Topological manifolds of dimension ≠4 have a [Lipschitz structure](http://en.wikipedia.org/wiki/Lipschitz_continuity#Lipschitz_manifolds). [Ed: Is this "well-known"? Is it obvious? Can somebody give a reference?] Does this imply the following result?



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> 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.
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In other words, can two manifolds (of dimension ≠4) be homeomorphic without being [bi-Lipschitz homeomorphic](http://en.wikipedia.org/wiki/Lipschitz_continuity#Lipschitz_manifolds)?



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


