Bourbaki

Proposition 5

Tag 0A32, permanent. Cite it as [Bourbaki, Tag 0A32].

Lie Groups and Lie Algebras, chapter II, § 2. Free Lie algebras

Let gg be a Lie algebra and PP a submodule of gg. We define the submodules PnP_n of gg by the formulae P1=PP_1 = P and Pn+1=[P,Pn]P_{n+1} = [P, P_n] for n≥1n \geq 1. Then

[Pm,Pn]⊂Pm+n, [P_m, P_n] \subset P_{m+n},
Pn=∑p=1n−1[Pp,Pn−p]for n≥2. P_n = \sum_{p=1}^{n-1} [P_p, P_{n-p}] \quad \text{for } n \geq 2.
We prove (18) by induction on mm. The case m=1m = 1 is obvious. By the Jacobi identity,
[[P,Pm],Pn]⊂[Pm,[P,Pn]]+[P,[Pm,Pn]], [[P, P_m], P_n] \subset [P_m, [P, P_n]] + [P, [P_m, P_n]],
that is
[Pm+1,Pn]⊂[Pm,Pn+1]+[P,[Pm,Pn]]. [P_{m+1}, P_n] \subset [P_m, P_{n+1}] + [P, [P_m, P_n]].
The induction hypothesis implies that [Pm,Pn+1]⊂Pm+n+1[P_m, P_{n+1}] \subset P_{m+n+1} and [Pm,Pn]⊂Pm+n[P_m, P_n] \subset P_{m+n}, whence
[Pm+1,Pn]⊂Pm+n+1+[P,Pm+n]=Pm+n+1. [P_{m+1}, P_n] \subset P_{m+n+1} + [P, P_{m+n}] = P_{m+n+1}.
By formula (18), Pn⊃∑p=1n−1[Pp,Pn−p]⊃[P1,Pn−1]=PnP_n \supset \sum_{p=1}^{n-1} [P_p, P_{n-p}] \supset [P_1, P_{n-1}] = P_n, whence (19).

When we take P=gP = g, the sequence (Pn)(P_n) is the lower central series (Cng\mathscr{C}^n g) of gg (Chapter I, § 1, no. 5). Hence:

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

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Transcribed from the printed volume. Not translated.