Universal enveloping algebra of Malcev Lie algebra associated to nilpotent group

Question

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 $0$.

Answer 1

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.

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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.

See for instance the original paper of Milnor-Moore or the book Tensor Categories by Etingof et al.