Is a complex line bundle over a compact Riemann surface topologically trivial iff it is holomorphically trivial? If so, how does one demonstrate that, and if not, what is a counterexample?
closed as offtopic by abx, paul garrett, José FigueroaO'Farrill, Ariyan Javanpeykar, Qiaochu Yuan Feb 3 '15 at 17:30
This question appears to be offtopic. The users who voted to close gave this specific reason:
 "This question does not appear to be about research level mathematics within the scope defined in the help center." – abx, paul garrett, José FigueroaO'Farrill, Ariyan Javanpeykar, Qiaochu Yuan

3$\begingroup$ Yes, over any surface of positive genus, there is a whole torus of nontrivial line bundles of degree $0$. This question is more suited for math.stackexchange. $\endgroup$ – Alex Degtyarev Feb 3 '15 at 13:11

4$\begingroup$ Why don't you try to read any introductory book on Riemann surfaces before asking such a question? $\endgroup$ – abx Feb 3 '15 at 13:22
There are holomorphic line bundles over a compact Riemann surface $X$ that are topologically trivial, yet not holomorphically trivial. To see this, note that smooth complex line bundles are classified by a complete invariant, called the degree. By contrast, we have the Picard group $Pic(X)$ of isomorphism classes of holomorphic line bundles on $X$.
One always has a surjective group morphism $Pic(X)\rightarrow\mathbb{Z}$, defined by taking degrees of holomorphic line bundles. In general (ie. for positive genus), this map is not an isomorphism. Its kernel therefore contains smoothly (hence topologically) trivial complex line bundles that are not holomorphically trivial.

1$\begingroup$ Just to clarify the adverb "in general", let me point out the trivial and only counterexample: What you say is OK, unless $X$ is Riemann sphere, in which case the kernel of $Pic(X)\to\mathbf Z$ is zero. $\endgroup$ – ACL Feb 4 '15 at 0:39

$\begingroup$ Yes, "in general" means genus $>0$ in this case. I'll record this above to clarify. $\endgroup$ – Peter Crooks Feb 4 '15 at 0:40