Bourbaki

Proposition 1

Tag 0A4M, permanent. Cite it as [Bourbaki, Tag 0A4M].

Lie Groups and Lie Algebras, chapter II, § 4. Central filtrations

(i) Let α,β\alpha, \beta be in R\mathbf{R}. There exists a biadditive mapping

ϕαβ:grα(G)×grβ(G)→grα+β(G) \phi_{\alpha \beta} : \mathrm{gr}_\alpha(G) \times \mathrm{gr}_\beta(G) \to \mathrm{gr}_{\alpha + \beta}(G)

which maps (xGα+,yGβ+)(xG_\alpha^+, yG_\beta^+) onto (x,y)Gα+β+(x, y)G_{\alpha+\beta}^+.

(ii) Let ϕ\phi be the biadditive mapping of gr(G)×gr(G)\mathrm{gr}(G) \times \mathrm{gr}(G) into gr(G)\mathrm{gr}(G) whose restriction to grα(G)×grβ(G)\mathrm{gr}_\alpha(G) \times \mathrm{gr}_\beta(G) is ϕαβ\phi_{\alpha\beta} for every ordered pair (α,β)(\alpha, \beta). The mapping ϕ\phi gives gr(G)\mathrm{gr}(G) a Lie Z\mathbf{Z}-algebra structure.

(i) Recall the identity

(xx′,y)=(x,y)x′(x′,y)(xx', y) = (x, y)x'(x', y)

for x,x′,yx, x', y in GG (Algebra, Chapter I, § 6, no. 2, formula (4 bis)).

For x∈Gαx \in G_\alpha and y∈Gβy \in G_\beta, the class modulo Gα+β+G_{\alpha+\beta}^+ of the element (x,y)(x, y) of Gα+βG_{\alpha+\beta} will be denoted by f(x,y)f(x, y). For aa in Gα+βG_{\alpha+\beta} and x′x' in GG, a−1.ax′=(a,x′)∈Gα+β+a^{-1}.ax' = (a, x') \in G_{\alpha+\beta}^+; in particular f(x,y)f(x, y) is equal to the class modulo Gα+β+G_{\alpha+\beta}^+ of (x,y)x′(x, y)x'. Formula (15) therefore implies

f(xx′,y)=f(x,y)f(x′,y).f(xx', y) = f(x, y)f(x', y).

Now (y,x)=(x,y)−1(y, x) = (x, y)^{-1}, whence

f(y,x)=f(x,y)−1.f(y, x) = f(x, y)^{-1}.

From (16) and (17) we deduce

f(x,yy′)=f(x,y)f(x,y′).f(x, yy') = f(x, y)f(x, y').

We have to prove that the mapping f:Gα×Gβ→grα+β(G)f : G_\alpha \times G_\beta \to \mathrm{gr}_{\alpha+\beta}(G) defines on taking quotients a mapping ϕαβ:grα(G)×grβ(G)→grα+β(G)\phi_{\alpha\beta} : \mathrm{gr}_\alpha(G) \times \mathrm{gr}_\beta(G) \to \mathrm{gr}_{\alpha+\beta}(G). By (16) and (18) it suffices to prove that f(x,y)=0f(x, y) = 0 if x∈Gα+x \in G_\alpha^+ or y∈Gβ+y \in G_\beta^+, which follows from (13').

(ii) As (x,x)=e(x, x) = e, it follows from (17) that ϕ\phi is an alternating Z\mathbf{Z}-bilinear mapping. Hence it remains to prove that, for u∈grα(G)u \in \mathrm{gr}_\alpha(G), v∈grβ(G)v \in \mathrm{gr}_\beta(G) and w∈grγ(G)w \in \mathrm{gr}_\gamma(G),

ϕ(u,ϕ(v,w))+ϕ(v,ϕ(w,u))+ϕ(w,ϕ(u,v))=0.\phi(u, \phi(v, w)) + \phi(v, \phi(w, u)) + \phi(w, \phi(u, v)) = 0.

Let x∈Gα,y∈Gβx \in G_\alpha, y \in G_\beta and z∈Gγz \in G_\gamma be elements representing respectively u,vu, v and ww. We know that xyx^y and xx are two elements of GαG_\alpha which are congruent modulo Gα+G_\alpha^+ and hence xyx^y is a representative of uu in GαG_\alpha; as (y,z)(y, z) is a representative of ϕ(v,w)\phi(v, w) in Gβ+γG_{\beta+\gamma}, we see that (xy,(y,z))(x^y, (y, z)) is a representative of ϕ(u,ϕ(v,w))\phi(u, \phi(v, w)) in Gα+β+γG_{\alpha+\beta+\gamma}. By cyclic permutation, we see that (yz,(z,x))(y^z, (z, x)) and (zx,(x,y))(z^x, (x, y)) represent respectively ϕ(v,ϕ(w,u))\phi(v, \phi(w, u)) and ϕ(w,ϕ(u,v))\phi(w, \phi(u, v)) in Gα+β+γG_{\alpha+\beta+\gamma}. Relation (19) is then a consequence of the following identity (Algebra, Chapter I, § 6, no. 2, formula (15)):

(xy,(y,z)).(yz,(z,x)).(zx,(x,y))=e.(x^y, (y, z)).(y^z, (z, x)).(z^x, (x, y)) = e.

The Lie algebra gr(G)\mathrm{gr}(G) over Z\mathbf{Z} defined in Proposition 1 is called the graded Lie algebra associated with the filtered group GG.

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

Cited by

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Cites

Transcribed from the printed volume. Not translated.