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Let $G$ be a finitely generated group, $S$ a fixed symmetric generating set and $B(n)$ the ball of radius $n$ about the identity with respect to the word length induced by $S$ on $G$.

Fix $k\geq1$ and denote by $\zeta_k(G,d_S)$ the infimum over $n\geq1$ of $\frac{|B(nk+k)|}{|B(nk)|}$.

Observe that:

  1. $\zeta_k(G,d_S)=1$ for all $k$ (well, $k=1$ is enough) implies that $G$ has sub-exponential growth.
  2. If $G$ has polynomial growth, then $\zeta_k(G,d_S)=1$, for all $k$ (Gromov + Pansu - by the way, is there a direct proof of this fact, without using such a big theorems?).

What happens in the middle? More formally:

Question: What can we say about $\zeta_k(G,d_S)$ when $G$ has intermediate growth? Is it always $1$? Is it always $>1$? Can be both?

Update: The answer has been provided by Martin Kassabov below: the condition $\zeta_k(G)=1$ is equivalent for $G$ to have sub-exponential growth.

Thanks in advance,

Valerio

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  • $\begingroup$ Nice question. My impression is that there is no simpler proof of 2. You can get some way beyond polynomial growth by quoting Shalom and Tao. $\endgroup$
    – Ben Green
    Commented Nov 10, 2011 at 11:02
  • $\begingroup$ See also mathoverflow.net/questions/36126/… $\endgroup$
    – Andreas Thom
    Commented Nov 10, 2011 at 16:06
  • $\begingroup$ Thank you Andreas, very interesting, even if it seems that doesn't help. $\endgroup$
    – Valerio Capraro
    Commented Nov 10, 2011 at 17:55

2 Answers 2

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If $G$ has sub exponential growth one has that $\lim_s \sqrt[s]{B(s)}=1$ if you assume that $\zeta_k(G) = c >1$ then by an easy induction we have $B(nk) > K c^n$ which implies that $$ \limsup_s \sqrt[s]{B(s)} > \limsup_n \sqrt[nk]{B(nk)} > \limsup_n \sqrt[nk]{kc^n} = \sqrt[k]{c} > 1 $$

Therefore $\zeta_k(G) \leq 1$, but it is clear that $\zeta_k(G)\geq 1$, i.e $\zeta_k(G)=1$.

I.e. for any group of sub exponential growth $\zeta_k(G) =1$.

The same is true if you replace the infimum in the the definition of $\zeta_k(G)$ with limsup, but the argument is more involved and uses that sub-multiplicaticity estimates.

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  • $\begingroup$ I take the last comment back -- using submultiplicativity one can only show that $lim_k \sqrt[k]{\bar zeta_k(G)} =1 $ where $\bar \zeta_k(G) = \limsup_n \frac{B(kn+k)}{B(kn)}$. $\endgroup$
    – kassabov
    Commented Nov 10, 2011 at 22:14
  • $\begingroup$ It seems good. This is nice, also because gives a direct proof in the case of polynomial growth, when all people with whom I talked was convinced of the necessity to use both Gromov and Pansu theorem. $\endgroup$
    – Valerio Capraro
    Commented Nov 10, 2011 at 22:45
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    $\begingroup$ When making my comment above I was addressing the question for "supremum" instead of "infimum". I feel one does need Gromov/Pansu to do that. $\endgroup$
    – Ben Green
    Commented Nov 11, 2011 at 10:02
  • $\begingroup$ @martin: do you think one use your corollary 1.3 to exibit at least a group of intermediate growth such that $\overline\zeta(G)=1$? $\endgroup$
    – Valerio Capraro
    Commented Nov 11, 2011 at 19:50
  • $\begingroup$ @valerio: I think most (if not all) groups of intermediate growth satisfy $\bar \zeta_k=1$. Our result with Igor is not strong enough to get the value of $\bar \zeta$ but my guess is that it is also 1. I feel that there are no groups (of intermediate growth) with $\bar \zeta >1$. $\endgroup$
    – kassabov
    Commented Nov 11, 2011 at 23:41
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Martin Kassabov just gave a seminar this Tuesday in Royal Holloway about joint work with Igor Pak, see http://arxiv.org/PS_cache/arxiv/pdf/1108/1108.0262v1.pdf. I think it might be useful to answer your question, but you will have to check the details.

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  • $\begingroup$ it seems to be very useful. In particular, corollary 1.3. Tomorrow I will have a look at the details. Thank you very much. $\endgroup$
    – Valerio Capraro
    Commented Nov 10, 2011 at 18:14
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    $\begingroup$ Yiftach: Thanks for advertising my work, but I think it has nothing to do with question $\endgroup$
    – kassabov
    Commented Nov 10, 2011 at 22:11
  • $\begingroup$ As long as I spelled your name correctly... Anyhow, your results, even if do not answer this questions directly, are interesting and relevant to the topic. $\endgroup$
    – Yiftach Barnea
    Commented Nov 11, 2011 at 11:58

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