## BI: Embedding hyperbolic spaces in products of trees

Lately, I’m having fun trying to quasi-isometrically embed metric spaces of all sorts in products of trees, like in this paper with David Hume (!). [EDIT: and this paper with John MacKay] This hobby started when David told me about this paper and related ones, where it is shown that nice hyperbolic spaces quasi-isometrically embed in the product of finitely many trees, the minimal number of trees being the dimension of the boundary plus one. Admitting such an embedding has nice consequences, for example having finite asymptotic dimension (also finite asymptotic Assouad-Nagata dimension, which is a stronger requirement). David is interested in such embeddings because another consequence is that your space has Hilbert compression exponent 1, which means that it gets very close to admitting a quasi-isometric embedding into a Hilbert space. Computer scientists are interested in this, apparently… Also, such embeddings give you information about the dimension of the asymptotic cones.

Ok, back to hyperbolic spaces. Disclaimer: I’m not sure that the authors think about their construction in the same way that I do… This is the general plan: “fill” your given hyperbolic space with trees, where each tree is assigned a colour so that trees with the same colour have “very little in common”, and finally “put together” trees of the same colour to form a bigger tree. In order to illustrate the meaning of the words in quote, let us consider $\mathbb{H}^2$, regarded as the universal cover of the genus 2 surface. Both $S_1$ and $S_2$ have free fundamental group, and their universal covers are fattened trees, ie., roughly, they contain a core tree and they retract onto it. So, you have a bunch of fattened trees covering $\mathbb{H}^2$, and you can colour them red or blue according to which subsurface they cover. Also, you have lipschitz retractions of $\mathbb{H}^2$ onto each of those trees: we can assume that the boundary curve of $S_1,S_2$ is a geodesic, so that all the fattened trees we are considering are convex in $\mathbb{H}^2$, and the retractions are closest point projections. Let me try to convince you with a picture that that the projection of a fattened tree onto another one with the same colour has bounded diameter, the projection of a red fattened tree onto another red fattened tree is depicted in black: Now, you can put together the core trees of all the fattened trees with the same colour to form trees $T_{red},T_{blue}$: just start with any core tree $T$, consider a nearby tree $T'$ of the same colour and glue an edge connecting, roughly, the projection of $T'$ on $T$ to that of $T$ on $T'$. Keep going.  You have a natural coarsely lipschitz map $\phi:\mathbb{H}^2\to T_{red}\times T_{blue}$. We have to understand why this map doesn’t reduce distances too much. This is not hard to see, if you have a geodesic $\gamma$ in $\mathbb{H}^2$ you can consider all its intersections with the fattened trees. Each of those subgeodesics  contribute at least its length minus some constant to the distance between the image of the endpoints of $\gamma$ through $\phi$. The constant is offset by the length of one of  the edges we’ve added.

This is it for $\mathbb{H}^2$, how should one proceed for other spaces? The idea is to construct the required trees by constructing the corresponding Cantor set in the boundary, the (quasi-)trees then being made of all bi-infinite geodesics connecting points in the Cantor set.

In order to construct the Cantor sets one first constructs a sequence of coloured coverings $\mathcal{U}_i$ satisfying a certain list of properties, here are some of the properties. Given one of those coverings, elements with the same colour are disjoint,  and elements in $\mathcal{U}_{i+1}$ are either contained or disjoint from any given set with the  same colour in $\mathcal{U}_i$. The diameters of the elements $\mathcal{U}_i$ tends to $0$ as $i\to\infty$. Here are some of the green sets of $\mathcal{U}_i$, $\mathcal{U}_{i+1}$, $\mathcal{U}_{i+2}$. The required Cantor sets arise in the following way. Start with an element $U$ of some $\mathcal{U}_i$. Then consider all elements in $\mathcal{U}_{i+1}$ with the same colour that are contained in $U$. Keep doing this, and take the intersection of all the sets you have considered in the process.

The properties I listed earlier don’t suffice to carry out the construction. One needs a quantitative control on the distance between elements of $\mathcal{U}_i$ with the same colour, meaning that it should be at least some constant times the max of the diameter of the elements of the cover. One can turn this quantitative control in the boundary into the required quantitative control in the interior. Ok, so you are left with finding coverings with the required properties. Indeed, if you’re familiar with the covering dimension, you’ve already spotted that the minimal number of colours needed is at least the topological dimension of the boundary plus one. Indeed, when dropping the quantitative control requirement, the existence of a sequence of coverings with the said properties and $n+1$ colours is equivalent to (and very close to being the usual definition of) having covering dimension at most $n$, say for compact metric spaces. One might have to use more colours than the covering dimension plus one, a priori, but this is enough in the case of boundaries of hyperbolic groups, as shown here. In any case, if you don’t care about the optimal number of trees, the fact that one can perform the construction with finitely many colours follows from the fact that boundaries of hyperbolic groups are doubling, i.e. balls of radius $r$ can be covered by at most $N$ balls of radius $r/2$, for some $N$ that does not depend on the ball.

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