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!