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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!

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

2 Answers 2

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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.

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  • $\begingroup$ Thanks a lot! This is very neat! $\endgroup$
    – abrewer
    Jun 17, 2019 at 18:39
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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 $\left<g,h\right>$ 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 $\left<g\right>$ one argues type-wise: If $g$ is elliptic, then $\left<g\right>$ 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 $\left<g\right>$ 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.

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