Timeline for Cohomology groups of an intersection
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Oct 4, 2010 at 5:15 | answer | added | Torsten Ekedahl | timeline score: 1 | |
| Oct 4, 2010 at 2:43 | history | edited | Ben Webster♦ | CC BY-SA 2.5 |
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| Oct 4, 2010 at 0:25 | history | edited | Charlie Frohman | CC BY-SA 2.5 |
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| Oct 3, 2010 at 23:04 | comment | added | Charlie Frohman | For instance, on $S^1\times S^1$ take a $(p,q)$ curve and and $(r,s)$ curve that intersect each other efficiently, and its true. In $(S^2)^{2n}$ if you take $P$ and $Q$ copies of $(S^2)^n$ embedded as components of the {\em small} diagonal where you set $n$ coordinates in pairs equal, will have this property. | |
| Oct 3, 2010 at 21:22 | comment | added | shenghao | Excuse me, Torsten, can you tell me precisely what the Mayer-Vietoris sequence is in the case of closed submanifolds (or give a reference)? Do we use the cohom of the ambient space or the union of the two submanifolds in the sequence? Thanks. | |
| Oct 3, 2010 at 20:40 | comment | added | Torsten Ekedahl | When the closed subspaces are well-behaved we have a Mayer-Vietoris also for closed subspaces. This is the case for closed submanifolds. | |
| Oct 3, 2010 at 20:25 | comment | added | shenghao | Don't we need open subsets (or sets whose interior....) for Mayer-Vietoris sequence? To Charlie: the coordinate ring can be interpreted as coherent cohomology. So maybe we don't expect to have analog for singular cohomology. | |
| Oct 3, 2010 at 15:03 | comment | added | Torsten Ekedahl | The cohomology of the intersection fits into a Meyer-Vietoris sequence which shows that the proposed formula is essentially never true. | |
| Oct 3, 2010 at 14:46 | comment | added | shenghao | Take X to be the complex affine plane C^2, and P and Q are two smooth curves of high genuses, so that H^1(P) or H^1(Q) is non-trivial. Then the righthand-side is non-trivial in degree 1, but P\cap Q is discrete (assume P and Q are in general position). So I guess maybe there is no "simple" conditions to make it true. Might be true in some special cases. | |
| Oct 3, 2010 at 14:31 | history | asked | Charlie Frohman | CC BY-SA 2.5 |