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A well-known property on groups acting on trees is:

Theorem: Let $T$ be a tree and $g,h \in \mathrm{Isom}(T)$ two elliptic isometries. If $\mathrm{Fix}(g) \cap \mathrm{Fix}(h) = \emptyset$ then the product $gh$ is a loxodromic isometry.

I am pretty sure that a similar statement exists for $\delta$-hyperbolic spaces, where $\mathrm{Fix}$ is replaced with $\mathrm{Fix}_{\delta}$, but I did not succeed in remembering in which article I saw it...

Does somebody know a reference?

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  • $\begingroup$ The condition $\mathrm{Fix}(g)\cap\mathrm{Fix}(h)=\emptyset$ should be exchanged to something like $$2{\cdot}|{\rm Fix}(g)-{\rm Fix}(h)|_X\approx|{\rm Fix}(g)-h{\cdot}{\rm Fix}(g)|_X.$$ (I do not know a ref.) $\endgroup$
    – Anton Petrunin
    Jul 19, 2015 at 13:29
  • $\begingroup$ Dear Anton - Your condition does not work, because it is not symmetric in $g$ and $h$. For example, consider the $(2,p,\infty)$ triangle group. Let $g$ be the element of order $p$ and let $h$ be the element of order $2$. The point is that the "broken geodesic" needs to be more-or-less straight at all of its corners (not just at half of them). $\endgroup$
    – Sam Nead
    Jul 19, 2015 at 21:15
  • $\begingroup$ Also, if $g = h$ then your equality is satisfied, as both sides are zero... $\endgroup$
    – Sam Nead
    Jul 19, 2015 at 21:22
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    $\begingroup$ Dear Seirios - Could I ask you to spell out the intended definition of ${\rm Fix}_\delta$? $\endgroup$
    – Sam Nead
    Jul 20, 2015 at 18:45
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    $\begingroup$ For me, $\mathrm{Fix}_{\delta}(g) = \{ x \in X \mid d(x,gx) \leq \delta \}$. $\endgroup$
    – Seirios
    Jul 28, 2015 at 9:05

2 Answers 2

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Check Lemme 2.3 in chapter 9 of "Geometrie et theorie des groupes" by Coornaert, Delzant and Papadopoulos. It implies that your conclusion holds provided that the quasi-fixed point sets are sufficiently far apart.

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  • $\begingroup$ I don't find the implication obvious, but maybe I am missing the key point. How do you apply this lemma? $\endgroup$
    – Seirios
    Jul 23, 2015 at 8:56
  • $\begingroup$ Take $x$ to be the midpoint of a geodesic joining the quasi-fixed points sets of $g$ and $h$. If the sets are sufficientely far apart, then $|g(x)-x|$ and $|h(x)-x|$ are large (say $>20\delta$) as they are roughly double the distance between $x$ and the respecitve quasi-fixed point set. Moreover $2(g(x).h(x))_x+6\delta$ is small (say $<10\delta$) as $x$ is close to the geodesic between $g(x)$ and $h(x)$. Thus your claim follows using this choice of $x$. $\endgroup$
    – Richard Weidmann
    Jul 28, 2015 at 6:25
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This is false. Consider the triangle group $\Delta = \Delta(p,q,r)$ with presentation $$\langle a, b, c \mid a^p = b^q = c^r = abc = 1 \rangle.$$ So the product of $a$ and $b$ is elliptic. However, as $p, q, r \to \infty$ the fundamental domain (a triangle) has side-lengths tending to infinity.

If you replace the $\delta$ in ${\rm Fix}_\delta(a)$ by a distance depending on the order of $a$, you should win.

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  • $\begingroup$ I don't remember the exact statement, but I know there exists a similar statement. By the way, thank you for the example. $\endgroup$
    – Seirios
    Jul 19, 2015 at 12:57
  • $\begingroup$ As Richard's answer indicates, you seem to have misinterpreted $\mathrm{Fix}_\delta$ to mean a neighbourhood of the fixed point set, whereas it should actually be the approximate fixed point set. $\endgroup$
    – HJRW
    Jul 20, 2015 at 13:11
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    $\begingroup$ Ah - very good. That makes much more sense. Well, if this is the definition meant then that should be included in the original question??? $\endgroup$
    – Sam Nead
    Jul 20, 2015 at 18:08

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