In this note we give an example of a set $\W\subset \R^4$ such that $L^2(\W)$
admits an orthonormal basis of exponentials $\{\frac{1}{|\W |^{1/2}}e^{2\pi i
x, \xi}\}_{\xi\in\L}$ for some set $\L\subset\R^4$, but which does not tile
$\R^4$ by translations. This improves Tao's recent 5-dimensional example, and
shows that one direction of Fuglede's conjecture fails already in dimension 4.
Some common properties of translational tiles and spectral sets are also
proved.Comment: 6 page

Matolcsi, Mate

English

arXiv

In this note we modify a recent example of Tao and give an example of a set Omega subset of R-4 such that L-2(Omega) admits an orthonormal basis of exponentials {1/vertical bar Omega vertical bar 1/2 e(2 pi i )}(xi epsilon Lambda) for some set Lambda subset of R-4, but which does not tile R-4 by translations. This shows that one direction of Fuglede's conjecture fails already in dimension 4. Some common properties of translational tiles and spectral sets are also proved

Matolcsi, Máté

Repository of the Academy's Library

arXiv:math/0611936v1  [math.CA]  30 Nov 2006FUGLEDE’S CONJECTURE FAILS IN DIMENSION 4MA´TE´ MATOLCSIAbstract. In this note we give an example of a set Ω ⊂ R4 suchthat L2(Ω) admits an orthonormal basis of exponentials { 1|Ω|1/2e2pii〈x,ξ〉}ξ∈Λfor some set Λ ⊂ R4, but which does not tile R4 by translations.This improves Tao’s recent 5-dimensional example, and shows thatone direction of Fuglede’s conjecture fails already in dimension 4.Some common properties of translational tiles and spectral sets arealso proved.2000 Mathematics Subject Classification. Primary 42B99, Sec-ondary 20K01.Keywords and phrases. Translational tiles, spectral sets, Fu-glede’s conjecture, Hadamard matrices.1. IntroductionLet Ω ⊂ Rd be a Lebesgue measureable set of finite non-zero mea-sure. A set Λ ⊂ Rd is said to be a spectrum of Ω if the functions{ 1|Ω|1/2e2pii〈x,ξ〉}ξ∈Λ form an orthonormal basis of L2(Ω) (here |Ω| de-notes the Lebesgue measure of Ω). If such Λ exists, Ω is said to bespectral. The set Ω is said to be a (translational) tile if it is possibleto tile Rd with a family of translates {t + Ω : t ∈ Σ} of Ω (ignoringoverlaps and gaps of measure zero).Tiles and spectral sets are connected by Fuglede’s famousConjecture 1.1. [1]A set Ω ⊂ Rd of finite, non-zero Lebesgue measureis a tile if and only if it is spectral.Fuglede [1] proved the conjecture in the special case when the spec-trum Λ or the translation set Σ is assumed to be a lattice. The generalcase of the conjecture was open for nearly 30 years, until last year Tao[13] showed an example to disprove one direction of the conjecture in5 and higher dimensions. Namely, he gave an example of a spectralset in R5 which is not a tile. The aim of this note is to modify Tao’sexample and prove that the conjecture fails already in dimension 4. Weemphasize that Propositions 2.1 and 2.5 are only new in the general-ity formulated below, and credit should be given to [13], where special12 MA´TE´ MATOLCSIcases are proved with the same idea. The essentially new results of thepaper are contained in Propositions 2.2, 2.3 and Theorem 3.1.We will restrict our attention to sets of the form of finite union ofunit cubes placed at points with integer coordinates, i.e.(1) Ω = ∪t∈Γ{t+ [0, 1)d},where Γ ⊂ Zd is finite (this means, essentially, that we will be consid-ering the tiling and spectral properties of sets in Zd). Besides giving acounterexample in dimension 4, we also prove some interesting proper-ties that are shared by spectral sets and tiles of the form (1).To asses our present knowledge concerning this topic it is interestingto note that Fuglede’s conjecture was recently proved in 2 dimensionsfor convex bodies in [5]. Even for finite union of intervals the conjectureis still open in dimesion 1. Some partial results concerning this caseappeared in [10], [11]. In arbitirary dimensions certain classes of non-tiling sets are already known to admit no spectrum, either. Resultsin this direction are contained in [4], [7] and [8]. For further recentresults related to Fuglede’s conjecture see [3], [12] [2] and [6]. Therecent survey [9] gives an overview not only of the Fuglede conjecturebut also of some recent results in the theory of translational tilings.Note, also, that one direction of the conjecture is still open in alldimensions, namely we have no examples of tiles which do not admita spectrum.2. Common properties of tiling and spectral setsIn this section we examine some natural properties of spectral setsand tiles of the form (1).We will need the following notations and definitions.The family of spectral sets (resp. tiles) of the form (1) will be denotedby Sd (resp. T d).A k×k complex matrixH is called a Hadamard matrix if all entries ofH have absolute value 1, and HH∗ = kI (where I denotes the identitymatrix). This means that the rows (and also the columns) of H forman orthogonal basis of Ck. A log-Hadamard matrix is any real squarematrix (hi,j)ki,j=1 such that the matrix (e2piihi,j )ki,j=1 is Hadamard.For a given finite set {t1, . . . tk} ⊂ Zd let T denote the d× k matrixwith jth column tj . (With a slight abuse of notation we will denote theset itself by T as well.) The set T = {t1, . . . tk} is called spectral if thereexists a k×d real matrix L such that LT is log-Hadamard (in the usualterminology of harmonic analysis it is also required that the entries ofL belong to the interval [0, 1) but the definition above will be moreFUGLEDE’S CONJECTURE FAILS IN DIMENSION 4 3convenient for present purposes). The (rows of the) matrix L is the(not necessarily unique) spectrum of T . The set T = {t1, . . . tk} ⊂ Zdis called a tile if it is possible to tile Zd with a family of translates ofT . The family of spectral sets (resp. tiles) of Zd will be denoted by Sd0(resp. T d0 ).The set T = {t1, . . . tk} ⊂ Zd is called m-spectral for some positiveinteger m if it admits a spectrum L with all entries being multiplesof 1/m (which, essentially, means that T is spectral in the group Zdm).The set T is called an m-tile if it is possible to tile the cube [0, m)dwith translates of T mod m. (This means that T tiles the group Zdm.)The family of m-spectral sets (resp. m-tiles) of Zd will be denoted bySdm (resp. Tdm).It is obvious that Sdm ⊂ Sd0 and Tdm ⊂ Td0 , and it is also clear that ifT ∈ T d0 then T +[0, 1)d ∈ T d. It is also well-known that if T ∈ Sd0 withspectrum L = {l1, . . . lk} then Ω = T + [0, 1)d ∈ Sd with spectrum Λ =L+Zd (the orthogonality of the functions e2pii〈λ,x〉 is an easy calculation,while the completeness can be seen by checking the Parseval identityfor functions of the form e2pii〈s,x〉 and noting that the linear span ofthese functions is dense in L2(Ω)).Tao’s example [13] shows that Sd0 6= Td0 and Sd 6= T d for d ≥ 5,therefore Fuglede’s conjecture is false, in general. Nevertheless, it isinteresting to see that Sd0 and Td0 share certain properties. Namely, itis reasonable to look for ’natural’ properties of T d0 and check whetherthe same holds for Sd0 , and vica versa. The first result in this directionestablishes the ’composition property’ for Sd0 and Td0 . This property isalready used in Tao’s example (in less generality, but the idea is thesame).Proposition 2.1. Assume that the set T = {t1, . . . tk} ⊂ Zd is an m-tile (resp. m-spectral) and the set S = {s1, . . . sr} ⊂ Zd is an n-tile(resp. n-spectral). Then the set Γ := T + mS is an mn-tile (resp.mn-spectral).Proof. This property is ’natural’ for tiles, therefore we prove only thespectral part of the proposition.Let L := { l1m, . . . lkm} be the spectrum of T , and Q := { q1n, . . . qrn} bethe spectrum of S. We claim that the spectrum of Γ is Λ = L+ Qm.The cardinality of Λ and Γ is the same (mn), therefore it is enoughto check orthogonality of the functions e2pii〈λ,x〉.To see this write λi1 =li1m+qi1nm, λi2 =li2m+qi2nm, and4 MA´TE´ MATOLCSI(2)∑x∈Γe2pii〈λi1−λi2 ,x〉 =∑t∈T∑s∈Se2pii〈li1−li2m+qi1−qi2nm,t+ms〉.Notice that if qi1 6= qi2 then the second summation gives 0 for allfixed t ∈ T (because of the spectral property of Q with respect to S),therefore the whole sum is 0. On the other hand, if qi1 = qi2 thenthe second summation gives re2pii〈li1−li2m,t〉, and the whole sum equals 0because of the spectral property of L with respect to T . The next two propositions are not directly used in our 4-dimensionalcounterexample below, but they are of independent interest. First weexamine a ’natural’ ascendence property of spectral sets.Proposition 2.2. Assume that for a given set T = {t1, . . . tk} ⊂ Zdthere exists a d1×d matrix L1 with integer entries for which the columnsof L1T form a spectral (resp. m-spectral) set T1 in Zd1 . Then T itselfis spectral (resp. m-spectral).Proof. If L denotes a spectrum of T1 then LL1 is clearly a spectrum ofT , because (LL1)T = L(L1T ).It may be a little surprising that the same property holds also fortiles.Proposition 2.3. Assume that for a given set T = {t1, . . . tk} ⊂ Zdthere exists a d1×d matrix L1 with integer entries for which the columnsof L1T form a tile (resp. m-tile) T1 in Zd1. Then T itself is a tile (resp.m-tile).Proof. We prove the statement for tiles, and remark that the case ofm-tiles is settled exactly the same way.Note first that the elements of T1 must be different from each otherbecause of the tiling condition. Let Σ1 denote the translation set cor-responding to T1, i.e. Σ1 + T1 = Zd1 . Define Σ := L−11 [Σ1] = {σ ∈ Zd :L1σ ∈ Σ1}. We claim that Σ is a good translation set for T .To see this, we check first that the translated copies of T are disjoint.Assume therefore that σ1+ti = σ2+tj for some σ1, σ2 ∈ Σ and ti, tj ∈ T .Applying the transformation L1 we get L1σ1+L1ti = L1σ2+L1tj , andin view of the tiling condition on T1 we conclude that L1ti = L1tj . Thisimplies ti = tj , and hence σ1 = σ2.Next we check that the translates of T cover the whole of Zd. Letz ∈ Zd arbitrary, and let z1 = L1z ∈ Zd1 . Then, by assumption,z1 = σ1 + t1 with some σ1 ∈ Σ1 and t1 ∈ T1. Take the inverse imageFUGLEDE’S CONJECTURE FAILS IN DIMENSION 4 5of t1, i.e. the element ti ∈ T for which L1ti = t1. Then σ := z − ti iscontained in Σ because L1(z − ti) = z1 − t1 = σ1 ∈ Σ1. Let us mention an interesting consequence of this proposition.Corollary 2.4. Let T = {t1, . . . tk} ⊂ Zd be a linearly independent setof vectors (over R). Then T is a tile in Zd.Proof. By the assumption of linear independence we can delete d − krows of the matrix T , so that the remaining k-dimensional columnsPT = {Pt1, . . . P tk} span Rk. (The deletion of rows corresponds tomultiplication by a k × d matrix P with entries 1’s and 0’s.) In viewof Proposition 2.3, it is enough to show that PT is a tile in Zk.Consider the column vectors k = (0, 1, . . . k − 1)∗ (the ∗ in the ex-ponent denotes transposition) and j := (det PT ) · PT ∗−1k. Noticethat the matrix (det PT ) ·PT ∗−1 (and hence the vector j) has integerentries, and the 1-dimensional set j∗ ·PT = (detPT ) k∗ obviously tilesZ. Hence, Proposition 2.3 applies, and PT is a tile in Zk. Finally, we examine a property of ’non-tiling’ sets which played anessential role in the counterexample of [13] (we give a slightly general-ized version here).Proposition 2.5. Assume that the set T = {t1, . . . tk} ⊂ [0, m)d is notan m-tile. Then for large n the set Sn := T +(m · [0, n)d)is not a tile.Proof. The proof proceeds along the same lines as in [13], and we in-clude it only for completeness.Assume, for contradiction, that Sn tiles Zd with some translationset Σ. Take a cube Cl = [0, l)d, where l is much larger than n. LetΣl := {σ ∈ Σ : (σ + Sn) ∩ Cl 6= ∅}. Note that #Sn = knd. We have#Σl ≤(l+2mn)dknd, because all Σl-translates of Sn are contained in thecube (−mn, l +mn)d.Let A denote the annulus A := [−m,mn+m)d−[m,mn−m)d. Then#A (≈ 4dm(mn)d−1) ≤ 5dm(mn)d−1, if n is large enough compared tom. Hence, Σl + A cannot cover the cube Cl−m := [0, l −m)d because(#Σl)(#A) ≤ (l + 2mn)d(5dm(mn)d−1knd)< (l − m)d, if the numbersn, l are chosen so that n is sufficiently large compared to m, and l issufficiently large compared to n.Take a point x ∈ Cl−m not covered by Σl + A. Consider the cubeCxm := x+[0, m)d. This cube is fully inside Cl, therefore if any translateσ + Sn intersects Cxm then σ necessarliy belongs to Σl. The point x isnot covered by the annulus σ+A, therefore Cxm is contained in the cubeσ+[0, mn)d. Let S denote the set T +m ·Zd. In view of what has been6 MA´TE´ MATOLCSIsaid, we have (σ+Sn)∩Cxm = (σ+S)∩Cxm = (x+[0, m)d)∩(σ+T+mZd).The mod m reduction of this set is exactly the translate σ+T mod m.Hence, the tiling of the cube Cxm by Σ-translates of Sn contradicts theassumption that T is not an m-tile. It would be interesting to see whether the corresponding result holdsalso for spectral sets.3. Counterexample in dimension 4We can now show that one direction of Fuglede’s conjecture failsalready in dimension 4.Theorem 3.1. For any d ≥ 4 there exists a finite union of unit cubesΩ = ∪kj=1{sj + [0, 1)d} in Rd, such that Ω is spectral but not a tile.Proof. Take the matrixK :=0 0 0 0 0 00 0 1 1 2 20 1 0 2 2 10 1 2 0 1 20 2 2 1 0 10 2 1 2 1 0and observe that 13K is log-Hadamard (and corresponds to the Hadamardmatrix H given in [13]).Note that K is of rank 4 mod 3, therefore it is possible to make amod 3 decomposition K = L · T , where L is a 6× 4 and T is a 4× 6matrix. We give a specific mod 3 decomposition for convenience:L :=0 0 0 00 1 1 21 0 2 21 2 0 12 2 1 02 1 2 1and T :=0 1 0 0 0 20 0 1 0 0 20 0 0 1 0 20 0 0 0 1 2 .Note, however, that infinitely many decompositions exist, and eachdecomposition corresponds to a counterexample as explained below.(We also remark that the original example of [13], although not ex-plicitely stated in the paper, corresponds to the decomposition K =K · I, where I denotes the identity matrix.)Forgetting about mod 3 calculations and regarding L and T as realmatrices we see that K ′ := 13L ·T is a log-Hadamard matrix. ThereforeFUGLEDE’S CONJECTURE FAILS IN DIMENSION 4 7the set T ∈ Z4 above is 3-spectral, but obviously not a 3-tile becausethe number of elements of T (i.e. 6) does not divide 34. Therefore, forlarge n, the set S := T + 3[0, n)4 is 3n-spectral, but not a tile in viewof Propositions 2.1 and 2.5.This means that the set Ω := ∪s∈S{s+[0, 1)4} is spectral in R4 (thisfact was already mentioned in the discussion preceding Proposition2.1). The fact that Ω does not tile R4 follows as in Proposition 2.5.Namely, in a large cube Cl we find a point x which is not covered by anytranslate of the annulus A, and note that any translate of Ω is eitherdisjoint from the cube x+ [0, 3)4 or it covers exactly 6 unit volumes ofit. Therefore x + [0, 3)4 cannot be covered by translates of Ω becauseof obvious divisibility reasons.References[1] B. Fuglede, Commuting self-adjoint partial differential operators and a grouptheoretic problem, J. Funct. Anal. 16(1974), 101-121.[2] B. Fuglede, Orthogonal exponentials on the ball, Expo. Math. 19(2001), 267-272.[3] A. Iosevich, S. Pedersen, Spectral and tiling properties of the unit cube, Inter.Math. Res. Notices 16(1998), 819-828.[4] A. Iosevich, N. Katz, T. Tao, Convex bodies with a point of curvature do nothave Fourier bases, Amer. J. Math. 123(2001), 115-120.[5] A. Iosevich, N. Katz, T. Tao, The Fuglede conjecture holds for convex planardomains, Math. Res. Letters, 10(2003), 559-570.[6] A. Iosevich, M. Rudnev, A combinatorial approach to orthogonal exponentials,Inter. Math. Res. Notices 49(2003)1-12.[7] M. Kolountzakis, Non-symmetric convex domains have no basis of exponentials,Illinois J. Math. 44(2000), 542-550.[8] M. Kolountzakis, M. Papadimitrakis, A class of non-convex polytops which ad-mit no orthogonal basis of exponentials, Illinois J. Math. 46(2002), 4, 1227-1232.[9] M. Kolountzakis, The study of translational tiling with Fourier analysis, Pro-ceedings of the Milano Conference on Fourier Analysis and Convexity, to appear.[10] I. Laba, Fuglede’s conjecture for a union of two intervals, Proc. Amer. Math.Soc. 129(2001), 2965-2972.[11] I Laba, The spectral set conjecture and multiplicative properties of roots ofpolynomials, J. London Math. Soc. 65(2002), 661-671.[12] J.C. Lagarias, P.W. Shor, Y. Wang, Orthonormal bases for exponentials forthe n-cube, Duke Math. J. 103(2000), 1, 25-37.[13] T. Tao, Fuglede’s conjecture is false in 5 and higher dimensions, Math. Res.Letters, to appear.Alfre´d Re´nyi Institute of Mathematics, Hungarian Academy of Sci-ences POB 127 H-1364 Budapest, Hungary Tel: (+361) 483-8302, Fax:(+361) 483-8333E-mail address : matomate@renyi.hu

A class of non-convex polytops which admit no orthogonal basis of exponentials,

A combinatorial approach to orthogonal exponentials,

Commuting self-adjoint partial diﬀerential operators and a group theoretic problem,

Convex bodies with a point of curvature do not have Fourier bases,

Fuglede’s conjecture for a union of two intervals,

Fuglede’s conjecture is false in 5 and higher dimensions,

Non-symmetric convex domains have no basis of exponentials,

Orthogonal exponentials on the ball,

Orthonormal bases for exponentials for the n-cube,

Spectral and tiling properties of the unit cube,

The Fuglede conjecture holds for convex planar domains,

The spectral set conjecture and multiplicative properties of roots of polynomials,

The study of translational tiling with Fourier analysis,

Fuglede`s conjecture fails in dimension 4

Fuglede's conjecture fails in dimension 4

Fuglede's conjecture fails in dimension 4

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