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Let $G$ be a group generated by $a,b$ (for the sake of simplicity). Consider the element $$S=a+b+a^{-1}+b^{-1}\in{\mathbb C}[G],$$ which may also be interpreted as an operator in $l^2(G)$ (by left regular representation). Obviously $\|S\|\le 4$, and by the old result of Kesten $\|S\|=4$ iff $G$ is amenable.

Now, suppose that $G$ is not amenable, that is, $\|S\|<4$. Is there an eigenvector in $l^2(G)$ corresponding to the eigenvalue $\|S\|$? (I do not know the answer even for a free group.)

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  • $\begingroup$ What is $\parallel S\parallel$ in the case of the free group? $\endgroup$
    – user1688
    Commented Jul 4, 2018 at 10:31
  • $\begingroup$ $S$ can be viewed as the adjacency operator of the Cayley graph. $\endgroup$
    – user1688
    Commented Jul 4, 2018 at 11:06
  • $\begingroup$ Corbennick@ For a free group, $\|S\|=2\sqrt{3}$ (which is the minimal possible value). $\endgroup$
    – Alex Gavrilov
    Commented Jul 4, 2018 at 11:34
  • $\begingroup$ You might want to look at the paper "Spectral Analysis for Adjacency Operators on Graphs" by Mantoiu et al. $\endgroup$
    – user1688
    Commented Jul 4, 2018 at 11:47
  • $\begingroup$ The answer is always negative for infinite groups. This is a property of the random walk on the group. I remember that a result in the book of Woess implies that $\tau(S^n)/\|S\|^n$ converges zero. However, the weak limit is (or contains) the spectral projection at the point $\|S\|$. $\endgroup$
    – Andreas Thom
    Commented Jul 8, 2018 at 7:17

1 Answer 1

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Linnell proved in his paper "Zero divisors and $\ell^2(G)$ C. R. Acad. Sci. Paris Sér. I Math. 315 (1992), no. 1, 49-53." that all elements of the group ring of a right orderable groups, containing Abelian and free groups, are nonzero divisors, i.e. if $0\neq\alpha\in\mathbb C[G]$, then for all $0\neq\beta\in\ell^2(G)$ we have $\alpha\beta\neq0\neq\beta\alpha$. So, in particular, the answer to your question in the case of right orderable groups is negative.

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