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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?

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migrated from mathoverflow.net Jul 25 '13 at 19:30

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$X$ is not necessarily algebraic. Of course, it is bilipschitz equivalent to a algebraic one, and bilipschitz equivalences respect volume growth. –  user72694 Jul 25 '13 at 12:45
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You need more hypotheses: $\mathbb{C}$ minus a closed disk has the upper half plane as universal cover. –  Richard Kent Jul 25 '13 at 13:01
    
Since the curve has no boundary, the fundamental group acts indiscretely on the boundary, so it is not in an abelian subgroup of rank 1. Abelian subgroups of higher rank are indiscrete, but the fundamental group must be discrete. –  Will Sawin Jul 25 '13 at 15:05
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5 Answers 5

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 $F_g$.

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Here, $\Sigma_g$ is of course the closed surface of genus $g$. Since non-closed surfaces have free fundamental groups, the arguments in those cases are even easier. –  HJRW Jul 25 '13 at 13:23
    
I am not sure about the whole premise of the question: If OP regards computation of $\pi_1(\Sigma_g)$ as "difficult", he/she will probably find computation for the surface with nonempty boundary, handlebody, or a graph, "difficult" as well. –  Misha Jul 25 '13 at 13:52
    
Misha - I agree. I was just hoping that the OP would find this helpful. –  HJRW Jul 25 '13 at 20:38
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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 $r\to\infty$.

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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.

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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.

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Of course Richard Kent's objection holds here as well, and so one must list out those $S$ which are quotients of abelian subgroups and forbid them from consideration. –  Lee Mosher Jul 25 '13 at 14:39
    
One could also state this in fewer words using just $H$ itself, allowing consideration of the "point at infinity" as one possible fixed point for an abelian subgroup. –  Lee Mosher Jul 25 '13 at 14:44
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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.

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