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Fixing a dimension $n \ge 4$, is the class of closed hyperbolic $n$-manifolds recursively enumerable?

Since hyperbolic manifolds are triangulable I can reformulate this in the following more explicit way: say $M_1, \ldots, M_m,\ldots$ is an enumeration of all triangulations of $n$-manifolds. Is there a Turing machine which outputs a sequence $i_1, \ldots, i_m, \ldots$ such that each $M_{i_k}$ is hyperbolic, every hyperbolic $n$-manifold occurs as one of the $M_{i_k}$ and no two of them are homeomorphic to each other?

This problem reduces to that of deciding whether a triangulated manifold is hyperbolic, since the homeomorphism problem for hyperbolic manifolds is decidable. This is well-known to be possible when $n$ is $2$ or $3$ but i haven't found any references for $n \ge 4$.

Added later: actually it seems that it is not possible to list all triangulations of manifolds in dimensions $6$ and higher. The question still makes sense for n=4, 5. While it may not be possible to get a complete list of all triangulation of all smooth manifolds it is possible (as outlined by HJRW in the comments) to get a list of triangulations which include all smooth manifolds at least once. So the question makes sense for all dimensions again.

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    $\begingroup$ You could compute the fundamental group, then the isolated points of the $\mathrm{SO}_0(n,1)$ character variety. You can then probably check completeness using the developing map and your triangulation. Each should be a "finite" check, though it will take thousands of years to do the first handful... $\endgroup$
    – Toffee
    Jan 5, 2022 at 13:38
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    $\begingroup$ @Toffee: You seem to be suggesting that the Casson--Manning algorithm in 3 dimensions (arxiv.org/abs/math/0102154) also works in higher dimensions. That may well be true, but probably requires more than a couple of lines of justification! $\endgroup$
    – HJRW
    Jan 5, 2022 at 14:44
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    $\begingroup$ I'm curious about the added dimension restriction. Is this a theorem? It sounds like a plausible obstruction in the topological category, but since hyperbolic manifolds are in particular smooth and hence have a unique PL structure, it doesn't seem like it should be a problem. $\endgroup$
    – HJRW
    Jan 5, 2022 at 14:50
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    $\begingroup$ @JeanRaimbault: I have something fairly naive in mind too! Assume we have such a list in dimension $n-1$. Then, in particular, we can list some triangulations of spheres in dimension $n-1$, as discussed above (invoking, among other things, the Poincaré conjecture). Now, in dimension $n$, list all finite simplicial complexes $K$, and for each such $K$ in parallel check to see if every link is on our list of $(n-1)$-dimensional spheres…. $\endgroup$
    – HJRW
    Jan 5, 2022 at 16:57
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    $\begingroup$ … Output $K$ if every link is on the list. Inducting on $n$, the result is some list of triangulations of $n$-dimensional manifolds. I think it’s big enough to include all the smooth ones. $\endgroup$
    – HJRW
    Jan 5, 2022 at 16:58

1 Answer 1

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The class of closed hyperbolic manifolds is recursively enumerable. I’ll describe a terrible algorithm which nevertheless gives an enumeration.

A couple of basic facts: a hyperbolic $n$-manifold $M$ admits a triangulation by geodesic simplices, and the representation of the fundamental group into $PO(n,1)$ may be conjugated to have matrices with algebraic entries. The former follows from taking a dirichlet domain and subdividing, the latter from Mostow rigidity.

By taking the preimage of the triangulation in $\mathbb{H}^n$ (realized eg as one sheet of a hyperboloid in $\mathbb{R}^{n,1}$), we have a triangulation of $\mathbb{H}^n$ invariant under the action of $\pi_1(M)$ acting by algebraic matrices. We now perturb the vertices of the triangulation equivariantly to be points with algebraic coordinates (this is why we use simplices rather than more complicated polyhedra, since they are stable under small perturbations).

We may recover $M$ from a finite amount of this information: choose one simplex from each orbit, and for each pair of faces that are glued together, there will be a matrix in $PO(n,1)$ with algebraic entries gluing the faces together by an isometry.

Conversely, if we have a collection of simplices in $\mathbb{H}^n$ with algebraic coordinates, and the faces are paired by isometries, then we may check that they give rise to an $n$-manifold using the Poincaré polyhedron theorem. For each codimension two face of the simplices, we check that the angles about the faces sum to $2\pi$. This is possible since all of the coordinates of the vertices are algebraic. What I have in mind here is a version of Poincaré’s theorem due to Seifert; see Rob Riley’s paper.

To enumerate hyperbolic n-manifolds, recursively enumerate simplices in $\mathbb{H}^n$ with algebraic coordinates, together with pairings between the faces. Throw out the ones that don’t satisfy Poincaré’s theorem. Then throw out repeats eg by using Sela’s algorithm to eliminate manifolds with the same fundamental group.

(Let me know if you’d like more details on any aspect of this algorithm, I realize it’s very sketchy. )

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    $\begingroup$ That's a pretty clear sketch (though hardly practical, as you mention). It seems clear that one can enumerate algebraic points on an affine quadric, and Riley's paper gives a good degree of detail regarding verification of Poincaré's hypotheses. $\endgroup$ Jun 9, 2022 at 11:53
  • $\begingroup$ @JeanRaimbault Attempting a practical implementation, there are many things one could try to improve. For example, enumerate triangulations first like you suggested. Even though the sphere recognition (and hence manifold recognition) problem is not solvable, the space of hyperbolic manifold metrics on a simplicial complex ought to be semialgebraic, and hence computable. Thus one could search through pseudomanifold triangulations first and try to find a hyperbolic manifold structure. Also the finite volume case should be algorithmic too. $\endgroup$
    – Ian Agol
    Jun 10, 2022 at 0:49

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