aa r X i v : . [ m a t h . N T ] J un Contemporary Mathematics
On spectral sets of integers
Dorin Ervin Dutkay and Isabelle Kraus
Abstract.
Based on tiles and on the Coven-Meyerowitz property, we present some exam-ples and some general constructions of spectral subsets of integers.
Contents
1. Introduction 12. One prime power 53. Szab´o’s examples 114. Some general constructions 175. Appendix 22References 24
1. Introduction
Definition . Let G be a locally compact abelian group and let b G be its Pontryagindual group. A finite subset A of G is called spectral (in G ) if there exists a subset Λ of b G with A such that(1.1) 1 A X a ∈ A ϕ ( a ) ϕ ′ ( a ) = δ ϕϕ ′ , ( ϕ, ϕ ′ ∈ Λ)In this case, Λ is called a spectrum for A (in the group b G ).It is easy to check that the spectral property can be rephrased in the following ways. Mathematics Subject Classification.
Key words and phrases. spectral set, tiling, cyclotomic polynomial, Coven-Meyerowitz property. c (cid:13) Proposition . Let A be a finite subset of a locally compact abelian group and Λ afinite subset of b G with A . The following statements are equivalent: (i) Λ is a spectrum for A . (ii) The matrix (1.2) 1 √ A ( ϕ ( a )) a ∈ A,ϕ ∈ Λ is unitary. (iii) For every a, a ′ ∈ A , (1.3) 1 X ϕ ∈ Λ ϕ ( a − a ′ ) = δ aa ′ . We will be interested mainly in the spectral subsets of Z and the spectral subsets of Z N .Since the dual group of Z is the group T = { z ∈ C : | z | = 1 } which can be identified with[0 , A of Z is spectral if and only if there exists a finite subset Λ in [0 , R ) such that A and the matrix1 √ A (cid:0) e πiaλ (cid:1) a ∈ A,λ ∈ Λ is unitary.Since the dual group of Z N is Z N , a subset A of Z N is spectral if and only if there existsa subset Λ of Z N such that A = √ A (cid:0) e πiaλ/N (cid:1) a ∈ A,λ ∈ Λ is unitary. Definition . For two subsets A and B of Z , we write A ⊕ B to indicate that for each c ∈ A + B there are unique numbers a ∈ A and b ∈ B such that a + b = c .For a set A of non-negative integers we denote by A ( x ) the associated polynomial A ( x ) = X a ∈ A x a . A subset A of Z is called a tile if there exists a subset C of Z such that A ⊕ C = Z .In 1974 [ Fug74 ], Fuglede proposed a conjecture that states that Lebesgue measurablespectral sets in R n coincide with sets that tile R n . The conjecture was disproved by Tao[ Tao04 ] in dimensions five and higher and later in dimensions three and higher [
FMM06,
N SPECTRAL SETS OF INTEGERS 3
KM06b, KM06a, Mat05 ]. All these counterexamples were based on some constructionsin finite groups, so the Fuglede conjecture fails for groups of the form Z N × Z N × Z N .However, the conjecture is still open at this moment in dimensions one and two. It is knownthat the Fuglede conjecture in R , under some additional hypotheses, can be reduced to theFuglede conjecture for Z , see [ DL14 ]. Conjecture . [Fuglede’s conjecture for Z ] A finite subset of Z is spectral if andonly if it is a tile. A basic result (see [
New77, CM99 ]) shows that every tiling set C is periodic , i.e., thereexists N ∈ N such that C + N = C . If B is any set consisting of one representative from C for each class modulo N , then C = B ⊕ N Z and so A ⊕ ( B ⊕ N Z ) = Z , and therefore A ⊕ B is a complete set of representatives modulo N . Proposition . [ CM99 ] Let N be a positive integer and A , B sets of non-negativeintegers. The following statements are equivalent: (i) A ⊕ ( B ⊕ N Z ) = Z . (ii) A ⊕ B is a complete set of representatives of Z N . In other words A ⊕ B = Z N ,where addition is understood modulo N . (iii) A ( x ) B ( x ) ≡ x + · · · + x N − mod( x N − . (iv) A (1) B (1) = N and for every factor t > of N , the cyclotomic polynomial Φ t ( x ) divides A ( x ) or B ( x ) . Thus, tiles for Z coincide with tiles for the groups Z N .In [ CM99 ], Coven and Meyerowitz found a sufficient condition for a subset of Z to be atile, formulated in terms of cyclotomic polynomials. Theorem . [ CM99 ] Let A be a finite set of non-negative integers with correspondingpolynomial A ( x ) = P a ∈ A x a . Let S A be the set of prime powers s such that the cyclotomicpolynomial Φ s ( x ) divides A ( x ) . Consider the following conditions on A ( x ) . (T1) A (1) = Q s ∈ S A Φ s (1) . (T2) If s , . . . , s m ∈ S A are powers of distinct primes, then the cyclotomic polynomial Φ s ...s m ( x ) divides A ( x ) .If A ( x ) satisfies (T1) and (T2), then A tiles the integers with period N := lcm( S A ) . Thetiling set B can be obtained as follows: define B ( x ) = Q Φ s ( x t ( s ) ) , where the product is takenover all prime power factors s of N which are not in S A , and t ( s ) is the largest factor of DORIN ERVIN DUTKAY AND ISABELLE KRAUS N relatively prime to s . Then B ( x ) is the polynomial associated to a set of non-negativeintegers B . Definition . A finite set A of non-negative integers is said to satisfy the Coven-Meyerowitz property (or the CM-property), if it satisfies conditions (T1) and (T2) in Theorem1.6. We call the tiling set B in Theorem 1.6, the Coven-Meyerowitz (CM) tiling set associatedto A , and we denote it by B = CM( A ).The converse of the Coven-Meyerowitz theorem also seems to be true, but at the moment,it is just a conjecture. Coven and Meyerowitz showed that tiles satisfy the (T1) property. Theorem . [ CM99 ] Let A be a finite set of non-negative integers with correspondingpolynomial A ( x ) = P a ∈ A x a and let S A be the set of prime powers s such that the cyclotomicpolynomial Φ s ( x ) divides A ( x ) . If A tiles the integers, then (T1) holds. Also, they proved that tiles with a cardinality that has only one or two prime factorssatisfy the CM-property.
Theorem . [ CM99 ] Let A be a finite set of non-negative integers with correspondingpolynomial A ( x ) = P a ∈ A x a such that A has at most two prime factors and A tiles Z .Then A satisfies (T2), and therefore it has the CM-property. Remark . Note that the CM-tiling set B or C S = B ⊕ lcm( S A ) Z does not dependon A , it depends only on S = S A . Also, the proof shows that if A satisfies (T1) and (T2),then it has a universal tiling of period lcm( S A ), which is a tiling set for all the sets A ′ with S A ′ = S A .Note also that B, C S ⊂ p Z for every prime factor p ∈ S A since p is a factor of N andevery divisor Φ s ( x t ( s ) ) of B ( x ) is a polynomial in x p . This is because either s = p α +1 with α ≥
1, and then we use Proposition 5.4(iii), or s is a power of a prime different than p andthen t ( s ) is a multiple of p , being the largest factor of lcm( S A ) relatively prime to s .Later, Laba proved that sets with the CM-property are spectral. Theorem . [ Lab02 ] Let A be set of non-negative integers that satisfies the CM-property. Then A is a spectral set. A spectrum for A can be obtained as follows: considerthe set Λ A of all numbers of the form X s ∈ S A k s s , N SPECTRAL SETS OF INTEGERS 5 where k s ∈ { , , . . . , p − } if s = p α ∈ S A , with p prime. Definition . With the notations as in Theorems 1.6 and 1.11, for N := lcm( S A ),we denote by L A := Λ A and we call L A the Laba spectrum of A .Combining the results of Coven-Meyerowitz and Laba, Dutkay and Haussermann showedthat if a set has the CM-property, then the tiling sets and the spectra are in a nice comple-mentary relation. Theorem . [ DH15 ] Let A be a finite set of non-negative integers with the CM-property. Let N = lcm( S A ) , and let B = CM( A ) be its CM-tiling set. Then B has theCM-property, and if L A , L B are the corresponding Laba spectra, then ( N · L A ) ⊕ ( N · L B ) = Z N . Many examples of tiles are found in the literature. Many fewer spectral sets are known.In this paper, we gather some of the examples of tiling sets in the literature and show thatthey have the CM-property and explicitly describe the tiling sets and the spectra. In Section2, we describe the tiles with cardinality of a prime power. In Section 3, we describe Szab´o’sexamples and show that they have the CM-property and describe the tiling sets and spectra.In Section 4, we present some general constructions of spectral sets, tiling sets, and sets withthe CM-property.
2. One prime power
Theorem . Let A be a set of non-negative integers with cardinality p n , where p isprime and n ∈ N . Then the set A tiles the integers if and only if there exist integers ≤ α < · · · < α n and for each ≤ k ≤ n , and each ≤ i , . . . , i k − ≤ p − there exists acomplete set of representatives modulo p , { a i ,...,i k − ,i k : 0 ≤ i k ≤ p − } , a i ,...,i k − , = 0 , suchthat the set A is congruent modulo p α n to the set (2.1) A ′ = ( n X k =1 p α k − a i ,...,i k : 0 ≤ i , . . . , i n ≤ n − ) . In this case S A = { p α , . . . , p α n } , the CM-tiling set is (2.2) B = CM( A ) = X j =0 ,...,αn − j = α − ,...,α n − b j p j : 0 ≤ b j ≤ p − , ≤ j ≤ α n − , j = α − , . . . , α n − . DORIN ERVIN DUTKAY AND ISABELLE KRAUS
The Laba spectra of A and B are, respectively (2.3) L A = ( n X i =1 k i p α i : 0 ≤ k i ≤ p − , ≤ i ≤ n ) (2.4) L B = X j =1 ,...,αn j = α ,...,α n k j p j : 0 ≤ k j ≤ p − , ≤ j ≤ α n , j = α , . . . , α n Remark . Let us explain a bit more the structure of the set A ′ . Think of the base p decomposition of a number. For the set A ′ , we only use the digits corresponding to positions α − , α − , . . . , α n −
1. The rest of the digits are 0. In position α − p , { a i : 0 ≤ i ≤ p − } with a = 0. Once the first digit a i ischosen for the digit in position α −
1, we use another complete set of representatives modulo p , { a i ,i : 0 ≤ i ≤ p − } , with a i , = 0. Note that, this complete set of representatives isallowed to be different for different choices of i .For 1 ≤ k ≤ n , once the digits a i , a i ,i , . . . , a i ,...,i k − have been chosen for positions α − , α − , . . . , α k − − α k − { a i ,...,i k : 0 ≤ i k ≤ p − } , with a i ,...,i k − , = 0.We will need some results from [ CM99 ]. Definition . Let S be a set of powers of at most two primes. Define T S to bethe collection of all subsets A of { , , . . . , lcm( S ) − } which tile the integers and satisfymin( A ) = 0 and S A = S . Note that T ∅ = { } because lcm( ∅ ) = 1. Lemma . Let S be a set of powers of at most two primes. A finite set A ′ with min( A ′ ) = 0 and S A ′ = S tiles the integers if and only if A ′ is congruent modulo lcm( S ) toa member of T S . Proof.
Let A be an element of T S and A ′ ≡ A (mod lcm( S )). Let N = lcm( S ). Since A ′ ∈ T S , there exists a set B such that A ′ ⊕ B = Z N . Then, since A ≡ A ′ (mod N ), it followsthat A ⊕ B = Z N .Conversely, if A ′ tiles the integers and S A = S , then by Lemma 2.5, A ′ has at mosttwo prime factors, so it has the CM-property, by Theorem 1.9. Therefore, it has a tilingset of period lcm( S ), B ⊕ lcm( S ) Z , by Remark 1.10. Let A be the set obtained from A ′ byreducing modulo lcm( S ). Then min( A ) = 0 and A ⊂ { , , . . . , lcm( S ) − } . Also, A has the N SPECTRAL SETS OF INTEGERS 7 same tiling set B ⊕ lcm( S ) Z . Then, by Lemma 5.3, S A ′ = S A as the complement of S B inthe set of all prime power factors of lcm( S ). (cid:3) Lemma . Let A be a finite set of non-negative integers which is a tile. Then A hasat most two prime factors if and only if S A consists of powers of at most two primes. Proof.
Since A is a tile, it satisfies the (T1) property, by Theorem 1.8. So, usingProposition 5.4(iv), A = Y s ∈ S A Φ s (1) = Y p α ∈ S A p. Thus A has at most two prime factors if and only if S A consists of powers of at most twoprimes. (cid:3) Lemma . [ CM99 ] Suppose S contains powers of only one prime p . Let S = { p α : p α +1 ∈ S, α ≥ } . (i) If p S then T S = { pA : A ∈ T S } . (ii) If p ∈ S then T S = (cid:8) ∪ p − i =0 ( { a i } ⊕ pA i ) : A i ∈ T S , a = 0 , { a , a , . . . , a p − } a completeset of representatives modulo p and every { a i } ⊕ pA i ⊂ { , , . . . , lcm( S ) − } (cid:9) . Proposition . [ CM99 ] Let p be a prime number. Then (i) The only member of T ∅ is { } . (ii) For α ≥ , the only member of T { p α +1 } is p α { , , . . . , p − } . Theorem . Let p be a prime number. Let S = { p α , p α , . . . , p α n } with ≤ α <α < · · · < α n } . The following statements are equivalent: (i) A ∈ T S . (ii) For ≤ k ≤ n − , there exist numbers a i ,...,i k i , . . . , i k = 0 , . . . , p − with thefollowing properties (a) The set { a i : 0 ≤ i ≤ p − } is a complete set of representatives modulo p , a = 0 , (b) For each ≤ k ≤ n − , and each i , . . . , i k − in { , . . . , p − } , the set { a i ,...,i k − ,i k : 0 ≤ i k ≤ p − } is a complete set of representatives modulo p , a i ,...,i k − , = 0 , DORIN ERVIN DUTKAY AND ISABELLE KRAUS (c)
For each ≤ k ≤ n − a i ,...,i k + p α k +1 − α k a i ,...,i k +1 + · · · + p α n − − α k a i ,...,i n − ≤ p α n − α k − . (d)(2.6) A = (cid:8) p α − a i + p α − a i ,i + · · · + p α n − − a i ,...,i n − + p α n − j : 0 ≤ i , . . . , i n , j ≤ p − (cid:9) . Proof.
We prove the equivalence of (i) and (ii) by induction on n . For n = 1, the resultfollows from Proposition 2.7. Assume now the statements are equivalent for n and take S = { p α , . . . , p α n +1 } . Using Lemma 2.6(i), we have that A ∈ T S if and only if A = p α − A with A ∈ T S where S = { p, p α − α +1 , . . . , p α n +1 − α +1 } . Using Lemma 2.6(ii), we have that A ∈ T S if and only if there exists a complete set of representatives modulo p , { a i : 0 ≤ i ≤ p − } , a = 0 and, for each 0 ≤ i ≤ p −
1, a set A i ∈ T S ′ , where S ′ = { p α − α , . . . , p α n +1 − α } suchthat a i + pA i ⊂ { , . . . , lcm( S ) − } = { , . . . , p α n +1 − α +1 − } , and A = ∪ p − i =0 ( { a i } + pA i ) . Using the induction hypothesis for the set S ′ , we get that, for each 0 ≤ i ≤ p − A i must be of the form A i = (cid:8) p α − α − a i ,i + p α − α − a i ,i ,i + · · · + p α n − α − a i ,...,i n + p α n +1 − α − j : 0 ≤ i , . . . , i n , j ≤ p − (cid:9) , where for each 0 ≤ i , . . . , i k − ≤ p −
1, the set { a i ,...,i k : 0 ≤ i k ≤ p − } is a complete setof representatives modulo p and a i ,...,i k − , = 0. Also a i ,i ,...,i k + p ( α k +1 − α ) − ( α k − α ) a i ,...,i k +1 + · · · + p ( α n − α ) − ( α k − α ) ≤ p ( α n +1 − α ) − ( α k − α ) − k ≥
2. We must also have a i + p α − α a i ,i + . . . p α n − α a i ,...,i n + p α n +1 − α ( p − ≤ p α n +1 − α +1 , and this implies (c) for k = 1.Then A = p α − ∪ p − i =0 ( { a i } + pA i ) , and (d) follows. (cid:3) N SPECTRAL SETS OF INTEGERS 9
Proof of Theorem 2.1.
Assume that A and A ′ have the given form. We show that A ′ ⊕ B = Z p αn . Note that A ′ = p n and B = p α n − n so A ′ · B = p α n . By Lemma 5.2,it is enough to show that ( A ′ − A ′ ) ∩ ( B − B ) = { } in Z p αn . If we pick an element in theintersection, it can be written in both ways as n X k =1 p α k − ( a i ,...,i k − a i ′ ,...,i ′ k ) = X j =0 ,...,αn − j = α − ,...,α n − p j ( b j − b ′ j ) . Take the first index l such that i l = i ′ l . Then the left-hand side is divisible by p α l − butnot by p α l . Since p α l − does not appear on the right hand side, it follows, by contradiction,that both sides are equal to 0.For the converse, if A is a tile then the result follows from Theorem 2.8.It remains to check that the CM-tiling set and the Laba spectra are those given in(2.2),(2.3) and (2.4). By Lemma 5.3, we have that S A and S B are complementary, so S B = (cid:8) p j : j ∈ { , . . . , α n } , j = α , . . . , α n (cid:9) . Then the CM-tiling set is defined by the polynomial Y j =1 ...αn j = α ,...,α n Φ j ( x ) = Y j =1 ...αn j = α ,...,α n (1 + x p j − + x p j − + · · · + x ( p − p j − )= X ≤ bj ≤ p − ≤ j ≤ α n ,j = α ,...,α n x P ≤ j ≤ αn,j = α ,...,αn b j x pj − . This implies (2.2). Since we have the form of S B , (2.3) and (2.4) follow immediately. (cid:3) Remark . In [
New77 ], Newman classifies the finite sets of integers which tile Z whenthe number of elements in the set is a prime power. The tiling condition is stated in Theorem2.10. In Proposition 2.11, we determine the relation between the numbers e ij in Newman’spaper and the numbers α i and the set S A in Theorem 2.1. Theorem . [ New77 ] Let a , a , . . . , a k be distinct integers with k = p α , p a prime, α a positive integer. For each pair a i , a j , i = j , we denote by p e ij the highest power of p whichdivides a i − a j . The set { a , a , . . . , a k } is a tile if and only if there are at most α distinct e ij . Proposition . Let A = { a , . . . , a k } be a set of non-negative integers which tile Z ,with A = p n , p a prime, n a positive integer. Then S A = { p e ij +1 : 1 ≤ i, j ≤ p n } , where e ij denotes the highest power of p which divides a i − a j , as in Theorem 2.10. Proof.
Let S A = { p α , . . . , p α n } . Since A is a tile, we have by Theorem 2.1 that A iscongruent modulo p α n to the set A ′ = ( n X k =1 p α k − a i ,...,i k : 0 ≤ i , . . . , i n ≤ n − ) . Take a, a ′ ∈ A . Then a = n X k =1 p α k − a i ,...,i k + p α n m,a ′ = n X k =1 p α k − a i ′ ,...,i k ′ + p α n m ′ . So a − a ′ = n X k =1 ( a i ,...,i k − a i ′ ,...,i k ′ ) p α k − + p α n ( m − m ′ ) . If i = i ′ , . . . , i n = i n ′ , then a = a ′ . So then m = m ′ . If there exists a k such that i k = i k ′ ,then take the smallest such k . Let e be the largest power of p such that p e divides a − a ′ .Then a − a ′ = p α k − ( a i ,...,i k − ,i k − a i ′ ,...,i k − ′ ,i k ′ ) + n X j = k +1 ( a i ,...,i j − a i ′ ,...,i j ′ ) p α j − + p α n ( m − m ′ ) . Since { a i ,...,i k − ,l : 0 ≤ l ≤ p − } is a complete set of representatives modulo p , this meansthat a − a ′ is divisible by p α k − , but not p α k . Therefore, e = α k −
1. Relabeling a = a i and a ′ = a j , we get that e ij = α k −
1. Doing this for all a i , a j ∈ A , we get that the set of all e ij is { α − , . . . , α n − } and the result follows. (cid:3) For the case when the cardinality of a tile has two prime factors, to describe the structureof the such tiles, one can use the following lemma from [
CM99 ]. Lemma . Suppose S contains powers of both the primes p and q . Let S = { p α : p α +1 ∈ S } ∪ { q β : q β ∈ S } , S ′ = { p α : p α ∈ S } ∪ { q β : q β +1 ∈ S } . (i) If p ∈ S , then T S = (cid:8) ∪ p − i =0 ( { a i } ⊕ pA i ) : A i ∈ T S , a = 0 , { a , a , . . . , a p − } a complete N SPECTRAL SETS OF INTEGERS 11 set of representatives modulo p and every { a i } ⊕ pA i ⊂ { , , . . . , lcm( S ) − } (cid:9) . (ii) If q ∈ S , then T S = (cid:8) ∪ q − i =0 ( { a i } ⊕ qA i ) : A i ∈ T S ′ , a = 0 , { a , a , . . . , a q − } a completeset of representatives modulo q and every { a i } ⊕ qA i ⊂ { , , . . . , lcm( S ) − } (cid:9) . (iii) If p, q S , then A ⊂ p Z or A ⊂ q Z and { A ∈ T S : A ⊂ p Z } = { pA : A ∈ T S } , { A ∈ T S : A ⊂ q Z } = { qA : A ∈ T S ′ } .
3. Szab´o’s examples
In [
Sza85 ], Szab´o constructed a class of examples to give a negative answer to twoquestions due to K. Corr´adi and A.D. Sands respectively:If G is a finite abelian group and G = A ⊕ A is one of its normed factorizations, inthe sense that both factors contain the zero, must one of the factors contain some propersubgroup of G ?If G is a finite abelian group and G = A ⊕ A is a factorization, must one of the factorsbe contained in some proper subgroup of G ?We will present here Szab´o’s examples and prove that they have the CM-property andfind their tiling sets and spectra. We begin with a general proposition which encapsulatesthe core ideas in Szab´o’s examples. Proposition . Let G be a finite abelian group. Suppose we have a factorization A ⊕ B ′ = G . Let B , . . . , B r be disjoint subsets of B ′ and g , . . . , g r in G with the propertythat (3.1) A + B i = A + B i + g i for all i. Define (3.2) B = B ′ ∪ r [ i =1 ( B i + g i ) ! \ r [ i =1 B i ! . Then A ⊕ B = G . Proof.
The sets A + B i are disjoint. Indeed, if not, then there are a, a ′ ∈ A , b i ∈ B i , b j ∈ B j , i = j such that a + b i = a ′ + b j . Because of the unique factorization, a = a ′ and b i = b j , which contradicts the fact that B i and B j are disjoint. We have the disjoint union G = ( A + ( B ′ \ ∪ i B i )) ∪ [ i ( A + B i ) = ( A + ( B ′ \ ∪ i B i )) ∪ [ i ( A + B i + g i ) . Take b = b ′ in B and we want to prove that A + b is disjoint from A + b ′ . If b or b ′ is in B ′ \ ∪ i B i , this is clear from the hypothesis. If b and b ′ lie in different sets B i + g i this isagain clear since the union above is disjoint. If b, b ′ ∈ B i + g i , then b = c + g i , b ′ = c ′ + g i with c, c ′ ∈ B i ⊂ B ′ , and since b = b ′ , we have c = c ′ . Therefore A + c is disjoint from A + c ′ and the same is true for A + b = A + c + g i and A + b ′ = A + c ′ + g i . Thus A ⊕ B = G . (cid:3) Szab´o constructed his examples in Z m × · · · × Z m r . We note here that working in thegroup Z m × . . . Z m r with m , . . . , m r relatively prime is equivalent to working in Z m ...m r because of the following isomorphism. Proposition . Let m , . . . , m r be relatively prime non-negative integers. Let m = m . . . m r . The map Ψ : Z m × · · · × Z m r → Z m , (3.3) Ψ( k , . . . , k r ) = r X i =1 k i mm i is an isomorphism. Proof.
It is clear that Ψ is a morphism. We check that it is injective. If Ψ( k , . . . , k r ) =0 then P ri =1 k i mm i = 0. Since m/m i is divisible by m j for all i = j , then k j mm j ≡ m j ).Since mm j is relatively prime to m j , it is invertible in Z m j so k j ≡ m j ). Thus( k , . . . , k r ) = (0 , , . . . , (cid:3) Example . [ Sza85 ] Let G be the direct product of the cyclic groups of orders m , . . . , m r and generators g , . . . , g r respectively. Assume r ≥
3. We can think of G as Z m × · · · × Z m r and we can pick the generators g = (1 , , . . . , , . . . , g r = (0 , . . . , , g = ( q, . . . ,
0) where q is some numberwhich is relatively prime to m . Let π be a permutation of the set { , . . . , r } that does nothave cycles of length 1 or 2. (Szab´o assumes π is cyclic, and since r ≥
3, our condition issatisfied, but we do not need π to be cyclic.)If g ∈ G \ { } and m is a positive integer which is less than or equal to the order of g ,we denote by [ g ] m = { , g, g, . . . , ( m − g } .Assume now m i = u i v i where u i , v i are integers greater than one. N SPECTRAL SETS OF INTEGERS 13
Obviously G = r X i =1 [ g i ] m i = r X i =1 ([ g i ] u i + [ u i g i ] v i ) = A ⊕ B ′ , where A = r X i =1 [ g i ] u i and B ′ = r X i =1 [ u i g i ] v i . Then Szab´o picks B i = [ u i g i ] v i + u π ( i ) g π ( i ) and g i = g i , as in Proposition 3.1. An easycheck shows that the sets B i are disjoint (here is where we need π to have only cycles oflength ≥
3) and A + B i + g i = A + B i (the main property used here is that [ g i ] u i +[ u i g i ] v i + g i =[ g i ] m i + g i = [ g i ] m i = [ g i ] u i + [ u i g i ] v i . Thus, the properties in Proposition 3.1 are satisfied andwith B = B ′ ∪ r [ i =1 ([ u i g i ] v i + u π ( i ) g π ( i ) + g i ) ! \ r [ i =1 ([ u i g i ] v i + u π ( i ) g π ( i ) ) ! , we have a new factorization A ⊕ B = G .Next, we construct spectra for the sets A, B ′ and B . For this, we regard G as wementioned before: G = Z m × · · · × Z m r and g = (1 , , . . . , , . . . , g r = (0 , . . . , , A = r X i =1 [ g i ] u i = { ( k , . . . , k r ) : 0 ≤ k i ≤ u i − i } ,B ′ = r X i =1 [ u i g i ] v i = { (( u k , . . . , u r k r ) : 0 ≤ k i ≤ v i − } ,B = B ′ ∪ r [ i =1 ([ u i g i ] v i + u π ( i ) g π ( i ) + g i ) ! \ r [ i =1 ([ u i g i ] v i + u π ( i ) g π ( i ) ) ! = { (( u k , . . . , u r k r ) : 0 ≤ k i ≤ v i − } ∪ r [ i =1 (cid:8) (0 , . . . , , k i u i + 1 , , . . . , , u π ( i ) , , . . . ,
0) : 0 ≤ k i ≤ v i − i (cid:9) \ r [ i =1 (cid:8) (0 , . . . , , k i u i , , . . . , , u π ( i ) , , . . . ,
0) : 0 ≤ k i ≤ v i − i (cid:9) , where k i u i + 1 and k i u i are on position i and u π ( i ) is on position π ( i ). Proposition . The set A has a spectrum (in G ) Λ A = { ( v j , . . . , v r j r ) : 0 ≤ j i ≤ u i − for all i } . The sets B ′ and B have spectrum (in G ) Λ B = { ( j , . . . , j r ) : 0 ≤ j i ≤ v i − for all i } . Also Λ A ⊕ Λ B = G . Proof.
For the set A , we check that { , v i , v i , . . . , ( u i − v i } in Z m i has spectrum { , , . . . , u i − } . Indeed u i − X j =0 e πi ( k − k ′ ) jviuivi = u i δ kk ′ for all k, k ′ ∈ { , , . . . , u i − } . Then, the result follows from Proposition 5.1. For B ′ , a similar argument can be used.For B we will use a lemma: Lemma . Given ( k , . . . , k r ) ∈ Z m × · · · × Z m r , if one of the k i is a non-zero multipleof u i , then v − X l =0 · · · v r − X l r =0 e πi l k m . . . e πi lrkrmr = 0 . Proof.
The sum splits into the product of the sums P v i − l i =0 e πi likiuivi , and if k i = ku i ,0 < k < v i then we further get P v i − l i =0 e πi likvi = 0. (cid:3) Now, take two distinct points b = ( b , . . . , b r ) = b ′ = ( b ′ , . . . , b ′ r ) in B . We want to provethat(3.4) X ( l ,...,l r ) ∈ Λ B e πi l b − b ′ m . . . e πi lr ( br − b ′ r ) mr = 0 . We denote B = { ( u k , . . . , u r k r ) : 0 ≤ k i ≤ v i − } \ r [ i =1 (cid:8) (0 , . . . , , k i u i , , . . . , , u π ( i ) , , . . . ,
0) : 0 ≤ k i ≤ v i − i (cid:9) ,B i = (cid:8) (0 , . . . , , k i u i , , . . . , , u π ( i ) , , . . . ,
0) : 0 ≤ k i ≤ v i − i (cid:9) , ˜ B i = (cid:8) (0 , . . . , , k i u i + 1 , , . . . , , u π ( i ) , , . . . ,
0) : 0 ≤ k i ≤ v i − i (cid:9) . So B = B ∪ ∪ ri =1 ˜ B i , disjoint union. N SPECTRAL SETS OF INTEGERS 15 If b, b ′ ∈ B , then the result follows from the fact that Λ B is a spectrum for B ′ , so B ′ is a spectrum for Λ B . If b ∈ B and b ′ ∈ ˜ B i , then b ′ is of the form b ′ = (0 , . . . , , k ′ i u i +1 , , . . . , , u π ( i ) , , . . . ,
0) and b is of the form b = ( k u , . . . , k r u r ).If one of the k , k , . . . , k i − , k i +1 , . . . , k π ( i ) − , k π ( i )+1 , . . . , k r is non-zero, then we applyLemma 3.5 to b − b ′ and obtain (3.4).If all these are zero, then, if k π ( i ) = 1, again we use Lemma 3.5 and obtain (3.4). If k π ( i ) = 1, then that means that b ∈ B i , a contradiction.If b ′ ∈ ˜ B i , and b is in a set ˜ B j with j = i , then b is of the form b = (0 , . . . , , k j u j +1 , , . . . , , u π ( j ) , , . . . , π ( i ) = j then π ( j ) = i (becasue π has no cycles of length 2),and π ( j ) = π ( i ). Therefore, applying Lemma 3.5 to b − b ′ , using the π ( j ) component, we get(3.4).If π ( i ) = j , then since π ( i ) = π ( j ), we can use the π ( i ) component in Lemma 3.5 for b − b ′ and again obtain (3.4).Finally, if b, b ′ ∈ ˜ B i , then b − b ′ ∈ ( B ′ − B ′ ) \ { } and the result follows, since Λ B is aspectrum for B ′ .The factorization Λ A ⊕ Λ B = G is obvious. (cid:3) Note that the factorization Λ A ⊕ Λ B = G is completely analogous to the factorization A ⊕ B ′ = G ; the roles of u i and v i are interchanged. Therefore, we can perform the sametype of operations on Λ A as we did on B ′ and get new factorizations and spectra.So, let us consider a permutation σ of { , . . . , r } which has only cycles of length ≥ ′ A = Λ A ∪ r [ i =1 ([ v i g i ] u i + v σ ( i ) g σ ( i ) ) + g σ ( i ) ! \ r [ i =1 ([ v i g i ] u i + v σ ( i ) g σ ( i ) ) ! . Proposition . Λ ′ A is a spectrum of A and Λ ′ A ⊕ Λ B = G . Proposition . Let m = m . . . m r . With the isomorphism Φ in Proposition 3.2, theimage of the sets A, B ′ , B, Λ A , Λ ′ A and Λ B in Z m have the CM-property. Proof.
We have Ψ( A ) = ( r X i =1 k i mm i : 0 ≤ k i ≤ u i − ) . We use the notation Q n ( x ) = 1 + x + · · · + x n − . We have A ( x ) = r Y i =1 Q u i (cid:16) x mmi (cid:17) . Then, with Proposition 5.4(ii) and (vi) (since u i and mm i are mutually prime), we get A ( x ) = Ψ( A )( x ) = r Y i =1 Y d | u i ,d> Φ u i (cid:16) x mmi (cid:17) = r Y i =1 Y d | u i ,d> Y t | mmi Φ dt ( x ) . If p α , α > p α ( x ) divides A ( x ), then there exist i ∈{ , . . . , r } , d > d | u i and t | mm i such that p α = dt . Since d and t are relatively prime and d >
1, we must have p α = d . Thus, S A consists of the prime powers that divide one of the u i . Take p α , . . . , p α n n in S A . Then p α divides one of the u i . By relabeling, let us assume p α , . . . , p α j j divide u i and p α j +1 j +1 , . . . , p α n n do not, hence they divide some u i j +1 , . . . , u i n , respec-tively, different than u i . Then d = p α . . . p α j j > d | u i . Also t = p α j +1 j +1 . . . p α n n divides mm i therefore Φ dt ( x ) divides A ( x ) and so A has the CM-property.For the sets B ′ and B , first note, from Lemma 5.3, that S B = S B ′ consists of primepowers p α , α > m but do not divide any of the u i .Take s , . . . , s n prime powers in S B . We will show that, for s = s . . . s n , Φ s ( x ) does notdivide A ( x ), so, by Proposition 1.5, it has to divide B ( x ) and B ′ ( x ), which will mean that B and B ′ satisfy the CM-property.Suppose not, so Φ s ( x ) divides A ( x ), then there exists i ∈ { , . . . , r } , d > d | u i and t | mm i such that s = dt . Each of the s k divides exactly one of the m j . By relabeling, suppose s , . . . , s j are the prime powers that divide m i and s j +1 , . . . , s n do not. Then d = s . . . s j .But, since d | u i it means that s , . . . , s j divide u i , a contradiction. Therefore, none of the s k divide m i so all of them divide mm i . But then, since s = dt , it follows that d = 1, again acontradiction. Thus, Φ s ( x ) does not divide A ( x ) so it has to divide B ( x ) and B ′ ( x ), whichimplies that B and B ′ have the CM-property.To show that the Λ sets have the CM-property is completely analogous. (cid:3) Example . This example will show that it is possible for the cyclotomic polynomialsto have multiplicity in a tile. They can appear with multiplicity in A ( x ) and they can appearin both A ( x ) and B ( x ). However, from Lemma 5.3, we know that this is not possible forΦ s ( x ) when s is a prime power. N SPECTRAL SETS OF INTEGERS 17
Take m = 4, m = 9, u = v = 2, u = v = 3. ThenΨ( A ) = { a + 4 b : a ∈ { , } , b ∈ { , , }} , Ψ( B ′ ) = { a + 12 b : a ∈ { , } , b ∈ { , , }} . As in the proof of Proposition 3.7, we have A ( x ) = (1 + x )(1 + x + x ) = Φ ( x )Φ ( x )Φ ( x )Φ ( x )Φ ( x )Φ ( x ) . Also, with Lemma 5.4, B ′ ( x ) = (1 + x )(1 + x + x ) = Q ( x ) Q ( x ) = Q (( x ) ) Q (( x ) )= Φ ( x )Φ ( x )Φ ( x )Φ ( x )Φ ( x )Φ ( x ) = Φ ( x )Φ ( x )Φ ( x )Φ ( x )Φ ( x )Φ ( x ) .
4. Some general constructions
Proposition . Let A be a set of non-negative integers. If A is a complete set ofrepresentatives modulo N , then A has the CM-property. Proof.
We have A ⊕ { } = Z N . By Proposition 1.5, there is an integer polynomial suchthat(4.1) A ( x ) = 1 + x + · · · + x N − + ( x N − Q ( x ) . Since A is a tile, it satisfies (T1), by Theorem 1.8. So N = A = Y s ∈ S A Φ s (1) . Since Φ s (1) = p if s = p α (by Proposition 5.4), it follows that the prime numbers that appearin the prime powers in S A divide N and also, all prime powers that divide N must appearin S A , otherwise N = A > Y s ∈ S A Φ s (1) . Thus, S A consists of all prime powers that divide N so (T2) is also satisfied, because if s , . . . , s n are powers of distinct primes in S A then s = s . . . s n divides N so Φ s ( x ) divides A ( x ), by (4.1). (cid:3) Proposition . Let A be a finite set of non-negative integers which has a spectrum Λ in N Z . If r is relatively prime to N , then r Λ is a spectrum for A . Proof.
Let λN = λ ′ N in Λ with λ, λ ′ ∈ Z . Then A ( e πi λ − λ ′ N ) = 0. Let λ − λ ′ N = ks with k, s ∈ Z , ( k, s ) = 1. Then Φ s ( z ) divides A ( z ) since it is the minimal polynomial for e πi ks . Since r is relatively prime to N , it is also relatively prime to s ( s divides N ). Then e πi krs isalso a primitive root of order s so Φ s ( e πi krs ) = 0 so A ( e π λ − λ ′ N r ) = A ( e πi krs ) = 0. This showsthat r Λ is a spectrum for A . (cid:3) For tiles, there is an analogous but more powerful result, due to Tijdeman [
Tij95 ]. Theorem . [Tijdeman’s theorem] If A ⊕ B = Z N and r is relatively prime to A ,then rA ⊕ B = Z N . Question.
Is Proposition 4.2 true if r is merely relatively prime to A ? This would bea dual of Tijdeman’s theorem. Theorem . Let A be a tile in Z N , A ⊕ B = Z N , let M be some postive integer and,for each a ∈ A , let A a be a tile in Z M with a common tiling set C , A a ⊕ C = Z M . Then theset ˜ A = ∪ a ∈ A ( { a } ⊕ N A a ) is a tile in Z NM with tiling set ˜ B = B ⊕ N C . If, in addition, thesets A and A a , a ∈ A satisfy the CM-property, then the set ˜ A satisfies the CM-property. Proof.
Take x in Z NM . It can be written uniquely as x = k + N k with k ∈ Z N and k ∈ Z M . Since A ⊕ B = Z N , k can be written uniquely as k = a + b , with a ∈ A and b ∈ B . Since A a ⊕ C = Z M , k can be written uniquely as k = a ′ + c with a ′ ∈ A a and c ∈ C . Thus x = ( a + N a ′ ) + ( b + N c ).Assume now that the sets A and A a , a ∈ A satisfy the CM-condition. Since ˜ A is a tile,by Theorem 1.8, ˜ A satisfies the (T1) property. To check the (T2) property, first we have tocompute the set S ˜ A .Note first that(4.2) ˜ A ( x ) = X a ∈ A x a A a ( x N ) . Since A a ⊕ C = Z M for all a , it follows the the sets A a have the same cardinality, and,from Lemma 5.3, that S A a is the complement of S C in the set of all prime power factors of M . Therefore, the sets S A a are all equal. By Lemma 5.5, it follows that the sets S NA a areall equal, and the sets N A a satisfy the CM-property.We will prove that(4.3) S ˜ A = S A ∪ S NA a disjoint union.If s is a prime power in S A , then s divides N (by Lemma 5.3) and if ω = e πi/s then ω N = 1. Thus ( N A a )( ω ) = A a ( ω N ) = A a (1) = A a = 0 so s is not in S NA a . So the sets S A and S NA a are disjoint. N SPECTRAL SETS OF INTEGERS 19
With (4.2), we have ˜ A ( ω ) = X a ∈ A ω a A a (1) = A a · A ( ω ) = 0 . Thus s ∈ S ˜ A .If s is a prime power in S NA a , then Φ s ( x ) divides ( N A a )( x ) = A a ( x N ) for all a ∈ A , andby (4.2), it follows that Φ s ( x ) divides ˜ A ( x ), so s ∈ S ˜ A . This proves that S A ∪ S NA a ⊂ S ˜ A .Since the sets ˜ A , A and N A a are tiles, they satisfy the (T1) property. Therefore, we have A = Y s ∈ S ˜ A Φ s (1) ≥ Y s ∈ S A Φ s (1) Y s ∈ S NAa Φ s (1) = A · N A a ) = A. Thus, we have equality in the inequality, and with Proposition 5.4, since Φ s (1) > s , it follows that we cannot have more elements in S ˜ A , so (4.3) is satisfied.Now take s , . . . , s n powers of distinct primes in S ˜ A . If all the s i are in S A , then since A satisfies the (T2) property, it follows that for s = s . . . s n , Φ s ( x ) divides A ( x ). Also, in thiscase, s divides N , so if ω = e πi/s then ω N = 1. Using (4.2), we obtain that ˜ A ( ω ) = 0 soΦ s ( x ) divide ˜ A ( x ).If all the s i are in S NA a , then since N A a satisfies the (T2) property, we get that Φ s ( x )divides ( N A a )( x ) = A a ( x N ), for all a ∈ A , and using (4.2), we obtain that Φ s ( x ) divides˜ A ( x ).Now assume s , . . . , s j are in S A (and hence they divide N ) and s j +1 , . . . , s n are in S NA a .Let s ′ = s j +1 . . . s n . We can factor N as N = N N N where N contains all the primefactors that appear in s , . . . , s j , N contains all the prime factors of N that appear in s ′ and N contains all the prime factors of N that do not appear in s , . . . , s n , and N , N , N are mutually prime. (The numbers s . . . s j and s ′ are relatively prime, because the s i arepowers of distinct primes.)Then Ns ...s n = N N N s ...s j s ′ can be reduced to k k s ′′ N where k = N s ...s j is an integer whichcontains only prime factors that appear in s , . . . , s j , N s ′ = k s ′′ , with k , s ′′ relatively primeintegers, and k contains only prime factors that appear in s ′ .Then we also have Ns ′ = N N N s ′ = N k s ′′ N , with s ′ and N k N relatively prime, so if ω ′ = e πi/s then ω ′ N = e πi Ns ′ is a primitive root of unity of order s ′′ . Since N A a satisfies the(T2) property, we get that Φ s ′ ( x ) divides ( N A a )( x ) so A a ( ω ′ N ) = 0 which means that Φ s ′′ ( x )divides A a ( x ). Then we also have Ns ′ = N N N s ′ = N k s ′′ N , with s ′ and N k N relativelyprime, so if ω ′ = e πi/s then ω ′ N = e πi Ns ′ is a primitive root of unity of order s ′′ . Since N A a satisfies the (T2) property, we get that Φ s ′ ( x ) divides ( N A a )( x ) so A a ( ω ′ N ) = 0 which meansthat Φ s ′′ ( x ) divides A a ( x ).Since k k N is also relatively prime with s ′′ , it follows that, for ω = e πi/s , ω N = e πi k k N s ′′ is also a primitive root of unity of order s ′′ . Therefore, ( N A a )( ω ) = A a ( ω N ) = 0 whichmeans that Φ s ( x ) divides ( N A a )( x ) = A a ( x N ), for all a ∈ A . From (4.2), it follows thatΦ s ( x ) divides ˜ A ( x ).Thus, ˜ A ( x ) satisfies the (T2) property. (cid:3) Remark . We see, from Lemma 2.12, that, in the case when A is a tile and A hasat most two prime factors p and q , then reducing modulo lcm( S A ), either A is contained in p Z or q Z , or it is of the form given in Theorem 4.4. We show in the next example that thisis not always the case. Example . Consider Szab´o’s example, with m = 2 , m = 3 , m = 5 , so G = Z × Z × Z which is isomorphic to Z by the isomorphism in Proposition 3.2. Let u = v = 2, u = v = 3, u = v = 5 and the permutation π of { , , } , π (1) = 2, π (2) = 3, π (3) = 1. Then A = { , } × { , , } × { , , , , } ⊂ Z × Z × Z ,B ′ = { , } × { , , } × { , , , , } ⊂ Z × Z × Z . To construct the set B we replace the subset B = { , } × { } × { } of B ′ with theset ˜ B = B + g = { , } × { } × { } , the set B = { } × { , , } × { } with the set˜ B = B + g = { } × { , , } × { } , and the set B = { } × { } × { , , , , } withthe set ˜ B = B + g = { } × { } × { , , , , } . so B = B ′ \ ( B ∪ B ∪ B ) ∪ ( ˜ B ∪ ˜ B ∪ ˜ B ) . The isomorphism in Proposition 3.2 is given by Z × Z × Z ∋ ( a , a , a ) a · · + a · · + a · · ∈ Z . We have that Ψ( B ), which has the CM-property and tiles Z , according to Proposition3.7, does not have the form in Theorem 4.4.Note that if a set ˜ A is of the form in Theorem 4.4, i.e.,˜ A = ∪ a ∈ a ( { a } ⊕ N A a ) , N SPECTRAL SETS OF INTEGERS 21 then, for k ∈ Z N , if k ∈ A , then { a ∈ ˜ A : a ≡ k (mod N ) } = A a = A/ C , a constantwhich does not depend on k , and which divides A , and it is 0 if k A . We say that ˜ A is equidistributed mod N .We will see that B is not equidistributed mod 2. Indeed, let B := B ′ \ ( B ∪ B ∪ B ).Then, Ψ( B ) ≡ , Ψ( ˜ B ) ≡ , Ψ( ˜ B ) ≡ , Ψ( ˜ B ) ≡ , Ψ( B ) ≡ , Ψ( ˜ B ) ≡ , Ψ( ˜ B ) ≡ , Ψ( ˜ B ) ≡ , Ψ( B ) ≡ , Ψ( ˜ B ) ≡ , Ψ( ˜ B ) ≡ , Ψ( ˜ B ) ≡ . Thus, Ψ( B ) ≡ B ) has 20 elements, a number which does not divide B ) = 30. This means that Ψ( B ) cannot have the form in Theorem 4.4.We have Ψ( ˜ B ) = { , } , Ψ( ˜ B ) = { , , } , Ψ( ˜ B ) = { , , , , } ,Ψ( B ) = { , , , , , , , , , , , , , , , , , , , } . Thus Ψ( B ) = { , , , , , , , , , , , , , , , , , , , , , , , , , , , , , } . Proposition . Let A be a finite set of non-negative integers which has a spectrum Λ ⊂ N Z . Suppose { A a : a ∈ A } are finite sets of non-negative integers that have a commonspectrum Λ . Then the set ∪ a ∈ A ( { a } ⊕ N A a ) is spectral with spectrum Λ + N Λ . Proof.
Let ( λ , λ ) = ( λ ′ , λ ′ ) in Λ × Λ . Since N ( λ − λ ′ ) ∈ Z , we have X a ∈ A X b ∈ A a e πi ( a + Nb )( λ − λ ′ + N ( λ − λ ′ )) = X a ∈ A e πia ( λ − λ ′ + N ( λ − λ ′ )) X b ∈ A a e πib ( λ − λ ′ ) . Since Λ is a spectrum for all sets A a , we get that, if λ = λ ′ , the last sum is 0, for all a ∈ A .If λ = λ ′ , then the above sum becomes X a ∈ A e πia ( λ − λ ′ ) A a = X a ∈ A e πia ( λ − λ ′ ) and since Λ is a spectrum for A , this sum is 0 if λ = λ ′ . (cid:3) Corollary . Let A be a spectral set in Z N with spectrum Λ . Suppose { A a : a ∈ A } are subsets of Z M that have a common spectrum Λ in Z M . Then the set ∪ a ∈ A ( { a } ⊕ N A a ) is spectral in Z NM with spectrum M Λ + Λ . Next, we present a result which generalizes and refines a result from Terence Tao’s blog,due to I. Laba.
Theorem . Suppose A ⊕ B = Z N and A and B are relatively prime. Then A consists of a single representative from each class modulo N/ B , and B consists of asingle representative from each class modulo N/ A . In other words, A is a complete setof representatives modulo N/ B . In particular, A and B have the CM-property and arespectral. Proof.
Let r = B . Since A and r are relatively prime, by Theorem 4.3, rA ⊕ B = Z N . So B ⊕ ( rA ⊕ N Z ) = Z . Then, by Lemma 5.6, if B ′ = { b ∈ B : b ≡ r ) } , we have B ′ ⊕ ( A ⊕ N/r Z ) = Z and B ′ = B/r = 1. Thus A ⊕ N/r Z = Z , which means that A consists of a single representative from each class modulo N/r . By symmetry, the statementis also true for B . (cid:3) Remark . If A is a complete set of representatives modulo k and N is a multiple of k , N = kl , then A always tiles Z N with B = { ki : i = 0 , . . . , l − } which is a subgroup of Z N . Note however that the tile B does not have to be a subgroup of Z N , even when A = k and B = l are mutually prime. For example, { , , , } ⊕ { , , } = Z and neither ofthe two sets is a subgroup of Z .
5. Appendix
Proposition . If A , . . . , A n are finite spectral sets in G , . . . , G n respectively, withcorresponding spectra Λ , . . . , Λ n , then A ×· · ·× A n is spectral in G ×· · ·× G n with spectrum Λ × · · · × Λ n . Proof.
Let ( ϕ , . . . , ϕ n ) , ( ϕ ′ , . . . , ϕ ′ n ) ∈ Λ × · · · × Λ n . Then X ( a ,...,a n ) ∈ A ×···× A n ϕ ( a ) . . . ϕ n ( a n ) ϕ ′ ( a ) . . . ϕ ′ n ( a n )= X a ∈ A ϕ ( a ) ϕ ′ ( a ) ! . . . X a n ∈ A n ϕ n ( a n ) ϕ ′ n ( a n ) ! = A δ ϕ ϕ ′ . . . A n δ ϕ n ϕ ′ n . (cid:3) Lemma . Let A and B be subsets of Z N such that A · B = N . Then A ⊕ B = Z N if and only if ( A − A ) ∩ ( B − B ) = { } . N SPECTRAL SETS OF INTEGERS 23
Proof.
The direct implication is clear. For the converse, define the map A × B → Z N ,( a, b ) a + b . The condition implies that the map is injective. Since the two sets have thesame cardinality N , the map is also surjective, so A ⊕ B = Z N . (cid:3) Lemma . [ CM99 ] Let A ( x ) and B ( x ) be polynomials with coefficients 0 and 1, n = A (1) B (1) , and R the set of prime power factors of N . If Φ t ( x ) divides A ( x ) or B ( x ) forevery factor t > of N , then (i) A (1) = Q s ∈ S A Φ s (1) and B (1) = Q s ∈ S B Φ s (1) . (ii) S A and S B are disjoint sets whose union is R . Proposition . [ CM99 ] Let p be a prime. (i) A polynomial P ( x ) ∈ Z [ x ] is divisible by Φ s ( x ) if and only if P ( ω ) = 0 for aprimitive s -th root of unity ω . (ii) 1 + x + · · · + x s − = Q t> ,t | s Φ t ( x ) . (iii) Φ p ( x ) = 1 + x + · · · + x p − and Φ p α +1 ( x ) = Φ p ( x p α ) . (iv) Φ s (1) = if s = 1 q if s is a power of a prime q otherwise. (v) Φ s ( x p ) = ( Φ ps ( x ) if p is a factor of s Φ s ( x )Φ ps ( x ) if p is not a factor s. (vi) If s and t are relatively prime, then Φ s ( x t ) = Q r | t Φ rs ( x ) . Lemma . [ CM99 ] Let k > and let A = kA be a finite set of non-negative integers. (i) A tiles the integers if and only if A tiles the integers. (ii) If p is prime, then S pA = { p α +1 : p α ∈ S A } ∪ { q β ∈ S A : q prime , q = p } . (iii) A ( x ) satisfies (T1) if and only if A ( x ) satisfies (T1). (iv) A ( x ) satisfies (T2) if and only if A ( x ) satisfies (T2). Lemma . [ CM99 ] Suppose A ⊕ C = Z , where A is a finite set of non-negative integers, k > and C ⊂ k Z . For i = 0 , , . . . , k − , let A i = { a ∈ A : A ≡ i (mod k ) } , a i = min( A i ) ,and A i = { a − a i : a ∈ A i } /k . Then (i) A ( x ) = x a A ( x k ) + x a A ( x k ) + · · · + x a k − A k − ( x k ) . (ii) Every A i ⊕ C/k = Z . (iii) The elements of A are equally distributed modulo k , i.e., every A i = ( A ) /k . (iv) S A = S A = · · · = S A k − . (v) When k is prime S A = { k } ∪ S kA and if every A i ( x ) satisfies (T2), then A ( x ) satisfies (T2).Acknowledgments This work was partially supported by a grant from the Simons Foundation(
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N SPECTRAL SETS OF INTEGERS 25 [Dorin Ervin Dutkay] University of Central Florida, Department of Mathematics, 4000Central Florida Blvd., P.O. Box 161364, Orlando, FL 32816-1364, U.S.A.,
E-mail address : [email protected] [Isabelle Kraus] University of Central Florida, Department of Mathematics, 4393 An-dromeda Loop N., Orlando, FL 32816-1364, U.S.A.,
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