# Some questions about the Malcev completion

Let $G$ be an abstract group. The Malcev completion $\widehat{G}$ of $G$ (over $\mathbb{Q}$) is the set of group-like elements in the complete Hopf algebra $\widehat{\mathbb{Q}[G]} = \lim_n \mathbb{Q}[G]/I^n, I = \langle g-1 \,|\, g\in G\rangle.$ The nilpotent completion of $G$ is also defined by $G^{\mathrm{nil}}=\lim_n G/G^{(n)}, G^{(1)}=G, G^{(n+1)}= (G,G^{(n)}).$ These limit definitions induce two natural filtrations (both denoted by $F$) on Malcev and nilpotent completion. Also in both cases $\forall m,n, \,(F_n,F_m)\subset F_{n+m}$ and so there exists a natural structure of Lie algebras on the direct sum of graded quotients: $E\widehat{G} = \oplus F_n \widehat{G}/F_{n+1}\widehat{G}, EG^{\mathrm{nil}}= \oplus F_n G^{\mathrm{nil}}/F_{n+1}G^{\mathrm{nil}}$.

I have three questions on these concepts:

1- What is the relation between these two completion processes? I know (By theorems in Chapter 8 of Fresse's book: Homotopy of operads and Grothendieck-Teichmuller groups) in the case of finitely generated free groups, $E\widehat{G} \simeq EG^{\mathrm{nil}}\otimes \mathbb{Q}.$ Does this hold in general? If not, does there exist simple sufficient conditions for this to be true?

2-It's evident that $\widehat{G}^{(n)}\subset F_n(\widehat{G})$. Does there exist examples of $G$ for which this inclusion is strict for some $n$?

3-Is $\widehat{G}=\widehat{\widehat{G}}$ in general?

Thanks!

• Do you have reference for the "complete Hopf algebra" $\widehat{\mathbb{Q}[G]} = \lim_n \mathbb{Q}[G]/I^n$ ? I suppose it is not a standard Hopf algebra with comultiplication $\Delta : \mathcal{H}\rightarrow \mathcal{H}\otimes_\mathbb{Q} \mathcal{H}$, no ? – Duchamp Gérard H. E. Aug 24 '15 at 7:02
• @DuchampGérardH.E. $\mathbb{Q}[G]$ has a standard comultiplication defined by $\Delta(g) = g\otimes g, g\in G$ which respects the filtration by the powers of the augmentation ideal $I$. So this induces a coproduct on the $I$-adic completion. cf. chapters 7 and 8 in Fresse's book on homotopy of operads. – Mostafa Aug 24 '15 at 7:14
• I think the coproduct is $\Delta : \mathcal{H}\rightarrow \mathcal{H}\hat{\otimes}\mathcal{H}$ – Duchamp Gérard H. E. Aug 24 '15 at 10:20
• @DuchampGérardH.E. You are right, in fact $\widehat{\mathbb{Q}[G]}$ is a Hopf algebra in the category of complete filtered modules. – Mostafa Aug 24 '15 at 11:14
• OK, thanks, I got the chapter "Hopf algebras" in Fresse's book. – Duchamp Gérard H. E. Aug 24 '15 at 13:32

A nice exposition of all of this can be found in section 12,1 of the book of Chmutov-Duzhin-Mostovoy (http://www.pdmi.ras.ru/~duzhin/papers/cdbook/).

Assume for the sake of safety that $G$ is finitely generated. The series $G^{(n)}$ is characterized by the property that $G/G^{(n)}$ is the largest quotient of $G$ which is nilpotent of class $n$. There is another series $G_n$ characterized by the fact that the successive quotient are the largest torsion free quotient of $G$ which are nilpotent of class $n$. More or less by definition: $$G_n=\{x \in G\ |\ \exists k, x^k \in G^{(n)}\}$$

Now one can show that $G_n$ coincides with the filtration induced on $G$ by the augmentation ideal of $\mathbb{Q}[G]$, i.e. $G_n=(1+I^n)\cap G$.

The moral is that the difference between $G^{nil}$ and $\hat{G}$ is about torsion, hence disapears after tensoring with $\mathbb{Q}$. So:

1) is always true, it's a theorem of Quillen

I'm pretty confident that 2) is always an equality since you already completed everything

3) is definitely true: $\hat{G}$ is characterized by the property that the map $G\rightarrow \hat{G}$ is universal among maps from $G$ to pro-unipotent groups. Hence the pair $(\hat{G},id:\hat{G}\rightarrow \hat{G})$ realizes $\hat{G}$ as its own Malcev completion.

• A long time has passed since this thread was started, but nevertheless: where (as in bibliographical reference) can one find the result of Quillen which you mention at 1)? – ΑΘΩ Jun 14 at 6:20
• It should be in "On the associated graded ring of a group ring" (sciencedirect.com/science/article/pii/0021869368900690) – Adrien Jun 15 at 11:57