Let $G$ be a f.g. group and $d$ be a word metric w.r.t. a symmetric generating set. For $g\in G$, define $|g|:=d(g,e)$, where $e$ is the group identity. For $k\in\mathbb N$, put $$n_k:=\#\{g\in G: |g|\leq k\}\quad\text{and}\quad m_k:=\#\{g\in G: |g|=k\}$$ In which groups $\lim_k\frac{m_k}{n_k}=0$?

I don't know if it is helpful or not; all groups I work with, are torsion free.

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    $\begingroup$ Notice that $m_k=n_k-n_{k-1}$, so a (not so good) answer is whenever $n_{k-1}/n_k$ converges to 1. $\endgroup$ – M. Dus Sep 25 '18 at 7:58
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    $\begingroup$ This might be related to the Følner condition and thus to amenability. $\endgroup$ – ThiKu Sep 25 '18 at 8:03
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    $\begingroup$ In order to have a chance of having this $G$ has to be amenable. Moreover, the balls have to constitute a sequence of Foelner sets. This is certainly the case for virtually nilpotent groups, but other than that things are a bit more sketchy. $\endgroup$ – duh Sep 25 '18 at 9:25
  • $\begingroup$ About the nilpotent case, see also arxiv.org/abs/math/0506362 and the references therein. $\endgroup$ – YCor Sep 25 '18 at 20:01

The answer was basically given by @duh in his/her comment. The group satisfies $\frac{m_k}{n_k}\to 0$ if and only if the balls are Foelner sequences. In particular, the group has to be amenable.

Now the question becomes which amenable groups satisfy that balls are Foelner sequences ?

As suggested by @duh, virtually nilpotent groups do satisfy this condition, but you do not need to mention Foelner sequences. We know that virtually nilpotent have polynomial growth. Usually, this is stated as $a_1k^d-b_1\leq n_k\leq a_2k^d+b_2$ for some $a_1,b_1,a_2,b_2$, where $d$ is the homogeneous dimension of the group. However, Pansu improved this result, showing that $n_k\sim ck^d$ for some constant $c$. In particular, $\frac{n_k}{n_{k-1}}$ converges to 1 and since $m_k=n_k-n_{k-1}$, we indeed have that $\frac{m_k}{n_k}$ converges to 0.

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    $\begingroup$ As a reference, all this is related to Problem 34.A (Page 206) in De La Harpe's Book, Topics in geometric group theory. $\endgroup$ – Mustafa Gokhan Benli Sep 25 '18 at 13:42
  • $\begingroup$ With a bit a researching: if the growth is exponential, then $n_{k+1}/n_k$ cannot converge to 1, see mathoverflow.net/questions/43964/…. There are also examples where $n_{k+1}/n_k$ does not converge, see mathoverflow.net/questions/36126/… $\endgroup$ – M. Dus Sep 25 '18 at 14:51
  • $\begingroup$ The balls form a Folner set for groups of intermediate growth $\endgroup$ – Benjamin Steinberg Sep 25 '18 at 16:41
  • $\begingroup$ @BenjaminSteinberg Thank you for this comment. Do you have a reference ? $\endgroup$ – M. Dus Sep 25 '18 at 17:03
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    $\begingroup$ @M.Dus, I actually oversimplified. What is known is that if a group $G$ has intermediate growth then some subsequence of balls give Folner sets. But is unknown if you can use all of them. I don't know an exact reference for this but it is I think implicit in Chou's proof that groups of intermediate growth are amenable but not elementary amenable. $\endgroup$ – Benjamin Steinberg Sep 25 '18 at 18:14

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