Let $X$ be a smooth projective variety over a field $k$. Let $h^{p,q}=dim_k H^q(X,\Omega_{X/k}^p)$ be the Hodge numbers.

If $k$ is of char $0$, by Lefschetz principle, we always have Hodge symmetry, i.e. $h^{p,q}=h^{q,p}$, for the following reason: we can regard $X$ as a base change of a smooth projective variety $X_0$ over a finitely generated field $k_0$, but fix an arbitrary embedding $k_0\rightarrow \mathbb{C}$, by base change to $\mathbb{C}$, the flat base change theorem and Hodge symmetry for complex smooth projective variety will give us the reason.

Do we always have Hodge symmetry for char $p$ ? If we do, how do we prove it?

My motivation might be very trivial for experts...I was computing the Hodge numbers for abelian varieties. Let X be an abelian variety over $k$ of dim $g$, then I want to compute all the hodge numbers (for any character). I crucially need Hodge symmetry:

The relative differential is trivial: $\Omega_{X/k}=\mathcal{O}_X^{\oplus g}$. If we have Hodge symmetry, then the Hodge numbers $h^{p,q}=dim H^q(X,\Omega^p)=dim H^q(X,\mathcal{O}_X^{\oplus \binom {g}{p}})= \binom{g}{p} dim H^q(X, \mathcal{O}_X)=\binom{g}{p} h^{q,0}=\binom{g}{p}h^{0,q}=\binom{g}{p}\binom{g}{q} h^{0,0}=\binom{g}{p}\binom{g}{q}$.

Is there any proof that doesn't depend on Hodge symmetry? Any references are welcome!

  • 1
    $\begingroup$ Sorry to disappoint but there is no Hodge symmetry in positive characteristic. I am not if it holds for abelian varieties. I suggest you narrow down your question to them. $\endgroup$
    – Bugs Bunny
    May 27, 2021 at 6:28
  • $\begingroup$ Thanks for your comment! I wonder if there is any explicit counterexample $\endgroup$
    – Yuan Yang
    May 27, 2021 at 6:32
  • 4
    $\begingroup$ Hodge symmetry fails in general; the first example is due to Serre (see e.g. section 1 of 2001.02787 for a modern version of that argument, although the original is also very readable). For abelian varieties it's still ok; see for example §13, Corollary 2 in Mumford's book (plus the computation that $\Omega_A \cong \mathcal O_A^g$). $\endgroup$ May 27, 2021 at 6:35
  • $\begingroup$ @R.vanDobbendeBruyn Oh my god! I was answered by the original author $\endgroup$
    – Yuan Yang
    May 27, 2021 at 6:48
  • $\begingroup$ @R.vanDobbendeBruyn Appreciate that!Respect!this question confused me for a long time. It's wonderful for me to have those references $\endgroup$
    – Yuan Yang
    May 27, 2021 at 7:16

1 Answer 1


Hodge symmetry fails in positive characteristic in general; see Serre's Mexico paper [Ser58, Prop. 16] (for a more modern/conceptual version of that argument, see e.g. [vDdB21, Prop. 1.4]).

However, it is true for abelian varieties; see for example [Mum08, §13, Cor. 2]. As you note, this boils down to the computation that $$h^i(A,\mathcal O_A) = {g \choose i}.$$


[vDdB21] R. van Dobben de Bruyn, The Hodge ring of varieties in positive characteristic. Algebra Number Theory 15.3, p. 729–745 (2021).

[Mum08] D. Mumford, Abelian varieties. Second edition, corrected reprint. With appendices by C. P. Ramanujam and Yuri Manin. New Delhi: Hindustan Book Agency/distrib. by American Mathematical Society (AMS); Bombay: Tata Institute of Fundamental Research (ISBN 978-81-85931-86-9), 2008. ZBL1177.14001.

[Ser58] J.-P. Serre, Sur la topologie des variétés algébriques en caractéristique $p$. Sympos. internac. Topología algebraica 24–53 (1958). ZBL0098.13103.


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