Smooth bijection between non-diffeomorphic smooth manifolds? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T01:04:49Z http://mathoverflow.net/feeds/question/41984 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41984/smooth-bijection-between-non-diffeomorphic-smooth-manifolds Smooth bijection between non-diffeomorphic smooth manifolds? D. Savitt 2010-10-13T06:15:09Z 2010-10-13T08:32:01Z <p>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 <i>are</i> 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?)</p> <p>(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?)</p> http://mathoverflow.net/questions/41984/smooth-bijection-between-non-diffeomorphic-smooth-manifolds/41993#41993 Answer by Greg Kuperberg for Smooth bijection between non-diffeomorphic smooth manifolds? Greg Kuperberg 2010-10-13T08:32:01Z 2010-10-13T08:32:01Z <p>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.</p> <p>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.</p> <p>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.</p>