# Holomorphic maps into ruled surfaces

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.

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.