# Top Dolbeault cohomology group vs top De Rham cohomology group of a compact complex manifold

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...

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.
• 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... Nov 7, 2021 at 3:08