On a compact complex surface, every holomorphic 1-form is closed

Given a compact complex manifold $$X$$ with two dimension.Then every holomorphic 1-form $$\omega\in H^{0}(X,\Omega^{1})$$ satisfies $$d\omega=0$$. I want to know how to solve this problem by a simple method.

• Do you mean that $X$ is a Riemann surface, or that it has complex dimension $2$? Aug 17, 2020 at 8:10
• @Mindlack complex dimension 2. Aug 17, 2020 at 8:20

When $$X$$ is compact Kahler, then followed by the Hodge theorem, $$H^0(X,\Omega^1)\cong H^{1,0}(X)$$ are all represented by harmonic forms, in particular, every holomorphic $$1$$-forms are harmonic, so closed.

In general, let's work with $$X$$ a compact complex surface. Let $$\omega$$ a holomorphic 1-form on $$X$$, then by Stokes theorem,

$$\int_{X}d\omega\wedge d\bar{\omega}=\int_Xd(\omega\wedge d \bar{\omega})=0.\tag{1}\label{1}$$

On the other hand, write $$d\omega=fdz_1\wedge dz_2$$ locally with $$f$$ holomorphic, so $$d\omega\wedge d\bar{\omega}=-|f|^2dz_1\wedge d\bar{z}_1\wedge dz_2\wedge d\bar{z}_2=4|f|^2dx_1\wedge dy_1\wedge dx_2\wedge dy_2,$$

with $$z_j=x_j+iy_j$$, $$j=1,2$$. Now condition $$(\ref{1})$$ implies $$f=0$$, so $$d\omega=0$$.