Assume we have a homeomoprhism $\phi:M\rightarrow M$, where $M$ is a topological manifold which admits at least one smooth structure.

Is it always possible to construct a smooth structure on $M$ w.r.t to it $\phi$ will be a diffeomorphism?

Of course when there is little freedom in defining the smooth structure the answer is clearly no in general: Take the classic example of $M=\mathbb{R}$, $\phi(x)= x^3$. $\phi$ is a homeomorphism, but its inverse is not smooth w.r.t to the standard smooth strucutre on $\mathbb{R}$. (which is the only one that exists on $\mathbb{R}$). Note that $\phi$ itself is smooth.

Update: As pointed out by Andy Putman I was wrong to think there is no smooth structure making $\phi$ a diffeomorphism. (Such a structure in fact exsits, is clearly different from the standard one, although diffeomorphic to it).

I guess in most cases the answer will be negative. However, I would still like to find some sufficient or necessary conditions on $M$ and $\phi$ that ensure such a smooth structure exists. (In particular , is it always possible to ensure that at least one of $\phi$,$\phi^{-1}$ is smooth?).

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    $\begingroup$ Just to bring out the main theme of my answer, if you fix a smooth structure (as you can in many cases, for instance in low dimensions), you are asking for conditions that ensure that a homeomorphism is topologically conjugate to a diffeomorphism. There is no easy answer to this -- it depends in a delicate way on the dynamics of the homeomorphism. For example, the $5$-sphere has a unique smooth structure, and I have no idea what possible form an answer to this question would take there. $\endgroup$ – Andy Putman Apr 27 '15 at 19:05

Let me first answer your last question in the negative: there exist homeomorphisms $f:M \rightarrow M$ of smoothable manifolds $M$ such that neither $f$ nor $f^{-1}$ are smooth with respect to any smooth structure on $M$. In fact, we can take $M = S^3$. Bing has constructed homeomorphisms $f:S^3 \rightarrow S^3$ such that $f \circ f = \text{id}$ and such that the fixed set of $f$ is the Alexander horned sphere. See

Bing, R. H. A homeomorphism between the 3-sphere and the sum of two solid horned spheres. Ann. of Math. (2) 56, (1952). 354–362.

The map $f$ (and hence the map $f^{-1} = f$) is not smooth with respect to any smooth structure on $S^3$ because of a theorem of P. Smith that says that the fixed set of any periodic diffeomorphism of $S^3$ is a smoothly embedded surface or circle. This is the beginning of a long story that culminated in the proof of the Smith conjecture, which says that any periodic diffeomorphism of $S^3$ is smoothly conjugate to an element of the orthogonal group. This was one of the early triumphs of Thurston's work; for a detailed account of it, see the book

The Smith conjecture. Papers presented at the symposium held at Columbia University, New York, 1979. Edited by John W. Morgan and Hyman Bass. Pure and Applied Mathematics, 112. Academic Press, Inc., Orlando, FL, 1984. xv+243 pp. ISBN: 0-12-506980-4

This brings me to my second point. It is FALSE that the map $g:\mathbb{R} \rightarrow \mathbb{R}$ defined via $g(x) = x^3$ is not a diffeomorphism with respect to any smooth structure on $\mathbb{R}$. It is true that $\mathbb{R}$ has a unique smooth structure (as does $S^3$ above), but this means something weaker than what you are claiming it means. Namely, if $X$ and $Y$ are any two smoothings of $\mathbb{R}$, then there exists a diffeomorphism $X \rightarrow Y$; however, this diffeomorphism might not be the identity.

Here is a sketch of how you construct the appropriate smooth structure. Constructing this smooth structure is equivalent to constructing a homeomorphism $\psi:\mathbb{R} \rightarrow \mathbb{R}$ such that $\psi \circ g \circ \psi^{-1}$ is a smooth map; the desired smooth structure is then obtained by pulling back the standard smooth structure on $\mathbb{R}$ via $\psi$. Here are some observations about $g$:

  1. It has three fixed points, namely $-1$ and $0$ and $1$.

  2. On the interval $(-\infty,-1)$, it shifts points to the left.

  3. On the interval $(-1,0)$, it shifts points to the right.

  4. On the interval $(0,1)$, it shifts points to the left.

  5. On the interval $(1,\infty)$, it shifts points to the right.

It is easy to construct a diffeomorphism $h: \mathbb{R} \rightarrow \mathbb{R}$ with the above properties (since $0$ will be an attracting fixed point of $h$, the derivative of $h$ at $0$ will be a number $\alpha$ satisfying $0 < \alpha < 1$). But it is then an easy exercise to see that $h$ will be topologically conjugate to $g$.

  • $\begingroup$ I think it's probably a little easier to show $f$ is not conjugate to a diffeomorphism. Namely any smooth involution has a fixed point set which is locally a smooth submanifold, but if an Alexander horned sphere was locally a smooth submanifold (for some smooth structure) then it would have a normal bundle (which is trivial as it is orientable), and $\pi_1(S^3)$ would be isomorphic to the free product of the two components of the complement of the horned sphere by Van-Kampen. $\endgroup$ – PVAL Apr 27 '15 at 21:45
  • $\begingroup$ @PVAL: I don't think it is true in general that the fixed point set of a diffeomorphism is a manifold. For instance, I think that there exists a diffeomorphism $f$ of $\mathbb{R}$ whose fixed set is the usual middle third Cantor set (just take the identity function and perturb it on each interval in the complement such that it stays smooth bijective with nonzero derivative and such that the derivative tends towards one on the boundary of the interval). $\endgroup$ – Andy Putman Apr 27 '15 at 21:50
  • $\begingroup$ You're right. I spoke incorrectly. It is in fact true that that any smooth INVOLUTION has fixed point set which is locally a smooth submanifold (see here math.stackexchange.com/questions/238939/…). $\endgroup$ – PVAL Apr 27 '15 at 21:51
  • $\begingroup$ @PVAL: Yes, I believe this for involutions, which does give an easier proof (though I think the stuff about the Smith conjecture is useful to show how subtle this kind of stuff is). $\endgroup$ – Andy Putman Apr 27 '15 at 21:54

I will address your first question: Is it always possible to construct a smooth structure on M w.r.t to it $\phi$ will be a diffeomorphism?

It is not always possible, even with a change of smooth structure, as this example in dimension 4 shows. Take M to be a K3 surface, a simply-connected spin manifold with signature -16 and $b_2 = 22$. Donaldson (Donaldson, Polynomial invariants for smooth four-manifolds, Topology 29 (1990)) showed that there is an automorphism $\phi$ of the intersection form that is not realized by a self-difffeomorphism of M. By Freedman's work, it is realized by a self-homeomorphism. The key property of $\phi$ is that it reverses the orientation of $H_2^+(M)$. (It takes a little work to make sense of this statement.)

By itself, this does not answer your question because you want to allow different smooth structures. But Donaldson's proof relies on the fact that one of his polynomial invariants is nonzero for $M$, plus the fact that the sign of this invariant is determined by an orientation of $H_2^+(M)$ (a so-called homology orientation). These ingredients can equally well be supplied by Seiberg-Witten theory. The crucial point then is a theorem of Morgan-Szabó (Homotopy K3 surfaces and mod 2 Seiberg-Witten invariants, Mathematical Research Letters, 4, (1997) 17-21), to the effect that any smooth 4-manifold homology equivalent to the K3 surface has an odd (and hence non-vanishing!) Seiberg-Witten invariant. Thus Donaldson's argument would apply to any smooth structure on $M$.

There are presumably examples of this in higher dimensions as well; I would bet that the answer is always negative for simply connected manifolds of dimension at least 5, for which one has a reasonable idea of all of the smooth structures. You might like to have a look at the old paper of Kirby-Scharlemann, A curious category which equals TOP, which treats related issues. You can find it on Kirby's web page.


Just a slight addition to Danny Ruberman's answer: more example in the same vein are discussed in David Gay's note "4-manifolds which are homeomorphic but not diffeomorphic".


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