I have a question on the exact definition of branch points (on Riemann surfaces) I have the following definition::

Let $f:X \rightarrow Y$ be a holomorphic function between Riemann surfaces. $x \in X$ is a branch point, if the multiplicity, with which $f$ takes the value $f(x)$ in $x$ is $>1$. The mmultiplicity can be computed by expressing $f$ as a function between open susets of $\mathbb{C}$ by using charts.

In other words, branch points are the points where $f$ has no neighborhood on which it is injective.

For example, look at the function

$f: \mathbb{P}^1 \rightarrow \mathbb{P}^1, z \mapsto z^2 + \dfrac{1}{z^2}$

I computed the branch points to be $1, 0, \infty$ by constructing the function with the charts.

Now, I found this exercise on the internet, where they computed the points $\pm 2$ and $\infty$ as branch points.

Did I make a mistake or is there another definition for branch points?


Note that $f(-z)=f(z), f(1/z)=f(z)$ so certainly if $z_0$ is a branch point $\pm 1/z_0, -z_0$ are other branch points.

Also $f(-z)=f(z)$ implies that $0$ is a branch point, as around $0$ $f(-a)=f(a)$. Hence $0, \infty $ are branch points.

For seeking others, one can make stupid computations $f'(z)=2z-{2\over z^3}$, and $f'(z)=0$ iff $z^4=1$, ie $z\in \{1,-1,i,-i\}$. Conclusion 6 branch points$\{0, \infty,1,-1,i,-i\}$.

  • $\begingroup$ oh my bad, I've computed the root of 1 to be only 1... $\endgroup$ – User1 Feb 22 at 10:58
  • $\begingroup$ Ok, but thank you very much, this means that there is another definition for branch points then the one I know $\endgroup$ – User1 Feb 22 at 10:59
  • $\begingroup$ No it is the same $f'(x_0) = 0$ iff $x_0$ is a branch point iff $f$ is not locally injective near $x_0$ ($f'(x_0)\not = 0$ implies that locally one can choose $f$ as a coordinate (implicit function theorem). $\endgroup$ – Thomas Feb 22 at 11:11
  • $\begingroup$ oh no I didn't mean your definition, I get that, I meant the one I found where the branch points were $\pm 2$. $\endgroup$ – User1 Feb 22 at 11:28
  • $\begingroup$ @User1 That is correct, they're probably using this definition. Then $0, \pm 1, \pm i, \infty$ are ramification points (each of index $2$), and the corresponding $f(z)$ gives three branch points $\pm 2, \infty$. $\endgroup$ – Maxim Mar 9 at 17:20

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