1
$\begingroup$

Let $B$ be a finite set in $\mathbf{P}^1(\mathbf{C})$. Let $G$ be the fundamental group of $\mathbf{P}^1(\mathbf{C}) - B$.

We can view $G$ as a subgroup of $\mathrm{SL}_2(\mathbf{R})$.

Why is $G$ Fuchsian of finite volume?

That is, why is $G$ a discrete subgroup of finite volume?

  • 0
    *How* do you view $G$ as a subgroup of $\mathrm{SL}_2(\mathbb R)$? $G$ is a finitely generated free group, and there are *many* free subgroups inside $\mathrm{SL}_2(\mathbb R)$...2011-11-04

1 Answers 1

3

Firstly, you must assume that $B$ contains at least three elements. In this case, the surface $S$ you are considering (a finitely punctured sphere) is hyperbolic; that is, there is a holomorphic universal covering map $\pi:H^U\to S$, where $H^u$ is the upper half plane. Since the automorphism group of $H^u$ is precisely $\operatorname{SL}_2(\mathbb{R})$, you can think of the fundamental group, via the group of deck transformation, as such a subgroup.

Discreteness is trivial since the set of deck transformations is discrete.

Finite volume just means (if I do not misunderstand your question) that the hyperbolic metric on $S$ has finite volume. For this, you just need to check that the hyperbolic metric near a puncture has finite volume: recall that the metric near a puncture looks like $ \frac{|dz|}{|z|\cdot|\log|z||},$ if we choose local coordinates so that the puncture is at zero.

Alternatively, think of a fundamental domain for the universal cover in $H^u$, this will be bounded by finitely many geodesics, which form cusps on the boundary. Since the density of the hyperbolic metric in $H^u$ is just $1/\operatorname{Im}(z)$, you can easily check that each such cusp will have finite volume.