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For a group $G$ with a non-elementary general type action by isometries on a Gromov hyperbolic geodesic space $(X,d)$, it is well known that you can construct free subgroups of $G$ via the ping pong argument with (sufficiently large powers of) two hyperbolic elements with distinct fixed points in $\partial X$.

In this question I ask for a reference for the case of ping pong with two parabolic isometries. This is perhaps well known in the case of $X=\mathbb{H}^2$, but I am interested in general Gromov hyperbolic spaces. I think the following should be true, but ideally I would like to cite it. Does anyone know of such a reference?

Let $G$ be a group acting by isometries (not necessarily properly) on a Gromov hyperbolic geodesic metric space $(X,d)$. Fix a basepoint $o\in X$.

Suppose

  1. that $f,g\in G$ both induce parabolic isometries on $(X,d)$,
  2. the fixed points of $f$ and $g$ in $\partial X$ are distinct, and
  3. for any $B>0$ there exists $N>0$ such that for all $n>N$ we have $d(o,f^n o),d(o,g^n o) >B$.

Then for sufficiently large $M$ we have that $f^M$ and $g^M$ generate a free group $F$ of rank two, and, any non-trivial $h\in F$ is either hyperbolic on $(X,d)$ or $h$ is conjugate (in $F$) to a power of $f$ or $g$.

A reference would be ideal and greatly appreciated. Thank you.

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    $\begingroup$ 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. $\endgroup$
    – YCor
    May 4, 2021 at 10:51
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    $\begingroup$ 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 $g^M$. $\endgroup$
    – Richard Weidmann
    May 5, 2021 at 19:33
  • $\begingroup$ 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 $\endgroup$
    – NWMT
    May 26, 2021 at 11:35
  • $\begingroup$ 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. $\endgroup$
    – NWMT
    May 27, 2021 at 11:26
  • $\begingroup$ 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. $\endgroup$
    – NWMT
    May 27, 2021 at 11:40

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