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 $K$, 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 $E_1$. However, Hodge symmetry can definitely fail. A standard example is the non-archimedean Hopf surface $X = \mathbf{A}^2_{K} \smallsetminus \{ (0,0) \} / p^{\mathbf{Z}}$ where $p^n$ acts through diagonal multiplication. By a fun direct calculation, one checks that $H^0(X,\Omega^1_X)=0$ and $H^1(X,\mathcal{O}_X) = K$, 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 $X$ satisfies (*) if it admits a formal model $\mathfrak{X}$ over $\mathcal{O}_K$ 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 $X$ be a smooth proper rigid space satisfying (*). Then $h^{1,0}(X) = h^{0,1}(X)$.

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 $D$ be the complex disk, with $D^\times =D \smallsetminus \{0 \}$ the punctured disk. Let $f:Y \to D$ be a proper map of complex analytic spaces. Suppose that $f^{-1}(D^\times) \to D^\times$ is a submersion, and that the central fiber $Y_0=f^{-1}(0)$ is the analytification of a projective (and not necessarily smooth) algebraic variety. Then for all $t \in D^\times$ with $|t| \ll 1$, the fiber $Y_t$ satisfies Hodge symmetry and Hodge-de Rham degeneration.

Of course, the analogy is that $\mathfrak{X} \to \mathrm{Spf} \mathcal{O}_K$ is analogous to $Y \to D$, and $X$ is analogous to the “nearby” fibers $Y_t$ with $0<|t| \ll 1$.

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.