A busy few days: we had a run of interesting visitors this week in Madison, including Thomas Lam, who gave a beautiful talk about total positivity (a subfield of algebraic combinatorics, not a self-help philosophy); Melanie Matchett Wood, who explained how to parametrize binary forms of degree n in the Bhargava style, not only over Z but over an arbitrary base scheme (which is to say, not really in the Bhargava style!); and Davesh Maulik, who showed us how one can rather miraculously count rational curves on a single K3 by counting rational curves on a suitably chosen one-parameter family of K3s, and then “dividing” by the Noether-Lefschetz theory attached to the family. Very agreeably for number theorists, a key point is the product formulae of Borcherds, which provide modular forms on the moduli space of K3s whose zeroes and poles are supported on the countable union of subvarieties where the Picard number jumps upwards from its generic value.
This led to an amusing conversation at lunch about countable unions of subvarieties. Here’s a remark: if A is an abelian variety over the complex numbers, it’s completely obvious that A(C) contains some non-torsion points; the torsion locus is a countable union of varieties of strictly lower dimension (in this case 0) and thus can’t cover A(C). On the other hand, if A is over Fpbar, every point of A(Fpbar) is defined over some finite field, and thus all these points are torsion. The case of Qbar is intermediate in difficulty; indeed, there are nontorsion points on every abelian variety over Qbar, but this is not, in some sense, by “pure thought” — one might, for instance, use the argument that torsion points have height 0 but that there are plainly points of arbitrarily large height on A(Qbar). This uses some actual theorems, not just a comparison of cardinalities. Similarly, one can ask: are there elliptic curves over an algebraically closed field k with End(E/k) = Z? When k = C, the answer is obviously yes. When k = Fpbar, the answer is no, thanks to Frobenius. And when k is Qbar, the answer is again no, but maybe one has to use a bit more — for instance, that a CM elliptic curve over a number field has potentially good reduction everywhere.
In general, it’s pretty hard to see whether a countable union of subvarieties of X/Qbar covers all the Qbar-points! Here are two well-known open questions in this vein.
- Does there exist, for every g > 3, an abelian variety over Qbar which is not isogenous to the Jacobian of any smooth genus-g curve? This is an old question of Oort. Surely the answer is yes; but many people, including me, have puzzled over it without seeing how to make a proof. (Variant that just occurred to me: is there a one-parameter family of abelian varieties over C of dimension greater than 4 in which infinitely many fibers are isogenous to Jacobians, but the generic fiber is not?)
- If X is a K3 surface over Qbar, does there exist a point of X(Qbar) not contained in any rational curve on X? Bogomolov and Tschinkel have conjectured (or at least suggested) that the answer is no; see question 9 on the list of open questions (.pdf) from last month’s Banff conference on the arithmetic of K3s. They can prove that the answer is no when Qbar is replaced by Fpbar, and the K3 is a Kummer surface.
Oscar Villareal, in his 2005 Ph.D. thesis, proved that when A/Qbar is a semiabelian variety, the orbit of a proper closed subvariety X of A under End(A) doesn’t cover all of A(Qbar). Another, easier example appears in my paper with Elsholtz, Hall, and Kowalski, where one sees that there are lots of abelian varieties over Qbar in any reasonable one-parameter family which are simple; that is, they avoid the countable union of subvarieties parametrizing abelian varieties isogenous to products. In fact, one can show there are lots of such examples not only over Qbar, but over Q; and this is very closely related to why this problem is so much easier. It’s not at all hard to show that there are families of abelian varieties whose generic fiber is simple (or non-isogenous to a Jacobian, or whatever.) And an abelian variety over Q (as opposed to Qbar) should really be thought of as a family of abelian varieties, whose base has fundamental group Gal(Qbar/Q). It’s easy to check that if the image of Gal(Qbar/Q) in its action on the Tate modules of A is big (i.e. if the “family” of abelian varieties has “big mondromy”) it’s impossible for A to be non-simple, even over Qbar. But if we were weird beings who knew about Qbar but not its subfield Q, I think the theorem might be hard to prove! This point of view doesn’t say anything interesting, as far as I can see, for the two open questions above.
If there’s an overall story hiding here, I sort of hope it would look something like this. Arbitrary countable unions of closed subvarieties can certainly cover all the Qbar points of a variety X; just send a hypersurface through each point! But one might hope for some well-defined class of “small” countable unions that automatically miss “most” points of X(Qbar). More precisely, you could imagine a topology on X(Qbar) in which the closed sets consisted of both the usual Zariski-closed sets and certain countable unions of those; and in which algebraic morphisms were still continuous, of course. Natural candidates for “new closed sets” would be loci where certain cycle spaces jump in dimension — for instance, if M is a moduli space of n-dimensional varieties X, then the algebraic part of H^n(X x X) jumps precisely when X has an “extra correspondence” not possessed by the general fiber. So, for instance, the union of all the modular curves on a Hilbert modular surface would be closed in this sense. So would the locus of pairs (A,X) on Agbar x Mgbar where A is isogenous to Jac(X); and the projection of this locus down to Agbar is (apart from the difference between Mgbar and Mg) the countable union contemplated by Oort.
The hope would then be that every closed set S in the mystery topology misses most of X(Qbar); more precisely, one might hope that for every sufficiently large d, the probability that a point in X(Qbar) with field of definition of degree d and absolute height at most H lies in S decreases as H grows. Is that ridiculous?