# Can all countable $CAT(0)$ cube complexes be isometrically embedded in $l^1(\mathbb{N},\mathbb{R})$?

In this paper (theorem 2), Chepoi & Hagen say

There exists an infinite $$CAT(0)$$ cube complex $$X$$ with constant maximum degree which cannot be isometrically embedded into a Cartesian product of a finite number of trees, i.e., the chromatic number of its crossing graph is infinite.

so that you can't hope to isometrically embed it in a $$\mathbb{R}^n$$ with $$l^1$$ metric. But what about $$l^1(\mathbb{N})$$:

Question 1: Under what conditions does a $$CAT(0)$$ cube complex, with the "polyhedral complex" metric induced by the $$l^1$$ metric on cubes, embed in $$l^1(\mathbb N)$$ ?

Question 2: Same question but only about embedding the $$0$$ skeleton of said complex.

My reason for the question is the following: It is known that the $$0$$-skeleton of a $$CAT(0)$$ cube complex is a median space, and intuitively it makes sense since it looks a lot like a nice subset of $$\mathbb{Z}^n$$ for some $$n$$ big enough, which is itself median. I'm therefore wondering if you could prove median-ness of this $$0$$-skeleton by embedding your complex in $$l^1(\mathbb{N})$$, and then arguing that the embedding is actually submedian, or whatever allows you to argue median-ness back for the skeleton.

EDIT: added the obvious "countable" condition. Then an obvious embedding would be to enumerate the vertices and send them to the "basis" of $$l^1$$ … I wonder if/where that would break.

• @YCor: I made the question explicit. I hope this is precise enough now. – ouimerci Jan 26 at 14:35
• Chepoi-Hagen stable link: doi.org/10.1016/j.jctb.2013.04.003 – YCor Jan 26 at 15:58

All CAT(0) cube complexes $$C$$ with $$\ell^1$$-metric embed isometrically into $$\ell^1$$. If the set of vertices is countable, one can choose $$\ell^1$$ of a countable set.
Indeed, say that a subset $$B$$ of the vertex set $$V_C$$ of $$C$$ is (totally) convex if it contains vertices of all geodesic paths between any two elements of $$B$$. Call it $$B$$ "biconvex" if both $$B$$ and its complement in $$V_C$$ are convex. (It's usually called "halfspace" but I don't think it's a good choice of terminology.) Let $$\mathcal{B}$$ be the set of biconvex subset. For every oriented edge $$(x,y)$$, there exists a unique $$B=B_{x,y}\in\mathcal{B}$$ such that $$y\in B$$ and $$x\notin B$$ (namely $$B_{x,y}=\{z\in V_C:zy\le zx\}$$). It is known that all biconvex subsets have this form. This shows that the cardinal of $$\mathcal{B}$$ is bounded above by the cardinal of $$V_C$$ (if infinite).
An isometric embedding $$f$$ of $$C$$ into $$\ell^1(\mathcal{B})$$ consists in the following: for $$x\in C$$, let $$\mathcal{B}(x)$$ be the set of biconvex subsets containing $$x$$. Fix a vertex $$x_0$$; map $$x\in V_C$$ to $$f(x)=1_{\mathcal{B}(x)}-1_{\mathcal{B}(x_0)}$$. Here we endow $$\mathcal{B}$$ with the atomic measure for which singletons have measure $$1/2$$, so that $$f$$ is an isometric embedding: indeed for all $$x,y\in V_C$$, each of $$\mathcal{B}(x)\smallsetminus \mathcal{B}(y)$$ and $$\mathcal{B}(y)\smallsetminus \mathcal{B}(x)$$ have exactly $$xy$$ elements. It is not hard to check that $$f$$ has a canonical affine extension to cells.
• Mmh, that's a nice argument. I assume for vertices $x,y$, $xy$ means the distance from $x$ to $y$ ? Also, is that a standard kind of argument ? – ouimerci Jan 26 at 15:12