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I am trying to compare various definitions and theorems I have seen recently concerning Fuchsian groups. Some of these seem to contradict each other, so I was hoping to get some clarification.

First, a Fuchsian group is a discrete subgroup of $\text{PSL}_2(\mathbb R)$, which we can view as a group of transformations of the upper half-plane $\mathbb H$ that acts discontinuously. A lattice is a Fuchsian group with finite covolume. In other words, it is a Fuchsian group that has a fundamental domain in $\mathbb H$ with finite hyperbolic area (and then any two fundamental domains have the same hyperbolic area).

My confusion arises with the notion of a Fuchsian group of the first kind. Every definition I have seen for this term is roughly the same. A Fuchsian group $\Gamma$ is said to be of the first kind if every point in $\mathbb R\cup\{\infty\}$ (the boundary of $\mathbb H$) is a limit point of the orbit $\Gamma z$ for some $z\in\mathbb H$. Here, the notion of "limit point" is with respect to the topology on the Riemann sphere $\mathbb C_\infty$.

Now, in the book "Introduction to the Spectral Theory of Automorphic Forms" by Iwaniec, the following theorem is stated:

Theorem: Every Fuchsian group of the first kind has a finite number of generators and fundamental domain of finite volume.

So this leads me to believe that a Fuchsian group of the first kind is necessarily a lattice. However, this Wikipedia article of Fuchsian groups (in the section on Limit Sets) defines a Fuchsian group of the first kind and then states "This happens if the quotient space $\mathbb H/\Gamma$ has finite volume, but there are Fuchsian groups of the first kind of infinite covolume." So Wikipedia is telling me that every lattice is of the first kind, but not conversely.

Any clarification on these matters would be greatly appreciated. In addition, if the Wikipedia article is correct, it might be helpful to have an example (provided it is not too complicated) of a Fuchsian group of the first kind that is not a lattice. Furthermore, is the other part of the above theorem (that a Fuchsian group of the first kind is finitely-generated) true? Thanks in advance.

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3 Answers 3

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The claim that Iwaniec makes is simply false; the mistake originates in Lemma 8 in:

C. L. Siegel, Discontinuous groups. Ann. of Math. (2) 44 (1943), 674--689.

The mistake in the proof is that Siegel incorrectly assumes that every Fuchsian group has a finitely-sided fundamental polygon. (I looked closely in the paper, Siegel's definition of a Fuchsian group only requires discreteness, no assumptions on fundamental polygons.)

The correct result is:

Theorem. The following are equivalent for a Fuchsian group $\Gamma< PSL(2,R)$ of the first kind:

  1. $\Gamma$ is finitely generated.

  2. $\Gamma$ is a lattice in $PSL(2,R)$, i.e. ${\mathbb H}^2/\Gamma$ has finite area.

  3. One (equivalently every) fundamental polygon of $\Gamma$ has only finitely many sides.

See for instance Theorem 10.1.2 in

A.F.Beardon, "Geometry of Discrete Groups", Springer Verlag, 1983.

The example that Lee Mosher gave you is definitely valid although a proof that the uniformizing group $\Gamma$ is of the 1st kind would take some nontrivial effort involving extremal length considerations. An easier example is the following.

Take a torsion-free Fuchsian subgroup $\Gamma_1< PSL(2,R)$ such that ${\mathbb H}^2/\Gamma$ has finite area. Take a nontrivial normal subgroup $\Gamma_2< \Gamma_1$ of infinite index. It is a general fact about general Fuchsian groups that the limit set of a nontrivial normal subgroup $\Gamma_2\triangleleft \Gamma_1$ equals the limit set of $\Gamma_1$. Since our group $\Gamma_1$ is of the first kind, so is the normal subgroup $\Gamma_2$. On the other hand, area is multiplicative under isometric coverings between Riemannian surfaces: If $S_2\to S_1$ is an isometric covering of degree $d$ of Riemannian surfaces, then $$ Area(S_2)=d Area(S_1). $$ (The same holds for manifolds in all dimensions, but the area would mean volume.) In our case, the degree of the covering map $$ {\mathbb H}^2/\Gamma_2\to {\mathbb H}^2/\Gamma_1 $$ equals the index $|\Gamma_1: \Gamma_2|=\infty$. Hence, $$ Area({\mathbb H}^2/\Gamma_2)=\infty. $$

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I'll write an answer that avoids the "first kind" and "second kind" terminology (for three reasons: (1) I don't ever see these terms used nowadays; (2) I never could keep them straight anyway; (3) I would probably write my answer incorrectly if I attempted to use them).

I'll describe a Fuchsian group $\Gamma$ which is not finitely generated such that $\Lambda(\Gamma) = \mathbb{R} \cup \{\infty\}$.

If $\Gamma$ is a Fuchsian group and if $\Lambda(\Gamma) \ne \mathbb{R} \cup \{\infty\}$ then there is an interval $I \subset \mathbb{R} \cup \{\infty\}$ whose endpoints are in $\Lambda(\Gamma)$ but whose interior is disjoint from $\Lambda(\Gamma)$. Let $\gamma$ be the geodesic in $\mathbb{H}$ with the same endpoints as $\Lambda$, let $P\subset \mathbb{H}$ be the half-plane with finite boundary $\gamma$ and infinite boundary $I$, and let $S_I < \Gamma$ be the infinite cyclic subgroup that stabilizes $I$, and $\gamma$, and $P$. In this situation, the infinite quotient cylinder $P / S_I$ embeds in the Riemann surface $\mathbb{H} / \Gamma$ and is a neighborhood of an end of the topological space $\mathbb{H} / \Gamma$, so that space has an isolated end. So it suffices to describe a Riemann surface with no isolated end and whose fundamental group is not finitely generated.

For this purpose, simply take $S^2 - C$ where $C$ is a Cantor set. Restricting the conformal structure on $S^2$ to get a conformal structure on $S^2 - C$, apply the uniformization theorem to get a Fuchsian group $\Gamma$ such that $\mathbb{H} / \Gamma$ is conformally equivalent to $S^2 - C$.

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  • $\begingroup$ I don't agree with your reason (1), though I wholeheartedly agree with your reason (2). $\endgroup$
    – Kimball
    Apr 1, 2018 at 12:51
  • $\begingroup$ The sphere has positive curvature, is the universal cover of $S^2-C$ indeed $\mathbb H$? And if $\mathbb H/\Gamma$ is only conformally equivalent to $S^2-C$, does it still have finite volume? $\endgroup$ Jul 4, 2018 at 13:13
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It appears that there are simply two definitions, and it seems (but I'm not an expert, and I welcome reactions) that the stronger one is the older one:

  1. A Fuchsian group of the first kind is a lattice, that is, a discrete subgroup with finite covolume.
  2. A fuchsian group of the first kind is a discrete subgroup whose limit set is $P^1(\mathbb R)$.

The first condition implies the second.

Definition 1 can be found in Takeuchi K., A characterization of arithmetic Fuchsian groups (1975).

Definition 2 is used in Iwaniec (2002), at Wikipedia (currently), in Katok, S., Fuchsian groups (1992)

My guess is that Iwaniec stated the equivalence between 1 and 2 without proof or reference, possibly due to the confusion that arises from these nonequivalent definitions.

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