There is a well known correspondence between line bundles over curves and divisors. For each line bundle, consider a rational section, take poles and zeros and we have a corresponding
divisor (up to linear equivalence ). But what if there is no such section ? For example,
consider a line bundle over $\mathbb{P}^1$ with transition function $e^{1/z}$ with $z \neq 0,\infty$. What is the degree of this line bundle?
[Math] Complex line bundle over curves
ag.algebraic-geometrycomplex-geometry
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Best Answer
As several people have pointed out, your example has degree $0$. Another way to see this is to observe that given a section, the number of zeros minus poles in both hemispheres is a difference of two winding numbers. This would work out to $(1/2\pi i)\int_\gamma d\log g_{12}$, where $\gamma$ is the equator and $g_{12}$ is the transition function (as in Henri's answer). In fancier terms, this is the first Chern number. In your example, this works out to $0$ (again). For $\mathbb{P}^1$, the degree is the sole invariant.