Let $G$ be non-amenable finitely generated group.

1) Is it true that there exists a sequence $S(n)$ of sets which generate $G$ and such that

$\frac{1}{|S(n)|}||\sum_{g\in S(n)} \lambda(g)||\rightarrow 0$ when $n\rightarrow \infty$.

2) The same as (1), but $S(n)$ is finite subset of $G$.

Here $\lambda:G\rightarrow B(l^2(G))$ is left regular representation of $G$.

Also 1) is reformulation of 2).

Edit: here are some discussions on the question.

  • 1
    $\begingroup$ It must be the left regular representation, and the norm must be the operator norm in $l^{2}(G)$, right? $\endgroup$ – Jon Bannon Nov 4 '10 at 16:46
  • 1
    $\begingroup$ The questions, assuming Jon is right as to the notation, are equivalent: 1) obviously implies 2), and if you have a sequence $S(n)$ satisfying 2), then fix a finite generating set $T$ and consider the family $S'(n):=S(n)\cup T$. Simple argument using triangle inequality shows that $S'(n)$ give you the answer to the first question. I find the question interesting but I won't upvote unless you explain your notation explicitly. $\endgroup$ – Łukasz Grabowski Nov 4 '10 at 17:45
  • 1
    $\begingroup$ In both 1 and 2, S(n) is finite set Of course 1 and 2 are equivalent, I wanted to specify how 2 is related to generators of G in general. The given norm is the norm of the left regular representation, computed on $l^2(G)$. Of course if one assumes that $S(n)$ is given with "multiplicities", then the answer for the question is positive. $\endgroup$ – Kate Juschenko Nov 4 '10 at 21:25
  • 1
    $\begingroup$ @Kate: It's readable, and I think a really nice question! $\endgroup$ – Jon Bannon Nov 5 '10 at 12:29
  • 2
    $\begingroup$ I was thinking more along the lines of this old result of H. Kesten: ams.org/mathscinet-getitem?mr=112053 (at least in the case where we assume each S(n) to be symmetric) $\endgroup$ – Yemon Choi Nov 6 '10 at 0:10

It was proved by Nagnibeda-Smirnova and myself (http://arxiv.org/abs/1206.2183) that if a group $\Gamma$ contains an infinite normal subgroup $N$ such that $\Gamma/N$ is not amenable then the question above for $\Gamma$ is true. This gave many examples of groups which are non-amenable and does not contain $\mathbb{F}_2$ as a subgroup, such as Burnside and Golod-Shafarevych groups. As an easy consequence: if $\Gamma$ is not amenable then for $\Gamma\times \mathbb{Z}$ the question is true.

The question in the complete generality was finally solved by Andreas Thom (http://arxiv.org/abs/1306.1767)

There are applications to percolation theory (due to Nagnibeda-Pak) and to operator space analog of von Neumann's conjecture (due to Pisier) listed in the citations above.

To my current knowledge (please correct me here), the question itself should be contributed either to Gilles Pisier or to Nagnibeda-Pak.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.