3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.