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.

  • $\begingroup$ This is a local calculation that's in any number of texts (basically it just boils down to the First Fundamental Theorem of Calculus). Nothing to do with Stokes's Theorem. Nothing to do with Riemann surfaces. $\endgroup$ May 1, 2017 at 16:55
  • $\begingroup$ I'm also a bit confused about that Stoke's part. In the end I decided to go with 'local examination will show us that $dF=\omega$. I couldn't ask my supervisor soon enough so that I would have a second chance to understand his concerns. $\endgroup$
    – VadaVad
    May 2, 2017 at 20:58
  • $\begingroup$ I have this clearly written out for differential forms in my multivariable mathematics textbook (and the argument appears pretty carefully done in my YouTube lectures linked in my profile). $\endgroup$ May 2, 2017 at 21:42

1 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. $dF=\omega$.

  • $\begingroup$ Excuse me, is this "$a(d/dt|_{t=1})$" a typo on the fifth line? So you define a 'path' on to $X$ whose pushforward at $x=\gamma(1)$ is $v$. The basic idea is to use 'everybodys favorite' $D \int_0^t k(s) ds=k(t)$? $\endgroup$
    – VadaVad
    Apr 29, 2017 at 15:11
  • $\begingroup$ @VadaVad yes it was a typo, now corrected. And indeed, the basic idea is that. $\endgroup$
    – user8268
    Apr 29, 2017 at 18:00

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