Is the infinite regular tree $T_infty$ quasi-isometric to the $n$-regular tree $T_n$?












2












$begingroup$


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







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$endgroup$












  • $begingroup$
    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
    Jan 10 at 19:40










  • $begingroup$
    @YCor: Yes, it is an equivalent notion.
    $endgroup$
    – Moishe Cohen
    Jan 10 at 21:07
















2












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
    Jan 10 at 19:40










  • $begingroup$
    @YCor: Yes, it is an equivalent notion.
    $endgroup$
    – Moishe Cohen
    Jan 10 at 21:07














2












2








2





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 7 at 12:52









CKayCKay

132




132












  • $begingroup$
    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
    Jan 10 at 19:40










  • $begingroup$
    @YCor: Yes, it is an equivalent notion.
    $endgroup$
    – Moishe Cohen
    Jan 10 at 21:07


















  • $begingroup$
    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
    Jan 10 at 19:40










  • $begingroup$
    @YCor: Yes, it is an equivalent notion.
    $endgroup$
    – Moishe Cohen
    Jan 10 at 21:07
















$begingroup$
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
Jan 10 at 19:40




$begingroup$
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
Jan 10 at 19:40












$begingroup$
@YCor: Yes, it is an equivalent notion.
$endgroup$
– Moishe Cohen
Jan 10 at 21:07




$begingroup$
@YCor: Yes, it is an equivalent notion.
$endgroup$
– Moishe Cohen
Jan 10 at 21:07










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.






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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.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 13:52









        Moishe CohenMoishe Cohen

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