7
$\begingroup$

Suppose that $X$ is a (projective) K3 surface over a field $k$. A polarization of $X$ is an element $\lambda\in Pic_X(k)$ that is represented over an algebraic closure $\overline{k}$ by an ample line bundle $L$ over $X_{\overline{k}}$. Given such a $\lambda$, we can consider its self-intersection number $(\lambda,\lambda)\in 2\mathbb{Z}$. Suppose that $k$ has finite characteristic $p>2$. Is it possible to find a finite extension $k'/k$ and a polarization $\lambda$ of $X_{k'}$ such that $(\lambda,\lambda)$ is co-prime to $p$?

This question has a negative answer for polarizations of abelian varieties (see BCnrd's answer here), so I'm not optimistic that things are any better for K3 surfaces. On the other hand, one has an affirmative answer to the analogue for polarizations of abelian varieties, if one is allowed to modify the abelian variety up to isogeny.

So here's my backup question: Is there an analogous notion of an 'isogeny' of K3 surfaces that might help here? At the least, two K3 surfaces that are `isogenous' should have isometric Neron-Severi lattices (up to tensoring with $\mathbb{Q}$) and also isomorphic $l$-adic realizations, compatible with the cycle class map from the Neron-Severi lattices.

Improve this question
$\endgroup$
5
  • $\begingroup$ I haven't thought this through carefully (my new years resolution is to limit mathoverflow activity), but suppose $k=\bar k$ and you took a general point $(X,\lambda)$ in the moduli space of polarized K3's with $\lambda^2=2p$. I suspect that you should be able to argue that $\lambda$ generates $Pic(X)$, so you would have counterexample. $\endgroup$
    – Donu Arapura
    Commented Jan 3, 2012 at 19:48
  • $\begingroup$ @Keerthi: Are you assuming that $k$ is a finite field or is it just assumed to be of characteristic $>2$? $\endgroup$
    – naf
    Commented Jan 4, 2012 at 7:36
  • $\begingroup$ Are things different for finite fields? I don't want to use the Tate conjecture. $\endgroup$
    – Keerthi Madapusi
    Commented Jan 4, 2012 at 14:42
  • $\begingroup$ If $k$ is finite then $Pic(X)$ is expected to have rank at least two. It's not clear, at least to me, what to expect in this case. $\endgroup$
    – naf
    Commented Jan 4, 2012 at 15:38
  • $\begingroup$ There is indeed a notion of isogeny for K3 surfaces; see the paper of Mukai "On the moduli space of bundles on a K3 surface I" and the preprint "Fourier-Mukai partners of K3 surfaces in positive characteristic" by Lieblich and Olsson (arxiv.org/abs/1112.5114) $\endgroup$
    – naf
    Commented Jan 4, 2012 at 15:51

2 Answers 2

Reset to default
10
$\begingroup$

To expand my comment, suppose $k=\bar k$ has characteristic $p=3$. Let $X$ be a generic complete intersection in $\mathbb{P}^4$ of multidegree $(2,3)$. Apply Deligne's version of Noether-Lefschetz (SGA 7, exp XIX) to conclude that $Pic(X)$ is generated by $\mathcal{O}_X(1)$. This implies that any polarization has degree divisible by $6$. (There goes my resolution.)

Improve this answer
$\endgroup$
5
  • 2
    $\begingroup$ 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... $\endgroup$
    – M P
    Commented Jan 4, 2012 at 0:24
  • 3
    $\begingroup$ @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.... $\endgroup$
    – Remke Kloosterman
    Commented Jan 4, 2012 at 10:11
  • 2
    $\begingroup$ ... 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. $\endgroup$
    – Remke Kloosterman
    Commented Jan 4, 2012 at 10:17
  • $\begingroup$ Yes, agreed. I should have said that $k$ should be transcendental. $\endgroup$
    – Donu Arapura
    Commented Jan 4, 2012 at 13:07
  • $\begingroup$ Just to make sure, if $k$ is finite, then what is our expectation? Can we say something without appealing to the Tate conjecture? $\endgroup$
    – Keerthi Madapusi
    Commented Jan 4, 2012 at 14:44
3
$\begingroup$

Let $S$ be a smooth projective surface and $m:=\gcd(\deg \lambda)$ where $\lambda$ runs through the ample cone. Let $m'$ be the gcd of the entries of the intersection matrix of $S$

Then $m$ equals either $m'$ or $2m'$. The fact that $m'$ divides $m$ is obvious. To prove that $m$ divides $2m'$, note that there exist an effective divisor $D$ on $m$ such that at least one of $gcd(D^2,H^2)=m$ or $gcd(D.H,H^2)=m$ holds.

In particular, there exist infinitely many positive $n$ such that $\gcd(H^2, (nH+D)^2)=\gcd(H^2,2n(H.D)+D^2)\in \{m,2m\}$.

For $n$ sufficiently large one has that $D+nH$ is ample.

For a $K3$ surface over a finite field it is believed that the geometric Picard number is always even. So if you want to produce a counterexample you might try to construct a complete intersection of degree 3,2 with geometric Picard number 2, such that the second generator of the Picard group has genus 1 modulo 3, and such that the intersection number of this curve with $O(1)$ is divisible by 3.

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .