Timeline for Polarizations of K3 surfaces over finite fields
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 4, 2012 at 14:44 | comment | added | Keerthi Madapusi | Just to make sure, if $k$ is finite, then what is our expectation? Can we say something without appealing to the Tate conjecture? | |
| Jan 4, 2012 at 13:07 | comment | added | Donu Arapura | Yes, agreed. I should have said that $k$ should be transcendental. | |
| Jan 4, 2012 at 10:17 | comment | added | Remke Kloosterman | ... if $k$ is the algebraic closure of a prime field you might have to exclude all the $k$-rational points of the parameter space. If $k$ is a finite field this phenomena actually happens (see the previous comments). In characteristic zero there are several results suggesting that the complement of the countable union of the subvarieties contains a $\mathbb{Q}$-rational point. | |
| Jan 4, 2012 at 10:11 | comment | added | Remke Kloosterman | @MP: If you believe the Tate conjecture (proven for most K3 surfaces over finite fields) then for a surface over a finite field the geometric Picard number has the same parity as the second betti number. The point is that $\overline{\mathbb{F}_p}$ is too small to apply Deligne's version of Noether-Lefschetz. @Donu: You need to assume that $k$ is transcendental over its prime field. From Deligne's definition of generic it follows that there is a countable union of subvarieties in the parameter space, where the conclusion $\rho=1$ does not hold.... | |
| Jan 4, 2012 at 0:24 | comment | added | M P | I thought that over the algebraic closure of a finite field a K3 surface always had even rank, but I do not have SGA7 available... | |
| Jan 3, 2012 at 21:06 | vote | accept | Keerthi Madapusi | ||
| Jan 3, 2012 at 21:06 | |||||
| Jan 3, 2012 at 20:08 | history | answered | Donu Arapura | CC BY-SA 3.0 |