On spectral sets of integers

# On spectral sets of integers

Dorin Ervin Dutkay [Dorin Ervin Dutkay] University of Central Florida
Department of Mathematics
4000 Central Florida Blvd.
P.O. Box 161364
Orlando, FL 32816-1364
U.S.A.
and  Isabelle Kraus [Isabelle Kraus] University of Central Florida
Department of Mathematics
4393 Andromeda Loop N.
Orlando, FL 32816-1364
U.S.A.
###### Abstract.

Based on tiles and on the Coven-Meyerowitz property, we present some examples and some general constructions of spectral subsets of integers.

###### Key words and phrases:
spectral set, tiling, cyclotomic polynomial, Coven-Meyerowitz property
05B45,65T50

## 1. Introduction

###### Definition 1.1.

Let be a locally compact abelian group and let be its Pontryagin dual group. A finite subset of is called spectral (in ) if there exists a subset of with such that

 (1.1) 1#A∑a∈Aφ(a)¯¯¯¯φ′(a)=δφφ′,(φ,φ′∈Λ)

In this case, is called a spectrum for (in the group ).

It is easy to check that the spectral property can be rephrased in the following ways.

###### Proposition 1.2.

Let be a finite subset of a locally compact abelian group and a finite subset of with . The following statements are equivalent:

1. is a spectrum for .

2. The matrix

 (1.2) 1√#A(φ(a))a∈A,φ∈Λ

is unitary.

3. For every ,

 (1.3) 1#Λ∑φ∈Λφ(a−a′)=δaa′.

We will be interested mainly in the spectral subsets of and the spectral subsets of . Since the dual group of is the group which can be identified with , a finite subset of is spectral if and only if there exists a finite subset in (or ) such that and the matrix

 1√#A(e2πiaλ)a∈A,λ∈Λ

is unitary.

Since the dual group of is , a subset of is spectral if and only if there exists a subset of such that and the matrix

 1√#A(e2πiaλ/N)a∈A,λ∈Λ

is unitary.

###### Definition 1.3.

For two subsets and of , we write to indicate that for each there are unique numbers and such that .

For a set of non-negative integers we denote by the associated polynomial

 A(x)=∑a∈Axa.

A subset of is called a tile if there exists a subset of such that .

In 1974 [Fug74], Fuglede proposed a conjecture that states that Lebesgue measurable spectral sets in coincide with sets that tile . The conjecture was disproved by Tao [Tao04] in dimensions five and higher and later in dimensions three and higher [FMM06, KM06b, KM06a, Mat05]. All these counterexamples were based on some constructions in finite groups, so the Fuglede conjecture fails for groups of the form . However, the conjecture is still open at this moment in dimensions one and two. It is known that the Fuglede conjecture in , under some additional hypotheses, can be reduced to the Fuglede conjecture for , see [DL14].

###### Conjecture 1.4.

[Fuglede’s conjecture for ] A finite subset of is spectral if and only if it is a tile.

A basic result (see [New77, CM99]) shows that every tiling set is periodic, i.e., there exists such that . If is any set consisting of one representative from for each class modulo , then and so , and therefore is a complete set of representatives modulo .

###### Proposition 1.5.

[CM99] Let be a positive integer and , sets of non-negative integers. The following statements are equivalent:

1. .

2. is a complete set of representatives of . In other words , where addition is understood modulo .

3. .

4. and for every factor of , the cyclotomic polynomial divides or .

Thus, tiles for coincide with tiles for the groups .

In [CM99], Coven and Meyerowitz found a sufficient condition for a subset of to be a tile, formulated in terms of cyclotomic polynomials.

###### Theorem 1.6.

[CM99] Let be a finite set of non-negative integers with corresponding polynomial . Let be the set of prime powers such that the cyclotomic polynomial divides . Consider the following conditions on .

1. .

2. If are powers of distinct primes, then the cyclotomic polynomial divides .

If satisfies (T1) and (T2), then tiles the integers with period . The tiling set can be obtained as follows: define , where the product is taken over all prime power factors of which are not in , and is the largest factor of relatively prime to . Then is the polynomial associated to a set of non-negative integers .

###### Definition 1.7.

A finite set of non-negative integers is said to satisfy the Coven-Meyerowitz property (or the CM-property), if it satisfies conditions (T1) and (T2) in Theorem 1.6. We call the tiling set in Theorem 1.6, the Coven-Meyerowitz (CM) tiling set associated to , and we denote it by .

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 1.8.

[CM99] Let be a finite set of non-negative integers with corresponding polynomial and let be the set of prime powers such that the cyclotomic polynomial divides . If tiles the integers, then (T1) holds.

Also, they proved that tiles with a cardinality that has only one or two prime factors satisfy the CM-property.

###### Theorem 1.9.

[CM99] Let be a finite set of non-negative integers with corresponding polynomial such that has at most two prime factors and tiles . Then satisfies (T2), and therefore it has the CM-property.

###### Remark 1.10.

Note that the CM-tiling set or does not depend on , it depends only on . Also, the proof shows that if satisfies (T1) and (T2), then it has a universal tiling of period , which is a tiling set for all the sets with .

Note also that for every prime factor since is a factor of and every divisor of is a polynomial in . This is because either with , and then we use Proposition 5.4(iii), or is a power of a prime different than and then is a multiple of , being the largest factor of relatively prime to .

Later, Łaba proved that sets with the CM-property are spectral.

###### Theorem 1.11.

[Łab02] Let be set of non-negative integers that satisfies the CM-property. Then is a spectral set. A spectrum for can be obtained as follows: consider the set of all numbers of the form

 ∑s∈SAkss,

where if , with prime.

###### Definition 1.12.

With the notations as in Theorems 1.6 and 1.11, for , we denote by and we call the Łaba spectrum of .

Combining the results of Coven-Meyerowitz and Łaba, Dutkay and Haussermann showed that if a set has the CM-property, then the tiling sets and the spectra are in a nice complementary relation.

###### Theorem 1.13.

[DH15] Let be a finite set of non-negative integers with the CM-property. Let , and let be its CM-tiling set. Then has the CM-property, and if , are the corresponding Łaba spectra, then .

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 that they have the CM-property and explicitly describe the tiling sets and the spectra. In Section 2, we describe the tiles with cardinality of a prime power. In Section 3, we describe Szabó’s examples 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 with the CM-property.

## 2. One prime power

###### Theorem 2.1.

Let be a set of non-negative integers with cardinality , where is prime and . Then the set tiles the integers if and only if there exist integers and for each , and each there exists a complete set of representatives modulo , , , such that the set is congruent modulo to the set

 (2.1) A′={n∑k=1pαk−1ai1,…,ik:0≤i1,…,in≤n−1}.

In this case , the CM-tiling set is

 (2.2) B=\textupCM(A)=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩∑j=0,…,αn−1j≠α1−1,…,αn−1bjpj:0≤bj≤p−1,0≤j≤αn−1,j≠α1−1,…,αn−1⎫⎪ ⎪ ⎪⎬⎪ ⎪ ⎪⎭.

The Łaba spectra of and are, respectively

 (2.3) \textup\LA={n∑i=1kipαi:0≤ki≤p−1,1≤i≤n}
 (2.4) \textup\LB=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩∑j=1,…,αnj≠α1,…,αnkjpj:0≤kj≤p−1,1≤j≤αn,j≠α1,…,αn⎫⎪ ⎪ ⎪⎬⎪ ⎪ ⎪⎭
###### Remark 2.2.

Let us explain a bit more the structure of the set . Think of the base decomposition of a number. For the set , we only use the digits corresponding to positions . The rest of the digits are 0. In position we use a complete set of representatives modulo , with . Once the first digit is chosen for the digit in position , we use another complete set of representatives modulo , , with . Note that, this complete set of representatives is allowed to be different for different choices of .

For , once the digits have been chosen for positions respectively, for the digit in position we pick a complete set of representatives, , with .

We will need some results from [CM99].

###### Definition 2.3.

Let be a set of powers of at most two primes. Define to be the collection of all subsets of which tile the integers and satisfy and . Note that because .

###### Lemma 2.4.

Let be a set of powers of at most two primes. A finite set with and tiles the integers if and only if is congruent modulo to a member of .

###### Proof.

Let be an element of and . Let . Since , there exists a set such that . Then, since , it follows that .

Conversely, if tiles the integers and , then by Lemma 2.5, has at most two prime factors, so it has the CM-property, by Theorem 1.9. Therefore, it has a tiling set of period , , by Remark 1.10. Let be the set obtained from by reducing modulo . Then and . Also, has the same tiling set . Then, by Lemma 5.3, as the complement of in the set of all prime power factors of . ∎

###### Lemma 2.5.

Let be a finite set of non-negative integers which is a tile. Then has at most two prime factors if and only if consists of powers of at most two primes.

###### Proof.

Since is a tile, it satisfies the (T1) property, by Theorem 1.8. So, using Proposition 5.4(iv),

 #A=∏s∈SAΦs(1)=∏pα∈SAp.

Thus has at most two prime factors if and only if consists of powers of at most two primes. ∎

###### Lemma 2.6.

[CM99] Suppose contains powers of only one prime . Let .

1. If then

 TS={p¯¯¯¯A:¯¯¯¯A∈T¯¯¯S}.
2. If then

 TS={∪p−1i=0({ai}⊕p¯¯¯¯Ai):¯¯¯¯Ai∈T¯¯¯S,a0=0,{a0,a1,…,ap−1} a complete
 set of representatives modulo p and every {ai}⊕p¯¯¯¯Ai⊂{0,1,…,\textuplcm(S)−1}}.
###### Proposition 2.7.

[CM99] Let be a prime number. Then

1. The only member of is .

2. For , the only member of is .

###### Theorem 2.8.

Let be a prime number. Let with . The following statements are equivalent:

1. .

2. For , there exist numbers with the following properties

1. The set is a complete set of representatives modulo , ,

2. For each , and each in , the set is a complete set of representatives modulo , ,

3. For each

 (2.5) ai1,…,ik+pαk+1−αkai1,…,ik+1+⋯+pαn−1−αkai1,…,in−1≤pαn−αk−1.
4.  (2.6) A={pα1−1ai1+pα2−1ai1,i2+⋯+pαn−1−1ai1,…,in−1+pαn−1j:0≤i1,…,in,j≤p−1}.
###### Proof.

We prove the equivalence of (i) and (ii) by induction on . For , the result follows from Proposition 2.7. Assume now the statements are equivalent for and take . Using Lemma 2.6(i), we have that if and only if with where . Using Lemma 2.6(ii), we have that if and only if there exists a complete set of representatives modulo , , and, for each , a set , where such that , and

 ¯¯¯¯A=∪p−1i1=0({ai1}+p¯¯¯¯Ai1).

Using the induction hypothesis for the set , we get that, for each the set must be of the form

 ¯¯¯¯Ai1={pα2−α1−1ai1,i2+pα3−α1−1ai1,i2,i3+⋯+pαn−α1−1ai1,…,in
 +pαn+1−α1−1j:0≤i2,…,in,j≤p−1},

where for each , the set is a complete set of representatives modulo and . Also

 ai1,i2,…,ik+p(αk+1−α1)−(αk−α1)ai1,…,ik+1+⋯+p(αn−α1)−(αk−α1)≤p(αn+1−α1)−(αk−α1)−1

and this implies (c) for . We must also have

 ai1+pα2−α1ai1,i2+…pαn−α1ai1,…,in+pαn+1−α1(p−1)≤pαn+1−α1+1,

and this implies (c) for .

Then

 A=pα1−1∪p−1i1=0({ai1}+p¯¯¯¯Ai1),

and (d) follows.

###### Proof of Theorem 2.1.

Assume that and have the given form. We show that . Note that and so . By Lemma 5.2, it is enough to show that in . If we pick an element in the intersection, it can be written in both ways as

 n∑k=1pαk−1(ai1,…,ik−ai′1,…,i′k)=∑j=0,…,αn−1j≠α1−1,…,αn−1pj(bj−b′j).

Take the first index such that . Then the left-hand side is divisible by but not by . Since does not appear on the right hand side, it follows, by contradiction, that both sides are equal to 0.

For the converse, if is a tile then the result follows from Theorem 2.8.

It remains to check that the CM-tiling set and the Łaba spectra are those given in (2.2),(2.3) and (2.4). By Lemma 5.3, we have that and are complementary, so

 SB={pj:j∈{1,…,αn},j≠α1,…,αn}.

Then the CM-tiling set is defined by the polynomial

 ∏j=1…αnj≠α1,…,αnΦj(x)=∏j=1…αnj≠α1,…,αn(1+xpj−1+x2pj−1+⋯+x(p−1)pj−1)
 =∑0≤bj≤p−11≤j≤αn,j≠α1,…,αnx∑1≤j≤αn,j≠α1,…,αnbjxpj−1.

This implies (2.2). Since we have the form of , (2.3) and (2.4) follow immediately.

###### Remark 2.9.

In [New77], Newman classifies the finite sets of integers which tile when the number of elements in the set is a prime power. The tiling condition is stated in Theorem 2.10. In Proposition 2.11, we determine the relation between the numbers in Newman’s paper and the numbers and the set in Theorem 2.1.

###### Theorem 2.10.

[New77] Let be distinct integers with , a prime, a positive integer. For each pair , we denote by the highest power of which divides . The set is a tile if and only if there are at most distinct .

###### Proposition 2.11.

Let be a set of non-negative integers which tile , with , a prime, a positive integer. Then , where denotes the highest power of which divides , as in Theorem 2.10.

###### Proof.

Let . Since is a tile, we have by Theorem 2.1 that is congruent modulo to the set

 A′={n∑k=1pαk−1ai1,…,ik:0≤i1,…,in≤n−1}.

Take . Then

 a=n∑k=1pαk−1ai1,…,ik+pαnm,
 a′=n∑k=1pαk−1ai1′,…,ik′+pαnm′.

So

 a−a′=n∑k=1(ai1,…,ik−ai1′,…,ik′)pαk−1+pαn(m−m′).

If , then . So then . If there exists a such that , then take the smallest such . Let be the largest power of such that divides . Then

 a−a′=pαk−1(ai1,…,ik−1,ik−ai1′,…,ik−1′,ik′)+n∑j=k+1(ai1,…,ij−ai1′,…,ij′)pαj−1+pαn(m−m′).

Since is a complete set of representatives modulo , this means that is divisible by , but not . Therefore, . Relabeling and , we get that . Doing this for all , we get that the set of all is and the result follows. ∎

For the case when the cardinality of a tile has two prime factors, to describe the structure of the such tiles, one can use the following lemma from [CM99].

###### Lemma 2.12.

Suppose contains powers of both the primes and . Let

 ¯¯¯¯S={pα:pα+1∈S}∪{qβ:qβ∈S},¯¯¯¯S′={pα:pα∈S}∪{qβ:qβ+1∈S}.
1. If , then

 TS={∪p−1i=0({ai}⊕p¯¯¯¯Ai):¯¯¯¯Ai∈T¯¯¯S,a0=0,{a0,a1,…,ap−1} a complete
 set of representatives modulo p and every {ai}⊕p¯¯¯¯Ai⊂{0,1,…,\textuplcm(S)−1}}.
2. If , then

 TS={∪q−1i=0({ai}⊕q¯¯¯¯Ai):¯¯¯¯Ai∈T¯¯¯S′,a0=0,{a0,a1,…,aq−1} a complete
 set of representatives modulo q and every {ai}⊕q¯¯¯¯Ai⊂{0,1,…,\textuplcm(S)−1}}.
3. If , then or and

 {A∈TS:A⊂pZ}={p¯¯¯¯A:¯¯¯¯A∈T¯¯¯S},{A∈TS:A⊂qZ}={q¯¯¯¯A:¯¯¯¯A∈T¯¯¯S′}.

## 3. Szabó’s examples

In [Sza85], Szabó constructed a class of examples to give a negative answer to two questions due to K. Corrádi and A.D. Sands respectively:

If is a finite abelian group and is one of its normed factorizations, in the sense that both factors contain the zero, must one of the factors contain some proper subgroup of ?

If is a finite abelian group and is a factorization, must one of the factors be contained in some proper subgroup of ?

We will present here Szabó’s examples and prove that they have the CM-property and find their tiling sets and spectra. We begin with a general proposition which encapsulates the core ideas in Szabó’s examples.

###### Proposition 3.1.

Let be a finite abelian group. Suppose we have a factorization . Let be disjoint subsets of and in with the property that

 (3.1) A+Bi=A+Bi+gi for all i.

Define

 (3.2) B=B′∪(r⋃i=1(Bi+gi))∖(r⋃i=1Bi).

Then .

###### Proof.

The sets are disjoint. Indeed, if not, then there are , , , such that . Because of the unique factorization, and , which contradicts the fact that and are disjoint. We have the disjoint union

 G=(A+(B′∖∪iBi))∪⋃i(A+Bi)=(A+(B′∖∪iBi))∪⋃i(A+Bi+gi).

Take in and we want to prove that is disjoint from . If or is in , this is clear from the hypothesis. If and lie in different sets this is again clear since the union above is disjoint. If , then , with , and since , we have . Therefore is disjoint from and the same is true for and . Thus . ∎

Szabó constructed his examples in . We note here that working in the group with relatively prime is equivalent to working in because of the following isomorphism.

###### Proposition 3.2.

Let be relatively prime non-negative integers. Let . The map ,

 (3.3) Ψ(k1,…,kr)=r∑i=1kimmi

is an isomorphism.

###### Proof.

It is clear that is a morphism. We check that it is injective. If then . Since is divisible by for all , then . Since is relatively prime to , it is invertible in so . Thus . Thus, is injective. Since the two sets have the same cardinality, is also surjective. ∎

###### Example 3.3.

[Sza85] Let be the direct product of the cyclic groups of orders and generators respectively. Assume . We can think of as and we can pick the generators . But we can also pick other generators, for example where is some number which is relatively prime to . Let be a permutation of the set that does not have cycles of length 1 or 2. (Szabó assumes is cyclic, and since , our condition is satisfied, but we do not need to be cyclic.)

If and is a positive integer which is less than or equal to the order of , we denote by .

Assume now where are integers greater than one.

Obviously

 G=r∑i=1[gi]mi=r∑i=1([gi]ui+[uigi]vi)=A⊕B′,

where

 A=r∑i=1[gi]ui and B′=r∑i=1[uigi]vi.

Then Szabó picks and , as in Proposition 3.1. An easy check shows that the sets