Can a non-amenable group have a 'centrally invariant mean'?

2012-05-23T08:30:22Z

Let $G$ be a countable, discrete group, and $f\in\ell^\infty(G)$. Let me say that $f$ has a centrally invariant mean 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 $g\in G$.

Question 1: Assume that every $f$ has a centrally invariant mean. Must $G$ be amenable?

Since the centrally invariant mean 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

Question 2: 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 $g$?

Remark. Positive answer to Q2 implies positive answer to Q1.

Proof. 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 http://arxiv.org/abs/1106.3127, Theorem 1.3).

Thanks in advance for any help,

Valerio

Can a non-amenable group have a 'centrally invariant mean'? - MathOverflow

Can a non-amenable group have a 'centrally invariant mean'?