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

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.

• The analogous fact is not true (in general) in characteristic $p > 0$. So any algebraic proof should at least use the fact that char $\mathbb{C}$ = 0. I don't know of any such proof.
– jmc
Dec 2, 2019 at 11:47

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 $$H^{qp}\stackrel{L^{n-i}}{\cong} H^{n-i+q,n-i+p}=H^{n-p,n-q}$$
• Thank you very much. I don't know anything about l-adic 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? Dec 2, 2019 at 16:36