Fuglede’s conjecture for a union of two intervals

I. Łaba ^{∗}

Department of Mathematics Princeton University Princeton, NJ 08544 U.S.A.

September 18, 2000

Abstract

We prove that a union of two intervals in R is a spectral set if and only if it tiles R by translations. Mathematics Subject Classification: 42A99.

1 The results

A Borel set Ω ⊂ R^{n} of positive measure is said to tile R^{n} by translations if there is a discrete set T ⊂ R^{n} such that, up to sets of measure 0, the sets Ω + t, t ∈ T, are disjoint and ⋃_{t∈T} (Ω + t) = R^{n}. We may rescale Ω so that |Ω| = 1. We say that Λ = {λ_{k} : k ∈ Z} ⊂ R^{n} is a spectrum for Ω if:

{e^{2πiλ_{k}·x}}_{k∈Z} is an orthonormal basis for L^{2}(Ω).  (1.1)

A spectral set is a domain Ω ⊂ R^{n} such that (1.1) holds for some Λ.

Fuglede [2] conjectured that a domain Ω ⊂ R^{n} is a spectral set if and only if it tiles R^{n} by translations, and proved this conjecture under the assumption that either Λ or T is a lattice. The conjecture is related to the question of the existence of commuting self-adjoint extensions of the operators −i\frac{∂}{∂x_{j}}, j = 1, …, n [2], [7], [17]; other relations between the tiling and spectral properties of subsets of R^{n} have been conjectured and, in some cases, proved, see [6], [8], [9], [11], [12], [15].

Recently there has been significant progress on the special case of the conjecture when Ω is assumed to be convex [10], [3], [4], and in particular the 2-dimensional convex case appears to be nearly resolved [5]. The non-convex case is considerably more complicated, and is not understood even in dimension 1. The strongest results yet in that direction seem to be those of Lagarias and Wang [14], [15], who proved that all tilings of R by a bounded region must be periodic, and that the corresponding translation sets are rational up to affine transformations. This in turn leads to a structure theorem for bounded tiles. It was also observed in [15] that the “tiling implies spectrum” part of Fuglede’s conjecture for compact sets in R would follow from a conjecture of Tijdeman [20] concerning factorization of finite cyclic groups; however, Tijdeman’s conjecture is now known to fail without additional assumptions (see [13] for a discussion). See also [16], [1] for partial results on the related problem of characterizing all tilings of Z by a finite set, and [15], [18] for a classification of domains in R^{n} which have L + Z^{n} as a spectrum for some finite set L. Another recent result [19] is that sets which tile [0, ∞) by translations are spectral sets.

∗Current address: Department of Mathematics, University of British Columbia, Vancouver, Canada V6T 1Z2

The purpose of the present article is to address the following special case of Fuglede’s conjecture in one dimension. Let Ω = I_{1} ∪ I_{2}, where I_{1}, I_{2} are disjoint intervals of non-zero length. By scaling, translation, and symmetric reflection, we may assume that

Ω = (0, r) ∪ (a, a + 1 − r), 0 < r ≤ \frac{1}{2}, a ≥ r.  (1.2)

Our first theorem characterizes all Ω’s of the form (1.2) which are spectral sets.

Theorem 1.1 Suppose that Λ is a spectrum for Ω, 0 ∈ Λ. Then at least one of the following holds:

(i) a − r ∈ Z and Λ = Z;

(ii) r = \frac{1}{2}, a = \frac{n}{2} for some n ∈ Z, and Λ = 2Z ⋃(\frac{p}{n} + 2Z) for some odd integer p. Conversely, if Ω, Λ satisfy (1.2) and if either (i) or (ii) holds, then Λ is a spectrum for Ω.

As a corollary, we prove that Fuglede’s conjecture holds for a union of two intervals.

Theorem 1.2 Let Ω ⊂ R be a union of two disjoint intervals, |Ω| = 1. Then Ω has a spectrum if and only if it tiles R by translations.

Theorem 1.2 follows easily from Theorem 1.1. We may assume that Ω is as in (1.2). Suppose that Λ is a spectrum for Ω; without loss of generality we may assume that 0 ∈ Λ. Then by Theorem 1.1 one of the conclusions (i), (ii) must hold, and in each of these cases Ω tiles R by translations. Conversely, if Ω tiles R by translations, by Proposition 2.1 Ω must satisfy Theorem 1.1(i) or (ii), hence by Theorem 1.1 again it is a spectral set.

Theorem 1.1 will be proved as follows. Suppose that Λ = {λ_{k} : k ∈ Z}is a spectrum for Ω; we may assume that λ_{0} = 0. Let λ_{kk′} = λ_{k} − λ_{k′}, Λ − Λ = {λ_{kk′} : k, k′ ∈ Z}, and

Z_{Ω} = {0} ∪ {λ ∈ R : χ̂_{Ω}(λ) = 0}.  (1.3)

Then the functions e^{2πiλ_{k}x} are mutually orthogonal in L^{2}(Ω), hence Λ ⊂ Λ − Λ ⊂ Z_{Ω}. This will lead to a number of restrictions on the possible values of λ_{k}. Next, let

ϕ_{λ}(x) = χ_{(0,r)}e^{2πiλx},  (1.4)

where χ_{(0,r)} denotes the characteristic function of (0, r). By Parseval’s formula, the Fourier coefficients c_{k} = ∫_{0}^{r} e^{2πi(λ−λ_{k})x}dx of ϕ_{λ} satisfy

∑_{k∈Z} c_{k}^{2} = ∥χ_{(0,r)}e^{2πiλx}∥_{L^{2}(Ω)}^{2} = r.  (1.5)

Given that the λ_{k}’s are subject to the orthogonality restrictions mentioned above, we will find that there are not enough λ_{k}’s for (1.5) to hold, unless the conditions of Theorem 1.1 are satisfied.

The author is grateful to Alex Iosevich for helpful conversations about spectral sets and Fuglede’s conjecture, and to Jeffrey Lagarias and Yang Wang for bringing the references [13] and [19] to her attention.

2 Tiling implies spectrum

Proposition 2.1 If Ω as in (1.2) tiles R by translations, it must satisfy (i) or (ii) of Theorem 1.1.

Proof. Suppose that R may be tiled by translates of Ω. Assume first that r = \frac{1}{2}. Then any copy of Ω used in the tiling has a “gap” of length a − r = a − \frac{1}{2}, which must be covered by non-overlapping intervals of length \frac{1}{2}, hence a ∈ \frac{1}{2}Z as in Theorem 1.1(ii).

Assume now that 0 < r < \frac{1}{2}. Let I_{1} = (0, r), I_{2} = (a, a+1−r). We will prove that translates of I_{1} and I_{2} must alternate in any tiling T of R by translates of Ω; this implies immediately that a − r ∈ Z as in Theorem 1.1(i).

• If T contained two consecutive translates (τ, τ + r) and (τ + r, τ + 2r) of I_{1}, it would also contain the matching translates (τ + a, τ + a + 1 − r) and (τ + a + r, τ + a + 1) of I_{2}, which is impossible since the latter two intervals overlap.

• Suppose now that T contains two consecutive translates (τ + a, τ + a + 1 − r) and (τ + a + 1−r, τ +a+2−2r) of I_{2}. Then T must also contain the matching translates I′_{1} = (τ, τ +r) and I′′_{1} = (τ + 1 − r, τ + 2 − 2r) of I_{1}. The gap between I′_{1} and I′′_{1} has length 1 − 2r, which is strictly less than 1 − r = |I_{2}|, so that I′_{1} must be followed by another translate of I_{1}. But this has just been shown to be impossible.

Next, we prove the second part of Theorem 1.1. This easy result appears to have been known to several authors, see e.g., the examples in [2], [8], [15]. Since we will rely on it later on in the proof of the “hard” part of the theorem, we include the short proof.

Proposition 2.2 If Λ and Ω are as in Theorem 1.1(i) or (ii), then Λ is a spectrum for Ω.

Proof. If (i) holds, then Ω is a fundamental domain for Z, and, consequently, Λ = Z is a spectrum [2]. Suppose now that (ii) holds. For any function f on Ω, we define functions f_{+}, f_{−}:

f_{+}(x) = \frac{1}{2}(f(x) + f(x′)), f_{−}(x) = \frac{1}{2}(f(x) − f(x′)), x ∈ Ω,

where x′ = x + a if x ∈ (0, \frac{1}{2}), and x′ = x − a if x ∈ (a, a + \frac{1}{2}). Then:

f(x) = f_{+}(x) + f_{−}(x), f_{+}(x) = f_{+}(x′), f_{−}(x) = −f_{−}(x′).

It therefore suffices to prove that:

g(x) = ∑_{k∈Z} c_{k}e^{4kπix} for any g(x) ∈ L^{2}(Ω) such that g(x) = g(x′),  (2.1)

h(x) = ∑_{k∈Z} c′_{k}e^{(4k+ \frac{2p}{n})πix} for any h(x) ∈ L^{2}(Ω) such that h(x) = −h(x′).  (2.2)

Since e^{4kπix}, k ∈ Z, is a spectrum for (0, \frac{1}{2}), we have:

g(x) = ∑_{k∈Z} c_{k}e^{4kπix}, h(x) = e^{\frac{2p}{n} πix} ∑_{k∈Z} c′_{k}e^{4kπix}, x ∈ (0, \frac{1}{2}).

(2.1) follows immediately by periodicity. From the second equation above we find that (2.2) holds for all x ∈ (0, \frac{1}{2}), and that for such x:

e^{\frac{2p}{n} πi(x+a)} ∑_{k∈Z} c′_{k}e^{4kπi(x+a)} = −e^{\frac{2p}{n} πix} ∑_{k∈Z} c′_{k}e^{4kπix} = −h(x) = h(x + a),

where we used that \frac{2p}{n}a = p is odd. Hence (2.2) holds also for x ∈ (a, a + \frac{1}{2}). This ends the proof of Proposition 2.2.

3 Orthogonality

We now begin the proof of the first part of Theorem 1.1. Throughout the rest of the paper, Ω is assumed to satisfy (1.2), Λ = {λ_{k} : k ∈ Z} is a spectrum for Ω, λ_{0} = 0, λ_{kk′} = λ_{k} − λ_{k′}, Λ − Λ = {λ_{kk′} : k, k′ ∈ Z}, and Z_{Ω} is defined by (1.3).

Lemma 3.1 Z_{Ω} = Z_{1} ∪ Z_{2} ∪ Z_{3}, where:

Z_{1} = \{λ ∈ R : λa ∈ Z + \frac{1}{2}, λ(2r − 1) ∈ Z\},
Z_{2} = {λ ∈ Z : λr ∈ Z},
Z_{3} = {λ ∈ Z : λ(a − r) ∈ Z}.

Proof. Suppose that λ ≠ 0, λ ∈ Z_{Ω}. Then:

∫_{Ω} e^{2πiλx}dx = e^{2πiλr} − 1 + e^{2πiλ(a+1−r)} − e^{2πiλa} = 0.

All solutions to z_{1} + z_{2} + z_{3} + 1 = 0, |z_{i}| = 1, must be of the form {z_{1}, z_{2}, z_{3}} = {−1, z_{∗}, −z_{∗}}. Hence λ ∈ Z_{Ω} if and only if one of the following holds.

• e^{2πiλa} = −1 and e^{2πiλr} + e^{2πiλ(a+1−r)} = 0, hence λ ∈ Z_{1};

• e^{2πiλr} = 1 and e^{2πiλ(1−r)} = 1, hence λ ∈ Z_{2}; • e^{2πiλ(a+1−r)} = 1 and e^{2πiλa} = e^{2πiλr}, hence λ ∈ Z_{3}.

Observe that Z_{2}, Z_{3} are additive subgroups of Z.

Lemma 3.2 At least one of the following holds:

Λ ⊂ Z_{1} ∪ Z_{2},  (3.1)
Λ ⊂ Z_{1} ∪ Z_{3}.  (3.2)

Proof. By Lemma 3.1, Λ ⊂ Λ − Λ ⊂ Z_{Ω} ⊂ Z_{1} ∪ Z_{2} ∪ Z_{3}. If Z_{2} ⊂ Z_{3}, (3.2) holds. Suppose therefore that there is a λ_{i} ∈ Z_{2} \ Z_{3}. It suffices to prove that for any λ_{j} ∈ Z_{3} we must have λ_{j} ∈ Z_{1} or λ_{j} ∈ Z_{2}.

Let λ_{j} ∈ Z_{3}, then λ_{ij} = λ_{i} − λ_{j} ∈ Z_{Ω} by orthogonality. By Lemma 3.1, λ_{ij} ∈ Z_{1} ∪ Z_{2} ∪ Z_{3}. If λ_{ij} ∈ Z_{2}, then λ_{j} ∈ Z_{2} and we are done, and if λ_{ij} ∈ Z_{3}, then λ_{i} ∈ Z_{3}, which contradicts the above assumption on λ_{i}. Assume therefore that λ_{ij} ∈ Z_{1}. Then:

λ_{ij} ∈ Z, λ_{ij}a ∈ Z + \frac{1}{2}, λ_{ij}(2r − 1) ∈ Z,

hence

2λ_{j}r = 2λ_{i}r − λ_{ij}(2r − 1) − λ_{ij} ∈ Z.

If λ_{j}r ∈ Z, then λ_{j} ∈ Z_{2}; if λ_{j}r ∈ Z + \frac{1}{2}, then λ_{j}a ∈ Z + \frac{1}{2} by the definition of Z_{3} and λ_{j}(2r − 1) ∈ Z, so that λ_{j} ∈ Z_{1}.

Lemma 3.3 (i) Λ ⊂ Z_{2} is not possible; (ii) Λ ⊂ Z_{3} is possible only if a − r ∈ Z and Λ = Z_{3} = Z.

Proof. Suppose that Λ ⊂ Z_{i} for i = 2 or 3. Since Z_{i} is an additive subgroup of Z, we must have Z_{i} = pZ for some integer p > 0. Furthermore, if there was a λ ∈ pZ \ Λ, we would have λ_{k} − λ ∈ pZ and e^{2πiλx} would be orthogonal to e^{2πiλ_{k}x} for all λ_{k} ∈ Λ, which would contradict (1.1). Hence Λ = Z_{i} = pZ. We also observe that, if p were ≥ 2, any function of the form f(x) = ∑_{k∈Z} c_{k}e^{2πiλ_{k}x} would be periodic with period \frac{1}{p} ≤ \frac{1}{2}, which again would contradict (1.1). Thus Λ = Z_{i} = Z.

If i = 2, this is not possible, since nr cannot be an integer for all n ∈ Z if r ≤ \frac{1}{2}. If i = 3, we obtain that n(a − r) ∈ Z for all n ∈ Z; letting n = 1, we find that a − r ∈ Z.

If Ω, Λ are as in Lemma 3.3(ii), then Theorem 1.1(i) is satisfied and we are done. Thus we may assume throughout the sequel that:

Λ ⊄ Z_{2}, Λ ⊄ Z_{3}.  (3.3)

Lemma 3.4 If (3.3) holds, then Λ ⊂ Z_{1} ∪ (Z_{2} ∩ Z_{3}).

Proof. By Lemma 3.2, it suffices to prove that:

if Λ ∩ (Z_{1} \ Z_{2}) ≠ ∅, then Λ ∩ Z_{2} ⊂ Λ ∩ Z_{3};  (3.4)
if Λ ∩ (Z_{1} \ Z_{3}) ≠ ∅, then Λ ∩ Z_{3} ⊂ Λ ∩ Z_{2};  (3.5)

We will only prove (3.4), since the proof of (3.5) is almost identical. Suppose that λ_{i} ∈ Z_{1} \ Z_{2}, and let λ_{j} ∈ Z_{2}. By Lemma 3.1, λ_{ij} belongs to at least one of Z_{1}, Z_{2}, Z_{3}. We may not have λ_{ij} ∈ Z_{2}, since then λ_{i} would also be in Z_{2}. Thus we only need consider the following two cases.

• Let λ_{ij} ∈ Z_{1}. Then λ_{i}a, λ_{ij}a ∈ Z + \frac{1}{2}, hence λ_{j}a ∈ Z and λ_{j} ∈ Z_{2} ∩ Z_{3}.

• Assume now that λ_{ij} ∈ Z_{3}. Then λ_{i} ∈ Z, hence 2λ_{i}r ∈ Z. Moreover, λ_{i}r ∈ Z would imply λ_{i} ∈ Z_{2}, hence λ_{i}r ∈ Z + \frac{1}{2}. It follows that λ_{i}(a − r) ∈ Z; since also λ_{ij}(a − r) ∈ Z, we obtain that λ_{j}(a − r) ∈ Z and λ_{j} ∈ Z_{2} ∩ Z_{3}.

Lemma 3.5 Assume (3.3). Then:

(i) Λ − Λ ⊂ Z_{1} ∪ (Z_{2} ∩ Z_{3});

(ii) Λ ∩ Z_{1} ⊂ λ_{∗} + r^{−1}Z for some λ_{∗} ∈ R.

Proof. For k ∈ Z, let Λ_{k} = Λ − λ_{k} = {λ_{jk} : j ∈ Z}. Then Λ_{k} is also a spectrum for Ω, 0 ∈ Λ_{k}, hence all of the results obtained so far apply with Λ replaced by Λ_{k}. Thus (i) follows from Lemmas 3.3 and 3.4.

To prove (ii), it suffices to verify that λ_{ij}r ∈ Z whenever λ_{i}, λ_{j} ∈ Z_{1}. Indeed, if λ_{i}, λ_{j} ∈ Z_{1}, then λ_{ij}a ∈ Z, hence λ_{ij} ∉ Z_{1} and, by (i), λ_{ij} ∈ Z_{2} ∩ Z_{3}. But this implies that λ_{ij}r ∈ Z.

4 Completeness

Fix j, n ∈ Z, and consider the function ϕ_{λ} defined by (1.4) with λ = λ_{j} − nr^{−1}. The Fourier coefficients of ϕ_{λ} are:

c_{k} = ∫_{0}^{r} e^{2πi(λ−λ_{k})x}dx = ∫_{0}^{r} e^{2πi(λ_{jk}−nr^{−1})x}dx,

hence c_{k} = r if λ_{jk} = nr^{−1}, and:

c_{k} = \frac{1}{2πi(λ_{jk} − nr^{−1})}(e^{2πi(λ_{jk}r−n)} − 1), λ_{jk} ≠ nr^{−1}.  (4.1)

Define α_{jk} = λ_{jk}r. Plugging (4.1) into (1.5), we obtain that for all j ∈ Z:

\frac{1}{r} = 1 + ∑_{k:α_{jk}∉Z} \frac{1}{4π^{2}α_{jk}^{2}} |e^{2πiα_{jk}} − 1|^{2},  (4.2)

and for all n, j ∈ Z:

\frac{1}{r} = δ_{n,j} + ∑_{k:α_{jk}∉Z} \frac{1}{4π^{2}(α_{jk} − n)^{2}} |e^{2πi(α_{jk}−n)} − 1|^{2},  (4.3)

where δ_{n,j} = 1 if there is a k ∈ Z such that α_{jk} = n, and δ_{n,j} = 0 otherwise.

We define the equivalence relation between the indices k, k′:

k ∼ k′ ⇔ α_{kk′} ∈ Z,

and denote by A_{1}, A_{2}, …, A_{m}, … ⊂ Z the (non-empty and disjoint) equivalence classes with respect to this relation. Hence k, k′ belong to the same A_{m} if and only if α_{kk′} ∈ Z; in particular,

A_{m} := {α_{k} : k ∈ A_{m}} ⊂ β_{m} + Z

for some β_{m} ∈ [0, 1).

Lemma 4.1 Let M denote the number of distinct and non-empty A_{m}’s. Then:

M ≥ r^{−1}.  (4.4)

Moreover, if one of the A_{m}’s skips a number (i.e., A_{m} ≠ β_{m} + Z), then M ≥ r^{−1} + 1.

Proof. For each m, m′, let β_{mm′} = β_{m} − β_{m′}; note that β_{mm′} ≠ 0 if m ≠ m′. Fix m′ and j ∈ A_{m′}, then (4.2) may be rewritten as:

\frac{1}{r} = 1 + ∑_{m≠m′} S_{mm′},  (4.5)

where:

S_{mm′} = ∑_{k∈A_{m}} \frac{1}{4π^{2}α_{jk}^{2}} |e^{2πiβ_{mm′}} − 1|^{2}.

Clearly:

S_{mm′} ≤ S̃(β_{mm′}),  (4.6)

where:

S̃(β) = ∑_{k∈Z} \frac{1}{4π^{2}(β + k)^{2}} |e^{2πiβ} − 1|^{2}.  (4.7)

Hence (4.4) follows from (4.5) and Lemma 4.2 below.

Suppose now that A_{m′} skips a number. Then we may find j ∈ A_{m′} and n ∈ Z such that δ_{n,j} = 0, and (4.4) may be improved to M ≥ 1 + r^{−1} by using (4.3) instead of (4.2).

Lemma 4.2 Let S̃(β) be as in (4.7), then S̃(β) = 1 for all 0 < β < 1.

Proof. By Proposition 2.2, Λ = 2Z ∪ (\frac{p}{n} + 2Z), where n ∈ Z and p is an odd integer, is a spectrum for Ω = (0, \frac{1}{2}) ∪ (\frac{n}{2}, \frac{n+1}{2}). Plugging this back into (4.2) we obtain that:

1 = ∑_{k∈Z} \frac{1}{4π^{2}(β + k)^{2}} |e^{2πiβ} − 1|^{2}

for β = \frac{p}{2n}. However, the set of β of this form is dense in R, hence by continuity the lemma holds for all β ∈ (0, 1).

5 Conclusion

Proof of Theorem 1.1. If Λ is as in Lemma 3.3(ii), then Theorem 1.1(i) is satisfied. We may therefore assume that (3.3) holds. From Lemma 3.5 we have:

Λ − Λ ⊂ Z_{1} ∪ (Z_{2} ∩ Z_{3}), Z_{2} ∩ Z_{3} ⊂ r^{−1}Z, Z_{1} ⊂ (λ_{∗} + r^{−1}Z),  (5.1)

for some λ_{∗} ∈ R, hence M ≤ 2 (using the notation of Section 4). However, by Lemma 4.1 M ≥ r^{−1} ≥ 2, and this may be improved to M ≥ 3 if one of the A_{m}’s skips a number. Therefore we must have r = \frac{1}{2}, and:

Λ − Λ = 2Z ∪ (λ_{∗} + 2Z), Z_{2} ∩ Z_{3} = 2Z, Z_{1} = λ_{∗} + 2Z.  (5.2)

Pick λ_{ij}, λ_{kl} ∈ Z_{1} such that λ_{ij} − λ_{kl} = 2. From the definition of Z_{1} we have λ_{ij}a, λ_{kl}a ∈ Z + \frac{1}{2}, hence:

2a = \frac{a}{r} = λ_{ij}a − λ_{kl}a ∈ Z,

so that a = \frac{n}{2} for some n ∈ Z. Finally, we have λ_{∗}a = \frac{1}{2}nλ_{∗} ∈ Z + \frac{1}{2}, hence λ_{∗}n = p for some odd integer p. Thus Ω and Λ satisfy (ii) of Theorem 1.1.

References

[1] E. M. Coven, A. Meyerowitz: Tiling the integers with translates of one finite set, J. Algebra 212 (1999), 161–174.

[2] B. Fuglede: Commuting self-adjoint partial differential operators and a group theoretic problem, J. Funct. Anal. 16 (1974), 101–121.

[3] A. Iosevich, N. H. Katz, S. Pedersen: Fourier bases and a distance problem of Erdös, Math. Res. Lett. 6 (1999), 251–255.

[4] A. Iosevich, N. H. Katz, T. Tao: Convex bodies with a point of curvature do not have Fourier bases, Amer. J. Math, to appear.

[5] A. Iosevich, N. H. Katz, T. Tao: preprint in preparation.

[6] A. Iosevich, S. Pedersen: Spectral and tiling properties of the unit cube, Internat. Math. Res. Notices 16 (1998), 819–828.

[7] P. Jorgensen: Spectral theory of finite volume domains in R^{n}, Adv. Math. 44 (1982), 105–120.

[8] P. Jorgensen, S. Pedersen: Spectral theory for Borel sets in R^{n} of finite measure, J. Funct. Anal. 107 (1992), 72–104.

[9] P. Jorgensen, S. Pedersen: Spectral pairs in Cartesian coordinates, J. Fourier Anal. Appl. 5 (1999), 285–302.

[10] M. Kolountzakis: Non-symmetric convex domains have no basis of exponentials, Illinois J. Math., to appear.

[11] M. Kolountzakis: Packing, tiling, orthogonality, and completeness, Bull. London Math. Soc. 32 (2000), 589–599.

[12] J. C. Lagarias, J. A. Reed, Y. Wang: Orthonormal bases of exponentials for the n-cube, Duke Math. J. 103 (2000), 25–37.

[13] J. C. Lagarias, S. Szabó: Universal spectra and Tijdeman’s conjecture on factorization of cyclic groups, J. Fourier Anal. Appl., to appear.

[14] J. C. Lagarias, Y. Wang: Tiling the line with translates of one tile, Invent. Math. 124 (1996), 341–365.

[15] J. C. Lagarias, Y. Wang: Spectral sets and factorizations of finite abelian groups, J. Funct. Anal. 145 (1997), 73–98.

[16] D. J. Newman: Tesselation of integers, J. Number Theory 9 (1977), 107–111.

[17] S. Pedersen: Spectral theory of commuting self-adjoint partial differential operators, J. Funct. Anal. 73 (1987), 122–134.

[18] S. Pedersen: Spectral sets whose spectrum is a lattice with a base, J. Funct. Anal. 141 (1996), 496–509.

[19] S. Pedersen, Y. Wang: Universal spectra, universal tiling sets and the spectral set conjecture, Scand. J. Math., to appear.

[20] R. Tijdeman: Decomposition of the integers as a direct sum of two subsets, London Math. Soc. Lecture Note Ser., vol.215, Cambridge Univ. Press, 1995, pp. 261–276.