# Is the group generated by two loxodromic isometries with a fixed point in common cocompact?

If you have two distinct loxodromic isometries of the hyperbolic plane $\gamma_1, \gamma_2$ such that they have a fixed point in common. For simplicity let's take the half plane model and let the fixed point in common be $\infty$. Is the group $\langle \gamma_1,\gamma_2\rangle$ cocompact? Is it even discrete? I can show that if the translation length $|\gamma_1| = |\gamma_2|$ then the group contains a parabolic element, so either it's not cocompact, or it's not proper. How horrible can this group be? I'd like to be able to prove that it's not cocompact, but I'm starting to suspect that it might be.

• I would guess that the group is not discrete. The idea is to think of the rays fixed by the isometries, take a point, push it out to a different ray with one isometry, then pull it back with the other one. Doing this, in general, you may be able to show that the orbit of a point returns arbitrarily close to it? – Neal Nov 14 '14 at 4:18
• If I'm understanding what you're saying correctly what you've described is actually a parabolic isometry. Provided that the translation lengths of the two isometries are the same. I'm still trying to explore what happens when the translation lengths are different. The pushing and pulling gives your a new hyperbolic isometry, but with a translation length that is determined by the ratio of the original two. Even if it isn't discrete though I don't think that helps me much with understanding if it's cocompact or not. – Devin Murray Nov 14 '14 at 17:24
• What is your definition of cocompact? Are you using a definition which applies to the indiscrete case as well as to the discrete case? – Lee Mosher Nov 17 '14 at 13:37
• That's a good question. I've been using the definition that says that H/gamma is compact, but I've also been playing with whether or not it's co-bounded as well. I'd be thrilled to be able to show either. – Devin Murray Nov 20 '14 at 23:42

## 1 Answer

The group $\Gamma = \langle \gamma_1,\gamma_2\rangle$ is not discrete.

To prove this, consider the "height map" $\mathbb{H}^2 \to \mathbb{R}$ defined by $f(x,y) = \ln(y)$. The inverse images of points under this map are the horocycles $y = \text{(constant)}$ all based at $\infty$, each of which is preserved by both $\gamma_1$ and $\gamma_2$, and therefore there is an induced action of $\Gamma$ on $\mathbb{R}$, the height displacement''. Let $\delta_i$ denote the height displacement of $\gamma_i$. The "height displacement" subgroup $\langle\delta_1,\delta_2\rangle$ of $\mathbb{R}$ may be discrete (infinite cyclic) or indiscrete (abelian of rank 2). In either case, given $\epsilon>0$ there exist nonzero integers $m_1,m_2$ such that the absolute value of $\Delta_y = m_1\delta_1 + m_2 \delta_2$ is less than $\epsilon/2$ (in the case that the height displacement group is cyclic, one can arrange that $\Delta_y=0$).

Consider now the points $p = 0+1i$ in $\mathbb{H^2}$ and $q = \gamma_1^{m_1} \gamma_2^{m_2}(p)$, which are in the same orbit. Their vertical displacement is $\Delta_y < \epsilon/2$. Let $\Delta_x$ be the difference of their $x$-coordinates, which might still be large. As $k \to +\infty$ the points $\gamma_1^k(p)$, $\gamma_2^k(p)$ are connected by concatenation of a horizontal segment whose length $<C \Delta_x e^{-k}$ goes to zero, followed by a vertical segment of length $\Delta_y <\epsilon/2$, and so $d(\gamma_1^k(p),\gamma_2^k(q)) <\epsilon$ for large enough $k$. This shows that $\Gamma$ is not discrete.

On the other hand, $\Gamma$ is cocompact in the strong sense that there exists a compact subset $B \subset \mathbb{H}^2$ whose translates $\{\gamma B \mid \gamma \in \Gamma\}$ cover $\mathbb{H}^2$. One can choose this set $B$ to be a "horobrick" whose appearance is an $xy$-rectangle in the upper half plane, whose two vertical sides are hyperbolic geodesic segments and whose two horizontal sides are horocylic segments. Simply choose $B$ to have bottom left corner containing $p$ and top right corner containing $\gamma_2(q)$ (assuming $q$ has larger $x$-coordinate than $p$).

• Your proof of the non-discreteness is much cleaner than the one I arrived at. By any chance is the $q$ you defined supposed to be $\gamma^{m_1}_1\gamma^{m_2}_2(p)$? And the distance we are measuring as $k \to \infty$ is between $\gamma_1^k(p)$ and $\gamma_2^k(q)$ is it not? I might be misunderstanding something here if that is not the case. – Devin Murray Nov 25 '14 at 17:36
• @DevinMurray: Yes, those were typos. Thanks for the corrections. – Lee Mosher Nov 26 '14 at 0:11
• How do you guarantee the existence of ${m_1}$ and ${m_2}$? Thank you. – WhySee Mar 23 '18 at 19:08