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.

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

2 Answers 2

Reset to default
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.

Improve this answer
$\endgroup$
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.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ Isn't it about compact families only? $\endgroup$
    – Sasha
    Commented Nov 22, 2012 at 10:23
  • $\begingroup$ Sorry, yes I was assuming that was what was wanted! $\endgroup$
    – Jonny Evans
    Commented 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
    Commented Nov 22, 2012 at 13:01
  • $\begingroup$ Sorry - edited to reflect comments. $\endgroup$
    – Jonny Evans
    Commented Nov 22, 2012 at 16:21

You must log in to answer this question.

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