# Divisor of meromorphic section of point bundle over a Riemann surface

Let $X$ be a compact connected Riemann surface (not $\mathbb{P}^1$), $p\in X$ be a point on it. Let $L$ be the holomorphic line bundle associated to the divisor $D=p$. By construction $L$ comes with a holomorphic section $\sigma$ vanishing exactly at $p$; since

$$H^0(X,\mathcal{O}(L)) \cong \mathcal{L}(D)=\{ f \in \mathcal{M}(X) \ | \ (f)+p \geq 0 \} \cong \mathbb{C}$$

any other holomorphic section is a scalar multiple of $\sigma$, hence vanishes exactly at $p$.

Does this generalize to meromorphic sections, i.e. does any meromorphic section vanish at $p$? I know that any meromorphic section will differ from $\sigma$ by a meromorphic function, i.e. $\sigma ' = f \sigma$. A priori $f$ could have poles in $p$, allowing $\sigma '$ to be non zero at $p$. Of course not every 0-degree divisor is principal, so maybe there are restrictions which do not allow me to move the zero at $p$? Does this depend on the point I pick?

Thanks in advance for any comment.

• Well, at least $D = 2p$ would solve this. But $D = p$ depends on the cocycles... – user40276 Nov 20 '14 at 22:36

There is a very subtle but very serious misconception (which is never addressed in books) around this question , namely:

No meromorphic section on $X$ can have a simple zero at $p$ and be holomorphic at every $q\neq p$, since $X\ncong \mathbb P^1$.
But does this not contradict the fact that there exists $\sigma \in H^0(X,\mathcal O(p))$ vanishing only at $p$, with order $1$ ?
No it doesn't: the subtle point is that $\sigma$ vanishes at $p$ if seen as a section of $\mathcal O(p)$, and not as a usual meromorphic function.
The link between the two is given by the formula $$ord_p^{\mathcal O(p)}(\sigma)=ord_p(\sigma)+1$$ $$ord_r^{\mathcal O(p)}(\sigma)=ord_r(\sigma) \; \operatorname {for}\; r\neq p$$

For $q\neq p$ the vector bundle $\mathcal O(q)$ is ample, so that for large enough $N$ the global sections of $\mathcal O(p+Nq)$ generate all fibers of that last bundle.
In particular there is a holomorphic section $\tau \in H^0(X,\mathcal O(p+Nq))$ not vanishing at $p$ as a section of the vector bundle , i.e. $ord_p^{\mathcal O(p+Nq)}(\tau)=0$
That very section $\tau$ can also be seen as a meromorphic section of $\mathcal O(p)$ with $$ord_p^{\mathcal O(p)}(\tau)=ord_p^{\mathcal O(p+Nq)}(\tau)=0$$ In other words $\tau$ can be seen as a meromorphic section of $\mathcal O(p)$ not vanishing at $p$ as a meromorphic section of $\mathcal O(p)$.
This answers your question "does any meromorphic section vanish at $p$?" in the negative.