Branch points and Ramification points of a meromorphic map between Riemann Surfaces

Question

Let be $f(z)=\frac{z^3}{(1-z^2)}$ be considered as a meromorphic function on the Riemann Sphere $\mathbb C_{\infty}.$ Consider the affiliated holomoprhic map $F:\mathbb C_{\infty}\rightarrow \mathbb C_{\infty}$.

Now I want to determine all the branch points and ramification points of $F$. I also want to determine the degree deg($F$) of $F$.

Note: I am using Rick Miranda's "Algebraic Curves and Riemann Surfaces".

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The ramification point $p\in \mathbb C_{\infty}$ is a point with $mult_p(F)\geq 2$.

For poles and zeros its quite easy to determine whether they are ramification points or not..

I found for example the following point:

• $p=0$ because its a zero of $f$ and so $mult_p(F)=ord_p(F)=3\geq2$

The poles $+1$ and $-1$ are no ramification points sich their order are just $-1$. Also $\infty$ is no ramification points since

$$f(1/z)=\frac{1}{z(z^2-1)}$$

has a pole of order just $1$ in zero ($ord_1(F)=-1$).

But I think I am not done. How can I check the points which are neither poles nor zeros?

Thanks in advance!:)

Answer 1

Use Lemma 4.4 on page 45. Your task is to write your map $F: \mathbb C_\infty \to \mathbb C_\infty$ in local coordinates. We know the Riemann sphere can be covered with two coordinate charts, say $U_1$ and $U_2$. First we think about the maps $F: F^{-1}(U_i) \to U_i$ for $i=1, 2$. However, $F^{-1}(U_i)$ may not be a local coordinate, but $U_j \cap F^{-1}(U_i)$ is.

So in local coordinates from...

• $F^{-1}(U_1) \cap U_1 \to U_1$, $F$ looks like $\frac{z^3}{1-z^2}$
• $F^{-1}(U_2) \cap U_1 \to U_2$, $F$ looks like $1/\frac{z^3}{1-z^2}$
• $F^{-1}(U_1) \cap U_2 \to U_1$, $F$ looks like $\frac{(1/z)^3}{1-(1/z)^2}$
• $F^{-1}(U_2) \cap U_2 \to U_2$, $F$ looks like $1/\frac{(1/z)^3}{1-(1/z)^2}$

Now you can simplify all of this, take the derivatives, and apply the lemma.