Easy explanation of non-abelianness of hyperbolic curves

Question

I'm looking for easy proofs (or just an easy proof) of the following statement:

Let X be a hyperbolic Riemann surface, i.e., $X$ is a Riemann surface and the universal covering of $X$ is the complex upper half plane. Then the fundamental group of X is non-abelian.

One could resort to several different proofs:

1. Compute the fundamental group. This is a standard computation. (This is too "difficult" though.)

2. Use that X is algebraic and the complex upper half plane isn't. Therefore, its fundamental group is an infinite subgroup of PSL$_2(\mathbf R)$.

3. Anything else?

Answer 1

The simplest method may be to observe that there exist surjections onto 'easier' non-abelian groups. For instance, the inclusion of $\Sigma_g$ into a handlebody of genus $g$ induces an epimorphism

$\pi_1\Sigma_g\to F_g$

where the latter is the free group of rank $g$. Of course, the existence of any non-abelian $g$-generator group implies the non-abelian-ness of $F_g$.

Answer 2

If the fundamental group were abelian, it would have polynomial growth (it is easy to show that this concept is well-defined since the group is finitely generated). But fundamental groups of compact hyperbolic surfaces have exponential growth. This is just as easily deduced from the exponential growth of the volume of a hyperbolic ball of radius $r$ as $r\to\infty$.

Answer 3

The Gauss Bonnet theorem implies such a compact Riemann surface has negative curvature, hence negative euler characteristic. is that enough for your purposes, i.e. that it is a doughnut with ≥ 2 handles?

Or maybe this enables HJRW's answer.

Answer 4

Since you seem willing to accept that $\pi_1 S$ acts on the upper half plane $H$, perhaps you will accept that it acts on the Poincare disc $D$ via conjugation by a conformal isomorphism $H \leftrightarrow D$, and that the following counts as "easy": every abelian subgroup of $PSL(2,R)$ acting on the Poincare disc fixes a point in the closure of the Poincare disc, but the action of $\pi_1 S$ fixes no such point.

Answer 5

A hyperbolic compact Riemann surface $X$ with fundamental group $\pi$ is a $K(\pi, 1)$. It follows that $\pi$ has cohomological dimension $2$. If $\pi$ were abelian, then it would be a finite direct product of cyclic groups. Finite cyclic groups have infinite cohomological dimension, so if $\pi$ were abelian then it would have to be a finite direct product of copies of $\mathbb{Z}$, and the only such direct product with cohomological dimension $2$ is $\mathbb{Z} \times \mathbb{Z}$; moreover, $X$ would have to be homotopy equivalent to a torus. But tori have Euler characteristic $0$ and $X$ necessarily has negative Euler characteristic; contradiction.