# What is meromorphic differentials like on Riemann Sphere? [closed]

There is a proposition that every meromorphic differential on Riemann Sphere (or $$\mathbb{P}^1 = \mathbb{C} \cup \{ \infty \}$$) can be written as $$f dz$$ where $$f$$ is a meromorphic function on $$\mathbb{P}^1$$, and that $$f dz$$ is meromorphic if and only if $$f$$ is holomorphic on $$\mathbb{C}$$ and $$z^2 f(z)$$ tends to a finite limit as $$z \rightarrow \infty$$.

This is Exercise 6.3. on Page 178 of Frances Kirwan's Complex algebraic curves. I know how to prove the former half of the proposition (every meromorphic differential can be written as $$f dz$$), but I cannot find a proof of the second half. In fact, I'm not sure whether the propostion is true, because I seem to find a meromorphic differential which is not holomorphic on $$\mathbb{C}$$ (e.g. $$1/z$$ $$dz$$).

Does anyone can answer my question?

• Is it not a typo? Surely $fdz is holomorphic iff those conditions are satisfied. – Sam Gunningham Apr 25 at 14:32 • I'm not sure whether there is a typo in the original book. Additionally, it seems that there is no non-zero holomorphic differentials on$\mathbb{P}^1$? – bojohnzhang Apr 25 at 14:36 • As you have observed there are meromorphic differentials on$\P^1$which are not holomorphic on$\C$(like$dz/z$). So it seems that there is a typo in the original. Also, you can probably prove that the statement is true after replacing meromorphic by holomorphic. Then, indeed, you can show that there are no non-zero functions satisfying that condition... – Sam Gunningham Apr 25 at 14:39 • Maybe. Well anyway thanks for your help! – bojohnzhang Apr 25 at 14:54 • Oh, I got it. The exercise following this one is "To deduce that there are no holomorphic differentials on$\mathbb{P}^1\$", and this is just a proof of it (I know another proof previously, so I didn't get what the writer means). Well, thank you very much! – bojohnzhang Apr 25 at 15:06

$$f(z)dz$$ is holomorphic if and only if $$f(z)$$ is holomorphic on $$\mathbb{C}$$ and $$z^2f(z)$$ tends to a finite limit as $$z \rightarrow \infty$$.
This is easy to prove because $$f(z_1)dz_1=-f(1/z_2)/z_2^2 dz_2$$ when changing local coordinate, and $$\lim\limits_{z_1\to\infty}f(z_1)z_1^2 = \lim\limits_{z_2\to 0}f(1/z_2)/z_2^2$$
In fact, there is no non-zero holomorphic function $$f(z)$$ that satisfies the condition above, so there is no non-zero holomorphic differential on $$\mathbb{P}^1$$.