Timeline for Relative flasqueness?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 21, 2014 at 8:14 | vote | accept | Vladimir Baranovsky | ||
| Feb 20, 2014 at 23:48 | comment | added | Alex Degtyarev | Happens all the time :) BTW, your definition looks pretty much like flasqueness of the restriction (in the topological sense :) | |
| Feb 20, 2014 at 23:25 | comment | added | Piotr Achinger | @AlexDegtyarev Of course! I confused the two $f^*$. | |
| Feb 20, 2014 at 23:08 | comment | added | Alex Degtyarev | @PiotrAchinger The restriction in your example has lots of sections; it's just not a coherent sheaf. By restriction I mean what you get before tensoring by $\mathcal{O}$. (Flasqueness isn't quite an algebraic-geometric notion.) | |
| Feb 20, 2014 at 22:56 | comment | added | Piotr Achinger | @AlexDegtyarev in algebraic geometry, the stalks of $R^i \pi_*$ are not always $H^i(fiber)$. For example, take $i=0$ and $X =$ blowup of a surface $S$ at a point and $F=\mathcal{O}(-1)$. Then at that point, the fiber is $\mathbb{P}^1$ and so $F|_{fiber}$ has no sections, but $\pi_* \mathcal{O}(-1)$ is the ideal sheaf of the point. | |
| Feb 20, 2014 at 22:51 | answer | added | Piotr Achinger | timeline score: 5 | |
| Feb 20, 2014 at 22:30 | comment | added | Alex Degtyarev | One possible answer is in the title of your question: since the stalks of $R^i\pi_*$ are $H^i(\text{fiber})$, it would suffice to require that the restrictions to fibers are flasque. | |
| Feb 20, 2014 at 22:01 | history | asked | Vladimir Baranovsky | CC BY-SA 3.0 |