Let $\Sigma$ be a compact Riemann surfaces. Let $L \to \Sigma$ be a holomorphic line bundle. This gives rise to a ruled surface $\mathbb{P}(\underline{\mathbb{C}} \oplus L) \to \Sigma$, where $\mathbb{P}$ denotes the fiber-wise projectivization and $\underline{\mathbb{C}}:= \Sigma \times \mathbb{C} \to \Sigma$ is the trivial line bundle.

Now, we can take any meromorphic section of $s$ of $L$, which then gives rise to a holomorphic section $u$ of the ruled suface via the following construction: As long as $z$ is no pole of $s$, we define the section $u$ to be $$ u(z)=[1:s(z)] $$ Note that this section is holomrophic if we take the chart of $\mathbb{CP}^1$ not containing the point $[0:1]$.
If $p$ is a pole, we have locally that $s$ contains no other pole or zero and we can write locally $$ s(z)=z^{-m}g(z) $$ for some $m \in \mathbb{N}$ and $g(z)$ zero- and polefree. Then we define $$ u(z)=z^{-m}[1:g(z)], $$ which is holomorhpic, if we take the chart of $\mathbb{CP}^1$ not containing $[0:1]$.

So meromorphic sections of $L$ give rise to holomorphic sections of $\mathbb{P}(\underline{\mathbb{C}} \oplus L)$. My question now is: Does every meromorphic section of this ruled surface arise in that way?
I have the feeling, wether it does work or does not work, relies somehow on the compactness of $\Sigma$. There is the know result on the complex plane (non-compact), for example, that every meromorphic function on $\mathbb{C}$ must not be rational (take $e^z$), however, on its compactification, it must be rational. Furthermore, my knowledge in algebraic geometry is a a bit lacking, so the question might even yield a trivial answer.


1 Answer 1


Yes, you're right on both counts - every section of this ruled surface arises via a meromorphic section of $ \mathcal{L} $ and you should read some algebraic geometry to make it easy for yourself.

The sections of the ruled surface correspond to morphisms $ \Sigma \rightarrow \mathbb{P}( \mathcal{O}_{\Sigma} \oplus \mathcal{L}) $ over $ \Sigma $ and these in turn are given by line bundle quotients $ \mathcal{O}_{\Sigma} \oplus \mathcal{L}^{\vee} \rightarrow \mathcal{N} \rightarrow 0 $, where $ \mathcal{N} $ is a line bundle on $ \Sigma $. Such quotients are given by two morphisms $ \mathcal{O}_{\Sigma} \rightarrow \mathcal{N} $ and $ \mathcal{L}^{\vee} \rightarrow \mathcal{N} $ i.e. two global sections $ s_1 \in H^0(\Sigma, \mathcal{N}) $ and $ s_2 \in H^0(\Sigma, \mathcal{N} \otimes \mathcal{L}) $ which do not simultaneously vanish on $ \Sigma $. Then $ s = s_2/s_1 $ is a meromorphic section of $ \mathcal{L} $.
Conversely, if $ s $ is a meromorphic section of $ \mathcal{L} $ with divisor of poles being $ P $, we multiply by the canonical section $ 1_P \in H^0(\Sigma, \mathcal{O}_{\Sigma}(P)) $ to get a holomorphic section $ s. 1_P \in H^0(\Sigma, \mathcal{L} \otimes \mathcal{O}_{\Sigma}(P)) $ and because we've killed exactly the poles, $ 1_P $ and $ s. 1_P $ do not simultaneously vanish on $ \Sigma $. The upshot is that we get a line bundle quotient $ \mathcal{O}_{\Sigma} \oplus \mathcal{L}^{\vee} \rightarrow \mathcal{O}_{\Sigma}(P) \rightarrow 0 $ and these two constructions are clearly inverse operations.

  • $\begingroup$ Thanks for your answer and yes, I am currently trying to self-study a bit of algebraic geometry. I really like your argument though, as I can follow it :-) $\endgroup$
    – F. Conrad
    Commented Feb 5, 2023 at 11:43
  • $\begingroup$ no problem. There was a typo : “1_P (not s) and s.1_P don’t vanish simultaneously” I have corrected it. $\endgroup$ Commented Feb 5, 2023 at 11:49

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