Martin Olsson and I have a new paper about derived Torelli theorems for K3 surfaces. It is a piece of our gradual attempt to recast Torelli theorems in terms of a combination of the derived category and the Chow theory. While we previously showed a very coarse kind of Torelli statement – a certain kind of derived equivalence that respects a filtration on Chow theory implies the existence of an isomorphism – we are now able to show that there is an isomorphism that acts on the cohomology in the same way as the derived equivalence. (There is a slight subtlety involving the filtration-perserving condition, but this is precisely analogous to the condition in the Torelli theorem about preserving ample classes.) Our original proof ultimately used analysis, but the new one is purely algebraic.

The basic idea

The Hodge structure of a variety is a semi-linear object that mixes linear algebra together with various filtrations. Here’s another such gadget: the derived category , together with data about the support of perfect complexes. This can be interpreted in various ways. Here are two examples.

  1. One could record the support of complexes , which means for example that a complex with adjacent cohomologies that have large support will itself also have large support, even if the rest of the cohomologies are supported on small loci. So here the package would be together with the collection of subcategories parametrizing perfect complexes with support of dimension at most .
  2. One could record the image of in the Chow theory via the Chern character, in which case the support data attached to allows things like enormous adjacent support to cancel. Here the semi-linear package is together with the map together with the codimension filtration on .

The idea is that if we start with and , and we have an equivalence that preserves filtrations (in one of the sense above or another reasonable one), then and should be isomorphic. A stronger hope is that at least some of these filtration-preserving equivalences should actually come from isomorphisms themselves, at least after passing to some cohomology theory. That is to say, equivalences are given by kernels on , so they induce correspondences between all kinds of cohomology theories (etale, crystalline, etc.), and we can at least ask for an isomorphism that induces the same cohomological correspondences.

The original proof for K3 surfaces

In our original paper, we carefully lifted the filtered equivalence to characteristic and then used the filtration-preserving condition to reduce to the classical complex Torelli theorem. In the process, we lost some control over how the equivalence was acting on various cohomology theories. (Now that I think about it again, I think we just didn’t try. I would guess that a relatively minor adjustment to our methods should yield the general result of our new paper, but it would still involve lifting to characteristic and invoking the classical Torelli theorem.)

This was really fun, but it also meant that in trying to lay foundations for a Torelli-type theory in positive characteristic, we were in the uncomfortable position of invoking the analytic theory over the complex numbers. This limits the amount of insight into the true meaning of the result that comes from the proof. (I.e., maybe this derived category plus filtered thingamajig idea is really a red herring, and all we did was record a new consequence of the analytic theory.)

The new proof for K3 surfaces

In the new paper, we follow a suggestion of F. Charles to use degeneration to the supersingular case in place of deforming to the complex case. In fact, we are able to completely eliminate the use of Hodge theory in the situations of interest, giving us a purely algebraic proof. (Important note: the supersingular case relies heavily on the deep work of Ogus, so we don’t get a transparent proof!)

The delicate points here involve tracing the action on cohomology as we slide derived equivalences around over the moduli space of polarized K3 surfaces in positive characteristic. This also involves knowing things about the geometry of the moduli space of K3 surfaces; if one is not careful, one falls into an analytic trap, but there are algebraic proofs of the relevant facts, even if some of them happen in characteristic 0.

As an outcome, one gets a purely algebraic proof that any Fourier-Mukai partner of a K3 surface is a moduli space of stable sheaves. Look ma, no Hodge theory! Update: I now realize that Charles relies on analytic results for his birational boundedness statement as part of his second proof of the Tate conjecture for K3s (via a Zarhin-type trick), so there is still a morsel of analysis involved, since we need the Tate conjecture to get the geometry of the moduli space of K3s right. Interesting!

The strongest possible filtration condition

Perhaps it is worth pointing out what happens if one imposes the strongest possible filtration-preservation condition (number 1 in the list above). This condition is very strong. In this case, it seems pretty likely that the equivalence itself is given by composing an isomorphism , a twist by an invertible sheaf on , and a shift in the derived category.

Among other things, my student Dan Bragg has been thinking a bit about this. I think Dan has worked out all of the details for arbitrary smooth and , but I don’t want to put words in his mouth (so I’ll just say “I think”).

What happens next?

Martin, Dan, and I have been thinking about the situation for hypersurfaces (including what happens to Donagi’s result in positive characteristic). But really the interesting question is whether this holds for all (smooth projective) varieties or not, when one is not using the strongest possible filtration condition. Here is a list of vague questions.

  1. What is the closest one can get to condition 2 above and still have universal truth?
  2. What classes of varieties will be Torellish with respect to condition 2, or with respect to some other similar condition?
  3. Is there a hierarchy of conditions with different Torellish classes of varieties?

By “Torellish” I mean exactly what you think I mean.