Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is there an explicit bound on the number of tetrahedra needed to triangulate a hyperbolic 3-manifold of volume $V$? Or more generally a hyperbolic $n$-manifold of volume $V$?

share|improve this question
There can't be, as there's only finitely many manifolds triangulable with $n$ tetrahedra, and there's infinitely many hyperbolic manifolds of volume $\leq V$ provided $V$ is the volume of, say, the figure-8 knot complement -- there's infinitely many hyperbolic manifolds you get by filling in the boundary torus, and the volume is less than whatever the figure-8 complement volume is. –  Ryan Budney Sep 8 '10 at 18:28
The Wikipedia page with the relevant theorem: en.wikipedia.org/wiki/Hyperbolic_Dehn_surgery –  Ryan Budney Sep 8 '10 at 18:30
Er, I suppose such a bound in principal exists it just can't be a "nice" bound. There's only finitely many hyperbolic manifolds of a given volume, so the bound is the max of the minimal number of tetrahedra needed to triangulate those manifolds. That's not an informative bound. The argument above basically says bound, as a function of $V$ is awfully complicated and certainly not an increasing function in $V$ –  Ryan Budney Sep 8 '10 at 18:35
So the problem is that any such bound function $f(V)$ would "know" the volumes of all the cusped hyperbolic $3$-manifolds as that's where it would blow up. But this is the kind of information that (at present) is not readily available. Basic questions like "is this real number $x$ the volume of a hyperbolic $3$-manifold?" are difficult to answer. –  Ryan Budney Sep 8 '10 at 18:41
As far as I know, no one has computed an explicit constant (as pointed out, in dimension 3, one could only hope for a triangulation of the thick part). Breslin has some results: ams.org/mathscinet-getitem?mr=2507575 –  Ian Agol Sep 9 '10 at 6:12

1 Answer 1

up vote 18 down vote accepted

A couple of things are true: 1. If you have any Riemannian manifold of bounded infinitesimal geometry (curvature pinched above and below), its thick part, where the injectivity radius $> \epsilon$, can be triangulated with a number of simplices bounded by a constant times volume, where the constant depends on the curvature bounds and the dimension. I don't personally know the constant even for hyperbolic 3-manifolds, but I think there are people who can produce explicit bounds. This is basically a consequence of the compactness of the set of manifolds of bounded infinitesimal geometry and injectivity radius bounded below, together with the fact that all smooth manifolds admit a smooth triangulation, and that any smooth triangulation of a closed subset can be extended.

  1. For hyperbolic 3-manifolds, if you allow "spun triangulations" where some tetrahedra are allowed to have missing vertices that spiral infinitely around a short closed geodesic, then there is a similar bound, the number is less than some constant times volume. To do it: first triangulate the thick part leaving a boundary torus, then make cones on the boundary triangles that spiral around a short geodesic.

The answers are the same whether you're asking for a geodesic triangulation of a hyperbolic manifold, or any smooth triangulation.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.