29
$\begingroup$

The "textbook" example of a smooth bijection between smooth manifolds that is not a diffeomorphism is the map $\mathbb{R} \rightarrow \mathbb{R}$ sending $x \mapsto x^3$. However, in this example, the source and target manifolds are diffeomorphic -- just not by the given map. Is there an example of a smooth bijection $X \rightarrow Y$ of smooth manifolds where $X,Y$ are not diffeomorphic at all? (and if so, what?)

(For instance, is it possible to arrange a smooth bijection from a sphere to an exotic sphere, failing to be a diffeomorphism because of the existence of critical points? or do homeomorphisms between different smooth structures on spheres fail to be everywhere smooth in some catastrophic way?)

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ I guess the word "manifold" forbids me from just rolling the half-open interval around the circle, and perhaps also from sticking continuum-many discrete points on the line.... $\endgroup$
    – Theo Johnson-Freyd
    Oct 13, 2010 at 8:00
  • $\begingroup$ Alternately, maybe you don't want to rest that much on the word "manifold", and instead mean to ask for smooth homeomorphisms that are not diffeomorphisms? $\endgroup$
    – Theo Johnson-Freyd
    Oct 13, 2010 at 8:01
  • $\begingroup$ Hi Theo, yes, "manifold" with (what I think of as) the default meaning -- second countable, without boundary, etc.... $\endgroup$
    – D. Savitt
    Oct 13, 2010 at 8:07
  • $\begingroup$ (and of course a smooth homeomorphism would be even better, but I'd be happy with just a smooth bijection) $\endgroup$
    – D. Savitt
    Oct 13, 2010 at 8:09
  • 2
    $\begingroup$ In the compact case, every continuous bijection is a homeomorphism. $\endgroup$
    – Greg Kuperberg
    Oct 13, 2010 at 8:36

1 Answer 1

Reset to default
23
$\begingroup$

Every smooth manifold has a smooth triangulation, which yields a pseudofunctor from the category of smooth manifolds to the category of PL manifolds. (There is no actual functor; that would be crazy.) If two smooth manifolds are PL isomorphic, then the answer is yes. You can start with the PL isomorphism, and then build a homeomorphism that follows it and that has the property that all derivatives vanish in all directions perpendicular to every simplex. You can build the homeomorphism by induction from the $k$-skeleton to the $(k+1)$-skeleton using bump functions.

The PL Poincaré conjecture is true in dimensions other than 4, so all exotic spheres in the same dimension $n \ge 5$ are PL homeomorphic. (High-dimensional examples of exotic spheres start in dimension 7, it was calculated.) In dimension 4, by contrast, every PL manifold has a unique smooth structure, and it is not known whether there are any exotic spheres.

On the other hand, if the manifolds are homeomorphic but not even PL homeomorphic, then I don't know. It is known that every manifold of dimension $n \ge 5$ has a unique Lipschitz structure, but I do not know a Lipschitz version of the above argument. On the positive side, passing from smooth to Lipschitz is an actual functor, so the answer to a modified question, is there a Lipschitz-smooth homeomorphism, is yes, and you can even make it bi-Lipschitz.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Perhaps a side issue, but I don't understand what you mean by pseudofunctor here. To me, a pseudofunctor is a type of map between 2-category or bicategory structures, but I am not aware of any interesting 2-category structure on the category of PL manifolds. $\endgroup$
    – Todd Trimble ♦
    Oct 13, 2010 at 13:43
  • 3
    $\begingroup$ It looks like I used the wrong word; I just didn't check the definition. What I meant is a "mock functor", a partial functor defined for objects and isomorphisms that has no reasonable behavior for general morphisms. And even compositions of isomorphisms are somewhat problematic. $\endgroup$
    – Greg Kuperberg
    Oct 13, 2010 at 14:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.