The kernel of all invariant means - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T04:03:45Z http://mathoverflow.net/feeds/question/104981 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104981/the-kernel-of-all-invariant-means The kernel of all invariant means Antoine 2012-08-18T12:08:29Z 2012-08-18T12:08:29Z <p>Let $G$ be discrete group which is amenable (i.e. it admits an left-invariant mean, i.e. a continuous positive normalised linear functional on $m:\ell^\infty(G) \to \mathbb{R}$ such that $\forall g \in G, m \circ \lambda_g = m$ where $\lambda_g : \ell^\infty(G) \to \ell^\infty(G)$ is the left-regular action of $G$). Let $\mathcal{M}$ be the set of all invariant means on $G$.</p> <p>Let $c_0(G)$ be the space of bounded functions decreasing to $0$ at $\infty$ (i.e. $f \in c_0(G)$ if $f$ is in the closure of the space of fintily supported functions, i.e. $f \in \ell^\infty(G)$ and $\forall \epsilon >0$ there exists a finite set $F \subset G$ with $\|f\|_{\ell^\infty(G \setminus F)} \leq \epsilon$).</p> <p>It's not difficult to see that $c_0(G)$ is included in the kernel of all (left- or right- or bi-)invariant mean, provided $G$ is infinite. When $G$ is finite, there is only one such mean, and it is trivial to compute the kernel. However the cardinality of $\mathcal{M}$ is otherwise quite big (if I remember correctly, it is uncountable for $G=\mathbb{Z}$).</p> <p>$\textbf{Question}$: Assuming $G$ is infinite, is there a description/characterisation of the elements of $\ell^\infty(G)$ which belong to the kernel of all the invariant means, i.e. $\cap_{m \in \mathcal{M}} \ker m$?</p>