Action of a group on a Riemann surface

Question

If $G$ is a finite group of order $n$, and acting on a compact Riemann surface $X_{g'}$ ($g'$ is genus), then $X_{g'}/G$ is another compact Riemann surface, $X_g$, of genus $g$. Also then $X_{g'} \rightarrow X_g$ is a branched covering. Are the regular covering space of $X_g$ lie between $X_{g'}$ and $X_g$?

The numbers $g,g',n$ are related by

$2g-2=|G|.{(2g'-2)+|G|.\sum (1-\frac{1}{n_x})}$,

where $n_x$ is the branching order at branch point $x\in X$. Can one suggest a topological proof of this relation? (I want to look at the problem of symmetries of Riemann surface, in topological point of view.)

Answer 1

Another derivation can be given using the Lefschetz fixed point formula. Let $X$ be a Riemann surface, $G$ be a finite group acting faithfully on $X$. All cohomology groups are with complex coefficients. Then $G$ acts on $H^i(X)$, and the Lefschetz number of $f$ is

$$L_f:= \sum_{i=0}^{2} (-1)^i Tr (f|_{H^i (X)})$$.

You need two rather difficult results, which are very useful in the study of group actions.

1. The first is the Lefschetz fixed point formula (it is nicely discussed in the books by Bredon (Topology and Geometry) and also Greenberg-Harper (Lectures on Algebraic topology)). As far as I remember, there is also a proof in Farkas-Kra (for surfaces, which is what you want). The fixed point formula asserts that for $f \neq id$, $L_f$ is the number $|F_f|$ of fixed points of $f$.

2. Then we need the knowledge that $H^i(X/G)=H^i(X)^G$ (the space of invariants). This is not easy. You can use the Hodge presentation of cohomology ($X \to X/G$ is holomorphic and it has to be injective in cohomology by looking what it does to a holomorphic $1$-form; any $G$-invariant holomorphic $1$-form on $X$ descends to one on $X/G$). If you are comfortable with homological algebra, and you can find an alternative proof in a more general setting here:

Euler characteristic of orbifolds

Now we start the real proof. From elementary representation theory of finite groups, you deduce that

$$|G|\chi(X/G)=|G|\sum_{i} (-1)^i dim (H^i (X)^G) = \sum_{f \in G} L_f.$$

Using the fixed point formula and the obvious identity $L_{id}= \chi(X)$), write

$$\sum_{f \in G} L_f = L_{id} + \sum_{f \neq id} L_f = \chi(X) + \sum_{f \neq id} |F_f|.$$

Let $S \subset X$ be the union of the fixed point sets of all $id \neq f \in G$. This is finite, and let $G_p$ be the stabilizer subgroup at $p \in S$. A simple counting gives

$$\sum_{f \neq id} |F_f| = \sum_{p \in S} (|G_p|-1).$$

Hence

$$|G|\chi(X/G)=\chi(X) + \sum_{p \in S} (|G_p|-1).$$

Reinterprete this formula in terms of branching numbers and you are done.

Answer 2

Set $X:=X_{g'}$ and $Y=X_g$, and let $X^0$ and $Y^0$ be the open sets where the covering $X \to Y$ is unramified. If $e$ denotes the topological Euler number, we have

$e(X^0)=n \cdot e(Y^0)$,

$e(Y)= e(Y^0)+ \sum_x 1$,

$e(X)= e(X^0) + \sum_x \frac{n}{n_x}$,

where the sums extend over the points with non-trivial branching order.

Using these relations, we obtain

$e(X)=n \cdot e(Y)+ n \cdot \sum (\frac{1}{n_x}-1)$,

and since

$e(X)=2-2g(X), \quad e(Y)=2-2g(Y)$

we are done.