## BI: Thickness

Thick metric spaces were defined in this paper by Jason Behrstock, Cornelia Druţu and Lee Mosher as an obstruction for a group (actually, a metric space) to be (properly) relatively hyerbolic.
Indeed, the notion of thickness provides a very interesting way of thinking about certain families of groups, I’ll give some examples later.
Throughout the post “is not relatively hyperbolic” will mean “is not hyperbolic relative to any collection of proper subgroups” (in this context, proper is equivalent to infinite index).
Ok, let’s start with noticing an easy obstruction to relative hyperbolicity: if the group $G$ has one asymptotic cone without cut-points then it is not relatively hyperbolic. If you’re scared of asymptotic cones, another way of formulating this property is the following: one can draw the following picture in the group for arbitrarily large $R$ and some large enough $C$. If you’re not happy yet, just think of products of infinite diameter spaces, like $\mathbb{R}^2$.
Groups with the aforementioned property are called unconstricted, or thick of order 0. There is also a uniform version of this property, that applies to families of metric spaces. I won’t spell this out, I trust you’re smart enough to figure it out.
We are ready to define inductively thickness of order at most $n+1$. A metric space is thick of order at most $n+1$ if it is covered by a family of subspaces which is uniformly thick of order $n$ and with the following further property. For any pair of subspaces $A, B$ there is a sequence $A=A_0,\dots,A_n=B$ so that $diam(A_i\cap A_{i+1})=\infty$.

Here is my favorite way of thinking about thick metric spaces. Your metric space is thick of order $n+1$ if it’s built up starting from a certain family of subspaces, which are thick of order $n$. In order to obtain the your metric space back you also need a certain graph whose vertices are the given subspaces, and such that adjacent vertices correspond to pairs of subspaces that have infinite diameter intersection.
It’s time for some examples (of thickness of order 1) that will hopefully clarify this graph thing. I’ll start with the example with the simplest geometry among those I’ll mention (IMHO), even though it’s probably the least popular one: graph manifold groups. Indeed, many other fundamental groups of graphs of groups can be treated in the same way. If you don’t know what they are, you can think of graph manifolds as 3-manifolds built up in the following way. You start with a bunch of surfaces with boundary, you take the product of each of those with $S^1$ and glue along boundary components swapping the surface factor and the $S^1$ factor. Sorry for the picture, I couldn’t do any better. What I just described to you is actually the class of flip graph manifolds. Each fundamental group of a grah manifold is quasi-isometric to that of a flip graph manifold. It’s easy to see that flip graph manifolds are $CAT(0)$, by the way.
The universal cover of a surface with boundary is a “fattened tree”, as the surface contains a graph as a deformation retract. So, the universal cover of a (flip) graph manifold is built up from fattened trees times $\mathbb{R}$ glued together along copies of $\mathbb{R}^2$. The graph encoding this gluings is the Bass-Serre tree. You should now see the thick structure of order 1.

Next example: mapping class groups. The mapping class group of an orientable surface $S$ is the group of self-diffeomorphisms preserving the orientation of $S$, up to isotopy (or homotopy if you prefer, it makes no difference).
If you’re thinking “so what?”, let me mention that all closed 3-manifolds can be obtained gluing two copies of a handlebody, i.e. the “thickening” of a graph as suggested picture below, and the resulting manifold only depends on the isotopy class of the diffeomorphism you’re using to glue, that is to say the corresponding elements of the mapping class group. But you’re still thinking “so what?”, I’m afraid. Suppose that the closed surface $S$ has genus at least 2. One can form a graph, called curve complex, whose vertices are (isotopy classes) of simple closed curves, and there is an edge connecting two vertices if the corresponding curves are disjoint. The curve complex is connected. Also, it has infinite diameter and most remarkably it is hyperbolic.
The collection of subsets providing the thick structure on the mapping class group $MCG(S)$ of $S$ is the family of all left cosets of stabilizers of (isotopy classes of) simple closed curves in $S$. Same objects, different description: put in the collection, for all pairs $\gamma, \delta$ of simple closed curves, the subset of $MCG(S)$ mapping $\gamma$ to $\delta$. Stabilizers of simple closed curves have infinite center (and are not virtually cyclic), and so they are unconstricted of order 0. (The center is generated by a Dehn twist along the curve, indeed there is a central extension $1\to\mathbb{Z}\to Stab(c)\to MCG(S\backslash c)\to 1$, where $c$ is the curve.) The graph encoding the gluings is (or, can be taken to be) the curve complex.
So, you can think about $MCG(S)$ in the following way. Above each vertex of the curve complex you have a copy the stabilizer of the corresponding curve, and the intersection between subsets sitting above adjacent vertices is a copy of the stabilizer of the union of the curves corresponding to the vertices.

As a final note, let me mention that many right angled Artin groups (AKA partially commmutative groups) and $Out(F_n)$ for $n\geq 3$ are thick of order 1 as well. In the case of RAAGs the thick structure is given by left cosets of maximal sub-RAAGs that split as direct products. I might come back on this sooner or later…

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