Stabilizer of a hyperpalne and cocompactness

Question

If $G$ acts cocompactly on a CAT($0$) cube complex $X$, I wonder why every hyperplane is acted on cocompactly by its stabilizer?

I have no idea where to start.

Answer 1

I'm going to make up some terminology for "mid-cubes" (I cannot remember the standard terminology).

And then I'm going to list some simple facts.

• Each $n$-cube $C \subset X$ has exactly $n$ "mid-cubes", each of dimension $n-1$.
• For each hyperplane $H$, the intersection $C \cap H$ is either empty or a mid-cube of $C$. Furthermore, $H$ is a union of mid-cubes.
• For each $n$-cube $C$ and each mid-cube $M$, there exists a unique hyperplane $H$ such that $m=C \cap H$.
• Since $G$ acts cocompactly on $X$, the set of $G$-orbits of cubes is finite.
• And since each cube has finitely many mid-cubes, the set of $G$-orbits of mid-cubes is also finite.

Now let $H$ be a hyperplane. It is a union of mid-cubes. From the list of properties above, it follows that there exists a finite set of mid-cubes $M_1,...,M_k$ in $H$ having the final property:

• For each mid-cube $M \subset H$ there exists a unique $i \in \{1,\ldots,k\}$ and an element $g \in G$ such that $g \cdot M_i = M$. Furthermore, $g(H)=H$ (because $H$ is the unique hyperplane containing $M$ and $H$ is also the unique hyperplane containing $M_i$).

It follows that the $\text{Stab}(H)$-orbit of the compact set $M_1 \cup \cdots \cup M_k$ is equal to $H$. And that is the definition of the statement "$\text{Stab}(H)$ acts cocompactly on $H$".