Invariant means on an ideal

Abstract:

Let $G$ be a compact abelian group and $Q$ an invariant ideal of ${L^\infty }(G)$. Let ${M_Q}$ be the set of invariant means $\nu$ on ${L^\infty }(G)$ that are zero on $Q$, that is $\nu ({\chi _A}) = 1$ for ${\chi _A} \in Q$. We show that ${M_Q}$ is very large in the sense that a nonempty ${G_\delta }$ subset of ${M_Q}$ must contain a copy of $\beta {\mathbf {N}}$. Let ${E_Q}$ be the set of extreme points of ${M_Q}$. We show that its closure is very small in the sense that it contains no nonempty ${G_\delta }$ of ${M_Q}$. We also show that ${E_Q}$ is topologically very irregular in the sense that it contains no nonempty ${G_\delta }$ of its closure. The proofs are based on delicate constructions which rely on combinatorial type properties of abelian groups. Assume now that $G$ is locally compact, noncompact, nondiscrete and countable at infinity. Let $M$ be the set of invariant means on ${L^\infty }(G)$ and ${M_t}$, the set of topologically invariant means. We show that ${M_t}$ is very small in $M$. More precisely, each nonempty ${G_\delta }$ subset of $M$ contains a $\nu$ such that $\nu (f) = 1$ for some $f \in C(G)$]> with $0 \leqslant f \leqslant 1$ and the support of $f$ has a finite measure. Under continuum hypothesis, we also show that there exists points in ${M_t}$ which are extremal in $M$ (but, in general, ${M_t}$ is not a face of $M$, that is, not all the extreme points of ${M_t}$ are extremal in $M$).