In complex geometry, the most interesting class of complex manifolds is probably the Kahler class. In the non-archimedean world, say over a fixed p-adic base field , the analogue of a compact complex manifold is a smooth proper rigid analytic space. In some ways, these are already surprisingly “close” to being Kahler – in particular, the Hodge-de Rham spectral sequence of such a space always degenerates at . However, Hodge symmetry can definitely fail. A standard example is the non-archimedean Hopf surface where acts through diagonal multiplication. By a fun direct calculation, one checks that and , so Hodge symmetry fails in degree one.
We now see a natural question: is there is some non-archimedean analogue of the Kahler condition which restores Hodge symmetry? Two years ago, Shizhang Li hit upon the following candiate condition:
A smooth proper rigid space satisfies (*) if it admits a formal model over whose special fiber is projective (as opposed to merely proper).
Using fantastic ideas due to Shizhang, we managed to prove the following suggestive result.
Theorem. Let be a smooth proper rigid space satisfying (*). Then .
Of course, one can then guess that (*) implies Hodge symmetry in all degrees. This speculation seems to have caught the imagination of others in the field, but until recently I personally regarded it as not much more than wishful thinking. However, my perspective completely changed a month ago, when I learned from Shizhang that, according to Robert Friedman, the archimedean analogue of “(*) implies Hodge symmetry” is a theorem! More precisely, we have the following result:
Theorem. Let be the complex disk, with the punctured disk. Let be a proper map of complex analytic spaces. Suppose that is a submersion, and that the central fiber is the analytification of a projective (and not necessarily smooth) algebraic variety. Then for all with , the fiber satisfies Hodge symmetry and Hodge-de Rham degeneration.
Of course, the analogy is that is analogous to , and is analogous to the “nearby” fibers with .
The proof of this theorem uses the full power of mixed Hodge theory. In fact the claim about Hodge-de Rham degeneration is exactly Corollary 11.24 in the book of Peters-Steenbrink. Hodge symmetry is even more subtle, and the argument for this doesn’t seem to be written down anywhere; Friedman explained it to Shizhang, who explained it to me, but the details entailed such a horrible explosion of gradings, filtrations, and multi-indices that I can’t hope to reproduce it here.
Anyway, I’m now completely convinced that Shizhang’s condition (*) implies Hodge symmetry in all degrees, and that this is truly the “right” p-adic analogue of the Kahler condition.