Good evening, I can not prove the following result:
Let $\omega $ be a meromorphic 1-form on $ \mathbb {C} _ {\infty} = \mathbb {C} \cup \infty $ such that $ \omega_{|\mathbb{C}} = f (z) dz $. Show that f is ratio of polynomial functions. Any suggestions on how to develop the demonstration? thanks.
On
As suggest in the comment, on another cover $\mathbb C$ (which corresponds to $\mathbb C\setminus\{0\} \cup \{\infty\}$ with coordinates $\tilde z$), write $$\omega = \tilde f(\tilde z)$$
On the intersection of two coordinate $\mathbb C\setminus \{0\}$, we have $z = 1/\tilde z$, then
$$d\tilde z = d(1/ z) = -\frac{1}{ z^2} d z . $$
Write $f(z) = \sum_{n=k}^\infty a_n z^n$ and $\tilde f(\tilde z) = \sum_{n=m}^\infty b_n \tilde z^n$ for some $k, m\in \mathbb Z$, then
$$ \sum_{n=k}^\infty a_n z^n dz = f(z) dz = \tilde f(\tilde z) d\tilde z = \sum_{n=m}^\infty b_n \tilde z^n d\tilde z = \sum_{n=m}^\infty b_n z^{-n} \bigg(-\frac{1}{z^2}\bigg)dz$$
$$ \Rightarrow \sum_{n=k}^\infty a_n z^n = \sum_{n=m}^\infty (-b_n) z^{-n-2}$$ Then compare coefficients to see that $a_n = b_n = 0$ for large enough $n$.
Here is a more conceptual proof. Consider first the behaviour of $f$ at infinity, when we change the coordinates to $\frac{1}{z}$ we get $\omega=-\frac{1}{z^2}f(\frac{1}{z})dz$. Now $\frac{1}{z^2}f(\frac{1}{z})$ must have a pole or a finite value at $z=0$. So we can say that there is a $k$ such that $z^kf(z)$ is bounded at infinity. Now this function has finite number of poles. By subtracting the principal parts of these poles we get an entire function, bounded at infinity, so it must be constant. This then shows that $f$ is a rational function.