Let $G$ be an amenable group of exponential growth and let $S$ be a finite symmetric generating set. For each $k$, let $B_{k}$ be the closed ball of radius $k$ about the identity element in the corresponding Cayley graph of $G$ and let $b_{k} = |B_{k}|$. If $\lim b_{k+1}/b_{k}$ exists, then $\lim b_{k+1}/b_{k} = \lim b_{k}^{1/k} > 1$ and this easily implies that no subsequence of the $B_{k}$ forms a Folner sequence for $G$. But is this also true for those amenable groups of exponential growth for which $\lim b_{k+1}/b_{k}$ does not exist?

  • $\begingroup$ @Simon: I do not know any group for which the limit $\lim b_{k+1}/b_{k}$ does not exist. $\endgroup$ – Mark Sapir Oct 28 '10 at 15:21
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    $\begingroup$ @Mark: see mathoverflow.net/questions/36126/… $\endgroup$ – Andreas Thom Oct 28 '10 at 15:56
  • $\begingroup$ Just for completeness, Andreas Thom points to an answer where a very simple generating subset for $C_2\ast C_3\simeq PSL_2(\mathbf{Z})$ fails to have convergent $(b_{k+1}/b_k)$. (Of course this group is not amenable, so $\liminf b_{k+1}/b_k>1$ anyway.) $\endgroup$ – YCor Mar 28 '19 at 8:29
  • $\begingroup$ Also I guess that the main intended question is rather whether there's an amenable group of exponential growth for which $\liminf b_{k+1}/b_k=1$. [Rephrasing already made observations, exponential growth implies $\limsup b_{k+1}/b_k>1$.] $\endgroup$ – YCor Mar 28 '19 at 8:32
  • $\begingroup$ @YCor In fact, the example works not just for $C_2 * C_3$ but for any free product of two finite groups $A *B$ with the generating sets being $S = (A \setminus \{e\}) \cup (B \setminus \{e\})$ as long as those two groups have distinct cardinality. $\endgroup$ – ARG Mar 28 '19 at 14:48

I'm fairly sure [from personal communications] this question is (widely) open in general [and a negative answer would come as a surprise to many]. But the only groups I know of where it is known that $\lim \frac{b_{k+1}}{b_k}$ do not exist are the ones described here (as linked by Andreas Thom). Now, for these groups it is clear the answer is yes.

It turns out that, for the generating sets given there, these groups have "pinched" exponential growth. So as a consolation (or perhaps a tiny step of progress to answer the question) here is a partial answer for groups (and generating sets) that have pinched exponential growth.

Remark: for examples of such pairs and as well as groups where one generating set has pinched exponential growth while the other does not, see at the end of this answer.

Notation: Let $B(n)$ be the ball of radius $n$ and $b_n = |B(n)|$. Let $S(n) = B(n) \setminus B(n-1)$ (with $S(0) = \lbrace e_G \rbrace$) be the spheres and $s_n = |S(n)|$.

Definition: Say a pair $(G,S)$ (where $S$ is a finite generating set of the group $G$) has pinched exponential growth if there are constants $K<L \in\mathbb{R}$ and $M \in \mathbb{R}$ so that $K \mathrm{exp}(Mn) \leq b_n \leq L \mathrm{exp}(Mn)$.

Proposition: If $\mathrm{Cay}(G,S)$ has pinched exponential growth, then $\liminf \frac{b_{k+1}}{b_k} >1$.

Let $g = \lim_n b_n^{1/n} = \inf_n b_n^{1/n}$. Now, note there are infinitely many $n_i$ so that, for $n \leq n_i$, $s_{n_i} \geq s_n$ (otherwise $s_n$ is bounded and $b_n$ grows at most linearly). Clearly, $s_{n_i} \leq b_{n_i} \leq (n_i+1) s_{n_i}$, so $$ g = \liminf b_{n_i}^{1/n_i} \geq \liminf_i s_{n_i}^{1/n_i}. $$ But $b_n \leq \sum_{j=0}^n s_j$ so $$g = \liminf_i b_{n_i}^{1/n_i} \leq \liminf (n_i+1)^{1/n_i} s_{n_i}^{1/n_i} = \liminf |S(n_i)|^{1/n_i}. $$ Thus far we have $$g = \liminf |S(n_i)|^{1/n_i}.$$ However, $n \mapsto s_n$ is also sub-multiplicative, i.e. $s_{n+m} \leq s_n s_m$. Hence $$g = \lim_n s_n^{1/n} = \inf_n s_n^{1/n}.$$ The equality with the $\inf$ implies $s_n/g^n \geq 1$.

Next note that pinched exponential growth is equivalent to $\limsup b_n/g^n = L < \infty$ (for some $L$). Putting this together $$ \begin{array}{rll} \displaystyle \liminf \frac{s_{n+1}}{b_n} & \displaystyle = g \liminf \frac{s_{n+1}}{g^{n+1}} \cdot \frac{g^n}{b_n} \\ & \displaystyle \geq g \liminf \frac{s_{n+1}}{g^{n+1}} \bigg( \limsup \frac{b_n}{g^n} \bigg)^{-1} & \displaystyle \geq g /L \end{array} $$ This implies that $$ \frac{ |B(n+1)|}{|B(n)|} \geq 1 + \frac{g}{L} $$ as desired.

Three remarks on pinched exponential growth:

1- the property depends on the generating set. If $(G,S)$ has pinched exponential growth. Then $G \times \mathbb{Z}$ with "summed" generating set $\lbrace (s,0) \mid s \in S \rbrace \cup \lbrace (e_G,\pm 1) \rbrace$ has also pinched exponential growth. With the "product" generating set $\lbrace (s,\epsilon) \mid s \in S , \epsilon = \pm 1 \rbrace$ it does not have pinched growth. Also, $G \times G$ with the product generating set has pinched exponential growth, but with the summed generating set it does not.

2- with the "usual" generating sets, solvable Baumslag-Solitar and some lamplighters groups have pinched exponential growth (this can be seen on their growth series). For the growth series of $BS(1,n)$ see Collins, Edjvet and Gill, Growth series of the group $\langle x,y \mid x^{-1}y x=y^\ell \rangle$, Arch. Math. 62:1--11, 1994. For the growth series of lamplighters of the form $(\oplus_{i \in \mathbb{Z}} H) \rtimes \mathbb{Z}$) where the growth series of $H$ is known (i.e. $H$ is finite, or Abelian, or $BS(1,n)$, or a lamplighter or ...) see Johnson, Rational growth of wreath products, in "Groups St Andrews 1989" volume 2 309--315.

3- If the growth series is rational then having pinched exponential growth is the same as having exactly one root on the convergence radius. Groups with two roots on their convergence radius are known to exist, but I don't know if there is a group (with rational growth series and) 3 roots on the convergence radius.

  • $\begingroup$ I don't understand what you mean by pinched exponential growth. For it to make sense, you need to add "for every generating subset" or "for some generating subset", or replace "a group" with "a pair $(G,S)$ of a group with a generating subset"... $\endgroup$ – YCor Mar 28 '19 at 8:36
  • $\begingroup$ thanks, obviously as I mentioned at the end this property is highly dependent on the generating set. $\endgroup$ – ARG Mar 28 '19 at 9:06

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