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.

I am trying to understand an example of a line bundle over a Riemann surface; as it is very terse and short, I have lots of trouble. It is written in block-quotes below, and I ask questions as I go.

The canonical bundle $K$ over a Riemann surface $M$ is the cotangent bundle, or the bundle of holomorphic $1$-forms. Suppose we have local coordinates $z$ and $w$ with $$ w(z) = \phi_\beta \circ \phi_\alpha^{-1} (z) $$ a function of $z$ on the overlap.

Question 1: What is meant by "with $w(z) = \phi_\beta \circ \phi_\alpha^{-1} (z)$ a function of $z$ on the overlap"? Why is it a function of $z$? I don't understand this equation at all.

The $1$-forms $dz$ and $dw$ give local trivializations of the canonical bundle, and on the overlap $$ dw = w'dz. $$

Question 2: how are $dz$ and $dw$ are local trivializations? Isn't $dz$ at a point a map from the tangent space to $\mathbb{C}$? And how is this relation obtained?

Therefore the transition functions are $dw/dz$, where $w = \phi_\beta \circ \phi_\alpha^{-1}$.

Question 3: What does "dw/dz" even mean?

share|improve this question
add comment

1 Answer

up vote 5 down vote accepted

There are many abuses of language in this subject and in order to help you I'll describe a few things completely rigorously.

Let $(U,\phi_\alpha)$ and $(V,\phi_\beta)$ be two charts (=local coordinates) of your Riemann surface $X$ at $P\in X$, that is $P\in U\cap V$.
The overlap is $\phi_\alpha(U\cap V)\subset \mathbb C$ and you have a holomorphic isomorphism $w: \phi_\alpha(U\cap V)\to \phi_\beta(U\cap V):z\mapsto w(z)$ which answers Question 1.

If you define $z_0=z(P)$, you can compute the derivative $w'(z_0)=\frac {dw}{dz}(z_0)\in \mathbb C$ and this answers Question 3.

Finally, since a holomorphic function $\mathbb \phi$ defined in a neighbourhood of $P$ has a differential which is a linear form $d\phi(P):T_P(X)\to \mathbb C$, you get two linear forms $d{\phi_\alpha} (P),d{\phi_\beta} (P):T_P(X)\to \mathbb C$.
Are they equal? Not at all! They are related by $$d{\phi_\beta} (P)=w'(z_0)\cdot d{\phi_\alpha}(P) \in T_P^*(X)$$ an equality in the fiber of the cotangent bundle at $P$ which answers (I hope!) your Question 2.

share|improve this answer
    
Yes, thank you! This has made things a lot clearer! –  Rick Jan 30 '12 at 22:45
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.