Hodge diamonds that cannot be realised

In Paulsen–Schreieder [PS19] and vDdB–Paulsen [DBP20], the authors/we show that any block of numbers

    \[\left(\begin{array}{ccccc} & & h^{n,n} & & \\ & \iddots & & \ddots & \\ h^{n,0} & & & & h^{0,n} \\ & \ddots & & \iddots & \\ & & h^{0,0} & & \end{array}\right) \in \big(\mathbf Z/m\big)^{(n+1)^2}\]

satisfying h^{0,0} = 1, h^{p,q} = h^{n-p,n-q}, and h^{p,q} = h^{q,p} (characteristic 0 only) can be realised as the modulo m reduction of a Hodge diamond of a smooth projective variety.

While preparing for a talk on [DBP20], I came up with the following easy example of a Hodge diamond that cannot be realised integrally, while not obviously violating any of the conditions (symmetry, nonnegativity, hard Lefschetz, …).

Lemma. There is no smooth projective variety (in any characteristic) whose Hodge diamond is

    \[\begin{array}{ccccc} & & 1 & & \\ & 1 & & 1 & \\ 0 & & 1 & & 0 \\ & 1 & & 1 & \\ & & 1.\!\! & & \end{array}\]

Proof. If \operatorname{char} k > 0, we have \sum_{p+q = i}h^{p,q} \geq h_{\operatorname{dR}}^i \geq h_{\operatorname{cris}}^i, with equality for all i if and only if the Hodge–de Rham spectral sequence degenerates and H_{\operatorname{cris}}^i is torsion-free for all i. Because H^2_{\operatorname{cris}} contains an ample class, we must have equality on h^2, hence everywhere because of how spectral sequences and universal coefficients work.

Thus, in any characteristic, we conclude that h^1_{\operatorname{Weil}}(X) = 2, so \dim \mathbf{Pic}_X^0 = 1 and the same for \dim \mathbf{Alb}_X. Thus, X \to \mathbf{Alb}_X is a fibration, so a fibre and a relatively ample divisor are linearly independent in the Néron–Severi group, contradicting the assumption h^{1,1}(X) = 1. \qedsymbol

Remark. In characteristic zero, the Hodge diamonds

    \[\begin{array}{ccccc} & & 1 & & \\ & a & & a & \\ 0 & & 1 & & 0 \\ & a & & a & \\ & & 1 & & \end{array}\]

cannot occur for any a \geq 1, by essentially the same argument. Indeed, the only thing left to prove is that the image X \to \mathbf{Alb}_X cannot be a surface. If it were, then X would have a global 2-form; see e.g. [Beau96, Lemma V.18].

This argument does not work in positive characteristic due to the possibility of an inseparable Albanese map. It seems to follow from Bombieri–Mumford’s classification of surfaces in positive characteristic that the above Hodge diamond does not occur in positive characteristic either, but the analysis is a little intricate.

Remark. On the other hand, the nearly identical Hodge diamond

    \[\begin{array}{ccccc} & & 1 & & \\ & a & & a & \\ 0 & & 2 & & 0 \\ & a & & a & \\ & & 1 & & \end{array}\]

is realised by C \times \mathbf P^1, where C is a curve of genus a. This is some evidence that the full inverse Hodge problem is very difficult, and I do not expect a full classification of which Hodge diamonds are possible (even for surfaces this might be out of reach).


[Beau96] A. Beauville, Complex algebraic surfaces. London Mathematical Society Student Texts 34 (1996).

[DBP20] R. van Dobben de Bruyn and M. Paulsen, The construction problem for Hodge numbers modulo an integer in positive characteristic. Forum Math. Sigma (to appear).

[PS19] M. Paulsen and S. Schreieder, The construction problem for Hodge numbers modulo an integer. Algebra Number Theory 13.10, p. 2427–2434 (2019).

Rings that are localisations of each other

This is a post about an answer I gave on MathOverflow in 2016. Most people who have ever clicked on my profile will probably have seen it.

Question. If A and B are rings that are localisations of each other, are they necessarily isomorphic?

In other words, does the category of rings whose morphisms are localisations form a partial order?

In my previous post, I explained why k[x] and k[x,x^{-1}] are not isomorphic, even as rings. With this example in mind, it’s tempting to try the following:

Example. Let k be a field, and let K = k(x_1, x_2, \ldots). Let

    \[A = K[x_0,x_{-1},\ldots]\]

be an infinite-dimensional polynomial ring over K, and let

    \[B = A\left[\frac{1}{x_0}\right].\]

Then B is a localisation of A, and we can localise B further to obtain the ring


isomorphic to A by shifting all the indices by 1. To see that A and B are not isomorphic as rings, note that A^\times \cup \{0\} is closed under addition, and the same is not true in B. \qed

Is there a moral to this story? Not sure. Maybe the lesson is to do mathematics your own stupid way, because the weird arguments you come up with yourself may help you solve other problems in the future. The process is more important than the outcome.