Compact surface with closed 1-form

Question

Why a compact surface with closed a nowhere vanishes complex 1-form, giving $T^{2}$?

Answer 1

Let $X$ a compact Riemann surface and $\omega$ a non-zero holomorphic one-form having no zeros (that is in every local chart $\omega = f(z)dz$ with $f$ locally analytic with no zero).

Let $$F(p) = \int_{p_0}^p \omega$$ (in local chart $F(p) =F(a)+ \int_a^p f(z)dz$)

Let $U$ be the universal cover of $X$ : that is $U$ is the Riemann surface obtained from the set of curves $\Gamma : p_0 \to p$ in $X$ modulo homotopy.

$U$ is topologically a simply connected domain.

Locally $F'$ doesn't vanish so $F$ is locally biholomorphic, since $U,F(U)$ are simply connected domains then $F$ is biholomorphic $U \to F(U)$.

Pick some closed-loops $\gamma_1,\ldots,\gamma_J$ generating the fundamental group $\pi_1(X)$.

Let $L = \{ \sum_{j=1}^J c_j \int_{\gamma_j} \omega, c_j \in \Bbb{Z}\}$.

This makes clear $F(U) = \Bbb{C}$.

The injectivity of $F : U \to \Bbb{C}$ means $L$ must be a lattice.

And hence $F$ is biholomorphic $X \to \Bbb{C}/L$.