4
$\begingroup$

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".


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!:)

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks for the answer. Why should I consider so many maps? I thought its enough to take two of the above maps such that their domains cover the whole surface. $\endgroup$ – Marc Jul 24 '16 at 16:32
  • $\begingroup$ @Marc The setup of the lemma (above the statement) requires that we write the map in local coordinates. You need to map into a coordinate chart. If you just consider the first and third maps, the problem is that the first map doesn't map into a coordinate chart (because it hits $0$ and $\infty$ at $0$ and $\pm 1$). $\endgroup$ – hwong557 Jul 24 '16 at 16:54
  • $\begingroup$ @Marc I've revised my answer above to hopefully clarify this point. $\endgroup$ – hwong557 Jul 24 '16 at 17:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.