Quotient of space and a group of maps, Riemann surfaces

Question

I've been attempting to study Riemann surfaces, and I have continuously run into this notion which eludes me. I see people write things like $ \mathbb H / <z\mapsto z+1>$ or $\mathbb D / PSL$. I know what all these things are individually, but I don't understand how someone takes a quotient of a space and some maps (as opposed to say, treating the space as a group and taking the quotient over a normal subgroup). The end result of this is hopefully understanding how people come up with the "isomorphism maps" that are discussed in this context.

Can someone spell this out a little bit for me? The whole issue feels opaque to me.

Answer 1

Say $G$ is a group of bijections of the set $X$. For $x,y\in X$ say $x\sim y$ if $x=gy$ for some $g\in G$. Then $\sim$ is an equivalence relation on $X$, and $X/G$ is the same thing as the set $X/\sim$ of equivalences classes.

Answer 2

Suppose a group $G$ acts (on the left) on a set $X$, so we have a map \begin{align*} G \times X &\to X\\ (g,x) &\mapsto gx \, . \end{align*} Given $x \in X$, consider its orbit $Gx = \{gx : g \in G\}$. These orbits partition $X$, so we can define the quotient $G \backslash X = \{Gx : x \in X\}$ as the set of orbits. (I write $G \backslash X$ rather than $X/G$ because $G$ is acting on the left.)

If $X$ is a topological space and $G$ is a topological group and acting continuously on $X$, i.e., the map \begin{align*} G \times X &\to X\\ (g,x) &\mapsto gx \end{align*} is continuous, then we can upgrade $G \backslash X$ to a topological space by giving it the quotient topology as follows. We have a quotient map \begin{align*} \pi: X &\to G \backslash X\\ x &\mapsto Gx \end{align*} that sends $x$ to its orbit, and we define the topology on $G \backslash X$ to be the strongest topology such that $\pi$ is continuous. That is, $U \subseteq G \backslash X$ is open iff $\pi^{-1}(U) \subseteq X$ is open.

Let's look at a geometric-flavored example. The group $\text{PSL}_2(\mathbb{R})$ acts on the upper half-plane by Möbius transformations: \begin{align*} \text{PSL}_2(\mathbb{R}) \times \mathcal{H} &\to \mathcal{H}\\ \begin{pmatrix} a & b\\ c & d \end{pmatrix} \cdot z &= \frac{az + b}{cz + d} \, . \end{align*} Then $$ \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix} \cdot z = z + 1 $$ so it acts as $T(z) = z+1$. I claim that $\langle T \rangle \backslash \mathcal{H}$ is homeomorphic to an infinite cylinder. To see this, we find a fundamental domain for the action, i.e., a set of representative points in $\mathcal{H}$ so that each orbit has exactly one representative. In this case $\{z \in \mathcal{H} : 0 \leq \Re(z) < 1\}$, is such a set. Since $T$ identifies the left and right edges of this domain, then we get a cylinder as the quotient $\langle T \rangle \backslash \mathcal{H}$.

For another example, take the subgroup $\text{PSL}_2(\mathbb{Z}) \leq \text{PSL}_2(\mathbb{R})$ and form $\text{PSL}_2(\mathbb{Z})\backslash \mathcal{H}$. This has a nice fundamental domain as shown here. This yields a funny looking quotient that is homeomorphic to a sphere minus a point.