# Is every compact smooth Riemannian manifold bilipschitz equivalent to a finite simplicial complex?

Let $$M$$ be a compact smooth Riemannian manifold. Then it admits a triangulation, i.e. a finite simplicial complex $$K$$ which is homeomorphic to $$M$$. Any such simplicial complex carries a natural metric - the path metric which concides with the usual metric on each simplex. Can we always choose $$K$$ so that with this path metric, it is bilipschitz equivalent to $$M$$ with the metric coming from its Riemannian structure?

• Of course some homeomorphisms are not lipschitz, but the triangulation theorem gives you a piecewise smooth triangulation. So it's bilipschitz when restricted to each simplex, you just have to worry about the seams. I'm not sure if that's enough. But the triangulation theorem gives you something even stronger: a PL (piecewise linear) manifold structure on the simplicial complex, ie, the link of every simplex is PL manifold triangulation of the sphere. This is very good for induction. So I claim that every triangulation produced by this theorem is bilipschitz. – Ben Wieland Apr 27 '19 at 19:32