Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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
2  
The transition function is holomorphic at $\infty$ so the line bundle is trivial. –  Torsten Ekedahl Jul 20 '11 at 6:33
5  
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
1  
Dear Georges, yes I forgot to say that it and its inverse are holomorphic at $\infty$. –  Torsten Ekedahl Jul 20 '11 at 14:15
add comment

2 Answers

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
add comment

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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.