Let $X$ be a hyperelliptic Riemann surface, and let $J$ be the hyperelliptic involution. Then consider the quotient surface $X/ < J > ,$ my question is whether $X/ < J > $ is a Riemann surface? 


On the standard Riemann surface textbook, the answer is yes, $X/< J >$ is the Riemann sphere $S^{2}$. More generally, let $H$ be a subgroup of the automorphism group of a Riemann surface $\Sigma,$ then $\Sigma/H$ is also a Riemann surface, and the quotient map is holomorphic, and the fixed points of $H$ are the branch points of the quotient map.


However, when we consider the problem in another way, the above mentioned $X/< J >$ should be an orbifold. The fixed points of $J$ are the singular points of orbifold. An orbifold is not a manifold, and if $X/< J >$ is even not a manifold, how could it be a Riemann surface? We know that only when an group $G$ act freely and properly discontinuously on a manifold $M$, the quotient space $M/G$ is a manifold. Here the question is that the action of automorphism group of a Riemann surface on itself has the fixed points. 


A simple example is the American football model of an orbifold. Let $\tau$ be the $\pi$ rotation around $z$-axis, then we can get an orbifold with two singular points located at north pole and south pole of the sphere. However, under the Riemann surface view, $S^{2}$ is a Riemann surface, and $\tau$ is a holomorphic involution with two fixed points, so $S^{2}/<\tau>$ should be a Riemann surface. It seems we get a contradiction here. Where is the mistake or confusion in my above statement?



Quotient Surface of A Hyperelliptic Involution
Let $X$ be the compact Riemann surface and $H$ a finite subgroup of $G=Aut(X)$. Then you can think of two different constructions.


1. One is taking the quotient in the category of complex manifolds $X/H$ (which corresponds to the [GIT quotient](http://en.wikipedia.org/wiki/Geometric_invariant_theory) $X//H$ in the algebraic category). Quotients (of complex manifolds by complex Lie groups) don't always exist, so you have to *prove* that there actually exists a complex manifold $X/H$ which has the desired quotient properties (which is done in your textbook).
2. The other construction is considering the quotient *orbifold* $[X/H]$. In this case, since $H$ is finite, it has always a meaning, and is -indeed- a 1-dimensional complex-analytic orbifold.


The relation between the two constructions is that (in this case of smooth Riemann surfaces and finite groups) you can think of $[X/H]$ as the Riemann surface $X/H$ decorated, at the ramification points of the quotient map $\pi:X\to X/H$, with the stabilizers. 


You say:



> 
> "We know that only when an group G act freely and properly discontinuously on a manifold $M$, the quotient space $M/G$ is a manifold. Here the question is that the action of automorphism group of a Riemann surface on itself has the fixed points"
> 
> 
> 


I think the correct statement would be that when $G$ acts freely and properly discontinuously there exists a quotient $M/G$ in the category of manifolds, and the quotient orbifold $[M/G]$ (which is the quotient in the category of orbifolds) *happens to be* a manifold. *But* there are cases, like the case of $M=\mathbb{C}\mathbb{P}^1$ and $G=${$1,-1$}, in which the action is *not* free, nevertheless the quotient manifold $M/G$ *does exist* (in this case it's $\mathbb{C}\mathbb{P}^1$ itself) but *is different* from the quotient orbifold $[M/G]$ (in this case $[\mathbb{C}\mathbb{P}^1/H]$ is the Riemann sphere decorated at $0$ and $\infty$ with copies of $\mathbb{Z}/2\mathbb{Z}$).


