Let $G$ be a finitely generated amenable group acting transitively on an amenable Schreier graph $S$. Is it possible for $S$ to have uncountably many ends? An amenable graph with uncountably many ends might be something like a slowly growing tree. I guess the answer is positive, but I'd like to see a simple example.
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$\begingroup$ Your first sentence is confusing to me: if $G$ is a group with a finite generating set and $H$ a subgroup, there is an action of $G$ on $G/H$, a ("Schreier") graph structure on $G/H$, but the action of $G$ on $G/H$ does not preserve the graph structure except in a few rare exceptions. Here are a two distinct interpretations of the question: 1) does there exist a f.g. amenable group with a Schreier graph with uncountably many ends? 2) does there exist a transitive action of a f.g. amenable group on a graph with infinitely many ends?... Could you clarify? $\endgroup$– YCorOct 18, 2014 at 23:21
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$\begingroup$ @YCor: I think of $S$ as being a graph with edges (and possibly loops) labelled by generators of $G$, and $G$ acting by isometries (by moving the vertices along the labelled arrows). Does that clarify what I mean here? $\endgroup$– Michal KotowskiOct 18, 2014 at 23:25
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1$\begingroup$ No, it seems you make a confusion. The action of $G$ on $G/H$ endowed with the Schreier graph metric is not by isometries in general. $\endgroup$– YCorOct 18, 2014 at 23:31
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$\begingroup$ Sorry, you are right, of course the action is not by isometries, I've edited the question. So the action doesn't preserve the graph structure. $\endgroup$– Michal KotowskiOct 18, 2014 at 23:34
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2$\begingroup$ Because you can artificially make the graph amenable by taking the pointed union with a bi-infinite line, adding an extra-generator $q$. Since then $[G,q]$ is contained in the set of finitely supporting permutations, you deduce that the new acting group $\langle G,q\rangle$ is amenable (if $G$ is amenable). Hence if you have an amenable group with a Schreier graph with uncountably many ends, you also get an amenable group with an amenable Schreier graph with uncountably many ends. $\endgroup$– YCorOct 18, 2014 at 23:55
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