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May 27, 2021 at 11:40 comment added NWMT I'm reading a bit more closely and the classification of isometries requires $X$ to be proper, but if you already know your isometries are parabolic, then I think a metric argument will work.
May 27, 2021 at 11:26 comment added NWMT I'm not sure such a result exists for arbitrary (i.e. non proper) Gromov hyperbolic spaces. There exists an argument that uses only metric properties of $\partial X$, but the source I have, Ghys, Étienne. "Sur les groupes hyperboliques d'après Mikhael Gromov." Progr. Math. 83 (1990), assumes $X$ is proper.
May 26, 2021 at 11:35 comment added NWMT Are you requiring that $X$ is a proper metric space or are you allowing something more exotic, like an $\mathbb R$-tree? If $X$ is a proper metric space (i.e. locally compact) then $\partial X$ is compact and metrizable and it's just a standard ping pong argument using disjoint neighbourhoods of the limit points, the fact that their closures are compact, and the definition of a parabolic element here: en.wikipedia.org/wiki/Convergence_group
May 5, 2021 at 19:33 comment added Richard Weidmann Not a real answer. Lemme 2.3 of the book of Coornaert, Delzant and Papadopoulos gives a criterion when the product of two non-hyperbolic isometries is hyperbolic. In you case this should prove that for sufficiently large $M$ the product $f^{\pm M}g^{\pm M}$ is hyperbolic. With a little bit of work you should be able to make this argument work for any freely reduced product in $f^M$ and $g^M$.
May 4, 2021 at 10:51 comment added YCor It seems to me that (3) follows from (1), in an arbitrary proper Gromov-hyperbolic space (in a proper metric space $X$ with isometry $f$, $(d(o,f^n(o)))_n$ is either bounded or tends to infinity.
May 4, 2021 at 10:49 history edited YCor
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May 4, 2021 at 10:44 history asked user203667 CC BY-SA 4.0