# Line bundles minus a point are trivial

Let $X$ be a Riemann surface and $\pi: L \rightarrow X$ be a line bundle. If I delete a point $p$ in $X$, then does the line bundle over $X\setminus\{p\}$ become trivial?

Intuitively, I want to say "no". By simply taking local trivialisations $\phi_i: \pi^{-1}(U_i \setminus \{p\}) \rightarrow U_i \setminus \{p\} \times \mathbb{C}$, I would think that you would obtain a nontrivial bundle at the end. But perhaps, I am missing some sort of Riemann Roch argument.

• If you look in the topological/smooth category, $H^2(X \backslash p, \mathbb Z) = 0$ since $X \backslash p$ is not compact, and since smooth functions/continuous functions form an acyclic sheaf, the exponential exact sequence give you $H^1(X \backslash p, F^*) = 0$ where $F^*$ is the sheaf of continous/smooth nonvanishing function which precisely classify line bundles. So answer is "yes, $L$ will become trivial" also in the topological/smooth category. (If I am not mistaken). – user171326 Mar 16 '17 at 21:32
• @GeorgesElencwajg and indeed I made a stupid mistake (the trivialization doesn't need to be meromorphic near $p$) – user8268 Mar 16 '17 at 21:40
• Dear @N.H.: no you are not mistaken and actually every topological (or smooth) vector bundle of any rank on any non-compact Riemann surface is trivial. However if $X$ is compact it carries a unique algebraic structure and if the genus of $X$ is non-zero there exists an algebraic line bundle $L$ on $X$ and a point $p\in X$ such that $L$ is not algebraically trivial on $X\setminus \{p\}$ (but of course $L$ is holomorphically trivial there). – Georges Elencwajg Mar 16 '17 at 21:40
• Dear @user8268: No problem, I have removed my comment. – Georges Elencwajg Mar 16 '17 at 21:43
• @N.H. Let $X$ have positive genus, choose two points $p\neq q\in X$ and let $L=\mathcal O(q)$. If $\mathcal O(q)\vert (X\setminus \{p\})$ were trivial, there would exist a regular function $f\in \mathcal O(X\setminus \{p\})$ with $div(f)=1.q$. But then $f$ would extend to a rational function $\overline f \in Rat(X)$, so that necessarily $div(\overline f)=1.q-1.p$ (the divisor of a rational function on $X$ has degree zero). This is absurd because the two different points $p,q\in X$ cannot be linearly equivalent on $X$ (recall $g(X)\gt 0)$. So $L\vert X\setminus \{p\}$ is not trivial. – Georges Elencwajg Mar 16 '17 at 22:32

Yes, every holomorphic line bundle on $X$ becomes trivial on $X\setminus\{p\}$.
(Independently of whether $X$ is compact or not, $X\setminus\{p\}$ is definitely not compact so that you can apply theorem 30.4 to your $L$ restricted to $X\setminus\{p\}$)