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.

  • 2
    $\begingroup$ 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. $\endgroup$
    – jmc
    Commented Dec 2, 2019 at 11:47

1 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.

  • 1
    $\begingroup$ One possible reference for this is paragraph 4.4 of Fontaine--Messing, p-adic periods and p-adic étale cohomology. $\endgroup$
    – pbelmans
    Commented Dec 2, 2019 at 15:24
  • $\begingroup$ 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? $\endgroup$ 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. 107-126. $\endgroup$
    – abx
    Commented 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.