Let $\Gamma$ be a group generated by symmetric finite set $S$ and acting on $X$. The Schreier graph of the action is the graph with vertex set $X$ and $(x,y)$ is an edge if there is $s\in S$ such that $x=sy$.

Does there exists a faithful transitive action of the Thompson group F on a discrete set $X$ such that the Schreier graph of this action do not contain a binary tree?

  • $\begingroup$ What do you mean by ``contain"? As a subgraph or coarsely embedded? $\endgroup$ – user6976 May 10 '13 at 1:06
  • $\begingroup$ For the applications the best would be to know that it does not coarsely contain. But firstly, I want to know about containment as a subgraph only (I suspect that there is no such action). $\endgroup$ – Kate Juschenko May 10 '13 at 1:30
  • $\begingroup$ The subgraph interpretation is more difficult because that depends on the generating set. $\endgroup$ – user6976 May 10 '13 at 2:33
  • $\begingroup$ I am thinking about "avoiding" of the free semigroup and this can not be done for known to me actions. But yes, you are right for the general situation. $\endgroup$ – Kate Juschenko May 10 '13 at 3:15
  • $\begingroup$ Free subsemigroups don't give you trees as subgraphs, only coarsely embedded trees. $\endgroup$ – user6976 May 10 '13 at 7:00

The answer to the question as stated is yes, there is no binary tree as a subgraph for the standard action with standard generators (but there is a binary tree for some generating set). It follows from Proposition 1 in this paper of Savchuk http://arxiv.org/pdf/0803.0043.pdf and can be checked directly also.

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