# Holomorphic coordinates on Riemann surfaces

I have a big problem understanding the meaning of holomorphic coordinates on Riemann surfaces, especially in relation to 1-forms.

Holomorphic coordinates on a Riemann surface $$X$$ is an open set $$U \subset X$$ and a complex chart

$$\varphi:U \rightarrow \varphi(U)=:V \subset \mathbb{C}$$

As far as clear. Now you can define a differentiable function $$f:U \rightarrow \mathbb{C}$$ by: for every open $$W \subset U$$ and each chart $$\psi:W \rightarrow \psi(W)$$ you have $$f \circ \psi^{-1}: \psi(W) \rightarrow \mathbb{C}$$ is differentiable.

Now we defined Residues on Riemann surfaces in the following way:

Let $$X$$ be a Riemann surface, $$U \subset X$$ open, $$a \in X$$ and $$\alpha$$ a holomorphic 1-form on $$U \setminus \{a\}$$. Let $$\alpha=f \ dz$$ be the local expression for $$\alpha$$ in some hol. coordinate $$z = \varphi:V \rightarrow \varphi(V) \subset \mathbb{C}$$, where $$V$$ is a neighborhood of $$a$$ and $$z(a)=0$$. Then $$Res_0(f \circ \varphi^{-1})$$ is coordinate independent. As a consequence we can define $$Res_a(\alpha):= Res_0(f \circ \varphi^{-1})$$. \ Proof: The right-hand side is precisely $$c_{-1}$$ in the Laurent expansion $$f(z) = \sum\limits_{\nu= - \infty}^{\infty} c_{\nu} z^{\nu}$$.

I'm not even able to understand the following: If $$z:U \rightarrow V$$ is a local coordinate, look at $$f(z)$$, but if $$f(z)$$ is meant to be the composition, it doesn't make any sense for me since $$z(U) \subset \mathbb{C}$$ and $$f$$ is defined on $$U$$ ...

I am really hoping tha someone can lead me the way to understand this...

• I don't know where you saw "look at $f(z)$", but whoever said it was being careless. Most likely by $f(z)$, they actually mean the map $f\circ z^{-1}$. More verbosely, when they called $z$ a local coordinate, they actually mean that there is some chart $\phi$ that they will not explicitly mention, and by means of this chart, they are identifying $f$ with $f\circ \phi^{-1}$. $z$ in then just a variable over the complex numbers, and by $f(z)$ they mean $f(\phi^{-1}(z))$. – Paul Sinclair Feb 20 at 1:17
• And what about the differential form $f \ dz$, where $z$ is a local coordinate, how to understand this? – User1 Feb 20 at 9:38
• Thanks btw, tha already helped me, it seems like no big deal but since it was never properly defined I was confused. – User1 Feb 20 at 9:39
• $f\,dz$ is $(f\circ \phi^{-1})dz$ on $\phi(U)$. – Paul Sinclair Feb 20 at 17:01
• but what does $dz$ mean here? – User1 Feb 20 at 18:47