1
$\begingroup$

Is it true that any family of K3 surfaces over $\mathbb{C}$ whose Picard number is constant is isotrivial? Here isotrivial means locally analytically trivial.

Speculation: Let $\mathcal{M}$ be the moduli space of K3 surfaces over $\mathbb{C}$. It is known that for any point $[S]\in \mathcal{M}$ its open neighborhood $U$ contains a K3 surface $[T]$ such that $Pic(S)>Pic(T)$ and such points are dense. So basically I am asking whether I can take a family avoiding these points or not.

Thank you in advance.

$\endgroup$
1
  • $\begingroup$ You can take arbitrary family and remove the points where the Picard number jumps. $\endgroup$
    – Sasha
    Nov 22, 2012 at 10:24

2 Answers 2

3
$\begingroup$

The answer to your question is yes. This is proven in a paper by Oguiso.

http://arxiv.org/abs/math/0011258

There is a slightly more general criterion for the density of Hodge loci which appears for instance in this survey of Voisin (section 3.2)

http://www.math.polytechnique.fr/~voisin/Articlesweb/hodgeloci.pdf

and is attributed to Mark Green.

$\endgroup$
2
$\begingroup$

I think the answer to your question is yes:

http://arxiv.org/abs/alg-geom/9701013

Edit: As Sasha points out, this is only for compact families.

$\endgroup$
4
  • 1
    $\begingroup$ Isn't it about compact families only? $\endgroup$
    – Sasha
    Nov 22, 2012 at 10:23
  • $\begingroup$ Sorry, yes I was assuming that was what was wanted! $\endgroup$ Nov 22, 2012 at 12:18
  • 3
    $\begingroup$ The result holds more generally even if the base isn't compact; by a result of green, the jumping locus is analytically dense so you can't just remove the points. $\endgroup$
    – PRL
    Nov 22, 2012 at 13:01
  • $\begingroup$ Sorry - edited to reflect comments. $\endgroup$ Nov 22, 2012 at 16:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.