I cannot reconcile two simple facts for a compact complex manifold $X$ with $dim_{\mathbb{C}}=n$. Let $A^{p,q}(X)$ be differential forms of degree $(p,q)$ on $X$.

One the one hand, $$H^{2n}_{DR} (X)= A^{n,n}(X)/d\left(A^{n-1,n}(X) + A^{n,n-1}(X)\right) = A^{n,n}(X)/\left(\partial(A^{n-1,n}(X)) + \bar{\partial}(A^{n,n-1})\right)$$ and this group is isomorphic to $\mathbb{C}$ via integration of forms and Stokes' theorem.

On the other hand, $$H^n(X,K_X)\stackrel{Dolbeault}{====} A^{n,n}(X)/ \bar{\partial}(A^{n,n-1}(X))$$ is also said to be isomorphic to $\mathbb{C}$ via integration and before Serre's duality is established (p. 135 of Voisin's "Hodge Theory and Complex Algebraic Geometry I"; in fact, the above fact is used to prove Serre's duality in the book.)

Without the $\partial\bar{\partial}$-lemma (that requires $X$ to be Kahler) I don't see how to show that $$\partial \omega=d \omega, \quad \omega \in A^{n-1, n}(X)$$ is zero in Dolbeault cohomology (since it clearly integrates to zero by Stokes' theorem). And yet it should be true by Serre's duality that does not require the Kahler condition...


1 Answer 1


One can look at the dual statement: since $H^{2n}(X)\cong H^0(X)$ by Poincaré duality and $H^{n,n}(X)\cong H^{0,0}(X)$ by Hodge star. Then the fact that $H^0(X)\cong H^{0,0}(X)$ is reflected by $df=0\Leftrightarrow \bar\partial f=0$ on a compact complex manifold. However, I do not know how to explicitly show $d$-exact is equivalent to $\bar\partial$-exact in top forms.

  • 1
    $\begingroup$ In fact, this question bothered me so much that I asked a few of my friends about it. None of them came up with an answer yet. $\endgroup$
    – lEm
    Nov 6, 2021 at 5:59
  • $\begingroup$ In their textbooks, Huybrechts and Demailly seem to not use $H^n(X,K_X) \simeq \mathbb{C}$ in the proof of Serre's duality, maybe for a reason... $\endgroup$
    – Bananeen
    Nov 7, 2021 at 3:08

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