# Do we have Hodge symmetry for char $p$?

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!

• 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. May 27, 2021 at 6:28
• Thanks for your comment! I wonder if there is any explicit counterexample May 27, 2021 at 6:32
• 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 $\Omega_A \cong \mathcal O_A^g$). May 27, 2021 at 6:35
• @R.vanDobbendeBruyn Oh my god! I was answered by the original author May 27, 2021 at 6:48
• @R.vanDobbendeBruyn Appreciate that！Respect！this question confused me for a long time. It's wonderful for me to have those references May 27, 2021 at 7:16

However, it is true for abelian varieties; see for example [Mum08, §13, Cor. 2]. As you note, this boils down to the computation that $$h^i(A,\mathcal O_A) = {g \choose i}.$$
[Ser58] J.-P. Serre, Sur la topologie des variétés algébriques en caractéristique $$p$$. Sympos. internac. Topología algebraica 24–53 (1958). ZBL0098.13103.