In [this paper](https://www.sciencedirect.com/science/article/pii/S0095895613000300) (theorem 2), Chepoi & Hagen say



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> 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.
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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.~~



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


