Can a non-amenable group have a 'centrally invariant mean'? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T13:23:58Z http://mathoverflow.net/feeds/question/97728 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/97728/can-a-non-amenable-group-have-a-centrally-invariant-mean Can a non-amenable group have a 'centrally invariant mean'? Valerio Capraro 2012-05-23T08:30:22Z 2012-05-23T22:17:09Z <p>Let $G$ be a countable, discrete group, and $f\in\ell^\infty(G)$. Let me say that $f$ has a <em>centrally invariant mean</em> if there is a finitely additive probability measure $\mu$ on $G\times G$ such that $\int f(xgy)d\mu(x,y)=\int f(xy)d\mu(x,y)$, for all $g\in G$.</p> <blockquote> <p><strong>Question 1:</strong> Assume that every $f$ has a centrally invariant mean. Must $G$ be amenable?</p> </blockquote> <p>Since the <em>centrally invariant mean</em> may depend on $f$, this may seem a too weak condition to imply amenability (and maybe it is). However, let's consider the following apparently more plausible</p> <blockquote> <p><strong>Question 2:</strong> Assume $f$ has a centrally invariant mean. Does there exist another measure $\sigma$ on $G\times G$ such that $\int f(xyg)d\sigma(x,y)$ does not depend on $g$?</p> </blockquote> <p><strong>Remark.</strong> Positive answer to Q2 implies positive answer to Q1.</p> <p><em>Proof</em>. Let $\sigma$ as in Q2 and let $\sigma_\alpha$ be a net of countably additive probability measures on $G\times G$ which converges to $\sigma$ in the weak* topology. Define a net of countably additive probability measures on $G$ by setting $\mu_\alpha(x)=\sum_y\sigma_\alpha(xy^{-1},y)$. Let $\mu$ be a weak* limit of the net $\mu_\alpha$. A straightforward computation shows that $\int f(xg)d\mu(x)$ does not depend on $g$. Therefore every $f$ has a left invariant measure and this implies amenability, by the recent theorem of Justin Moore (see <a href="http://arxiv.org/abs/1106.3127" rel="nofollow">http://arxiv.org/abs/1106.3127</a>, Theorem 1.3).</p> <p>Thanks in advance for any help,</p> <p>Valerio</p>