non-proper parabolic isometries of hyperbolic spaces












4














In his seminal paper on hyperbolic groups (see Section 8.1) Gromov defines an isometry $f$ of a hyperbolic space $X$ to be parabolic if the orbit of any point $xin X$ under the action of $langle frangle$ has a unique accumulation point $a$ in the boundary $partial X$. He then goes on to call $f$ proper if for any sequence $(z_i)$ with $lim |z_i|=infty$ the sequence $(f^{z_i}(x))$ represents $a$. I don't think proper parabolic isometries are discussed in later sections of the paper.



In the book of Ghys, de la Harpe et al (see Chapitre 8, Theorem 17 and Lemma 18) it is shown that any parabolic isometry of a proper hyperbolic space is proper. The proof uses the assumption of properness in an essential way.



My question is: Are there non-proper hyperbolic spaces that have non-proper parabolic isometries.






gr.group-theory gt.geometric-topology mg.metric-geometry geometric-group-theory







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  • Actually every isometry of a nonempty proper metric space is either bounded (some/every orbits is bounded) or proper (in either the topological or metric sense). This follows from the fact that in a locally compact group, every cyclic subgroup either has compact closure, or is (closed and infinite).
    – YCor
    8 hours ago
















4














In his seminal paper on hyperbolic groups (see Section 8.1) Gromov defines an isometry $f$ of a hyperbolic space $X$ to be parabolic if the orbit of any point $xin X$ under the action of $langle frangle$ has a unique accumulation point $a$ in the boundary $partial X$. He then goes on to call $f$ proper if for any sequence $(z_i)$ with $lim |z_i|=infty$ the sequence $(f^{z_i}(x))$ represents $a$. I don't think proper parabolic isometries are discussed in later sections of the paper.



In the book of Ghys, de la Harpe et al (see Chapitre 8, Theorem 17 and Lemma 18) it is shown that any parabolic isometry of a proper hyperbolic space is proper. The proof uses the assumption of properness in an essential way.



My question is: Are there non-proper hyperbolic spaces that have non-proper parabolic isometries.






gr.group-theory gt.geometric-topology mg.metric-geometry geometric-group-theory







share|cite|improve this question
























  • Actually every isometry of a nonempty proper metric space is either bounded (some/every orbits is bounded) or proper (in either the topological or metric sense). This follows from the fact that in a locally compact group, every cyclic subgroup either has compact closure, or is (closed and infinite).
    – YCor
    8 hours ago














4












4








4







In his seminal paper on hyperbolic groups (see Section 8.1) Gromov defines an isometry $f$ of a hyperbolic space $X$ to be parabolic if the orbit of any point $xin X$ under the action of $langle frangle$ has a unique accumulation point $a$ in the boundary $partial X$. He then goes on to call $f$ proper if for any sequence $(z_i)$ with $lim |z_i|=infty$ the sequence $(f^{z_i}(x))$ represents $a$. I don't think proper parabolic isometries are discussed in later sections of the paper.



In the book of Ghys, de la Harpe et al (see Chapitre 8, Theorem 17 and Lemma 18) it is shown that any parabolic isometry of a proper hyperbolic space is proper. The proof uses the assumption of properness in an essential way.



My question is: Are there non-proper hyperbolic spaces that have non-proper parabolic isometries.






gr.group-theory gt.geometric-topology mg.metric-geometry geometric-group-theory







share|cite|improve this question















In his seminal paper on hyperbolic groups (see Section 8.1) Gromov defines an isometry $f$ of a hyperbolic space $X$ to be parabolic if the orbit of any point $xin X$ under the action of $langle frangle$ has a unique accumulation point $a$ in the boundary $partial X$. He then goes on to call $f$ proper if for any sequence $(z_i)$ with $lim |z_i|=infty$ the sequence $(f^{z_i}(x))$ represents $a$. I don't think proper parabolic isometries are discussed in later sections of the paper.



In the book of Ghys, de la Harpe et al (see Chapitre 8, Theorem 17 and Lemma 18) it is shown that any parabolic isometry of a proper hyperbolic space is proper. The proof uses the assumption of properness in an essential way.



My question is: Are there non-proper hyperbolic spaces that have non-proper parabolic isometries.






gr.group-theory gt.geometric-topology mg.metric-geometry geometric-group-theory




gr.group-theory gt.geometric-topology mg.metric-geometry geometric-group-theory






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edited 9 hours ago









YCor

27.2k480132




27.2k480132










asked 10 hours ago









Richard WeidmannRichard Weidmann

36118




36118












  • Actually every isometry of a nonempty proper metric space is either bounded (some/every orbits is bounded) or proper (in either the topological or metric sense). This follows from the fact that in a locally compact group, every cyclic subgroup either has compact closure, or is (closed and infinite).
    – YCor
    8 hours ago


















  • Actually every isometry of a nonempty proper metric space is either bounded (some/every orbits is bounded) or proper (in either the topological or metric sense). This follows from the fact that in a locally compact group, every cyclic subgroup either has compact closure, or is (closed and infinite).
    – YCor
    8 hours ago
















Actually every isometry of a nonempty proper metric space is either bounded (some/every orbits is bounded) or proper (in either the topological or metric sense). This follows from the fact that in a locally compact group, every cyclic subgroup either has compact closure, or is (closed and infinite).
– YCor
8 hours ago




Actually every isometry of a nonempty proper metric space is either bounded (some/every orbits is bounded) or proper (in either the topological or metric sense). This follows from the fact that in a locally compact group, every cyclic subgroup either has compact closure, or is (closed and infinite).
– YCor
8 hours ago










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Yes. Let me call this "metrically proper" to avoid ambiguity.



Indeed, Edelstein (On non-expansive mappings of Banach spaces, Proc.
Cambridge Philos. Soc. 60 (1964), 439-447) observed that there infinite-dimensional Hilbert spaces admit unbounded isometries that are not metrically proper. This is not Gromov-hyperbolic, but every Hilbert space $V$ embeds (as a horosphere) into a CAT($-1$) space $mathbf{H}^infty(V)$, called infinite hyperbolic space (here "hyperbolic" is meant in the sense of Poincaré/Lobachevski), and this embedding is equivariant for isometries (each isometry has a canonical isometric extension). This embedding is not isometric (infinitesimally it is), but still it is proper. So a non-proper unbounded isometry of $V$ extends to a non-proper isometry of $mathbf{H}^infty(V)$.






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  • Thank you, Yves.
    – Richard Weidmann
    9 hours ago











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Yes. Let me call this "metrically proper" to avoid ambiguity.



Indeed, Edelstein (On non-expansive mappings of Banach spaces, Proc.
Cambridge Philos. Soc. 60 (1964), 439-447) observed that there infinite-dimensional Hilbert spaces admit unbounded isometries that are not metrically proper. This is not Gromov-hyperbolic, but every Hilbert space $V$ embeds (as a horosphere) into a CAT($-1$) space $mathbf{H}^infty(V)$, called infinite hyperbolic space (here "hyperbolic" is meant in the sense of Poincaré/Lobachevski), and this embedding is equivariant for isometries (each isometry has a canonical isometric extension). This embedding is not isometric (infinitesimally it is), but still it is proper. So a non-proper unbounded isometry of $V$ extends to a non-proper isometry of $mathbf{H}^infty(V)$.






share|cite|improve this answer





















  • Thank you, Yves.
    – Richard Weidmann
    9 hours ago
















4














Yes. Let me call this "metrically proper" to avoid ambiguity.



Indeed, Edelstein (On non-expansive mappings of Banach spaces, Proc.
Cambridge Philos. Soc. 60 (1964), 439-447) observed that there infinite-dimensional Hilbert spaces admit unbounded isometries that are not metrically proper. This is not Gromov-hyperbolic, but every Hilbert space $V$ embeds (as a horosphere) into a CAT($-1$) space $mathbf{H}^infty(V)$, called infinite hyperbolic space (here "hyperbolic" is meant in the sense of Poincaré/Lobachevski), and this embedding is equivariant for isometries (each isometry has a canonical isometric extension). This embedding is not isometric (infinitesimally it is), but still it is proper. So a non-proper unbounded isometry of $V$ extends to a non-proper isometry of $mathbf{H}^infty(V)$.






share|cite|improve this answer





















  • Thank you, Yves.
    – Richard Weidmann
    9 hours ago














4












4








4






Yes. Let me call this "metrically proper" to avoid ambiguity.



Indeed, Edelstein (On non-expansive mappings of Banach spaces, Proc.
Cambridge Philos. Soc. 60 (1964), 439-447) observed that there infinite-dimensional Hilbert spaces admit unbounded isometries that are not metrically proper. This is not Gromov-hyperbolic, but every Hilbert space $V$ embeds (as a horosphere) into a CAT($-1$) space $mathbf{H}^infty(V)$, called infinite hyperbolic space (here "hyperbolic" is meant in the sense of Poincaré/Lobachevski), and this embedding is equivariant for isometries (each isometry has a canonical isometric extension). This embedding is not isometric (infinitesimally it is), but still it is proper. So a non-proper unbounded isometry of $V$ extends to a non-proper isometry of $mathbf{H}^infty(V)$.






share|cite|improve this answer












Yes. Let me call this "metrically proper" to avoid ambiguity.



Indeed, Edelstein (On non-expansive mappings of Banach spaces, Proc.
Cambridge Philos. Soc. 60 (1964), 439-447) observed that there infinite-dimensional Hilbert spaces admit unbounded isometries that are not metrically proper. This is not Gromov-hyperbolic, but every Hilbert space $V$ embeds (as a horosphere) into a CAT($-1$) space $mathbf{H}^infty(V)$, called infinite hyperbolic space (here "hyperbolic" is meant in the sense of Poincaré/Lobachevski), and this embedding is equivariant for isometries (each isometry has a canonical isometric extension). This embedding is not isometric (infinitesimally it is), but still it is proper. So a non-proper unbounded isometry of $V$ extends to a non-proper isometry of $mathbf{H}^infty(V)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 9 hours ago









YCorYCor

27.2k480132




27.2k480132












  • Thank you, Yves.
    – Richard Weidmann
    9 hours ago


















  • Thank you, Yves.
    – Richard Weidmann
    9 hours ago
















Thank you, Yves.
– Richard Weidmann
9 hours ago




Thank you, Yves.
– Richard Weidmann
9 hours ago


















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