2
$\begingroup$

Given a group $G$ acting on a $CAT(0)$ complex $X$ by isometries can $G$ contain a divisible element, i.e. an element $g\in G$ such that $\forall n\in\mathbb N$ there is $h\in G$ such that $g=h^n$.

$\endgroup$
  • $\begingroup$ You need to add the condition that the action is properly discontinuous and cocompact, as required by the definition of a CAT(0) group, otherwise there are easy counter-examples. $\endgroup$ – Moishe Kohan Sep 19 '16 at 19:00
1
$\begingroup$

Let $G$ be a group acting geometrically on a CAT(0) space $X$.

  • Show that there exists some $N \geq 0$ such that any torsion element of $G$ has order $\leq N$.
  • Deduce that $G$ cannot contain a divisible element of finite order.
  • Let $g \in G$ be an infinite order element. In particular, it must be loxodromic, ie., there exists a bi-infinite geodesic $\gamma$ on which it acts by translation. If $g=h^n$, then $h$ also acts on $\gamma$ by translation. Moreover, $\ell(g)=\ell(h)^n$, where $\ell(\cdot)$ denotes the translation length.
  • Show that, if $g$ is a divisible element and if we fix a point $x \in \gamma$, then there exist infinitely many elements $h \in G$ satifying $d(x,hx) \leq 1$. Hence a contradiction with the fact that $G$ acts geometrically on $X$.

In fact, the argument essentially holds in a more general setting, for instance for semihyperbolic groups.

$\endgroup$
  • $\begingroup$ why if $g$ has to be an infinite order element it has to be loxodromic? could it not be parabolic? $\endgroup$ – Spotty Oct 16 '16 at 19:14
  • $\begingroup$ If a group acts cocompactly on a CAT(0) space, then it contains no parabolic. This statement should be available in Bridson and Haefliger's book; I should be able to find a precise reference if needed. $\endgroup$ – Seirios Oct 17 '16 at 7:59

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.