**Question 1.** Does every CAT(0) space embed isometrically inside an integral of $\mathbb{R}$-trees?

Here an *integral* of $\mathbb{R}$ trees means the set of functions from a measure space $\mathcal{F}$ to a measurable field of based $\mathbb{R}$-trees over $\mathcal{F}$, so that the squared distance from the basepoint is integrable, as described in Proposition 44 and Remark 45 of https://dx.doi.org/10.1090/S0894-0347-06-00525-X .

(Originally the question referred to $\ell^2$ products of $\mathbb{R}$-trees, but Nicolas Monod below gave an argument that these do not suffice, and suggested the generalization.)

Really I'm most interested in the finite version of this question (where we also don't need to talk about $\mathbb{R}$-trees):

**Question 2.** Does every $k$-element subset of a CAT(0) space embed isometrically in a (finite) product of metric trees?

To go from Question 2 to Question 1, one would presumably attempt a limiting approximation argument, although it's a little complicated, since an approximation for $k$ points doesn't necessarily have an obvious relation to the approximation for $k+1$ points.

If you consider the set of possible squared-distances between $k$ points in any CAT(0) space, you get a subset $MC0_k$ of $\mathbb{R}^{\binom{k}{2}}$. This is clearly closed under scaling, and the fact that a product of CAT(0) spaces is still CAT(0) implies that $MC0_k$ is a convex subset. On the other hand, you could look at $MT_k$, the set of possible squared-distances between $k$ points in a tre. Question 2 is then asking whether the convex hull of $MT_k$ is $MC0_k$. This reformulation makes it clear that, if the answer to Question 2 is affirmative, you would need at most $\binom{k}{2}$ different trees to realize any $k$-element subset of a CAT(0) space.

For $k=4$, the answer to Question 2 appears to be affirmative. Petrunin gives an elegant characterization of which squared-distances between 4 points can be realized in a CAT(0) space: https://arxiv.org/abs/1411.5329 Let me sketch the argument.

If the four distances are realized in $\mathbb{E}^3$, we are done. Otherwise, the four distances can (generically) be realized by a space $X_0$ obtained by gluing together three triangles around a vertex, so that the angle sum around the vertex is greater than $2\pi$. Then consider the space $X_1$ obtained by deleting one of these triangles. This is again CAT(0), and all distances are the same except between one pair of points. The distances in $X_0$ can be realized as a convex combination of he distances in $X_1$ and the distances in another space $X_3$ where that changed length is shortened until the three triangles embed in $\mathbb{E}^2$.

You can proceed in a similar way, deleting the triangles one at a time, until you reduce $X_0$ to a convex combination of a tripod and spaces that embed in $\mathcal{E}^2$.

It looks like one could also push this through for $k=5$, although there are many more cases to consider.

*(Moved the question about extreme points here: What are the extremal CAT(0) metrics? )*

This is a version of this old question: Length inequalities in trees and CAT(0) spaces However, I started off asking the earlier question in a slightly confusing dual form, so I wanted to restate it with a more easily-understood question first.

*(Edited to refer to $\mathbb{R}$-trees)*

Nicolas Monod gave an elegant argument against Question 1 using products. I accepted his answer, but edited the question to refer to integrals instead, which is the natural generalization.

(Or, if you prefer finitary questions, Question 2 is still open, and Nicolas' argument against embedding in a product of trees doesn't really give any evidence against it.)

quasi-isometricallyembeds into a product of trees: arxiv.org/abs/math/0509355 . $\endgroup$2more comments