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?



Does a Riemannian manifold have a triangulation with quantitative bounds?
A series of papers has been published (or rather is being published) on this topic, staring with Stability of Delaunay-type structures for manifolds (see <http://dl.acm.org/citation.cfm?id=2261284>) by Boissonnat, Dyer, and Ghosh. I think the paper in the series you might find most useful is Delaunay triangulation of manifolds (see <https://arxiv.org/abs/1311.0117>). 


