Timeline for Cohomology of a sheaf of functions locally constant along a foliation
Current License: CC BY-SA 2.5
19 events
| when toggle format | what | by | license | comment | |
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| Jun 25, 2014 at 9:16 | answer | added | user54349 | timeline score: 2 | |
| Nov 28, 2011 at 21:18 | answer | added | R.S. | timeline score: 0 | |
| May 19, 2010 at 20:33 | vote | accept | Dmitri Panov | ||
| Feb 27, 2010 at 20:18 | answer | added | Dmitri Panov | timeline score: 4 | |
| Feb 27, 2010 at 18:57 | answer | added | José Figueroa-O'Farrill | timeline score: 4 | |
| Feb 27, 2010 at 17:02 | comment | added | David Ben-Zvi | I don't think it matters if the leaves are simply connected, as long as the sheaf is equivariant for the equivalence relation (ie is the constant sheaf along the fibers). In general (if F is just locally constant) I agree you'll get a sheaf (not a local system I don't think since the topology of leaves jumps) which is the (group algebra) of relative $\pi_1$ and you'll get a module over this, but at this point I'm not sure working downstairs gains you anything (ie you're basically saying, a sheaf on the fibers of a fibration - ie pushing forward the stack of sheaves rather than descending) | |
| Feb 27, 2010 at 16:57 | history | edited | Pete L. Clark | CC BY-SA 2.5 |
edited title
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| Feb 27, 2010 at 16:38 | comment | added | Chris Schommer-Pries | (cont). In the case that the leaves are not simply connected (eg. $M = S^1$ with a single leaf), then it is more complicated. I think that you get something like a local system on the quotient and then $\mathcal{F}$ is a module or something like that for this local system. I'm not sure. | |
| Feb 27, 2010 at 16:35 | comment | added | Chris Schommer-Pries | Right. After playing a bit more, I think that the following is plausible: If the leaves are simply connected (so that on each leaf locally constant sheaf = constant sheaf), then the sheaf is pulled back from some sheaf on the (bad) quotient space. This should be true in Dmitri's $S^3$ example above. I think that the Cech cohomlogy on M will then be the same as the sheaf cohomology on the quotient (Since the quotient can be bad, and this can be any sheaf, this could be really realy exotic). | |
| Feb 27, 2010 at 16:27 | comment | added | Dmitri Panov | Chris, for you first question: In the case M^n=S^1, the sheaf F that we get is just the sheaf of locally constant functions on S^1, and its cohomology are surelly not the same as cohomology of the functions of the quotinet, which is a point. For your second question: Personnaly I am horrified by non-Hausdrof spaces, so I would not be able to answer your question. But it is quite possible that you are right, that every locally constant sheaf on M^n comes from "the quotinet", it sounds plausible... | |
| Feb 27, 2010 at 16:26 | comment | added | David Ben-Zvi | Chris - I think the answer is yes, for formal reasons: passing to sheaves takes colimits of spaces to limits, which in this case will mean sheaves on the quotient are the same as sheaves on M equivariant for our equivalence relation (which the given F is). Likewise its sheaf cohomology will be calculated by (derived functor of) invariant sections upstairs. (Unless I'm confused, which is likely, we're using something much weaker than descent here, just (co)continuity of spaces --> sheaves: I wouldn't know what kind of descent theorems held in this kind of setting..) | |
| Feb 27, 2010 at 16:10 | comment | added | Chris Schommer-Pries | Dmitri, is there an example of an interesting non-trivial sheaf on $S^3$ which is locally constant along the fibers of the foliation (in your example)? Are there any that fail to come from the (bad) quotient space? | |
| Feb 27, 2010 at 16:08 | comment | added | Chris Schommer-Pries | I realize that the quotient can be horrific, but you can still ask the question. Locally constant sheaves still make sense on bad spaces (locally the sheafification of the constant presheaves). I'm wondering if the cohomology group on M is just the usual sheaf cohomology of the bad quotient space. | |
| Feb 27, 2010 at 15:46 | history | edited | Dmitri Panov | CC BY-SA 2.5 |
added 205 characters in body; added 4 characters in body
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| Feb 27, 2010 at 15:42 | comment | added | algori | Chris -- there isn't a reasonable quotient space in general. | |
| Feb 27, 2010 at 15:42 | comment | added | Dmitri Panov | Chris, notice, that the clouse of a leaf ot F can easily have dimension larger then the dimesnion of the leaf. So the quotient can be non Hasudorf.... Akhil, I don't know how to answer your question... | |
| Feb 27, 2010 at 15:36 | comment | added | Chris Schommer-Pries | Is such a sheaf pulled back from the quotient space $M/F$? | |
| Feb 27, 2010 at 15:31 | comment | added | Akhil Mathew | Does this have anything to do with the semicontinuity theorem for cohomology along fibers? Just curious. | |
| Feb 27, 2010 at 15:06 | history | asked | Dmitri Panov | CC BY-SA 2.5 |