5-point CAT(0) spaces after Tetsu Toyoda

We give another proof of Toyoda's theorem that describes 5-point subpaces in CAT(0) length spaces


Introduction
The CAT(0) comparison is a certain inequality for 6 distances between 4 points in a metric space. The following descriptions, the so-called (2+2)-comparison, is the most standard, we refer to [2,3] for other definitions and their equivalences.
Given a quadruple of points p, q, x, y in a metric space X, consider two model triangles 1 [pxỹ] =△(pxy) and [qxỹ] =△(qxy) with common side [xỹ].x yz pq If the inequality |p − q| X |p −z| + |z −q| holds for any pointz ∈ [xỹ], then we say that the quadruple p, q, x, y satisfies CAT(0) comparison; here |p − q| X denotes the distance from p to q in X.
If CAT(0) comparison holds for any quadruple (and any of its relabeling) in a metric space X, then we say that X is CAT(0).
It is not hard to check that if a quadruple of points satisfies CAT(0) comparison for all relabeling, then it admits a distance-preserving inclusion into a length CAT(0) space. The following theorem generalizes this statement to 5-point metric spaces.
1.1. Toyoda's theorem. Let P be a 5-point metric space that satisfies CAT(0) comparison. Then P admits a distance-preserving inclusion into a length CAT(0) space X.
Moreover, X can be chosen to be a subcomplex of a 4-simplex such that (1) each simplex in X has Euclidean metric and (2) the inclusion maps the 5 points on P to the vertexes of the simplex. the anonymous referee for help.
The first author was partially supported by RFBR grant 20-01-00070, the second author was partially supported by NSF grant DMS-2005279.

5-point arrays in 3-space
Denote by A the space of all 5 point arrays in R 3 that is nondegenerate in the following sense: (1) all 5 points do not lie on one plane and (2) no three points lie on one line. Note that A is connected.
A 5 point array x 1 , . . . , x 5 ∈ R 3 defines an affine map from a 4-simplex to R 3 . Fix an orientation of the 4-simplex and consider the induced orientations on its 5 facets. Each facet may be mapped in an orientation-preserving, degenerate, or orientation-reversing way. For each array consider the triple of integers (n + , n 0 , n − ), where n + , n 0 , and n − denote the number of orientationpreserving, degenerate, or orientation-reversing facets respectively.
Clearly n + + n 0 + n − = 5 and since all 5 points cannot lie in one plane, we have that n + 1, n − 1, and n 0 1. Therefore, the value m = n − − n + can take an integer value between −3 and 3; in this case, we say that an array belongs to A m .
It defines a subdivision of A into 7 subsets A −3 , . . . , A 3 with combinatorial configuration as on the diagram; quadruples in one plane are marked in gray and the triple (n + , n 0 , n − ) is written below.
Every two quadrilaterals in the array have 3 common points that define a plane. If the remaining two points lie on opposite sides from the plane, then the corresponding facets have the same orientation; if they lie on one side, then the orientations are opposite. Therefore, the 7 subsets A −3 , . . . A 3 can be described in the following way: a tetrahedron with preserved orientation and one point inside. A −2 a tetrahedron with preserved orientation and one point on a facet. A −1 a double triangular pyramid formed by two tetrahedrons with preserved orientation.
A 0 a pyramid over a convex quadrilateral A 1 a double triangular pyramid formed by two tetrahedrons with reversed orientation.
A 2 a tetrahedron with reversed orientation and one point on a facet. A 3 a tetrahedron with reversed orientation and one point inside. Note that the complement A\A 0 has two connected components formed by Observe that each array in A − has at least 3 positively oriented facets and each array in A + has at least 3 negatively oriented facets.

Associated form
In this section we recall some facts about the so-called associated form introduced in [6]; it is a quadratic form W x on R n−1 associated to a given n-point array x = (x 1 , . . . , x n ) in a metric space X.
Construction. Let △ be the standard simplex △ in R n−1 ; that is, the first (n − 1) of its vertices v 1 , . . . , v n form the standard basis on R n−1 , and v n = 0.
Recall that |a−b| X denotes the distance between points a and b in the metric for all i and j. Note that this identity defines W x uniquely. The constructed quadratic form W x will be called the form associated to the point array x.
Note that an array x = (x 1 , . . . , x n ) in a metric space X is isometric to an array in Euclidean space if and only if W x (v) 0 for any v ∈ R n−1 .
In particular, the condition W x 0 for a triple x = (x 1 , x 2 , x 3 ) means that all three triangle inequalities for the distances between x 1 , x 2 , and x 3 hold. For an n-point array, it implies that W x (v) 0 for any vector v in a plane spanned by a triple v i , v j , v k . In particular, we get the following: 3.1. Observation. Let W x be a form on R n−1 associated with a point array for any nonzero vector v ∈ L. Then the projections of any 3 vertices of △ to the quotient space R n−1 /L are not collinear.
CAT(0) condition. Consider a point array x with 4 points. From 3.1, it follows that W x is nonnegative on every plane parallel to a face of the tetrahedron △. In particular, W x can have at most one negative eigenvalue.
Assume W x (w) < 0 for some w ∈ R 3 . From 3.1, the line L w spanned by w is transversal to each of 4 planes parallel to a face of △.
Consider the projection of △ along L w to a transversal plane. The projection of the 4 vertices of △ lie in general position; that is, no three of them lie on one line. Therefore, we can see one of two combinatorial pictures shown on the diagram. Since the set of lines L w with W x (w) < 0 is connected, the combinatorics of the picture does not depend on the choice of w.

Claim.
If CAT(0) comparison holds in X, then the diagram on the right cannot appear.
(The converse holds as well, but we will not need it.) Proof. Suppose we see the picture on the right. Let In the plane spanned by [v 2 , v 4 ] and w, the vector w is timelike. Therefore we have the following reversed triangle inequality: here we use shortcut |a − b| = W (a − b).
Note that the triangles The claim implies the following: 3.3. Observation. Suppose a metric on x = (x 1 , . . . , x n ) satisfies CAT(0) comparison and W x is its associated form on R n−1 . Assume that L is a subspace of R n−1 such that W x (v) < 0 for any nonzero vector v ∈ L. Then if the projections of 4 vertices of △ to the quotient space R n−1 /L lies in one plane, then its projection looks like the picture on the left; that is, one of the points lies in the triangle formed by the remaining three points.
3.4. Corollary. Suppose a metric on x = (x 1 , . . . , x 5 ) satisfies CAT(0) comparison and W x is its associated form on R 4 . Assume that L is a subspace of R 4 such that W x (v) < 0 for any nonzero vector v ∈ L. Then dim L 1.
Moreover, if dim L = 1, then the projections of the vertices of △ to the quotient space R 3 = R 4 /L belong to A\A 0 (defined in the previous section).
Let W be a quadratic form on R 4 . Suppose that W has exactly one negative eigenvalue. Choose future and past cones C + and C − for W ; that is, C + and C − are connected components of the set Let K be a convex body in R 4 ; denote by Σ the surface of K. A point p lies on the upper side of Σ (briefly p ∈ Σ + ) if there is a spacelike hyperplane in R 4 that supports Σ at p from above; more precisely if the Minkowski sum {p} + C + does not intersect K.
Similarly, we define the lower side of Σ denoted by Σ − . Note that Σ + and Σ − might have common points. The subsets Σ + and Σ − are spacelike; in particular, the length of any Lipschitz curve in these subsets can be defined and it leads to induced intrinsic pseudometrics on Σ + and Σ − . Abusing notation, we will not distinguish a pseudometric space and the corresponding metric space.

Lemma.
Let Σ be the surface of a convex set K in R 4 and C ± be the future and past cones for a quadratic form W . Then the upper and lower sides Σ + and Σ − of Σ equipped with the induced intrinsic metric are CAT(0) length spaces.
Moreover, if a line segment [pq] in R 4 lies on Σ ± , then [pq] is a minimizing geodesic in Σ ± ; that is, This lemma is essentially stated by Anatolii Milka [5, Theorem 4]; we give a sketch of alternative proof based on smooth approximation.
Sketch. We can assume that W is nondegenerate; that is, after a linear change of coordinates it is the standard form on R 3,1 . If not, then there is a W -preserving projection of R 4 to a W -nondegenerate subspace; apply this projection and note that this subspace is isometric a subspace of R 3,1 .
Assume S is a smooth strictly spacelike hypersurface in R 3,1 with convex epigraph. By Gauss formula, S has nonpositive sectional curvature.
Suppose a strictly spacelike hyperplane Π cuts from S a disc D. Recall that Liberman's lemma [5,Theorem 3] implies that time coordinate is convex on any geodesic in S. We may assume that time is vanishing on Π; therefore, by the lemma, D has a convex set in S. Therefore the Cartan-Hadamard theorem [3]

Consider the Minkowski sum
it has a convex spacelike boundary ∂K − . Choose a strictly spacelike hyperplane Π that lies above K. Denote by D the subset of ∂K − below Π. Let us equip D with induced intrinsic pseudometric. By construction Σ − is isometric to ι(D). It follows that Σ − is CAT(0). Now suppose a line segment [pq] in R 4 lies on Σ − . Choose a supporting hyperplane Π at the midpoint of [pq]. Choose time coordinate that vanish on Π; by Liberman's lemma, every shortest path in Σ − between p and q has to lie on Π; that is, the intersection Σ − ∩ Π is a convex subset of Σ − . Therefore [pq] is convex in Σ − which implies the second statement.

Sublemma.
Let u and v be two lightlike vectors in R 3,1 . Suppose that the union of two half-lines s → p + s·u and t → q + t·v for s, t 0 is a spacelike set. Then the function (s, t) → |(p + s·u) − (q + t·v)| is nondecreasing in both arguments, where |w| := w, w for a spacelike vector w.
Proof. Since u and v are lightlike, u, u = v, v = 0. Since the union of two half-lines is spacelike, (p + s·u) − (q + t·v) is spacelike for any s, t 0. It follows that for any s, t 0. Therefore Whence the result.
Assume v is a nonzero vector in R 4 and p ∈ Σ. We say that p lies on the upper side of Σ with respect to v (briefly p ∈ Σ + (v)) if p + t·v / ∈ K for any t > 0. Correspondingly, p lies on the lower side of Σ with respect to v (briefly p ∈ Σ − (v)) if p + t·v / ∈ K for any t < 0.

Observation.
Let K be a compact convex set in R 4 and C ± be the future and past cones for a quadratic form W . Then the upper (lower) side of the boundary surface Σ of K can be described as the intersection of the upper (respectively lower) sides of Σ with respect to all vectors v ∈ C + ; that is,

Proof assembling
Proof of Toyoda's theorem. Let {x 1 , . . . , x 5 } be the points in P . Choose a 5-simplex △ in R 4 ; denote by W the form associated with the point array (x 1 , . . . , x 5 ).
If W 0, then P admits a distance preserving embedding into Euclidean 4-space, so one can take the convex hull of its image as X.
Suppose W (v) < 0 for some v ∈ R 4 . Since P is CAT(0), 3.4 implies that W has exactly one negative eigenvalue. Moreover, if a line L is spanned by a vector v such that W (v) < 0, then the projection of the vertices of the simplex to R 3 = R 4 /L belongs to A\A 0 .
The space of such lines L is connected. By 2.1, we can assume that all the projections belong to A − . That is, we can choose timelike orientation such that for any v ∈ C + the lower part Σ − (v) of Σ = ∂△ has at least 3 facets of △.
Since all edges of △ lie in Σ − , the inclusion P ֒→ Σ − is distance preserving. Whence we can take X = Σ − .
Finally, observe that in each case X is a subcomplex of △ that includes all edges and has a model metric on each simplex.

Remarks
Let us recall the definition of graph comparison given by Vladimir Zolotov and the authors [4] and use it to formulate a few related questions.
Let Γ be a graph with vertices v 1 , . . . , v n . A metric space X is said to meet the Γ-comparison if for any set of points in X labeled by vertices of Γ there is a model configurationṽ 1 , . . . ,ṽ n in the Hilbert space H such that if v j is adjacent to v j , then The C 4 -comparison (for the 4-cycle C 4 on the diagram) defines CAT(0) comparison. Tetsu Toyoda have shown that C 4 -comparison imlies graph comparisons for all cycles C n [9]; remakably, the metric space is not assumed to be intrinsic. The O 3 -comparison (for the octahedron graph O 3 on the diagram) defines another comparison. Since O 3 contains C 4 as an induced subgraph, we get that O 3 -comparison is stronger than C 4 -comparison.
6.1. Open question. Is it true that octahedron-comparison holds in any 6 points in a length CAT(0) space? And, assuming the answer is affirmative, what about the converse: is it true that any 6-point metric space that satisfies octahedron-comparison admits a distance preserving embedding in a length CAT(0) space?
The analogous questions for spaces with nonnegative curvature in the sense of Alexandrov (briefly CBB(0)) are open as well. The CBB(0) comparison is equivalent to the 3-tree comparison (for the tripod-tree shown first on the following diagram). It turns out that any length CBB(0) space satisfies the . . .

3-tree
4-tree 5-tree 6-tree 2(2)-tree 3(1)-tree comparison for the other trees on the diagram; it is formed by an infinite family of star-shaped trees and two trees with 6 vertices [1,4]. (The 4-tree comparison (the second tree on the diagram) is equivalent to the so-called (4+1)-point comparison in the terminology of [1].) We expect that this comparison provides a necessary and sufficient condition for 5-point sets. Namely, we expect an affirmative answer to the following stronger question.
6.2. Question. Suppose a 5-point metric space P satisfies the 4-tree comparison. Is it true that P admits a distance preserving embedding into a length CBB(0) space?
Finally, let us mention a related question about a 6-point condition.
6.3. Question. Suppose a 6-point metric space P satisfies the 5-tree, 2(2)tree, and 3(1)-tree comparisons. Is it true that P admits a distance preserving embedding into a length CBB(0) space?