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?
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3$\begingroup$ Yes, over any surface of positive genus, there is a whole torus of non-trivial line bundles of degree $0$. This question is more suited for math.stackexchange. $\endgroup$– Alex DegtyarevCommented Feb 3, 2015 at 13:11
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4$\begingroup$ Why don't you try to read any introductory book on Riemann surfaces before asking such a question? $\endgroup$– abxCommented Feb 3, 2015 at 13:22
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$\begingroup$ It is a tragedy that Stack Exchange closes questions like this one that are perfectly reasonable to ask. $\endgroup$– Daniel AsimovCommented Mar 11, 2023 at 19:05
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1 Answer
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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.
answered Feb 3, 2015 at 13:06
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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. $\endgroup$– ACLCommented Feb 4, 2015 at 0:39
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$\begingroup$ Yes, "in general" means genus $>0$ in this case. I'll record this above to clarify. $\endgroup$ Commented Feb 4, 2015 at 0:40