# Does a Riemannian manifold have a triangulation with quantitative bounds?

Suppose that $M$ is a closed Riemannian manifold with bounded geometry, i.e., curvature between $-1$ and $1$ and injectivity radius at least $1$. Since $M$ is a smooth manifold, it has a triangulation. Does it necessarily have a triangulation that is "nice" with respect to the metric?

For instance, is there an $\epsilon>0$ such that for any such $M$, there is a triangulation of $M$ whose simplices are all homeomorphic to the standard simplex by a map $f$ such that $f$ and $f^{-1}$ are both $\epsilon^{-1}$-Lipschitz?

The method I'm familiar with for constructing a triangulation is to embed $M$ in $\mathbb{R}^n$, construct a fine net of points in $M$, construct the Delaunay triangulation of those points, then project back to the manifold to get a triangulation. But this isn't very quantitative -- it depends on the embedding, and even if the embedding is nice, an unlucky choice of points will lead to some bad simplices. Is there a better way?

• Robert: There are two references which come close to what you want, but neither one does it. One is in Epstein et al book on word processing in groups (but only a sketch) and another is W. Breslin "Thick triangulations of hyperbolic n-manifolds". With extra work Breslin's arguments do generalize to the variable curvature case. Lastly, for most purposes, instead of a triangulation one can use "map to nerve", which is explained with all details in K. Grove and P. Petersen, Bounding homotopy types by geometry, Annals (1988). Jan 22, 2016 at 0:02
• One more sketch is in O. Attie, Quasi-isometry classification of some manifolds of bounded geometry, Math. Z. 216 (1994), 501–527. Jan 22, 2016 at 0:03
• Thank you for the references! I think I can see how to complete them. Usually, I would use a map to a nerve, but I ran into a problem where a true partition was necessary. Jan 22, 2016 at 2:41
• Robert: Let me know if and when you have a complete proof. In the GGT book we currently list this as an open problem, but with a reasonable path to a solution. Jan 22, 2016 at 18:54
• Could the almost-linear co-ordinates introduced by Jost and Karcher provide a means to proving this? You can find an English explanation of these co-ordinates in the following lecture notes: Jost, Jürgen Harmonic mappings between Riemannian manifolds. Proceedings of the Centre for Mathematical Analysis, Australian National University, 4. (I might be able to find my copy of it, if you think this would be useful) Jan 22, 2016 at 19:55

• If $(M^m,g)$ is of bounded geometry (in the $C^\infty$-sense, say), then there exists $\epsilon_0 > 0$ such that for that for any $0<\epsilon <\epsilon_)$ there is a countable cover of $M$ by geodesic balls $B_{\epsilon}(x_j)$, $\bigcup B_{\epsilon}(x_j) = M$, such that the cover of $M$ by the balls $B_{2\epsilon}(x_j)$ with double radius and the same centers is still uniformly locally finite.
Now choose points $y_k$ such that in each non-empty intersection of $B_{\epsilon}(x_j)$ they span an $m$-simplex, which is diffeomorphic to the standard $m$-simplex by any of the relevant exponential maps $\exp_{x_j}$. Then the maps you are looking for are just Riemannian exponential mappings which satisfy your assumptions by the properties of bounded geometry.
• This is good, but I think there are some constraints that need to be dealt with. For instance, if $m+1$ points are nearly coplanar in one chart, then in another chart, the orientation of that simplex could flip. So you need some bounds to make sure that the triangulation stays a triangulation between overlapping patches. I think this is roughly what Breslin does in his paper; he shows that, if you take small enough patches, then there is some set of points with well-defined Delaunay triangulation and limited "badness". Jan 22, 2016 at 16:46