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 $0$.


1 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.

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.