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?

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

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