Le Conte de la Mesure sur les Complexes Cubiques CAT(0)

A useful way to think of CAT(0) cube complexes is via this functorial construction. Begin with a set $Y$ together with a collection of (nonempty) two-sided (walls) partitions $\mathcal{P}\subset 2^{Y}$. Two sided here means that if $h\in\mathcal{P}$ then $Y\setminus h\in\mathcal{P}$. The wall may be thought of as the pair $\{h,Y\setminus h\}$.

Next, create a graph by declaring that each maximal non-empty intersection of sets from $\mathcal{P}$ is a vertex. We connect two vertices if their corresponding defining intersections differ by the choice of exactly one side of one partition from $\mathcal{P}$. In Figure 2, you can see a choice of 5 partitions. Each region is not empty. The regions labeled $A$ and $U_{1}$ differ by the choice of one partition and therefore, their associated vertices $a$ and $u_{1}$ respectively are connected by an edge.

Once we have done this, we may not have a CAT(0) cube complex. However, we may take the cubical completion (as in Figure 2, where we must add the unlabled “back” vertex to creat a 3-cube) to annul positive curvature. We note that this can always be done, except that distances may become infinite (if there are infinitely many partitions separating two regions, i.e. the finite interval condition is not satisfied) or the dimension may become infinite (if there are families of pairwise transverse partitions of unbounded cardinality, see Section 0.3 for the definition of transverse).

Conversely, start with (the vertex set of) a CAT(0) cube complex $X$. Each edge in $X$ belongs to an equivalence class generated by “being parallel across a square”. The compliment of that parallelism class has two sides, and the vertices that belong to each side give the two-sided partition of the vertex set (see Figure 3).

Finally, if we apply the previous construction to this collection of two-sided partitions, we get a CAT(0) cube complex, which is cannonically isomorphic to (the vertex set of) $X$.

A halfspace of a CAT(0) cube complex is one side of a two-sided partition, as discussed above. We have established that the vertex set, denoted by $X$, is cannonically determined by the halfspace structure $\mathfrak{H}\subset 2^{X}$. By mapping a vertex in $X$ to the collection of halfspaces that contain it, we obtain an isometric injection:

Here the extended metric on $2^{\mathfrak{H}}$ is given by half the cardinality of the symmetric difference $d(S,T)=\frac{1}{2}(S\triangle T)\in[0,\infty]$. While this metric is useful, we will rely on the standard topology on $2^{\mathfrak{H}}\cong\mathrm{Map}(\mathfrak{H}\to\{0,1\})$. This is given by declaring the basic open sets to be cylinder sets, which are themeselves determined by specifying values in finitely many coordinates. Equivalently, this is the topology of pointwise convergence of maps $\mathfrak{H}\to\{0,1\}$. With this topology, $2^{\mathfrak{H}}$ is compact. We note that the induced topology may be different than the metric topology on $X$. This is directly comparable to the weak-$*$ topology and respectively the metric topology on a Hilbert space.

Having found an injection of $X$ into the compact space $2^{\mathfrak{H}}$, we may take the closure in the image and this defines the Roller compactification. By removing the image of $X$ inside this closure, we are left with the Roller boundary.

Note that once we obtain the closure of $X$ in $2^{\mathfrak{H}}$, there is a cannonical extension of the halfspaces as partitions of $X$ to $\overline{X}$. Those are the basic clopen sets for our totally disconnected topology on $\overline{X}$.

The following is then immediate from the functorial construction, and can be thought of as forgetfulness.

We shall see in Section Lemma how and when we can find a cannonical section to this map. The example in Figure 4 is given by taking $\mathfrak{H}^{\prime}=\mathfrak{H}\setminus\mathcal{B}$ and a valence consideration shows that an isometric section is impossible.

The Roller boundary of a tree is, as a set, the visual boundary, but the topology is different when it is not (large-scale) locally finite. Consider the tree given by identifying at $0$ infinitely many copies of $[0,\infty]$, as in Figure 5. It is not difficult to see that $\infty_{n}\to 0$ since all halfspaces eventually contain $0$.

The Roller compactification has an important multiplicative property:

Combining this, with the discussion for trees, we obtain the Roller compactification of $\mathbb{Z}^{2}$, as in Figure 6, or for $F_{2}\times\mathbb{Z}$ as in Figure 7.

Given a pair of halfspaces $h,k\in\mathfrak{H}$ such that $k\neq h,h^{*}$ then either all of the pairwise intersections $h\cap k,h\cap k^{*},h^{*}\cap k,h^{*}\cap k^{*}$ are not empty (and in this case, we say that $h$ and $k$ are transverse, and write $h\pitchfork k$) or one of the following other cases hold (see Figure 8):

It is straightforward to verify that if $\mathfrak{H}=\mathfrak{H}_{1}\sqcup\mathfrak{H}_{2}$ is a nontrivial disjoint decomposition into nonempty involution invariant, pairwise transverse sets and $X$, $X_{1}$, and $X_{2}$ are the assocaited CAT(0) cube complexes respectively then there is a cannonical isomorphism $X\cong X_{1}\times X_{2}$. We say that $X$ is irreducible if it is not isomorphic to a product.

A pair of halfspaces $h,k\in\mathfrak{H}$, with $h\neq k^{*}$ are said to be facing if $h^{*}\cap k^{*}=\varnothing$. This corresponds to the lower right corner in Figure 8. A triple of halfspaces $h,k,\ell\in\mathfrak{H}$ are said to be a facing triple if they are pairwise facing.

Let $\mathcal{I}\subset\overline{X}$ be an arbitrary subset. The (involution invariant) collection of halfspaces that separate points in $\mathcal{I}$ is

Similarly, we denote the collection of halfspaces that contain $\mathcal{I}$ as $\mathfrak{H}_{\mathcal{I}}^{+}$, and the collection of compliments of halfspaces that contain $\mathcal{I}$, namely the halfspaces that trivially intersect $\mathcal{I}$ is denoted by $\mathfrak{H}_{\mathcal{I}}^{-}$, meaning that $\mathfrak{H}_{\mathcal{I}}^{-}=(\mathfrak{H}_{\mathcal{I}}^{+})^{*}$. We have established the following decomposition:

We say that, $\mathcal{I}$ is an interval if there exists $x,y\in\overline{X}$ such that $\mathcal{I}=\underset{h\in\mathfrak{H}_{\{x,y\}}^{+}}{\overset{}{\cap}}h$. In this case we write $\mathcal{I}=\mathcal{I}(x,y)$ and say that it is the interval between $x$ and $y$. We also call $x$ and $y$ endpoints of $\mathcal{I}$. Clearly, $\mathfrak{H}_{\mathcal{I}(x,y)}$ does not contain a facing triple.

The structure of intervals in an arbitrary CAT(0) cube complex can be quite exotic; please see the examples in Figures 9 and 10 or come up with your own!

Nevertheless, the following theorem of [BCG⁺09], which relies on Dilworth’s Theorem for partially ordered sets, shows that intervals are not too wild.

The following is an important property of halfspaces in a CAT(0) cube complex. The first item is the standard Helly property for halfspaces. The second follows from the first by applying the finite intersection property to the collection of sets $\mathcal{S}$, each of which is clopen in the compact space $\overline{X}$. Comparing it with the classical Helly property for convex sets in euclidean space, we may consider it as telling us that, in a sense, CAT(0) cube complexes are of Helly-dimension 1.

Lifting Decompositions

Let $X$ be a CAT(0) cube complex with associated halfspace structure $\mathfrak{H}$. We shall say $\mathcal{S}\subset\mathfrak{H}$ is consistent if it satisfies the following two properties:

1. (1)

   If $h\in\mathcal{S}$ then $h^{*}\notin\mathcal{S}$.

2. (2)

   If $h,k\in\mathfrak{H}$, with $h\subset k$, and $h\in\mathcal{S}$ then $k\in\mathcal{S}$.

We note that, according to the functorial construction, if $\mathcal{S}$ is consistent and $\mathcal{S}\sqcup\mathcal{S}^{*}=\mathfrak{H}$ then $\mathcal{S}$ is the collection of halfspaces containing a single point.

Consider a probability measure $\mu\in\textrm{Prob}(\overline{X})$ and the associated collection of halfspaces $\mathfrak{H}_{\mu}^{+}:=\{h\in\mathfrak{H}:\mu(h)>1/2\}$. It is straightforward to verify that $\mathfrak{H}_{\mu}^{+}$ is consistent and that the collection of halfspaces $\mathfrak{H}_{\mu}=\{h\in\mathfrak{H}:\mu(h)=1/2\}$ does not contain any facing triples. Applying the Lifting Decomposition to $\mathfrak{H}_{\mu}^{+}$, we get that $\underset{h\in\mathfrak{H}^{+}_{\mu}}{\overset{}{\cap}}\,h$ is isomorphic to $\overline{X}(\mathfrak{H}_{\mu})$.

Let $x,y,z\in\overline{X}$ and consider the associated probability measure $\mu_{(x,y,z)}=\frac{1}{3}(\delta_{x}+\delta_{y}+\delta_{z})$. A simple parity argument shows that if $h\in\mathfrak{H}$ then $\mu_{(x,y,z)}(h)\neq\frac{1}{2}$. Therefore, we have that $\underset{h\in\mathfrak{H}_{\mu}^{+}}{\overset{}{\cap}}h$ is a single point, which we shall call the median of the triple and denote it by $m(x,y,z)$.

While this is not the standard definition of the median, it fits nicely within our context. We note that several of the natural properties of the median (e.g. invariance under permutation of the points) are immediate, including the property that

Proof of the Main Theorem

Suppose $\Gamma$ is amenable acting on $X$. Since $\overline{X}$ is compact and metrizable, there must be a $\Gamma$-invariant probability measure $\mu\in\textrm{Prob}(\overline{X})$. By the Lifting Decomposition and the previous lemma, it follows that $\underset{h\in\mathfrak{H}_{\mu}^{+}}{\overset{\,}{\cap}}h$ is a $\Gamma$-invariant interval.

Suppose now that $X$ is of finite dimension $D$ and that $\Gamma$ is not necessarily amenable but preserves an interval $\mathcal{I}\subset\overline{X}$. By Corollary 0.4 (see also [Fer18, Corollary 2.9]), the number of end points on which $\mathcal{I}$ is an interval is $2^{D^{\prime}}$ for some $0\leqslant{D^{\prime}}\leqslant D$ and we identify these with $\left({\mathbb{Z}/2\mathbb{Z}}\right)^{D^{\prime}}\cong\{0,1\}^{D^{\prime}}$. By [CS11, Proposition 2.6] we have that $\text{Aut}\,(\{0,1\}^{D^{\prime}})\cong\{0,1\}^{D^{\prime}}\rtimes\mathrm{Sym}({D^{\prime}})$. We may therefore project $\Gamma\to\text{Aut}\,(\{0,1\}^{D^{\prime}})$, and without loss of generality, assume $\Gamma\leqslant\text{Aut}\,(\{0,1\}^{D^{\prime}})$.

Let $\Gamma_{0}=\Gamma\cap\{0,1\}^{D^{\prime}}\lhd\Gamma$ and note that $|\Gamma_{0}|=2^{N}$ for some $0\leqslant N\leqslant{D^{\prime}}$. Fix a choice of right $\Gamma_{0}$-coset representatives $S\subset\Gamma$, with trivial $\{0,1\}^{D^{\prime}}$-coordinate. Let $\mathcal{O}\in\{0,1\}^{D^{\prime}}$ be defined component-wise by $\mathcal{O}_{i}\equiv 0$, for $i\in\{1,\dots,D^{\prime}\}$. Note that $\mathcal{O}$ is fixed by $\mathrm{Sym}({D^{\prime}})$ and has trivial stabilizer in $\Gamma_{0}$. Let $(\mathcal{O},\alpha)\in S$. Then $\Gamma_{0}(\mathcal{O},\alpha).\mathcal{O}=\Gamma_{0}.\mathcal{O}$. We have shown that

Therefore, by the orbit stabilizer theorem, we have