Timeline for Number of points of the nilpotent cone over a finite field and its cohomology
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 5, 2018 at 10:43 | answer | added | Dr. Evil | timeline score: 1 | |
| May 28, 2018 at 22:23 | comment | added | user74900 | @WillSawin I see now, thank you for the explanation. | |
| May 28, 2018 at 20:29 | comment | added | Will Sawin | @AknazarKazhymurat Being a rational homology manifold in this context is a local condition on the dualizing sheaf, I believe, and doesn't have to do with homology necessarily. | |
| May 28, 2018 at 17:16 | comment | added | user74900 | @WillSawin 'If it's a rational homology...' --- did you mean dual to homology, not cohomology? | |
| May 27, 2018 at 16:52 | answer | added | Jim Humphreys | timeline score: 7 | |
| May 27, 2018 at 14:03 | comment | added | Geordie Williamson | If I remember correctly precisely these questions are addressed in "Partial resolutions of nilpotent varieties" by Borho-MacPherson. In any case it would be worth having a look at this paper. | |
| May 27, 2018 at 9:13 | vote | accept | Zhiyu | ||
| May 27, 2018 at 6:51 | answer | added | Will Sawin | timeline score: 11 | |
| May 26, 2018 at 21:08 | comment | added | Zhiyu | @WillSawin So you finish by induction on dimension? Could you provide some details on "Sufficient to show....This is equivalent....exceptional divisor. "? As for any projective variety X we can produce an affine cone whose number of rational points is related to number of X in a simple way, so isn't the vanishing result a restriction on X? | |
| May 26, 2018 at 19:05 | comment | added | Will Sawin | One can compute the ordinary cohomology in every characteristic directly from the fact that it is an affine cone. In characteristic zero, it follows from the fact that affine cones are contractible. In characteristic $p$, one can reason like this (don't know if it's the simplest argument): Sufficient to show the cohomology relative to the origin is zero. This is equivalent to the cohomology of the blow-up at the origin relative to the exceptional divisor. But the blow-up at the origin is an $\mathbb A^1$-bundle on the exceptional divisor, so they have the same cohomology. | |
| May 26, 2018 at 18:32 | comment | added | Zhiyu | @WillSawin Thanks, how do you compute its cohomology in char 0 ? As base change fails in general non-smooth case, betti number of special fiber and generic fiber might be different. | |
| May 26, 2018 at 18:19 | comment | added | Will Sawin | If it's a rational homology manifold, then its cohomology with compact supports is dual to its cohomology, which is one-dimensional in degree zero and zero-dimensional in all higher degrees. Hence the compactly supported cohomology lives in degree $2 ( \dim G - \operatorname{rank} G)$. If this is also true in characteristic $p$, this implies that the number of points is $q^{ \dim G - \operatorname{rank} G}$. | |
| May 26, 2018 at 16:06 | comment | added | Zhiyu | @WillSawin Thank you, I am not familiar with the positive characterestic case. I am mostly interested in Q2 and compute the case $n=2$. | |
| May 26, 2018 at 13:27 | history | edited | Zhiyu | CC BY-SA 4.0 |
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| May 26, 2018 at 13:26 | comment | added | Will Sawin | You only have trace formula for the compactly supported etale cohomology in general, not the intersection cohomology. However in the rationally smooth case the etale cohomology with compact supports agrees with the intersection cohomology with compact supports. (I don't think there is a difference between rationally smooth in the usual and intersection cohomology senses.) | |
| May 26, 2018 at 12:15 | history | asked | Zhiyu | CC BY-SA 4.0 |