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Let $G$ be a group acting on a locally finite tree $T$. Then the boundary $\partial T$ is a Cantor set on which $G$ acts by homeomorphisms (indeed by quasi-isometries under a suitable metric). However, even if $G$ is the full automorphism group of $T$, we can't get the full quasi-isometry group of $\partial T$, as the tree structure puts further restrictions on the action.

What ways are known of identifying those group actions by homeomorphisms on the Cantor set which arise as actions on the boundary of a locally finite tree? Is there a nice algebraic description of $\mathrm{Aut}(T)$ as a subgroup of $\mathrm{Homeo}(\partial T)$?

Edit: Here is an example of a sufficient criterion to show what I mean. Suppose that $G$ acts on the Cantor set $X$ by homeomorphisms, with finite orbits on the clopen subsets. Then one can produce a tree $T$ such that $G$ acts on $T$ with a global fixed point and $\partial T$ is $G$-homeomorphic to $X$. So an action with finite orbits on the clopen subsets is automatically 'arboreal' (we do not care about which tree appears, just that it exists).

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you should clarify: the boundary of T is bounded, so a quasi-isometry is the same as a self-bijection (not necessarily continuous), so this is not very interesting to consider. –  YCor May 18 '12 at 14:05
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There are certainly necessary conditions, such as each infinite element $g$ which does not preserve any Borel probability measure on $\partial T$ must have 2 fixed points and act with north-south dynamics. This corresponds to the fact that any group element acting on a tree has a fixed point (elliptic) or has non-zero translation length (hyperbolic), in which case it preserves an axis. –  Ian Agol May 18 '12 at 14:14
    
@Agol: Every isometry preserves a Borel probability measure on the boundary :) –  YCor May 18 '12 at 22:09
    
@Yves: I see, I probably should have added some more conditions on the measure. I think the condition could be changed to either an element lies in a compact subgroup of $Aut(\partial{\mathcal{T}})$ or it acts with north-south dynamics. –  Ian Agol May 19 '12 at 5:49
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The term for that behavior is "bilipschitz" (of course, "lipschitz" means just bounded above). However, what metric on $\partial T$ do you intend to use? The boundary Cantor set of a tree $T$ has many metrics, some more and some less related to the tree $T$ itself. –  Lee Mosher May 21 '12 at 12:45

3 Answers 3

up vote 10 down vote accepted

Besides our paper that Alessandro mentions, Sageev, Whyte and I also have a paper "Maximally Symmetric Trees" which comes pretty close to doing what you ask. For certain trees it gives nice descriptions of $Aut(T)$ in terms of the representation into the quasi-isometry group $QI(T)$ (which, by the paper "Un groupe hyperbolique est determine par son bord" of Paulin, is characterized as the quasiconformal subgroup of $Homeo(\partial T)$ in the appropriate sense).

Although our paper applies only indirectly to all the trees $T$ relevant to your question, it applies very directly to all those trees which are "maximally symmetric", meaning that any continuous, proper, cocompact monomorphism from $Aut(T)$ into $Aut(T')$ for some other tree $T'$ is an isomorphism. Furthermore, the paper describes an explicit way to "expand" any tree $T$ to a maximally symmetric tree $T'$ (equivalently, to embed $Aut(T)$ continuously, properly, and cocompactly in $Aut(T')$), and it gives an explicit enumeration of maximally symmetric trees. The main theorem says that $T$ is maximally symmetric if and only if $Aut(T)$ is a maximal uniform cobounded subgroup of $QI(T)$.

In all of this, I should clarify by saying that the trees in question are always finite valence trees on which the automorphism group acts cocompactly.

ADDED LATER: I'll admit that this characterization of $Aut(T)$ is more geometric than algebraic, but it at least has well-known analogues in quasi-isometric geometry. For instance it is the analogue of the fact that $Isom(H^3)$, the isometry group of hyperbolic 3-space, is a maximal uniform cobounded subgroup of $QI(H^3)$ (the latter being identified with $Conf(S^2)$, the group of conformal homeomorphisms of the 2-sphere at infinity). The difference is that up to conjugacy there is only one maximal uniform cobounded subgroup of $QI(H^3)$, whereas maximally symmetric trees give countably many different conjugacy classes of maximal uniform cobounded subgroups of $QI(T)$.

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Thanks, I will have to have a look at these papers. –  Colin Reid May 21 '12 at 12:38

You may wish to take a look at "Quasi-actions on trees. I." by Mosher, Sageev and Whyte. They consider cobounded quasi-actions by quasi-isometries with bounded constants on trees with bounded valence and show that they are conjugate to actual isometric actions on trees. As noted in Theorem 5 of the same paper, such actions correspond to uniformly quasiconformal actions on the boundary which are cocompact on the space of triples.

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I'll focus on potential topological characterizations of the action. As I mentioned in the comments above, every element of $G$ will either lie in a maximal compact subgroup of $Aut(\partial{T})$, or will have north-south dynamics (a hyperbolic element).

If the group $G$ acts discretely, cocompactly on $\mathcal{T}$, then there's Bowditch's characterization that this is equivalent to $G$ acting properly dicontinuously and cocompactly on the triple points in $\partial{T}$. In fact, in this case $G$ will be a virtually free hyperbolic group, and a (finite) graph of finite groups.

I believe that $Aut(\mathcal{T})$ is closed in $Aut(\partial{\mathcal{T}})$, with the induced topology. In this case, one may assume that $G\leq Aut(\mathcal{T})$ is closed, otherwise taking its closure $\overline{G}$ in $Aut(\partial{\mathcal{T}})$ will correspond to taking its closure in $Aut(\mathcal{T})$, and if $\overline{G}\leq Aut(\mathcal{T})$, then $G\leq Aut(\mathcal{T})$. So assume now that $G$ is closed in $Aut(\partial{\mathcal{T}})$.

If $G$ is a compact subgroup of $Aut(\partial{\mathcal{T}})$, then I think it should be a profinite group, which acts on a tree (elliptically with a global fixed point). However, I'm not sure exactly how to tell if the action of $G$ on $\partial{\mathcal{T}}$ corresponds to this group action. Let's assume from now on that $G$ is not compact.

If $G$ does not act cocompactly on $\mathcal{T}$, then consider the limit set $\Lambda(G)\subset \partial{\mathcal{T}}$, which is the set of accumulation points of $Gx\subset \partial{\mathcal{T}}$ for any $x\in\partial{\mathcal{T}}$. One may realize this as the closure of the limit points of hyperbolic elements of $G$. Then $G$ should act cocompactly on the convex hull of $\Lambda(G)=\mathcal{R}$ inside of $\mathcal{T}$. So one may replace $\mathcal{T}$ with $\mathcal{R}$, and $\partial{\mathcal{T}}$ with $\partial{\mathcal{R}}=\Lambda(G)$.

So now assume that $G$ is a closed noncompact subgroup of $Aut(\partial{\mathcal{T}})$, and $\Lambda(G)=\partial{\mathcal{T}}$. Now, I believe that there should be a generalization of Bowditch's theorem. $G$ should be hyperbolic as a topological group, generated by a compact subset. I think this should be equivalent to $G$ acting properly cocompactly on the triple point space of $\partial{\mathcal{T}}$, but I don't know if this is proven (or even correct). Compare to the action of $Isom(\mathbb{H}^n)$ on $(\partial\mathbb{H}^n)^3-\Delta$, which is proper and cocompact.

There is another possible topological characterization in terms of actions on spaces of walls, but this is really just an encoding of the tree on $\partial{T}$ in terms of how each edge partitions $\partial{T}$ into pairs of clopen sets. If $G$ is closed, but noncompact and nondiscrete, then one might be able to encode the tree by the maximal compact subgroups of $G$ and their intersections, essentially Bass-Serre theory. But I haven't thought this through.

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@Ian: In this degree of generality, you will not get compactly generated groups. For instance, take an infinitely generated subgroup $G$ of a f.g. free group $F$ and let $G$ act on the tree which is the Cayley graph of $F$. So, I think you have to assume that $G$ acts on the set of triples in the Cantor set cocompactly, but not a priori discretely. –  Misha May 19 '12 at 14:48
    
Right, I see, I should only consider cocompact actions. –  Ian Agol May 20 '12 at 1:44

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