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 $V$?



Is there an explicit bound on the number of tetrahedra needed to triangulate a hyperbolic 3-manifold of volume V?
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.


