Holomorphic and Harmonic $1-$ Forms on Riemann Surfaces Let $\mathbb{S}$ be a Riemann Surface.
We say that a differential form $\omega$, on the Riemann Surface $\mathbb{S}$, is a Harmonic $1-$ Form if it and its conjugate $\omega^\star$ are both closed, i.e., $d\omega=0$ and $d\omega^\star=0$.
We say that a differential form $v$, on the  Riemann Surface $\mathbb{S}$, is a Holomorphic Differential Form if it can be locally represented as $dv=f(z)dz$ where $f$ is a holomorphic function.
Now, the question is:
What is the relationship between Harmonic $1-$ Forms and Holomorphic Differential Forms?
 A: Note that there is a completely different and much more common use of the term "Harmonic form" in Hodge theory. I'm pointing this out, because this is what confused me initially.
Anyhow, let $\omega$ be a complex differential form on $\mathbb{S}$. Harmonicity and Holomorphicity are both local conditions, so we may pick coordinates $z=(x,y)$ and expand $\omega=fdz+gd\overline{z}$ with smooth functions $f,g$ in these coordinates. Then, $\omega^{\ast}=i(fdz-gd\overline{z})$. If $\omega$ is harmonic, $\omega$ and $\omega^{\ast}$ are closed, whence $\frac{1}{2}(i\omega+\omega^{\ast})=fdz$ is closed. But $d(fdz)=\frac{\partial f}{\partial\overline{z}}d\overline{z}\wedge dz$, so that $fdz$ is closed iff $f$ is holomorphic. Thus, the $(1,0)$-part of a harmonic $1$-form is a holomorphic $1$-form. Conversely, if $\omega=fdz$ is holomorphic, it is closed, but then $\omega^{\ast}=ifdz$ is closed as well, so that $\omega$ is harmonic. In total, holomorphic $1$-forms are precisely the $(1,0)$-parts of harmonic $1$-forms.
