Few distance sets in spaces and product spaces
Abstract
Kusner asked if points is the maximum number of points in such that the distance between any two points is . We present an improvement to the best known upper bound when is large in terms of , as well as a generalization of the bound to -distance sets. We also study equilateral sets in the sums of Euclidean spaces, deriving upper bounds on the size of an equilateral set for when , is even, and for any .
1 Introduction
1.1 Background
A classic exercise in linear algebra asks for the maximum number of points in such that the pairwise distances take only two values. One can associate a polynomial to each point in the set such that the polynomials are linearly independent. Then, one can show that the polynomials all lie in a subspace of dimension . Since the number of linearly independent vectors cannot exceed the dimension of the subspace, the cardinality of the set is at most . This beautiful argument was found by Larman, Rogers, and Seidel [10] in 1977, illustrating the power of algebraic techniques in combinatorics.
We can ask the much more general question: “In a metric space , what is the maximum number of points such that the pairwise distances take only values?” We use , or just if , to denote the answer to this question (by convention, we do not count as a distance). A set of points satisfying this question’s conditions, i.e., the cardinality of the set is , is called an -distance set. A -distance set is also known as an equilateral set. Also, we typically restrict the metric space to a normed space, so that the specific distances used do not matter. Thus, we will always assume that the largest of the distances is .
The posed question has been studied on many different spaces. The most famous result would be the upper bound on an -distance set in -dimensional Euclidean space, found by Bannai, Bannai, and Stanton [3]. This result was also discovered independently by Blokhuis [4]. Another important case is when all points lie on the -dimensional sphere. There is strong motivation for this problem as it has many applications in coding theory and design theory [6]. In particular, having tight upper bounds can help us find extremal configurations which satisfy unique properties.
Similar results have been obtained in the hyperbolic space [4], the Hamming space [12], and the Johnson space [12]. Not as much is known in an arbitrary finite-dimensional normed space, known as a Minkowski space, other than Petty’s [13] general bound of , where . Swanepoel [17] then conjectured that for a Minkowski space with dimension and proved it for the case. We should mention that equilateral sets in Minkowski spaces have been applied in differential geometry, where they are used to find minimal surfaces [11].
1.2 Definitions
In our paper, we investigate this problem on with the norm, as well as on the sum of Euclidean spaces. For a point and a , the norm is defined to be
and the norm is defined to be
The norm is the well-known “taxicab” norm, and the norm is the standard Euclidean norm. Throughout the paper, we write instead of to emphasize that we are using the Euclidean norm. We also use to emphasize that we are in -dimensional Euclidean space.
For Euclidean spaces , we define their sum as the product space equipped with the norm
where for . When , the norm is
Then, for any , we let the distance between two points and be and use to describe this space.
Although our notation for the norm in spaces and in sums is the same, the norm being used should be clear from context.
1.3 Previous Work and Our Results
We first study -distance sets in equipped with the norm. This space is denoted by . The two most famous questions pertaining to this problem are Kusner’s [7] conjectures on equilateral sets.
Conjecture 1 (Kusner).
.
Conjecture 2 (Kusner).
for .
For Conjecture 1, note that the set , where is the -th standard basis vector, is equilateral in , so . Currently, the best known upper bound is due to Alon and Pudlák [1]. It is also known that Conjecture 1 holds for (Bandelt, Chepoi, and Laurent [2]) and (Koolen, Laurent, and Schrijver [9]).
As for Conjecture 2, note that the set is equilateral for a suitable choice of , so . For and large enough, Swanepoel [19] actually showed that , disproving Conjecture 2 in this case. The first nontrivial upper bound of was found by Smyth [15] using an approximation argument. This result was later improved by Alon and Pudlák [1] and is currently the best known upper bound on for arbitrary and .
Theorem 1 (Alon and Pudlák).
For every , we have , where one may take for an absolute .
Our first result is an improvement of Theorem 1 when is large in terms of . One can check that , so Theorem 2 is indeed an improvement.
Theorem 2.
Let be the constant from Theorem 1. If and , then .
We should mention that when satisfies other special properties, Theorem 1 can be strengthened. If is an even integer, Swanepoel [19] used a linear independence argument to show that
If is an odd integer, Alon and Pudlák’s argument for extends to , giving the bound in this case.
Our next result is a generalization of Theorem 1 to -distance sets. As far as we know, the only literature on -distance sets in is by Smyth [16]. Our theorem below is strictly stronger than Conjecture 5 in [16].
Theorem 3.
If is a positive integer and is a real number satisfying , then for a constant depending on and .
Our next three results are on equilateral sets in the sum of Euclidean spaces. As far as we know, this problem has not been well studied in this space. Swanepoel [20] showed that for normed spaces and , where is the finite Borsuk number. However, explicit bounds are still unknown when and are Euclidean spaces and when we take an sum instead of an sum. Our first result in this area almost completely resolves the problem for . Note that we have the obvious lower bound by taking the Cartesian product of the two equilateral sets. This lower bound is known to meet the upper bound when [20].
Theorem 4.
Let and be Euclidean spaces. Then, .
Our second result in this area provides an upper bound when is even.
Theorem 5.
Let and be Euclidean spaces, and let be an even integer. Then, .
Finally, we extend Alon and Pudlák’s [1] result on equilateral sets in to certain product spaces. We present an upper bound on the sum of Euclidean spaces for any . Observe that Theorem 1 is a special case of our theorem below when .
Theorem 6.
Let be Euclidean spaces and set . If , then for a constant depending on and .
2 Preliminaries
In this section, we present two famous results that we use later. The first is the Rank Lemma, which allows us to estimate the rank of a symmetric matrix. The second is Jackson’s Theorem, a celebrated result from approximation theory. This theorem allows us to approximate by a polynomial with sufficiently small error.
2.1 The Rank Lemma
Lemma 1 (Rank Lemma).
For a real symmetric nonzero matrix ,
We will frequently make use of the following corollary.
Corollary.
Let be a real symmetric matrix with and for all . Then
Choosing gives .
2.2 Jackson’s Theorem
Theorem 7 (Jackson, [8]).
For any and positive integer , there exists a polynomial with degree at most such that
where is an absolute constant, denotes the modulus of continuity of , and uses falling factorial notation.
From Theorem 7, we can recover the following lemma. This lemma was first given and proved in [14] and later also transmitted in [18].
Lemma 2.
For any and , there exists a polynomial with degree at most such that
(1) |
where .
We will always assume that the polynomial in Lemma 2 is even and that . If is not even, we can take the even part of . If , we can take the polynomial as this only increases the error term by a factor of .
3 Bound on equilateral sets in for large
We start with an important lemma about the norm.
Lemma 3.
Suppose . Then for any ,
Proof.
The left hand inequality is a well-known property of the norm. The right hand inequality is the Power Mean Inequality, which can be proven using Jensen’s Inequality. ∎
Suppose our equilateral set is . Let be the closest even integer to , rounding up if is odd. The main idea is that by Lemma 3, we can approximate by . If is large enough in terms of , the error term is sufficiently small. From there, it suffices to bound the size of an approximately-equilateral set in , which can be done with a linear algebra argument.
Proof of Theorem 2.
We use the notation explained above. There are two cases to consider depending on whether is greater than or less than .
Case 1. We have is odd, so .
From our bound on and the fact that ,
where the first inequality holds by Theorem 1 and the last inequality by the inequality valid for . Assuming without loss of generality that , we may rearrange the result above into
Because is the smallest even integer greater than , we have
(2) |
Now, embed the points into . Since , Lemma 3 implies
(3) |
for all . Consider the functions given by , and let be the matrix with . Clearly, , and from (2) and (3),
for all . Since is symmetric, Lemma 1 tells us that . For the upper bound on the rank, note that every lies in the span of the set of polynomials
The -th row vector of is . Hence, every row vector belongs to the subspace spanned by , which has dimension at most . It follows that . Combining the upper and lower bound, we have
as desired.
Case 2. We have is even, so .
Since , . So, similar to Case 3, we have
Rearranging the result gives us
Because is the largest even integer less than or equal to , we have
(4) |
Embed the points into . This time, Lemma 3 implies
(5) |
We define and like in Case 3. Again, , and from (4) and (5),
for all . Applying Lemma 1, . So, similar to Case 3, we have the bound . The final result is
Having considered all cases, the proof is complete. ∎
4 Bound on -distance sets in
Smyth [14], see also [16], first combined Jackson’s theorem with the invertibility of diagonally dominated matrices to prove for . Alon and Pudlák [1] then improved this bound by substituting the rank lemma for the invertibility lemma to achieve Theorem 1. We would like to extend this method of proof to the -distance case. However, as pointed out by Smyth [16], if one of the distances is arbitrarily small, we need arbitrarily high degrees of approximation. So, we want to impose a lower bound on our distances.
Consider a two-distance set with distances and , where is very small. Intuitively, this set should look like several “clusters” of points, such that the distance between each cluster is . So, if we look globally, this set looks like an equilateral set with distance , but if we look locally at each cluster, we have an equilateral set with distance . This means that we have essentially reduced the problem from one about two-distance sets to one about equilateral sets. We carry this intuition to -distance sets and formalize this argument with an induction.
Proof of Theorem 3.
We use strong induction on . The base case is , which was proven by Alon and Pudlák [1]. For the inductive step, assume that the statement holds true for all . We prove that it is true for .
Let be a -distance set in . Let our distances be . There are two cases to consider depending on whether the distances are lower bounded or not.
Case 1. The smallest distance is less than .
There exists an index such that . Let be a maximal -distance set using the distances . Every point is a distance at most from some point in , otherwise can be added to , contrary to the maximality of . So, if we draw a closed ball of radius around every point in , every point in lies within some ball. These balls are disjoint from the condition . Furthermore, by the triangle inequality, within each ball, the distance between any two points is at most , and thus is at most . It follows that within every ball, we have a -distance set, where , using of the distances from . This implies the bound
Applying the inductive hypothesis gives us
as desired. The first inequality follows by analyzing the function
on the interval . Its first derivative is
so is increasing on and hence on . The second inequality follows by analyzing the related function
on the interval . It is symmetric about , and its first derivative is
Since , is decreasing on and increasing on . Thus, it is maximised at and .
Case 2. The smallest distance is at least .
Suppose consists of the points . For convenience, let . Fix as the constant from Lemma 2. Define as
Then, let be a positive integer satisfying
which is possible since .
Lemma 2 allows us to pick an even polynomial with and degree at most such that
Consider the functions given by
and their polynomial approximations
Let be the matrix given by . First, since , for all . We now estimate for . Expand
and
where denotes the -th elementary symmetric polynomial in . For convenience, we define and . Applying the triangle inequality with (1), we have
Since , recall that . Combining this with the fact that implies that is positive. Now, we are ready to upper bound . Since for , we have . Thus
The first inequality is achieved by the triangle inequality, the factorization , and the upper bound on . The second inequality follows by the AM-GM inequality. The last inequality holds because by our choice of and , whence .
Now, since , we can lower bound with
On the other hand, since , we can upper bound with
This gives us . Putting everything together, we have
from our choice of and .
Recall that is even, so the matrix is symmetric. Lemma 1 then tells us that . We now find an upper bound for the rank.
Note that the polynomials belong to the span of the set
This set consists of polynomials. Thus, all the belong to the span of a set of at most polynomials. The -th row vector of is . Hence, every row vector belongs to the subspace spanned by
which has dimension at most . This implies .
The upper and lower bounds combine to give . Using the upper bound on and the condition , we can rearrange the inequality to obtain
This completes the induction. ∎
5 Bound on equilateral sets in product spaces
Proof of Theorem 4.
Write for each point , where and . Let be our equilateral set with cardinality . Consider the functions defined by
for all . Note that for all , so the are linearly independent.
We can expand as
So, the are all spanned by the following set of polynomials
This implies the bound .
We will now prove that the set of polynomials is linearly independent. Assume for the sake of contradiction that we have a dependence
(6) |
The left hand side of (6) is a polynomial in the that is identically zero. Thus, extracting the coefficient of , we have
(7) |
Extracting the coefficient of , for the appropriate , we have
(8) |
Extracting the coefficient of and applying (7), we have
(9) |
Now, plug into (6), multiply both sides by , and sum over all .
Applying (7), (8), and (9) implies for all . It easily follows that all other coefficients are zero, as desired.
Now, we know that . This rearranges into . ∎
6 Bound on equilateral sets in product spaces for even
Proof of Theorem 5.
Let be an equilateral set in with points. For every , define the function by
for all so that for all . It follows that the polynomials are linearly independent. Now, we can expand as
Utilizing multi-index notation, we can expand
The sum is taken over all non-negative integers and non-negative integer vectors with entries such that , where is the sum of the components of . Similarly, we can expand
We want to count the number of monomials in each polynomial. Since is a polynomial in and is a polynomial in , the two sets of monomials are disjoint, except for the constant monomial. Let us consider first. By choosing , we must count all the monomials with degree at most . There are of these. If the degree is greater than , say , we only need to count the monomials formed when and . Hence, the total number of monomials in is
Similarly, there are monomials in . In total, has monomials (we subtract as the constant monomial is counted twice). This gives an upper bound on .
By finding a larger linearly independent set of polynomials, a trick first used by Blokhuis [5], we can lower this bound to . We prove that the set of polynomials
where are integer vectors with entries and sums of components respectively, is linearly independent. The details are very similar to those in Blokhuis’s [4] bound for the -distance set in .
Suppose we have a dependence
(10) |
The main claim is the following lemma.
Lemma 4.
For all with ,
Similarly, for all with ,
Proof.
We only prove the first statement. The argument for the second statement is identical.
Suppose . Consider the part of the left hand side of (10) that is homogeneous in with degree . Note that the monomials do not contribute to this, so we only have to look at the part from the . Using our expansion of above, the part of homogeneous in the variables with degree is
The left hand side of (10) is a polynomial in the which is identically zero. Thus, we have
Now, substitute , multiply by , and sum over all .
This is a sum of squares where all coefficients have the same sign. So,
Plugging in proves the lemma. ∎
Remark.
This proof can be easily extended to the sum of Euclidean spaces. If we consider the product space , our bound is just .
7 Bound on equilateral sets in product spaces for
Proof of Theorem 6.
Let be an equilateral set in , and write where for .
Let be the constant from Lemma 2. Take and set to be a positive integer satisfying
By Lemma 2, there exists an even polynomial with and degree at most that approximates . Thus, for and ,
Next, for define the functions by
Let be the matrix given by . Since , we have . For ,
with the last inequality following from our conditions on and . Because is symmetric, we can apply Lemma 1 to get .
Now, we look for an upper bound on the rank. We can write any point as
Because is even, all terms in the expansion of have integer exponent, i.e., is actually a polynomial. Additionally, since has degree at most , each is in the span of the set
where each set consists of all the monomials with degree less than formed by . Thus, the are spanned by at most polynomials. Because every row vector of , each of the form , belongs to the subspace spanned by the set , we have
where we let and use the well-known bound for . Now, recall that and , so . Thus,
Combining this inequality with the condition yields
as desired. ∎
8 Acknowledgements
The authors would like to thank the MIT PRIMES program for making this research possible. The authors would also like to thank Larry Guth for introducing them to the problem, as well as Yufei Zhao for some helpful suggestions.
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