Suppose that a finite group $G$ acts on a Riemman Surface $X$. Suppose that $\{p_1,\ldots,p_r\}\subset X$ represent a complete set of inequivalent branch points under the action of $G$. Let $G_j$ denote the stabilizer of $p_j$ (if $p_j$ is interchanged by another point in the same orbit, then the associated group is conjugated to $G_j$) and let $m_j$ denote the order of $G_j$ (or any conjugated). If $g$ and $\gamma$ denotes the genus of $X$ and $X/G$ respectively, then the Riemann-Hurwitz Formula tells us that:
$g=|G| \cdot (\gamma-1)+1+\frac{|G|}{2}\sum_{j=1}^{r}{(1-\frac{1}{m_j})}$
I tried to use this formula on an specific example to see if I understood the concepts. Let $X=\{x^4+y^4=z^4 \}\subset \mathbb{P}^2$; $G=\mathbb{Z}_4 \times \mathbb{Z}_4$. Consider the following action on $X$ defined on the generators
- $(1,0) \cdot (x:y:z)=(ix:y:z)$
- $(0,1) \cdot (x:y:z)=(x:iy:z)$
I already proved that there are only two orbits given by:
- $O_1=\{(1:0:1),(i:0:1),(-1:0:1),(-i:0:1) \}$
where the stabilizer is given by $G_1=0 \times \mathbb{Z}_4$
- $O_2=\{(0:1:1), (0:i:1), (0:-1:1), (0:-i:1) \}$
where the stabilizer is given by $G_2=\mathbb{Z}_4 \times 0$
If $p_1=(1:0:1)$ and $p_2=(0:1:1)$ then $\{p_1,p_2\}\subset X$ is a complete set of non equivalent branch points. I want to compute $\gamma$ using the above formula. First we need to compute the genus of $X$ this can be done with the genus degree formula and it follows that $g=3$. If we replace the values $g=3, |G|=16 , m_1=m_2=4$ on the above formula we get that $\gamma= \frac{3}{8}$ which is clearly nonsense.
This is why I´m asking for help to understand this concepts.
I'll change your example slightly to make it more manageable, because $\mathbb{Z}/4\times \mathbb{Z}/4$ has 14 nontrivial subgroups. So lets do it instead with $\mathbb{Z}/2\times \mathbb{Z}/2$, acting via:
$(1,0).[a:b:c]=[-a:b:c]$
$(0,1).[a:b:c]=[a:-b:c]$
$(1,1).[a:b:c]=[a:b:-c]$
We are thinking of $X\rightarrow X/G$ as nearly a covering space, and Riemann Hurwitz tells us we can compare the euler characteristics by dividing by the degree of the map (the order of the group, in this case 4), and correcting the points with nontrivial ramification. Under a group action, nontrivial ramification arises from points with nontrivial stabilisers (nontrivial branch points downstairs), hence your formula in those terms, so lets compute the points of $X$ with nontrivial stabilisers.
The (nontrivial) stabiliser group could be either cyclic, generated by $(1,0),(0,1)$ or $(1,1)$, or it could be all of $\mathbb{Z}/2\times \mathbb{Z}/2$.
First, if a point is fixed by $(1,0)$, it is either of the form $[1:0:0]$ or $[0:a:b]$. We see then by symmetry that the only points (of $\mathbb{P}^2$) fixed by all of $\mathbb{Z}/2\times \mathbb{Z}/2$ are $[1:0:0],[0:1:0]$ and $[0:0:1]$, and none of these lie on our curve. So the points with nontrivial stabiliser will be the points on our curve that have a single coordinate zero, in other words, the intersection of the $X=0$, $Y=0$, $Z=0$ lines with our curve. We can directly work out what these are, but to save space, lets just note that each line has $4$ points of intersection, either by direct computation or Bezouts theorem. So all together, we have $12$ points with nontrivial stabiliser, being $\mathbb{Z}/2$, so $6$ branch points of $X/G$.
Thus, Riemann Hurtwitz tells us that:
$3=4(\gamma-1)+1+2\cdot 6(1-0.5)$
So the genus $\gamma$ of the quotient is therefore $0$, so we have a complex projective line as our quotient. As a bonus, since the only quotients of $\mathbb{P}^1$ are $\mathbb{P}^1$, and your example is a quotient of this one, we see that the genus of your quotient is also $0$. In general, we need to check all the points with nontrivial stabilisers, so there is a lot of potential work when the group has some complexity. The reason your counts were off was that there are a lot more points with nontrivial stabiliser in your example than the ones you listed.