Function defined as integral of 1-form on Riemann surface

Question

Suppose $X$ is compact simply connected Riemann surface, $\omega$ is non-vanishing 1-form on $X$. Choose basepoint $z_0$ and define $F\colon X \to \mathbb{C}$ as $$ F(x) = \int_{[z_0,x]} \omega .$$ Here $[z_0,x]$ is any path from $z_0$ to $x$. This is well defined because $X$ is simply connected.

I claim that $dF = \omega$, but my supervisor insists that this isn't so clear. His reasonings are

• When evaluating difference quotient of $dF_p ([\gamma])$ we will eventually need Stokes or something similar.
• We need to choose suitable $\gamma$.
• $dF_p$ is $TX_p \to T\mathbb{C}_{F(p)}$ and $\omega$ is $TX_p \to \mathbb{C}$

The third point throws me a bit off. If it helps, then $\omega$ is actually $p^* (\alpha)$ where $p\colon X \to Y$ is a covering map and $\alpha$ is non-vanishing 1-form on $Y$.

I would appreciate either a good reference or some good arguments to cover this up.

Answer 1

Here is a possible a formal argument (I guess it can be simplified):

If $v\in T_xX$ is any tangent vector, choose a smooth map $\gamma:(-2,2)\to X$ s.t. $\gamma(0)=z_0$, $\gamma(1)=x$, and $\gamma_*(d/dt|_{t=1})=v$. Define $f:(-2,2)\to\Bbb C$ as $f(x)=\int_0^x\gamma^*\omega$. By the definition of $F$ you have $f=\gamma^*F$, hence $df=\gamma^*dF$, hence $\frac{df}{dt}(1)=dF(v)$. If $\gamma^*\omega=g(t)dt$, Newton-Leibniz says $\frac{df}{dt}(1)=g(1)=(\gamma^*\omega)(d/dt|_{t=1})=\omega(v)$. Hence $dF(v)=\omega(v)$ for every $v$, i.e. $dF=\omega$.