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This question arose in the comments of A question about groups of intermediate growth. I think it might be interesting to put it more in evidence.

Let $G$ be a f.g. group with a fixed symmetric set of generators $S$ and denote by $B(n)$ the ball of radius $n$ about the identity w.r.t. the word metric induced by $S$.

Fix an integer $k\geq1$ and define $\overline\zeta_k(G)=\lim\sup_n\frac{|B(nk+k)|}{|B(nk)|}$.

Observe that

  1. If $G$ has polynomial growth, then $\overline\zeta_k(G)=1$, for all $k$.
  2. If $\overline\zeta_k(G)=1$ for all $k$, then $G$ has sub-exponential growth.

General question: What can we say about $\overline\zeta_k(G)$ if $G$ has intermediate growth?

Martin Kassabov, in the comment to my question, suspects that it should be always (or most of the times) equals to $1$, but I cannot find even a single examples of a group of intermediate growth for which it is equal to $1$. I have to say that my knowledge about groups of intermediate growth is very little and I just tried to apply Corollary 1.3 in http://arxiv.org/PS_cache/arxiv/pdf/1108/1108.0262v1.pdf, but, as already observed by Martin, it is not strong enough to give an example of groups of intermediate growth whose $\overline\zeta_k(G)=1$.

Particular question: Is there an example of group of intermediate growth for which $\overline\zeta_k(X)=1$, for all $k$?

Thanks in advance,

Valerio

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    $\begingroup$ there is not "easy proof" of the observation 1, and for observation 2 it is enough to have it for some $k$. $\endgroup$
    – kassabov
    Commented Nov 15, 2011 at 16:41
  • $\begingroup$ Yes, I was sure about the second point... $\endgroup$
    – Valerio Capraro
    Commented Nov 15, 2011 at 19:32

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It seems that if $\bar \zeta_k > 1$ for some $k$ then $\bar \zeta_k > 1$ for all $k$. Also it is clear $\lim_k \sqrt[k]{\bar \zeta_k} = 1$.

My guess is that most known groups of intermediate growth satisfy $\bar \zeta_k=1$, but proving this require very carefull estimates for the size of the balls. Notice that until recently the growth type on any group of initermediate growth have not be computed, which requires just a "crude" estimates of the growth type. You can see the recent papers of Bartholdi and Eschler: https://arxiv.org/abs/1110.3650 and https://arxiv.org/abs/1011.5266, where they have computed the growth type of many groups, but I fell that their estimates are far from what you need to get $\bar\zeta_k=1$.

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  • $\begingroup$ the second claim (about the limit) is clear and follows from your argument in the other question. Can you please give me some details about the first claim: if $\overline\zeta_k >1$ for some $k$, then $\overline\zeta_k>1$, for all $k$? Thank you very much in advance. $\endgroup$
    – Valerio Capraro
    Commented Nov 15, 2011 at 22:27
  • $\begingroup$ Using that $B(n)$ is an increasing function one can easily see that $\sqrt{\zeta_k} \leq \zeta_l$ for $k<l$ -- if the ratio $B(n+k)/B(n)$ is large for some $n$ we can find $m$ such that $m < n < n+k < m +l$ which implies that $B(m+l)/B(m)$ is large. However it might not be possible to ensure that $m$ is divisible by $l$, in this case we can pick $m$ between $n$ and $n+k$ and one of the ratios $B(m+l)/B(m)$ and $B(m)/B(m-l)$ will be large. $\endgroup$
    – kassabov
    Commented Nov 16, 2011 at 11:20
  • $\begingroup$ Again using that $B(n)$ is an increasing we have $\zeta_k \leq \zeta_{nk} \leq (\zeta_k)^n$ -- just break the interval of size $nk$ into $n$ intervals of size $k$. These inequalities give that if one $\zeta_k =1$ then all others are also equal to $1$. $\endgroup$
    – kassabov
    Commented Nov 16, 2011 at 11:22

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