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ON FUGLEDE’S CONJECTURE FOR THREEINTERVALS

DEBASHISH BOSE, C.P. ANIL KUMAR, R. KRISHNAN,AND SHOBHA MADAN

Abstract. In this paper, we prove the Tiling implies Spectral partof Fuglede’s cojecture for the three interval case. Then we prove theconverse Spectral implies Tiling in the case of three equal intervalsand also in the case where the intervals have lengths 1/2, 1/4, 1/4.Next, we consider a set Ω ⊂ R, which is a union of n intervals. IfΩ is a spectral set, we prove a structure theorem for the spectrumprovided the spectrum is assumed to be contained in some lattice.The method of this proof has some implications on the Spectral

implies Tiling part of Fuglede’s conjecture for three intervals. Inthe final step in the proof, we need a symbolic computation us-ing Mathematica. Finally with one additional assumption we canconclude that the Spectral implies Tiling holds.

1. Introduction

Let Ω be a Lebesgue measurable subset of R with finite positivemeasure. For λ ∈ R, let

eλ(x) = e2πiλx, x ∈ R

Ω is said to be a spectral set if there exists a subset Λ ⊂ R such thatthe set EΛ = eλ : λ ∈ Λ is an orthonormal basis for the Hilbert spaceL2(Ω), and then the pair (Ω,Λ) is called a spectral pair.We say that Ω as above tiles R by translations if there exists a subset

T ⊂ R, such that the sets Ω + t : t ∈ T forms a partition a.e. ofR. Here Ω + t = x + t : x ∈ Ω. These definitions clearly extend toRd, d > 1.

Fuglede’s Conjecture. Let Ω be a measurable set in Rd withfinite positive measure. Then Ω is spectral if and only if Ω tiles Rd bytranslations.

Fuglede proved this conjecture under the additional assumption thatthe spectrum, or the tiling set is a lattice [F]. In recent years there hasbeen a lot of activity on this problem. It is now known that in this

2000 Mathematics Subject Classification. Primary: 42A99.1

2DEBASHISH BOSE, C.P. ANIL KUMAR, R. KRISHNAN, AND SHOBHA MADAN

generality, the conjecture is false in both directions if the dimensiond ≥ 3 [T], [KM1], [KM2] and [M], [FR], [FMM]. However the conjectureis still open in all dimensions d ≥ 3 under the additional hypothesisthat the set Ω is a convex set. For convex sets, the conjecture is trivialfor d = 1 and for d = 2, it was proved in [K1] and [IKT]. In dimension 1,the problem is related to number theoretic questions, and many partialresults are known (see, for example, [KL], [L2]. [LW], [PW]).

In this paper we restrict ourselves to one dimension and to the casewhen the set Ω is a union of three intervals. This work was inspired bythe paper [L1], by I. Laba, where Fuglede’s conjecture is proved in thecase that Ω is a union of two intervals.

In section 2, we prove that for Ω a union of three intervals, ”Tilingimplies Spectral”. This proof, though somewhat long uses elementaryarguments. First we are able to identify the different cases to be con-sidered and then we can analyse the tiling patterns in each case toprove that Ω is spectral.

In section 3, we consider two particular cases where Ω = A∪B ∪C,and either |A| = |B| = |C| = 1/3, or |A| = 1/2, |B| = |C| = 1/4, andprove ”Spectral implies Tiling” in each case. Here, orthogonality condi-tions impose restrictions on the end-points of the intervals, and we canuse a powerful theorem on tesselation of integers due to Newman[N].

In Section 4 we briefly digress to the the n-interval case. We assumethat Ω = ∪n

j=1[aj , aj + rj),∑n

j=1 rj = 1 is spectral. Then if a spectrum

Λ contains an arithmetic progression (AP) of length 2n + 1, we provethat Λ must contain the complete AP. Since by [L] the spectrum haspositive asymptotic density, the above condition on Λ holds if Λ iscontained in some lattice, (using [S]), and in this case we also get thatΛ is rational. We remark that all known spectra lie in a lattice. Inthis paper, we relax this hypothesis somewhat; it suffices to assumethat Λ − Λ is δ-separated, then Λ contains APs of arbitrary length(Proposition 4.5).

We return to three intervals in Section 5, with the assumption thatΛ−Λ is δ-separated. It then turns out that either (i) Ω/Z ≈ [0, 1] or (ii)Ω/2Z ≈ [0, 1/2]∪ [n/2, (n+1)/2], for some n or (iii) the case is that ofequal intervals (not necessarily 3 equal intervals!). The first two casesare that of [F] and [L1] respectively. The third case is rather complex,and we have had to use symbolic computation on Mathematica in anattempt to resolve Fuglede’s conjecture for three intervals (of coursewith our assumption on the spectrum). The analysis is given in Section6.

ON FUGLEDE’S CONJECTURE FOR THREE INTERVALS 3

2. Tiling implies Spectral

Theorem 2.1. Let Ω = A ∪ B ∪ C, |A| + |B| + |C| = 1. If Ω tiles R

by translations, then Ω is a spectral set.

For the proof of this theorem we adopt a notation for convenience:Whenever we need to keep track of the intervals A, B, C as being partof a translate of Ω, we will write Ωt = Ω+ t = At ∪Bt ∪ Ct.

We begin with a simple lemma.

Lemma 2.2. Suppose Ω tiles R. Suppose that in some tiling by Ω, AAoccurs, then either |A| = 1

2or |A| = |B| = |C| = 1

3

Proof. Suppose not all three intervals are equal, and that for somes, t ∈ T , AsAt, occurs in some tiling of R by Ω. Then |A| ≥ |B|, |C|. IfBB also occurs, then |B| ≥ |A|, so |A| = |B| < 1/2. Now CC cannotalso occur (for then |A| = |B| = |C|). The gap between Cs and Ct

equals 0 < |A| − |C| < |A|, so it cannot be filled at all. Hence if AAoccurs, neither BB nor CC can also occur in that tiling. Now the gapbetween Bs and Bt is of length |A| − |B| < |A|, so it can be filled onlyby a C. It follows that |A| − |B| = |C| and so |A| = 1

2.

This leads us to consider the following cases:

Case 1: |A|, |B|, |C| 6= 12and not all are equal:

(1a): No two of |A|, |B|, |C| are equal(1b): |A| = |B| 6= |C|

Case 2: |A| = 12

(2a): |B| 6= |C|(2b): |B| = |C| = 1

4

Case 3: |A| = |B| = |C| = 13.

Proof of Theorem: We will prove that Tiling implies Spectral, ineach of the above cases.

Case(1a) : |A|, |B|, |C| distinct and not equal to 1/2.We claim that in this case the tiling pattern has to be of the form

−−−ABC|ABC|ABC −−−

or−−−ACB|ACB|ACB −−−

ie. Ω tiles R by Z, hence is spectral [F]. Our claim will be provedif we show that no three consecutive intervals appear as XYX , since

4DEBASHISH BOSE, C.P. ANIL KUMAR, R. KRISHNAN, AND SHOBHA MADAN

we already know that XX cannot appear. So, suppose AtBAs occurs.Then the gap between the Ct and Cs is of length |A| + |B| − |C| <|A| + |B| and so cannot be filled by AB, (nor by AA, BB), henceequals either |A| or |B|. But then, either |B| = |C| or |A| = |C|, acontradiction.

Case(1b): |A| = |B| 6= |C|, |C| 6= 1/2.Of course none of AA,BB,CC can occur in the tiling. We show

below that ABA and BAB cannot occur: Suppose AsBAt occurs, thenthe gap between Cs and Ct has length |A|+|B|−|C| = 4|A|−1 and hasto be filled by A’s and B’s. Let 4|A| − 1 = m|A|. Clearly m < 4, andwe check easily that this leads to a contradiction (m = 0 ⇒ |C| = 1/2;m = 1 ⇒ |A| = |B| = |C| = 1/3, m = 2 ⇒ |A| = |B| = 1/2 andm = 3 ⇒ |A| = 1).

This means that in any tiling, the gap between two consecutive C’sis filled by at most two of A and B. In other words, C is translatedby at most 1; hence also A and B. But if none of the translates isgreater than 1, then CAC, CBC, ACA, and BCB are excluded, sothat the the tiling pattern must be either −−−ABCABCABC −−−or −−−ACBACBACB −−−, ie. Ω tiles R by Z.

Case (2a) : |A| = 1/2, |B| 6= |C|By Lemma 2.2 , BB and CC cannot occur in the tiling. Next we

show that none of BAB,ABA,CAC,ACA can occur. Note that BABoccurs iff ABA occurs. Suppose AsBAt occurs. The gap between Cs

and Ct is of length |A| + |B| − |C| < |A| + |B|. This gap has to befilled by a single A or a single B; and this would imply |B| = |C| or|A| = |C| = 1/2. Similarly CAC and ACA cannot occur.

Next, suppose that there is a string BsAA...AABt in the tiling with nconsecutive A’s. The gap between Cs and Ct is of length n/2+|B|−|C|and has to be filled by A’s and B’s. But B’s can occur just afterthe Cs and just before the Ct, otherwise ABA will occur. If no Boccurs, then |B| = |C| and if a single B occurs then |C| = 0 or 1/2.Hence there exists a string − − −CBAA − − − ABC − − so thatn2+ |B| − |C| = 2|B|+ m

2, for some integer m, which means m = n− 1.

We can repeat the argument n times to to get that BAB must occursomewhere, which is not possible. Similarly CAA−−− AC does notoccur.

Thus the only possibilities are tiling patterns of the form :

(1) −−−AA−−−A |BC |BC −−−|BC|AA−−−ABC −−−(2) −−−AA−−−A |BC |BC −−−|BC|BA−−−ACB−−−

ON FUGLEDE’S CONJECTURE FOR THREE INTERVALS 5

and (1) ,(2) with B and C interchanged. We show that (2) and (2)cannot happen. Suppose (2) occurs; consider the part AsBC − − −BCBAt and the Cs, Ct corresponding to these A’s. The gap is oflength n

2+ 1

2+ |B| − |C| and has to be filled by A’s and B’s. But this

leads to a contradiction as shown above. So finally the tiling pattern

has to be of the form (1) or (1).We write Ω = [0, 1/2) ∪ [b, b + r) ∪ [c, c + 1/2 − r). It is easy to see

that the above tiling pattern for Ω, say by a tiling set T implies thatboth Ω1 = [0, 1/2)∪ [b, b+ 1/2) and Ω2 = [0, 1/2) ∪ [c− r, c− r + 1/2)tile R by the same tiling set T . Using the result for two intervals [L1],this implies that b = n

2, c− r = k

2with n, k ∈ Z . Then c− b− r = k−n

2.

Observe that n is a period for the tiling set, but n/2 is not (since once[0, n

2) is filled, the tiling upto [0, n) is completely determined ). So if

k0 is the period of the tiling then k0|n, but k0 6 |n2. Hence if 2j |n and

2j+1 6 |n the same is true for k0. Hence nk0

∈ 2Z + 1. Also k0|k, andkk0

∈ 2Z+ 1. Hence k0|m, where m = c− b− r.

We show finally that the set Λ = 2Z∪ (2Z+ 1k0) is a spectrum for Ω

(we have taken p = nk0). To check orthogonality, note that

∫

Ω

e2πiλξdξ =

∫ 1/2

0

+

∫ b+r

b

+

∫ c+1/2−r

c

e2πiλξdξ

=

∫ 1/2

0

e2πiλξdξ +

∫ b+r

b

e2πiλξdξ +

∫ b+1/2+m

b+r+m

e2πiλξdξ

=

∫ 1/2

0

+

∫ b+r

b

e2πiλξdξ + e2πiλm∫ b+1/2

b+r

e2πiλξdξ

=

∫

Ω1

e2πiλξdξ if λ ∈ Λ = 0

since k0|m and since Λ is a spectrum for Ω1

As in [L1], it is easy to see that Λ is complete.Case 2b is a special case of 4 equal intervals, and Case 3 is 3 equal

intervals. In each of these cases tiling implies spectral follows by moregeneral results [L2]. However, this case can be handled by using atheorem due to Newman [N], and we do this in the next section.

6DEBASHISH BOSE, C.P. ANIL KUMAR, R. KRISHNAN, AND SHOBHA MADAN

3. The Equal Interval Cases

We will prove both implications of Fuglede’s conjecture for the setsΩ3 = [0, 1] ∪ [a, a+ 1] ∪ [b, b+ 1] and Ω4 = [0, 2] ∪ [a, a+ 1] ∪ [b, b+ 1].An essential ingredient of our proofs will be the following theorem ontesselation of integers.

Theorem 3.1. (Newman) Let a1, a2, ..., ak be distinct integers with k =pα, p a prime and α a positive integer. For each pair ai, aj; i 6= j, let eijdenote the largest integer so that peij/(ai−aj). Then the set a1, a2, ...aktesselates iff there are at most α distinct eij’s.

Tiling implies Spectral: Case 3

Suppose Ω3 tiles R by translations. It is easy to see that a, b ∈ Z. Weapply Newman’s Theorem to the integers 0, a, b, so that p = 3, α = 1.Let a = 3jn and b = 3km with n, m not divisible by 3. Our hypothesisimplies that j = k and that m − n is not divisible by 3. It followsthat if a = 3j(3r + 1), then b = 3j(3s + 2), r, s ∈ Z. But then wecan check easily that the set Λ = Z ∪ (Z + 1

3j+1 ) ∪ (Z + 23j+1 ) is a

spectrum for Ω3. To check completeness, suppose that f ∈ L2(Ω)satisfies

⟨f, e2πiλ.

⟩= 0 ∀λ ∈ Λ. Define 1−periodic functions by

(1)

f#1 (x)

f#2 (x)

f#3 (x)

=

1 1 11 ω ω2

1 ω2 ω

f(x)f(x+ a)f(x+ b)

where x ∈ [0, 1]. Then check that⟨f, e2πiλ.

⟩= 0, ∀λ ∈ Λ iff⟨

f#j , e

2πik.⟩= 0, ∀k ∈ Z, j = 1, 2, 3 iff f#

j = 0, j = 1, 2, 3 iff f = 0.

Tiling implies Spectral: Case 2b

Assume that Ω4 tiles R by translations. Again it is easy to see thata, b ∈ Z. This time we use Newman’s theorem for the integers 0, 1, a, bwith p = 2, α = 2. Let a = 2j(2n+ 1), b = 2k(2m+ 1).

1. If j 6= 0, k 6= 0, the set of ei,j ’s contains 0, j, k. so j = k, butthen a− b = 2j.2(n−m), the set still has cardinality 3, and thereforecannot tile.2. Suppose that j = 0, k 6= 0. Then consider the four integers 0, 1, a =2n + 1, b = 2k(2m + 1) which tesselate Z. If n = 2l−1r with l ≥ 1andr an odd integer, the set of ei,j’s is the set 0, k, l. Hence k = l.

ON FUGLEDE’S CONJECTURE FOR THREE INTERVALS 7

3. Let j, k = 0. We see then that there are at least three distinct ei,j ’s,so that this case is not possible.We conclude that if Ω4 tiles, then a = 2lr + 1, b = 2ls, where

r, s ∈ 2Z+ 1 and then it is easy to see that the set Λ = Z∪Z+ 12l+1 is

a spectrum.

Proposition 3.2. Let Ω = A∪B ∪C; |A| = |B| = |C| = 1/3. SupposeΩ is a spectral set. Then Ω tiles R by translations.

Proof. Suppose Ω = [0, 13] ∪ [a, a + 1/3] ∪ [b, b + 1/3]. Without loss of

generality, let 0 ∈ Λ ⊂ Λ− Λ ⊂ ZΩ

χΩ(λ) = e2πiλ3 − 1 + e2πiλ(a+

13) − e2πiλa + e2πiλ(b+

13) − e2πiλb

= (e2πiλ3 − 1)(1 + e2πiλa + e2πiλb)

Hence, if λ ∈ ZΩ, then either λ ∈ 3Z or 1 + e2πiλa + e2πiλb = 0Let

Z1Ω = λ : e2πiλa = ω, e2πiλb = ω2

Z2Ω = λ : e2πiλa = ω2, e2πiλb = ω

We collect some easy facts in the following lemma:

Lemma. Let λ1 ∈ Z1Ω, λ2 ∈ Z2

Ω, then the following hold:

(1) −λ1, 2λ1 ∈ Z2Ω and −λ2, 2λ2 ∈ Z1

Ω

(2) λ1 − λ2 ∈ Z2Ω and −λ1 + λ2 ∈ Z1

Ω,(3) λ1 + 3λ1Z ⊂ Z1

Ω, λ2 + 3λ2Z ⊂ Z2Ω ; also λ2 + 3λ1Z ⊂ Z2

Ω λ1 +3λ2Z ⊂ Z1

Ω.(4) Let α be the smallest positive number in Z1

Ω. Then Z1Ω =

α + 3αZ, Z2Ω = 2α + 3αZ

Hence, we have

ZΩ = 3Z ∪ Z1Ω ∪ Z2

Ω = 3Z ∪ (α + 3αZ) ∪ (2α + 3αZ)

Note that elements of 3Z are mutually orthogonal, but this set alonecannot be the full spectrum.Now consider two elements λ, λ′ ∈ Z1

Ω. Then λ − λ′ = 3α(n −m), n,m ∈ Z and also e2πi(λ−λ′)a = 1. Thus the only way that twoelements of Z1

Ω can be orthogonal is if λ − λ′ ∈ 3Z. We also get thata, b ∈ Z/3. Now since αa ∈ Z+1/3, we must have α ∈ Q, say α =

pq

with (p, q) = 1 . We conclude that

(λ1 + 3Z) ⊆ (α + 3αZ), (2λ1 + 3Z) ⊆ (2α + 3αZ),

and infact,Λ ⊆ 3Z ∪ (λ1 + 3Z) ∪ (2λ1 + 3Z) ⊆ ZΩ

8DEBASHISH BOSE, C.P. ANIL KUMAR, R. KRISHNAN, AND SHOBHA MADAN

Putting λ1 = α + 3αk, for some k ∈ Z, we see easily that3pqZ + 3Z ⊆ 3p

qZ. But this is possible only if p = 1. It follows that

a = 3n+13q, b = 3m+2

3q. We end the proof by a simple application of

Newman’s Theorem to the three integers 0, (3n+ 1)q, (3m+ 2)q, withp = 3, α = 1, to conclude that Ω3 tiles R.

Proposition 3.3. Let Ω = A ∪ B ∪ C; |A| = 12, |B| = 1

4, |C| = 1

4.

Suppose Ω is a spectral set. Then Ω tiles R by translations.

Proof. Let Ω3 = [0, 2] ∪ [a, a + 1] ∪ [b, b + 1] be a spectral set. ByTheorem of [PD] [Pedersen-Jorgensen] a, b ∈ Z, so that Ω3 = [0, 2] ∪[M,M + 1] ∪ [N,N + 1].

χΩ(λ) = (e2πiλ − 1)(1 + e2πiλ + e2πiλM + e2πiλN)

HenceZΩ = Z ∪ Z1

Ω ∪ Z2Ω ∪ Z3

Ω,

where

Z1Ω = λ : 1 + e2πiλ = 0, e2πiλN + e2πiλM = 0

Z2Ω = λ : 1 + e2πiλN = 0, e2πiλM + e2πiλ = 0

Z3Ω = λ : 1 + e2πiλM = 0, e2πiλN + e2πiλ = 0

We collect some facts, which are easy to verify:

(1) λ ∈ Z1Ω iff λ ∈ Z+ 1/2 and M −N is odd. We let N = 2n+ 1

and M = 2m. Then either Z1Ω = φ or Z1

Ω = Z+ 1/2.(2) If Z1

Ω 6= φ, then Z+ 1/2 ⊆ Z2Ω.

(3) If λ, λ′ ∈ Z3Ω are such that eλ and eλ′ are mutually orthogonal,

then λ− λ′ ∈ Z/2.(4) Λ ∩ Z1

Ω 6= φ. Hence Λ = Z/2 ∪ Λ ∩ Z3Ω

(5) λ ∈ Λ ∩ Z3Ω iff 2λn, 2λm ∈ Z+ 1/2.

From these facts we deduce that if j is the largest integer such that2j|n, this is also true for m.Finally we see that if Ω is spectral, then N = 2j+1r + 1, b = 2j+1s,

with r, s ∈ 2Z+ 1. Newman’s theorem once again says that the set ofintegers 0, 1, 2j+1r + 1, 2j+1s tesselate Z.

ON FUGLEDE’S CONJECTURE FOR THREE INTERVALS 9

4. Structure of the Spectrum for n intervals

Let Ω = ∪nj=1[aj , aj + rj),

∑nj=1 rj = 1. In this section we assume

that Ω is a spectral set with spectrum Λ. In one dimension, all knownexamples of spectra are rational and periodic, though it is not knownwhether it has to be so. We will prove below that if, we assume that Λis a subset of some discrete lattice L, then Λ is rational. A consequenceof this hypothesis is that then Λ contains arbitrarily long arithmeticprogressions (Szemeredi’s theorem[S]). Clearly we may assume that 0 ∈Λ. Then 0 ∈ Λ ⊂ Λ− Λ ⊂ ZΩ, where ZΩ = t ∈ R : χΩ(t) = 0 ∪ 0.

Proposition 4.1. If ZΩ contains an arithmetic progression of length2n+ 1, say 0, d, 2d, · · · , 2nd then

(1) the complete arithmetic progression dZ ⊂ ZΩ,(2) d ∈ Z, and(3) Ω d-tiles R.

Proof. Note that if t ∈ ZΩ, then∑n

j=1[e2πit(aj+rj) − e2πitaj ] = 0

The hypothesis says that χΩ(ld) = 0; l = 1, 2, ..., 2n, hence

∑n

j=1[e2πild(aj+rj) − e2πildaj ] = 0, l = 1, 2, ...2n

We write ζ2j = e2πidaj ; ζ2j−1 = e2πid(aj+rj); j = 1, 2, ..., n;, then theabove system of equations can be rewritten as

(2)

ζ1 − ζ2 + ζ3 − ζ4 + · · ·+ ζ2n−1 − ζ2n = 0ζ21 − ζ22 + ζ23 − ζ24 + · · ·+ ζ22n−1 − ζ22n = 0

...ζ2n1 − ζ2n2 + ζ2n3 − ζ2n4 + · · ·+ ζ2n2n−1 − ζ2n2n = 0

Equivalently,

(3)

1 1 · · · 1 1ζ1 ζ2 · · · ζ2n−1 ζ2n...

.... . .

......

ζ2n−21 ζ2n−2

2 · · · ζ2n−22n−1 ζ2n−2

2n

ζ2n−11 ζ2n−1

2 · · · ζ2n−12n−1 ζ2n−1

2n

ζ1−ζ2...

ζ2n−1

−ζ2n

=

00...00

Since (ζ1,−ζ2, .....− ζ2n)T 6= 0, we have

10DEBASHISH BOSE, C.P. ANIL KUMAR, R. KRISHNAN, AND SHOBHA MADAN

(4)

2n∏

1=i<j

(ζi − ζj) = 0

Next, we need a lemma. We will say (ζi, ζj) is a solution of (4) ifζi − ζj = 0, for some i 6= j. The solution of (4) is good if i + j is oddand is bad if i+ j is even.

Lemma 4.2. If (4) holds, then there exists a good solution of (4)

Proof. Suppose not. Then all solutions of (4) are bad. Since the so-lution set of (4) is non-empty, without loss of generality let ζ1 = ζ3.Then the system reduces to

(5)

1 1 · · · 1 1ζ2 ζ3 · · · ζ2n−1 ζ2n...

.... . .

......

ζ2n−32 ζ2n−3

3 · · · ζ2n−32n−1 ζ2n−3

2n

ζ2n−22 ζ2n−2

3 · · · ζ2n−22n−1 ζ2n−2

2n

−ζ22ζ3...

ζ2n−1

−ζ2n

=

00...00

hence

(6)2n∏

2=i<j

(ζi − ζj) = 0

Note that any solution(good) of (6) is a solution(good) of (4). Thusif (6) has a good solution that would lead to a contradiction.So (6) does not have any good solution. Repeating this 2n− 2 times

we would be left with a 2 × 2 Vandermonde matrix which is singularas

(7)

(1 1ζi ζj

)(nζi−nζj

)=

(00

)

where i is odd and j is even. This implies that ζi = ζj with i + jodd. This is a contradiction to our assumption .

Now we can complete the Proof of Proposition 4.1

Proof. Observe that whenever we get a good solution of (2.3), (2.2)reduces to an (n−2)×(n−2) Vandermonde matrix. Then by the samearguments the reduced matrix would have a good solution. Repeatingthis process n times, we get a partition of ζi into n distinct pairs

ON FUGLEDE’S CONJECTURE FOR THREE INTERVALS 11

(ζi, ζj) such that each (ζi, ζj) is a good solution of (4). But then ζki =ζkj , ∀k ∈ Z Hence by relabelling, if necessary,

(8) χΩ(kd) =

n∑

j=1

(ζk2j−1 − ζk2j) = 0 ∀ k ∈ Z

Thus dZ ⊂ ZΩ. Now consider

(9) F (x) =∑

k∈Z

χΩ (x+ k/d) , x ∈ [0, 1/d)

Thus F is 1dperiodic and integer valued thus

(10) F (ld) = d∑

k∈Z

∫ 1d

0

1Ω(x+k

d)e−2πildxdx

= dχΩ(ld) = d δl,0

Thus F (t) = d a.e. so d ∈ Z and Ω d-tiles the real line.

Using Proposition 2.1, we now prove the corresponding result for thespectrum.

Proposition 4.3. Suppose Ω is spectral and Λ a spectrum with 0 ∈ Λ.If for some a, d ∈ R, a, a + d, ..., a+ 2nd ∈ Λ, then a+ dZ ⊆ Λ.

Proof. Since

a, a+ d, ..., a+ 2nd ∈ Λ

Shifting Λ by a we get Λ1 = Λ− a is a spectrum for Ω and

0, d, ..., 2nd ∈ Λ1 ⊂ Λ1 − Λ1 ⊂ ZΩ

Thus surely dZ ⊂ ZΩ by Proposition 4.1. Now let λ ∈ Λ1.Then

−λ, d− λ, 2d− λ, ..., 2nd− λ ∈ ZΩ

Let

ξ2j = e−2πiλaj , ξ2j−1 = e−2πiλ(aj+rj); j = 1, ..., n

ζ2j = e2πidaj , ζ2j−1 = e2πid(aj+rj); j = 1, ..., n

Thus we have

(11) ξ1ζk1 − ξ2ζ

k2 + ...+ ξ2n−1ζ

k2n−1 − ξ2nζ

k2n = 0 for k = 0, ..., 2n.

Since the ζi′s can be partitioned into n disjoint pairs (ζi, ζj) such that

ζi = ζj and i + j odd, without loss of generality, let ζ2j−1 = ζ2j , j =1, 2, ..., n. Thus (11) can be written as

12DEBASHISH BOSE, C.P. ANIL KUMAR, R. KRISHNAN, AND SHOBHA MADAN

(12)

1 1 · · · 1ζ1 ζ3 · · · ζ2n−1...

.... . .

...ζn−11 ζn−1

3 · · · ζn−12n−1

ξ1 − ξ2ξ3 − ξ4

...ξ2n−1 − ξ2n

=

00...0

Now if [ξ1 − ξ2, ξ3 − ξ4, ..., ξ2n−1 − ξ2n]T is the trivial solution i.e.

ξ2j−1 − ξ2j = 0, ∀ j = 1, ..., n, then ∀ k ∈ Z,

χΩ(kd− λ) = ξ1ζk1 − ξ2ζ

k2 + · · ·+ ξ2n−1ζ

k2n−1 − ξ2nζ

k2n

= ζk1 (ξ1 − ξ2) + · · ·+ ζk2n−1(ξ2n−1 − ξ2n) = 0

Thus dZ− λ ∈ ZΩ.

If however [ξ1− ξ2, ξ3− ξ4, ..., ξ2n−1− ξ2n]T is not the trivial solution,

then ζ2l+1 = ζ2k+1 for some l, k ∈ 1, ..., n; l 6= kRemoving all the redundant variables we get a non-singular Vander-

monde matrix satisfying

(13)

1 1 · · · 1η1 η3 · · · η2k−1...

.... . .

...ηk−11 ηk−1

3 · · · ηk−12k−1

∑1∑3...∑2k−1

=

00...0

where ∑k=

∑

j:ζ2j−1=ηk

ξ2j−1 − ξ2j

and such that the above matrix (ηkj ) is non-singular. Then each of the∑i = 0, i = 1, · · · , k. But then once again

χΩ(pd− λ) = ηp1∑

1+ ηp3

∑3+ · · ·+ ηp2k−1

∑2k−1

= 0 ∀p ∈ Z

Thus dZ − λ ∈ ZΩ. We already have dZ ⊆ ZΩ and now we have seenif λ ∈ Λ1, then dZ− λ ∈ ZΩ. Thus dZ ⊆ Λ1, hence a+ dZ ⊂ Λ.

With the above Proposition it is clear that we need to explore con-ditions under which the spectrum, when it exists, contains Arithmeticprogressions. Note that Λ has positive asymptotic density, so that ifΛ is a subset of a lattice, the well-known theorem of Szemeredi[S],guarantees the existence of APs of arbitrary lengths. In the followinglemma, we are able to relax this condition to a local condition.

ON FUGLEDE’S CONJECTURE FOR THREE INTERVALS 13

Lemma 4.4. . If Λ is a spectrum, such that Γ = Λ−Λ is δ-separated,then Λ contains a complete arithmetic progression. Further, Λ is con-tained in a lattice with a base.

Proof. : The δ-separability of Γ means that for some δ1 > 0

inf|γj − γk| : γj, γk ∈ Γ, γj 6= γk = δ1 > 0

Let δ = δ1/2 and consider the map from Λ to the lattice δZ given by

ψ(λ) = [λ

δ]δ

Then the subset ψ(Λ) has positive density, hence by [S], given any N,there exists an arithmentic progression of length N , say

ψ(λ1), ψ(λ2), ..., ψ(λN),

with ψ(λj+1) = ψ(λj)+dδ, d ∈ Z, j = 1, 2, ...N−1. But then λj+1−λjmust lie in the interval (d − δ, d + δ) for each j = 1, 2, ...N − 1. Sincethis interval has length δ1, all the above elements must be the same (inany interval of length δ1, there can be at most one element of Λ − Λ).But this means that the λj’s are in arithmetic progression. Now usingProposition 4.3, we get that Λ contains the complete AP dZ.Let Λs := λn+1 − λn|λn ∈ Λ be the set of successive differences of

spectral sequences, the spectral gaps. We will show that Λs is finite.Λs ⊆ Λ − Λ ⊆ ZΩ. As Ω is measurable and of finite measure thereexists a neighbourhood around 0 which does not intersect ZΩ. Thus Λs

is bounded below .But Λs is bounded above [see IP]. So by the compactness of Ω we

get that Λs is finite as χΩ can be extended analytically to the entirecomplex plane and zeros of an entire function are isolated.

Let Λs = r1, r2, ..., rk. The set of solutions for∑k

i=1 airi = dwith ai ∈ N ∪ 0 is finite. Thus Λ ⊆ 0, b1, ..., bl + dZ for somel. ie Λ is contained in a lattice with a base. So we have dZ ⊆ Λ ⊆0, b1, ..., bl+ dZ

We end this section by showing that the hypothesis Λ−Λ ⊆ L givesmore information on the spectrum

Theorem 4.5. If Ω as above is spectral, with spectrum Λ such that0 ∈ Λ ⊂ Λ− Λ ⊂ L, L a lattice, then Λ is rational.

Proof. : Let L = θZ, and suppose Λ ⊆ θZ. Since Λ is a spectrum ithas asymptotic density 1. By Szemeredi’s theorem Λ contains arbitrar-ily long arithmetic progressions, and so in particular, of length 2n+1.

14DEBASHISH BOSE, C.P. ANIL KUMAR, R. KRISHNAN, AND SHOBHA MADAN

Without loss of generality, suppose

0, Kθ, ..., 2nKθ ∈ Λ ⊆ ZΩ

where K ∈ Z. We already know that if Λ contains an arithmeticprogression of length 2n then the complete arithmetic progression is inΛ and the common difference d ∈ N. Hence Kθ ∈ Z, which means thatθ ∈ Q. So Λ ⊆ Q.

5. Three Intervals: Spectral implies Tiling

Consider now the particular case of three intervals. Let Ω = [0, r1)∪[a2, a1 + r2) ∪ [a3 + r3), r1 + r2 + r3 = 1. Suppose that Ω is spectraland that its spectrum Λ contains an arithmetic progression of length 6and common difference d. By translating if necessary we may assumethat dZ ⊂ Λ, d ∈ Z, and so by Proposition 4.1, Ω d−tiles R. Ifd = 1, Λ = Z, and it is well-known that then Ω tiles R by Z. Ifd 6= 1, letλ ∈ Λ \ dZ. Put

ζ2j−1 = e2πid(aj+rj), ζ2j = e2πidaj

ξ2j−1 = e2πiλ(aj+rj), ξ2j = e2πiλaj

We know that among the ζj ’s there are good pairs. After reindexingsuppose ζ2j−1 = ζ2j . Then by orthonormality we have

(14)

1 1 1ζ1 ζ3 ζ5ζ21 ζ23 ζ25

ξ1 − ξ2ξ3 − ξ4ξ5 − ξ6

=

000

Let A denote the linear transformation represented by the abovematrix. We consider three cases:

Case 1. Rank A=3. In this case, ξ1 = ξ2, ξ3 = ξ4 and ξ5 = ξ6. Butthen it follows that Λ is a group, and also that Λ ⊆ Z. But Λ hasdensity 1, we must have Λ = Z.

Case 2. Rank A=2. In this case, without loss of generality , assumethat ζ1 = ζ5, then

(15)

(1 1ζ1 ζ3

)(ξ1 − ξ2 + ξ5 − ξ6

ξ3 − ξ4

)=

(00

)

Hence

ξ1 − ξ2 + ξ5 − ξ6 = 0 and ξ3 − ξ4 = 0

ON FUGLEDE’S CONJECTURE FOR THREE INTERVALS 15

It follows that in this case

ZΩ = ZΩ(1) ∪ ZΩ(2) ∪ ZΩ(3)

where

ZΩ(1) = λ : ξ3 = ξ4, ξ1 = ξ2, ξ5 = ξ6

ZΩ(2) = λ : ξ3 = ξ4, ξ1 = ξ6, ξ2 = ξ5

ZΩ(3) = λ : ξ3 = ξ4, ξ1 = −ξ5, ξ2 = −ξ6

andΛ− Λ ⊂ ZΩ

We proceed as in ([L1]) to conclude that

(1) Either Λ ⊂ ZΩ(1) ∪ ZΩ(3) or Λ ⊂ ZΩ(2) ∪ ZΩ(3),(2) Λ ⊂ Λ− Λ ⊂ ZΩ(3) ∪ (ZΩ(1) ∩ ZΩ(2))(3) ZΩ(1) ∩ ZΩ(2) = KZ, since this set is a subgroup of Z, and(4) Λ ∩ ZΩ(3)− Λ ∩ ZΩ(3) ∈ ZΩ(1) ∩ ZΩ(2) \ ZΩ(3)

It follows that Λ∩ZΩ(3) ⊆ β + kZ. Then by density considerations,we get that Λ = kZ ∪ (kZ + β), with k = 2. Then Spectral impliesTiling follows from [P] and [PW].

Case 3. Rank A=1In this case, ζ1 = ζ3 = ζ5 = ζ2 = ζ4 = ζ6. Then

e2πida1 = e2πid(a1+r1) = e2πida2 = e2πid(a2+r2) = e2πida3 = e2πid(a3+r3)

Taking a1 = 0 we get

a1 = 0, a2 =l2d, a3 =

l3d; r1 =

k1d, r2 =

k2d, r3 =

k3d

where l2, l3, k1, k2, k3 ∈ Z and k1 + k2 + k3 = d.It follows from this that we are in the case of d equal intervals, in

three groups. But then the spectrum is periodic and of the form

Λ = L+ dZ

If d = 1 then Λ = Z, so there is nothing to prove. If d = 2, thenr1, r2, r3 ≥ 1/2, which is not possible. hence, we may assume thatd ≥ 3. Now if λ ∈ Λ,

e2πiλa1 − 1 + e2πiλ(a2+r2) − e2πiλa2 + e2πiλ(a3+r3) − e2πiλa3 = 0

If we assume further that Λ ⊆ Q, this is a case of six roots of unitysumming to 0, say α1+· · ·+α6 = 0. Poonen and Rubinstein [PR] ( haveclassified all minimal vanishing sums of roots of unity α1+ · · ·+αn = 0of weight n ≤ 12 (see also[LL]). There are three possible ways in whichsix roots of unity (α1, α2, ..., α6) can sum up to zero:

16DEBASHISH BOSE, C.P. ANIL KUMAR, R. KRISHNAN, AND SHOBHA MADAN

(T1) ( a 2 sub-sum = 0). ασ(1)+ασ(2) = ασ(3)+ασ(4) = ασ(5)+ασ(6) =0.

(T2) (a 3 sub-sum = 0). ασ(1)+ασ(2)+ασ(3) = ασ(4)+ασ(5)+ασ(6) = 0.(T3) (no sub-sum = 0). ασ(n) = ρn; n = 1, · · · , 4; ασ(5) = −ω, ασ(6) =

−ω2 (after normalizing ) where ρ is a fifth root of unity and ωis a cube root of unity.

Here σ is some element of S6, the group of permutations of 6 objects.It turns out that there are now many possibilities and we use a

symbolic computation explained in the next section.

6. A symbolic computation using Mathematica

We begin with the setting of Case 3 in the previous section, whereRankA = 1. Let d be the smallest positive integer in ZΩ such thatdZ ⊆ Λ, and let Λ = L+dZ = ∪d

j=1λj+dZ, with λ1 = 0. We also have

d ≥ 3. Observe that if d′ ∈ ZΩ is such that e2πid′aj = e2πid

′(aj+rj) = 1for all j ∈ 1, 2, 3, then d′ ∈ dZ.

We now introduce some notation and explain the analysis behindthe computation carried out. First we map ZΩ into C6; λ → vλ =(e2πiλ(a1+r1),−e2πiλa1 , e2πiλ(a2+r2),−e2πiλa2 , e2πiλ(a3+r3),−e2πiλa3). In par-ticular v0 = (1,−1, 1,−1, 1,−1)

Define a conjugate bilinear form onC6 as follows.For v = (x1, x2, ..., x6),w = (y1, y2, ..., y6), (skew dot product)

SDP (v, w) = x1y1 − x2y2 + x3y3 − x4y4 + x5y5 − x6y6

Note that SDP (vλ, v0) = 0 ∀λ ∈ ZΩ. Let

G(v, w) = (x1y1,−x2y2, x3y3,−x4y4, x5y5,−x6y6)

We will say that a vector vλ is of Type1, Type2 or Type3 if it satisfies(T1), (T2) or (T3) respectively, listed at the end of the last section.

We make some observations

(1) SDP (vλ, v0) = 0, ∀λ ∈ ZΩ.(2) G(vλ1 , vλ2) = vλ1−λ2 .(3) If λ ⊂ Q, then all components of vλ, λ ∈ λ are roots of unity.(4) Since a1 = 0, the second coordinate in the image of ZΩ in C6 is

always −1.(5) The image of Λ in C6 consists of precisely d elements corre-

sponding to the different cosets of dZ (for, if vλ1 = vλ2 , thenG(vλ1 , vλ2) = v0, so λ1 − λ2 ∈ dZ.

ON FUGLEDE’S CONJECTURE FOR THREE INTERVALS 17

The computation is done under the following assumption:

Assumption: For λ1, λ2 ∈ Λ, vλ1−λ2 is of Type 1 if and onlyif λ1 − λ2 ∈ dZThrough the symbolic computation we shall investigate the maxi-

mum possible cardinality of a set λ1, λ2, ..., λd such that λi−λj ∈ ZΩ

and such that vλi−λjis a vector of either Type 2 or Type 3.

The first case of this investigation is analyzed below.

Case 1. vλ1 , vλ2 are both Type 2 vectors. First we make some

observations.1. if (α1, α2, ..., α6) are 6 roots of unity such that their sum is zero

and if ασ(1) + ασ(2) + ασ(3) = 0 thenασ(1)

ασ(2),

ασ(2)

ασ(3),

ασ(3)

ασ(1)are powers of ω

where ω3 = 1.

2. if (α1, α2, ..., α6) is as above and has no subsum zero i.e. it is aType3 vector, then it has to be a permutation of (xρ, xρ2, xρ3, xρ4,−xω,− xω2) where x is some root of unity. So there cannot be two pairsamong these six elements whose ratios are powers of −ω.

3. If the vector (±ω∗,±ω∗,±ω∗,±ω∗,±ω∗,±ω∗) where ω∗ is somepower of ω, with exactly 3 positive signs and 3 negative signs has sumzero then it has to be of Type1, because Type2 would imply that it isa permutation of (1, ω, ω2,−1,−ω,−ω2) which is a Type1 vector.

4. For a Type2 vector all elements in a 3 subsum adding up to zerohave the same sign.

Let u[x] = (1, ω, ω2, x, xω, xω2) and u[y] = (1, ω, ω2, y, yω, yω2) betwo vectors of Type2 where x and y are roots of unity. Take theconjugate SDP of these two vectors by permuting and conjugating thesecond vector u[y]. The terms in the conjugate SDP, ignoring the signs,will fall into one of the following four categories upto a permutation ofthe first 3 elements and last 3 elements.

1. (ω∗, ω∗, ω∗, xyω∗, xyω∗, xyω∗)2. (yω∗, yω∗, yω∗, xω∗, xω∗, xω∗)3. (ω∗, ω∗, yω∗, xω∗, xyω∗, xyω∗)4. (ω∗, yω∗, yω∗, xω∗, xω∗, xyω∗)

Here ω∗ represents some power of ω. The first case arises when wetake conjugate SDP of u[x] and permuted u[y] which is of the type(ω∗, ω∗, ω∗, yω∗, yω∗, yω∗). The second case is similar. The third casearises when we take the conjugate SDP of u[x] and permuted u[y] which

18DEBASHISH BOSE, C.P. ANIL KUMAR, R. KRISHNAN, AND SHOBHA MADAN

is of the type (ω∗, ω∗, yω∗, ω∗, yω∗, yω∗) upto a permutation of the first3 elements and last 3 elements. Finally the fourth case is again similar.

In all cases, after putting in the signs, they are not of type 3, as thereare atleast two pairs whose ratios are powers of −ω.

Now we need to consider only two cases.

Case 1. (ω∗, ω∗, ω∗, xyω∗, xyω∗, xyω∗)If there is a 3 subsum being zero, after the signs are put in appro-

priately, which involves ω∗ and xyω∗ then the ratio (xyω∗/ω∗) = xy isa power of −ω. Hence the set (ω∗, ω∗, ω∗, xyω∗, xyω∗, xyω∗) becomes(±ω∗,±ω∗,±ω∗,±ω∗,±ω∗,±ω∗) with exactly 3 positive signs and 3negative signs. So the terms in this SDP form a Type1 vector. Hencethis possibility is not considered. So if there is a 3 subsum being zerothen it should be ω∗ + ω∗ + ω∗ = 0. and xyω∗ + xyω∗ + xyω∗ = 0.The 3 pluses occur with first 3 elements and 3 minuses occur withthe last 3 elements or viceversa. Once u[x] is fixed with these signs(1, ω, ω2, y, yω, yω2) can be permuted in the first 3 and last 3 elements.In this case x and y can be any root of unity other than −1,−ω,−ω2.

Case 3. (ω∗, ω∗, yω∗, xω∗, xyω∗, xyω∗)If there is a 3 subsum being zero after the signs are put in appropri-

ately, then the 3 subsum which involves ω∗ has to involve one of theterms yω∗, xω∗, xyω∗. So either x is a power of −ω or y is a powerof −ω or xy is a power of −ω. If xy is a power of −ω then bothyω∗, xω∗ will also be involved in a 3 subsum which has ω∗. So both xand y are powers of −ω. Hence the set (ω∗, ω∗, yω∗, xω∗, xyω∗, xyω∗)becomes (±ω∗,±ω∗,±ω∗,±ω∗,±ω∗,±ω∗) with exactly 3 positive signsand 3 negative signs. So the terms in this SDP form a Type1 vector.Hence this possibility is not considered. Hence either x is a power of−ω or y is a power of −ω but not both.

If x is a power of −ω then x is a power of ω as u[x] is a Type2vector and the 3 subsum would be ω∗ + ω∗ + ω∗ = 0. and yω∗ +yω∗ + yω∗ = 0. The 3 pluses occur with elements which involve y and3 minuses occur with the other 3 elements or viceversa. So, keepingu[y] = (1, ω, ω2, y, yω, yω2) fixed, the 3 pluses occur with first 3 ele-ments and 3 minuses occur with the last three elements or viceversa.The vector u[x] = (1, ω, ω2, x, xω, xω2) with x = ω∗ is itself a permu-tation of (1, ω, ω2, 1, ω, ω2). Now u[x] has to be permuted in such away that the terms in the SDP which involve y have to be a permu-tation of ±(y, yω, yω2). Hence permuted u[x] has to be of the form

ON FUGLEDE’S CONJECTURE FOR THREE INTERVALS 19

(σ(1), σ(omega), σ(ω2), µ(1), µ(ω), µ(ω2)) where σ and µ are permuta-tions of (1, ω, ω2). In this case y can be any root of unity other than−1,−ω,−ω2.

The analysis when y is a power of −ω is identical.So all the cases where the two Type2 vectors u[x] and u[y] can have

conjugate SDP zero and the terms in the SDP forms a Type2 vector,reduce to the case where the 3 pluses occur at the first 3 positions and3 minuses occur at the last 3 positions or viceversa, with u[x] fixed andu[y] permuted among the first 3 positions and last 3 positions or whereall the permuted y terms in u[y] occur in the first 3 positions and otherpermuted 3 terms occur at the last 3 positions, as given below

(16)

+1 +1 +1 −1 −1 −11 ω ω2 x xω xω2

σ(1) σ(ω) σ(ω2) yµ(1) yµ(ω) yµ(ω2)

It follows that if the spectrum contains only vectors of Type1 andType2, then d ≤ 3. This follows from the computation given at theend ofhttp://www.imsc.res.in/ ∼ rkrishnan/FugledeComputation.html -Type2withType2.nb

Now in the case when Λ contains one vector of Type3, Two casescan arise

1. Λ contains only vectors of Type1 and Type3, then from the Math-ematica symbolic computation, available in the filehttp://www.imsc.res.in/∼rkrishnan/FugledeComputation.html -Type3withType3.nb we conclude again the d = 3.

2. Lastly if Λ contains vectors of Type 3 as well as of Type 2, thenthere can be at most one coset coming from each type. The details ofthis computation are available athttp://www.imsc.res.in/∼rkrishnan/FugledeComputation.html -Type3withType2.nb and vwithuandv1.nb

We conclude that d = 3 in all the above cases, which reduces tothe case of three equal intervals, for which we have already proved theconjecture.

We have thus proved the spectral implies tiling part of Fuglede’sconjecture for three intervals under two assumptions, namely (a) Λ iscontained in a lattice, and (b) (Λ− Λ)∩ Type1 = dZ.

20DEBASHISH BOSE, C.P. ANIL KUMAR, R. KRISHNAN, AND SHOBHA MADAN

Final Remark. Note that the additional assumptions made on thespectrum are used to reduce the Rank A = 1 case to the case of threeequal intervals. However, without any additional assumptions on Λ,this case still corresponds to an equal interval case grouped together inthree bunches.

Acknowledgement. The first author would like to thank Biswaran-jan Behera for useful discussions during the initial stages of this work.

References

[FMM] Farkas, B., Mattolcsi, M. and Mora, P., On Fuglede’s Conjecture and theExistence of Universal Spectra, J. Fourier Anal. Appl. 12(2006)483-494.

[FR] Farkas, B. and Revesz, Sz.Gy. Tiles with no spectra in dimension 4, Math.Scand. 98(2006), 44-52.

[F] Fuglede, B., Commuting self-adjoint partial differential operators and agroup theoretic problem, J. Funct. Anal. 16(1974)101-121.

[IKT] Iosevich, A.,Katz, N.H., Tao,T., Fuglede conjecture holds for convex bodiesin the plane. Mat. Res. Letters 10(2003)559-570.

[IP] Iosevich, A. and Pedersen, S., How large are the spectral gaps?, Pacific J.Math. 192(2000),307-314.

[JP] Jorgensen, P. and Pedersen, S., Fractal Geometry and Stochastics, Progressin Probability, 17(1995)151-219.

[K1] Kolountzakis, M., Non-symmetric convex domains have no basis of expo-nentials, Ill. J. Math. 44(2000), 542-550.

[KL] Kolountzakis, M.and Laba,I., Tiling and spectral properties of near-cubicdomains, Studia Math. 160 (2004), 287–299.

[KM1] Kolountzakis, M. and Matolcsi, M., Tiles with no spectra. Forum Math.18(2006) 519-528.

[KM2] Kolountzakis, M. and Matolcsi, M., Complex Hadamard matrices and thespectral set conjecture, Collect. Math. Vol. Extra(2006) 281-291.

[L1] Laba, I., Fuglede’s conjecture for a union of two intervals, Proc. AMS.121(2001) 2965-2972.

[L2] Laba, I., The spectral set conjecture and multiplicative properties of rootsof polynomials. J. London Math. Soc 65(2002)661-671.

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

[LL] Lam, T.Y. and Leung, K.K., On vanishing sums of roots of unity. FiniteFields Appl. 2 (1996), no. 4, 422–438.

[L] Landau, H., Necessary density conditions for sampling and interpolation ofcertain entire functions, Acta Math., 117(1967) 37-52.

[M] Matolsci., Fuglede’s conjecture fails in dimension 4. Proc. Amer. Math. Soc.133 (2005), 3021–3026.

[N] Newman, D.J., Tesselation of integers, J. Number Theory 9 (1977) 107-111.[P] Pedersen, S., Spectral sets whose spectrum is a lattice with a base. J. Funct.

Anal. 141(1996),496-509.

ON FUGLEDE’S CONJECTURE FOR THREE INTERVALS 21

[PW] Pedersen, S. and Wang, Y., Universal spectra, universal tiling sets and thespetral set conjecture, Math. Scand. 88(2001)246-256.

[PR] Poonen, B. and Rubinstein, M., The number of intersection points madeby the diagonals of a regular polygon. SIAM J. Discrete Math. 11(1998)135-156.

[S] Szemeredi,E., On sets of integers containing no k elements in arithmeticprogression. Acta Arith. 27 (1975), 199–245.

[T] Tao, T., Fuglede’s conjecture is false in 5 and higher dimensions. Math.Res. Lett. 11 (2004), no. 2-3, 251–258.

Debashish Bose:India

E-mail address : debashishb@wientech.com

C.P. Anil Kumar: Infosys, Bangalore, India

E-mail address : anilkumar p@infosys.com

R. Krishnan: Institute of Mathematical Sciences, Chennai-600113,

India

E-mail address : rkrishnan@imsc.res.in

Indian Institute of Technology, Kanpur-208116, India

E-mail address : madan@iitk.ac.in