# Non-cocompact action on CAT(0) cube complex and hyperplane stabilizers

A hyperplane of a cube complex $$X$$ is a connected component of taking an $$(n-1)$$-cube for each midcube of $$X$$ and identifiying midcubes along faces of adjacent $$n$$-cubes of $$X$$.

If a group $$G$$ acts on a finite dimensional CAT(0) cube complex not cocompactly (with perhaps extra conditions), must there be an infinite number of orbits of hyperplanes $$G \hat{\mathfrak{h}}_{1}, G \hat{\mathfrak{h}}_{2}, \ldots$$ such that the stabilizers of these hyperplanes $$Stab(\hat{\mathfrak{h}_{i}}) \neq \{ 1 \}$$ are non-trivial?

• Let $\mathbf{Z}$ act on $\mathbf{Z}^2$ (the latter viewed as CAT(0) cube complex) by powers of $(m,n)\mapsto (m+1,n+1)$. Then it's not cocompact while there are only two hyperplane orbits. – YCor Aug 2 '19 at 11:49
• I also don't understand the non-trivial stabilizer stipulation. Say $G$ itself is trivial (and $X$ is non-compact), then all stabilizers are trivial, i.e. the set of orbits such that the stabilizers are non-trivial is empty, quite the opposite of being infinite. (Of course, the stabilizers may all be trivial in more elaborate examples, such as YCor's.) – Victor Protsak Aug 2 '19 at 13:46

Proposition: (Sageev) Let $$G$$ be a finitely generated group acting on a CAT(0) cube complex $$X$$. Then there exists a $$G$$-invariant convex subcomplex $$Y \subset X$$ containing only finitely many $$G$$-orbits of hyperplanes.
Sketch of proof. Let $$s_1, \ldots, s_n$$ be generators of $$G$$ and $$x_0 \in X$$ a vertex. Set $$Y$$ as the convex hull of the orbit $$G \cdot x_0$$. The hyperplanes of $$Y$$ are exactly the hyperplanes of $$X$$ separating two vertices of $$G \cdot x_0$$. Let $$J$$ be such a hyperplane. So there exist $$g,h \in G$$ such that $$J$$ separates $$gx_0$$ and $$hx_0$$. By translating, we may suppose that $$g=1$$. Write $$h$$ as a word of generators $$r_1 \cdots r_k$$. By looking at a path $$[x_0,r_1x_0] \cup [r_1x_0,r_1r_2 x_0] \cup \cdots \cup [r_1 \cdots r_{k-1}x_0, r_1 \cdots r_{k-1}r_k x_0]$$ from $$x_0$$ and $$hx_0$$, we find that there exists some $$i$$ such that $$J$$ separates $$r_1 \cdots r_ix_0$$ and $$r_1 \cdots r_ir_{i+1} x_0$$. Up to translating $$J$$, we may suppose that $$J$$ separates $$x_0$$ and $$r_{i+1}x_0$$.
In other words, any hyperplane of $$Y$$ has a translate which separates $$x_0$$ from $$s x_0$$ for some generator $$s$$. It follows that there exist only finitely many orbits of hyperplanes in $$Y$$. $$\square$$