Does every locally compact connected homogeneous metric space admit a vertex-transitive 'grid'?Is every locally compact connected homogeneous metric space a manifold cross a continuum?Uniform Embedding into Euclidean SpaceIs there a compact connected Hausdorff space in which every non-empty $G_delta$ set has non-empty interior?Coordinate chart of concave functions near a regular point in Alexandrov spacesCovering numbers of uniformly bounded subsets of Gromov-Hausdorff spaceRandom walk uniformly hitting a compact setFunction as sum of distances over a connected, compact metric space
Does every locally compact connected homogeneous metric space admit a vertex-transitive 'grid'?
Is every locally compact connected homogeneous metric space a manifold cross a continuum?Uniform Embedding into Euclidean SpaceIs there a compact connected Hausdorff space in which every non-empty $G_delta$ set has non-empty interior?Coordinate chart of concave functions near a regular point in Alexandrov spacesCovering numbers of uniformly bounded subsets of Gromov-Hausdorff spaceRandom walk uniformly hitting a compact setFunction as sum of distances over a connected, compact metric space
$begingroup$
This is a followup to this easier version of this question on MSE, which Lee Mosher answered in the positive in the special case that $X$ is a hyperbolic space. It's also vaguely related to this question.
Suppose that $(X,d)$ is a locally compact connected homogeneous metric space, where by homogeneous I mean that for any $x_0,x_1 in X$ there exists an isometry $f:Xrightarrow X$ such that $f(x_0)=x_1$. Does there always exist a set $Ysubseteq X$ with the following properties?
$Y$ is uniformly discrete, i.e. there is an $varepsilon > 0$ such that for any distinct $x,yin Y$, $d(x,y) > varepsilon$.
$Y$ uniformly compactly covers $X$, i.e. for some $x in Y$ there is a compact set $K ni x$ such that translates of $K$ under $f in mathrmAut(X) : f(Y) = Y$ cover all of $X$.
$Y$ is vertex-transitive, i.e. for any $x,yin Y$ there is a isometry $f:X rightarrow X$ such that $y=f(x)$ and $f(Y)=Y$.
EDIT: I realized I hadn't captured what I wanted with the second bullet point.
gt.geometric-topology mg.metric-geometry topological-groups tiling
edited 10 hours ago
YCor
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asked 11 hours ago
James HansonJames Hanson
2,0287 silver badges26 bronze badges
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add a comment
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$begingroup$
This is a followup to this easier version of this question on MSE, which Lee Mosher answered in the positive in the special case that $X$ is a hyperbolic space. It's also vaguely related to this question.
Suppose that $(X,d)$ is a locally compact connected homogeneous metric space, where by homogeneous I mean that for any $x_0,x_1 in X$ there exists an isometry $f:Xrightarrow X$ such that $f(x_0)=x_1$. Does there always exist a set $Ysubseteq X$ with the following properties?
$Y$ is uniformly discrete, i.e. there is an $varepsilon > 0$ such that for any distinct $x,yin Y$, $d(x,y) > varepsilon$.
$Y$ uniformly compactly covers $X$, i.e. for some $x in Y$ there is a compact set $K ni x$ such that translates of $K$ under $f in mathrmAut(X) : f(Y) = Y$ cover all of $X$.
$Y$ is vertex-transitive, i.e. for any $x,yin Y$ there is a isometry $f:X rightarrow X$ such that $y=f(x)$ and $f(Y)=Y$.
EDIT: I realized I hadn't captured what I wanted with the second bullet point.
gt.geometric-topology mg.metric-geometry topological-groups tiling
edited 10 hours ago
YCor
31.8k4 gold badges96 silver badges148 bronze badges
asked 11 hours ago
James HansonJames Hanson
2,0287 silver badges26 bronze badges
$endgroup$
$begingroup$
Your assumptions are equivalent to that there exists a cocompact lattice in the isometry group of $X$. (The setwise-stabilizer of $Y$ in $Isom(X)$.) This is already false for some 3-dimensional homogeneous manifolds (solvable simply connected Lie groups).
$endgroup$
– Misha
5 hours ago
add a comment
|
$begingroup$
This is a followup to this easier version of this question on MSE, which Lee Mosher answered in the positive in the special case that $X$ is a hyperbolic space. It's also vaguely related to this question.
Suppose that $(X,d)$ is a locally compact connected homogeneous metric space, where by homogeneous I mean that for any $x_0,x_1 in X$ there exists an isometry $f:Xrightarrow X$ such that $f(x_0)=x_1$. Does there always exist a set $Ysubseteq X$ with the following properties?
$Y$ is uniformly discrete, i.e. there is an $varepsilon > 0$ such that for any distinct $x,yin Y$, $d(x,y) > varepsilon$.
$Y$ uniformly compactly covers $X$, i.e. for some $x in Y$ there is a compact set $K ni x$ such that translates of $K$ under $f in mathrmAut(X) : f(Y) = Y$ cover all of $X$.
$Y$ is vertex-transitive, i.e. for any $x,yin Y$ there is a isometry $f:X rightarrow X$ such that $y=f(x)$ and $f(Y)=Y$.
EDIT: I realized I hadn't captured what I wanted with the second bullet point.
gt.geometric-topology mg.metric-geometry topological-groups tiling
edited 10 hours ago
YCor
31.8k4 gold badges96 silver badges148 bronze badges
asked 11 hours ago
James HansonJames Hanson
2,0287 silver badges26 bronze badges
$endgroup$
This is a followup to this easier version of this question on MSE, which Lee Mosher answered in the positive in the special case that $X$ is a hyperbolic space. It's also vaguely related to this question.
Suppose that $(X,d)$ is a locally compact connected homogeneous metric space, where by homogeneous I mean that for any $x_0,x_1 in X$ there exists an isometry $f:Xrightarrow X$ such that $f(x_0)=x_1$. Does there always exist a set $Ysubseteq X$ with the following properties?
$Y$ is uniformly discrete, i.e. there is an $varepsilon > 0$ such that for any distinct $x,yin Y$, $d(x,y) > varepsilon$.
$Y$ uniformly compactly covers $X$, i.e. for some $x in Y$ there is a compact set $K ni x$ such that translates of $K$ under $f in mathrmAut(X) : f(Y) = Y$ cover all of $X$.
$Y$ is vertex-transitive, i.e. for any $x,yin Y$ there is a isometry $f:X rightarrow X$ such that $y=f(x)$ and $f(Y)=Y$.
EDIT: I realized I hadn't captured what I wanted with the second bullet point.
gt.geometric-topology mg.metric-geometry topological-groups tiling
gt.geometric-topology mg.metric-geometry topological-groups tiling
edited 10 hours ago
YCor
31.8k4 gold badges96 silver badges148 bronze badges
asked 11 hours ago
James HansonJames Hanson
2,0287 silver badges26 bronze badges
edited 10 hours ago
YCor
31.8k4 gold badges96 silver badges148 bronze badges
asked 11 hours ago
James HansonJames Hanson
2,0287 silver badges26 bronze badges
edited 10 hours ago
YCor
31.8k4 gold badges96 silver badges148 bronze badges
edited 10 hours ago
YCor
31.8k4 gold badges96 silver badges148 bronze badges
edited 10 hours ago
YCor
31.8k4 gold badges96 silver badges148 bronze badges
31.8k4 gold badges96 silver badges148 bronze badges
asked 11 hours ago
James HansonJames Hanson
2,0287 silver badges26 bronze badges
asked 11 hours ago
James HansonJames Hanson
2,0287 silver badges26 bronze badges
asked 11 hours ago
James HansonJames Hanson
2,0287 silver badges26 bronze badges
2,0287 silver badges26 bronze badges
$begingroup$
Your assumptions are equivalent to that there exists a cocompact lattice in the isometry group of $X$. (The setwise-stabilizer of $Y$ in $Isom(X)$.) This is already false for some 3-dimensional homogeneous manifolds (solvable simply connected Lie groups).
$endgroup$
– Misha
5 hours ago
add a comment
|
$begingroup$
Your assumptions are equivalent to that there exists a cocompact lattice in the isometry group of $X$. (The setwise-stabilizer of $Y$ in $Isom(X)$.) This is already false for some 3-dimensional homogeneous manifolds (solvable simply connected Lie groups).
$endgroup$
– Misha
5 hours ago
$begingroup$
Your assumptions are equivalent to that there exists a cocompact lattice in the isometry group of $X$. (The setwise-stabilizer of $Y$ in $Isom(X)$.) This is already false for some 3-dimensional homogeneous manifolds (solvable simply connected Lie groups).
$endgroup$
– Misha
5 hours ago
$begingroup$
Your assumptions are equivalent to that there exists a cocompact lattice in the isometry group of $X$. (The setwise-stabilizer of $Y$ in $Isom(X)$.) This is already false for some 3-dimensional homogeneous manifolds (solvable simply connected Lie groups).
$endgroup$
– Misha
5 hours ago
add a comment
|
1 Answer
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$begingroup$
The answer is no. A quick answer can be done as follows:
(1) Pansu proved (1989) that two Carnot Lie groups are quasi-isometric if and only if they are isomorphic.
(2) There exists continuum many non-isomorphic 7-dimensional Carnot Lie groups.
If $Y$ is a proper, uniformly discrete and isometry-transitive metric space, then $Y$ is QI to its isometry group $G$, which is locally compact. Assuming it QI to a nilpotent Lie group in addition implies (by results of Gromov/Losert/Trofimov) that $G$, modulo a compact normal subgroup, is discrete and virtually nilpotent. There are only countably many QI classes. Hence at least one of the examples in (2) yields (for every choice of left-invariant Riemannian metric) a negative answer to your question.
There is an alternative using hyperbolicity rather polynomial growth, consisting of a continuum family of negatively curved homogeneous 3-folds. That they're not QI is also a result of Pansu. That a totally locally compact group QI to it has to be compact-by-discrete is an immediate consequence of the 2-dimensional Hilbert-Smith conjecture (which is an old theorem).
answered 10 hours ago
YCorYCor
31.8k4 gold badges96 silver badges148 bronze badges
$endgroup$
$begingroup$
I have a question regarding the sentence"If $Y$ is proper, uniformly discrete...". Is this a fact about Carnot Lie groups, Lie groups in general, homogeneous manifolds in general, or locally compact homogeneous metric spaces in general?
$endgroup$
– James Hanson
10 hours ago
$begingroup$
Changed "it's" to "$Y$". It's about arbitrary metric spaces.
$endgroup$
– YCor
9 hours ago
add a comment
|
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$begingroup$
The answer is no. A quick answer can be done as follows:
(1) Pansu proved (1989) that two Carnot Lie groups are quasi-isometric if and only if they are isomorphic.
(2) There exists continuum many non-isomorphic 7-dimensional Carnot Lie groups.
If $Y$ is a proper, uniformly discrete and isometry-transitive metric space, then $Y$ is QI to its isometry group $G$, which is locally compact. Assuming it QI to a nilpotent Lie group in addition implies (by results of Gromov/Losert/Trofimov) that $G$, modulo a compact normal subgroup, is discrete and virtually nilpotent. There are only countably many QI classes. Hence at least one of the examples in (2) yields (for every choice of left-invariant Riemannian metric) a negative answer to your question.
There is an alternative using hyperbolicity rather polynomial growth, consisting of a continuum family of negatively curved homogeneous 3-folds. That they're not QI is also a result of Pansu. That a totally locally compact group QI to it has to be compact-by-discrete is an immediate consequence of the 2-dimensional Hilbert-Smith conjecture (which is an old theorem).
answered 10 hours ago
YCorYCor
31.8k4 gold badges96 silver badges148 bronze badges
$endgroup$
$begingroup$
I have a question regarding the sentence"If $Y$ is proper, uniformly discrete...". Is this a fact about Carnot Lie groups, Lie groups in general, homogeneous manifolds in general, or locally compact homogeneous metric spaces in general?
$endgroup$
– James Hanson
10 hours ago
$begingroup$
Changed "it's" to "$Y$". It's about arbitrary metric spaces.
$endgroup$
– YCor
9 hours ago
add a comment
|
$begingroup$
The answer is no. A quick answer can be done as follows:
(1) Pansu proved (1989) that two Carnot Lie groups are quasi-isometric if and only if they are isomorphic.
(2) There exists continuum many non-isomorphic 7-dimensional Carnot Lie groups.
If $Y$ is a proper, uniformly discrete and isometry-transitive metric space, then $Y$ is QI to its isometry group $G$, which is locally compact. Assuming it QI to a nilpotent Lie group in addition implies (by results of Gromov/Losert/Trofimov) that $G$, modulo a compact normal subgroup, is discrete and virtually nilpotent. There are only countably many QI classes. Hence at least one of the examples in (2) yields (for every choice of left-invariant Riemannian metric) a negative answer to your question.
There is an alternative using hyperbolicity rather polynomial growth, consisting of a continuum family of negatively curved homogeneous 3-folds. That they're not QI is also a result of Pansu. That a totally locally compact group QI to it has to be compact-by-discrete is an immediate consequence of the 2-dimensional Hilbert-Smith conjecture (which is an old theorem).
answered 10 hours ago
YCorYCor
31.8k4 gold badges96 silver badges148 bronze badges
$endgroup$
$begingroup$
I have a question regarding the sentence"If $Y$ is proper, uniformly discrete...". Is this a fact about Carnot Lie groups, Lie groups in general, homogeneous manifolds in general, or locally compact homogeneous metric spaces in general?
$endgroup$
– James Hanson
10 hours ago
$begingroup$
Changed "it's" to "$Y$". It's about arbitrary metric spaces.
$endgroup$
– YCor
9 hours ago
add a comment
|
$begingroup$
The answer is no. A quick answer can be done as follows:
(1) Pansu proved (1989) that two Carnot Lie groups are quasi-isometric if and only if they are isomorphic.
(2) There exists continuum many non-isomorphic 7-dimensional Carnot Lie groups.
If $Y$ is a proper, uniformly discrete and isometry-transitive metric space, then $Y$ is QI to its isometry group $G$, which is locally compact. Assuming it QI to a nilpotent Lie group in addition implies (by results of Gromov/Losert/Trofimov) that $G$, modulo a compact normal subgroup, is discrete and virtually nilpotent. There are only countably many QI classes. Hence at least one of the examples in (2) yields (for every choice of left-invariant Riemannian metric) a negative answer to your question.
There is an alternative using hyperbolicity rather polynomial growth, consisting of a continuum family of negatively curved homogeneous 3-folds. That they're not QI is also a result of Pansu. That a totally locally compact group QI to it has to be compact-by-discrete is an immediate consequence of the 2-dimensional Hilbert-Smith conjecture (which is an old theorem).
answered 10 hours ago
YCorYCor
31.8k4 gold badges96 silver badges148 bronze badges
$endgroup$
The answer is no. A quick answer can be done as follows:
(1) Pansu proved (1989) that two Carnot Lie groups are quasi-isometric if and only if they are isomorphic.
(2) There exists continuum many non-isomorphic 7-dimensional Carnot Lie groups.
If $Y$ is a proper, uniformly discrete and isometry-transitive metric space, then $Y$ is QI to its isometry group $G$, which is locally compact. Assuming it QI to a nilpotent Lie group in addition implies (by results of Gromov/Losert/Trofimov) that $G$, modulo a compact normal subgroup, is discrete and virtually nilpotent. There are only countably many QI classes. Hence at least one of the examples in (2) yields (for every choice of left-invariant Riemannian metric) a negative answer to your question.
There is an alternative using hyperbolicity rather polynomial growth, consisting of a continuum family of negatively curved homogeneous 3-folds. That they're not QI is also a result of Pansu. That a totally locally compact group QI to it has to be compact-by-discrete is an immediate consequence of the 2-dimensional Hilbert-Smith conjecture (which is an old theorem).
answered 10 hours ago
YCorYCor
31.8k4 gold badges96 silver badges148 bronze badges
edited 9 hours ago
answered 10 hours ago
YCorYCor
31.8k4 gold badges96 silver badges148 bronze badges
answered 10 hours ago
YCorYCor
31.8k4 gold badges96 silver badges148 bronze badges
answered 10 hours ago
YCorYCor
31.8k4 gold badges96 silver badges148 bronze badges
31.8k4 gold badges96 silver badges148 bronze badges
$begingroup$
I have a question regarding the sentence"If $Y$ is proper, uniformly discrete...". Is this a fact about Carnot Lie groups, Lie groups in general, homogeneous manifolds in general, or locally compact homogeneous metric spaces in general?
$endgroup$
– James Hanson
10 hours ago
$begingroup$
Changed "it's" to "$Y$". It's about arbitrary metric spaces.
$endgroup$
– YCor
9 hours ago
add a comment
|
$begingroup$
I have a question regarding the sentence"If $Y$ is proper, uniformly discrete...". Is this a fact about Carnot Lie groups, Lie groups in general, homogeneous manifolds in general, or locally compact homogeneous metric spaces in general?
$endgroup$
– James Hanson
10 hours ago
$begingroup$
Changed "it's" to "$Y$". It's about arbitrary metric spaces.
$endgroup$
– YCor
9 hours ago
$begingroup$
I have a question regarding the sentence"If $Y$ is proper, uniformly discrete...". Is this a fact about Carnot Lie groups, Lie groups in general, homogeneous manifolds in general, or locally compact homogeneous metric spaces in general?
$endgroup$
– James Hanson
10 hours ago
$begingroup$
I have a question regarding the sentence"If $Y$ is proper, uniformly discrete...". Is this a fact about Carnot Lie groups, Lie groups in general, homogeneous manifolds in general, or locally compact homogeneous metric spaces in general?
$endgroup$
– James Hanson
10 hours ago
$begingroup$
Changed "it's" to "$Y$". It's about arbitrary metric spaces.
$endgroup$
– YCor
9 hours ago
$begingroup$
Changed "it's" to "$Y$". It's about arbitrary metric spaces.
$endgroup$
– YCor
9 hours ago
add a comment
|
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$begingroup$
Your assumptions are equivalent to that there exists a cocompact lattice in the isometry group of $X$. (The setwise-stabilizer of $Y$ in $Isom(X)$.) This is already false for some 3-dimensional homogeneous manifolds (solvable simply connected Lie groups).
$endgroup$
– Misha
5 hours ago