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Let $X$ be a $\delta$-hyperbolic geodesic space. Then we have the following classification of isometries on $X$:

Theorem: Let $g$ be an isometry on $X$. Then, exactely one of the following case holds: (i) $g$ is elliptic: for every $x \in X$, $\{ g^n \cdot x : n \in \mathbb{Z} \}$ is bounded; (ii) $g$ is parabolic: the action induced by $g$ on the boundary $\partial X$ has exactly one fixed point; (iii) $g$ is hyperbolic: the action induced by $g$ on the boundary $\partial X$ has exactly two fixed point.

Now let $G$ be a non elementary (ie. neither finite nor virtually $\mathbb{Z}$) hyperbolic group. If $X$ is a Cayley graph of $G$ then $G$ has a natural isometric action on $X$ and:

Theorem: $G$ has no parabolic isometry.

I would like to generelize this theorem for other hyperbolic space $X$. For example, let $G$ be a non elementary hyperbolic group acting freely on a hyperbolic geodesic space $X$ by isometry; can $G$ contain a parabolic isometry? (Feel free to modify the hypotheses.)

Edit 1: We can add the following requirement (where $X$ is $\delta$-hyperbolic): for all $x \in X$, $B(x,n_0\delta) \cap G \cdot x$ contains at most $C_0$ points, where $n_0$ and $C_0$ doesn't depend on $x$.

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Yes, there can be parabolic isometries in your sense. Examples can be obtained by "cone constructions", e.g. as follows:

Let $\mathbb{H}^2$ be the upper half-plane and let $X = (0,\infty) \times \mathbb{H}^2$ with the Riemannian metric $(dt)^2 + e^{-2t}\frac{\lvert dz\rvert^2}{y^2}$, which is a $\operatorname{CAT}(-1)$-space and hence geodesic and $\delta$-hyperbolic. Then $\operatorname{PSL}(2,\mathbb{R})$ acts freely by isometries on $X$ while it fixes a point in the boundary.

Any non-elementary Gromov-hyperbolic subgroup of $\operatorname{SL}(2,\mathbb{R})$ will yield an example.

A rather silly sufficient condition is to assume properness and cocompactness of the action, because the Svarc-Milnor lemma reduces this (up to quasi-isometric equivalence) to the case you already know.

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  • $\begingroup$ Thank you for your example; I added a possible condition, I hope it is sufficient. $\endgroup$ – Seirios Feb 17 '13 at 12:41
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Gromov in his original 1987 book (Section 3.1) wrote a classification for arbitrary isometric group actions on hyperbolic spaces (with no further assumption) into 5 main classes. It goes at follows (the terminology is borrowed from here)

1: bounded: orbits are bounded

2: horocyclic: orbits are unbounded, $G$ acts with no hyperbolic isometry (hence there's a unique fixed boundary point)

3: lineal: there's a hyperbolic isometry and a fixed pair at infinity

4: focal: there's a hyperbolic isometry and a unique fixed point at infinity

5: general type: there's a hyperbolic isometry and no fixed point or pair at infinity.

According to Gromov, 1,2,3 are the elementary actions, i.e. the closure of an orbit in the boundary has at most 2 points, while in the non-elementary cases 4,5, this closure is uncountable.

If the action is cobounded, then case 2 (horocyclic action) can't occur. For the even more specific case of a discrete hyperbolic finitely generated group, case 4 (focal action) can't occur as well and Cases 1,3 are the virtually cyclic groups (finite and infinite).

On the other hand, all 5 cases occur for a group actions on trees, or on the hyperbolic plane (exercise).

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  • $\begingroup$ PS: the question is vaguely stated (especially if we're allowed to change the hypotheses). Anyway, a dense free subgroup of $SL_2(\mathbf{C})$ can probably act freely on the hyperbolic 3-space, with parabolic isometries. But I don't think hyperbolicity of the acting group, nor freeness of the action, are so relevant. The non-existence of parabolics is quite specific to cocompact proper actions of discrete groups. $\endgroup$ – YCor Feb 21 '13 at 21:12

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