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Suppose I have a smooth Riemannian manifold $X$ with induced distance function $d$, and a bi-Lipschitz (with respect to $d$) homeomorphism $$\phi: X \to X.$$

Under what circumstances could $\phi$ be smoothable to a diffeomorphism? By "smoothable" in this case I mean "homotopic to a diffeomorphism through bi-Lipschitz homeomorphisms" (this might not be standard, I suppose). Are there any clear obstructions?

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    $\begingroup$ What do you mean by "smoothable"? $\endgroup$
    – Igor Belegradek
    Commented Jul 19, 2020 at 20:38
  • $\begingroup$ @Igor Belegradek: I guess the OP means "approximated by $C^1$ diffeomorphisms", with uniform convergence? $\endgroup$
    – leo monsaingeon
    Commented Jul 19, 2020 at 21:08
  • $\begingroup$ Sorry, just edited the question to make it more clear what I mean! $\endgroup$
    – Rohil Prasad
    Commented Jul 19, 2020 at 22:58

2 Answers 2

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Any self-homeomorphism of a manifold of dimension $\neq 4$ is topologically isotopic to a bi-Lipschitz homeomorphism, see lemma 2.4 in Lipschitz and quasiconformal approximation of homeomorphism pairs by Jouni Luukkainen.

There are exotic spheres (e.g. in dimension $7$) that admit a self-homeomorphism that is not homotopic to a diffeomorphism, see here. This gives a bi-Lipschitz homeomorphism that is not homotopic to a diffeomorphism.

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  • $\begingroup$ This is interesting, thanks. Is there a counterexample in dimension 4? Perhaps one can appeal to Donaldson-Sullivans work on Lipschitz/quasi-conformal 4-manifolds... $\endgroup$
    – Rohil Prasad
    Commented Jul 20, 2020 at 16:13
  • $\begingroup$ @RohilPrasad: I do not know what happens in dimension 4. You now have all the references that I have, and should be able to explore it further. $\endgroup$
    – Igor Belegradek
    Commented Jul 20, 2020 at 16:27
  • $\begingroup$ I am not sure if the answer is correct. The link shows a self-homeomorphis, that is not isotopic to a diffeomorphism. This is because the manifold with reversed orientation are not diffeomorphic. I think the homeomorphism cannot be bi-Lipschitz due to uniqueness of the Lipschitz structure in dimensions $\neq 4$. $\endgroup$
    – Piotr Hajlasz
    Commented Sep 23, 2020 at 1:39
  • $\begingroup$ @PiotrHajlasz: I do not claim to understand Luukkainen's paper, but I think what I say is a formal consequence of his Lemma 2.4 for $A=\emptyset$ and $B=M$. $\endgroup$
    – Igor Belegradek
    Commented Sep 23, 2020 at 1:59
  • $\begingroup$ I am sorry, I was stupid. Exotic spheres are bi-Lipschitz homeomorphic exactly for a reason I mentioned: uniqueness of a Lipschitz structure. $\endgroup$
    – Piotr Hajlasz
    Commented Sep 23, 2020 at 3:03
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There is some interest in a related question in non-linear elasticity, specifically people there would consider a function "smoothable" if there is a close-by (in some norm applying both to function and its inverse) diffeomorphism.

In 2D there are some results with regards to this, see e.g. Smooth approximation of bi-Lipschitz orientation-preserving homeomorphisms by Danieri & Pratelli, who prove that there is a close diffeomorphism in some bi-Sobolev norm for domains in the plane (which if I am not mistaken should imply the same result at least for compact manifolds). But the proof uses a bi-Lipschitz extension theorem, so I am not sure if one can construct homotopies from that result easily and there seems to be no way to extend this to higher dimensions.

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