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.

  • 1
    $\begingroup$ Do you mean that $X$ is a Riemann surface, or that it has complex dimension $2$? $\endgroup$
    – Aphelli
    Aug 17, 2020 at 8:10
  • $\begingroup$ @Mindlack complex dimension 2. $\endgroup$
    – Kevin
    Aug 17, 2020 at 8:20

1 Answer 1


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 $d\omega=0$.


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