Suppose that $π : X → Y$ is a map of degree d between Riemann surfaces.
(a) For any point $p$ of $X$, explain why the ramification index $k_{p}$ cannot be bigger than $d$, i.e., that $k_{p} \leq d$.
(b) If $d = 2$ (a double cover) explain why a point $p$ of $X$ is either not a ramification point, or has ramification index exactly $2$.
(c) Again in the case that $d = 2$ explain why the number of branch points (on $Y$ ) is the same as the number of ramification points (on $X$).

For (a) $d=\sum k_{p_j}$, where $k_p$ is the ramification index of $p \in X$ so neither of the ramification indices can be greater than the degree. As a consequence of this, with $d=2$ the ramification index of a point can either be $1$ (in which case it is not a ramification point) or the ramification index could be $2$. I'm confused about part (c), is it because any ramification point in $X$ will have ramification index $2$ and image of a ramification point is a branch point?

  • $\begingroup$ I think you're just supposed to show why two ramification points cannot land on the same branch point. $\endgroup$ Apr 3, 2017 at 21:33
  • $\begingroup$ That makes sense! Thanks @ElizabethS.Q.Goodman $\endgroup$ Apr 3, 2017 at 21:49

1 Answer 1


Let $\pi\colon X \rightarrow Y$ is a proper non-constant holomorphic map between Riemann surfaces of degree $d\leq 3$. Denote the set of ramification points and branch points of $\pi$ by $R$ and $B$, respectively. Then the function $R \rightarrow B;x\mapsto \pi(x)$ is bijective.

It is clearly surjective. Now, assume for a contradiction that there exist distinct $x_1,x_2\in R$ such that $\pi(x_1)=\pi(x_2)$. Then $3\geq\sum_{p\in\pi^{-1}(\pi(x_1))}\operatorname{mult}_p(\pi)\geq 4$, which is absurd.


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