# A non-positively curved cube complex that admits a local isometric embedding into a Salvetti complex is special.

I am trying to prove the following:

"A non-positively curved cube complex $X$ that admits a local isometric embedding (that maps cubes to cubes) into the Salvetti complex of some right-angled Artin group is special."

By "special" we mean that hyperplanes are embedded and 2-sided and that they do not self-osculate or inter-osculate (hopefully, this is the usual definition).

I think that one might show that the Salvetti complex itself is special and then that the condition of $X$ being locally isometrically embedded pulls back the "specialness" property on $X$, but I cannot see how this last part should be done, and I would be glad to receive any hint.

## 1 Answer

The key point to notice is that a combinatorial map $f: X \to Y$ sends two edges dual to the same hyperplane either to a single edge or to two edges dual to the same hyperplane. Thus, if $f$ is moreover locally injective and if $e_1,e_2$ are two edges with a common endpoint and dual to the same hyperplane, then $f(e_1)$ and $f(e_2)$ are two distinct edges of $Y$ dual to the same hyperplane. As a consequence, $f$ sends self-intersecting hyperplanes to self-intersecting hyperplanes. The same argument essentially holds for the other pathological configurations of hyperplanes.

• How do I see that such a map $f$ maps the edges the way you described it? – noctusraid Aug 4 '16 at 15:17
• A local isometry is locally injective, so it sends a square to a square. A fortiori, two edges which are opposite sides of some square are sent to two edges which are opposite sides of some square as well. Now, because two edges are dual to the same hyperplane iff there exists a sequence of edges between them such that any two successive edges are opposite sides of some square, you should be able to conclude. – Seirios Aug 5 '16 at 5:46
• Yes I forgot about the "square to square", now it makes perfectly sense to me, thanks! – noctusraid Aug 5 '16 at 6:58