$SL(2,\mathbb{Z})$ is a lattice in $SL(2,\mathbb{R})$

Question

Definition (1). A subgroup $H$ of a topological group $G$ (with nice properties) is called a lattice if $H$ is discrete and $G/H$ has finite volume (with respect to the Haar measure).

Definition (2). Let $X$ be a topological space and let $G$ be a group acting on $X$. A fundamental domain, $\mathcal{F}$, for this action is a subset of $X$ which contains one element of each orbit.

I am trying to understand the proof of the following Theorem:

Theorem $SL(2,\mathbb{Z})$ is a lattice in $SL(2,\mathbb{R})$.

I am reading this note (or book). https://arxiv.org/pdf/math/0106063.pdf

Question (1). In example (1.3.7) at page 15, the author defines a fundamental domain $\mathcal{F}$ for the action of $SL(2,\mathbb{Z})$ on upper-half plane $\mathbb{H}$. Then he proves that the volume of $\mathcal{F}$ is finite. How does $vol(\mathcal{F})<\infty$ conclude that the volume of $SL(2,\mathbb{R})/SL(2,\mathbb{Z})$ is finite?

Question (2). Can someone explain how he proves that the $vol(\mathcal{F})$ is finite? It's not clear why the hyperbolic length is $(dx)/(2y)$ and the hyperbolic width is $(dy)/(2y)$.

Answer 1

The first answer comes from the definition of fundamental domain. As you said, $\mathcal{F}$ is a fundamental domain for the $SL(2,\Bbb Z)$ action on $\Bbb H$. This means that for each $z\in \Bbb H$ the $SL(2,\Bbb Z)-$orbit meets $\mathcal{F}$ at most at two points. More precisely, if $z$ lies in the interior of $\mathcal{F}$, then $\Big(SL(2,\Bbb Z)\cdot z\Big)\cap \mathcal{F}=\{z\}$. If $z$ lies on the boundary of $\mathcal{F}$, then $\Big(SL(2,\Bbb Z)\cdot z\Big)\cap \mathcal{F}$ consists of two points - both on the boundary - which are identified. The only exception is given by the point $i $ which lies on the boundary but its orbit meets $\mathcal{F}$ only once. The quotient space $SL(2,\Bbb R)/SL(2,\Bbb Z)$ is identified with the quotient space you get by gluing the identified points on the boundary of $\mathcal{F}$. Since $\textsf{vol}(\mathcal{F})$ is finite then also the volume of $SL(2,\Bbb R)/SL(2,\Bbb Z)$ has to be finite.

About question two: I give you another explanation different from the one provided in the notes. The point is that the hyperbolic area of a hyperbolic triangle $T$ - possibly with one or more vertices at infinity - is given by the formula $\mathcal{A}_{\Bbb H}(T)=\pi-\alpha-\beta-\gamma$ where $\alpha,\beta,\gamma$ are the inner angles of the triangle. For each vertex at infinity the inner angle is zero. This is more clear if you look at the situation on the disc model of the hyperbolic plane. Furthermore, any standard book of hyperbolic geometry of differential geometry have a proof of this result. In fact, it highlights one of the main differences between euclidean and hyperbolic geometry, so it isn't hard to find it. In your case, since one vertex lies at infinity, its inner angle is zero. The others are less than $\frac\pi2$ - you can prove it by using standard tools of euclidean geometry (because the magnitude of an angle is given by the magnitude of two tangent vectors at the vertex. Since the euclidean geometry and hyperbolic geometry are conformal, even if they are not isometric geometries, euclidean tools work as well.). Concluding, the hyperbolic area of that strip, actually a hyperbolic triangle with one vertex at infinity has finite area.

A couple of final remarks:

1. The fact that the infinite strip has finite area might seems something counterintuitive at the beginning. The hyperbolic plane is a Riemannian manifold, namely a differential manifold $\Bbb H$ endowed with the Riemann metric $ds^2=\frac{dx^2+dy^2}{y^2}$. The denominator $y^2$ is the reason why a triangle with one or more vertex at infinity has finite area. Indeed, in the euclidean sense the vertical lines of the domain $\mathcal{F}$ have the same distance in the euclidean geometry, but their distance decrease as soon an $y\to\infty$ in the hyperbolic geometry, that's the point. Not only, it is possible to show that the distance of the vertical lines in hyperbolic geometry is actually zero!! Nice, uh?
2. Second thing: unlike euclidean geometry, each hyperbolic triangle has finite area!! and two similar triangles are automatically isometric.