Various authors (including Bergeron and Venkatesh) have shown that the cohomology of certain arithmetic groups have a *lot* of torsion. For example, if is a co-compact arithmetic lattice in , and is an acyclic local system, then

The proof relies on the fact that the difference in ranks of and is one. As the invariant grows, one expects there to be less torsion. How much torsion should one expect in general? I’m not sure I have an answer, but the point of this post is that Poincare duality gives a non-trivial bound, at least if one restricts to covers up a -adic tower. Let be a semi-simple group over , Let , let be a maximal compact, let , let be a co-compact lattice, and let be an acyclic local system. Suppose that and . Then, for a fixed prime (for which is split) and varying , I claim that one has the inequality

An elementary exercise shows that is trivial as a local system for and large enough . The inequality above can then be reduced to the following claim: there is an inequality:

Assume otherwise. The main point is as follows: taking the inverse limit over all , we obtain modules over the Iwasawa algebra . This algebra, by results of Lazard and Venjakob, is essentially a regular local ring, in particular, it makes sense to talk about the dimension of modules over that ring. If the inequality above does not hold, then these modules will have small dimension, explicitly, co-dimension greater than . This is so small that Poincare duality will, Ouroboros like — swallow itself completely and collapse into nothingness. However, the only way that could happen is if there was nothing to start with, which is nonsense.

More mathematically, consider the completed homology groups

The homology groups may be computed by a complex of free -modules obtain by lifting an initial triangulation on the base. (Here one thinks of group cohomology as the cohomology of the associated arithmetic quotients, of course.) Poincare duality then explains what happens when one takes the dual of this sequence and considers the corresponding homology groups, namely, there is a spectral sequence:

This spectral sequence might be more familiar to some readers if one imagines to be a field, in which case the zeroth Ext group is a Hom and the higher Exts vanish, and one obtains the duality isomorphisms between homology and cohomology over a field. Or, if was the integers, then then zeroth Ext group is a Hom, the first Ext group is torsion, the higher Ext groups vanish, and one obtains the usual short exact sequence comparing the dual of homology to cohomology up to a torsion error term.) The dimension assumption we made implies that the limits are small as -modules, in particular that for all . The key here is a Theorem of Ardakov and Brown relating the size of the cohomology growth under towers to the codimension of the module. Yet putting this assumption into the spectral sequence shows that all terms with vanish, and hence that . Yet it is easy to see that

, and thus we have a contradiction.

In fact, this is the same argument that ME and I used to give lower bounds on torsion for -adic analytic covers of -manifolds. There is some slack where the argument can be improved – since one only needs vanishing for a triangular portion of the spectral sequence, you are in good shape if you have extra information about the lower rows. Of course, the * real* answer to the amount of mod p torsion in these towers (which is a different question to the original one of torsion over the integers) should be:

where was defined above.

**Edit:** In a previous version of this post, I confused the roles of and . For complex groups one has , and this is asymptotically the correct estimate for simple real groups. In general, one has , with the worse case, ironically, corresponding to (any number of copies of) . So you get a bound of the form:

Nice post. By the way, when you wrote you meant …

Dear Florian, you are completely correct. It has now been fixed.

I like the argument.

However, I think it is not so easy to relate the obtained lower bound to the index. In fact, the inequality n \geq 2d is not correct if G(R) has sufficiently many compact factors. But we can’t ignore the compact factors when calculating the index of the congruence subgroups, so I think the square root bound does not follow from the argument.

Thanks for the remark. I had in my notes, but the only examples I computed were for complex lie groups for which . Then when I went to prove the inequality , I “conveniently” replaced by for which the corresponding equality

isvalid (except, as you say, when is not compact).