Distance geometry for kissing spheres

Linear Algebra and its Applications 479 (2015) 185–201

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Linear Algebra and its Applications www.elsevier.com/locate/laa

Distance geometry for kissing spheres Hao Chen 1 Freie Universität Berlin, Institut für Mathematik, Arnimallee 2, 14195, Berlin, Germany

a r t i c l e

i n f o

Article history: Received 25 December 2013 Accepted 9 April 2015 Available online xxxx Submitted by R. Brualdi

a b s t r a c t A kissing sphere is a sphere that is tangent to a fixed reference ball. We develop in this paper a distance geometry for kissing spheres, which turns out to be a generalization of the classical Euclidean distance geometry. © 2015 Elsevier Inc. All rights reserved.

MSC: 51K05 15A83 52C17 Keywords: Distance geometry Distance matrix Cayley–Menger matrix Matrix completion Distance completion

1. Introduction Distance geometry studies the geometry based solely on knowledge of distances. A typical question of Euclidean distance geometry is like this: Are there three points A, B, C in a plane such that the distances are dE (A, B) = 3, dE (B, C) = 4 and dE (C, A) = 5? Here dE is the Euclidean distance. In classical Euclidean distance geometry, previous E-mail address: [email protected]. The author was supported by the Deutsche Forschungsgemeinschaft within the Research Training Group ‘Methods for Discrete Structures’ (GRK 1408). 1

http://dx.doi.org/10.1016/j.laa.2015.04.012 0024-3795/© 2015 Elsevier Inc. All rights reserved.

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works show that answers to such questions lie in the distance matrix and the Cayley– Menger matrix. This will be reviewed in detail in Section 2. A kissing sphere is a sphere that is tangent to a fixed reference ball. The name comes from the billiards term. The classical kissing number problem [1] asks for the maximal number of unit balls with pairwisely disjoint interiors that can touch a reference unit ball. However, in the present paper, we do not require the kissing spheres to have a fixed radius. Following the approaches of Euclidean distance geometry, we develop in this paper a distance geometry for kissing spheres, based on a Möbius invariant distance function, see Definition 3.2. A key observation is that the distance matrix for kissing spheres also plays the role of a Cayley–Menger matrix. It is then possible to adapt proof techniques from Euclidean distance geometry for our use. The central problem is whether there is a set of kissing spheres that realizes given distances. If distances are specified for every pair of kissing spheres, this is the embeddability problem. Our first main result, Theorem 4.1, asserts that the answer to an embeddability problem lies in the distance matrix in a similar way as in Euclidean distance geometry. If distances are not specified for some pairs of points, one would like to find values for these unknown distances so that the completed distances can be realized by a set of kissing spheres. This is the distance completion problem, another central problem in distance geometry. In general it is difficult to tell if a distance completion is possible. However, in Euclidean distance geometry, the existence of a solution is much easier to tell if the known distances form a chordal graph. By embedding the kissing spheres into Lorentz space, we come to the same conclusion for the distance completion problem for kissing spheres, which is our second main result, Theorem 4.4. Apart from obvious similarities between the results above, we also notice that the distance geometry for kissing spheres degenerates to Euclidean distance geometry in several ways. In this sense, the distance geometry developed in this paper generalizes Euclidean distance geometry. The paper is organized as follows. In Section 2, we review results of Euclidean distance geometry. We define in Section 3 the space of kissing spheres, together with a distance function that is invariant under Möbius transformations. Then we study the embeddability problem and distance completability problem in Section 4. Finally, we discuss in Section 5 the ball packing problem, which is in fact the motivation for this paper. 2. Preliminaries on Euclidean distance geometry In this section, we introduce distance geometry in a very general framework, and present some results from Euclidean distance geometry. Let I be a set. A non-negative function d : I × I → R≥0 is a distance function on I if d(i, j) = d(j, i) and d(i, i) = 0 for all i, j ∈ I. The pair (I, d) is called a distance space. So the Euclidean distance space (En , dE ) is the Euclidean space En equipped with the Euclidean distance dE . Note that the distance function is not necessarily a metric, as

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d(i, j) = 0 is allowed for i = j, and the triangular inequality is not required. Our distance function dK for kissing spheres, see Definition 3.2, is an example of non-metric distance function. Definition 2.1 (Isometrically embeddable). Let (I, d) and (I  , d ) be two distance spaces. We say that (I, d) is isometrically embeddable into (I  , d ) if there exists a map σ : I → I  such that d (σ(i), σ(j)) = d(i, j) for all i, j ∈ I. In this case, we call the map σ an isometric embedding. For the embeddability problems studied in distance geometry, the set I is often finite. In this case, we can assume that I = {0, . . . , k}, and write σi instead of σ(i) for i ∈ I. For a subset J ⊂ I, we denote by (J, d) the distance space consisting of J and the distance function d restricted to J. In Euclidean distance geometry, the embeddability problem asks: Given a finite distance space (I, d), is it isometrically embeddable into the Euclidean distance space (En , dE )? For a finite distance space (I, d), where I = {0, . . . , k}, there are two powerful tools for solving this problem. One is the distance matrix D(I, d), defined as the (k + 1) × (k + 1) matrix whose i, j entry is the squared distance d(i, j)2 for i, j ∈ I. The other is the Cayley–Menger matrix M (I, d), defined as the (k + 2) × (k + 2) matrix  M (I, d) =

D(I, d) eT

e 0

 ,

where e denotes the all-ones column vector of length k + 1. The following theorem combines some important results of Euclidean distance geometry (see [2–5]). Theorem 2.2. For a finite distance space (I, d), where d is not identically zero, consider the following statements: (i) (I, d) is isometrically embeddable into (En , dE ). (ii) The rank of the Cayley–Menger matrix M (I, d) is at most n + 2 and (−1)|J| det M (J, d) ≥ 0 for all J ⊆ I. (iii) The Cayley–Menger matrix M (I, d) has exactly one positive eigenvalue and at most n + 1 negative eigenvalues. (iv) The distance matrix D(I, d) has exactly one positive eigenvalue and at most n + 1 negative eigenvalues. Then, (i) ⇔ (ii) ⇔ (iii) ⇒ (iv).

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If we are not given a complete information about the distances, a natural question to ask is whether the distance completion problem has a solution. In the language of graph theory: Definition 2.3 (Distance completable). Let G = (V, E) be an undirected graph with a length function  : E → R≥0 . We say that (G, ) is (distance) completable in a distance space (I, d) if there is a distance function dV on the vertex set V such that (V, dV ) is isometrically embeddable into (I, d) and dV (u, v) = (u, v) for all (u, v) ∈ E. For a subgraph H ⊂ G, we denote by (H, ) the graph H equipped with the length function  restricted to H. In Euclidean distance geometry, the distance completability problem asks: Given a graph G = (V, E) and a length function  on G, is (G, ) completable in (En , dE )? We define two sets of length functions as follows: n CE (G) = { : E → R≥0 | (G, ) is completable in (En , dE )}, n KE (G) = { : E → R≥0 | for all cliques K ⊂ G, (K, ) is completable in (En , dE )}.

In general these two sets are not equal, but we have the following theorem: n n (G) = KE (G) if and only if G is chordal. Theorem 2.4 (Laurent [6]). CE

Here, a chordal graph is a graph without chordless cycles with more than three edges, or equivalently, every such cycle has an edge joining two vertices not adjacent in the cycle. Therefore, for a chordal graph, in order to tell for a length function whether it is completable in Euclidean distance space, we only need to check all the maximal cliques for the embeddability, which makes the distance completability problem much easier. 3. Distance space of kissing spheres 3.1. Kissing spheres We now define kissing spheres, the main object of study of this paper. For this, we ˆ n = En ∪ {∞}, with Cartesian work in the extended n-dimensional Euclidean space E coordinate system (x0 , . . . , xn−1 ). The sphere centered at o ∈ En with diameter φ > 0 is the set {x ∈ En | dE (x, o) = φ/2}. We regard (n − 1)-hyperplanes (with the adjoined infinity point) as spheres of infinite diameter, whose center o is at infinity. Let κ be a real number. A ball of curvature κ may be one of the following: • a set {x | dE (x, o) ≤ 1/κ} if κ > 0; • a set {x | dE (x, o) ≥ −1/κ} if κ < 0; • a closed half-space if κ = 0.

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ˆ 2 , as defined in Definition 3.1. Fig. 1. Three kissing spheres to the gray reference ball in E

In the first two cases, o ∈ En is the center of the ball. A sphere is said to be tangent ˆ n if t is the only element in their intersection. We call t the to a ball at a point t ∈ E tangency point, which can be at infinity if it involves a ball of curvature 0 and a sphere of infinite diameter. ˆ n as the reference ball. A kissing sphere Definition 3.1 (Kissing sphere). Fix a ball in E is a sphere tangent to the reference ball. Our main concern is the combinatorics, which means relations like tangency, intersection and disjointness between the kissing spheres. These are defined as follows. Two kissing spheres are tangent to each other if their intersection consists of a single point which is not on the boundary of the reference ball. Two kissing spheres intersect if their intersection consists of more than one point, or if they are tangent to the reference ball at the same tangency point. Two kissing spheres are disjoint if their intersection is empty. For example, in Fig. 1, A is tangent to B, B is disjoint from C, and C intersects A. In the following, we assume the reference ball to be the half-space x0 ≤ 0. If it is not the case, we can always find a Möbius transformation that sends the reference ball to the half-space without any change to the combinatorics of kissing spheres. Therefore we define a kissing sphere alternatively as follows. ˆ n is a sphere tangent Definition 3.1 (Kissing sphere, alternative). A kissing sphere in E to the half-space x0 ≤ 0. Such a kissing sphere must lie in the half-space x0 ≥ 0. Note that the tangency point can be at infinity. In this case, the kissing sphere is a hyperplane x0 = h > 0 (with the adjoined infinity point), and we say that it is a hyperplane at level h. ˆ n as defined in Definition 3.1 . For a We denote by Kn the set of kissing spheres in E n kissing sphere p ∈ K , as shown in Fig. 2 for n = 2, we denote by t(p) the tangency point on the (n − 1)-dimensional hyperplane x0 = 0, and by φ(p) ∈ R≥0 ∪ {∞} the diameter of p. Therefore if φ(p) < ∞, the pair (φ(p), t(p)) is the “north pole” of p, situated in the half-space x0 > 0.

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Fig. 2. A kissing sphere as defined in Definition 3.1 .

3.2. A distance function ˆn Möbius transformations are conformal diffeomorphisms from S n to S n . Viewing E n as S , Möbius transformations map spheres to spheres. They form a group called the Möbius group, denoted by Möb(n). Please note that, as in [7], we do not require Möbius transformations to preserve the orientation. Therefore reflections and inversions are also Möbius transformations. Let T ∈ Möb(n) be a Möbius transformation that preserves the half-space x0 ≤ 0. Then T maps kissing spheres to kissing spheres. For a sphere centered on the hyperplane x0 = 0, its image under T is also centered on x0 = 0. Therefore ˆ n−1 . the restriction of T on the hyperplane x0 = 0 is a Möbius transformation on E Conversely, by the Poincaré extension (see [8, Section 3.3]), any Möbius transformation ˆ n−1 can be naturally extended to a Möbius transformation on E ˆ n that preserves the on E half-space x0 ≤ 0. We thus define the action of a Möbius transformation T ∈ Möb(n − 1) on Kn to be the action of its Poincaré extension. We define a distance function dK on Kn as follows. Definition 3.2 (Distance function for kissing spheres). Let p, q ∈ Kn be two kissing spheres. If there is a Möbius transformation T preserving the half-space x0 ≤ 0 such that φ(T p) = φ(T q) = 1, then the distance between p and q is dK (p, q) := dE (t(T p), t(T q)) (see Fig. 3). If such a T does not exist, dK (p, q) := 0. The required transformation T does not exist if and only if t(p) = t(q) but p = q. Theorem 3.3. The distance function dK of Definition 3.2 is well defined. In other words, dK is independent of the choice of T . We need to prove that dK defined on Kn is invariant under the action of Möbius transformations in Möb(n − 1). Before the proof, we shall look at the effect that an

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Fig. 3. Definition of the distance, when T exists.

inversion preserving the half-space x0 ≤ 0 has on a kissing sphere. Let s be a sphere with radius r centered at a point o on the hyperplane x0 = 0. For a kissing sphere p ∈ Kn , we denote by ps the image of p under the inversion transform with respect to s. We have φ(ps ) =

r2 φ(p) dE (o, t(p))2

(1)

and dE (o, t(ps )) =

r2 , dE (o, t(p))

(2)

where dE (x, y) is the Euclidean distance on the hyperplane x0 = 0. The effect of such an inversion on p is then the same as the effect of a dilation of scale factor r2 /dE (o, t(p))2 . Notably, the scale factor does not depend on the diameter of p. Proof of Theorem 3.3. An explicit calculation is not necessary, but may help understanding the situation. Let p, q be two kissing spheres such that φ(p), φ(q) < ∞ and dE (t(p), t(q)) > 0. The infinite case and the degenerate case will be discussed later.  Choose a point o on the line segment t(p)t(q), such that dE (o, t(p))/dE (o, t(q)) = φ(p)/φ(q). That is,  dE (t(p), t(q)) φ(p)  dE (o, t(p)) =  , φ(p) + φ(q)  dE (t(p), t(q)) φ(q)  dE (o, t(q)) =  . φ(p) + φ(q) Let s be a sphere centered at o with radius dE (t(p), t(q))  r=  . φ(p) + φ(q) Then, by (1) and (2), we have φ(ps ) = φ(q s ) = 1, and dE (t(p), t(q)) dK (p, q) = dE (t(ps ), t(q s )) = dE (o, t(ps )) + dE (o, t(q s )) =  . φ(p)φ(q)

(3)

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A Möbius transformation is generated by reflections and inversions. For details, see [8, Definition 3.1.1], where inversions are interpreted as reflections with respect to a sphere, or [9, Theorem 3.8], where reflections are viewed as inversions with respect to a plane. Theorem 3.3 is obviously true for reflections. We shall study the inversions in detail. An inversion preserving the half-space x0 ≤ 0 must have its inversion sphere centered at a point o on the hyperplane x0 = 0. Let s be such an inversion sphere of radius r. By (2), we have dE (o, t(ps )) = r2 /dE (o, t(p)) and dE (o, t(q s )) = r2 /dE (o, t(q)). Thanks to the independence of the scale factor of the diameter, the triangle ot(p)t(q) and the triangle ot(q s )t(ps ) are similar, and dE (t(ps ), t(q s )) =

r2 dE (t(p), t(q)) . dE (o, t(p))dE (o, t(q))

We then have dE (t(ps ), t(q s )) dE (t(p), t(q)) dK (ps , q s ) =  =  = dK (p, q), s s φ(p )φ(q ) φ(p)φ(q) which proves the theorem. We now extend the calculation to the infinite case. Let p be a hyperplane at level h. Consider again a sphere s centered at a point o on x0 = 0 with radius r. Then φ(q s ) = r2 φ(q)/dE (o, t(q))2 , φ(ps ) = r2 /h, and dE (t(ps ), t(q s )) = r2 /dE (o, t(q)). Since the inversion preserves the distance dK , by (3), we have dE (t(ps ), t(q s )) dK (p, q) =  = φ(ps )φ(q s )

 h . φ(q)

(4)

Finally we study the degenerate case. That is, t(p) = t(q) but p = q. Since a Möbius transformation is bijective, it is impossible to transform p and q into spheres of the same diameter. According to the definition, dK (p, q) = 0. This is reasonable, since it’s the limit of (3) as dE (t(p), t(q)) tends to 0, or the limit of (4) as φ(q) tends to infinity. This is Möbius invariant since t(p) = t(q) holds under any Möbius transformation. 2 Remark. As mentioned in the preliminaries, dK is not a metric. More specifically, the triangle inequality is not satisfied in general, and there are p, q ∈ Kn such that p = q but dK (p, q) = 0. Theorem 3.4. The distance function dK reflects the combinatorics as follows: ⎧ >1 ⎪ ⎪ ⎨= 1 dK (p, q) <1 ⎪ ⎪ ⎩ =0

if if if if

p is disjoint from q; p is tangent to q; p intersects q; t(p) = t(q).

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We notice from (3) that the distance space for a set of kissing spheres is conformally Euclidean, as defined in the following (compare the conformal equivalence defined by Luo [10] and Bobenko et al. [11]): Definition 3.5 (Conformal embedding). Let (I, d) and (I  , d ) be two distance spaces. We say that (I, d) is conformally embeddable into (I  , d ) if there exist a map ξ : I → I  and a real valued function f : I → R≥0 , such that d (ξi , ξj ) = f (i)f (j)d(i, j) for all i, j ∈ I. We say that ξ is a conformal embedding with the conformal factor f . (I, d) and (I  , d ) are conformally equivalent if the conformal embedding is bijective. (I, d) is conformally Euclidean if it is conformally embeddable into the Euclidean distance space. In fact, if a finite distance space (I, d) is embeddable into (Kn , dK ), we can always choose an isometric embedding σ such that φ(σi ) < ∞ for all i ∈ I. We then recognize √ in (3) that t ◦ σ embeds (I, d) conformally into (En−1 , dE ), with conformal factor φ ◦ σ. Remark. If there exists an embedding σ such that φ(σi ) = 1 for all i ∈ I, then dK reduces to Euclidean distance. 4. Main results 4.1. Embeddability problem We now prove our first main theorem (compare Theorem 2.2). Theorem 4.1. Given a finite distance space (I, d), where d is not identically zero, the following statements are equivalent: (i) (I, d) is isometrically embeddable into (Kn , dK ). (ii) The rank of the distance matrix D(I, d) is at most n + 1 and (−1)|J| det D(J, d) ≤ 0 for all J ⊆ I. (iii) The distance matrix D(I, d) has exactly one positive eigenvalue and at most n negative eigenvalues. Our proof is inspired by the proofs in [12, Sect. 6.2]. It will use the notion of Schur

A B complement: Consider a block matrix C , where A and D are square matrices, and D

0 D is invertible. After a block Gaussian elimination, it becomes P0 D , where P = A − BD−1 C is called the Schur complement of D.

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Proof of Theorem 4.1. Assume that I = {0, · · · , k} and that for i, j ∈ I, d(i, j) = 0 if and only if i = j. So the off-diagonal entries are all positive. The degenerate case, where some off-diagonal entries vanish, will be discussed later. If (I, d) is isometrically embeddable into (Kn , dK ), we can choose an embedding σ : I → Kn such that σk is the hyperplane at level 1. This is always possible, because dK is invariant under the action of Möb(n − 1). We then write the distance matrix D(I, d) explicitly. It will be in the form of  D(I, d) =

D(I \ {k}, d) Φt

Φ 0

 ,

(5)

where Φ denotes a k×1 column matrix whose i-th entry is 1/φ(σi ) for 0 ≤ i < k. Consider the distance sub-matrix D({0, k}, d), one can easily verify that its Schur complement, denoted by P (I, d), is in the form of dE (t(σi ), t(σj ))2 − dE (t(σi ), t(σ0 ))2 − dE (t(σ0 ), t(σj ))2 φ(σi )φ(σj )

t(σi ) − t(σ0 ) t(σj ) − t(σ0 ) , = −2 φ(σi ) φ(σj )

P (I, d)ij =

for i, j ∈ I \ {0, k}, where · , · is the Euclidean inner product. Since I \ {k} is embedded by σ into Kn , xi = t(σi ) −t(σ0 ) are vectors in the (n −1)-dimensional hyperplane x0 = 0. Therefore, P (I, d) is negative semi-definite, whose rank is at most n − 1. Inversely, let P (I, d) be the Schur complement of the submatrix D({0, k}, d). If it is negative semi-definite with rank at most n − 1, it can be written in the form P (I, d)ij = −2xi , xj  for 0 < i, j < k, where xi are vectors in an (n − 1)-dimensional hyperplane. Without loss of generality, we may assume it to be the hyperplane x0 = 0. Then an embedding can be constructed by setting σk as the hyperplane at level 1, and φ(σi ) = 1/D(I, d)ik  xi /D(I, d)ik t(σi ) = 0

if i > 0 if i = 0

for 0 ≤ i < k. We have proved that (I, d) is isometrically embeddable into (Kn , dK ) if and only if P (I, d) is negative semi-definite of rank at most n −1. We also have the following relations for the Schur complement det D(I, d) = −

det P (I, d) φ(σ0 )2

(6a)

In D(I, d) = (1, 1, 0) + In P (I, d)

(6b)

rk D(I, d) = rk P (I, d) + 2

(6c)

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where rk M is the rank of M , and In M is the inertia of matrix M , which is a triple indicating (in order) the number of positive, negative and zero eigenvalues of M . Here we use the convention that the determinant of an empty matrix is 1. These relations allow us to express the negative semi-definiteness and the rank of P (I, d) by the distance matrices. A principal submatrix of P (I, d) is of the form P (J, d) for some subset J of I such that J ⊇ {0, k}. P (I, d) is negative semi-definite if and only if for any principal submatrix P (J, d), (−1)|J| det P (J, d) ≥ 0. Notice that the choice of {0, k} is arbitrary, since we can always apply a permutation on I, to bring any index to 0 or k. So we have (−1)|J| det P (J, d) ≥ 0 for all J ⊆ I (the case |J| ≤ 1 is trivial). Then Eqs. (6a) and (6c) prove (i) ⇔ (ii). P (I, d) is negative semi-definite if and only if all its eigenvalues are non-positive. Then Eqs. (6b) and (6c) prove (i) ⇔ (iii). For the degenerate case, suppose that the d(0, k) = 0 and keep other distances positive. Now the sub-distance-matrix D({0, k}, d) is a zero matrix, so we cannot argue by Schur complement. However, since d is not identically zero, we can always relabel the elements of I so that d(0, k) > 0. So the argument for the rank and the inertia is still valid. For a subset J ⊆ I, if 0 or k is not in J, the non-degenerate results applies. Otherwise, if {0, k} ⊆ J, the 0-th row (and column) of D(J, d) is a multiple of the k-th row (and column), so det D(J, d) = 0. Therefore, by induction, the theorem remains valid as long as the off-diagonal entries of D(I, d) are not all zero, as assumed in the theorem. 2 Eq. (5) shows that the distance matrix for kissing spheres is playing dual roles: It combines the power of the distance matrix and the Cayley–Menger matrix. Alternatively, this can be seen by comparing Theorem 2.2 and Theorem 4.1, Remark. We notice from (5) that if φ(σi ) = 1 for all i ∈ I \ {k}, then Φ = e and D(I, d) reduces to an Euclidean Cayley–Menger matrix. 4.2. Embedding into the lightcone ˆ n can be embedded into the Lorentz We now show that the space of kissing spheres in E n,1 space R , which will be useful later for the study of distance completability problem. The Lorentz space Rn,1 is an (n + 1)-dimensional real vector space with an indefinite inner product of signature (n, 1). Explicitly, with the coordinate system x = (x0 , . . . , xn−1 , t), the Lorentz inner product · , ·n,1 on Rn,1 is defined as x, x n,1 = x0 x0 + . . . + xn−1 xn−1 − tt for two vectors x and x in Rn,1 . A vector x is space-like (resp. light-like, time-like) if x, xn,1 is positive (resp. zero, negative). The lightcone, denoted by Ln , is the set of light-like vectors, i.e. Ln = {x | x, xn,1 = 0}. A vector x is future-directed (resp.

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past-directed) if its last coordinate t is positive (resp. negative). We denote by Ln+ the future-directed lightcone, that is, the set of future directed light-like vectors (the origin excluded). Let (I, d) be a distance space isometrically embeddable into (Kn , dK ). Since D(I, d) has exactly one positive eigenvalue and at most n negative eigenvalues, we can factor it into D(I, d) = Qt ΛQ, where Λ is an (n + 1) × (n + 1) diagonal matrix, with 1 as its last entry, and all other entries being −1. The columns of Q can be viewed as vectors in the Lorentz space Rn,1 , indexed by the elements of I. We can therefore write D(I, d)ij = −xi , xj n,1 for a system of vectors {xi }i∈I in Rn,1 . Since xi , xi n,1 = −D(I, d)ii = 0, all the vectors xi are on the lightcone Ln . Since −xi , xj n,1 =

1 xi − xj , xi − xj n,1 = D(I, d)ij ≥ 0 2

for all i = j, the difference between any two vectors cannot be time-like. Therefore, {xi }i∈I have to be either all future-directed, or all past-directed. The Lorentz inner product induces a distance function dL (x, x )2 := −x, x n,1 =

1 x − x , x − x n,1 ≥ 0 2

(7)

on the future-directed lightcone Ln+ . We can therefore view D(I, d) as the Lorentz distance matrix for a set of future-directed light-like vectors. We have just proved the following theorem: Theorem 4.2. (Kn , dK ) and (Ln+ , dL ) are isometric. That is, there is a bijective isometric embedding between (Kn , dK ) and (Ln+ , dL ). For example, let Kn∗ be the set of kissing spheres of finite diameter, the following map Ψ∗ from Kn∗ to Ln+ is an isometric embedding: √

Ψ∗ (p) =

 2

1 − t(p)22 , 2t(p), 1 + t(p)22 . 2φ(p)

(8)

It is easy to verify that Ψ(p), Ψ(p)n,1 = 0 and dK (p, q) = dL (Ψ(p), Ψ(q)). We can extend Ψ∗ to a map Ψ defined on Kn by setting √ Ψ(p) =

2 (−h, 0, . . . , 0, h) 2

if p is a hyperplane at level h. One verifies that Ψ is bijective. Remark. A continuous version of Theorem 4.2 can be found in [13], which asserts that a Riemann space is conformally Euclidean if and only if it can be embedded into the lightcone. A generalization for conformally flat Riemann manifolds can be found in [14].

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The correspondence stated by Theorem 4.2 is not new. If we regard the half-space x0 ≥ 0 as the Poincaré half-space model for the n-dimensional hyperbolic space, then the kissing spheres are the horospheres. It is classical in hyperbolic geometry that horospheres can be identified to vectors in the lightcone, see for instance [15]. Furthermore, in the projective model of Möbius geometry, points are mapped to light-like directions [9, Eq. (2.3)]. In fact, our isometric embedding maps the tangency points of kissing spheres to light-like directions in the same way, and use the vector lengths to distinguish different kissing spheres with the same tangency point. Note that the invariance of D(I, d) under the action of Möb(n −1) is reflected in Rn,1 as the invariance under Lorentz transformations that preserves the direction of time. Indeed, Möb(n − 1) is isomorphic to the orthochronous Lorentz group O+ (n, 1) [9, Corollary 3.3]. If the reference ball has a non-zero curvature κ, we can either make it zero by a Möbus transformation before applying the isometric embedding in (8), or directly apply the following embedding to a kissing sphere p: Ψ(p) =

 √2 2

+

√  2 (ˆt(p), 1), κφ(p)

(9)

where φ(p) is now the signed diameter, negative if the sphere surrounds the reference ball, and ˆt(p) refers to the unit direction vector for the tangency point, taking the center of the reference ball as the origin. The Lorentz distance dL (Ψ(p), Ψ(q)) can be used as a distance function for kissing spheres to a reference ball of non-zero curvature. A non-zero vector y and a real number c determine a hyperplane H = {x | x, yn,1 = c}. We call y a normal vector of H. A hyperplane is said to be space-like (resp. light-like, time-like) if its normal vector is time-like (reps. light-like, space-like). We observe two situations where distance geometry for kissing spheres reduces to Euclidean distance geometry. • Let I be a set of kissing spheres. Suppose that for all p ∈ I, Ψ(p) lies on the same lightlike hyperplane H. That is, H can be written in the form H = {x | −x, yn,1 = 1}, where y is a light-like vector. Then the kissing sphere Ψ−1 (y) is tangent to all the el√ √ ements of I. There is a Lorentz transformation L that sends y to (− 22 , 0, · · · , 0, 22 ). For all p ∈ I, Ψ−1 LΨ(p) is a unit kissing sphere (kissing spheres of unit diameter) since they are tangent to the hyperplane at level 1. Therefore D(I, d) degenerates to Euclidean distance matrix. This is the same situation as remarked at the end of Section 3. • If for every p ∈ I, Ψ(p) lies on a space-like hyperplane H, then there is a Lorentz transformation L that sends H to a time-constant hyperplane. The intersection of LH with the lightcone is an (n −1)-sphere, and dL becomes to Euclidean distance dE . One way to view this is to consider unit kissing spheres kissing a reference ball of finite radius. It turns out that dK , induced by (9), equals the Euclidean distance between their centers.

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Seidel also observed in [16, Theorem 4.5.3] that the distance matrix for points in Euclidean space can be written as the Gram matrix of light-like vectors on a time-constant hyperplane. 4.3. Distance completability problem We now study the distance completability problem in (Kn , dK ). The following are consequences of our results for the embeddability problem. Theorem 4.3. Let a graph G = (V, E) be given with a length function  : E → R≥0 . Then (G, ) is completable in (Kn , dK ) if and only if there is a non-negative symmetric matrix D satisfying (C1) (C2) (C3) (C4)

Duv = 0 if u = v. Duv = (u, v)2 if and only if (u, v) ∈ E. The rank of D is at most n + 1. D has exactly one positive eigenvalue.

D is in fact the distance matrix corresponding to a distance function realizing the given edge lengths. We call D a target matrix. This theorem transforms a distance completion problem to a matrix completion problem: Some entries of the matrix being given (C1 and C2), find the values for the other entries, so that the rank of matrix is low (C3). We refer to [17,18] for more about matrix completion. Comparing to the classical matrix completion problem, the condition (C4) is new. The target matrix is usually positive semi-definite, but here we need it to be indefinite. The result in the previous section can help on this problem. If (G, ) is completable in (Kn , dK ), then for every clique K of G, (K, ) is completable in (Kn , dK ). The inverse is in general not true. Define two sets as in Euclidean case, n CK (G) = { : E → R≥0 | (G, ) is completable in (Kn , dK )}, n (G) = { : E → R≥0 | for all cliques K ⊂ G, (K, ) is completable in (Kn , dK )}. KK

We now prove our second main theorem (compare Theorem 2.4) n n (G) = KK (G) if and only if G is chordal. Theorem 4.4. CK

For the proof, we will employ techniques from [6], with some adaption. Proof of the “only if” part of Theorem 4.4. If G is not chordal, consider a chordless cycle C of length at least 4, and pick an edge e0 ∈ C. Construct a length function  by setting (e) = 1 if e has exactly one end in C or if e = e0 , otherwise (e) = 0. Then n n  ∈ KK (G) but  ∈ / CK (G). 2

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The “if” part will be derived from the following. If two graphs each has a clique of the same size, the clique-sum glues them together by identifying that clique. In the language of mathematics, consider a graph G = (V, E) and two of its subgraphs G1 = (V1 , E1 ) and G2 = (V2 , E2 ). G is the clique-sum of G1 and G2 if V = V1 ∪ V2 and W = V1 ∩ V2 induces a clique in G, and there is no edge joining a vertex in V1 \ W and a vertex in V2 \ W . We now use Theorem 4.2 to prove the following lemma. n n (Gi ) = KK (Gi ) Lemma 4.5. Let G be a clique-sum of G1 (V1 , E1 ) and G2 (V2 , E2 ). If CK n n for i = 1, 2, then CK (G) = KK (G). n n n (G). Obviously,  ∈ KK (Gi ) = CK (Gi ), for i = 1, 2. Proof. Let  be an element in KK Since (G1 , ) and (G2 , ) are n-completable, let D1 and D2 be the corresponding target matrices. We can find a system of future-directed null vectors xu ∈ Ln+ for u ∈ V1 such that (D1 )uv = dL (xu , xv )2 for u, v ∈ V1 , and yu ∈ Ln+ for u ∈ V2 such that (D2 )uv = dL (yu , yv )2 for u, v ∈ V2 . On the common clique, for all u, v ∈ V1 ∩ V2 , we have dL (xu , xv ) = dL (yu , yv ) = (u, v). Therefore, there is a Lorentz transformation L, such that Lxu = yu for u ∈ V1 ∩ V2 . Now we construct a system of vectors by setting zu = Lxu for u ∈ V1 , and zu = yu for u ∈ V2 \ V1 . The matrix Duv = dL (zu , zv ) is a n target matrix for (G, ), therefore  ∈ CK (G). 2

Proof of the “if” part of Theorem 4.4. It follows from Lemma 4.5 since any chordal graph can be built up by clique-sums. 2 Therefore, for a chordal graph, in order to tell for a length function whether it is completable in (Kn , dK ), we only need to check all the maximum cliques for embeddability, which makes it much easier to tell if there exists a solution to the distance completion problem. 5. A discussion on ball packings This paper is actually motivated by the ball packing problem. A ball packing is a collection of balls with disjoint interiors. The tangency graph of a ball packing takes the balls as vertices, and the tangency relations as edges. For a given graph, the ball packing problem asks whether there is a ball packing whose tangency graph is isomorphic to the given graph. Disk packings in dimension two are well understood thanks to Koebe– Andreev–Thurston’s disk packing theorem, but a generalization to higher dimensions turns out to be very difficult. Some attempts can be found in [19–22]. Kissing spheres can be viewed as balls restricted to the surface of the reference ball. A packing of balls touching an n-dimensional reference ball can be regarded as a ball packing of “dimension n − 1/2”. It’s a special ball packing in which one ball touches every others. Our hope is that the study of kissing spheres may help understanding ball packings. For example, we may consider the kissing sphere packing problem, which asks

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whether a graph can be realized by the tangency relations of a collection of pairwise non-intersecting kissing spheres. Special kissing sphere packings have been studied by Szirmai [23,24] in the form of horosphere packing. Some kissing-sphere packable graphs have been investigated by Maehara and Noha in [25]. In this paper, kissing spheres are allowed to intersect each other, so we are not dealing with packing problems. However, the distance function (see Definition 3.2) is intentionally designed to indicate the relations between kissing spheres (intersection, tangency or disjointness). It is a natural idea to apply distance geometry to the packing problem. In the language of distance completion problem, consider a length function  = 1 on G. If G is kissing-sphere packable, then (G, ) is completable in (Kn , dK ). However, since no intersection is allowed in a packing, one is only allowed to complete distances with values ≥ 1, which makes the problem much more complicated. Back to the ball packing problem. Similar to our representation of kissing spheres by ˆ n by vectors on the light-cone, there is a conventional way [7,9,26] to represent balls in E vectors on the one-sheet hyperboloid (de Sitter space) dSn+1 = {x ∈ Rn+1,1 | x, xn+1,1 = 1}.

(10)

If we take the Lorentz inner product dL (x, x )2 := −x, x n,1 as a “squared distance” (even though it’s nothing like a squared distance), it reflects the relation between the corresponding spheres (tangency, disjointness or intersection) in a similar way as the isometric embedding in (8). Embedding problem for de Sitter space has been studied by Kokkendorff in [27] in term of Gram matrix. This paper provides a geometric picture for the set of kissing spheres. However, for an application on the packing problem, more investigations on discrete geometry in Lorentz space are expected. One study in this direction is Boyd–Maxwell ball packings, initiated by Boyd in [28] and Maxwell in [26], and recently revisited by Labbé and the author in [29]. Acknowledgements I would like to thank Raman Sanyal, Louis Theran, Karim Adiprasito, and Günter M. Ziegler for helpful discussions. I would like to give special thanks to Leo Liberti for his suggestion on terminology. References [1] J.H. Conway, N.J.A. Sloane, Sphere Packings, Lattices and Groups, 3rd ed., Grundlehren Math. Wiss., vol. 290, Springer-Verlag, New York, 1999. [2] K. Menger, Géométrie Générale, Mémor. Sci. Math., vol. 124, Gauthier-Villars, Paris, 1954. [3] T.L. Hayden, J. Wells, Approximation by matrices positive semidefinite on a subspace, Linear Algebra Appl. 109 (1988) 115–130. [4] R.L. Graham, P.M. Winkler, On isometric embeddings of graphs, Trans. Amer. Math. Soc. 288 (1985) 527–536.

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