I guess the last sentence should use a name other than $latex \Gamma_1$.

Everything except the last sentence shows that in the Teichmuller space of a hyperbolic orbifold, the set of guys that nontrivial cover another orbifold is not everything. So, given any orbifold with more than one cusp, there is a point in its Teichmuller space that is minimal with respect to coverings (so it’s Fuchsian group is maximal).

]]>Either way, it’s great to get some mathematical activity here, and your reply encourages me to post some more questions in the near future.

]]>The answer is no. Let $latex \Gamma_1$ be your fuchsian group, and let $latex X = \mathbb{H}^2/\Gamma_1$. There are only countably many nontrivial topological orbifold covers $latex X \to Y$. Any such cover induces a map $latex T(Y) \to T(X)$ between Teichmuller spaces.

If $latex \Gamma_1$ is a proper subgroup of any fuchsian group $latex \Gamma_2$, then $latex X$ will lie in the image of some $latex T(Y)$. Since there are only countable many of these maps $latex T(Y) \to T(X)$, the union of their images cannot be all of $latex T(X)$, by the Baire category theorem (since the dimension of $latex T(Y)$ is less than that of $latex T(X)$). Now just pick $\Gamma_1$ so that it does not lie in any of the images.

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