## Fuchsiana

Mar 23rd, 2007 by Eliot P. Brenner

Here’s a little puzzle about Fuchsian groups. [A collaborator came up with it and we’d like to use it to simplify a proof in a paper, but I think it’s a question that may have totally independent interest.] Is any Fuchsian group $latex \Gamma_1$ of the first kind with at least one cusp contained as a subgroup of finite index in another Fuchsian group $latex \Gamma$ such that $latex \Gamma$ has exactly one cusp?

If one is more comfortable with this terminlogy, “Fuchsian group of the first kind with at least one cusp” equals “nonuniform lattice in $latex \mathrm{PSL}(2,\mathbf{R})”$.

For example, obviously if $latex \Gamma_1$ is a congruence subgroup of any sort, then obvserving that $latex \Gamma_1\leq\mathrm{PSL}(2,\mathbf{Z})$ suffices. This gives me the feeling if there’s a counter-example to be found, it would be among non-arithmetic lattices, but I’ve never had any sort of handle on those.

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on 08 Apr 2007 at 3:58 pm1RichardHello. Goofing around wordpress, I stumbled on your page.

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.

on 08 Apr 2007 at 4:27 pm2heatkernelThanks for your reply! There are at least two points I don’t understand. The first may be just a matter of notation. You say “then $latex X$ is in the image of some $latex T(Y)$”, but then later pick a $latex \Gamma_1$ that lies outside all the images. Did you mean $latex \Gamma_1$ the first time? The second is that I don’t see why anything about your choice of $latex \Gamma_1$ guarantees that it has more than one cusp. In other words, it seems to me that you may have provided a proof that there exist maximal Fuchsian subgroups, but it isn’t clear that every (or indeed, any!) maximal Fuchsian subgroup has more than one cusp.

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.

on 08 Apr 2007 at 4:42 pm3RichardGlad to be of some help.

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).

on 08 Apr 2007 at 4:54 pm4heatkernelIt’s much clearer to me now. Many thanks.