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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?

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How do you know it is? –  Rasmus Nov 4 '11 at 8:25
    
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 $\mathrm{SL}_2(\mathbb R)$... –  Mariano Suárez-Alvarez Nov 4 '11 at 11:55
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1 Answer 1

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.

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