finite covering

Question

I am trying to understand the following example, which I came across in a research article. I am posting it as a question below.

$\bf{Question}$. Let $\Sigma$ be a curve of genus two with the automorphism group $G$, and $p_1$, $p_2$, and $p_3$ are three points on $\Sigma$. Let $a_{1}$, $b_{1}$, $a_{2}$, $b_{2}$ denote the generators of $\pi_{1}(\Sigma)$ and $\gamma_{1}$, $\gamma_{2}$, $\gamma_{3}$ are simple loops about the points $p_{1}$, $p_{2}$, and $p_{3}$. Consider the following map $f$ from $\pi_{1} (\Sigma \setminus p_{1}, p_{2}, p_{3})$ to $\mathbb{Z}_{3}$ given by $\gamma_i \rightarrow 1$ and $a_{i}, b_{i} \rightarrow 0$. What it means the covering (unramified) $\Pi_{f} :\Sigma' \rightarrow \Sigma$ associated to the stabilizer of $f$? How one can think of such covering map in terms of the automorphism group $G$ of $\Sigma$? What one can say about the degree of $\Pi_{f}$?

Answer 1

By Riemann Existence Theorem, the surjection $f \colon \pi_1(\Sigma \setminus p_1, p_2, p_3) \to \mathbb{Z}_3$ gives a Galois cover with Galois group $\mathbb{Z}_3$ branched only at the points $p_1, p_2, p_3$.

This cover is precisely your covering $\Pi_f \colon \Sigma' \to \Sigma$. Then $\deg \Pi_f = |\mathbb{Z}_3|=3$. (By the way, here I do not understand why you wrote that the cover is "unramified", which is false. Maybe you meant "ramified"?)

The Galois group of $\Pi_f$ acts naturally on $\Sigma'$, and the action is free outside the preimages of $p_1$, $p_2$, $p_3$. At these preimages the stabilizer is obviously $\mathbb{Z}_3$ itself, so the cover $\Pi_f$ is totally ramified over $p_1$, $p_2$, $p_3$.

Then the genus of $\Sigma'$ can be computed by Riemann-Hurwitz formula, obtaining $$2g(\Sigma')-2= |\mathbb{Z}_3| \bigg(2g(\Sigma)-2 + 3 \bigg(1-\frac{1}{3} \bigg) \bigg),$$ that is $g(\Sigma')=7$.