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If $f$ and $g$ are two hyperbolic isometries of the hyperbolic space $\mathcal H^n$, we know that $f$ has 2 fixed points on $\partial \mathcal H^n$ and similiarly for $g$.

Is it true that $f$ and $g$ generate a non-abelian free group if $\mathrm{fix}(f)\cap \mathrm{fix}(g) = \varnothing$ ? If so, could you provide a reference for this fact ?

paf
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1 Answers1

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That is not true.

A simple example is the following one: consider the hyperbolic isometry

\begin{equation*} \phi:= \left[ \begin{array}{l l} 4 &0\\ 0 &\frac{1}{4}\\ \end{array} \right] \in\text{PSL}(2,\mathbb{R})=\text{Isom}^+\left(\mathbb{H}^2\right) \end{equation*}

and the finite order elliptic element

\begin{equation*} \rho:= \left[ \begin{array}{l l} \frac{1}{2} &-\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} &\frac{1}{2}\\ \end{array} \right] \in\text{PSL}(2,\mathbb{R})=\text{Isom}^+\left(\mathbb{H}^2\right) \end{equation*}

with $\rho^6=1$. The product

\begin{equation*} \psi:=\phi\rho= \left[ \begin{array}{l l} 4 &0\\ 0 &\frac{1}{4}\\ \end{array} \right] \cdot \left[ \begin{array}{l l} \frac{1}{2} &-\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} &\frac{1}{2}\\ \end{array} \right] = \left[ \begin{array}{l l} 2 &-2\sqrt{3}\\ \frac{\sqrt{3}}{8} &\frac{1}{8}\\ \end{array} \right] \end{equation*}

is again hyperbolic as $\text{tr}(\phi\rho)=2+\frac{1}{8}>2$. Moreover it is easy to see that its fixed points are disjoint from those of $\phi$.

The subgroup $\Gamma:=\langle\phi,\psi\rangle$ cannot be a non-abelian free group as it contains the torsion element $\rho$.

Gabriele
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    Thanks. Do you know if there is a condition (necessary and/or sufficient) on $f$ and $g$ for which $\langle f,g\rangle$ is a non-abelian free group ? – paf Jul 26 '16 at 15:58
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    If you assume that $\text{Fix}(f)\cap\text{Fix}(g)=\emptyset$ and that $\langle f,g\rangle<\text{Isom}^+\left(\mathbb{H}^n\right)$ is discrete then for sufficiently large $p,q\in\mathbb{N}$ the subgroup $\langle f^p,g^q\rangle$ is free non-abelian. Again the sole discreteness is not a sufficient condition (if you want I can give you an easy example). – Gabriele Jul 27 '16 at 09:45
  • I'm interested in your example. Moreover, do you have a reference for the first fact ? – paf Jul 27 '16 at 10:08
  • Reference: you can see the section "word-hyperbolic groups" of the Wikipedia page on the Ping-pong Lemma https://en.wikipedia.org/wiki/Ping-pong_lemma – Gabriele Jul 27 '16 at 10:13
  • Example: consider the torsion element $\rho=\left[\begin{array}{l l} \omega &0\ 0 &\omega^{-1} \end{array}\right]\in\text{PSL}(2,\mathbb{C})$ where $\omega=\frac{1}{2}+\frac{\sqrt{3}}{2}i$ satisfies $\omega^6=1$. Consider then the hyperbolic isometry $\phi=\left[\begin{array}{l l} 5 &2i\ -2i &1 \end{array}\right]\in\text{PSL}(2,\mathbb{C})$. As before the product $\psi=\rho\phi$ is hyperbolic (compute the trace). The subgroup $\langle\psi,\phi\rangle$ is discrete (it lies in $\text{PSL}(2,\mathbb{Z}[\omega])$) and is not free since it contains $\rho$. – Gabriele Jul 27 '16 at 10:32
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    For the statement in the comment of @Gabriele regarding freeness of $\langle f^p, g^q \rangle$ for large $p,q$, the key words (besides "ping pong lemma") are "Tits alternative", named after the mathematician Jacques Tits. – Lee Mosher Jul 28 '16 at 22:57