Multiple lattice tiles and Riesz bases of exponentials

# Multiple lattice tiles and Riesz bases of exponentials

Mihail N. Kolountzakis M.K.: Department of Mathematics, University of Crete, GR-700 13, Iraklio, Greece
August 28, 2019
###### Abstract.

Suppose is a bounded and measurable set and is a lattice. Suppose also that tiles multiply, at level , when translated at the locations . This means that the -translates of cover almost every point of exactly times. We show here that there is a set of exponentials , , where is some countable subset of , which forms a Riesz basis of . This result was recently proved by Grepstad and Lev under the extra assumption that has boundary of measure , using methods from the theory of quasicrystals. Our approach is rather more elementary and is based almost entirely on linear algebra. The set of frequencies turns out to be a finite union of shifted copies of the dual lattice . It can be chosen knowing only and and is the same for all that tile multiply with .

Keywords: Riesz bases of exponentials; Tiling

Notation: We write . If is a set then is its indicator function. If is a lattice in then denotes the dual lattice.

## 1. Introduction

### 1.1. Riesz bases

In this paper we deal with the question of existence of a Riesz (unconditional) basis of exponentials

 et(x):=e(t⋅x)=e2πit⋅x,  t∈L,

for the space , where is a domain of finite Lebesgue measure and is a countable set. By Riesz basis we mean that every can be written uniquely in the form

 (1) f(x)=∑t∈Lat⋅e(t⋅x)

with the coefficients satisfying

 (2) C1∥f∥22≤∑t∈L|at|2≤C2∥f∥22,

for some positive and finite constants .

### 1.2. Orthogonal bases

One very special example of a Riesz basis occurs when the exponentials , can be chosen to be orthogonal and complete for . One can then choose and for (2) to hold as an equality. For instance, if is the unit cube in then one can take and obtain such an orthogonal basis of exponentials. This case, where an orthogonal basis of exponentials exists, is a very rigid situation though and many “reasonable” domains do not have such a basis (a ball is one example, or any other smooth convex body or any non-symmetric convex body).

The problem of which domains admit an orthogonal basis of exponentials has been studied intensively. The so called Fuglede or Spectral Set Conjecture  (claiming that for to have such a basis it is necessary and sufficient that it can tile space by translations) was eventually proved to be false in dimension at least 3 [15, 8, 1, 2], in both directions. Yet the conjecture may still be true in several important special cases such as convex bodies , and it generated many interesting results even after the disproof of its general validity (a rather dated account may be found in ).

It is expected that the existence of a Riesz basis for a domain is a much more general, and perhaps even generic, phenomenon, although proofs of existence of a Riesz basis for specific domains are still rather rare, especially in higher dimension [9, 10, 12].

### 1.3. Lattice tiles

One general class of domains for which an orthogonal basis of exponentials is known to exist is the class of lattice tiles. A domain is said to tile space when translated at the locations of the lattice (a discrete additive subgroup of containing linearly independent vectors) if

 (3) ∑t∈LχΩ(x−t)=1,  for almost all x∈Rd.

Intuitively this condition means that one can cover with the -translates of , with no overlaps, except for a set of measure zero (usually the translates of , for “nice” domains ).

It is not hard to see that when has finite and non-zero measure then the set has density equal to . If is a lattice then we call an almost fundamental domain of and . A fundamental domain of is any set which contains exactly one element of each coset mod , for instance a fundamental parallelepiped. There are of course many others, as indicated in Figure 1.3.

Every lattice tile by the lattice has an orthogonal basis of exponentials, namely those with frequencies , where is the dual lattice .

### 1.4. Multiple tiling by a lattice

We say that a domain tiles multiply when its translates cover space the same number of times, almost everyhwere.

###### Definition 1.1.

Let be measurable and be a countable set. We say that tiles when translated by at level if

 (4) ∑t∈LχΩ(x−t)=k,

for almost every . If we do not specify then we mean .

Multiple tiles are a much wider class of domains that level-one tiles. For instance, any centrally symmetric convex polygon in the plane whose vertices have integer coordinates tiles multiply by the lattice at some level . In contrast, only parallelograms or symmetric hexagons can tile at level one.

Another difference is the fact that if two disjoint domains and both tile multiply when translated at the locations then so does their union. In the case of multiple lattice tiling this operation gives essentially the totality of multiple tiles starting from level-one tiles, according to the following easy Lemma.

###### Lemma 1.

Suppose is a measurable set which tiles at level when translated by the lattice . Then we can write

 (5) Ω=Ω1∪⋯∪Ωk∪E,

where has measure and the are measurable, mutually disjoint and each is an almost fundamental domain of the lattice .

###### Proof.

Let be a measurable fundamental domain of , for instance one of its fundamental parallelepipeds. For almost every (call the exceptional set ) it follows from our tiling assumption that contains exactly points, which we denote by

 p1(x)

ordered according to the lexicographical ordering in . We also have that almost every point of belongs to exactly one such list.

Let then , for . In other words, for (almost) each one of the classes mod we distribute its occurences in into the sets . It is easy to see that the are disjoint and measurable and that they are almost fundamental domains of . ∎

### 1.5. Multiple lattice tiles have Riesz bases of exponentials

It is not true that domains that tile multiply by a lattice have an orthogonal basis of exponentials. For instance, it is known  that the only convex polygons that have such a basis are parallelograms and symmetric hexagons, yet every symmetric convex polygon with integer vertices is a multiple tile, a much wider class.

It is however true that multiple tiles have a Riesz basis of exponentials. The main result of this paper is the following theorem.

###### Theorem 1.

Suppose is bounded, measurable and tiles multiply at level with the lattice . Then there are vectors such that the exponentials

 (6) e((aj+λ∗)⋅x), j=1,2,…,k,λ∗∈Λ∗

form a Riesz basis for .

The vectors depend on and only, not on .

Theorem 1 was proved by Grepstad and Lev  with the additional topological assumption that the boundary has Lebesgue measure 0.

In  the result is proved following the method of [14, 13] on quasicrystals. Our approach is more elementary and almost entirely based on linear algebra. The authors of  have pointed out to me that there are similarities of the method in this paper and the methods in [10, 11, 12].

As an interesting corollary of Theorem 1 let us mention, as is done in , that, according to the recent result of , if is a centrally symmetric polytope in , whose codimension 1 faces are also centrally symmetric and whose vertices all have rational coordinates, then has a Riesz basis of exponentials.

###### Open Problem 1.

Is Theorem 1 still true if is of finite measure but unbounded?

## 2. Proof of the main result

The essence of the proof is contained in the following lemma.

###### Lemma 2.

Suppose is bounded, measurable and tiles multiply at level with the lattice . Then there exist vectors such that the following is true.

For any there are unique measurable functions such that

1. The are -periodic,

2. The are in of any almost fundamental domain of , and

3. We have the decomposition

 (7) f(x)=k∑j=1e(aj⋅x)fj(x),  for a.e.\ % x∈Ω.

Finally we have

 (8) C1∥f∥2L2(Ω)≤k∑j=1∥∥fj∥∥2L2(Ω)≤C2∥f∥2L2(Ω) ,

where depend only on and not on .

###### Proof.

Using Lemma 1 we can write as the disjoint union

 Ω=Ω1∪⋯∪Ωk,

where each is a measurable almost fundamental domain of . We can now define for and for almost every

 (9) ωj(x) as the unique point in Ωj s.t.\ x−ωj(x)∈Λ, and
 (10) λj(x)=x−ωj(x).

(The maps are clearly measurable and measure-preserving when restricted to a fundamental domain of .) Since the sought-after are to be -periodic it is enough to define them on and extend them to by their -periodicity. We may therefore rewrite our target decomposition (7) equivalently as follows.

 (11) For each x∈Ω1 and r=1,2,…,k: f(ωr(x))=k∑j=1e(aj⋅(x−λr(x)))fj(x).

We view (11) as a linear system

 (12) M˜F=F

whose right-hand side is the column vector

 F=(f(ω1(x)),f(ω2(x)),…,f(ωk(x)))⊤

and the unknowns form the column vector

 ˜F=(f1(x),f2(x),…,fk(x))⊤.

We have a different linear system for each and its matrix is with

 (13) Mr,j=Mr,j(x)=e(aj⋅(x−λr(x))),  r,j=1,2,…,k.

Factoring we can write this matrix as

 (14) M(x)=N(x)diag(e(a1⋅x),e(a2⋅x),…,e(ak⋅x)),

with the matrix given by

 Nr,j=Nr,j(x)=e(−aj⋅λr(x)),  r,j=1,2,…,k.

The key observation here is that when varying the number of different matrices that arise (the are fixed) is finite and depends on only. The reason for this is that the vectors are among the vectors in the bounded set , hence they take values in a finite set. (This is the only place where the boundedness of is used.)

Let us now see that the vectors can be chosen so that all the (finitely many) possible matrices are invertible. We have

 (15) detN(x)=∑π∈Sksgn(π)e(−k∑j=1aj⋅λπj(x)),

where denotes the permutation group on . By the definition of the vectors and the disjointness of the sets it follows that for each no two can be the same. View now the expression (15) as a function of the vector . Clearly it is a trigonometric polynomial and it is not identically zero as all the frequencies (for in the symmetric group )

 (16)

are distinct precisely because all the are distinct. Since the zero-set of any trigonometric polynomial (that is not identically zero) is a set of codimension at least 1 it follows that the vectors can be chosen so that all the matrices that arise are invertible.

Let now and consider the solution of the linear system (12) at that now takes the form

 (17) ˜F(x)=diag(e(−a1⋅x),e(−a2⋅x),…,e(−ak⋅x))N(x)−1F(x).

Since runs through a finite number of invertible matrices it follows that there are finite constants , independent of , such that for any we have

 (18) A1∥F(x)∥2ℓ2≤∥∥˜F(x)∥∥2ℓ2≤A2∥F(x)∥2ℓ2.

Integrating (18) over we obtain

 (19) A1∥f∥2L2(Ω)≤k∑j=1∥∥fj∥∥2L2(Ω1)≤A2∥f∥2L2(Ω).

This implies (8) with , . To show the uniqueness of the decomposition (7) observe that any such decomposition must satisfy the linear system (17), whose non-singularity has been ensured by our choice of the . ∎

We can now complete the proof of our main result.

###### Proof of Theorem 1.

Let . By Lemma 2 we can write as in (7). Since the are -periodic and are in of any almost fundamental domain of it follows that we can expand each in the frequencies of (the dual lattice of )

 (20) fj(x)=∑λ∗∈Λ∗fj,λ∗e(λ∗⋅x), j=1,2,…,k,

with

 (21)

since the exponentials , , form an orthogonal basis of .

The completeness of (6) follows from (7):

 (22) f(x)=k∑j=1∑λ∗∈Λ∗fj,λ∗e((aj+λ∗)⋅x).

The fact that (6) is a Riesz sequence follows from (8):

 1C2k∑j,λ∗∣∣fj,λ∗∣∣2≤∥∥ ∥∥∑j,λ∗fj,λ∗e((aj+λ∗)⋅x)∥∥ ∥∥2L2(Ω)≤1C1k∑j,λ∗∣∣fj,λ∗∣∣2.

As is clear from the proof above, the -tuples of vectors that appear in Theorem 1 are a generic choice: almost all -tuples will do. The exceptional set in is a set of lower dimension.

With a little more care one can see that one can choose the vectors to depend on and only and not on . In the proof of Lemma 2 the were chosen to ensure that the trigonometric polynomials (15) are all non-zero. Fix and and form the set of all polynomials of the form (15) which are not identically zero. This set of polynomials is countable and each such polynomial vanishes on a set of codimension at least 1 in . It follows that the union of their zero sets cannot possibly exhaust and we only have to choose the to avoid that union.

Thus there is a choice of that works for all of the same lattice. This proof does not give uniform values for the constants and in (8) though. ∎

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