Integral of $\omega\wedge\overline{\omega}$ on Riemann surface

Question

Let $X$ be a Riemann surface of genus $g$ and $\omega$ a meromorphic 1-form on it. I've read that if $\omega$ has just a simple pole in $x\in X$ (and is holomorphic on $X\setminus\{x\}$) then the integral $\int_X\omega\wedge\overline{\omega}$ can't be finite. I've thought about that but I can't figure out why. In particular, how should the integral be considered and why is it $\int_X\omega\wedge\overline{\omega}=\infty$?

Answer 1

To show that it is infinite, one need only to calculate the expression locally around the simple pole.

Locally on an open ball (identified with the unit ball $B$ in $\mathbb C$) we have

$$\omega = \left(\frac{1}{z} + g(z)\right) dz,$$

where $g$ is holomorphic (thus bounded). Since

$$\int_B \left(\frac 1z\right) \left(\frac 1{\bar z}\right) dz \wedge d\bar z = -2i \int_B \frac{1}{x^2 + y^2} dx\wedge dy = -4\pi i\int_0^1 \frac{1}{r} dr =\ \infty,$$

the total integral cannot be finite.