Timeline for Sheaves without global sections
Current License: CC BY-SA 2.5
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 27, 2010 at 23:30 | vote | accept | Bugs Bunny | ||
| Oct 27, 2010 at 0:14 | comment | added | Karl Schwede | That's of course true too, and another good point. | |
| Oct 26, 2010 at 17:19 | comment | added | David Treumann | But the condition that Frob is an isomorphism on H^i(X,O) is much weaker than being Frobenius split, like Torsten's example of a generic hypersurface. By now, does the opposite question look more interesting? On which varieties does every vector bundle have cohomology? | |
| Oct 26, 2010 at 16:29 | comment | added | Karl Schwede | Sasha, that brings up a good point. Most Frobenius split varieties in nature are log Fano, so that is close to the answer Sandor mentioned below. But they need not be (although they always must be log Calabi-Yau). So it seems that if you start in characteristic zero with a (log) Calabi-Yau, up to some conjectures, for infinitely many reduction to characteristic $p$, there is such a sheaf $M$ by the observation you and David made. It seems natural to ask if such a sheaf still exists in characteristic zero. | |
| Oct 26, 2010 at 4:39 | answer | added | Torsten Ekedahl | timeline score: 6 | |
| Oct 26, 2010 at 3:38 | answer | added | Sasha | timeline score: 43 | |
| Oct 26, 2010 at 3:15 | comment | added | Sasha | David, there is an important class of varieties, called ``Frobenius split varieties''. For these the morphism $O \to Frob_* O$ splits, so $O$ is a direct summand of the Frobenius pushforward. The complement summand (isomorphic to the cone of $O \to Frob_* O$) then has no derived global sections. | |
| Oct 26, 2010 at 1:36 | comment | added | Donu Arapura | Yes, I think so. If $X$ is a Kummer surface obtained from a product of supersingular elliptic curves $E_i$, I think that $H^2(X,O_X)\cong H^1(E_1,O_{E_1})\otimes H^1(E_2,O_{E_2})$ compatibly with $F$-action. | |
| Oct 26, 2010 at 0:31 | answer | added | Sándor Kovács | timeline score: 9 | |
| Oct 26, 2010 at 0:05 | answer | added | Daniel Erman | timeline score: 11 | |
| Oct 26, 2010 at 0:03 | comment | added | David Treumann | Good lord I've been trying to make precisely that computation for like 90 minutes. For a del Pezzo surface it works because H^i(X,O) vanishes for i > 0. Are there "supersingular K3s" on which Frob:H^2(X,O) --> H^2(X,O) vanishes? | |
| Oct 25, 2010 at 23:56 | comment | added | Donu Arapura | David, I think that this won't work if for example $X$ is supersingular elliptic curve. In this case, the map $H^1(X,O)\to H^1(X,F_*O)$ is $0$, so you would cohomology on the cone. | |
| Oct 25, 2010 at 23:39 | answer | added | Donu Arapura | timeline score: 10 | |
| Oct 25, 2010 at 22:41 | comment | added | David Treumann | In characteristic p, you can try the cone on the map O --> Frob_* O. I'm confused about whether this works. | |
| Oct 25, 2010 at 22:24 | history | asked | Bugs Bunny | CC BY-SA 2.5 |