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Let $f$ be a rational function on a compact connected Riemann surface $X$. The rational function $f$ induces a holomorphic map $\overline{f}:X\to \mathbf{P}^1(\mathbf{C})$.

Let $x$ be a point on the Riemann sphere $\mathbf{P}^1(\mathbf{C})$. How can I check that if $b$ is a branch point of $\overline{f}$ by looking at the derivative of $f$?

How does this work when $X=\mathbf{P}^1(\mathbf{C})$?

Harry
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1 Answers1

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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 $1/f$.

user8268
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  • Just a minor nitpick on terminology: If $b$ is in $X$, one should call it a ramification point right? The branch locus being the image of the ramification locus. – Harry Apr 02 '12 at 19:10
  • @Harry: you are of course right – user8268 Apr 02 '12 at 19:12
  • A little problem: $z=\infty$ is not branch point of $f(z)=\sqrt{z^2+1}$, but $(\frac{1}{f})'(0)=0$ – Andrews Jun 09 '19 at 01:54
  • @Andrews $z$ is not a local coordinate around $z=\infty$. We can use $w=1/z$, so now $1/f =w/\sqrt{1+w^2}$. And indeed $\frac{d(1/f)}{dw}(0)\neq 0$. No problem here. – user8268 Jun 09 '19 at 08:49
  • @user8268 Thanks for your time and kindly explanation, this contributes a lot to this question – Andrews Jun 09 '19 at 10:29
  • And sometimes, it'll be complicated to determine the remification index/branch number at $\infty$, like in these questions: Q1, Q2 – Andrews Jun 09 '19 at 10:37
  • Hi @Andrews, sorry I know it's been some time this post has been answered, but reading your comments I wonder how come that $\frac{1}{f}(w) = \frac{w}{\sqrt{1 + w^2}}$, and not just $\frac{1}{\sqrt{1 + w^2}}$ ? – Rhaena Apr 21 '21 at 07:58