(Split off from [Does every CAT(0) space embed in a measurable integral of $\mathbb{R}$-trees?](https://mathoverflow.net/questions/204710/does-every-cat0-space-embed-in-a-product-of-trees/204723) )


Fix an integer $k \ge 2$, and let
$MC0\_k \subset \mathbb{R}^{\binom{k}{2}}$ be the set of possible squared-distances between $k$ points (not necessarily distinct) in any CAT(0) space. This is clearly closed under scaling, and the fact that a product of CAT(0) spaces is a CAT(0) space implies that $MC0\_k$ is a convex set. What are its extreme rays?


(Recall that an *extreme* *point* in a convex set $C$ is a point that is not in the interior of any line segment in $C$. In the context of convex cones, as here, an *extreme ray* consists of points that aren't in the interior of a line segment that is not contained in a ray through the origin.)


One guess is that the extreme rays in $MC0\_k$ are trees with $k$ marked points, where there is at least one marked point at each vertex of valence $>3$. For $k=4$, this includes embeddings in $\mathbb{R}$ as well as tripods, with the central vertex marked.


Note that this is related to the (known-difficult) question of characterizing which lengths can appear as distances in a CAT(0) space, i.e., characterizing $MC0\_k$ in the notation above.



What are the extremal CAT(0) metrics?
Let me describe a 6-point counterexample.


Let $K$ be a 2-dimensional cone with total angle $\theta=2{\cdot}\pi+\varepsilon$, where $\varepsilon$ is small and positive (any $0<\varepsilon<\tfrac\pi2$ will do).
Note that $K$ is CAT(0).


Consider the following 6 points in $K$: the tip $p$ + and an orbit $\{x\_1,x\_2,x\_3,x\_4,x\_5\}$ of the rotation by angle $\tfrac\theta5$.


Suppose that these 6 points admit an embedding into a product of thees, say $L$.
Let $q$ be the image of $p$; denote by $\Sigma\_q$ its space of directions.
Note that any closed geodesic in $\Sigma\_q$ has length either $2{\cdot}\pi$ or at least $3{\cdot}\pi$. (The latter statement can be prove along the same lines as Gromov's flag condition.)


Denote by $y\_i$ the image of $x\_i$. For any $i$ (mod 5) there is a flat geodesic quadrilateral $qy\_{i-1}y\_iy\_{i+1}$ in $L$ that is isometric to the quadrilateral $px\_{i-1}x\_ix\_{i+1}$.
It follows that there is a closed geodesic in $\Sigma\_q$ of length $\theta$ --- a contradiction.


**Comments:**


* There is a similar example with 5 points --- take a plane convex quadrilateral with line segment attached at one vertex. By Reshetnyak theorem, it is a CAT(0) space. The vertices of the quadraliteral + the end of the segment form a 5-point subspace that cannot be embedded in a product of trees.
* Another related observation: [octahedron comparison holds for products of trees](https://arxiv.org/abs/2212.06445).
That is, if you choose six points in the product of trees and label them by vertices of an octahedron, then there is a configuration in the Euclidean space such that edges are not getting larger and the diagonals are not getting smaller.
It is unknow if it holds in general CAT(0) space.


