non-proper parabolic isometries of hyperbolic spaces
$begingroup$
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
edited Jan 8 at 20:00
YCor
27.5k481134
asked Jan 8 at 19:45
Richard WeidmannRichard Weidmann
36118
$endgroup$
add a comment |
$begingroup$
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
edited Jan 8 at 20:00
YCor
27.5k481134
asked Jan 8 at 19:45
Richard WeidmannRichard Weidmann
36118
$endgroup$
$begingroup$
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).
$endgroup$
– YCor
Jan 8 at 21:41
add a comment |
$begingroup$
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
edited Jan 8 at 20:00
YCor
27.5k481134
asked Jan 8 at 19:45
Richard WeidmannRichard Weidmann
36118
$endgroup$
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
edited Jan 8 at 20:00
YCor
27.5k481134
asked Jan 8 at 19:45
Richard WeidmannRichard Weidmann
36118
edited Jan 8 at 20:00
YCor
27.5k481134
asked Jan 8 at 19:45
Richard WeidmannRichard Weidmann
36118
edited Jan 8 at 20:00
YCor
27.5k481134
edited Jan 8 at 20:00
YCor
27.5k481134
edited Jan 8 at 20:00
YCor
27.5k481134
27.5k481134
asked Jan 8 at 19:45
Richard WeidmannRichard Weidmann
36118
asked Jan 8 at 19:45
Richard WeidmannRichard Weidmann
36118
asked Jan 8 at 19:45
Richard WeidmannRichard Weidmann
36118
36118
$begingroup$
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).
$endgroup$
– YCor
Jan 8 at 21:41
add a comment |
$begingroup$
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).
$endgroup$
– YCor
Jan 8 at 21:41
$begingroup$
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).
$endgroup$
– YCor
Jan 8 at 21:41
$begingroup$
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).
$endgroup$
– YCor
Jan 8 at 21:41
add a comment |
1 Answer
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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 $mathbf{H}^infty(V)$.
answered Jan 8 at 20:03
YCorYCor
27.5k481134
$endgroup$
$begingroup$
Thank you, Yves.
$endgroup$
– Richard Weidmann
Jan 8 at 20:21
add a comment |
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$begingroup$
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 $mathbf{H}^infty(V)$.
answered Jan 8 at 20:03
YCorYCor
27.5k481134
$endgroup$
$begingroup$
Thank you, Yves.
$endgroup$
– Richard Weidmann
Jan 8 at 20:21
add a comment |
$begingroup$
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 $mathbf{H}^infty(V)$.
answered Jan 8 at 20:03
YCorYCor
27.5k481134
$endgroup$
$begingroup$
Thank you, Yves.
$endgroup$
– Richard Weidmann
Jan 8 at 20:21
add a comment |
$begingroup$
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 $mathbf{H}^infty(V)$.
answered Jan 8 at 20:03
YCorYCor
27.5k481134
$endgroup$
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 $mathbf{H}^infty(V)$.
answered Jan 8 at 20:03
YCorYCor
27.5k481134
answered Jan 8 at 20:03
YCorYCor
27.5k481134
answered Jan 8 at 20:03
YCorYCor
27.5k481134
answered Jan 8 at 20:03
YCorYCor
27.5k481134
27.5k481134
$begingroup$
Thank you, Yves.
$endgroup$
– Richard Weidmann
Jan 8 at 20:21
add a comment |
$begingroup$
Thank you, Yves.
$endgroup$
– Richard Weidmann
Jan 8 at 20:21
$begingroup$
Thank you, Yves.
$endgroup$
– Richard Weidmann
Jan 8 at 20:21
$begingroup$
Thank you, Yves.
$endgroup$
– Richard Weidmann
Jan 8 at 20:21
add a comment |
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$begingroup$
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).
$endgroup$
– YCor
Jan 8 at 21:41