2
$\begingroup$

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!

$\endgroup$
5
  • 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

5
$\begingroup$

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 $$h^i(A,\mathcal O_A) = {g \choose i}.$$


References.

[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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.