Bourbaki

Exercise 26, § 4

Lie Groups and Lie Algebras, I, 98

Let GG be a group and let (Hn)n≥1(H_n)_{n \geq 1} be a decreasing sequence of subgroups of GG; suppose that H1=GH_1 = G and that if we write (x,y)=xyx−1y−1(x, y) = xyx^{-1}y^{-1}, the relations x∈Hi, y∈Hjx \in H_i,\ y \in H_j imply (x,y)∈Hi+j(x, y) \in H_{i+j}.

(a) Let Gi=Hi/Hi+1G_i = H_i/H_{i+1}; show that GiG_i is commutative and that the mapping x,y↦(x,y)x, y \mapsto (x, y) defines on passing to the quotient a Z\mathbf{Z}-bilinear mapping of Gi×GjG_i \times G_j into Gi+jG_{i+j}.

(b) Let gr⁡(G)=∑i=1∞Gi\operatorname{gr}(G) = \sum_{i=1}^\infty G_i and extend by linearity the mappings Gi×Gj→Gi+jG_i \times G_j \to G_{i+j} defined in (a) to a Z\mathbf{Z}-bilinear mapping of gr⁡(G)×gr⁡(G)\operatorname{gr}(G) \times \operatorname{gr}(G) into gr⁡(G)\operatorname{gr}(G). Show that gr⁡(G)\operatorname{gr}(G) is a Lie Z\mathbf{Z}-algebra with this mapping (to verify the Jacobi identity, use the following formula:

((x,y),zy).((y,z),xz).((z,x),yx)=ex,y,z in G, ((x, y), z^y) . ((y, z), x^z) . ((z, x), y^x) = e \qquad x, y, z \text{ in } G,

where xyx^y denotes yxy−1yxy^{-1} and ee denotes the identity element of GG.

(c) Suppose that there exists nn such that Hn={e}H_n = \{e\}. Show that gr⁡(G)\operatorname{gr}(G) is a nilpotent Lie Z\mathbf{Z}-algebra.

Solution

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Transcribed from the printed volume. Not translated.