any element of the Lie algebra L^((Ti)i∈I)=L^(I)\hat{L}((T_i)_{i \in I}) = \hat{L}(I). Such an element uu can be written uniquely as the sum of a summable family (uν)ν∈N(I)(u_\nu)_{\nu \in \mathbf{N}^{(I)}} where uν∈Lν(I)u_\nu \in L^\nu(I).

Suppose that II is finite. Let gg be a complete Hausdorff filtered Lie algebra such that g=⋃α>0gαg = \bigcup_{\alpha > 0} g_\alpha; let t=(ti)i∈It = (t_i)_{i \in I} be a family of elements of gg.

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

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