# Contacting Hyperplanes in a CAT(0) cube complex

I've been reading about CAT(0) cube complexes and have come across an assertion about contacting hyperplanes that I can not figure out.

First, we say that two hyperplanes $V, W$ cross if there is a $2$-cube whose distinct midcubes are contained in $V, W$ respectively. Two hyperplanes, $V, W$ osculate if they do not cross and there exist distinct $1$-cubes dual to $V,W$ which share a vertex. We say two hyperplanes contact if they cross or osculate, which is if $N(V) \cap N(W) \neq \emptyset$. (Where $N(V)$ is the carrier of the hyperplane $V$.)

(These definitions are from Mark Hagen's paper, Weak Hyperbolicity of Cube Complexes and Quasi-Arboreal Groups https://arxiv.org/pdf/1101.5191.pdf )

This paper states two hyperplanes $V,W$ contact if and only if no other hyperplane $U$ separates $V$ and $W$. The same statement is in other papers and in Hagen's thesis he actually defines contact as the separating condition and states that it is equivalent to the definition above. So it seems that there should be a straightforward proof, but in discussions with other graduate students we have not uncovered it.

This question has risen in the context of understanding the proof that the contact graph of a CAT(0) cube complex is quasi-isometric to a tree. The scenario we are worried about is roughly sketched in this picture. The orange hyperplanes are not separated by the blue (or any other hyperplane) but do not contact.

Any thoughts would be appreciated, thanks!

• Of course if there's $U$ between $V$ and $W$ then $V,W$ don't contact. So the question is: why (if true) if $V,W$ don't contact, they are separated by a hyperplane. I tend to believe it's indeed true (despite your picture) but have to think more. – YCor Mar 15 '17 at 17:38
• Yes! I forgot to mention that I do see that separating implies no contact. I am certainly inclined to believe the equivalence, and initially thought there should be a simple proof. My current thoughts are now to maybe see the second direction using Roller Duality. – ThePiper Mar 17 '17 at 3:28

## 1 Answer

The stronger statement is true that there is an edge path from $V$ to $W$ that only crosses hyperplanes separating $V$ and $W$. This is lemma 2.18(2) in https://arxiv.org/pdf/1212.1585.pdf. Note that the proof of claim 2.19 implicitly uses the fact that a vertex $x$ is incident to an edge crossing a hyperplane $H$ if and only if there is no hyperplane separating $x$ from $H$. This follows from lemma 2.17 of the aforementioned paper, but for a shorter more geometrical proof you can apply lemma 13.13 from http://math.hunter.cuny.edu/olgak/haglund_Wise.pdf with $Y=N(H)$ and $e$ on a shortest edge path from $H$ to $x$.