For each $\lambda\in[0,+\infty]_{{\bf Q}}:={\bf Q}_{\geq 0}\cup\infty$ we define a Lie subalgebra

$${\bf g}_{\lambda}=\left\{\sum_{n_{1},n_{2}}c_{m,n}R_{\alpha_{1}}^{-n_{1}}R_{\alpha_{2}}^{-n_{2}}\in{\bf g}\,|\,\,c_{n_{1},n_{2}}\in K,\,\,{n_{2}\over{n_{1}}}=\lambda\,\right\}\,\,.$$

Each ${\bf g}_{\lambda}$ is an abelian Lie algebra. It carries the induced filtration by Lie algebras ${\bf g}_{\lambda}^{\geq k}={\bf g}_{\lambda}\cap{\bf g}^{\geq k}$. Denote by $G_{\lambda}=\exp({\bf g}_{\lambda})$ the corresponding pro-nilpotent group.

Proof. Indeed, for the monomial $R_{\alpha_{1}}^{-n_{1}}R_{\alpha_{2}}^{-n_{2}}\in{\bf g}_{\lambda}$ which maps non-trivially to the quotient ${\bf g}_{\lambda}/{\bf g}^{\geq k}_{\lambda}$ we have: $n_{1}+n_{2}\leq k,n_{1}/n_{2}=\lambda$, where $n_{1},n_{2}$ are non-negative integers. There are finitely many such non-negative integers $n_{1}$ and $n_{2}$. $\blacksquare$

It follows from the Lemma that we have a natural isomorphism of vector spaces $\prod_{\lambda\in[0,+\infty]_{{\bf Q}}}{\bf g}_{\lambda}/{\bf g}_{\lambda}^{\geq k}\to{\bf g}/{\bf g}^{\geq k}$, hence the map

$$(f_{\lambda})_{\lambda\in[0,+\infty]_{{\bf Q}}}\mapsto\sum_{\lambda}f_{\lambda}=\sum_{i=1}^{N_{i}}f_{\lambda_{i}},\,\,\,\,\mbox{ where }f_{\lambda}\in{\bf g}_{\lambda}/{\bf g}_{\lambda}^{\geq k}\,\,\,\forall\lambda\in[0,+\infty]_{{\bf Q}}$$

is well-defined and gives rise (after taking the projective limit as $k\to+\infty$) to the isomorphism ${\bf g}\simeq\prod_{\lambda\in[0,+\infty]_{{\bf Q}}}{\bf g}_{\lambda}$.

In a similar way we define the map $\prod_{\to}:\prod_{\lambda\in[0,+\infty]_{{\bf Q}}}G_{\lambda}\to G$, the product is taken with respect to the natural order on ${\bf Q}$. Namely, for any $k\geq 1$ we define

$${\textstyle\prod_{\to}^{(k)}}:\prod_{i=1}^{N_{k}}G_{\lambda_{i}}/G_{\lambda_{i}}^{\geq k}\to G/G^{\geq k}\,,\,(g_{1},\dots,g_{N_{k}})\mapsto g_{1}\dots g_{N_{k}},\,\mbox{ for }g_{i}\in G_{\lambda_{i}}/G_{\lambda_{i}}^{\geq k}$$

and then set $\prod_{\to}:=\varprojlim_{k}\prod_{\to}^{(k)}\,$.

Proof. Let $k\geq 1$ be an integer. We claim that $\prod_{\to}^{(k)}$ is a bijection of sets (this implies the proposition by taking the projective limit as $k\to+\infty$). We will prove the bijection by induction in $k$ . Case $k=1$ is obvious because all the groups under considerations are trivial.

We would like to prove that $\prod_{\to}^{(k+1)}$ is a bijection assuming that $\prod_{\to}^{(k)}$ is a bijection. Let $h$ be an element of $G/G^{\geq k+1}$ and $\overline{h}$ its image in $G/G^{\geq k}$. By the induction assumption there exist unique $\overline{h}_{i}\in G_{\lambda_{i}}/G_{\lambda_{i}}^{\geq k+1},1\leq i\leq N_{k+1}$ such that $\overline{h}_{1}\dots\overline{h}_{N_{k+1}}=\overline{h}$. Let $h_{i},1\leq i\leq N_{k+1}$ be any liftings of $\overline{h}_{i}$ to $G_{i}/G_{i}^{\geq k}$. Then $h_{1}\dots h_{N_{k+1}}=h\pmod{G^{\geq k}}$, hence $c:=h_{1}\dots h_{N_{k+1}}h^{-1}$ belongs to $G^{\geq k}/G^{\geq k+1}\subset Center(G/G^{\geq k+1})$. The last inclusion holds because $[{\bf g},{\bf g}^{\geq k}]=[{\bf g}^{\geq 1},{\bf g}^{\geq k}]\subset{\bf g}^{\geq k+1}$.

Next we observe that the isomorphism of abelian Lie algebras

$$\bigoplus_{1\leq i\leq{N_{k+1}}}{\bf g}_{\lambda_{i}}^{\geq k}/{\bf g}_{\lambda_{i}}^{\geq k+1}\simeq{\bf g}^{\geq k}/{\bf g}^{\geq k+1}$$

implies an isomorphism of the corresponding abelian groups

$$\prod_{1\leq i\leq{N_{k+1}}}G_{i}^{\geq k}/G_{i}^{\geq k+1}\simeq G^{\geq k}/G^{\geq k+1}\,\,.$$

Hence we can write uniquely $c=c_{1}\dots c_{N_{k+1}}$, where $c_{i}\in G^{\geq k}/G^{\geq k+1}\subset Center(G/G^{\geq k+1})$. It follows that $\prod_{\to}^{(k+1)}\left((h_{i}c_{i}^{-1})\right)=h$. Also it is now clear that this decomposition of $h$ is unique. This concludes the proof. $\blacksquare$