# Parallel hyperplane to a hyperbolic isometry of a CAT(0)-cube complex

Consider a finite-dimensional geodesically complete CAT(0) cube complex $$X$$.

A hyperplane of $$X$$ is a convex subspace of $$X$$ that intersects the mid-point of edges of $$X$$ such that $$X - H$$ has two components.

A hyperbolic isometry $$\gamma$$ of $$X$$ is parallel to a hyperplane $$\hat{\mathfrak{h}}$$ if there exists an axis for $$\gamma$$ which is contained in a neighbourhood of $$\hat{\mathfrak{h}}$$.

Is every hyperbolic isometry of a geodesically complete finite dimensional CAT(0)-cube complex parallel to a hyperplane?

• The answer is negative even for simplicial trees. – AGenevois Aug 1 '19 at 13:04
• Please avoid creating a new account each time you're asking another question (cf mathoverflow.net/questions/337469/…) – YCor Aug 3 '19 at 10:53

As pointed out in the comments, the answer is negative. But we can prove even more:

Proposition: Let $$G$$ be a group acting properly and cocompactly on a CAT(0) cube complex $$X$$. Then there exist a $$G$$-invariant convex subcomplex $$Y \subset X$$ and an isometry $$g \in G$$ which not parallel to any of the hyperplanes of $$Y$$.

In other words, the question has a negative answer in most of the reasonnable situations.

Sketch of proof. Let $$Y \subset X$$ be a convex subcomplex on which $$G$$ acts essentially [CS, Proposition 3.5]. Decompose $$Y$$ as a Cartesian product $$Y_1 \times \cdots \times Y_n$$ of irreducible cube complexes and fix a finite-index subgroup $$\dot{G} \leq G$$ which preserves the product structure. Now look at the induced (cocompact) action $$\dot{G} \curvearrowright Y_i$$. As a consequence of [CS, Theorem 6.3], $$\dot{G}$$ contains a strongly contracting isometry of $$Y_i$$ (i.e., an isometry which skewers a pair of strongly separated hyperplanes).

Thus, $$\dot{G}$$ acts simultaneously on $$n$$ hyperbolic graphs, namely the contact graphs of $$Y_1, \ldots, Y_n$$, and $$\dot{G}$$ contains a loxodromic isometry for each action. By applying [CU], it follows that $$\dot{G}$$ contains an element inducing a loxodromic isometry simultaneously in all the contact graphs. Such an element, thought of as an isometry of $$Y$$, cannot be parallel to any hyperplane. $$\square$$

[CS] P.-E. Caprace and M. Sageev, Rank rigidity for CAT(0) cube complexes.

[CU] M. Clay and C. Uyanik, Simultaneous construction of hyperbolic isometries.