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I am not very familiar with F, but I know that it can be realized as a group of homeomorphisms of the boundary of the binary tree. I also know that F cannot be realized as a group of graph automorphisms of any regular rooted tree because F is not residually finite. However, if we topologize our trees with the path metric, can F be realized as a group of continuous prefix-preserving transformations of a regular rooted tree (where the transformations need not be injective or surjective)? If you know the answer, could you provide a reference? Thanks!

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    $\begingroup$ Thomson's group $F$ is a group. How do you plan to define the inverse of a transformations that is not injective or surjective? $\endgroup$
    – André Henriques
    Commented Aug 24, 2011 at 18:07
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    $\begingroup$ Consider, for example, two functions $a$ and $e$ from $\{1,2,3\}$ to $\{1,2,3\}$ defined by $a(1)=2, a(2)=1, a(3)=1$ and $e(1)=1,e(2)=2$, and $e(3)=2$. Then you can check that $a^2=e$, $e^2=e$, and $ea=ae=e$. So the semigroup generated by $a$ and $e$ is a group, and in fact it is the cyclic group of order 2. So you have a group defined by functions that are not injective or surjective. I'm just wondering if something like this can happen with F on a regular rooted tree, where the transformations are continuous and level-preserving. $\endgroup$
    – dan
    Commented Aug 24, 2011 at 19:54
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    $\begingroup$ So, just to clarify, I'm not looking for a group action of F, it would have to be a semigroup action. $\endgroup$
    – dan
    Commented Aug 24, 2011 at 19:59
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    $\begingroup$ The semigroup action would have to be invertible on the image of the identity element (like in your example) which should be a subtree by continuity. So if you restrict to the subspace which is the image of the identity, you would get bijections of the subtree. It might be more natural to consider maps of trees up to homotopy, such as in the action of Out(F_n) on outer space. $\endgroup$
    – Ian Agol
    Commented Aug 24, 2011 at 22:52

3 Answers 3

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A semigroup of level preserving transformations of a rooted tree is still residually finite. The levels are still finite and so the actions on the levels separate points into finite semigroups. Thus no such representation exists.

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    $\begingroup$ Thanks for the response Ben. The functions I care about just need to be prefix-preserving, but they need not be level-preserving. Once you jettison level-preserving then you no longer have residually finite. $\endgroup$
    – dan
    Commented Aug 24, 2011 at 21:05
  • $\begingroup$ @Dan, from your comment above I thought you wanted level-preserving. If you want prefix-preserving, then you can already realize Thompson's group F by automorphisms. It is shown in the article by Grigorchuk, Nekrashevych and Suschanskii that Thompson's groups F and V are groups generated by invertible asychronous automata. Asynchronous automata compute prefix-preserving maps that are not level-preserving. $\endgroup$
    – Benjamin Steinberg
    Commented Aug 24, 2011 at 21:32
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    $\begingroup$ I've read that paper, and I thought that the automaton they give generates F, but only as boundary transformations (the automaton seems to mimic the usual realization of F as homeomorphisms of the Cantor set if you consider the states as transformations of the boundary). The boundary transformations generated by the automaton are bijective, but if you think of that automaton as giving transformations of the tree itself, then the states are no longer bijective, and it's not clear to me what semigroup is generated by the states. Anyway, I'm only interested in the tree, not its boundary. $\endgroup$
    – dan
    Commented Aug 24, 2011 at 22:32
  • $\begingroup$ I haven't looked at that part of the paper in several years, so I don't remember exactly how the states act on the tree. I guess it must be an imitation of the partial action RW is speaking of but where instead of being undefined off a subtree, it does garbage. $\endgroup$
    – Benjamin Steinberg
    Commented Aug 25, 2011 at 0:20
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There is a way to get $F$ to act on the infinite binary tree bijectively, but I doubt it satisfies most of your other requirements. It basically does something sensible with the "missing finite subtree." I have only partly checked this out (meaning it seems to check for one of the two relations needed).

We let $T$ be the set of finite words (including the empty word) on the alphabet $\{0,1\}$. This is a binary tree by letting the left child of $u\in T$ be $u0$ and the right child of $u$ be $u1$. We define two permutations of $T$.

The permutation $x_0$ is determined by the following rules:

$\emptyset\rightarrow 1$; $0\rightarrow \emptyset$; $00u\rightarrow 0u$; $01u\rightarrow 10u$; $1u\rightarrow 11u$.

The permutation $x_1$ is determined by the following rules:

$\emptyset\rightarrow \emptyset$; $0u\rightarrow 0u$; $1\rightarrow 11$; $10\rightarrow 1$; $100u\rightarrow 10u$; $101u\rightarrow 110u$; $11u\rightarrow 111u$.

These are the usual rules for the action of $x_0$ and $x_1$ in $F$ on infinite words in $\{0,1\}$ restricted to finite words and extended to the few cases that the rules usually omit.

As I said, it checks for the relation:

$(x_1)^{x_0x_0} = (x_1)^{x_0x_1}$.

Here $a^b$ means $b^{-1}ab$ and the actions are to be composed from left to right (they are right actions).

The other relation that defines $F$ with the one above is

$(x_1)^{x_0x_0x_0} = (x_1)^{x_0x_0x_1}$.

If the second fails while the first succeeds, I will be stunned.

Assuming that the second relation checks out (not too hard, I am just too lazy), then these two permutations of $T$ generate a copy of $F$. On "most" of $T$, the action agrees with the usual action. How well this cooperates with what you want is for you to decide.

The definitions can be tinkered with a bit. I doubt that the relations can survive a lot of tinkering though.

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  • $\begingroup$ I should point out that there is nothing new here. If all finite words in 0,1 are made infinite by appending an infinite string of repetitions of the letter "t" at the end and 0<t<1 is declared, then the actions above preserve lex order. Perhaps more succinctly, the actions preserve the infix order on the nodes of T. Thus the above is just a coding of the usual action of F on [0,1]. T naturally codes the containment relation of subintervals of [0,1]. F breaks those relations. It is unlikely that a friendly F action on T exists. $\endgroup$
    – Matt Brin
    Commented Jan 17, 2012 at 13:32
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There is a so-called group of hierarchomorphisms $\mathsf{Hier}(T)$ of a homogeneous tree $T$ introduced by Neretin. It consists of homeomorphisms of the boundary $\partial T$ which can be extended to $T$, except for a finite subtree, and is, in a sense, similar to the group of diffeomorphisms of the circle. Thompson's group $F$ can be realized as a subgroup of $\mathsf{Hier}(T)$.

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  • $\begingroup$ I think one has to be slightly careful what you mean by the group structure in the following sense. The classical Higman-Scott construction of F via partial auomorphisms of a rooted tree involves first composing partial functions and then extending maximally. $\endgroup$
    – Benjamin Steinberg
    Commented Aug 24, 2011 at 20:16
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    $\begingroup$ Thanks for the response. I'm not familiar with this work, but I do need to consider the entire tree and I can't do away with any finite subtrees. $\endgroup$
    – dan
    Commented Aug 24, 2011 at 21:05

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