I am currently reading the book Geometry and Spectra of Compact Riemann Surfaces by Peter Buser. The author gives an example of a hyperbolic cylinder to demonstrate his ideas in pasting two hyperbolic structures.
First, the author constructs a hyperbolic cylinder with the following shape ($\subset \mathbb{H}$ for sure):
based on the equivalence relationship $$\sim:\gamma(t)=\gamma'(t)$$ where $\gamma,\gamma'$ are two geodesics.
Then the author writes that the following quotient is a cylinder $$C=S/\sim$$ where $S$ is the annulus.
Here is my question
The author then mentions that if $m(z)=\frac{b}{a}z$ and $\Gamma=\{m^k|k\in\mathbb{Z}\}$, then $\Gamma\backslash \mathbb{H}=C$. Why is this true? Can anyone "describe" $\Gamma\backslash\mathbb{H}$ for me?
I could picture that $C=S/\sim$ is a one-layer "twisted" cylinder, but $\Gamma\backslash \mathbb{H}$ is really confusing. I am confused by the elements in $\Gamma\backslash\mathbb{H}$. For example, $m^2\in\Gamma$ and $$m^2(\mathbb{H})=\frac{b^2}{a^2}\mathbb{H}\in\Gamma\backslash\mathbb{H}$$ Isn't this pasting the geodesic that goes through $a^2$ to the geodesic that goes through $b^2$?
I know there is something seriously wrong with my understanding, so please help me. Thanks in advance.
Something about $m$ in the statement made by the author looks off. The correct statement should probably be that if $m = \frac{b}{a}$ (no $z$) and $\Gamma = \{m^k \mid k \in \mathbb Z\}$ then the quotient spaces $\Gamma \setminus \mathbb H$ and $S / \sim$ are homeomorphic. What's more, the inclusion $S \hookrightarrow \mathbb H$ induces the homeomorphism.
Let me explain why this is true using basic quotient space material.
Each element of $\Gamma \setminus \mathbb H$ is a coset of the form $\Gamma z = \{m^k z \mid k \in \mathbb Z\}$, for some $z \in \mathbb H$. Furthermore, the intersection $\Gamma z \cap S$ satisfies one of two possibilities: it is a unique point, contained in the interior of $S$; or it is a pair of points, one of them on the inner semicircle of $S$ having the form $\gamma(t)$, and the other on the outer semicircle having the form $\gamma'(t)$ for the same value $t$, and $\gamma'(t) = m \gamma(t)$.
To put this another way, $\Gamma \setminus \mathbb H$ is a set of equivalence classes on the set $\mathbb H$, $S / \sim$ is a set of equivalence classes on the set $S$, and the intersection with $S$ of each $\Gamma \setminus \mathbb H$ equivalence class is equal to an $S / \sim$ equivalence class.
From this, you immediately obtain a one-to-one correspondence, i.e. a bijection $$\Gamma \setminus \mathbb H \leftrightarrow S / \sim $$ So far so good. You can see from this that the geodesic $\gamma$ is being pasted to the geodesic $\gamma'$ in both of these descriptions; you could write this as $$m(\gamma)=\gamma' $$ There is, of course, also another geodesic $m^2(\gamma)$, lying outside of $S$, and there is in fact an expanding sequence of geodesics $m^2(\gamma),m^3(\gamma),m^4(\gamma),\ldots$ lying further and further outside of $S$. There is also a shrinking sequence of geodesics $m^{-1}(\gamma)$, $m^{-2}(\gamma)$, $m^{-3}(\gamma)$ lying further and further inside of $S$.
Now this is more than just a bijection, it's actually a homeomorphism of quotient topologies. So the upshot is that since $S / \sim$ is homeomorphic to an open annulus (what you are calling a "twisted cylinder"), it follows that $\Gamma \setminus \mathbb H$ is also homeomorphic to an annulus. You'll need more advanced tools of quotient topologies to prove this, and you can find them for example in Munkres "Topology".