Periodicity of the spectrum in dimension one

# Periodicity of the spectrum in dimension one

Alex Iosevich A.I.: Department of Mathematics, 915 Hylan Building, University of Rochester, Rochester, NY 14627, U.S.A.  and  Mihail N. Kolountzakis M.K.: Department of Mathematics, University of Crete, Knossos Ave., GR-714 09, Iraklio, Greece
July 27, 2019
###### Abstract.

A bounded measurable set , of Lebesgue measure 1, in the real line is called spectral if there is a set of real numbers (“frequencies”) such that the exponential functions , , form a complete orthonormal system of . Such a set is called a spectrum of . In this note we prove that any spectrum of a bounded measurable set must be periodic.

A.I.: Supported by NSF grant DMS10-45404
M.K.: Supported by research grant No 3223 from the Univ. of Crete and by NSF grant DMS10-45404 and grants of the University of Rochester, whose hospitality is gratefully acknowledged

Keywords: Spectral sets; Fuglede’s Conjecture.

AMS Primary Classification: 42B99

## 1. Tilings, spectral sets and periodicity

### 1.1. Spectra of domains in Euclidean space and the Fuglede Conjecture

Let be a bounded measurable set and let us assume for simplicity that has Lebesgue measure 1. The concept of a spectrum of that we deal with in this paper may be interpreted as a way of using Fourier series for functions defined on with non-standard frequencies. It was introduced by Fuglede [4] who was studying a problem of Segal on the extendability of the partial differential operators (on )

 ∂∂x1,∂∂x2,…,∂∂xd

to commuting operators on all of .

###### Definition 1.1.

A set is called a spectrum of (and is said to be a spectral set) if the set of exponentials

 E(Λ)={eλ(x)=e2πiλ⋅x: λ∈Λ}

is a complete orthonormal set in .

(The inner product in is .)

It is easy to see (see, for instance, [9]) that the orthogonality of is equivalent to the packing condition

 (1) ∑λ∈Λ|ˆχΩ|2(x−λ)≤1,  a.e. (x),

as well as to the condition

 (2) Λ−Λ⊆{0}∪{ˆχΩ=0}.

The orthogonality and completeness of is in turn equivalent to the tiling condition

 (3) ∑λ∈Λ|ˆχΩ|2(x−λ)=1,  a.e. (x).

These equivalent conditions follow from the identity

 (4) ⟨eλ,eμ⟩=∫Ωeλ¯¯¯¯¯eμ=ˆχΩ(μ−λ)

and from the completeness of all the exponentials in . Condition (1) is roughly expressing the validity of Bessel’s inequality for the system of exponentials while condition (3) says that Bessel’s inequality holds as equality.

If is a spectrum of then so is any translate of but there may be other spectra as well.

Example: If is the cube of unit volume in then is a spectrum of . Let us remark here that there are spectra of which are very different from translates of the lattice [7, 15, 8].

In the one dimensional case, which will concern us in this paper, condition (2) implies that the set has gaps bounded below by a positive number, the smallest positive zero of . (Note that, since is a bounded set, the function can be defined for all complex and is an entire function. This guarantees that its zeros are a discrete set.)

The Fuglede or Spectral Set Conjecture. Research on spectral sets has been driven for many years by a conjecture of Fuglede [4] which stated that a set is spectral if and only if it is a translational tile. A set is a translational tile if we can translate copies of around and fill space without overlaps. More precisely there exists a set such that

 (5) ∑s∈SχΩ(x−s)=1,  a.e. (x).

One can extend the definition of translational tiling to functions from sets.

###### Definition 1.2.

We say that the nonnegative function tiles by translation with the set if

 ∑s∈Sf(x−s)=ℓ, for almost every x∈Rd,

where is a constant (the level of the tiling).

Thus the question of spectrality for a set is essentially a tiling question for the function (the power-spectrum). Taking into account the equivalent condition (3) one can now, more elegantly, restate the Fuglede Conjecture as the equivalence

 (6) χΩ tiles Rd by translation at level 1⟺|ˆχΩ|2 tiles Rd by translation at level 1.

In this form the conjectured equivalence is perhaps more justified. However this conjecture is now known to be false in both directions if [24, 19, 12, 13, 2, 3], but remains open in dimensions and and it is not out of the question that the conjecture is true if one restricts the domain to be convex. (It is known that the direction “tiling spectrality” is true in the case of convex domains; see for instance [9].) The equivalence (6) is also known, from the time of Fuglede’s paper [4], to be true if one adds the word lattice to both sides (that is, lattice tiles are the same as sets with a lattice spectrum).

### 1.2. Periodicity of spectra and tilings

The property of periodicity is a very important property for a tiling.

###### Definition 1.3.

A set is called (fully) periodic if there exists a lattice (a discrete subgroup of with linearly independent generators; the period lattice) such that for all . We call a translation tiling periodic if the set of translations is periodic.

As an example of the importance of periodicity for a tiling we mention its connection to decidability [22], a question to which the study of tilings has provided several examples and problems. Although the general problem of tiling (not restricting the motions to be translations) is undecidable, it is not hard to see that, when the assumption of periodicity is added, the problem becomes decidable. Roughly, if one knows a priori that a set admits periodic tilings, if it admits any, then the question “Does admit a tiling?” can be answered algorithmically by simultaneously enumerating all possible counterexamples to tiling (if a tiling does not exist then the obstacle will show up at some finite stage) as well as all possible tilings of finite regions. If a tiling does not exist then the first enumeration will produce a counterexample. Otherwise, if a tiling exists then a periodic tiling exists and one of the finite regions that can be tiled will show this periodicity and can therefore be extended to all space. More details of this argument can be found at [22].

The so-called Periodic Tiling Conjecture [5, 17] should be mentioned: if a set tiles by translations (at level 1) then it can also tile by a periodic set of translations. By the argument sketched above this conjecture implies decidability for whatever class of sets it holds true. For instance, when one considers finite discrete sets (tiling is defined analogously to the continuous case) then one could prove the decidability of the tiling question if one managed to first prove that any such set that tiles by translation can also tile by a periodic set of translations. Both questions are open for (but see [23, 26] for some special cases). For all translational tilings by finite subsets of are necessarily periodic [21] and the problem is decidable. Another class of tilings where the Periodic Tiling Conjecture holds is the case when is assumed to be a convex polytope in , for any [25, 20].

In dimension it is known [18, 16, 11] that all translational tilings by a bounded measurable set are necessarily periodic. More generally it is known that whenever is an integrable function on the real line which tiles the real line by translation with a set of translates then is of the form

 (7) S=J⋃j=1(αjZ+βj),

where the real numbers are necessarily commensurable (and is in that case periodic) if the tiling is indecomposable (cannot be made up by superimposing other tilings). But this result is not applicable to the periodicity of spectra as the power-spectrum is never of compact support when is bounded (a qualitative expression of the uncertainty principle).

The question of periodicity of one-dimensional spectra was explicitly raised in [14]. It was recently proved (first in [1] and then a simplified proof was given in [10]) that if is a finite union of intervals in the real line then any spectrum of is periodic. See also [17] where periodicity of spectra and of tilings plays an important role.

Theorem A [Bose and Madan [1]; Kolountzakis [10]]
If is a finite union of intervals of total length and is a spectrum of then there exists a positive integer such that .

Our purpose in this note is to improve Theorem 1.2 by removing the assumption that is a finite union of intervals.

###### Theorem 1.

Suppose that is a spectrum of , a bounded measurable set of measure 1. Then is periodic and any period is a positive integer.

The proof of Theorem 1 is given in §2.

###### Corollary 2.

If , a bounded measurable set of measure 1, is spectral then tiles the real line at some integer level when translated at the locations .

###### Proof.

Let is a spectrum of . By Theorem 1 we know that is a periodic set and let be one of its periods: . Then we have (the number of elements in each period must be in order for to have density 1, hence is an integer), and, by (2), this implies that for all nonzero . Hence tiles when translated at (see, e.g. [9]) at level . ∎

Theorem 1 is not true in dimension higher than 1. For instance, even when is as simple as a cube, it may have spectra that are not periodic [15, 7, 8].

## 2. Proof of periodicity for spectra in dimension 1

### 2.1. The spectrum as a double sequence of symbols

Because of (2) we have that the gap between any two elements of is bounded below by , the smallest positive zero of . Let us now observe that the gap between successive elements of is also bounded above by a constant that depends only on .

###### Lemma 1.

If is a bounded measurable set of measure 1 then there is a finite number such that if is any spectrum of then the gap between any two successive elements of is at most .

###### Proof.

Lemma 1 is essentially a special case of Lemma 2.3 of [11]. In that Lemma it is proved that if tiles the line with a set ,

 ∑a∈Af(x−a)=w,  for almost all x∈R, with w>0 % a constant,

then the set has asymptotic density equal to . This means that the ratio

 |A∩I||I|

tends to as the length of the interval tends to infinity. The convergence is uniform over the choice of the set and the location of the interval . 111 Inequality (2.4) speaks of , but none of the other quantities that appear in it depend on . This means that inequality (2.4) holds even if we take to be the number of elements of in any interval of length . In fact, one can prove that cannot be if is sufficiently large, depending on , without taking the limit in inequality (2.4) and without talking about asymptotic density. This uniformity of course implies that the maximum gap of is bounded by a quantity that depends on only.

Since is a tiling and we deduce that has gaps bounded above by a function of alone. ∎

Let now

 Z={ξ∈R: ˆχΩ(ξ)=0}

and define the finite set (as is discrete)

 (8) Σ=Z∩(0,Δ]={s1,s2,…,sk},

where is the quantity given by Corollary 1.

We now view the set as a finite set of symbols (alphabet) and consider the set of all bidirectional sequences of elements of equipped with the product topology. A sequence of elements of converges to if for all the double sequences and agree in the window for large enough . More precisely, for all there is such that for we have

 xnj=xj,  for −k≤j≤k.

is a metrizable compact space so that each sequence has a convergent subsequence. This is just another way of phrasing a diagonal argument that is somewhat more convenient to use. The proof below may of course be phrased avoiding topological notions altogether and replacing the convergence of each subsequence with a diagonal argument.

The space is the natural space in which to view a spectrum of , as the set is locally of finite complexity: because of (2) the difference of any two successive elements of can be only be an element of . By demanding, as we may, that is always in we can therefore represent any set with the sequence of its successive differences. More precisely, we map any set whose successive differences are in and which contains

 Λ={⋯<−λ2<−λ−1<λ0=0<λ1<λ2<⋯}

to the element given by

 Λn=λn+1−λn,   (n∈Z).

This correspondence is a bijection and we will use one or the other form of the set as it suits us.

### 2.2. Symbolic sequences determined by their values in a half-line

Suppose . We say that is determined by left half-lines if knowing an element of to the left of any index suffices to determine the element in the remaining positions to the right of , i.e. if for any and we have

 (xi=yi for i≤n)⟹(xi=yi for all i∈Z).

Determination of by right half-lines is defined analogously.

We similarly say that is determined by any window of size (a positive integer) if for any and any knowing for completely determines .

###### Theorem 3.

Suppose is a closed, shift-invariant set which is determined by left half-lines and by right half-lines. Then there is a finite number such that is determined by windows of size .

###### Proof.

It is enough to show that there is a finite window size such that whenever two elements of agree on a window of size then they necessarily agree at the first index to the right of that window. For in that case they necessarily agree at the entire right half-line to the right of the window and are by assumption equal elements of .

Assume this is not true. Then there are elements of , , which agree at some window of width but disagree at the first location to the right of that window. Using the shift invariance of we may assume that

 xn−n=yn−n, xn−n+1=yn−n+1, …, xn−1=yn−1  &  xn0≠yn0.

By the compactness of the space there are and a subsequence of ’s such that and . By the meaning of convergence in the space we have that the sequences and agree for all negative indices and disagree at . This contradicts the assumption that is determined by left half-lines. ∎

###### Theorem 4.

If is shift-invariant and is determined by windows of size then all elements of are periodic, and the period can be chosen to be at most .

###### Proof.

Fix . Since there are at most different window-contents of length , it follows that there are two indices , , such that

 xi=xj,xi+1=xj+1,…,xi+w−1=xj+w−1.

Writing for the left shift of (i.e. ) we have that and agree at the window . By assumption then , which is another way of saying that the sequence has period . ∎

### 2.3. Symbolic sequences with spectral gaps

Suppose is a spectrum of the bounded set of measure 1. Write , where is a unit point mass at point . It is well known (see, for instance, [9]) that the Fourier Transform of the tempered distribution is supported by 0 plus the zeros of the function

 (|ˆχΩ|2)∧=χΩ∗χ−Ω

which is a continuous function with value 1 at the origin. Therefore there is an interval , with , such that has a spectral gap:

 (9) suppˆδΛ∩(0,a)=∅.

With defined by (8) let consist of all sequences which correspond to sets with gaps from such that (9) holds. The set is obviously shift-invariant as shifting a sequence in corresponds to translation of the set and translation will not affect the support of .

###### Lemma 2.

The set is closed in .

###### Proof.

Suppose and and that . It is enough to show that as this is what it means for to have no support in and therefore . By definition of the Fourier Transform

 ˆδΛ(ϕ)=δΛ(ˆϕ)=∑λ∈Λˆϕ(λ)=∗limn→∞∑λ∈Λnˆϕ(λ)=limn→∞δΛn(ˆϕ)=limn→∞ˆδΛn(ϕ)=0.

The justification for the equality above is very easy given (a) the rapid decay of , and, (b) the fact that all have the same positive minimum gap.

Indeed, due to (a) and (b) we can find for any an such that

 ∣∣ ∣∣∑λ∈L & |λ|>Rˆϕ(λ)∣∣ ∣∣<ϵ  for L=Λ or L=Λn,

and also an such that for . It follows that for we have

 ∣∣ˆδΛ(ϕ)∣∣=∣∣ˆδΛ(ϕ)−ˆδΛn(ϕ)∣∣=∣∣ ∣∣∑λ∈Λ & |λ|>Rˆϕ(λ)−∑λ∈Λn & |λ|>Rˆϕ(λ)∣∣ ∣∣≤2ϵ.

This implies that as we had to show. ∎

###### Theorem 5.

The sequences in are determined by both left half-lines and right half-lines.

###### Proof.

Suppose that is not determined by left half-lines (the argument is similar for right half-lines). Then there are distinct such that for all negative integers . Both and have a spectral gap at and therefore so does their difference

 μ=δΛ1−δΛ2.

Notice that is supported in the half-line . Suppose . It follows from the rapid decay of that the measure

 ν=ˆψ⋅μ

is totally bounded and still has a spectral gap at the interval . But the measure is also supported in the half-line and by the F. and M. Riesz Theorem [6] its Fourier Transform is mutually absolutely continuous with respect to the Lebesgue measure on the line. But this is incompatible with the vanishing of in some interval. Therefore must be identically and, since , is otherwise arbitrary, it follows that , or , a contradiction. It follows that is indeed determined by left half-lines. ∎

### 2.4. Conclusion of the argument

By Lemma 2 and Theorem 5 the set defined above, right after (9), given is a closed shift-invariant subset of and its elements are determined by half-lines. By Theorem 3 there exists a finite number such that the elements of are determined by their values at any window of width . By Theorem 4 all elements of are therefore periodic sequences. Since all spectra of can also be viewed as elements of , the periodicity of any spectrum of follows from the periodicity of the sequence of its successive differences.

The fact that any period of is a positive integer is a consequence of the fact that has density 1: if is a period of this implies that there are exactly elements of in each interval hence is an integer.

Acknowledgment: We are grateful to Dorin Dutkay for pointing out an error in our initial proof of Lemma 1.

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