Proposition 7

(i) The family $(G(a_\lambda))$ is a central filtration on $G$ (Chapter II, § 4, no. 4, Definition 2). (ii) For $\lambda \in \mathbf{R}_+^*$, $G(a_\lambda) = \{ x \in G | \omega(x) \geq \lambda \}$, $G(a_\lambda^+) = \{ x \in G | \omega(x) > \lambda \}$. (ii) is obvious. We prove (i). Clearly $G(a_\lambda) = \bigcap_{\mu < \lambda} G(a_\mu)$ and $G = \bigcup_{\lambda > 0} G(a_\lambda)$. On the other hand, if $x \in G(a_\lambda)$ and $y \in G(a_\mu)$, then

by Proposition 6 applied with $a = a_\lambda, b = a_\mu, c = c' = a_{\lambda+\mu}$.

By Chapter II, § 4, no. 4, we can form the group $\mathrm{gr}(G)$ associated with the group $G$, with the central filtration $(G(a_\lambda))$. Writing $G_\lambda = G(a_\lambda)/G(a_\lambda^+)$ for all $\lambda > 0$, we obtain $\mathrm{gr}(G) = \bigoplus_{\lambda > 0} G_\lambda$. Recall (loc. cit., Proposition 1) that the commutator in $G$ allows us to define a bracket in $\mathrm{gr}(G)$ with which $\mathrm{gr}(G)$ is a Lie algebra, as follows: if $\bar{x} \in G_\lambda$ and $\bar{y} \in G_\mu$, choose a representative $x$ of $\bar{x}$ in $G(a_\lambda)$ and a representative $y$ of $\bar{y}$ in $G(a_\mu)$; then $[\bar{x}, \bar{y}]$ is the class of $x^{[-1]}y^{[-1]}x.y \in G(a_{\lambda+\mu})$ in $G_{\lambda+\mu}$. By Proposition 6, applied with $a = a_\lambda, b = a_\mu, c = a_{\lambda+\mu}, c' = a_{\lambda+\mu}^+$, we see that $[\bar{x}, \bar{y}]$ is also the class of $[x, y]$ in $G_{\lambda+\mu}$. Thus, when $G$ is considered as a Lie subalgebra of $L(G) = K^r$, filtered by the $G(a_\lambda)$, the associated graded Lie algebra (Chapter II, § 4, no. 3) is equal to $\mathrm{gr}(G)$.