About the growth rate of a group

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.

• 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. Sep 25, 2018 at 7:58
• This might be related to the Følner condition and thus to amenability. Sep 25, 2018 at 8:03
• 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.
– duh
Sep 25, 2018 at 9:25
• About the nilpotent case, see also arxiv.org/abs/math/0506362 and the references therein.
– YCor
Sep 25, 2018 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.
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.
• 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/… Sep 25, 2018 at 14:51
• @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. Sep 25, 2018 at 18:14