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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?

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    $\begingroup$ They can even be homology spheres or homology projective planes. $\endgroup$ Apr 18, 2023 at 14:43
  • $\begingroup$ Thank you! I see that my first question was a silly one then... $\endgroup$
    – TSU
    Apr 19, 2023 at 7:20
  • $\begingroup$ 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). $\endgroup$
    – YCor
    Jul 12, 2023 at 10:13

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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).

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

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