In a blog post on ldtopology, a recent arxiv posting of Lins-Lins is discussed. The main argument of that paper is difficult to algorithmically distinguish two 3-manifolds and to that end the authors provide the challenge of showing two specific manifolds are not homeomorphic.

Nathan Dunfield implemented a computation of the index 6 subgroups of the fundamental groups of the two manifolds. In the comments of the aforementioned blog post, he showed this computation distinguishes these two sets of subgroups by abelianizing the subgroups. In particular, this indicates the two manifolds are not homeomorphic.

My first question is, "Can this trick always be used?", or more specifically:

Let $M_1$ and $M_2$ be non-homeomorphic (finite volume) hyperbolic 3-manifolds and let $S_{i,k}$ be the set of degree $k$ covers of $M_i$. Finally, let $H_{i,k}$ be the sets of integral first homology groups for the manifolds in $S_{i,k}$. Is there a known pair of non-homeomorphic (finite volume) hyperbolic 3-manifolds $M_1$ and $M_2$ such that for all $n$, the sets of $H_{1,n}$ and $H_{2,n}$ are identical, i.e. there is a bijection between them?

As a second question:

In the absence of a pair of hyperbolic examples, is there a known pair of 3-manifolds with non-isomorphic fundamental groups with this property?

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    $\begingroup$ Neil: I do not think that with the present technology one can prove or disprove this for hyperbolic manifolds. For non-hyperbolic ones, all Sol-manifolds have isomorphic homology. $\endgroup$
    – Misha
    Apr 26, 2013 at 11:21
  • $\begingroup$ Misha: Thank you for bringing up Sol-manifolds. I wrote the post with integral first homology groups in mind, and so I edited the question to reflect this. Since integral first homology can be often used to distinguish Sol-manifolds, do you know of a pair of Sol-manifolds that would answer the (refined) second question? $\endgroup$ Apr 26, 2013 at 12:26
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    $\begingroup$ It's natural to include in your data the action of $\pi_1M_i$ on $H_{i,k}$, which factors through a finite quotient. If you do, then a positive answer to either question would imply that $\pi_1M_1$ and $\pi_1M_2$ have isomorphic profinite completions. For 3-manifolds with non-solvable fundamental group, this is Question 9.28 in our survey article on 3-manifold groups (arXiv:1205.0202v3). Without this data, it's not so clear to me, but it seems likely that the two questions are equivalent (ie you can reconstruct the action on homology). $\endgroup$
    – HJRW
    Apr 26, 2013 at 12:39
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    $\begingroup$ Neil - there exist pairs of Sol manifolds with isomorphic profinite completions. I suspect they will answer your question in the integral case. $\endgroup$
    – HJRW
    Apr 26, 2013 at 12:40
  • $\begingroup$ (See, for instance, Corollary 1.4 of L. Funar, 'Torus bundles not distinguished by TQFT invariants'.) $\endgroup$
    – HJRW
    Apr 26, 2013 at 12:41

1 Answer 1


To atone for my comment above, I'm going to write out the Sol example. A Sol lattice takes the form $\mathbb Z^2\rtimes \mathbb Z$, where the action is hyperbolic. The rank of the abelianization is always 1, though the amount of torsion varies. The profinite completion is $\hat{\mathbb Z}^2\rtimes \hat{\mathbb Z}$. We can consider the group $\mathbb Z^2$ together with the action as a module over $\mathbb Z[t,t^{-1}]$. One invariant is the characteristic polynomial $P(t)$. Fixing this, we can consider the group as a module over $\mathbb Z[t,t^{-1}]/P(t)$. For hyperbolic actions, this ring is an order in a real quadratic number field. Moreover, every real quadratic number field has a unit, and thus has an order of this form. We can give $\mathbb Z^2$ the structure of a rank 1 projective module over this ring; and if the ring has nontrivial class group, there are non-isomorphic ways to do this. So this gives rise to two Sol lattices that are not isomorphic, but are pretty similar, having the same characteristic polynomial. The obstruction to their isomorphism is global, so when we take profinite completions and are dealing with modules over local rings, where all projectives are free, the groups become isomorphic. Thus their finite index subgroups are in bijection preserving the quotients of corresponding groups; and preserving the abelianizations. The action on the finite quotient group on the homology of the subgroup is trivial on the torsion-free part, because it is always just $\mathbb Z$, and matches the corresponding action in the profinite group, so does not distinguish the two discrete groups.

Similarly, it seems to me that there are pretty good candidates for hyperbolic three-manifolds with isomorphic profinite completions. Then these would have a bijection between finite index subgroups preserving the abelianizations and quotient groups, answering the original question. However, the action of the finite quotients on the abelianizations might distinguish them, although that is a subtle invariant. The idea is to take a imaginary quadratic number field with appropriate class group and form nonisomorphic rank 2 projective modules. For the right choice of projective modules, the automorphism groups are not isomorphic as group schemes and, I think, as discrete groups. Over the number field all projective modules are free, so the groups of rational automorphisms are equal, and thus the lattices are commensurable. Over the ring of integers in a local field, all projective modules are free, so the congruence completions are isomorphic. The congruence subgroup property does not apply to these groups, but the congruence kernel is attached to the rational group, so pretty much the same for the two groups. Compatibility between the congruence completion and the congruence kernel seems to me to be the big problem in checking this candidate. Or maybe you could use two $(3,1)$ quadratic forms in the same genus.


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