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This question is motivated by a partial answer to another question on MO.

Given an infinite finitely generated group $G$ and a finite generating set $S$, let $b_n^S$ be the cardinality of the ball of radius $n$ [around the identity].

Say a pair $G$ has pinched exponential growth w.r.t. $S$ if there are constants $K<L \in \mathbb{R}_{> 0}$ and $M \in \mathbb{R}_{>0}$ so that $K \cdot \mathrm{exp}(Mn) \leq b_n^S \leq L \cdot \mathrm{exp}(Mn).$

Question: What are examples of amenable groups $G$ which have exponential growth, but there is no choice of generating set $S$, so that the group has pinched exponential growth (w.r.t. $S$)?

Remarks:

  • the question could be strengthened to require that the group does not have a rational growth series. (If there are more than one pole on the radius of convergence of the growth series, then the group does not have pinched exponential growth.)

  • the question could also be strengthened to require that the group be finitely presented.

  • an equivalent formulation of the question is: let $g(S) =\displaystyle \lim_{n \to \infty} \sqrt[n]{b_n^S}$. Is there an $S$ so that $\limsup_n \dfrac{ b_n^S}{g(S)^n} < \infty$?

[EDIT: Further remarks:

  • there would be a weaker form of the question where non-amenable groups could be considered. Non-amenable groups are however not interesting in this context (see linked question and comment below)

  • [complement to a comment below] If $S$ is a generating set, let $S' = S \cup \lbrace e \rbrace$. Assume $G$ has pinched exponential growth w.r.t. $S_G$ and $H$ has pinched growth w.r.t. to $S_H$. Consider the generating sets $S_+ = (S_G \times \lbrace e_H \rbrace) \cup (\lbrace e_G \rbrace \times S_H)$ and $S_\times = S_G' \times S_H'$. Then it's not too difficult to check that $G \times H$ has pinched growth w.r.t. $S_\times$ (because the balls are just product of the balls). It does not have however pinched growth w.r.t. $S_+$. (To see this note that the balls have the form $\cup_{i =0}^n \sigma_i(G) \times \sigma_{n-i}(H)$ where $\sigma_i$ are the sphere of radius $i$. Combined with the fact that the sphere are also pinched [see linked answer for details] gives the fact that the growth is not pinched)

  • [Still extension of that comment] Assume $G$ has pinched growth w.r.t. to $S_G$ and $H = \mathbb{Z}$ with $S_H = \lbrace 1;-1 \rbrace$ as generating. Then $G \times H$ has pinched growth w.r.t. $S_+$ but not w.r.t. $S_\times$. Indeed, the spheres w.r.t. $S_+$ have cardinality of the balls in $G$. So the cardinality of the balls w.r.t. $S_+$ can be bounded by geometric sums. On the other hand the balls w.r.t. $S_\times$ have the cardinality of a ball in $\mathbb{Z}$ times a ball in $G$ (i.e. bounded above and below by some function of the form $n exp(Mn)$). Hence not pinched. The case of $G=F_2$ is particularly easy to check (see comment below).

  • If $G$ has pinched growth, then $G \times \mathbb{Z}^2$ does not have pinched growth w.r.t. to either $S_\times$ or $S_+$. But that does not exclude another generating set with pinched growth.

  • Most "classical" amenable groups of exponential growth have a generating set with pinched growth (this includes the lamplighter group, solvable Baumslag-Solitar groups and metabelian polycyclic groups).

END of EDIT]

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  • $\begingroup$ Is there a particular reason to assume amenability? $\endgroup$
    – YCor
    Jan 4, 2021 at 17:55
  • $\begingroup$ Every group of exponential growth has "pinched" growth. $\endgroup$
    – markvs
    Jan 4, 2021 at 17:55
  • $\begingroup$ @dodd what do you have in mind? the classical thing is that $(b_n^S)^{1/n}$ converges. $\endgroup$
    – YCor
    Jan 4, 2021 at 18:01
  • $\begingroup$ @Ycor it's natural in the context of the linked question ("it never happens that balls in a group of exponential growth are a Folner sequence"). If it were so that amenable groups of exponential growth always have a generating set with pinched exponential growth, then one would at least have that for some generating set, balls are not Folner. I seriously doubt this is the case however... $\endgroup$
    – ARG
    Jan 4, 2021 at 19:31
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    $\begingroup$ @dodd that's clearly false. Look at the group $G = F_2 \times \mathbb{Z}$ where $F_2$ is the free group with the "product" generating set $\lbrace (a,0), (b,0), (a,1), (b,1),(e,1) \rbrace$ where $a,b \in F_2$ are a free generating set and $e \in F_2$ is the identity. Then a ball of radius in $G$ is made of a ball of radius $n$ in $F_2$ times a ball of radius $n$ in $\mathbb{Z}$. The cardinality is consequently $2 \cdot 3^n -1$ times $2n+1$. But $b_n = 2(2n +1)(3^n-1)$ is not pinched growth. $\endgroup$
    – ARG
    Jan 4, 2021 at 19:40

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