I know that for compact Kähler manifolds $M$ there is an isomorphism: $$ H^p(M, \Omega_M^q) = H^q(M, \Omega_M^p) $$ where $\Omega_M$ is the sheaf of holomorphic $1$forms. It is because $H^p(M, \Omega_M^q) = H^{p,q}_{\bar{\partial}}(M)=\mathcal{H}^{p,q}(M)$ the set of harmonic forms on $M$. We can then apply conjugation on $\mathcal{H}^{p,q}(M)$ to get $\mathcal{H}^{q,p}(M)$. So the Hodge numbers satisfy $h^{p,q} = h^{q,p}$. This means if I have a complex smooth projective scheme $X$ and take the sheaf of differentials $\Omega_X$, then we still have $h^{p,q} = h^{q,p}$ because analytification of $X$ is a compact Kähler manifold. I wonder if there is an algebraic argument for this fact.
1 Answer
I remember seeing such a proof in an article by Messing, who attributed it to Gabber. Let $X$ be a smooth projective variety of dimension $n$ over a field of characteristic $0$. Suppose that $p+q=i\le n$. Serre duality gives $h^{pq}= h^{np, nq}$. Now put this together with algebraic proofs of hard Lefschetz for $\ell$adic cohomology (Deligne); suitable comparison theorems, which implies HL for algebraic de Rham; the algebraic proof of degeneration of the Hodge to de Rham spectral sequence (Faltings, DeligneIllusie). Then we get an isomorphism $$ H^{qp}\stackrel{L^{ni}}{\cong} H^{ni+q,ni+p}=H^{np,nq}$$
If you consider all the prerequisites, it's no easier than the analytic proof.

1$\begingroup$ One possible reference for this is paragraph 4.4 of FontaineMessing, padic periods and padic étale cohomology. $\endgroup$– pbelmansCommented Dec 2, 2019 at 15:24

$\begingroup$ Thank you very much. I don't know anything about ladic cohomology yet, but in case of char 0 the $l$adic cohomology coincides with normal sheaf cohomology, then we have a pure algebraic proof of Hard Lefschetz. Is it true? $\endgroup$ Commented Dec 2, 2019 at 16:36

$\begingroup$ For a much more elementary proof, see §5 in Deligne's Théorème de Lefschetz et critères de dégénérescence de suites spectrales, Publ. IHES 35 (1968) p. 107126. $\endgroup$– abxCommented Dec 2, 2019 at 16:48

$\begingroup$ Thank you, it is also a good opportunity to learn more French :) $\endgroup$ Commented Dec 2, 2019 at 17:45