MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\Pi_{f}$?

share|cite|improve this question
Although you already have an answer, for future readers it might help to say which research article it was. – Julian Kuelshammer Jun 26 '13 at 6:31
up vote 2 down vote accepted

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 $g(\Sigma')=7$.

share|cite|improve this answer
Thanks a lot Francesco. – user34848 Jun 10 '13 at 21:35
You are welcome. But you meant "ramified", didn't you? – Francesco Polizzi Jun 10 '13 at 21:36
No, I am considering unramified covering $\Pi_{f}$ – user34848 Jun 10 '13 at 21:43
Sorry, but I really do not understand. The images of the loops $\gamma_i$ are non-trivial in $\mathbb{Z}_3$, so the map $\Pi_f$ is ramified at the points $p_i$, with stabilizer the whole of $\mathbb{Z}_3$. This is a very standard application of Riemann Existence Theorem. The cover over $\sigma \setminus p_1, p_2, p_3$ is étale, but the cover over $\Sigma$ is not. – Francesco Polizzi Jun 10 '13 at 21:48
Obviously I mean $\Sigma \setminus p_1, p_2, p_3$ – Francesco Polizzi Jun 10 '13 at 21:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.