## Gromov boundary of hyperbolic groups

May 16, 2011

As we have seen in pervious posts, the Cayley graphs of groups equipped with the word metric is a very special class of geodesic metric space – they are graphs that have tons of symmetries. Because of that symmetry, we can’t construct groups with any kind of Gromov boundary we want. In fact, there are only few possibilities and they look funny. In this post I want to introduce a result of Misha Kapovich and Bruce Kleiner that says:

Let $G$ be a hyperbolic group that’s not a semidirect product $H \ltimes N$ where $N$ is finite or virtually cyclic. (In those cases the boundary of $G$ can be obtained from the boundary of $H$\$).

Theorem: When $G$ has $1$-dimensional boundary, then the boundary is homeomorphic to either a Sierpinski carpet, a Menger curve or $S^1$.

OK. So what are those spaces? (don’t worry, I had no clue about what a ‘Menger curve’ is before reading this paper).

The Sierpinski carpet

(I believe most people have seen this one)

Start with the unit square, divide it into nine equal smaller squares, delete the middle one. Repeat the process to the eight remaining squares. and repeat… Of course we then take the infinite intersection to get a space with no interior.

Proposition: The Sierpinski carpet is (covering) 1-dimensional, connected, locally connected, has no local cut point (meaning we cannot make any open subset of it disconnected by removing a point).

Theorem: Any compact metrizable planar space satisfying the above property is a Sierpinski carpet.

The Menger curve

Now we go to $\mathbb{R}^3$, the Menger curve is the intersection of the Sierpinski carpet times the unit interval, one in each of the $x, y, z$ direction.

Equivalently, we may take the unit cube $[0,1]^3$, subtract the following seven smaller cubes in the middle: In the next stage, we delete the middle ‘cross’ from each of the remaining 20 cubes: Proceed, take intersection.

Proposition: The Menger curve is 1-dimensional, connected, locally connected, has no local cut point.

Note this is one dimensional because we can decompose the ‘curve’ to pieces of arbitrary small diameter by cutting along thin rectangular tubes, meaning if we take those pieces and slightly thicken them there is no triple intersections.

Theorem: Any compact metrizable nowhere planar (meaning no open set of it can be embedded in the plane) space satisfying the above property is a Menger curve.

Now we look at our theorem, infact the proof is merely a translation from the conditions on the group to topological properties of the boundary and then seeing the boundary as a topological space satisfies our universal properties.

A group being Gromov hyperbolic implies the boundary is compact metrizable.

No splitting over finite or virtually cyclic group implies the boundary is connected, locally connected and if it’s not $S^1$, then it has no local cut point.

Now what remains is to show, for groups, if the boundary is not planar then it’s nowhere planar. This is an easy argument using the fact that the group acts minimally on the boundary.

Please refer to first part of their paper for details and full proof of the theorem.

Remark: When study classical Polish-school topology, I never understood how on earth would one need all those universal properties (i.e. any xxx space is a xxx, usually comes with a long condition include ten or so items >.<). Now I see in fact such thing can be powerful. i.e. sometimes this allows us to actually get a grib on what does some completely unimaginable spaces actually look like!

Another wonderful example of this is the recent work of S. Hensel and P. Przytycki and the even more recent work of David Gabai which shows ending lamination spaces are Nobeling curves.

### 6 Responses to “Gromov boundary of hyperbolic groups”

1. alreadydone Says:

I guess the link of the “even more recent work” should be http://www.math.princeton.edu/facultypapers/Gabai/HighConnectivity0.913.pdf, but by now it is the same as the link of the “recent work”.

• 777 Says:

Oh yes… you are absolutely right! Thanks a lot for the careful reading~ (and I’m amazed that you actually found the correct paper…are you familiar with this work?)

• alreadydone Says:

Actually not. I just found the two links are the same, became curious and Googled with the keywords:) Although interested, I have not been engaged in research in this area. I’m only about to attend graduate school this year.

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