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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.

flag	asked Nov 22 at 8:08 Pooya 11●1

You can take arbitrary family and remove the points where the Picard number jumps. – Sasha Nov 22 at 10:24

HT · Answer 1 · 2012-11-22 13:44:46Z UTC

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.

Jonny Evans · Answer 2 · 2012-11-22 14:58:32Z UTC

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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.

edited Nov 22 at 14:58

answered Nov 22 at 9:36

Jonny Evans
2,149●1●8●25

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Isn't it about compact families only? – Sasha Nov 22 at 10:23

Sorry, yes I was assuming that was what was wanted! – Jonny Evans Nov 22 at 12:18

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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. – PRL Nov 22 at 13:01

Sorry - edited to reflect comments. – Jonny Evans Nov 22 at 16:21

Isotrivial K3 family and Picard number

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