# Can two Riemannian manifolds (dim≠4) be homeomorphic without being bi-Lipschitz homeomorphic?

Topological manifolds of dimension ≠4 have a Lipschitz structure. [Ed: Is this "well-known"? Is it obvious? Can somebody give a reference?] Does this imply the following result?

Assume M and N are smooth Riemannian manifold, with same dimension other than 4. If M homeomorphic to N, then M is bi-Lipschitz homeomorphic to N.

In other words, can two manifolds (of dimension ≠4) be homeomorphic without being bi-Lipschitz homeomorphic?

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Is it a question or a statement? – Anton Petrunin Mar 30 '10 at 22:41
Whatever it is, it doesn't seem to be at.algebraic-topology. – Reid Barton Mar 30 '10 at 22:44
This question needs massive improvement. Complete sentences and correct capitalization would help; glossing the terms would also be appropriate; motivation and background, here on MO, is generally a must-have. – Theo Johnson-Freyd Mar 30 '10 at 23:03
It's not that bad, Theo, surely? My reading of it is as a question about whether two (Riemannian) smooth manifolds of dimension $\ne 4$ can be homeomorphic without being bi-Lipshitz holomorphic. – Yemon Choi Mar 30 '10 at 23:08
Yes, but probably you should spell it with the "c" ;-) – JBL Mar 30 '10 at 23:49

If you don't assume compactness, then no. Silly example: $\mathbb R^1$ and $(0,1)$. Example with complete metrics: $\mathbb R^2$ and $\mathbb H^2$ (they have essentially different volume growths and hence are not bi-Lipschitz equivalent).
If $M$ and $N$ are closed, then yes, by Sullivan's uniqueness result pointed out to by Leonid Kovalev in comments (provided that the MR review is correct - I'm not an expert in any way and don't have access to the paper). The uniqueness means that for every two Lipschitz structures there is an isomorphism between them. And for Lipschitz structures defined by Riemannian metrics, isomorphisms are a locally bi-Lipschitz homeomorphisms of the metrics. By compactness, locally bi-Lipschitz implies bi-Lipschitz.