Few Distance Sets in ℓ p Spaces and ℓ p Product Spaces
aa r X i v : . [ m a t h . M G ] N ov FEW DISTANCE SETS IN ℓ p SPACES AND ℓ p PRODUCT SPACES
RICHARD CHEN, FENG GUI, JASON TANG, AND NATHAN XIONG
Abstract.
Kusner asked if n + 1 points is the maximum number of points in R n such that the ℓ p distance between any two points is 1. We present an improvement to the best known upper boundwhen p is large in terms of n , as well as a generalization of the bound to s -distance sets. We alsostudy equilateral sets in the ℓ p sums of Euclidean spaces, deriving upper bounds on the size of anequilateral set for when p = ∞ , p is even, and for any 1 ≤ p < ∞ . Introduction
Background
A classic exercise in linear algebra asks for the maximum number of points in R n such that thepairwise distances take only two values. One can associate a polynomial to each point in the setsuch that the polynomials are linearly independent. Then, one can show that the polynomials alllie in a subspace of dimension ( n + 1)( n + 4) /
2. Since the number of linearly independent vectorscannot exceed the dimension of the subspace, the cardinality of the set is at most ( n + 1)( n + 4) / X , what is the maximumnumber of points such that the pairwise distances take only s values?” We use e s ( X ), or just e ( X ) if s = 1, to denote the answer to this question (by convention, we do not count 0 as adistance). A set of points S ⊆ X satisfying this question’s conditions, i.e., the cardinality of theset { d ( x, y ) : x, y ∈ S, x = y } is s , is called an s -distance set . A 1-distance set is also known as an equilateral set . Also, we typically restrict the metric space to a normed space, so that the specificdistances used do not matter. Thus, we will always assume that the largest of the s distances is 1.The posed question has been studied on many different spaces. The most famous result wouldbe the upper bound (cid:0) n + ss (cid:1) on an s -distance set in n -dimensional Euclidean space, found by Bannai,Bannai, and Stanton [BBS83]. This result was also discovered independently by Blokhuis [Blo84a].Another important case is when all points lie on the n -dimensional sphere. There is strong motiva-tion for this problem as it has many applications in coding theory and design theory [DGS77]. Inparticular, having tight upper bounds can help us find extremal configurations which satisfy uniqueproperties.Similar results have been obtained in the hyperbolic space [Blo84a], the Hamming space [MN11],and the Johnson space [MN11]. Not as much is known in an arbitrary finite-dimensional normedspace, known as a Minkowski space , other than Petty’s [Pet71] general bound of e ( X ) ≤ n , where n = dim X . Swanepoel [Swa99] then conjectured that e s ( X ) ≤ ( s + 1) n for a Minkowski space X with dimension n and proved it for the n = 2 case. We should mention that equilateral sets in Minkowski spaces have been applied in differential geometry, where they are used to find minimalsurfaces [Mor92].1.2.
Definitions
In our paper, we investigate this problem on R n with the ℓ p norm, as well as on the ℓ p sum ofEuclidean spaces. For a point x = ( x , . . . , x n ) ∈ R n and a p ≥
1, the ℓ p norm is defined to be k x k p = n X i =1 | x i | p ! p , and the ℓ ∞ norm is defined to be k x k ∞ = max ≤ i ≤ n | x i | . The ℓ norm is the well-known “taxicab” norm, and the ℓ norm is the standard Euclidean norm.Throughout the paper, we write k·k instead of k·k to emphasize that we are using the Euclideannorm. We also use E n to emphasize that we are in n -dimensional Euclidean space.For Euclidean spaces E a , . . . , E a n , we define their ℓ p sum as the product space E a × · · · × E a n equipped with the norm k ( x , . . . , x n ) k p = n X i =1 k x i k p ! p , where x i ∈ E a i for i = 1 , . . . , n . We use E a ⊕ p · · · ⊕ p E a n to describe this space.When p = ∞ , the norm is just k ( x , . . . , x n ) k ∞ = max ≤ i ≤ n k x i k . Although our notation for the norm in ℓ p spaces and in ℓ p sums is the same, the norm beingused should be clear from context.1.3. Previous Work and Our Results
We first study s -distance sets in R n equipped with the ℓ p norm. This space is denoted by ℓ np = ( R n , k·k p ). The two most famous questions pertaining to this problem are Kusner’s [Guy83]conjectures on equilateral sets. Conjecture 1 (Kusner) . e ( ℓ n ) = 2 n . Conjecture 2 (Kusner) . e ( ℓ np ) = n + 1 for < p < ∞ . For Conjecture 1, note that the set {± e i : i = 1 , . . . , n } , where e i is the i -th standard basis vector,is equilateral in ℓ n , so e ( ℓ n ) ≥ n . Currently, the best known upper bound is e ( ℓ n ) ≤ cn log n dueto Alon and Pudl´ak [AP03]. It is also known that Conjecture 1 holds for n = 3 (Bandelt, Chepoi,and Laurent [BCL98]) and n = 4 (Koolen, Laurent, and Schrijver [KLS00]).As for Conjecture 2, note that the set { e , . . . , e n , λ P ni =1 e i } is equilateral for a suitable choiceof λ , so e ( ℓ np ) ≥ n + 1. For 1 < p < n large enough, Swanepoel [Swa04b] actually showedthat e ( ℓ np ) > n + 1, disproving Conjecture 2 in this case. The first nontrivial upper bound of e ( ℓ np ) ≤ c p n ( p +1) / ( p − was found by Smyth [Smy02] using an approximation argument. This result EW DISTANCE SETS IN ℓ p SPACES AND ℓ p PRODUCT SPACES 3 was later improved by Alon and Pudl´ak [AP03] and is currently the best known upper bound on e ( ℓ np ) for arbitrary n and p . Theorem 1 (Alon and Pudl´ak) . For every p ≥ , we have e ( ℓ np ) ≤ c p n (2 p +2) / (2 p − , where one maytake c p = cp for an absolute c > . Our first result is an improvement of Theorem 1 when p is large in terms of n . One can checkthat c >
2, so Theorem 2 is indeed an improvement.
Theorem 2.
Let c > be the constant from Theorem 1. If n > and p ≥ c ( n log n ) , then e ( ℓ np ) ≤ p + 1) n . We should mention that when p satisfies other special properties, Theorem 1 can be strengthened.If p is an even integer, Swanepoel [Swa04b] used a linear independence argument to show that e ( ℓ np ) ≤ ( ( p − n + 1 if p ≡ p n + 1 if p ≡ p is an odd integer, Alon and Pudl´ak’s argument for e ( ℓ n ) extends to e ( ℓ np ), giving the bound e ( ℓ np ) < c p n log n in this case.Our next result is a generalization of Theorem 1 to s -distance sets. As far as we know, the onlyliterature on s -distance sets in ℓ np is by Smyth [Smy13]. Our theorem below is strictly stronger thanConjecture 5 in [Smy13]. Theorem 3. If s is a positive integer and p is a real number satisfying p > s , then e s ( ℓ np ) ≤ c p,s n (2 ps +2 s ) / (2 p − s ) for a constant c p,s depending on p and s . Our next three results are on equilateral sets in the ℓ p sum of Euclidean spaces. As far aswe know, this problem has not been well studied in this space. Swanepoel [Swa18] showed that e ( X ⊕ ∞ Y ) ≤ e ( X ) b f ( Y ) for normed spaces X and Y , where b f is the finite Borsuk number.However, explicit bounds are still unknown when X and Y are Euclidean spaces and when we takean ℓ p sum instead of an ℓ ∞ sum. Our first result in this area almost completely resolves the problemfor E a ⊕ ∞ E b . Note that we have the obvious lower bound e ( E a ⊕ ∞ E b ) ≥ e ( E a ) e ( E b ) = ( a + 1)( b + 1)by taking the Cartesian product of the two equilateral sets. This lower bound actually meets theupper bound when a = 2 , Theorem 4.
Let E a and E b be Euclidean spaces. Then, e ( E a ⊕ ∞ E b ) ≤ ( a + 1)( b + 1) + 1 . Our second result in this area provides an upper bound when p is even. Theorem 5.
Let E a and E b be Euclidean spaces, and let p be an even integer. Then, e ( E a ⊕ p E b ) ≤ (cid:0) a + p/ a (cid:1) + (cid:0) b + p/ b (cid:1) . Finally, we extend Alon and Pudl´ak’s [AP03] result on equilateral sets in ℓ np to ℓ p product spaces.We present an upper bound on the ℓ p sum of n Euclidean spaces for any 1 ≤ p < ∞ . Observe thatTheorem 1 is a special case of our theorem below when a = · · · = a n = 1. Theorem 6.
Let E a , . . . , E a n be Euclidean spaces and set a = max ≤ i ≤ n a i . If p > a , then e ( E a ⊕ p · · · ⊕ p E a n ) ≤ c p,a n p +2 a p − a for a constant c p,a depending on p and a . RICHARD CHEN, FENG GUI, JASON TANG, AND NATHAN XIONG
Our paper is structured as follows. In Section 2, we introduce two important tools used in ourproofs. Then, we prove Theorems 2, 3, 4, 5, and 6 in Sections 3, 4, 5, 6, and 7 respectively.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 | x | p by apolynomial with sufficiently small error.2.1. The Rank LemmaLemma 1 (Rank Lemma) . For a real symmetric n × n nonzero matrix A , rank A ≥ ( P ni =1 a ii ) P ni,j =1 a ij . We will frequently make use of the following corollary.
Corollary.
Let A be a real symmetric n × n matrix with a ii = 1 and | a ij | ≤ ε for all i = j . Then rank A ≥ n n − ε . Choosing ε = n − / gives rank A ≥ n/ . Jackson’s TheoremTheorem 7 (Jackson, [Jac30]) . For any f ∈ C k [ − , and positive integer d , there exists a poly-nomial P with degree at most d such that | f ( x ) − P ( x ) | ≤ c k ( n + 1) k ω ( f ( k ) , /d ) for all x ∈ [ − , , where c > is an absolute constant, ω ( f, δ ) denotes the modulus of continuity of f , and ( n + 1) k =( n + 1) n · · · ( n − k + 2) uses falling factorial notation. From Theorem 7, we can recover the following lemma. For its proof, see [Swa04a].
Lemma 2.
For any p ≥ and d ≥ ⌈ p ⌉ , there exists a polynomial P with degree at most d such that (1) | P ( x ) − | x | p | ≤ B ( p ) d p for all x ∈ [ − , where B ( p ) = ( ⌈ p ⌉ p (1 + π / ⌈ p ⌉ ( p ) ⌈ p ⌉− ) / ⌈ p ⌉ ! . We will always assume that the polynomial P in Lemma 2 is even and that P (0) = 0. If P is noteven, we can take the even part of P . If P (0) = 0, we can take the polynomial Q ( x ) = P ( x ) − P (0)as this only increases the error term by a factor of 2. EW DISTANCE SETS IN ℓ p SPACES AND ℓ p PRODUCT SPACES 5 Bound on equilateral sets in ℓ np for large p We start with an important lemma about the ℓ p norm. Lemma 3.
Suppose ≤ p ≤ q . Then for any x ∈ R n , k x k q ≤ k x k p ≤ n p − q k x k q . Proof.
The left hand inequality is a well-known property of the ℓ p norm. The right hand inequalityis the Power Mean Inequality, which can be proven using Jensen’s Inequality. (cid:3) Suppose our equilateral set is { p , . . . , p m } . Let k be the closest even integer to p , rounding up if p is odd. The main idea is that by Lemma 3, we can approximate k·k p by k·k k . If p is large enoughin terms of n , the error term is sufficiently small. From there, it suffices to bound the size of anapproximately-equilateral set in ℓ nk , 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 de-pending on whether k is greater than or less than p . Case 1.
We have ⌊ p ⌋ is odd, so p < k .From our bound on p and the fact that p ≥ m ≤ cpn (2 p +2) / (2 p − ≤ cpn ≤ p (log n ) ≤ (1 − n − /p ) − , where the first inequality holds by Theorem 1. Assuming without loss of generality that m >
1, wemay rearrange the result above into − p log n (cid:18) − √ m (cid:19) ≥ . Because k is the smallest even integer greater than p , we have(2) p < k ≤ p + 1 ≤ p − p log n (cid:18) − √ m (cid:19) . Now, embed the m points into ℓ nk . Since p < k , Lemma 3 implies(3) n k − p ≤ k p i − p j k k ≤ RICHARD CHEN, FENG GUI, JASON TANG, AND NATHAN XIONG for all i = j . Consider the m functions f i : R n → R given by f i ( x ) = 1 − k p i − x k kk , and let A bethe matrix with a ij = f i ( p j ). Clearly, a ii = 1, and from (2) and (3), | a ij | ≤ − n − kp ≤ − n − (1 − log n (1 − / √ m )) = 1 √ m , for all i = j . Since A is symmetric, Lemma 1 tells us rank A ≥ m/
2. For the upper bound on therank, note that every f i lies in the span of the set of polynomials S = { , x , . . . , x n , x , . . . , x n , . . . , x k − , . . . , x k − n , n X t =1 x kt } . The i -th row vector of A is ( f i ( p ) , . . . , f i ( p m )). Hence, every row vector belongs to the subspacespanned by { ( f ( p ) , . . . , f ( p m )) : f ∈ S} , which has dimension at most |S| = ( k − n + 2 ≤ kn . Itfollows that rank A ≤ kn ≤ ( p + 1) n . Combining the upper and lower bound, we have m ≤ p + 1) n, as desired. Case 2.
We have ⌊ p ⌋ is even, so p ≥ k .Since c, n ≥ p >
4. So, similar to in Case 1, we have m ≤ cpn (2 p +2) / (2 p − ≤ cpn − /p ≤ n − /p · p (log n ) ≤ n − /p (1 − n − /p ) − = ( n /p − − . Rearranging the result gives us p log n (cid:18) √ m (cid:19) ≥ . Because k is the largest even integer less than or equal to p , we have(4) p ≥ k ≥ p − ≥ p − p log n (cid:18) √ m (cid:19) . Embed the m points into ℓ nk . This time, Lemma 3 implies(5) 1 ≤ k p i − p j k k ≤ n k − p . EW DISTANCE SETS IN ℓ p SPACES AND ℓ p PRODUCT SPACES 7
We define f i and A like in Case 1. Again, a ii = 1, and from (4) and (5), | a ij | ≤ n − kp − ≤ n − (1 − log n (1+1 / √ m )) −
1= 1 √ m , for all i = j . Applying Lemma 1, rank A ≥ m/
2. Similar to in Case 1, we have the boundrank A ≤ kn ≤ pn . The final result is m ≤ pn < p + 1) n. Having considered all cases, the proof is complete. (cid:3) Bound on s -distance sets in ℓ np We would like to extend Alon and Pudl´ak’s [AP03] idea of combining Jackson’s Theorem withthe rank lemma to the s -distance case. However, as pointed out by Smyth [Smy13], if one of thedistances is arbitrarily small, we need arbitrarily high degrees of approximation. So, we want toimpose a lower bound on our distances.Consider a two-distance set with distances 1 and a , where a is very small. Intuitively, this setshould look like several “clusters” of points, such that the distance between each cluster is 1. So, ifwe look globally, this set looks like an equilateral set with distance 1, but if we look locally at eachcluster, we have an equilateral set with distance a . This means that we have essentially reducedthe problem from one about two-distance sets to one about equilateral sets. We carry this intuitionto s -distance sets and formalize this argument with an induction. Proof of Theorem 3.
We use strong induction on s . The base case is s = 1, which was proven byAlon and Pudl´ak [AP03]. For the inductive step, assume that the statement holds true for all1 ≤ s < k . We prove that it is true for s = k .Let S be a k -distance set in ℓ np . Let our k distances be 1 = a > a > · · · > a k . There are twocases to consider depending on whether the k distances are lower bounded or not. Case 1.
The smallest distance a k is less than 2 − k .There exists an index 1 ≤ i < k such that a i > a i +1 . Let S ′ ⊆ S be a maximal i -distance setusing the distances a , . . . , a i . From the maximality of S ′ , every point p ∈ S \ S ′ is within a i +1 of some point in S ′ . Draw a closed ball of radius a i +1 around every point in S ′ . These balls aredisjoint from our condition. The condition also implies that within each ball, the distance betweenany two points is at most a i +1 . Thus, within every ball, we have a ( k − i )-distance set using thedistances a i +1 , . . . , a k . This implies the bound e k ( ℓ np ) ≤ e i ( ℓ np ) · e k − i ( ℓ np ) . Applying the inductive hypothesis gives us e k ( ℓ np ) ≤ c p,i n pi +2 i p − i · c p,k − i n p ( k − i )+2( k − i )2 p − ( k − i ) ≤ c p,k n pk +2 k p − k , RICHARD CHEN, FENG GUI, JASON TANG, AND NATHAN XIONG as desired. The second inequality follows by analyzing the function f ( x ) = 2 px + 2 x p − x + 2 p ( k − x ) + 2( k − x )2 p − ( k − x )on the interval [0 , k ]. It is symmetric about x = k , and its first derivative is f ′ ( x ) = 4 p ( p + 1)(2 p − x ) − p ( p + 1)(2 p − ( k − x )) . Since 2 p > k , f is decreasing on [0 , k ] and increasing on [ k , k ]. Thus, it is maximised at x = 0 and x = k . Case 2.
The smallest distance a k is at least 2 − k .Suppose S consists of the points { p , . . . , p m } . For convenience, let π = a p a p · · · a pk . Fix B ( p ) asthe constant from Lemma 2. Define c as c = max( B ( p ) k · pk − pk +2 k , (2 /p − − p ) . Then, let d be a positive integer satisfying cn √ m < d p < cn √ m, which is possible since c ≥ (2 /p − − p .Lemma 2 allows us to pick an even polynomial P with P (0) = 0 and degree at most d such that | P ( x ) − | x | p | ≤ B ( p ) d p for all x ∈ [ − , . Consider the m functions f i : R n → R given by f i ( x ) = 1 π k Y u =1 a pu − n X t =1 | x t − p it | p ! , and their polynomial approximations g i ( x ) = 1 π k Y u =1 a pu − n X t =1 P ( x t − p it ) ! . Let A be the m × m matrix given by a ij = g i ( p j ). First, since P (0) = 0, a ii = 1 for all i . We nowestimate a ij for i = j . Expand f i ( x ) = 1 π k X ℓ =0 ( − k + ℓ σ ( ℓ ) n X t =1 | x t − p it | p ! k − ℓ , and g i ( x ) = 1 π k X ℓ =0 ( − k + ℓ σ ( ℓ ) n X t =1 P ( x t − p it ) ! k − ℓ , EW DISTANCE SETS IN ℓ p SPACES AND ℓ p PRODUCT SPACES 9 where σ ( ℓ ) denotes the ℓ -th elementary symmetric polynomial in a p , . . . , a pk . For convenience, wedefine X ij = P nt =1 | p jt − p it | p and Y ij = P nt =1 P ( p jt − p it ). Applying the triangle inequality with(1), we have | X ij − Y ij | ≤ nB ( p ) d p . Since i = j , recall that X ij ≥ − k . Combining this with the fact that c > B ( p ) · k − implies that Y ij is positive. Now, we are ready to attack a ij . | a ij | = 1 π (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k X ℓ =0 ( − k + ℓ σ ( ℓ )( Y k − ℓij − X k − ℓij ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ π · nB ( p ) d p k − X ℓ =0 σ ( ℓ ) k − ℓ X r =1 | X k − ℓ − rij Y r − ij |≤ π · nB ( p ) d p k − X ℓ =0 σ ( ℓ ) k − ℓ X r =1 (cid:12)(cid:12)(cid:12)(cid:12) ( k − ℓ − r ) X ij + ( r − Y ij k − ℓ − (cid:12)(cid:12)(cid:12)(cid:12) k − ℓ − = 1 π · nB ( p ) d p k − X ℓ =0 σ ( ℓ ) k − ℓ X r =1 (cid:12)(cid:12)(cid:12)(cid:12) ( r − Y ij − X ij ) k − ℓ − X ij (cid:12)(cid:12)(cid:12)(cid:12) k − ℓ − ≤ π · nB ( p ) d p k − X ℓ =0 σ ( ℓ )( k − ℓ ) (cid:18) nB ( p ) d p + 1 (cid:19) k − ℓ − ≤ π · nB ( p ) kd p · k k − X ℓ =0 σ ( ℓ ) . Now, since a u ≥ − k , we can lower bound π with π = a p a p · · · a pk ≥ (2 (1 − k ) p ) k = 2 pk − pk . On the other hand, since a t ≤
1, we can upper bound σ ( ℓ ) with σ ( ℓ ) = X ≤ j 2. Using the upper bound on d p andrearranging the inequality, we obtain m ≤ c p,k n (2 pk +2 k ) / (2 p − k ) . This completes the induction. (cid:3) Bound on equilateral sets in ℓ ∞ product spaces Proof of Theorem 4. Write x = ( e x , e x ) for each point x ∈ E a ⊕ ∞ E b , where e x ∈ E a and e x ∈ E b .Let S be our equilateral set with cardinality m . Consider the m functions f u : R a + b → R definedby f u ( x ) = (cid:16) − k e x − e u k (cid:17) (cid:16) − k e x − e u k (cid:17) , for all u ∈ S . Note that f u ( v ) = δ uv , so the f u are linearly independent.We can expand f u as f u ( x ) = − a X t =1 ( e x t − e u t ) ! − b X t =1 ( e x t − e u t ) ! = − k e u k − k e x k + 2 a X t =1 e x t e u t ! − k e u k − k e x k + 2 b X t =1 e x t e u t ! So, the f u are all spanned by the following set of ( a + 2)( b + 2) polynomials { , e x i , e x j , e x i e x j , k e x k , k e x k , k e x k e x i , k e x k e x j , k e x k k e x k : 1 ≤ i ≤ a, ≤ j ≤ b } . This implies the bound m ≤ ( a + 2)( b + 2).We will now prove that the set of polynomials { f u , , x k , k e x k : u ∈ S, ≤ k ≤ a + b } is linearlyindependent. Assume for the sake of contradiction that we have a dependence(6) X u ∈ S α u f u + a + b X k =1 β k x k + γ k e x k + δ = 0 . EW DISTANCE SETS IN ℓ p SPACES AND ℓ p PRODUCT SPACES 11 The left hand side of (6) is a polynomial in the x k that is identically zero. Thus, extracting thecoefficient of k e x k k e x k , we have(7) X u ∈ S α u = 0 . Extracting the coefficient of k e x r k x k , for the appropriate r ∈ { , } , we have(8) X u ∈ S α u u k = 0 . Extracting the coefficient of k e x k and applying (7), we have(9) X u ∈ S α u k e u k = 0 . Now, plug u into (6), multiply both sides by α u , and sum over all u ∈ S . X u ∈ S α u + a + b X k =1 β k X u ∈ S α u u k + γ X u ∈ S α u k e u k + δ X u ∈ S α u = 0 . Applying (7), (8), and (9) implies α u = 0 for all u ∈ S . It easily follows that all other coefficientsare zero, as desired.Now, we know that m + a + b + 2 ≤ ( a + 2)( b + 2). This rearranges into m ≤ ( a + 1)( b + 1) + 1. (cid:3) Bound on equilateral sets in ℓ p product spaces for even p Proof of Theorem 5. Let S be an equilateral set in E a ⊕ p E b with m points. For every u ∈ S , definethe function f u : R a + b → R by f u ( x ) = 1 − k e x − e u k p − k e x − e u k p for all x = ( f x , f x ) ∈ E a ⊕ p E b so that f u ( v ) = δ uv . It follows that the m polynomials are linearlyindependent. Now, we can expand f u as f u ( x ) = 1 − a X t =1 ( e x t − e u t ) ! p/ − b X t =1 ( e x t − e u t ) ! p/ = 1 − k e x k + k e u k − a X t =1 e x t e u t ! p/ − k e x k + k e u k − b X t =1 e x t e u t ! p/ = 1 − e f u ( x ) − e f u ( x )Utilizing multi-index notation, we can expand e f u ( x ) = X ε,gε + γ ≤ p/ ( − γ (cid:18) γg (cid:19)(cid:18) p/ ε, γ, p/ − ε − γ (cid:19)e u g k e u k p − ε − γ e x g k e x k ε . The sum is taken over all non-negative integers ε and non-negative integer vectors g with a entriessuch that ε + γ ≤ p/ 2, where γ is the sum of the components of g . Similarly, we can expand e f u ( x ) = X ε,gε + γ ≤ p/ ( − γ (cid:18) γg (cid:19)(cid:18) p/ ε, γ, p/ − ε − γ (cid:19)e u g k e u k p − ε − γ e x g k e x k ε . We want to count the number of monomials in each polynomial. Since e f u is a polynomial in e x , . . . , e x a and e f u is a polynomial in e x , . . . , e x b , the two sets of monomials are disjoint, exceptfor the constant monomial. Let us consider e f u first. By choosing ε = 0, we must count all themonomials with degree at most p/ 2. There are (cid:0) a + p/ a (cid:1) of these. If the degree is greater than p/ p/ c , we only need to count the monomials formed when ε = c and γ = p/ − c . Hence, thetotal number of monomials in e f u is (cid:18) a + p/ a (cid:19) + (cid:18) a + p/ − a − (cid:19) + (cid:18) a + p/ − a − (cid:19) + · · · + (cid:18) a − a − (cid:19) = (cid:18) a + p/ a (cid:19) + (cid:18) a + p/ − a (cid:19) . Similarly, there are (cid:0) b + p/ b (cid:1) + (cid:0) b + p/ − b (cid:1) monomials in e f u . In total, f u has (cid:0) a + p/ a (cid:1) + (cid:0) a + p/ − a (cid:1) + (cid:0) b + p/ b (cid:1) + (cid:0) b + p/ − b (cid:1) − m .By finding a larger linearly independent set of polynomials, a trick first used by Blokhuis [Blo84b],we can lower this bound to (cid:0) a + p/ a (cid:1) + (cid:0) b + p/ b (cid:1) . We prove that the set of polynomials { f u , e x m , e x n , u ∈ S, < µ < p/ , < ν < p/ } is linearly independent. The details are very similar to those in Blokhuis’s [Blo84a] bound for the s -distance set in R n .Suppose we have a dependence(10) X u ∈ S a u f u ( x ) + X <µ
Lemma 4. For all m with µ < p/ , X u ∈ S a u e u m = 0 . Similarly, for all n with ν < p/ , X u ∈ S a u e u n = 0 . Proof. We only prove the first statement. The argument for the second statement is identical.Suppose µ = t < p/ 2. Consider the part of the left hand side of (10) that is homogeneous in e x , . . . , e x a with degree p − t . Note that the monomials e x m do not contribute to this, so we onlyhave to look at the part from the f u . Using our expansion of f u above, the part of f u homogeneous EW DISTANCE SETS IN ℓ p SPACES AND ℓ p PRODUCT SPACES 13 in the a variables with degree p − t is − X ε,g ε + γ = p − tε + γ ≤ p/ ( − γ (cid:18) γg (cid:19)(cid:18) p/ ε, γ, p/ − ε − γ (cid:19)e u g k e u k p − ε − γ e x g k e x k ε . The left hand side of (10) is a polynomial in the x i which is identically zero. Thus, we have X ε,g ε + γ = p − tε + γ ≤ p/ ( − γ (cid:18) γg (cid:19)(cid:18) p/ ε, γ, p/ − ε − γ (cid:19) X u ∈ S a u e u g k e u k p − ε − γ e x g k e x k ε = 0 . Now, substitute x = v , multiply by a v k e v k t − p , and sum over all v ∈ S . X ε,g ε + γ = p − tε + γ ≤ p/ ( − γ (cid:18) γg (cid:19)(cid:18) p/ ε, γ, p/ − ε − γ (cid:19) X u ∈ S a u k e u k p − ε − γ e u g ! = 0 . This is a sum of squares where all coefficients have the same sign. So, X u ∈ S a u k e u k p − ε − γ e u g = 0 . Plugging in γ = t proves the lemma. (cid:3) Now, plug in u into (10), multiply by a u and sum over all u ∈ S , X u ∈ S a u + X <µ
This proof can be easily extended to the ℓ p sum of n Euclidean spaces. If we consider theproduct space E a ⊕ p · · · ⊕ p E a n , our bound is just (cid:0) a + p/ a (cid:1) + · · · + (cid:0) a n + p/ a n (cid:1) .7. Bound on equilateral sets in ℓ p product spaces for ≤ p < ∞ Proof of Theorem 6. Let { p , . . . , p m } be an equilateral set in E a ⊕ p · · · ⊕ p E a n , and write p i =( e p i , . . . , e p in ) where e p ik ∈ E a k for k = 1 , . . . , n .Let B ( p ) be the constant from Lemma 2. Take c = max( B ( p ) , (2 /p − − p ) and set d to be apositive integer satisfying cn √ m < d p < cn √ m. By Lemma 2, there exists an even polynomial P with P (0) = 0 and degree at most d that approx-imates | x | p . Thus, for i = j and k = 1 , . . . , n , (cid:12)(cid:12) P ( (cid:13)(cid:13) e p ik − e p jk (cid:13)(cid:13) ) − (cid:13)(cid:13) e p ik − e p jk (cid:13)(cid:13) p (cid:12)(cid:12) ≤ B ( p ) d p . Next, for i = 1 , . . . , m define the functions f i : R a + ··· + a n → R by f i ( x ) = 1 − n X k =1 P ( k e p ik − e x k k ) . Let M be the m × m matrix given by m ij = f i ( p j ). Since P (0) = 0, we have m ii = 1. For i = j , | m ij | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 (cid:13)(cid:13) e p ik − e p j k (cid:13)(cid:13) p − n X k =1 P ( (cid:13)(cid:13) e p ik − e p jk (cid:13)(cid:13) ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ n X k =1 (cid:12)(cid:12)(cid:13)(cid:13) e p ik − e p j k (cid:13)(cid:13) p − P ( (cid:13)(cid:13) e p ik − e p jk (cid:13)(cid:13) ) (cid:12)(cid:12) ≤ nB ( p ) d p < √ m , with the last inequality following from our conditions on c and d . Because M is symmetric, we canapply Lemma 1 to get rank M ≥ m/ x ∈ E a ⊕ p · · · ⊕ p E a n as x = ( e x (1)1 , e x (2)1 , . . . , e x ( a )1 , e x (1)2 , e x (2)2 , . . . , e x ( a )2 , . . . , e x (1) n , e x (2) n , . . . , e x ( a n ) n ) . Because P is even, all terms in the expansion of P ( k e p ik − e x k k ) have integer exponent, i.e., f i isactually a polynomial. Additionally, since P has degree at most d , each f i is in the span of the set S = { , n X k =1 k e x k k d } ∪ S ∪ S ∪ · · · ∪ S n , where each set S i consists of all the monomials with degree less than d formed by e x (1) i , . . . , e x ( a i ) i .Thus, the f i are spanned by at most 2 + P nk =1 (cid:0) a k + d − a k (cid:1) − n polynomials. Because every rowvector of M , each of the form ( f i ( p ) , . . . , f i ( p m )), belongs to the subspace spanned by the set { ( f ( p ) , . . . , f ( p m )) : f ∈ S} , we haverank M ≤ n X k =1 (cid:18) a k + d − a k (cid:19) − n ≤ n X k =1 (cid:18) a k + d − a k (cid:19) ≤ n (cid:18) a + d − a (cid:19) < ne a (cid:18) da (cid:19) a , EW DISTANCE SETS IN ℓ p SPACES AND ℓ p PRODUCT SPACES 15 where we let a = max ≤ i ≤ n a i . Now, recall that d ≥ B ( p ) /p ≥ p and 2 p > a , so d > a/ 2. 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In New Trends in Intuitive Geometry ,pages 407–458. Springer-Verlag Berlin Heidelberg, 2018. Lexington High School, Lexington, MA 02421, USA Email address : Department of Mathematics, Massachusetts Institute of Technology, Cambridge, MA 02139, USA Email address : [email protected] Belmont High School, Belmont, MA 02478, USA Email address : [email protected] Phillips Academy Andover, Andover, MA 01810, USA Email address ::