# Show that a space of holomorphic sections of a line bundle is isomorphic to the space of meromorphic functions on a Riemann surface.

Let $$\Sigma$$ be a Riemann surface, $$x \in \Sigma$$ and let $$z: U \rightarrow D \subset \mathbb{C}$$ be a local coordinate system centered at $$x$$. For every $$k \in \mathbb{Z}$$, a holomorphic line bundle $$L_{kx}$$ on $$\Sigma$$ is defined by gluing the trivial vector bundles $$(\Sigma \backslash \{x\}) \times \mathbb{C} \rightarrow \Sigma$$ and $$U \times \mathbb{C} \rightarrow U$$ via the holomorphic transition function $$(U \backslash \{x\}) \times \mathbb{C} \rightarrow (U \backslash \{x\}) \times \mathbb{C}, \ (p, v) \mapsto (p, z(p)^k v)$$

Can someone give me some hints how to prove the following claims ?

The space of global holomorphic sections of $$L_{kx}$$ is isomorphic to the space of meromorphic functions on $$\Sigma$$, holomorphic outside of $$x$$ and has a pole of order at most $$k$$ if $$k \geq 0$$ and a zero of order $$k$$ if $$k < 0$$.

So, given a section $$(s_0,s_1)$$, we can define a meromorphic function $$s:\Sigma\to\C$$ given by $$s(p)=\begin{cases}s_0(p)&\text{if }p\neq x\\s_1(p)z(p)^{-k}&\text{if }p=x\end{cases}$$ which is visibly holomorphic outside of $$x$$. $$z(x)=0$$ and $$s_1$$ is holomorphic at $$x$$ (possibly vanishing there). If $$k\ge0$$, then $$s$$ has a pole or order $$\le k$$ at $$x$$, and if $$k<0$$, then $$s$$ has a zero of order $$\ge|k|$$ at $$x$$. In either case, we can say that $$\ord_x(s)\ge-k$$.
Conversely, given a meromorphic function $$s:\Sigma\to\C$$ which is holomorphic on $$\Sigma\sm\{x\}$$, but with $$\ord_x(s)\ge-k$$, we can construct a section $$(s_0,s_1)$$ of $$L_{kx}$$ by setting $$s_0(p)=s(p)$$ and $$s_1(p)=s(p)z(p)^k$$. Note that $$\ord_{s_1}(x)=\ord_s(x)+k\ord_z(x)=\ord_s(x)+k\ge0,$$ so $$s_1$$ is holomorphic, and $$(s_0,s_1)$$ really gives a holomorphic section.
• Why do you assume $z$ is biholomorphic at $x$ ? I think given a line bundle [$X = \bigcup U_j$ open cover and $g_{ij}$ holomorphic on $U_i \cap U_j$] and a non-zero global section [$(f_i)$: $f_i$ holomorphic on $U_i$ and $f_j = f_i g_{ij}$ on $U_i \cap U_j$] the existence of $(f_i)$ implies the compatibility conditions for the $g_{ij}$ and it gives a global meromorphic function when for some $l$, each $g_{jl}$ extends to a meromorphic function on $U_j$ so that $f(p) = f_j(p)g_{jl}(p), p \in U_j$ is globally meromorphic on $X$ – reuns Aug 31 '19 at 18:59
• We're told in the beginning that $z$ is biholomorphic. In the original question, $z:U\to D\subset\mathbb C$ is said to be a local coordinate system centered at $x$; I take this to mean that $z$ is a chart in $\Sigma$'s atlas with $z(x)=0$. Unless I'm mistaken, this should, be definition, mean that $z$ is biholomorphic at $x$. I'm not sure what you are trying to get across with your point about when a section of a line bundle gives a global meromorphic function. Here, we only have one transition function $g_{01}(p)=z(p)^k$ defined on $U\setminus\{x\}$... – Niven Aug 31 '19 at 20:21
• ... I guess to keep my notation in line with yours, I should say our "one" transition function is $g_{10}(p)=z(p)^{-k}$ ($U_1=U$,$U_0=\Sigma\setminus\{x\}$). Then, for $j=0,1$, we see that $g_{i0}$ extends to a meromorphic function on $U_j$ and the meromorphic function $s(p)=s_j(p)g_{j0}(p)$ on $X$ you get from your discussion is the same at the one I wrote down. Maybe you were trying to say that you don't need $z$ to be biholomorphic to get a global meromorphic function on $X$. This is true, but you to get the right interpretation of sections of this bundle, you need that... – Niven Aug 31 '19 at 20:35
• $\operatorname{ord}_z(x)=1$ which follows from the fact that we know $z$ is biholomorphic here, and is not true for arbitrary transition functions. – Niven Aug 31 '19 at 20:38