Does anybody know the actions of Thompson group F which are not conjugate to the standard one?

Motivation is to find actions such that the Schreier graph of the action does not contain a binary tree.

I've decided to ask a separate question on the motivation here

  • $\begingroup$ Actions on what? Circle? $\endgroup$ – Misha May 4 '13 at 15:38
  • $\begingroup$ Misha, yes, on circle. But I am also interested in any actions of this group. $\endgroup$ – Kate Juschenko May 4 '13 at 16:41
  • $\begingroup$ Misha, I am interested in those actions, where the Schreier graph of the action does not contain a binary tree. Does this kind of actions are known? $\endgroup$ – Kate Juschenko May 4 '13 at 16:56
  • $\begingroup$ The question seems to concern transitive actions on discrete sets. Actually, I don't know if $F$ admits a Schreier graph of subexponential growth (or even with no bilipschitz binary tree) besides the ones factoring through the action of an abelian quotient of $F$. $\endgroup$ – YCor May 4 '13 at 16:58
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    $\begingroup$ Yves, PS: sorry, I am thinking that I write clearly, but in fact I don't... $\endgroup$ – Kate Juschenko May 6 '13 at 21:56

$F$ is bi-orderable, so if you care about actions by homeomorphisms on the line, you can pick up a bi-ordering and produce a dynamical realization: a faithful almost free action ( see Proposition 3.4 and Example 3.5 in this paper of Navas http://arxiv.org/abs/0710.2466 for a "non-standard" action of $F$). For the discrete case, the proof of Proposition 1.8 in the same paper might be an inspiration... All the bi-orderings of $F$ were described by Navas and Rivas here http://arxiv.org/abs/0808.1688 . (As a side remark, in one dimensional dynamics one usually consider two actions equivalent if they are semi-conjugate, rather than conjugate)

(this was meant to be just a comment, but don't have enough points...:)


Well, every action of $F$ corresponds to a subgroup $H\leq F$ in the standard way. Specifically, the "standard" action on the interval corresponds to the stabilizers of various points in the interval. The Schreier graphs of these actions have been studied here.

If you want other actions of Thompson's group $F$, you need to look at subgroups that aren't just the stabilizer of a point. If you want to avoid a binary tree in the Schreier graph, you want this subgroup to be "large" enough so that every conjugate intersects the $\langle x_0,x_1\rangle$ monoid. However, if you want the action to be faithful, the subgroup cannot be so large that it contains the commutator subgroup.

Of course, it's not clear how to make this concept of "large" precise. I suppose one possible definition is "does not stabilize any point in $(0,1)$".

With that in mind, here are some relatively "large" subgroups of Thompson's group $F$. I have no idea whether their Schreier graphs contain binary trees.

  1. Given any subset of the dyadics (or indeed any subset of the interval), one could consider the consider certain Cantor sets inside the interval, such as the set of points whose binary expansion subgroup of elements of $F$ that stabilize that subset. One interesting subset to look at might be $\{\ldots,\frac{1}{16},\frac18,\frac14,\frac12,\frac34,\frac78,\frac{15}{16},\ldots\}$. Another might be the set of dyadics of the form $k/4^n$, where $k$ is an odd integer. Finally, one can consider certain Cantor sets inside of $[0,1]$, such as the set of points whose binary expansion has a "$0$" in every odd-numbered position.

  2. There is a copy of $F_3$ inside of $F$, which can be described as follows:

    • Start by labeling the interval $[0,1]$ with the letter $A$.
    • Now, whenever you subdivide an $A$ interval, label the left half $A$ and the right half $B$
    • Finally, whenever you subdivide a $B$ interval, label both halves $A$.
    • Let $H$ be the subgroup of elements of $F$ that map linearly between the intervals of two dyadic subdivisions in a label-preserving way. Then $H$ is isomorphic to $F_3$.

    This construction can be generalized to give a copy of $F_n$ for any $n$.

  3. More generally, it is possible to find copies of many different diagram groups inside of $F$ by using different labelings. The copy of $F_3$ above corresponds to the diagram group for the monoid presentation $\langle A,B \mid A=AB, B=A^2\rangle$. See this paper by Guba and Sapir for some further examples.

  4. Another source of "large" subgroups comes from various laminations of the unit disk. Given any lamination, one can consider the subgroup consisting of all elements of $F$ that preserve the lamination. These groups have not been studied at all, though in a recent preprint Bradley Forrest and I have considered an analogous subgroup of Thompson's group $T$.

  • $\begingroup$ Thanks Jim for this list of actions. Currently I don't see if any of those does not contain binary tree and are faithful. $\endgroup$ – Kate Juschenko May 6 '13 at 0:20
  • $\begingroup$ @Kate: a homomorphism from $F$ to any group is either injective of factors through the abelianization. In other words, any action (on any set) of $F$ not factoring through $\mathbf{Z}^2$ is faithful. Still equivalently, any nontrivial normal subgroup of $F$ contains $[F,F]$. $\endgroup$ – YCor May 6 '13 at 11:20
  • $\begingroup$ @Kate: Using Yves's argument, it follows that all of the actions above are faithful. I have no idea whether the Schreier graphs contain binary trees, although if I had to guess, I would guess the Schreier graph for every faithful action of $F$ contains a binary tree. $\endgroup$ – Jim Belk May 6 '13 at 16:36
  • $\begingroup$ Yves, yes, this is clear to me... But one needs to check this for the specified actions, which is what I just did. $\endgroup$ – Kate Juschenko May 6 '13 at 21:47
  • $\begingroup$ Jim, this would be interesting to know, my intuition suggests the same, but I might be wrong. $\endgroup$ – Kate Juschenko May 6 '13 at 21:50

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