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I just want to confirm if I am right here. Can I say every Fuchsian group corresponds to a way to glue the hyperbolic plane to get a suface?

Then the Poincare polygon theorem means that, given a convex finitely sided polygon and side pairing with appropriate angle sums of vertex cycle, we can find a Fuchsian group such that this polygon is fundamental (and thus tessellate the corresponding surface).

I am not sure if this is correct because my intuition here is on complex plane (rather than on hyperbolic plane), where the Fuchsian group is $\langle +1, -1, +i, -i\rangle$ and the fundamental polygon can be the unit square. Then the Fuchsian group corresponds to a torus and the unit square forms a tessellation.

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  • $\begingroup$ It depends on what do you mean by "get a surface" (I am ignoring the "glue" part, you probably mean a quotient map): If all you are asking if you get a topological surface or a surface equipped with a path-metric, then, yes. If you want to get a surface (possibly with boundary) such that the quotient map from $H^2$ is a covering map, then the answer is no. Also, what exactly do you mean by a Fuchsian group? The definition which exist in the literature are not entirely consistent, all what they agree on is that it should be discrete. As of now, your question is unclear. $\endgroup$ – Moishe Kohan Jun 11 at 19:10
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Every complete Riemannian 2-manifold of constant curvature $-1$ has $H^2$ as a universal cover, and its deck transformation group acts freely and discretely on $H^2$ and is thus a Fuchsian group, but not every Fuchsian group acts freely, so not every quotient of $H^2$ by a Fuchsian group is a Riemannian manifold. For example, if $r$ is a point reflection across some point $p \in H^2$, then $G := \{1, r\}$ is a Fuchsian group, but $H^2/G$ has a cone point at $p$.

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