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 $1/f$.
- $z=\infty$ 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 $f(z)=(z^{2}+1)^{1/2}$?
- 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.
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 $z=\infty$.