# Does every cofinite Fuchsian group contain a hyperbolic element?

Let $$\Gamma\subset PSL_2(\mathbb{R})$$ be a cofinite Fuchsian group (e.g. a Fuchsian group with finite fundamental domain). Does $$\Gamma$$ necessarily contain a hyperbolic element?

At first, I tried to use the fact that $$tr(\gamma)>2$$ if $$\gamma\in \Gamma$$ is hyperbolic, but I failed at this. (Which does not mean it is not possible and if it were, I would appreciate the simplicity of this approach.)

Now, I thought one could use the following two facts

• A non-elementary Fuchsian group (the orbit $$\Gamma z$$ is infinite for all $$z\in \mathbb{H}$$) must contain a hyperbolic element.

• A Fuchsian group is elementary if it is either cyclic or generated by the Moebius transformations $$g(z)=kz$$ and $$h(z)=-\frac{1}{z}$$

If I wanted to use the above, I would need to show that the fundamental domain of both a cyclic group and the one generated by $$g$$ and $$h$$ are finite, I assume. However, I fail with that. Maybe somebody can help me there?

I appreciate any help - so if my thoughts are leading in the wrong direction, I am very happy to check out a new approach!

• Your definition of "elementary" (every orbit in $H^2$ is infinite) is wrong. Jun 15, 2019 at 22:31
• I defined non-elementary groups. Is it still wrong? Jun 17, 2019 at 18:36
• Yes, it is still wrong. Hint: Consider the cyclic group generated by $z\mapsto z+1$. :) Jun 17, 2019 at 19:00
• Ah, I forgot the boundaries! Does it become right if I turn $\mathbb{H}$ into $\mathbb{H}\cup\mathbb{R}\cup \infty$ then? Jun 18, 2019 at 18:12
• What becomes right? Your definition of elementary groups? Then yes, it is a correct definition. But Lee Mosher's proof then does not apply. Jun 18, 2019 at 19:53

You say that you know that a Fuchsian group $$\Gamma$$ contains a hyperbolic element if it is true that $$\Gamma z$$ is an infinite set for every $$z \in \Gamma$$.

Assuming this to be known, it's pretty straightforward to prove the following two implications:

$$\Gamma$$ is cofinite $$\implies$$ $$\Gamma$$ is infinite $$\implies \Gamma z$$ is infinite for every $$z \in \mathbb H^2$$.

The first implication should be pretty obvious. However, your definition of cofinite is somewhat vague: I'm not sure whether "finite fundamental domain" means "finite diameter" or "finite area" or something else. Nonetheless, whatever it means, it should follow immediately that $$\Gamma$$ is infinite, because if $$D$$ is a finite fundamental domain then, using the fact that $$\mathbb H^2$$ is not finite, and using the equation $$\mathbb H^2 = \cup_{g \in \Gamma} g \cdot D$$, it follows that $$\Gamma$$ is infinite. For instance, if $$D$$ has finite diameter then one uses that $$\mathbb H^2$$ has infinite diameter to conclude that $$\Gamma$$ is infinite; alternatively if $$D$$ has finite area then one uses that $$\mathbb H^2$$ has infinite area to conclude that $$\Gamma$$ is infinite.

For the second implication, we use the fact that a Fuchsian group, by definition, is a discrete group. From this it follows that for each point $$z \in \mathbb H^2$$, its stabilizer subgroup $$\text{Stab}(z) = \{g \in \Gamma \mid g \cdot z = z\}$$ is a finite group (if $$\text{Stab}(z)$$ were infinite then $$\text{Stab}(z)$$ would contain rotations centered at $$z$$ having arbitrarily small angle, contradicting discreteness). It follows that the index $$[\Gamma:\text{Stab}(z)]$$ is infinite.

Now apply the orbit stabilizer theorem, to conclude that the cardinality of the orbit set $$\Gamma z$$ equals the index of $$\text{Stab}(z)$$. So $$\Gamma z$$ is infinite.

• Thanks a lot! This is very neat! Jun 17, 2019 at 18:39

One can indeed prove that using the facts abrewer listed.

First: Since the area of any fundamental domain for $$\Gamma$$ depends only on the signature of $$\Gamma$$ (which is a topological invariant), it is preserved under conjugation with elements in $$\mathrm{PSL}_{2}(\mathbb{R})$$.

For the group $$\left$$ we conjugate by $$q:z\longmapsto\frac{z}{-z+1}.$$ This takes the fixed points of $$g$$, $$0$$ and $$\infty$$, to $$0$$ resp. $$-1$$. Hence, the axis of $$\widetilde{g}= qgq^{-1}$$ becomes the geodesic $$\gamma$$ from $$0$$ to $$-1$$. The fixed point of $$\widetilde{h}= qhq^{-1}$$ becomes $$-\tfrac{1}{2}+\tfrac{\mathrm{i}}{2}$$, which is the summit of $$\gamma(\mathbb{R})$$. Since $$\widetilde{h}$$ is again an involution, its fixed point is the summit of its isometric sphere $$I(\widetilde{h})$$. Hence, $$I(\widetilde{h})$$ and $$\gamma(\mathbb{R})$$ coincide. Since $$\gamma$$ is the axis of $$\widetilde{g}$$, it meets $$I(\widetilde{g})$$ and $$I(\widetilde{g}^{-1})$$ at right angles. This shows that $$\mathcal{F}=\mathbb{H}\setminus\left(\overline{I(\widetilde{h})}\cup\overline{I(\widetilde{g})}\cup\overline{I(\widetilde{g})^{-1}}\right)$$ is a Poincaré polyhedron and thus, a fundamental domain for the group generated by its side pairing transformations, which is $$\left<\widetilde{g},\widetilde{h}\right>$$. Now $$\mathcal{F}$$ is easily seen to have infinite area.

For the cyclic groups $$\left$$ one argues type-wise: If $$g$$ is elliptic, then $$\left$$ is finite. Hence, it cannot be cofinite, since there is no hope that a finite collection of finite area sets tessellates the infinite area space $$\mathbb{H}$$. If $$g$$ is hyperbolic or parabolic, one again conjugates so that no fixed point of $$\left$$ equals infinity. The isometric spheres of $$g^{2},g^{3},\dots$$ resp. of $$g^{-2},g^{-3},\dots$$ are easily seen to be contained in the interior of $$I(g)$$ resp. $$I(g^{-1})$$. Again, one obtains a Poincaré polyhedron of infinite area.

We conclude that a cofinite group cannot be elementary and thus, it must contain hyperbolic elements.