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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.

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  • $\begingroup$ 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). $\endgroup$ – user171326 Mar 16 '17 at 21:32
  • $\begingroup$ @GeorgesElencwajg and indeed I made a stupid mistake (the trivialization doesn't need to be meromorphic near $p$) $\endgroup$ – user8268 Mar 16 '17 at 21:40
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    $\begingroup$ 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). $\endgroup$ – Georges Elencwajg Mar 16 '17 at 21:40
  • $\begingroup$ Dear @user8268: No problem, I have removed my comment. $\endgroup$ – Georges Elencwajg Mar 16 '17 at 21:43
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    $\begingroup$ @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. $\endgroup$ – Georges Elencwajg Mar 16 '17 at 22:32
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Yes, every holomorphic line bundle on $X$ becomes trivial on $X\setminus\{p\}$.
Much more generally every holomorphic vector bundle on every non-compact Riemann surface is trivial: you can find this remarkable theorem as Theorem 30.4 (page 229) of Forster's Lectures on Riemann Surfaces.
(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 $X\setminus\{p\}$)

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  • $\begingroup$ Sorry for this very late reply. Thanks, Georges, N.H. and user8268 for the replies and discussion! This is my first post on stack exchange so I do appreciate the quick responses. $\endgroup$ – Marielle O. Mar 21 '17 at 10:28
  • $\begingroup$ Welcome to our wonderful site, dear MPNimbus. I hope we will have the pleasure to meet you here often. $\endgroup$ – Georges Elencwajg Mar 21 '17 at 10:32

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