K3 surfaces with good reduction away from finitely many places

Question

Let S be a finite set of primes in Q. What, if anything, do we know about K3 surfaces over Q with good reduction away from S? (To be more precise, I suppose I mean schemes over Spec Z[1/S] whose geometric fibers are (smooth) K3 surfaces, endowed with polarization of some fixed degree.) Are there only finitely many isomorphism classes, as would be the case for curves of fixed genus? If one doesn't know (or expect) finiteness, does one have an upper bound for the number of such K3 surfaces X/Q of bounded height?

Answer 1

Some thoughts.

There are no such varieties when S = 1. This is a consequence of a theorem of Fontaine, MR1274493 (Schémas propres et lisses sur Z).

I think that one should only expect finitely many such varieties for any fixed S. Let me give an argument that uses every possible conjecture I know. There may be an unconditional proof, but that would probably require knowing something about K3-surfaces.

I first want to claim that the ramification at primes q|S is "bounded" independently of X. The corresponding fact for elliptic curves will be that the power of the conductor for each q|N is bounded by 2 (if p > 3) or (if p = 2 or 3) by some fixed number I can't remember.

The most obvious argument along these lines is to consider the representation on inertia I_q acting on the p-adic etale cohomology groups H^2(X). These correspond to Galois representations with image in GL_22(Z_p). The argument I have in mind for elliptic curves works directly in this case, providing that one has "independence of p" statement for the Weil-Deligne representations at q (quick hint: the image of wild inertia divides the gcd of the orders of GL_22(F_p) over all primes p). This may require the existence of semi-stable models, which one certainly has for elliptic curves, but I don't know for K3-surfaces.

The next step is to use a Langlands-type conjecture. The p-adic representation V on H^2(X) may be reducible, but at least we know that each irreducible chunk will correspond to an irreducible Galois representation of Q into GL_n(Z_p) for some n (at most 22). Each of these, conjecturally, will correspond to a cuspidal automorphic form of fixed weight and level divisible only by q|S. Moreover, from the previous paragraph, the level will be bounded at q|S. Thus there will only be finitely many representations which can occur as H^2(X) for any K3-surface X/Z[1/S]. (Maybe I am assuming here that the Galois representation acting on H^2(X) is semi-simple --- let us do so, since this is a conjecture of Grothendieck and Serre.)

Finally, I want to deduce from any equality H^2(X) = H^2(X') that X is (essentially) X'. From the Tate conjecture we deduce the existence of correspondences X~~>X' and X'~~>X over Q whose composition induces an isomorphism on H^2(X) --- and now hopefully some knowledge of the geometry of K3 surfaces is enough to show that these sets of "isogenous" K3 surfaces form a finite set.

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EDIT:

As Buzzard points out, I obscured the fact in the last paragraph that some more arithmetic may be necessary. What I meant to say is that understanding isogeny classes of K3's over Q will first require understanding isogeny classes over C, and hopefully this second task will be the hard part.

As David points out, the Torelli theorem for K3 will surely be relevant here. I think there can be non-isomorphic isogenous K3s, however. If one takes an isogeny of abelian surfaces A->B then one can presumably promote this to an isogeny of the associated Kummer surfaces.

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EDIT:

Here is another thought. Deligne proves the Weil conjecture for K3 surfaces:

http://www.its.caltech.edu/~clyons/DeligneWeilK3trans.pdf

The philosophy is that there should be an inclusion of motives H^2(X) --> H^1(A) tensor H^1(A) for some abelian variety A (possibly of some huge but uniformly bounded dimension, like 2^19). It may be possible (conjecturally or otherwise) to reduce your question to the analogous statement for A, for which it is known. (Prop 6.5 is relevant here). It may well be possible to show that the variety A is defined over Z[1/2S], for example. I could make this edit more coherent but I'm off to lunch, so treat this as a thought fragment.

Answer 2

Yves Andre has proved the finiteness of the number of K3 surfaces over a number field $K$ with a polarisation of fixed degree $d$ and having good reduction (as a polarised variety) outside a fixed finite set of primes.

The reference is: "On the Shafarevich and Tate conjectures for hyper-Kähler varieties".Math. Ann. 305 (1996), no. 2, 205–248.

Answer 3

One classical trick that may be useful for reduction to the abelian scheme case is the Kuga-Satake construction which takes a weight 2 Hodge structure of K3 type to a weight 1 Hodge structure built out of its Clifford algebra. Although I've never read it, there is a paper of Rizov (http://arxiv.org/abs/math/0608497) which is supposed to make this a priori transcendental construction work over a general base (when you start with an honest family of polarized K3 surfaces). There may still be an issue about working up to isogeny, but at least this may let you shortcut the arithmetic discussion.

Answer 4

Here is a result about Kummer surfaces (one of the K's in K3 stands for Kummer) proved by Tetsushi Ito in his unpublished Master's thesis (Tokyo). I have a copy dated January 2001.

Theorem 5.2 . Let $K$ be a number field, $S$ a finite set of places of $K$ containing all archimedean places, and $d$ an integer. There are only finitely many Kummer surfaces over $K$, polarised of degree $d$, which have a rational point and which have good reduction outside $S$.

He uses Faltings' proof of the Shafarevich Conjecture for abelian varieties.