Volume of a Riemannian manifold and its relation to fundamental group

Question

I am reading a book (Mapping Class Group by Farb and Margalit) and it says (in a proof of one theorem):

If $S$ admits a hyperbolic metric (they define such a surface to be of finite area and complete) and we had $\pi_1(S)\cong \mathbb{Z}\ $ then the surface would have an infinite volume which is a contradiciton. Hence $\pi_1(S)$ is NOT isomorphic to $\mathbb{Z}$.

Questions:

1. Why $\pi_1(S)\cong \mathbb{Z}\ $ implies the volume of $S\ $ is infinite?

2. Is there a general relation between the fundamental group and the volume of a surface?

Can someone help me, please?

Answer 1

Hyperbolic surfaces are of the form $\mathbb{H}^2/\Gamma$, where $\Gamma$ is a Fuchsian group which is isomorphic to the fundamental group. If $\Gamma \cong \mathbb{Z}$, then it is generated by a single element $\gamma$. The order of $\gamma$ is infinite, so it cannot be elliptic. It is thus either parabolic (and hence conjugate to $\begin{pmatrix} 1 & \lambda \\ 0 & 1 \end{pmatrix}$ for some $\lambda$) or hyperbolic (and hence conjugate to $\begin{pmatrix} \lambda & 0 \\ 0 & 1/\lambda \end{pmatrix}$ for some $\lambda \neq 0$). It is an easy exercise to show that the quotient of $\mathbb{H}^2$ by the subgroup generated by $\begin{pmatrix} 1 & \lambda \\ 0 & 1 \end{pmatrix}$ has infinite volume, and similarly for $\begin{pmatrix} \lambda & 0 \\ 0 & 1/\lambda \end{pmatrix}$.

For compact surfaces with Riemannian metrics, the Gauss-Bonnet theorem determines the volume in terms of the Euler characteristic (and hence in terms of the fundamental group).

I recommend reading Katok's book "Fuchsian Groups" if you want to acquire comfort with the above types of arguments.

Answer 2

I will write what I understand so far:

The fundamental group of an orientable surface of genus $g$ with $b$ boundary components and $n$ punctures will be generated by $2g+b+n$ generators. The Euler characteristic in this case is $\chi=2-(2g+b+n)$ i.e $\chi=2-$ the number of generators. So we get $\chi=1.$ obviously $g=0$ and the either $b=1$ or $n=1$ i.e sphere with one boundary component OR sphere with a puncture .

If sphere with one boundary component: I do not know what to do because there is no guarantee that the boundary is totally geodesic.

If sphere with a puncture (this corresponds to the case of parabolic element), that is just the plane, and it has infinite area (No need for Gauss-Bonnet).

Please correct me if I made some mistakes.