# Universal enveloping algebra of Malcev Lie algebra associated to nilpotent group

Let $$G$$ be a finitely generated torsion-free nilpotent group. The Malcev completion of $$G$$ is a nilpotent Lie group $$N$$ into which $$G$$ embeds as a lattice. One way to construct this is to take the completion $$\widehat{\mathbb{R}[G]}$$ of the real group ring with respect to the augmentation ideal. This is a Hopf algebra, and $$N$$ is the set of group-like elements in it.

The Lie algebra of $$N$$ is the set of primitive elements $$P(\widehat{\mathbb{R}[G]})$$. Let $$R$$ be the subalgebra of $$\widehat{\mathbb{R}[G]}$$ generated by $$P(\widehat{\mathbb{R}[G]})$$. Question: is $$R$$ the universal enveloping algebra of $$P(\widehat{\mathbb{R}[G]})$$?

If I understand the setup correctly, Quillen proved in his paper "On the associated graded of a group ring" that the associated graded of $$R$$ is the universal enveloping algebra of the associated graded of $$N$$, but I don't know if that is true before we take the associated graded.

Of course, the field $$\mathbb{R}$$ is only playing a minor role here, and the natural question is to replace it by an arbitrary field of characteristic $$0$$.

Yes, equivalently, for $$A = \widehat{kG}$$ where $$k$$ is of characteristic zero, the map $$U = \widehat{U(\mathrm{Prim }A)} \to A$$ is injective. In fact, this holds whenever $$A$$ is complete with respect to its augmentation ideal.
Proof: let $$f: U \to A$$ be the natural map, and let $$J$$ be the augmentation ideal of $$U$$.
If $$f$$ is not injective, then there is a least $$n$$ such that $$J^n \cap ker(f) \neq 0$$. Let $$x \in J^n \cap ker(f)$$. Then $$y=\Delta(x) - x\otimes 1 - 1\otimes x \in \sum_{p=1}^{n-1} J^p \otimes J^{n-p}.$$ Since $$ker(f)$$ is a Hopf ideal, $$0 = (f\otimes f)(\Delta(x)) - f(x)\otimes 1 - 1 \otimes f(x) = (f \otimes f)(y)$$. But $$f\otimes f$$ is injective on $$\sum_{p=1}^{n-1} J^p \otimes J^{n-p}$$ by hypothesis on $$n$$, so $$y = 0$$ and $$x$$ is primitive. But the primitive elements of $$U$$ are exactly $$\mathrm{Prim}\, A$$, and the map $$\mathrm{Prim}\, A \to A$$ is injective. Hence $$x = 0$$, a contradiction.
In fact, more is true when $$A$$ is cocommutative.
Theorem (Milnor-Moore): if $$A$$ is a cocommutative Hopf algebra over a field $$k$$ of characteristic zero, complete with respect to its augmentation ideal $$I$$, then $$A \cong U = \widehat{U(\mathrm{Prim}\, A)}$$, the completion of the enveloping algebra of its primitive elements.