Take the 2-minute tour ×

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?

share|improve this question
The transition function is holomorphic at $\infty$ so the line bundle is trivial. –  Torsten Ekedahl Jul 20 '11 at 6:33
Dear Torsten, your argument is a little incomplete. After all, the non-trivial bundle $\mathcal O(k)$ for $k>0$ has as transition function $1/z^k$, which is holomorphic at $\infty$. The crucial point is that indeed $e^{1/z}$ is holomorphic but also that it has neither zeros nor poles on $\mathbb P^1\setminus 0$. –  Georges Elencwajg Jul 20 '11 at 8:59
Dear Georges, yes I forgot to say that it and its inverse are holomorphic at $\infty$. –  Torsten Ekedahl Jul 20 '11 at 14:15

2 Answers 2

Here is the way you get a non-vanishing holomorphic section which will trivialize your line bundle.

Denote $U_1=\{[x:y]; x\neq 0 \}$ and $U_2=\{[x:y]; y\neq 0\}$, this is a covering of $\mathbb P^1$. The transition function of the bundle is $g_{12}([x:y])=e^{y/x}$ (you could also take $e^{x/y}$, this wouldn't change the argument), defined on $U_1 \cap U_2$.

Then you may defined the following holomorphic functions: on $U_1$, you put $s_1([x:y]):=e^{y/x}$ and on $U_2$, $s_2([x:y]):=1$. Thus $s_1$ and $s_2$ are non-vanishing holomorphic functions, and on $U_1 \cap U_2$, you have $s_1=g_{12}s_2$ so that they form a section of our line bundle.

EDIT: In particular the degree of the line bundle is 0!

Moreover, on a Riemann surface (or more generally on any smooth projective complex variety), any line bundle admits a meromorphic section, so that the correspondence your are talking about still holds.

share|improve this 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.

share|improve this answer

Your Answer


