Timeline for Connected components of space of maps between two manifolds
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
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| May 27, 2017 at 8:03 | comment | added | user21574 | As additional comment: We can define always $\mathcal M_{g,k}(M,g,\beta):=(\bar\partial_J)^{-1}(0)$ where $\bar\partial_J: W^{l+1,p}(\Sigma, M,\beta)\times \mathcal M_{g,k}\to W^{l,p}(\Sigma, \Omega^{0,1}(\Sigma,u^*TM))$ and if we change the direction of map then we get Banach vector bundle | |
| May 4, 2010 at 21:33 | answer | added | Ady | timeline score: 2 | |
| May 2, 2010 at 15:46 | comment | added | macbeth | Thanks to all for useful comments. @Tim, yes, that is what I mean: since intuition (supported by the discussion here) is that the connected components are homotopy equivalence classes, and since homology doesn't determine homotopy, it should be false that the connected components are the homology classes. (But perfectly true that homology classes are unions of connected components.) I am feeling happy that here there be no pathology-dragons. | |
| May 2, 2010 at 14:57 | comment | added | Tim Perutz | Two remarks: first, homology classes of smooth mappings $M\to N$ do not usually determine homotopy classes, though they do, by the Hurewicz theorem, when $M=S^2$ and $N$ is simply connected (maybe that's what you mean about "unions of connected components"?). Second, life on the not-quite-continuous Sobolev borderline (e.g. $W^2_1$ on a surface) is very precarious, both for analysts and topologists. | |
| May 2, 2010 at 14:53 | comment | added | Paul | I think if you use the methods Dylan and Dan mention below you get that $\pi_0(W^{k,p})=\pi_0(C^r)=\pi_0(C^\infy)$. I don't know a good reference to this, but you could check Palais' book called "foundations of global non-linear analysis". Hirsch's book has results that relate $C^r$ and $C^\infty$ for all $r$, so maybe that and Sobolev embedding does the trick. | |
| May 2, 2010 at 8:13 | vote | accept | macbeth | ||
| May 2, 2010 at 6:54 | answer | added | Dan Ramras | timeline score: 5 | |
| May 2, 2010 at 4:38 | history | edited | Dylan Thurston |
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| May 2, 2010 at 3:27 | answer | added | Dylan Thurston | timeline score: 5 | |
| May 1, 2010 at 21:32 | comment | added | macbeth | Ah, of course you're right -- I forgot about the Sobolev estimates. In that case, should the McD-S remark about "connected components" be interpreted as meaning "unions of connected components"? | |
| May 1, 2010 at 21:04 | comment | added | Paul | I'm going to guess that $k$ and $p$ were chosen large enough so that $W^{k,p}\subset C^r$ for some $r\ge 0$, (by Sobolev), so that $W^{k,p}$ maps in the same path component are continuous and homotopic. | |
| May 1, 2010 at 20:27 | history | asked | macbeth | CC BY-SA 2.5 |