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?

  • 1
    $\begingroup$ You are missing the discreteness assumption. $\endgroup$ – 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 $X$.

a. Steps 1 and 2 show that the first case is impossible.

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.