Timeline for Rational homotopy type of a complement
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
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| May 15, 2010 at 14:58 | comment | added | Ryan Budney | I think he has an example of homotopy-equivalent simply-connected manifolds that are not homeomorphic such that certain configuration spaces in the manifolds are not homotopy-equivalent. I don't know if he's using rational homotopy techniques or not but I think it's likely. | |
| May 15, 2010 at 14:44 | comment | added | algori | Woops, it seems that quite a lot happened in this thread since I last looked at it (and probably in mathematics in the past years as well:)). Ryan, thanks, I'd be interested in hearing Salvatore's response. In the mean time, what exactly is the statement? Is it that the rational homotopy type of the configuration spaces is not determined by the rational homotopy type of the manifold? Or the same statement with "rational" replaced by "integral"? Or a mixture of both? | |
| May 15, 2010 at 13:54 | comment | added | Ryan Budney | Longoni's response to your question is "hasn't anything happened in mathematics in the past 10 years?" (joking) Salvatore says he'll eventually log into MathOverFlow and give a proper response. | |
| Apr 17, 2010 at 2:14 | comment | added | Oscar Randal-Williams | @Paul: The map f does not necessarily induce a map between the complements. | |
| Apr 16, 2010 at 23:54 | answer | added | Dev Sinha | timeline score: 2 | |
| Apr 16, 2010 at 23:51 | comment | added | Paul | If everything is simply connected, then it seems like Mayer-Vietoris and the 5-lemma shows $f$ induces an isomorphism on rational homology, and hence by Hurewicz/Whitehead it's a rational homotopy equivalence. I figured this was really a $\pi_1$ question. | |
| Apr 16, 2010 at 21:18 | comment | added | Ryan Budney | There's no knot examples to your revised question. Smale proved that even inequivalent smooth knots in $S^n$ (with codimension $>2$) have diffeomorphic complements. | |
| Apr 16, 2010 at 20:19 | comment | added | Somnath Basu | @Ryan - +1 for the "Longoni is a banker in Milano" statement! | |
| Apr 16, 2010 at 19:47 | comment | added | Ryan Budney | I think Salvatore ?might? have a result analogous to his paper with Longoni using simply-connected manifolds instead of lens spaces. The argument proceeds much the same -- the configuration spaces aren't homotopy-equivalent even though the underlying manifolds are. I'll ask him about it in person next week. I'll visit Longoni as well. I don't think Salvatore has written up that paper yet. Longoni is a banker in Milano. | |
| Apr 16, 2010 at 19:41 | history | edited | algori | CC BY-SA 2.5 |
added the simple-connectedness assumption
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| Apr 16, 2010 at 18:57 | comment | added | algori | Thanks, indeed! What if we assume everything simply connected? | |
| Apr 16, 2010 at 18:20 | comment | added | Ryan Budney | There is a homotopy equivalence $(S^3,K_1) \to (S^3,K_2)$ with $K_1$ the unknot, and $K_2$ the trefoil. | |
| Apr 16, 2010 at 18:01 | history | asked | algori | CC BY-SA 2.5 |