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If $\Gamma\subseteq PSL(2,\Bbb R)$ is a geometric finite (i.e. finitely generated; i.e. $\Gamma\backslash\mathbb{H}$ has finite volume) Fuchsian group which is not co-compact and has $\infty$ as a cusp, then why does the minimum \begin{align*} \operatorname{min}\{c>0\mid \begin{pmatrix} *&*\\ c&* \end{pmatrix}\in\Gamma\} \end{align*} exist? I've been reading that this has to do with the "standard polygon" for $\Gamma$ and that $c^{-1}$ is the radius of the largest isometric circle but I don't know the construction of this polygon neither did I find sources of the construction of this. Is the standard polygon connected to the Dirichlet domain (this I know)?

Thanks in advance for any hint.

Nightgap
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    This is Shimizu's Lemma. See for example Lemma 2.21 here: https://books.google.com/books?id=h77tCAAAQBAJ&pg=PA45&lpg=PA45&dq=shimizu%27s+lemma+fuchsian+group&source=bl&ots=x-pVWiP32g&sig=ACfU3U3GjUWTtFe9KNPaCdZ0UfWZHrnfCw&hl=en&sa=X&ved=2ahUKEwimlMTFv4jiAhUP-6wKHVNQA4gQ6AEwCHoECAkQAQ#v=onepage&q=shimizu's%20lemma%20fuchsian%20group&f=false – Grant Lakeland May 07 '19 at 03:48
  • Thank you very much. But why does this imply the existence of the minimum. Doesn't it only show that the minimum, if it exists, is greater than one? – Nightgap May 07 '19 at 07:55
  • It shows that the minimum is at least 1 when the parabolic fixing infinity is ( 1 1 | 0 1 ). The point is that if infinity is a cusp, then there must be an element of the form (1 * | 0 1 ), and the group can be conjugated in PSL(2,R) so that it is (1 1 | 0 1) and the Lemma can be applied. Then the inverse conjugation gives you the minimum for your group. – Grant Lakeland May 07 '19 at 13:56
  • Regarding the fundamental domain side, I think the right way to look at it is from the perspective of the Ford domain, which can be seen as a sort of "Dirichlet domain based at infinity" whose sides are isometric circles, along with two sides identified by the parabolic. Beardon's Geometry of Discrete Groups is a good reference for this. As you say, $c^{-1}$ is the radius of the isometric circle, and in a geometrically finite group we cannot have a sequence of radii going to infinity. – Grant Lakeland May 07 '19 at 15:03
  • "... geometrically finite group with $\infty$ as a cusp, we cannot have..." – Grant Lakeland May 07 '19 at 15:19

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