Timeline for Does a smooth homeomorphism of closed manifolds preserve cobordism fundamental class?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Jul 5, 2013 at 7:26 | answer | added | Oscar Randal-Williams | timeline score: 6 | |
| Jul 4, 2013 at 16:43 | comment | added | Fernando Muro | Now I see the problem for bordism! Thanks for your comments. | |
| Jul 4, 2013 at 14:37 | comment | added | Johannes Ebert | Last but not least: any proof attempt that does not refer to special structural properties of oriented cobordism is doomed to fail. There exists exotic spheres that are not spin cobordant to the standard sphere. This means that the result is wrong for spin bordism. | |
| Jul 4, 2013 at 14:35 | comment | added | Johannes Ebert | at $0$ is $0$. Composing with such a homeomorphism (in appropriate charts) creates a smooth homeomorphism between exotic spheres (it is of course not a diffeomorphism). | |
| Jul 4, 2013 at 14:34 | comment | added | Johannes Ebert | So, already the question whether the images of the fundamental classes in $\Omega_{\ast} (pt)$ are the same is nontrivial. 2: from undergraduate analysis, it is known that a smooth homeomorphism does not need to be a diffeomorphism. In fact, if I am not mistaken, there exist smooth homeomorphism between exotic spheres. Let $M$ be an exotic sphere. By the solution of the highdimensional Poincare conjecture, there is a homeomorphism $S^n \to M$ which is everywhere smooth, except in one point. Now pick a homeomorphism of $R^n$ that is smooth, regular outside the origin and whose $\infty$-jet | |
| Jul 4, 2013 at 14:28 | comment | added | Johannes Ebert | I think the question is difficult. Here are my reasons: 1. it is true that homeomorphic manifolds are oriented cobordant, even if they are not diffeomorphic. This requires, to my knowledge, two fields medal theorems: Thom and Wall did prove that two oriented smooth manifolds are cobordant if their Pontrjagin numbers and Stiefl-Whitney numbers agree. The Stiefel-Whitney numbers are homotopy invariants, because you can express then in terms of the Spivak normal fibration and the Steenrod operations. The Pontrjagin numbers are invariant under homeomorphisms by Novikov's famous result. | |
| Jul 4, 2013 at 14:22 | comment | added | Johannes Ebert | @Dylan: sorry for the allegation. | |
| Jul 4, 2013 at 11:31 | comment | added | Fernando Muro | Oriented bordism is a generalized cohomology theory. The n-dimensional group of a topological space X is the set of maps M -> X from oriented smooth manifolds M modulo cobordism. The fundamental class of a manifold is represented by the identity map. | |
| Jul 4, 2013 at 11:09 | comment | added | Andrew Stacey | Maybe I'm missing something (not being an expert on cobordism), but isn't the definition that $M$ and $N$ are cobordant if there is a manifold $W$ and a diffeomorphism $\partial W \cong M \amalg N$? If $M \cong N$ then we can take $W = M \times [0,1]$. But here we're asking for if $f \colon M \to N$ is a smooth homeomorphism not necessarily a diffeomorphism (eg $x \mapsto x^3$). So you'd need to adjust the cobordism to take that into account. One thing that would give it would be if whenever $f \colon M \to N$ was a smooth homeomorphism then there was a diff $g \colon M \to N$. | |
| Jul 4, 2013 at 10:50 | comment | added | Dylan Wilson | @JohannesEbert: I did not vote to close... | |
| Jul 4, 2013 at 10:37 | comment | added | Fernando Muro | All diffeomorphisms are smooth homeomorphisms, aren't they? Anyway, smoothness is irrelevant for thisquestion I think. The up votes and your comments make me think that I may be terribly mistaken, but so far I can't see how. | |
| Jul 4, 2013 at 10:10 | comment | added | Andy Putman | @Fernando : For a diffeomorphism, the inverse is also smooth. Not so with a smooth homeomorphism. | |
| Jul 4, 2013 at 8:52 | comment | added | Fernando Muro | A diffeomorphism is both smooth and a homeomorphism, so I think Dylan's answer it totally pertinent. Besides, I don't think he voted to close since I did and so far there's only one vote. I can't see a non-trivial part in the question, and I think it's completely answered by Dylan. | |
| Jul 4, 2013 at 7:42 | comment | added | Johannes Ebert | @Dylan: please read questions carefully before you vote to close. User35353 wrote ''smooth map which is also a homeomorphism'' and not ''diffeomorphism'' (in the latter case, the answer is indeed trivial). It is a nontrivial question and I would like to see a qualified answer (which I cannot provide). | |
| Jul 4, 2013 at 7:00 | review | Close votes | |||
| Jul 4, 2013 at 18:40 | |||||
| Jul 4, 2013 at 6:59 | comment | added | Dylan Wilson | The statement after "it is clear that" is false (take an orientation-reversing diffeomorphism, e.g.). Other than that, the statement is true, and also in cobordism where it is even trivially true if you define cobordism theory geometrically. | |
| Jul 4, 2013 at 4:37 | history | asked | John Pardon | CC BY-SA 3.0 |