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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?

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    $\begingroup$ 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. $\endgroup$ – YCor Aug 2 '19 at 11:49
  • $\begingroup$ 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.) $\endgroup$ – Victor Protsak Aug 2 '19 at 13:46
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Actually, having finitely many orbits of hyperplanes is quite common. For instance:

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 $Y$. $\square$

In other words, if we are looking at finitely generated groups only, we can always suppose that the cube complex contains only finitely many orbits of hyperplanes, without any assumption on the action.

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  • $\begingroup$ Actually, this proposition can be seen as consequence of an essentially trivial fact in the point of view of commensurating actions, see here (arXiv link). The proof of the commensurating fact is the first 5 lines of proof of Proposition 4B2. For the median graph (aka CAT(0) cube complex) application, see corollary 7E6. By the way Anthony, I knew this was known (as Caprace told me it was used several times in his paper with Sageev) but don't know an original reference, do you have one? $\endgroup$ – YCor Aug 2 '19 at 14:33
  • $\begingroup$ The argument I gave can be extracted from the proof of Theorem 5.1 in Sageev's thesis Ends of group pairs and non-positively curved cube complexes. $\endgroup$ – AGenevois Aug 2 '19 at 16:26

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