I promised to return to a more mathematical summary of the conference in Ventotene, and indeed I shall do so in the next two posts.

One of the themes of the conference was bounding the order of the torsion subgroup in arithmetic lattices. Tsachik Gelander gave a number of talks (in part) on the seven author paper. One nice result was a uniform bound of the shape

where ranges (say) over all lattices in for a fixed (The key result here is the uniformity — this result is much easier for covers of a fixed manifold.)

Two natural questions that came up (in conversation at least) during the conference are as follows:

- Can one do better in low degree?
- What is the true expectation for the size of this group for (say) congruence subgroups of

Let’s consider the first question. For (congruence) subgroups of , one can certainly say quite a bit more. For example, is essentially trivial, by the congruence subgroup property. However, in the stable range of cohomology (in particular, when the completed cohomology groups become stable), the groups are finite over and so contribute very little. One does, at least, have the following soft arguments for general groups.

** Proposition: ** Let be a semi-simple group over with -rank . Then is a torsion -module for

The proof is as follows: the boundary terms are also torsion, so it suffices to show that all the in the appropriate range are also torsion, where we consider Borel-Moore homology. Assume otherwise. Let From the spectral sequence we deduce that there is at least one such that Yet the homological dimension of is, by Borel-Serre, equal to and so all the homology in these degrees (and hence certainly the completed homology) vanishes.

One can do better in certain algebraic cases, where one can deduce vanishing of the completed cohomology in certain degrees by perfectoid technology (as in Corollary 4.2.3 of Scholze’s paper).

The answer to the second question, even conjecturally, is more mysterious. There are some speculations related to this question in Bergeron-Venkatesh. But it seems a little tricky to formulate a precise guess (for a good upper bound).