Timeline for Roots of line bundles in a family
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 15, 2022 at 7:37 | comment | added | Frank | In characteristic p you can have that line bundles become divisible by a power of p when specialised. This happens for supersingular K3 surfaces, but see also arxiv.org/pdf/0907.4781.pdf example 3.12. Actually Jason had also told me a few years ago of an example coming from the Moret-Bailly pencil of the Jacobian of a supersingular genus 2 curve. | |
| Jul 14, 2022 at 15:05 | comment | added | Cranium Clamp | abx and Jason Starr, thank you. If I understand correctly, the entire argument can be repeated for schemes over a field of char not dividing n by taking the exact sequence $ 0 \rightarrow Z/nZ \rightarrow \mathbb{G}_m \rightarrow \mathbb{G}_m \rightarrow 0 $ on a fiber in the etale topology, where the second map is multiplying by n. Is that correct? What happens if char k divides n, out of curiosity? | |
| Jul 14, 2022 at 14:29 | vote | accept | Cranium Clamp | ||
| Jul 14, 2022 at 14:00 | comment | added | abx | @Jason Starr: Right, this is more direct (and more elegant). Still the observation that having a $n$-th root can be checked on $c_1$ might be useful. | |
| Jul 14, 2022 at 11:05 | comment | added | Jason Starr | Actually, just the second part of this argument is sufficient (and thus also applies in characteristic $p$ prime to $n$). The obstruction to infinitesimally deforming $L$ is $n$ times the obstruction to infinitesimally deforming the root invertible sheaf inside $H^2(F,\mathcal{O}_F)$. This is a vector space over the ground field. Thus, if $p$ is prime to $n$, then if one obstruction vanishes, then both obstructions vanish. | |
| Jul 14, 2022 at 8:15 | history | answered | abx | CC BY-SA 4.0 |