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

Question

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.

Answer 1

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.

Answer 2

From the point of view of graph theory, it is well known that median graphs (i.e. one-skeleta of CAT(0) cube complexes) embeds isometrically in Hamming cubes. An approach I like uses restricted quotients, which have other nice applications.

Restricted quotients. Given a CAT(0) cube complex $X$ and a collection of hyperplanes $\mathcal{J}$, define the pseudo-metric $d_\mathcal{J} : X^{(0)} \times X^{(0)} \to \mathbb{N}$ as $$(x,y) \mapsto \text{number of hyperplanes in $\mathcal{J}$ separating $x,y$.}$$ The restricted quotient $X_\mathcal{J}$ is the cube complex obtained by cubulating the wallspace $(X, \mathcal{J})$. Such a cube complex can be defined in many different ways, but the key point to keep in mind is that its vertex-set is the quotient of $X^{(0)}$ by the equivalence relation: $x \sim_\mathcal{J} y$ if $d_\mathcal{J}(x,y)=0$.

For instance, start from the dual graph $\Gamma$ of $X \backslash \bigcup_{J\in \mathcal{J}} J$, i.e. the graph whose vertices are the connected components of $X \backslash \bigcup_{J\in \mathcal{J}} J$ and whose edges link two components if they are separated by a unique hyperplane. Then $X_\Gamma$ coincides with the cube complex obtained from $\Gamma$ by filling in all the subgraphs isomorphic to one-skeleta of $k$-cubes with $k$-cubes for every $k \geq 2$.

There is clearly a surjective map $\pi_\mathcal{J} : X \twoheadrightarrow X_\mathcal{J}$ (obtained by sending each vertex of $X$ to the vertex of $X_\mathcal{J}$ corresponding to the component that contains $x$). Moreover, $$d_{X_\mathcal{J}}(\pi_\mathcal{J}(x),\pi_\mathcal{J}(y)) = d_\mathcal{J}(x,y) \text{ for all vertices $x,y \in X$.}$$ (Here, my cube complexes are endowed with $\ell^1$-metrics.)

Constructing embeddings. Because the distance in $X$ between any two vertices coincides with the number of hyperplanes that separate them, it follows that, for every partition $\{ \mathcal{J}_i, \ i \in I\}$ of the hyperplanes of $X$, then $$\bigoplus\limits_{i \in I} \pi_{\mathcal{J}_i} : X \to \bigoplus\limits_{i \in I} X_{\mathcal{J}_i}$$ is an isometric embedding of $X$ into a product of CAT(0) cube complexes (namely, the restricted quotients).

Application 1: If our partition is $\{ \{J\}, \text{ $J$ hyperplane}\}$, then each restricted quotient is a single edge and one obtains an isometric embedding of $X$ into an infinite-dimensional cube $[0,1]^I$.

Application 2: Let $\Delta X$ denote the crossing graph of $X$, i.e. the graph whose vertices are the hyperplanes of $X$ and whose edges link two hyperplanes if they are transverse. Let $\chi$ denote the chromatic graph of $\Delta X$. By considering the partition of the hyperplanes of $X$ induced by a coloring of $\Delta X$ with $\chi$ colors, one obtains an isometric embedding of $X$ into a product of $\chi$ simplicial trees.