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?


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.

  • $\begingroup$ Yes, thank you! This has made things a lot clearer! $\endgroup$ – Rick Jan 30 '12 at 22:45

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