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?

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    $\begingroup$ Is it not a typo? Surely $fdz is holomorphic iff those conditions are satisfied. $\endgroup$ – Sam Gunningham Apr 25 at 14:32
  • $\begingroup$ 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 $\mathbb{P}^1$? $\endgroup$ – bojohnzhang Apr 25 at 14:36
  • $\begingroup$ 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... $\endgroup$ – Sam Gunningham Apr 25 at 14:39
  • $\begingroup$ Maybe. Well anyway thanks for your help! $\endgroup$ – bojohnzhang Apr 25 at 14:54
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    $\begingroup$ 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! $\endgroup$ – bojohnzhang Apr 25 at 15:06

The second half of the statment is not true, but if we replace "meromorphic" by holomorphic, it becomes a true proposition:

$f(z)dz$ 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 $\mathbb{P}^1$.

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