# Cohomology of cocompact lattices in hyperbolic spaces

I have a (maybe too naive) hope that cocompact torsion-free arithmetic lattices in hyperbolic spaces $$X \neq \mathbb{H}_\mathbb{R}^2$$ are uniquely determined by their cohomology with coefficients in $$\mathbb{Z}$$. I was wondering if there are any known counter examples to this claim, that is:

Question: Are there torsion free cocompact lattices of hyperbolic spaces (real, complex or quaternionic of dimension $$\geq 3$$) that are not isomorphic but have the same cohomolgy?

• They can even be homology spheres or homology projective planes. Apr 18, 2023 at 14:43
• Thank you! I see that my first question was a silly one then...
– TSU
Apr 19, 2023 at 7:20
• Maybe you want to also exclude a bit more (e.g. very small dimension?). Also you need to specify whether you mean cohomology as graded abelian group, or as graded algebra (which retains more information).
– YCor
Jul 12, 2023 at 10:13

It depends on what you mean with "the same cohomology". If you don't care about the multiplicative structure and work on $$\mathbb{Q}$$ or $$\mathbb{R}$$, then you are asking whether the collection of Betti numbers determine the lattice, right?

I would then say that the answer is no (and this goes in the same direction as the comment by Moishe Kohan): in PU(2,1), there are examples of cocompact torsion free lattices which have the same Betti numbers as $$\mathbb{CP}^2$$. The corresponding quotients of the unit ball of $$\mathbb{C}^{2}$$ are called "fake projective planes" (see the first example by Mumford, and the more recent classification by Prasad--Yeung).

There are several such lattices which are non-isomorphic, thus answering negatively your question (in the "weak" setting I described).

• OP considers $\mathbf{Z}$-coefficients, so keeps track of torsion therein. Thus this is more than just Betti numbers.
– YCor
Jul 12, 2023 at 10:14
• According to this paper (arxiv.org/pdf/1801.05291.pdf), there are two that have H_1 = Z/7Z. Jul 12, 2023 at 10:44
• Even with the multiplicative structure, integer (co)homology spheres (of which there are many with real hyperbolic structures) show that the answer is “no”.
– HJRW
Jul 13, 2023 at 7:42
• Yes, I suspected that the rational cohomology would be too weak, but it is always nice to see concrete examples where things go wrong
– TSU
Jul 31, 2023 at 12:20