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Let $G$ be a finitely generated group. The amenability of $G$ is equivalent to the existence of a certain "weak measure" on $G$. Is there such a characterization for residually amenable groups as well? That is:

Is the residual amenability of a discrete finitely generated group $G$ equivalent to the existence of a function $\mu \colon \mathcal{P}(G) \to \mathbb{R}$ satisfying certain properties?

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    $\begingroup$ Maybe in some artificial way, but I'm not aware of any useful statement of this kind. Moreover residual amenability, unlike amenability, does not pass to quotients, so this properties should not pass to push-forwards such as the existence of invariant means. $\endgroup$
    – YCor
    Commented Dec 28, 2014 at 19:46
  • $\begingroup$ @YCor Maybe if a group is residually amenable, then there is some way to induce a measure on it coming from the measures on the amenable images. If the group is residually finite then this is clearly possible. $\endgroup$
    – Pablo
    Commented Dec 28, 2014 at 19:57
  • $\begingroup$ Are you talking about Haar measure on the profinite completion? $\endgroup$
    – HJRW
    Commented Dec 28, 2014 at 21:05
  • $\begingroup$ @HJRW yes. I measure a set by taking the Haar measure of its closure inside the profinite completion. $\endgroup$
    – Pablo
    Commented Dec 29, 2014 at 7:47

1 Answer 1

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It is easy to see that a finitely generated group is residually amenable if and only if there exists an bi-invariant ultra-metric on $G$ and a finitely additive $G$-invariant measure on open (with respect to the metric) subsets of $G$.

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    $\begingroup$ (Note that Andreas probably assumes that the group is countable.) It's not easier for free groups than arbitrary groups. I expect that we consider a decreasing sequence of normal subgroups $(N_n)$ with $\bigcap N_n=1$ and $G/N_n$ amenable, then we say $d(g,h)=e^{-k}$ where $k=\sup\{m:g^{-1}h\in N_m\}$. (...) $\endgroup$
    – YCor
    Commented Dec 29, 2014 at 15:45
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    $\begingroup$ Then take a left $G$-invariant mean on $G/N_n$ and pull it back to a left $G$-invariant mean $m_n$ on $N_n$-invariant subsets of $G$. Then any limit point of the $m_n$ is a left $G$-invariant mean on subsets of $G$ that are $N_n$-invariant for some $n$, which are precisely open subsets for this metric. $\endgroup$
    – YCor
    Commented Dec 29, 2014 at 15:48
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    $\begingroup$ And conversely given these conditions, take $N_n$ to be the $e^{-n}$-ball, then $G/N_n$ is amenable and $\bigcap N_n=1$. Without the countability assumption this characterizes groups with a decreasing sequence $(N_n)$ of normal subgroups such that $\bigcap N_n=1$ and $G/N_n$ is amenable, or equivalently subgroups of countable products of amenable groups. (I need to think a little to find a residually amenable group -necessarily uncountable- not of this form) $\endgroup$
    – YCor
    Commented Dec 29, 2014 at 15:54
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    $\begingroup$ Here's one example of residually amenable group that's not "countably" residually amenable. Consider the infinite countable finitary alternating group $H$. Let $G=H^I=\prod_{i\in I}H_i$ where $I$ is an uncountable index set and $H_i$ is a copy of $H$. Then $H$ is locally finite, hence amenable; so $G$ is residually amenable; moreover $H$ is simple. Suppose by contradiction we have $(N_n)$ as above. For every $n$, let $I_n$ be the set of $i$ such that $H_i$ is not contained in $N_n$. $\endgroup$
    – YCor
    Commented Dec 29, 2014 at 16:27
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    $\begingroup$ Then $I=\bigcup I_n$ by injectivity of $G\to \prod G/N_n$. By simplicity of $H$, for every $i\in I_n$, $H_i$ is mapped injectively into $G/N_n$. Then since every nontrivial normal subgroup of $\prod_{i\in I_n}H_i$ intersects one of the factors, it follows that $G_n=\prod_{i\in I_n}H_i$ is mapped injectively into $G/N_n$. But since $H_i\simeq H$ contains all finite groups, if $I_n$ were infinite it would follow that $G_n$ would contain nonabelian free groups, contradicting the amenability of $G/N_n$. Hence each $I_n$ is finite and hence $I$ is countable, a contradiction. $\endgroup$
    – YCor
    Commented Dec 29, 2014 at 16:28

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