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 $k < 0$.

Thanks for your help.


1 Answer 1


$\newcommand{\C}{\mathbb C}\newcommand{\CP}{\mathbb C\mathbb P}\newcommand{\sm}{\setminus}\newcommand{\ord}{\operatorname{ord}}$A holomorphic section of $L_{kx}$ is given by a pair of functions $s_0:\Sigma\setminus\{x\}\to\C$ and $s_1:U\to\C$ that are compatible with the transition function. Here, being compatible with the transition function means that, for $p\in U\sm\{x\}$, $z(p)^ks_0(p)=s_1(p)$.

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.

These two constructions are clearly inverse to one another (and linear), so we have are desired isomorphism.

  • $\begingroup$ 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 $X$ $\endgroup$
    – reuns
    Commented Aug 31, 2019 at 18:59
  • $\begingroup$ 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 $U\setminus\{x\}$... $\endgroup$
    – Niven
    Commented Aug 31, 2019 at 20:21
  • $\begingroup$ ... 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... $\endgroup$
    – Niven
    Commented Aug 31, 2019 at 20:35
  • $\begingroup$ $\operatorname{ord}_z(x)=1$ which follows from the fact that we know $z$ is biholomorphic here, and is not true for arbitrary transition functions. $\endgroup$
    – Niven
    Commented Aug 31, 2019 at 20:38

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