Fix a field $k$. Given a discrete group $G$, the pro-algebraic completion (or Hochschild-Mostow completion) is the pro-algebraic group $A_{k}(G)$ with is universal with respect to finite dimensional representations of $G$. We usually omit the field $k$ from the notation when it is clear.

I have encountered pro-algebraic completions in a number of places, especially in relation to the schemetization problem, however there are next to no examples of pro-algebraic completions of interesting groups anywhere in the literature.

The only example that I have found is of the pro-algebraic completion of $\mathbb{Z}$, which is

$$A(\mathbb{Z})=\mathbb{G}_{a}(k)\times T\times \widehat{\mathbb{Z}},$$

where $T$ denotes denotes the pro-torus whose character group is $\mathrm{Hom}(\mathbb{Z},k^{\times})$, and $\widehat{\mathbb{Z}}$ denotes the pro-finite completion of the integers. This example, and its direct generalisation to abelian groups, was found in the following paper of Bass et.al.


Can anyone provide any other examples, or references containing examples, of pro-algebraic completions?

Tangentially, does anyone know of any methods to compute pro-algebraic completions?

I am particularly interested in understanding the pro-algebraic completions of the symmetric groups and the free groups.

  • 2
    $\begingroup$ I'm guessing that the pro-algebraic completion of $SL(n,\mathbb{Z})$, $n \geq 3$ ought to be known (or computable), since all of its representations to Lie groups (and presumably algebraic groups) are known via superrigidity. en.wikipedia.org/wiki/Superrigidity The free group might be trickier, since presumably the algebraic completion is at least as complicated as the profinite completion, which is essentially all rank $n$ finite groups (so is not well-understood in general). $\endgroup$
    – Ian Agol
    Mar 5, 2018 at 17:29

1 Answer 1


You ask specifically about the symmetric groups and the free groups. This is a bit funny, as a finite group forms its own proalgebraic completion (a slight generalization of this fact is Proposition 1 in Bass-Lubotzky-Magid-Mozes), while the proalgebraic completion of a non-abelian free group is clearly not a manageable object. Note that it follows that every group which virtually surjects onto a non-abelian free group, e.g a surface group or $\text{SL}_2(\mathbb{Z})$, has a huge completion.

As you mentioned, for f.g abelian groups it should be rather easy to work out the proalgebraic completion. I suppose this is also the case for f.g nilpotent groups: the completion should be the Malcev completion times a protorus times the profinite completion. I would start by thinking explicitly of the Heisenberg group to verify this guess.

As mentioned in Ian Agol's comment, Margulis Super-Rigidity provides a powerful tool for computing proalgebraic completions for higher rank arithmetic groups. This is also mentioned in the introduction of BLMM. In particular, when you also have the Congruence Subgroup Property you get very neat results, e.g the proalgebraic completion of $\text{SL}_n(\mathbb{Z})$ is $\text{SL}_n(\mathbb{C})\times \text{SL}_n(\hat{\mathbb{Z}})$ for $n\geq 3$.


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