Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Say I have a Line bundle induced by some Divisor $\mathfrak{L}(D)$.

And $D$ is given by the zero locus of some polynomial $P$. I actually know this polynomial explicitly.

Now, I also have a Connection and even this connection i know explicitly.

I've been trying to show, that this connection belongs to the above line bundle. To my mind, this should be possible, as the first chern class of the line bundle is nothing more than the Curvature form of the Connection.

Here is what I tried so far:

  • I tried computing the curvature form from the connection and then simply comparing it to the first chern class of $D$. This has proven difficult. Specifically, I don't know how to show that the Curvature form I find is of the class $c_1$ is in.

  • I tried using the fact that the transition functions of the line bundle are given by $g_{ab} = f_a/f_b$. And that a connection can be found by $D = d \log g_{ab}$. However, I don't really know how to construct these $f_a$ functions from my polynomial.

Is there another way? Or perhaps know how to solve the two problems above?

This must have been done by someone, but I can't find anything. Any examples are also highly appreciated! Thanks!

share|improve this question
    
Is $D$ linearly equivalent to $0$ so that $\mathcal{L}(D)\simeq \mathcal{O}_X$? In what way do you "know" the connection? I find it hard to be handed a connection without the data of the line bundle being used in the description which automatically tells you it belongs to that line bundle. –  Matt Jul 18 '12 at 17:23
    
Yes, some of my divisors are equivalent to 0. How can one use this? It is indeed a bit weird that I don't really know if the connection belongs to that line bundle or not, but sadly I only have hand-wavy arguments so far. And by "know the connection", I have an explicit formula for $D$. As well as for the Polynomial $P$ –  FMN Jul 18 '12 at 17:30
    
The actual problem arises from the fact that the Divisors, etc, all arise from a pure mathematical derivation of the problem. And the Connection thusfar has come from a physics perspective. Therefore both "languages" used are a bit incompatible up to exactly this point. I'm trying to understand better on how to make the connection –  FMN Jul 18 '12 at 17:32
    
Here's what is confusing me. To me a connection is a homomorphism $\nabla : \mathcal{L}\to \mathcal{L}\otimes \Omega^1$ that satisfies the Leibniz rule. There are other characterizations, like if you know it is flat then you can recover it from the local system or a representation of the fundamental group, but otherwise I don't know how to describe a connection without using $\mathcal{L}$ to do it. –  Matt Jul 19 '12 at 0:08
    
Basically I just have a 1-form given explicitly. It appeared in some physical equation and it was possible to solve for it. That's also my problem, I'm not even sure if it actually is a connection. That's why I was hoping to recover some information of the connection using Divisors –  FMN Jul 19 '12 at 8:09

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.