Quotient Surface of A Hyperelliptic Involution - MathOverflow

Quotient Surface of A Hyperelliptic Involution

2011-03-28T15:23:07Z

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?

Answer by Charles Matthews for Quotient Surface of A Hyperelliptic Involution

2011-03-28T15:40:07Z

If you think about J-equivariant holomorphic mappings, in the category of complex manifolds, this issue probably goes away. If there is a universal example of such a mapping to a manifold on which J acts trivially, that can be your quotient. If you step outside the category to some other category, why should the word "quotient" make sense in the same way? If you talk about "the" quotient space, you should be saying what structure you are talking about, also.

Answer by Qfwfq for Quotient Surface of A Hyperelliptic Involution

2011-03-28T15:44:11Z

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

One is taking the quotient in the category of complex manifolds $X/H$ (which corresponds to the GIT quotient $X//H$ 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).
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 $\mathbb{Z}/2\mathbb{Z}$).

Answer by Sebastian for Quotient Surface of A Hyperelliptic Involution

2011-03-30T09:51:20Z

Here is a very lowtech answer: Look at the $C$ plane, together with the $Z_2$-action $\sigma\colon z\mapsto-z.$ At a point $z\neq0$ you get an open neighbourhood $U$ with $u\cap\sigma(U)=\emptyset.$ Therefore you can just take $z$ as a coordinate on the image of $U$ under $\sigma$ in $C/\sigma$ as a nice coordinate. This "topological" quotient always works for finite groups away from the fixpoints, and you have the same coordinates (in some sense). The nice thing with Riemann surfaces is, that even if you are at a fix point, you get nice coordinates on the quotient, but they differ from the original one: in the above example, you should take $z^2$ as a local coordinate on a neighboorhood of the image of $0$ in the quotient. This makes sense, since any even function on $C$ (i.e. invariant under $\sigma$) is a function in $z^2.$

Answer by Nicolas Schmidt for Quotient Surface of A Hyperelliptic Involution

2012-02-14T23:32:51Z

I think the confusion stems from a common misconception, that orbifold points should somehow be considered as singular. The quotient in question - whether as a complex variety or as an orbifold - is not singular. In fact you will never be able to produce something singular by taking the quotient of a compact Riemann surface (i.e. a smooth complex algebraic curve) wrt the action of a finite group (because smoothness in dimension 1 is equivalent to normality, which is preserved by quotients under finite group actions).