Asymptotics of orbits on graphsGrowth rate of number of loops in a graphAsymptotics for forbidden subwords“Antipodal” maps on regular graphs?Average squared distance in $k$-regular graphsA question about expander graphsFinite vertex-transitive graphs that look like infinite vertex-transitive graphsLovász conjecture and 2-connected graphsHamming representability of finite graphsGraphs formed of vertices of distance $2$Reference on graphs such that contracting 2 non-adjacent vertices increases the Hadwiger number

Asymptotics of orbits on graphs


Growth rate of number of loops in a graphAsymptotics for forbidden subwords“Antipodal” maps on regular graphs?Average squared distance in $k$-regular graphsA question about expander graphsFinite vertex-transitive graphs that look like infinite vertex-transitive graphsLovász conjecture and 2-connected graphsHamming representability of finite graphsGraphs formed of vertices of distance $2$Reference on graphs such that contracting 2 non-adjacent vertices increases the Hadwiger number













3












$begingroup$


Let $X$ be a connected, locally finite graph with vertex set $V(X)$ and $G$ a group acting freely on $X$ such that $X/G$ is a finite graph. Fix a vertex $x$ and for $kinmathbb N$ set
$$
N(k)=# gin G: d(gx,x)le k,
$$

where $d$ is the vertex distance in the graph $X$.
Further set
$$
A(k)=#yin V(X):d(x,y)le k.
$$

Is it true that, as $ktoinfty$, the number $N(k)/A(k)$ tends to $#V(X/G)^-1$? If so, what error term estimates are known?










share|cite|improve this question











$endgroup$











  • $begingroup$
    Very interesting! Can you please add the reference or the source of inspiration for this problem?
    $endgroup$
    – SeF
    12 hours ago










  • $begingroup$
    It's kind of a graph analogue of lattice point counting.
    $endgroup$
    – Zero
    12 hours ago















3












$begingroup$


Let $X$ be a connected, locally finite graph with vertex set $V(X)$ and $G$ a group acting freely on $X$ such that $X/G$ is a finite graph. Fix a vertex $x$ and for $kinmathbb N$ set
$$
N(k)=# gin G: d(gx,x)le k,
$$

where $d$ is the vertex distance in the graph $X$.
Further set
$$
A(k)=#yin V(X):d(x,y)le k.
$$

Is it true that, as $ktoinfty$, the number $N(k)/A(k)$ tends to $#V(X/G)^-1$? If so, what error term estimates are known?










share|cite|improve this question











$endgroup$











  • $begingroup$
    Very interesting! Can you please add the reference or the source of inspiration for this problem?
    $endgroup$
    – SeF
    12 hours ago










  • $begingroup$
    It's kind of a graph analogue of lattice point counting.
    $endgroup$
    – Zero
    12 hours ago













3












3








3





$begingroup$


Let $X$ be a connected, locally finite graph with vertex set $V(X)$ and $G$ a group acting freely on $X$ such that $X/G$ is a finite graph. Fix a vertex $x$ and for $kinmathbb N$ set
$$
N(k)=# gin G: d(gx,x)le k,
$$

where $d$ is the vertex distance in the graph $X$.
Further set
$$
A(k)=#yin V(X):d(x,y)le k.
$$

Is it true that, as $ktoinfty$, the number $N(k)/A(k)$ tends to $#V(X/G)^-1$? If so, what error term estimates are known?










share|cite|improve this question











$endgroup$




Let $X$ be a connected, locally finite graph with vertex set $V(X)$ and $G$ a group acting freely on $X$ such that $X/G$ is a finite graph. Fix a vertex $x$ and for $kinmathbb N$ set
$$
N(k)=# gin G: d(gx,x)le k,
$$

where $d$ is the vertex distance in the graph $X$.
Further set
$$
A(k)=#yin V(X):d(x,y)le k.
$$

Is it true that, as $ktoinfty$, the number $N(k)/A(k)$ tends to $#V(X/G)^-1$? If so, what error term estimates are known?







graph-theory asymptotics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 12 hours ago







Zero

















asked 14 hours ago









ZeroZero

2567




2567











  • $begingroup$
    Very interesting! Can you please add the reference or the source of inspiration for this problem?
    $endgroup$
    – SeF
    12 hours ago










  • $begingroup$
    It's kind of a graph analogue of lattice point counting.
    $endgroup$
    – Zero
    12 hours ago
















  • $begingroup$
    Very interesting! Can you please add the reference or the source of inspiration for this problem?
    $endgroup$
    – SeF
    12 hours ago










  • $begingroup$
    It's kind of a graph analogue of lattice point counting.
    $endgroup$
    – Zero
    12 hours ago















$begingroup$
Very interesting! Can you please add the reference or the source of inspiration for this problem?
$endgroup$
– SeF
12 hours ago




$begingroup$
Very interesting! Can you please add the reference or the source of inspiration for this problem?
$endgroup$
– SeF
12 hours ago












$begingroup$
It's kind of a graph analogue of lattice point counting.
$endgroup$
– Zero
12 hours ago




$begingroup$
It's kind of a graph analogue of lattice point counting.
$endgroup$
– Zero
12 hours ago










1 Answer
1






active

oldest

votes


















5












$begingroup$

It is possible that the limit does not exist at all: Consider the free group on two generators acting on the $(4,2)$-biregular tree in the obvious way. This action is free and has 3 orbits (one containing all vertices of degree 4, and the other two containing "half" of the vertices of degree 2).



Let $x$ be a vertex of degree $4$. Then $N(k)$ is the number of vertices of degree 4 in $B_x(k)$, and $A(k)$ is the total number of vertices in $B_x(k)$. If we write $a_k$ and $b_k$ for the number of vertices at distance exactly $k$ from $x$ which have degree 4 or 2 respectively, we get $a_0 = 1$, and $b_2l+1 = a_2l+2 = 4cdot3^l$ and $b_2l = a_2l+1 = 0$ for $l geq 0$. Note that
$$fracN(k)A(k) = fracsum_i leq k a_isum_i leq k a_i + b_i$$
and if I'm not mistaken, plugging in the above values gives a limit of $frac 12$ for the subsequence of even $k$, and $frac 14$ for the subsequence of odd $k$.






share|cite|improve this answer









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    1 Answer
    1






    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    It is possible that the limit does not exist at all: Consider the free group on two generators acting on the $(4,2)$-biregular tree in the obvious way. This action is free and has 3 orbits (one containing all vertices of degree 4, and the other two containing "half" of the vertices of degree 2).



    Let $x$ be a vertex of degree $4$. Then $N(k)$ is the number of vertices of degree 4 in $B_x(k)$, and $A(k)$ is the total number of vertices in $B_x(k)$. If we write $a_k$ and $b_k$ for the number of vertices at distance exactly $k$ from $x$ which have degree 4 or 2 respectively, we get $a_0 = 1$, and $b_2l+1 = a_2l+2 = 4cdot3^l$ and $b_2l = a_2l+1 = 0$ for $l geq 0$. Note that
    $$fracN(k)A(k) = fracsum_i leq k a_isum_i leq k a_i + b_i$$
    and if I'm not mistaken, plugging in the above values gives a limit of $frac 12$ for the subsequence of even $k$, and $frac 14$ for the subsequence of odd $k$.






    share|cite|improve this answer









    $endgroup$

















      5












      $begingroup$

      It is possible that the limit does not exist at all: Consider the free group on two generators acting on the $(4,2)$-biregular tree in the obvious way. This action is free and has 3 orbits (one containing all vertices of degree 4, and the other two containing "half" of the vertices of degree 2).



      Let $x$ be a vertex of degree $4$. Then $N(k)$ is the number of vertices of degree 4 in $B_x(k)$, and $A(k)$ is the total number of vertices in $B_x(k)$. If we write $a_k$ and $b_k$ for the number of vertices at distance exactly $k$ from $x$ which have degree 4 or 2 respectively, we get $a_0 = 1$, and $b_2l+1 = a_2l+2 = 4cdot3^l$ and $b_2l = a_2l+1 = 0$ for $l geq 0$. Note that
      $$fracN(k)A(k) = fracsum_i leq k a_isum_i leq k a_i + b_i$$
      and if I'm not mistaken, plugging in the above values gives a limit of $frac 12$ for the subsequence of even $k$, and $frac 14$ for the subsequence of odd $k$.






      share|cite|improve this answer









      $endgroup$















        5












        5








        5





        $begingroup$

        It is possible that the limit does not exist at all: Consider the free group on two generators acting on the $(4,2)$-biregular tree in the obvious way. This action is free and has 3 orbits (one containing all vertices of degree 4, and the other two containing "half" of the vertices of degree 2).



        Let $x$ be a vertex of degree $4$. Then $N(k)$ is the number of vertices of degree 4 in $B_x(k)$, and $A(k)$ is the total number of vertices in $B_x(k)$. If we write $a_k$ and $b_k$ for the number of vertices at distance exactly $k$ from $x$ which have degree 4 or 2 respectively, we get $a_0 = 1$, and $b_2l+1 = a_2l+2 = 4cdot3^l$ and $b_2l = a_2l+1 = 0$ for $l geq 0$. Note that
        $$fracN(k)A(k) = fracsum_i leq k a_isum_i leq k a_i + b_i$$
        and if I'm not mistaken, plugging in the above values gives a limit of $frac 12$ for the subsequence of even $k$, and $frac 14$ for the subsequence of odd $k$.






        share|cite|improve this answer









        $endgroup$



        It is possible that the limit does not exist at all: Consider the free group on two generators acting on the $(4,2)$-biregular tree in the obvious way. This action is free and has 3 orbits (one containing all vertices of degree 4, and the other two containing "half" of the vertices of degree 2).



        Let $x$ be a vertex of degree $4$. Then $N(k)$ is the number of vertices of degree 4 in $B_x(k)$, and $A(k)$ is the total number of vertices in $B_x(k)$. If we write $a_k$ and $b_k$ for the number of vertices at distance exactly $k$ from $x$ which have degree 4 or 2 respectively, we get $a_0 = 1$, and $b_2l+1 = a_2l+2 = 4cdot3^l$ and $b_2l = a_2l+1 = 0$ for $l geq 0$. Note that
        $$fracN(k)A(k) = fracsum_i leq k a_isum_i leq k a_i + b_i$$
        and if I'm not mistaken, plugging in the above values gives a limit of $frac 12$ for the subsequence of even $k$, and $frac 14$ for the subsequence of odd $k$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 10 hours ago









        Florian LehnerFlorian Lehner

        54138




        54138



























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