What are the extremal CAT(0) metrics?

Question

(Split off from Does every CAT(0) space embed in a measurable integral of $\mathbb{R}$-trees? )

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.

Answer 1

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. 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.