# Almost nilpotent groups of hyperbolic isometries are cyclic

Let $$X$$ be a complete, simply connected, Riemannian manifold of negative curvature. Let $$\Gamma$$ be an almost nilpotent group of non-elliptic isometries containing a hyperbolic isometry $$\gamma$$. Then $$\Gamma$$ fixes a unique geodesic $$\ell$$ and is infinite cyclic.

It should follow from the fact that hyperbolic isometries fix a unique geodesic on each point of which they attain the minimum translation distance $$d_\gamma=\min_{x\in X}d(x,\gamma(x))$$.

I can prove the first part of the statement for nilpotent groups, because they have non-trivial center and if two isometries $$\gamma_1,\gamma_2$$ commute and one fixes a geodesic $$\ell=\gamma_1(\ell)$$, then also the other fixes it.

On the other hand I have some problem showing that $$\Gamma$$ is infinite cyclic. I was looking for counterexamples and came up with this: let $$\gamma_1$$ and $$\gamma_{\sqrt{2}}$$ two isometries of $$X$$ which translate the same geodesic $$\ell$$ of $$1$$ and $$\sqrt{2}$$. Their action on $$\ell$$ is the same of $$\mathbb{Z}^2$$. I know that there are no counterexamples, so I suspect that these two isometries generate a non-nilpotent group, and I would be able to detect it if I'd look at their action outside $$\ell$$.

So here are the questions: How can I prove the statement? Why the counterexample is in fact not a counterexample?

• You are missing the discreteness assumption. – Moishe Kohan Jan 9 at 21:38

In order to close this question. Yes, your example is a counter-example to the claim you are trying to prove. The key condition you are missing is that the group $$\Gamma$$ is discrete. Under this assumption the claim indeed holds. One proves this in three steps:
1. Suppose that $$g_1, g_2$$ are hyperbolic isometries which share exactly one fixed point $$p$$ in the sphere at infinity of $$X$$. Then $$[g_1, g_2]$$ is a parabolic isometry $$f$$ of $$X$$ fixing $$p$$.
2. If $$g$$ is a hyperbolic isometry of $$X$$ and $$f$$ is a parabolic isometry of $$X$$ such that $$g, f$$ have a common fixed point $$p$$ in the sphere at infinity of $$X$$, then either the sequence $$f_n:=g^n f g^{-n}$$ or $$f_{n}=g^{-n} f g^{n}$$ satisfies the property $$d(x, f_n(x))\to 0$$ for (any) $$x$$ on the axis of $$g$$. Thus, a discrete subgroup of isometries cannot contain two hyperbolic elements which share exactly one fixed points, cf. this question.
Step 3. If $$\Gamma$$ is almost nilpotent and contains non (nontrivial) elliptic elements then $$\Gamma$$ either has exactly one fixed point $$p$$ in the ideal boundary of $$X$$ or preserves a 2-point subset $$\{p, q\}$$ in this ideal boundary of $$X$$.
b. In the second case, $$\Gamma$$ will preserve the unique geodesic $$c\subset X$$ connecting $$p, q$$. Then, restricting $$\Gamma$$ to this geodesic we obtain an isomorphism of $$\Gamma$$ to a discrete isometry group of the real line. From this, you conclude that $$\Gamma$$ is cyclic (since it contains no elliptic elements).