Is there an explicit bound on the number of tetrahedra needed to triangulate a hyperbolic 3-manifold of volume V? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T11:29:05Z http://mathoverflow.net/feeds/question/38082 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38082/is-there-an-explicit-bound-on-the-number-of-tetrahedra-needed-to-triangulate-a-hy Is there an explicit bound on the number of tetrahedra needed to triangulate a hyperbolic 3-manifold of volume V? bb 2010-09-08T18:21:30Z 2010-09-08T20:16:43Z <p>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 $V$?</p> http://mathoverflow.net/questions/38082/is-there-an-explicit-bound-on-the-number-of-tetrahedra-needed-to-triangulate-a-hy/38099#38099 Answer by Bill Thurston for Is there an explicit bound on the number of tetrahedra needed to triangulate a hyperbolic 3-manifold of volume V? Bill Thurston 2010-09-08T20:16:43Z 2010-09-08T20:16:43Z <p>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.</p> <ol> <li>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.</li> </ol> <p>The answers are the same whether you're asking for a geodesic triangulation of a hyperbolic manifold, or any smooth triangulation.</p>