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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.

edited Apr 6 2011 at 19:37

asked Nov 4 2010 at 16:37

Kate Juschenko
3,814●5●27

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It must be the left regular representation, and the norm must be the operator norm in $l^{2}(G)$, right? – Jon Bannon Nov 4 2010 at 16:46

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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. – Łukasz Grabowski Nov 4 2010 at 17:45

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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. – Kate Juschenko Nov 4 2010 at 21:25

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@Kate: It's readable, and I think a really nice question! – Jon Bannon Nov 5 2010 at 12:29

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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) – Yemon Choi Nov 6 2010 at 0:10

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Amenability of groups

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