$h^{p,q} = h^{q,p}$ on complex smooth projective scheme

Question

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.

Answer 1

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 $h^{pq}= h^{n-p, n-q}$. 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, Deligne-Illusie). Then we get an isomorphism $$ H^{qp}\stackrel{L^{n-i}}{\cong} H^{n-i+q,n-i+p}=H^{n-p,n-q}$$

If you consider all the prerequisites, it's no easier than the analytic proof.