Timeline for What tools cannot work for orbifolds?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 12, 2013 at 15:41 | comment | added | André Henriques | Cohomology is a ring, and Poincare duality holds when you tensor everything with ℚ, so the intersection form certainly makes sense. | |
| Jul 12, 2013 at 14:34 | comment | added | Chris Gerig | Is there then a way to speak about intersection forms? I guess I'm just trying to see what can be salvaged; if you have any further elaborations on your answer then that would be great! | |
| Jul 11, 2013 at 15:10 | comment | added | André Henriques | Orbifold cohomology is a mix of manifold cohomolgoy (finite cohomological dimension) and group cohomology (infinite cohomological dimension). If you work with ℚ or ℝ coefficients, then the cohomology of finite groups vanishes, and you're back in a situation where orbifolds have finite dimensional cohomology and Poincare duality holds. | |
| Jul 11, 2013 at 12:55 | comment | added | Chris Gerig | Do you know what the crux of the issue is for integer coefficients? | |
| Jul 11, 2013 at 9:52 | comment | added | André Henriques | That's right. There is something called Chen-Ruan orbifold cohomology arxiv.org/abs/math/0004129 that satisfies Poincare duality. It is, however, a rather strange beast: the orbifold needs to be almost complex for these groups to be defined, and the grading is not by the integers, but only by the rationals! | |
| Jul 11, 2013 at 3:44 | comment | added | Chris Gerig | Can you comment on what makes it different from other orbifold cohomologies where it does work? (a google search shows me some results on orbifold Poincare duality). | |
| Jul 10, 2013 at 19:59 | history | answered | André Henriques | CC BY-SA 3.0 |