closed forms on a closed oriented surface

Question

I wanted to see if is true that every $(1,0)$-form on a closed oriented surface is closed? And also how does being closed implies this form being holomorphic?

Answer 1

Let $\alpha$ be a $(1,0)$-form on the Riemann surface $X$. $\alpha$ is closed iff $0=d\alpha = \partial \alpha + \bar \partial \alpha$.
$\partial \alpha$ is a $(2,0)$-form, but $\dim X = 1$, so $\partial \alpha= 0$.

Thus $\alpha$ is closed iff $\bar \partial \alpha=0$. But this means that $\alpha$ is a holomorphic $1$-form. So:

A $(1,0)$-form is closed iff it is holomorphic.

Keeping the above in mind, we easily can construct non-closed $(1,0)$-forms. For example, let $f$ be a smooth, but non-holomorphic function on the torus. Then $f dz$ is a $(1,0)$-form which is not closed.

Edit: Regarding the comment: Holomorphic $1$-forms are sections of the canonical bundle, i.e. elements of $H^0(X,\omega_X)$.
$\dim ^0(X,\omega_X)$ is called the geometric genus of $X$. It can be shown that it coincides with the topological notion of the genus. So only on $\mathbb CP^1$ there is no holomorphic $1$-form ($\neq 0$).