How to classify K3 surfaces over an arbitrary field k?
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asked Oct 19, 2009 at 19:28
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1$\begingroup$ Although I completely agree with David Lehavi's answer, if you work over $\mathbb{Z}_p$, probably you can also make sense of rigid analytic K3 surfaces that are not necessarily algebraic schemes over $\mathbb{Z}_p$ (just as there are complex analytic K3 surfaces that are not algebraic schemes over $\mathbb{C}$). $\endgroup$– Jason StarrFeb 13, 2013 at 22:12
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1 Answer
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The "standard" definition of a K3 surface is field independent (unless you are a physicist):
$p_g=1, q=0$, and trivial canonical class.
Some results:
- Mumford and Bombieri showed that you get (just as in the complex case) a 19 dimensional family of K3 surfaces for any degree (the 19 dimensional thingy is a deformation theory argument which is completely algebraic).
- Deligne showed that all the K3 surfaces in finite characteristics are reductions mod p.
What you obviously don't get is the fact that all these spaces sit together in a nice 20 dimensional complex ball. I also don't know if you can carry over any of the recent Kodaira dimension computation of these moduli (which are very analytic in nature).
Reference: Complex algebraic surfaces (Beauville): Chapter VIII and Appendix A.
answered Oct 19, 2009 at 22:10
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$\begingroup$ Field independent with the small twist that when people refer to K3 over k they usually mean algebraic surface, of which there are 19 parameters, while over C they mean a member of 20-dimensional family. $\endgroup$ Oct 20, 2009 at 4:11
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1$\begingroup$ @ilya - well, it depends if these people are algebraic or differential geometers I guess... $\endgroup$ Oct 20, 2009 at 5:46
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2$\begingroup$ it should be $p_g=1$ and $q=0$... $\endgroup$ Feb 12, 2013 at 1:45
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2$\begingroup$ + Yoyontzin: oops (and it stayed like this for the last 3.5 years) $\endgroup$ Feb 12, 2013 at 20:55