On almost-equidistant sets - II

A set in $\mathbb R^d$ is called almost-equidistant if for any three distinct points in the set, some two are at unit distance apart. We proved that an almost-equidistant set $V$ in $\mathbb R^d$ has $O(d)$ points in two cases: if the diameter of $V$ is at most $1$ or if $V$ is a subset of $(d-1)$-dimensional sphere of radius at most $\sqrt{1/2+O(d^{-2/3})}$. Also, we present a new proof of the result by Kupavskii, Mustafa and Swanepoel arXiv:1708.01590 that an almost-equidistant set in $\mathbb R^d$ has $O(d^{4/3})$ elements.


Introduction
A set in R d is called almost-equidistant if among any three points in the set, some two are at unit distance apart. The natural conjecture [12,Conjecture 7] claims that an almost-equidistant set in R d has O(d) points.
Using an elegant linear algebraic argument Rosenfeld [14] proved that an almostequidistant set on a (d − 1)-dimensional sphere of radius 1/ √ 2 has at most 2d points. Note that the set of the vertices of two unit (d − 1)-simplices lying on the same (d − 1)dimensional sphere of radius 1/ √ 2 is almost-equidistant. Bezdek and Lángi [2, Theorem 1] generalized Rosenfeld's approach and showed that an almost-equidistant set on a (d − 1)-dimensional sphere of radius < 1/ √ 2 has at most 2d + 2 points; this bound is tight because the vertices of two unit d-simplices inscribed in the same sphere form an almost-equidistant set. Balko, Pór, Scheucher, Swanepoel and Valtr [1] showed that an almost equidistant set has O(d 3/2 ) elements. This bound was improved by the author [12] to O(d 13/9 ). Recently, Kupavskii, Mustafa, Swanepoel [9] further improved to O(d 4/3 ). For more references we refer the interested readers to [1].
The first goal of the current article is to confirm the conjecture in two cases: for almostequidistant sets of diameter 1 (see Section 3) and for almost-equidistant sets lying on a (d − 1)-dimensional sphere of radius ≤ 1/2 + O(d −2/3 )) (see Section 4). The second aim is to give a new proof of the upper bound O(d 4/3 ) for the cardinality of an almostequidistant set in R d (see Section 5). Also, we will discuss several open problems related to almost-equidistant sets (see Section 6).

Preliminaries
Suppose that {v 1 , . . . , v n } ⊂ R d is an almost-equidistant set. Consider the matrix where J n is the n-by-n matrix of one's and I n is the identity matrix of size n. We need two simple facts proved in [12,Corollary 4 and Lemma 5]. We join them in the following lemma.
2) U has at most one eigenvalue > 1 and at least n − d − 2 eigenvalues equal to 1.
Suppose that {v 1 , . . . , v n } is an almost-equidistant set lying on the (d − 1)-dimensional sphere of radius 1/(2(1 − cos α)) with center in the origin of R d . Consider the matrix where 0 < α < π is a fixed angle. We omit the proof of the next lemma because it is similar to the proof of Lemma 1.
We will use the following lemma in next sections.
Lemma 3. Suppose that an n-by-n matrix W has n − k eigenvalues are equal to 1. Also, tr(W) = tr(W 3 ) = 0.
3. Assume that all eigenvalues of W but λ are ≤ 1. Then Proof. 1. Denote by λ 1 , . . . , λ k eigenvalues of W that are < 1. Then To finish the proof, we need Lemma 1 in [2]. For the sake of completeness we provide its proof here.
Here the equality is possible if x i are equal and therefore, x i ≥ 1.
Proof. Consider functions f, g : has the value of a tangent line to f (x) at x = 1 and the second point of the intersection of that tangent line and The equality case we leave as an exercise.
Assume that n > 2k. Introducing the notation l = n − 2k we can rewrite the first equality in (3) as: Finally, according to the second equality in (3) Assume that n ≥ 2k − 1. Introducing the notation l = n − 2k − 1 we can rewrite the first equality in (4) as: By Lemma 4, we obtain Moreover, if λ + λ ′ = 0 then (5) becomes a strict inequality (see Lemma 4). Since the second equality in (4), By Lemma 4, we obtain

Almost-equidistant diameter sets
A subset of R d is called an almost-equidistant diameter set if it is almost-equidistant and has diameter 1. The next theorem is about the maximal size of such sets.
Theorem 5. An almost-equidistant diameter set in R d has at most 2d + 4 points.
Proof. Suppose that the matrix U (see (1)) for an almost-equidistant diameter set V ⊆ R d . Clearly, the entries of U are non-positive.
Assume that U has an eigenvalue λ > 1. We need the following week form of the Perron-Frobenius theorem; see [5,11]  Theorem 6 (Perron-Frobenius theorem). If an n-by-n matrix has non-negative entries, then it has a non-negative real eigenvalue, which has maximum absolute value among all eigenvalues.

Almost-equidistant sets on small spheres
Proof. Assume that {v 1 , . . . , v n } ⊂ S is an almost-equidistant set and S has the center in the origin Consider the matrix U α (see (2)) for {v 1 , . . . , v n }. According to Lemma 2 there are two possible cases.
Therefore, it is enough to show that (7) holds. Without loss of generality we can assume that n i=1 v i = o, where o is the origin. By the definition of f (see (7)) we have Summing up the last inequality for i = 1, . . . , n we get Thus We will use the following theorem (see [4,Theorem 1]) several times.
2. If 1/m ≤ f /n < 1 then (11) and (12)  4. If f /n < 1/l then (11) and (12) imply This completes the proof. 6 6. Discussion 6.1. Almost-equidistant diameter sets. A graph (V, E) is called a diameter graph if its vertex set V ⊆ R d is a set of points of diameter 1 and a pair of vertices forms an edge if they are at unit distance apart. Of course, the set of vertices of two cliques in a diameter graph is an almost-equidistant diameter set. For instance, in [10, the last paragraph of Section 3] there is given an example of diameter graph in R d consisting of two cliques without common vertices such that they have d + 1 and ⌊ d+1 2 ⌋ vertices respectively. We believe that the vertex set of this diameter graph has the maximal size among almost-equidistant diameter sets in R d . There is the following conjecture [7, Conjecture 5.5] that arose in the context of study of cliques in diameter graphs.
Conjecture 13 (Schur). Let S 1 and S 2 be two unit simplices in R d forming a set of diameter 1 such that they have k and m vertices respectively. Then S 1 and S 2 share at least min{0, k + 2m − 2d − 2} vertices for k ≥ m.
6.2. Two-distant almost-equidistant sets. A subset of R d is called a two-distant set if there are only two distances formed by any two distinct points of the set. The following question seems to be interesting.

Problem 14.
What is the largest cardinality of a set in R d that is two-distant and almost-equidistant at the same time?
Let us consider a two-distant almost-equidistant set V ⊆ R d with distances 1 and a > 1 between points of V (the case a < 1 is not interesting because of Theorem 5). If the matrix U (see (1)) for V does not have an eigenvalue > 1 then Lemmas 1 and 3 imply |V | ≤ 2d + 4. Thus we can assume that U has an eigenvalue > 1. Note that the matrix U/(a 2 − 1) is an adjacency matrix of some triangle-free graph. By Lemma 1, Problem 14 is reduced to following question.
Problem 15. What is the minimal rank of the matrix A − λ 2 I n , where A is the adjacency matrix of some triangle-free graph on n vertices and λ 2 > 0 is its second largest eigenvalue?
It is worth pointing out that there are infinitely many triangle-free graphs on n vertices such that rank(A − λI) = O(n 3/4 ), where A is the adjacency matrix of the graph and λ is some real number; see [3, the proof of Theorem 5].