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In order to calculate genus of compact Riemann surface using Riemann-Hurwitz theorem, we have to determine the branch points first.

Question:

For holomorphic maps between $\Bbb{CP^1}$, is there a general way to decide if $\infty$ is a branch point? What about its ramification index($=$number of branches $-1$)?

This answer claims:

If $b\in X$ and $f(b)\neq\infty$ then $b$ is a branch point iff $f'(b)=0$ (derivative wrt. an arbitrary local coordinate; the ramification index is the maximal $k$ s.t. $f^{(k)}(b)=0$ (the number of branches meeting at $b$ is $k+1$). If $f(b)=\infty$, replace $f$ with $1/f$.

  1. $z=\infty$ is not branch point of $f(z)=\sqrt{z^2+1}$, while $(\frac{1}{f})'(0)=0$.

To see this, relevant links: (a) $z=\infty$ is not a branch point of $\sqrt{z^2+1}$ : what is the (compact) Riemann surface? (b) Is $z=\infty$ a branch point of $f(z)=(z^{2}+1)^{1/2}$?

  1. For $f(z)=z^2+\frac{1}{z^2}, f'(0)=\infty$, but $0$ is branch point of $f$.

Thanks for your time and patience.

Andrews
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    An holomorphic map $f : M \to \Bbb{CP^1}$ is a meromorphic function $F : M \to \Bbb{C}$, the multiplicity we want at $a \in M$ is, for $b \in \Bbb{CP}^1$ close to $f(a)$, the number of points close to $a$ mapped to $b$. If $a$ is a pole of $F$ this is the same as its order. If $a$ is a zero of $F$ this is the same as its order, if $a$ is not a pole this is the same as $1 \ +$ the order of the zero of $(F \circ \phi)'$ at $\phi^{-1}(a)$ with $\phi$ a local chart. Do you see why replacing $F$ by $1/F$ gives the same result ? – reuns Jun 09 '19 at 04:06
  • @reuns Is that because $F$ and $1/F$ have the same zero and same order of zero? – Andrews Jun 09 '19 at 10:26

1 Answers1

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As partially answered here and in comment:

For function $f(z)=\sqrt{z^2+1}$, $\infty$ is a pole of $f$, so when considering situation at $z=\infty$, $z$ is no longer a coordinate, we use $w=\frac{1}{z}$. The order of $f$ at $\infty$ is (by definition) exactly the order of zero for $\frac 1f$. And $\frac{d(1/f)}{dw}(0)\not=0$, so $\infty$ is not a branch point of $f$.

For function $f(z)=z^2+\frac{1}{z^2}$, $\infty$ is a pole of $f$, use $w=\frac 1z$. $0$ is pole of $f(w)=w^2+\frac{1}{w^2}$, $\frac{d(1/f)}{dw}(0)=0$, $\frac{d^2(1/f)}{dw^2}(0)\not=0$ therefore $\infty$ is branch point of $f$ with branch number $2$.

In general,

$(1)$ for $z_0\not=\infty$ and $z_0$ is not a pole

$z_0$ is branch point $\iff$ $f'(z_0)=0$ (from implicit function theorem, $f$ near $z_0$ is not injective)

ramification index is the maximal integer $k$ s.t. $f^{(k)}(z_0)=0$, the number of branches $z_0$ is $k+1$.

$(2)$ for $z_0\not=\infty$ but $z_0$ is a pole

$z_0$ is pole $\iff$ $z_0$ is zero of $1/f$.

ramification index of $z_0$ to $f$ = (order of zero $z_0$ to $1/f$) $-1$

= maximal integer $k$ s.t. $(1/f)^{(k)}(z_0)=0.$

$(3)$ for $z_0 = \infty$, $z$ is no longer local coordinate near $\infty$

take coordinate $w=\frac 1z$, take $g(w)=f(z)$, then $g(w)$ has same behavior near $w=0$ as $f(z)$ near $z=\infty$.

Andrews
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  • Hi ! Again sorry I know this post has been answerd some time ago, but there is one (probably simple) thing I'm not sure to catch: could you explain exactly which function you study to know if $\infty$ is a branch point or not ? Because I understand that we want to study a function around $0$ instead, since we can, however I'm really not sure wether we study $\frac{1}{f}(\frac{1}{z})$ or $\frac{1}{f}(z)$ or $f(\frac{1}{z})$ ? I hope you can enlighten me ^^ – Rhaena Apr 21 '21 at 08:06
  • It's yet again been some time since the previous comment, but in case it's still helpful: the goal is to make both the input and the output a finite value. So a very rough summary of the above answer: (1) If the input and output are both finite, you can just use $f(z)$. (2) If you have a pole at a finite point, you use $\frac 1f(z)$. (3.1) If you have a finite value at infinity, you use $f(\frac 1z)$. (3.2) If you have a pole at infinity, you use $\frac 1f(\frac 1z)$. – Jonathan Love Apr 12 '22 at 13:55