Is the infinite regular tree $T_infty$ quasi-isometric to the $n$-regular tree $T_n$?
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It is known that for $ngeq 3$ all $n$-regular trees $T_n$ are quasi-isometric to each other. This can e.g. be seen by using an edge contraction argument.
Is there also a quasi-isometry between the countably infinite regular tree $T_infty$ and some $T_n$?
trees geometric-group-theory
asked Jan 7 at 12:52
CKayCKay
132
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add a comment |
$begingroup$
It is known that for $ngeq 3$ all $n$-regular trees $T_n$ are quasi-isometric to each other. This can e.g. be seen by using an edge contraction argument.
Is there also a quasi-isometry between the countably infinite regular tree $T_infty$ and some $T_n$?
trees geometric-group-theory
asked Jan 7 at 12:52
CKayCKay
132
New contributor
CKay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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No, that would be impossible. The question is, what quasi-isometry invariants do you know? Do you know about growth? About ideal boundaries of Gromov-hyperbolic spaces? Either one can be used to prove non-existence of a quasiisometry.
$endgroup$
– Moishe Cohen
Jan 7 at 21:44
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As far as I know the QI-invariant 'growth' is only defined for finitely generated groups and can thus not be applied to $T_infty$ or is there a more general notion that I am missing? And isn't the boundary of all those trees a Cantor set? I do not yet see a contradiction to the existence of a QI.
$endgroup$
– CKay
Jan 8 at 12:10
1
$begingroup$
You can define growth in greater generality. The ideal boundaries of the trees of infinite valence will not be compact.
$endgroup$
– Moishe Cohen
Jan 8 at 13:29
1
$begingroup$
See §3Db in arxiv.org/abs/1403.3796: $T_n$ is coarsely proper while $T_infty$ is not. Coarsely proper seems to be equivalent to the notion of finite packing mentioned by @MoisheCohen. It's reasonable to say that a space that is not coarsely proper has growth $=infty$ (=greater than any function) and this is a coarse invariant.
$endgroup$
– YCor
2 days ago
$begingroup$
@YCor: Yes, it is an equivalent notion.
$endgroup$
– Moishe Cohen
2 days ago
add a comment |
$begingroup$
It is known that for $ngeq 3$ all $n$-regular trees $T_n$ are quasi-isometric to each other. This can e.g. be seen by using an edge contraction argument.
Is there also a quasi-isometry between the countably infinite regular tree $T_infty$ and some $T_n$?
trees geometric-group-theory
asked Jan 7 at 12:52
CKayCKay
132
New contributor
CKay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
It is known that for $ngeq 3$ all $n$-regular trees $T_n$ are quasi-isometric to each other. This can e.g. be seen by using an edge contraction argument.
Is there also a quasi-isometry between the countably infinite regular tree $T_infty$ and some $T_n$?
trees geometric-group-theory
trees geometric-group-theory
asked Jan 7 at 12:52
CKayCKay
132
New contributor
CKay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 7 at 12:52
CKayCKay
132
New contributor
CKay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 7 at 12:52
CKayCKay
132
New contributor
CKay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 7 at 12:52
CKayCKay
132
asked Jan 7 at 12:52
CKayCKay
132
132
New contributor
CKay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
CKay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
CKay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
No, that would be impossible. The question is, what quasi-isometry invariants do you know? Do you know about growth? About ideal boundaries of Gromov-hyperbolic spaces? Either one can be used to prove non-existence of a quasiisometry.
$endgroup$
– Moishe Cohen
Jan 7 at 21:44
$begingroup$
As far as I know the QI-invariant 'growth' is only defined for finitely generated groups and can thus not be applied to $T_infty$ or is there a more general notion that I am missing? And isn't the boundary of all those trees a Cantor set? I do not yet see a contradiction to the existence of a QI.
$endgroup$
– CKay
Jan 8 at 12:10
1
$begingroup$
You can define growth in greater generality. The ideal boundaries of the trees of infinite valence will not be compact.
$endgroup$
– Moishe Cohen
Jan 8 at 13:29
1
$begingroup$
See §3Db in arxiv.org/abs/1403.3796: $T_n$ is coarsely proper while $T_infty$ is not. Coarsely proper seems to be equivalent to the notion of finite packing mentioned by @MoisheCohen. It's reasonable to say that a space that is not coarsely proper has growth $=infty$ (=greater than any function) and this is a coarse invariant.
$endgroup$
– YCor
2 days ago
$begingroup$
@YCor: Yes, it is an equivalent notion.
$endgroup$
– Moishe Cohen
2 days ago
add a comment |
$begingroup$
No, that would be impossible. The question is, what quasi-isometry invariants do you know? Do you know about growth? About ideal boundaries of Gromov-hyperbolic spaces? Either one can be used to prove non-existence of a quasiisometry.
$endgroup$
– Moishe Cohen
Jan 7 at 21:44
$begingroup$
As far as I know the QI-invariant 'growth' is only defined for finitely generated groups and can thus not be applied to $T_infty$ or is there a more general notion that I am missing? And isn't the boundary of all those trees a Cantor set? I do not yet see a contradiction to the existence of a QI.
$endgroup$
– CKay
Jan 8 at 12:10
1
$begingroup$
You can define growth in greater generality. The ideal boundaries of the trees of infinite valence will not be compact.
$endgroup$
– Moishe Cohen
Jan 8 at 13:29
1
$begingroup$
See §3Db in arxiv.org/abs/1403.3796: $T_n$ is coarsely proper while $T_infty$ is not. Coarsely proper seems to be equivalent to the notion of finite packing mentioned by @MoisheCohen. It's reasonable to say that a space that is not coarsely proper has growth $=infty$ (=greater than any function) and this is a coarse invariant.
$endgroup$
– YCor
2 days ago
$begingroup$
@YCor: Yes, it is an equivalent notion.
$endgroup$
– Moishe Cohen
2 days ago
$begingroup$
No, that would be impossible. The question is, what quasi-isometry invariants do you know? Do you know about growth? About ideal boundaries of Gromov-hyperbolic spaces? Either one can be used to prove non-existence of a quasiisometry.
$endgroup$
– Moishe Cohen
Jan 7 at 21:44
$begingroup$
No, that would be impossible. The question is, what quasi-isometry invariants do you know? Do you know about growth? About ideal boundaries of Gromov-hyperbolic spaces? Either one can be used to prove non-existence of a quasiisometry.
$endgroup$
– Moishe Cohen
Jan 7 at 21:44
$begingroup$
As far as I know the QI-invariant 'growth' is only defined for finitely generated groups and can thus not be applied to $T_infty$ or is there a more general notion that I am missing? And isn't the boundary of all those trees a Cantor set? I do not yet see a contradiction to the existence of a QI.
$endgroup$
– CKay
Jan 8 at 12:10
$begingroup$
As far as I know the QI-invariant 'growth' is only defined for finitely generated groups and can thus not be applied to $T_infty$ or is there a more general notion that I am missing? And isn't the boundary of all those trees a Cantor set? I do not yet see a contradiction to the existence of a QI.
$endgroup$
– CKay
Jan 8 at 12:10
1
1
$begingroup$
You can define growth in greater generality. The ideal boundaries of the trees of infinite valence will not be compact.
$endgroup$
– Moishe Cohen
Jan 8 at 13:29
$begingroup$
You can define growth in greater generality. The ideal boundaries of the trees of infinite valence will not be compact.
$endgroup$
– Moishe Cohen
Jan 8 at 13:29
1
1
$begingroup$
See §3Db in arxiv.org/abs/1403.3796: $T_n$ is coarsely proper while $T_infty$ is not. Coarsely proper seems to be equivalent to the notion of finite packing mentioned by @MoisheCohen. It's reasonable to say that a space that is not coarsely proper has growth $=infty$ (=greater than any function) and this is a coarse invariant.
$endgroup$
– YCor
2 days ago
$begingroup$
See §3Db in arxiv.org/abs/1403.3796: $T_n$ is coarsely proper while $T_infty$ is not. Coarsely proper seems to be equivalent to the notion of finite packing mentioned by @MoisheCohen. It's reasonable to say that a space that is not coarsely proper has growth $=infty$ (=greater than any function) and this is a coarse invariant.
$endgroup$
– YCor
2 days ago
$begingroup$
@YCor: Yes, it is an equivalent notion.
$endgroup$
– Moishe Cohen
2 days ago
$begingroup$
@YCor: Yes, it is an equivalent notion.
$endgroup$
– Moishe Cohen
2 days ago
add a comment |
1 Answer
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$begingroup$
The easiest argument that $T_n$ and $T_infty$ are not quasiisometric is based on growth, more precisely, it is a fragment the same proof that shows that growth is quasiisometry-invariant:
Every finite valence graph has the "finite packing" property: There exists $rhoge 0$ such that for every $Rge rge rho$, every ball $B(x,R)$ contains only finitely many pairwise disjoint balls of the radius $r$. (One can even take $rho=0$ for graphs of finite valence.) In fact, if the valence is uniformly bounded like in your case, then the number of such $r$-balls is also uniformly bounded by a function of $R$ and $r$.
The finite packing property is easily seen to be quasiisometry-invariant: If $X_1, X_2$ are quasiisometric metric spaces and $X_1$ has finite packing property then $X_2$ does as well. The finite packing property is clearly false for regular trees of infinite valence.
answered Jan 8 at 13:52
Moishe CohenMoishe Cohen
46.1k342104
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$begingroup$
The easiest argument that $T_n$ and $T_infty$ are not quasiisometric is based on growth, more precisely, it is a fragment the same proof that shows that growth is quasiisometry-invariant:
Every finite valence graph has the "finite packing" property: There exists $rhoge 0$ such that for every $Rge rge rho$, every ball $B(x,R)$ contains only finitely many pairwise disjoint balls of the radius $r$. (One can even take $rho=0$ for graphs of finite valence.) In fact, if the valence is uniformly bounded like in your case, then the number of such $r$-balls is also uniformly bounded by a function of $R$ and $r$.
The finite packing property is easily seen to be quasiisometry-invariant: If $X_1, X_2$ are quasiisometric metric spaces and $X_1$ has finite packing property then $X_2$ does as well. The finite packing property is clearly false for regular trees of infinite valence.
answered Jan 8 at 13:52
Moishe CohenMoishe Cohen
46.1k342104
$endgroup$
add a comment |
$begingroup$
The easiest argument that $T_n$ and $T_infty$ are not quasiisometric is based on growth, more precisely, it is a fragment the same proof that shows that growth is quasiisometry-invariant:
Every finite valence graph has the "finite packing" property: There exists $rhoge 0$ such that for every $Rge rge rho$, every ball $B(x,R)$ contains only finitely many pairwise disjoint balls of the radius $r$. (One can even take $rho=0$ for graphs of finite valence.) In fact, if the valence is uniformly bounded like in your case, then the number of such $r$-balls is also uniformly bounded by a function of $R$ and $r$.
The finite packing property is easily seen to be quasiisometry-invariant: If $X_1, X_2$ are quasiisometric metric spaces and $X_1$ has finite packing property then $X_2$ does as well. The finite packing property is clearly false for regular trees of infinite valence.
answered Jan 8 at 13:52
Moishe CohenMoishe Cohen
46.1k342104
$endgroup$
add a comment |
$begingroup$
The easiest argument that $T_n$ and $T_infty$ are not quasiisometric is based on growth, more precisely, it is a fragment the same proof that shows that growth is quasiisometry-invariant:
Every finite valence graph has the "finite packing" property: There exists $rhoge 0$ such that for every $Rge rge rho$, every ball $B(x,R)$ contains only finitely many pairwise disjoint balls of the radius $r$. (One can even take $rho=0$ for graphs of finite valence.) In fact, if the valence is uniformly bounded like in your case, then the number of such $r$-balls is also uniformly bounded by a function of $R$ and $r$.
The finite packing property is easily seen to be quasiisometry-invariant: If $X_1, X_2$ are quasiisometric metric spaces and $X_1$ has finite packing property then $X_2$ does as well. The finite packing property is clearly false for regular trees of infinite valence.
answered Jan 8 at 13:52
Moishe CohenMoishe Cohen
46.1k342104
$endgroup$
The easiest argument that $T_n$ and $T_infty$ are not quasiisometric is based on growth, more precisely, it is a fragment the same proof that shows that growth is quasiisometry-invariant:
Every finite valence graph has the "finite packing" property: There exists $rhoge 0$ such that for every $Rge rge rho$, every ball $B(x,R)$ contains only finitely many pairwise disjoint balls of the radius $r$. (One can even take $rho=0$ for graphs of finite valence.) In fact, if the valence is uniformly bounded like in your case, then the number of such $r$-balls is also uniformly bounded by a function of $R$ and $r$.
The finite packing property is easily seen to be quasiisometry-invariant: If $X_1, X_2$ are quasiisometric metric spaces and $X_1$ has finite packing property then $X_2$ does as well. The finite packing property is clearly false for regular trees of infinite valence.
answered Jan 8 at 13:52
Moishe CohenMoishe Cohen
46.1k342104
answered Jan 8 at 13:52
Moishe CohenMoishe Cohen
46.1k342104
answered Jan 8 at 13:52
Moishe CohenMoishe Cohen
46.1k342104
answered Jan 8 at 13:52
Moishe CohenMoishe Cohen
46.1k342104
46.1k342104
add a comment |
add a comment |
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$begingroup$
No, that would be impossible. The question is, what quasi-isometry invariants do you know? Do you know about growth? About ideal boundaries of Gromov-hyperbolic spaces? Either one can be used to prove non-existence of a quasiisometry.
$endgroup$
– Moishe Cohen
Jan 7 at 21:44
$begingroup$
As far as I know the QI-invariant 'growth' is only defined for finitely generated groups and can thus not be applied to $T_infty$ or is there a more general notion that I am missing? And isn't the boundary of all those trees a Cantor set? I do not yet see a contradiction to the existence of a QI.
$endgroup$
– CKay
Jan 8 at 12:10
1
$begingroup$
You can define growth in greater generality. The ideal boundaries of the trees of infinite valence will not be compact.
$endgroup$
– Moishe Cohen
Jan 8 at 13:29
1
$begingroup$
See §3Db in arxiv.org/abs/1403.3796: $T_n$ is coarsely proper while $T_infty$ is not. Coarsely proper seems to be equivalent to the notion of finite packing mentioned by @MoisheCohen. It's reasonable to say that a space that is not coarsely proper has growth $=infty$ (=greater than any function) and this is a coarse invariant.
$endgroup$
– YCor
2 days ago
$begingroup$
@YCor: Yes, it is an equivalent notion.
$endgroup$
– Moishe Cohen
2 days ago