Timeline for Techniques for computing cup products in singular cohomology
Current License: CC BY-SA 3.0
9 events
when toggle format | what | by | license | comment | |
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May 14, 2011 at 21:54 | comment | added | Ryan Budney | I wouldn't say (1) is naive, since much of the work of completing the task is left un-done, in that you still have to perturb the homology classes to be transverse. This isn't always easily done. Morse theory can be useful in these situations. | |
May 14, 2011 at 19:28 | answer | added | Peter May | timeline score: 30 | |
May 11, 2011 at 23:41 | comment | added | Sergey Melikhov |
Strategy (1) does not need $X$ to be homotopy equivalent to a manifold. See R. Fenn, Techniques of geometric topology'', Chapter 1 and S. Buoncristiano, C. P. Rourke, B. J. Sanderson A geometric approach to homology theory'', Chapter 2.
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May 11, 2011 at 17:04 | comment | added | user332 | In my experience, any manifold that I understand well enough to be able to compute the de Rham algebra of is probably a manifold that I understand well enough to work out the intersection product on. Would you happen to know of an example which doesn't fit this description? | |
May 11, 2011 at 15:42 | comment | added | Donu Arapura | When I actually have to compute anything like this, I tend to use (1) or and sometimes $(\alpha,\beta)\mapsto \int\alpha\wedge \beta$ in de Rham. (2) is likely to lead to a mess. | |
May 11, 2011 at 14:29 | answer | added | Johannes Ebert | timeline score: 10 | |
May 11, 2011 at 13:24 | answer | added | James Cranch | timeline score: 8 | |
May 11, 2011 at 13:10 | answer | added | Jeff Strom | timeline score: 6 | |
May 11, 2011 at 12:45 | history | asked | user332 | CC BY-SA 3.0 |