Tiling and weak tiling in \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(\mathbb {Z}_p)^d$$\end{document}(Zp)d

We discuss the relation of tiling, weak tiling and spectral sets in finite abelian groups. In particular, in elementary p-groups \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(\mathbb {Z}_p)^d$$\end{document}(Zp)d, we introduce an averaging procedure that leads to a natural object of study: a 4-tuple of functions which can be regarded as a common generalization of tiles and spectral sets. We characterize such 4-tuples for \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$d=1, 2$$\end{document}d=1,2, and prove some partial results for \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$d=3$$\end{document}d=3.


Introduction
The concept of weak tiling was introduced recently in [15], in connection with Fuglede's conjecture for convex bodies in R d .In this note we will study the relation of tiling, weak tiling and spectral sets in finite abelian groups, with particular attention to elementary p-groups (Z p ) d .Our motivation is to give a new tool to prove the "spectral → tile" direction of Fuglede's conjecture in finite abelian groups.
We begin by recalling the necessary notions and fixing our notation.The cardinality of any finite set A will be denoted by |A|.We use the standard notation A ± B = {a ± b : a ∈ A, b ∈ B}, and k A = {ka : a ∈ A}, for any positive integer k.For any n, let Z n denote the cyclic group of order n.
Let G be a finite abelian group.A character is a homomorphism from G to the complex unit circle T. The dual group, denoted by G, is the collection of all characters of G.For a function f : G → C, the Fourier transform f : G → C is defined as

γ (x).
A set A ⊂ G is called spectral if the function space L 2 (A) admits an orthogonal basis consisting of characters restricted to A. Such an orthogonal basis of characters is called a spectrum of A. (The spectrum, if it exists, is not necessarily unique.) We say that a set A ⊂ G tiles G if there exists another set B ⊂ G such that each g ∈ G can be written uniquely as g = a + b, where a ∈ A, b ∈ B. We usually express this relation as A ⊕ B = G or, in the functional notation, 1 A * 1 B = 1 G , where * denotes convolution.The convolution of any two functions f , g : G → C is defined in the standard way as ( f * g)(x) = y∈G f (x − y)g(y).
For any function f : G → C the function f − is defined as f − (x) = f (−x).Some basic properties of the Fourier transform read as follows: and for real-valued even functions f = |G| f .Fuglede's conjecture [7] stated that a set A ⊂ G is spectral if and only if it tiles G.The conjecture was formulated explicitly in R d , but Fuglede already mentioned that the notions make sense in any locally compact abelian group G.In fact, the first counterexample by Tao [20] in R 5 was based on a counterexample in the finite group (Z 3 ) 5 .Since then, further counterexamples [5,12] have been constructed (all based on examples in finite groups) to both directions of the conjecture in R d , d ≥ 3. The conjecture (both directions of it) remains open for R and R 2 .Some further connections between the discrete and the continuous settings have been revealed by Dutkay and Lai in [2].
In this paper we will restrict our attention to finite abelian groups, and mostly to the case G = (Z p ) d with p being a prime.For finite abelian groups, several results have been discovered in recent years [3, 4, 6, 8-11, 13, 14, 16-19, 21].For the purposes of this note, we single out the results of [8] and [6,17].In [8] the authors prove both directions of Fuglede's conjecture in (Z p ) 2 .To the contrary, in [6] and [17] the authors prove (independently of each other) that for odd primes p there exist spectral sets in (Z p ) d , d ≥ 4, which do not tile the group, thus exhibiting a counterexample to the "spectral → tile" direction of Fuglede's conjecture in these groups.
In connection with Fuglede's conjecture, the notion of weak tiling was introduced in [15].For us, in the case of finite abelian groups, weak tiling can be formulated as follows.A set A ⊂ G tiles another set E ⊂ G weakly, if there exists a nonnegative function h : G → R such that 1 A * h = 1 E .We will soon see that it makes sense to introduce the further restrictions that h(0) = 1, the function h is positive definite (i.e.h ≥ 0), and restrict our attention to the case E = G.Definition 1.1 We say that a set A ⊂ G pd-tiles G weakly if there exists a nonnegative function h : In the terminology, "pd" stands for positive definite.Note that, due to positive definiteness, h is necessarily an even function, h(x) = h(−x).The assumption h(0) = 1 is essential, otherwise the constant function h = 1 |A| would provide a weak pd-tiling for any set A ⊂ G.Note also, as a comparison with the terminology in [15], that if A pd-tiles G weakly then A tiles its complement weakly.
A simple but important observation, also essentially contained in [15], is that if A pd-tiles G weakly with a function h, then The reason is that x = a − a and h(x) > 0 would imply Next, we show that a proper tiling always induces a weak pd-tiling of G.

Lemma 1.2 If A ⊕ B = G is a tiling, then A pd-tiles G weakly with h
Proof A⊕ B = G implies 1 A • 1 B = |G|δ 0 , so that the supports of the functions 1 A and 1 B are essentially disjoint (only intersect at 0).This implies that 1 We remark that the notation h 1 above (instead of using simply h) is for later convenience.
The essential observation in [15] is that if a set A is spectral, then it tiles its complement weakly.However, slightly more is true.

Lemma 1.3 Let G be a finite abelian group. If A ⊂ G is spectral, then A pd-tiles G weakly.
Proof Let S ⊂ G be a spectrum of A. Then |S| = |A| because the space L 2 (A) has dimension |A|, and S is a basis.Let |A| 2 1 S * 1 −S ≥ 0, so h 1 is nonnegative, normalized and positive definite, as required.Also, the weak tiling condition 1 A * h 1 = 1 G is most easily seen by taking Fourier transforms.
By this lemma, we can establish the "spectral → tile" direction of Fuglede's conjecture in a finite abelian group G, if we prove that any set that pd-tiles G weakly, actually tiles G properly.This motivates the following definition.Definition 1.4 Assume that a finite abelian group G has the property that whenever a set A pd-tiles G weakly then A tiles G properly.Then we call the group G pd-flat.
In the remainder of this note we will focus our attention on elementary p-groups G = (Z p ) d .As mentioned above, for odd primes and d ≥ 4, there exist spectral sets in G which do not tile G. Therefore, (Z p ) d is not pd-flat for d ≥ 4.
In Sect. 2 we introduce an averaging technique which leads to a natural generalization of spectral sets and tiles in (Z p ) d .Finally, in Sect. 3 we show that (Z p ) d is pd-flat for d = 1, 2, and we give some partial results and conjectures for d = 3.

Averaging
It is natural to go one step further with the generalization of tiling.
It turns out that this notion is very flexible, and for G = (Z p ) d it gives rise to a natural averaging procedure.In connection with this, we introduce a special class of functions.

Definition 2.2
We say that a function f : That is, a ray-type function is constant on any punctured line through the origin (but may have a different value at the origin).
We also need some information on the zeroes of the Fourier transform of the indicator function 1 A of a set A ⊂ (Z p ) d .We make the identification G = {0, 1, . . ., p−1} d , and the same for the dual group G = {0, 1, . . ., p − 1} d .It is sometimes useful to think of the elements of G as column vectors of length d, and elements of G as row vectors of length d.Also, in notation, we will use boldface letters for elements of G and G to indicate that they are vectors.With these identifications in mind, the action of a character t ∈ G on an element x ∈ G is given by e 2iπ t,x / p .For a function f : G → C, and t ∈ G, we have f (t) = x∈G f (x)e 2iπ t,x / p .In particular, for A ⊂ G, the Fourier transform of 1 A takes the form 1 A (t) = a∈A e 2iπ t,a / p .The important point here is that for any t = 0 we either have 1 A (kt) = 0 for all k = 1, 2, . . ., p − 1, or 1 A (kt) = 0 for all k = 1, 2, . . ., p − 1.This is well-known, and the reason is that 1 A (t) = 0 if and only if the sum a∈A e 2iπ t,a / p contains the same number of terms for each pth root of unity, in which case a∈A e 2iπ kt,a / p also contains the same number of terms of each pth root of unity.
Therefore, the zeros of the Fourier transform 1 A consist of punctured lines through the origin (for t = 0 a punctured line L is given by L = {t, 2t, . . ., (p − 1)t}; note that 0 / ∈ L, and we use the notation L = L ∪ {0}).The same is true for the zeroes of the Fourier transform of f We can now perform the first half of the averaging.

Lemma 2.3 Let G = (Z p ) d , and 1
, so the zeroes of these functions are the same punctured lines through the origin.
The assumption Due to the fact that the zeroes of 1 A and 1 k A coincide, and |A| = |k A|, we have 1 k A h 1 = |G|δ 0 , and therefore The function Finally, for the average |G|δ 0 is a functional pd-tiling since these properties hold for each f i .Also, by averaging, it is clear that f is ray-type, and the property supp f ∩ supp h 1 = {0} is inherited.
We can perform a second step of averaging for the function h 1 .Proof Recall from Lemma 2.3 that f is a ray-type function.We claim that f is also ray-type.To see this, note that f can be written in the form f = cδ 0 + i c i 1 L i for some lines L i (for convenience, we use proper lines in this decomposition, not punctured lines; the difference is absorbed in the constant c at the origin).The Fourier transform of 1 L i is p1 H i with the hyperplane H i being orthogonal to L i , and therefore the f = c1 G + p i 1 H i , which is clearly ray-type.
By Lemma 2.3 we have that f * h 1 = 1 G , which implies f h 1 = |G|δ 0 .This means that supp f ∩ supp h 1 = {0}.As f is ray-type, its support is a union of lines R i , and hence h 1 must be 0 on the punctured lines Ṙi .But h k (t) = h 1 (t/k), so h k is also 0 on Ṙi , and hence f h k = |G|δ 0 holds (for the value at zero we have Also, supp f ∩ supp h 1 = {0} by Lemma 2.3, and f is ray-type, so the support of h 1 is contained in rays where f = 0.This clearly implies that the support of h k is also contained in these rays, and therefore supp f ∩ supp h k = {0} After averaging, it is clear that h is ray-type, and the functional pd-tiling property f * h is preserved, as well as the condition supp f ∩ supp h = {0}.
We see from Lemmas 1.2 and 1.3 that A pd-tiles G weakly in both cases when A tiles G, or A is spectral in G.After applying Lemmas 2.3 and 2.4 we arrive at the following corollary.

Corollary 2.5 Let G = (Z p ) d , and assume A pd-tiles G weakly with 1
and the 4-tuple of functions ( f , h, f , h) satisfy the following properties: Proof Only 8 needs further explanation, and it follows by taking Fourier transform of the equation, noting that f = |G| f (and the same for h), and applying 5 and 3.
This corollary motivates the following definition.

Definition 2.6 A 4-tuple of functions ( f , h, f , h) is called a complementary 4-tuple
if they satisfy conditions (i)-(ix) of Corollary 2.5.(We do not necessarily require that f be constructed from a set A.) The notion of complementary 4-tuples is very appealing, because it puts the functions f , h and their Fourier transform f , h on equal footing.When studying spectral sets and tiles in (Z p ) d it is natural to try to characterize all complementary 4-tuples.We will do this for d = 1, 2 in the next section, and present some partial results for d = 3.
We also remark that an analogous averaging procedure can be carried out in cyclic groups Z n , leading to a similar notion of complementary 4-tuples in those groups.Studying these 4-tuples could lead to some insights concerning the Coven-Meyerowitz conjecture.This is subject to future research.

Complementary 4-tuples in (Z p ) d
In this section we analyze complementary 4-tuples in (Z p ) d , and prove that (Z p ) d is pd-flat for d = 1, 2, and also give some partial results for d = 3.
For convenience, we recall here what is known about Fuglede's conjecture in these groups.Both directions of the conjecture are true for d = 1, 2, with d = 1 being fairly trivial, and d = 2 being treated in [8].A simpler proof of the case d = 2 can be found in [10].For d = 3 it is known that all tiles are spectral [1], but it is not known whether all spectral sets are tiles.For d ≥ 4 (and p being odd) there are examples of spectral sets which do not tile the group [6,17] but the tile-spectral direction of the conjecture is open.
We now turn to the main results of this section.We emphasize here that a group G being pd-flat is formally stronger than the "spectral → tile" direction of Fuglede's conjecture in G.

is, every set A which pd-tiles G weakly, tiles G properly. Also, G is not pd-flat for d ≥ 4.
Proof Assume A pd-tiles G weakly.Then there exists a complementary 4-tuple ( f , h, f , h) with the properties listed in Corollary 2.5, and f is constructed from 1 A as in Corollary 2.5.
For d = 1 the support of any ray-type function is either {0} or G.If supp f = {0} then |A| = 1, and A tiles G trivially.If supp f = G then necessarily supp h = {0}, and h(0) = 1 implies h = δ 0 , and hence f must be the constant function 1.Therefore, |A| = p by (2), and hence A = G.
For d = 2 we must perform a case-by-case analysis of the function f .Let c = inf x∈G f (x) ≥ 0, and let L i be the lines in the support of f .Then f can be decomposed uniquely as f = c1 G + i d i 1 L i + mδ 0 with some d i ≥ 0 and m ∈ R (note that L i denotes the full line here, not the punctured line).
If c = 0 and m = 0, then the total mass of f is exactly p times the mass at 0, because this is true for all 1 L i .The mass at zero is f (0) = 1, and hence we have |A| = p, by equation (2).Also, c = 0 implies that f must be zero on some punctured line Ṙ, so A can have at most one element in each coset of the line R since A − A ⊂ supp ( f ).But |A| = p, so A must have exactly one element in each coset of R, and therefore A ⊕ R = G is a tiling.
Finally, we claim that c = 0 and m < 0 is not possible.Indeed, c = 0 means that f must be zero on some punctured line Ṙ, and therefore f = m < 0 on the orthogonal punctured line Ṙ⊥ , contradicting the nonnegativity of f .
For d ≥ 4 we know that there exist spectral sets in (Z p ) d which do not tile the group [6,17].Therefore, (Z p ) d cannot be pd-flat.
For d = 3, we conjecture that (Z p ) 3 is pd-flat, but unfortunately we cannot give a complete characterization of complementary 4-tuples, we can only prove some partial results in this direction.Therefore, the "spectral → tile" direction of Fuglede's conjecture remains open in this case.We have the following partial result.Proposition 3.2 Let G = (Z p ) 3 , and assume A ⊂ G pd-tiles G weakly with 1 A * h 1 = 1 G .Let ( f , h, f , h) be the complementary 4-tuple constructed in Corollary 2.5.Then either A tiles G properly, or the functions f , h, f , h all have the following "dispersive" property: for any 2-dimensional subspaces S ⊂ G, S ⊂ G the intersections supp f ∩ S, supp h ∩ S, supp f ∩ S and supp h ∩ S are all non-trivial (i.e.not equal to {0} or the whole plane).
Proof Exploiting the fact that all appearing functions are ray-type, we can represent each function in the following normalized form.Let us label the planes and lines through the origin as S 1 , S 2 , . . ., S p 2 + p+1 and L 1 , L 2 , . . ., L p 2 + p+1 .Then f can be represented uniquely as where the coefficients w, c i , d i ≥ 0, m ∈ R are defined by the following greedy algorithm: etc, and subsequently d 1 is the maximal value such that f − w1 G − ( We can represent f , h, f , h in this manner uniquely, after fixing the order of planes and lines in G and G. We will show that if any of f , h, f , h does not have the dispersive property (stated in the proposition), then A tiles G.This is done by a case-by-case analysis, where we will use the properties of complementary 4-tuples stated in Corollary 2.5.
Case I. Let us first consider the trivial case when the support of some of the appearing functions is the whole underlying group.Having dealt with Case I, we can assume for the rest of the argument that w = 0 in equation (3).Moreover, this also implies that m ≤ 0 in (3) because m > 0 would imply supp f = G.Altogether, we can conclude that the representation of f takes the form where c i , d i ≥ 0, and m ≤ 0. The same is true for the representations of h, f , h.Case II.Next, assume that the support of one of the appearing functions intersects a plane only at 0.
(a) Let supp f ∩ S = {0}.Let L ⊂ G denote the punctured line orthogonal to S. Then f = m ≤ 0 on L, and the nonnegativity of f implies m = 0. Also, the "empty" plane S intersects every other plane S i , so every S i has at least one "empty" line S ∩ S i , and hence the weight c i must be 0 in equation ( 4) for every i.Hence, f can be represented in the form f = i d i 1 L i , and therefore the total mass of f is p times the mass at 0 (as this is the case for every L i ), and hence |A| = p is implied by (2) and f (0) = 1.Furthermore, the fact that S is empty means that A − A ∩ S = {0}, so there is exactly one point of A on each coset of S. Therefore, A ⊕ S = G is a tiling.= p.This implies that x∈G h(x) = p 2 , and x∈G f (x) = p.This implies that in the decomposition (4) all coefficients c i = 0 and m = 0, otherwise the total mass of f would be greater than p times the mass at 0 (recall that c i ≥ and m ≤ 0).Therefore, f = i d i 1 L i .Also, as supp f = G, we have a line L ⊂ G such that f = 0 on L , which implies that f = 0 on the punctured plane S = L ⊥ .Finally, as |A| = p and A − A ∩ S = {0} we have that A ⊕ S = G is a tiling.
Case III.Finally, assume that the support of one of the appearing functions contains a whole plane.In this case, we can invoke 5 and 9 to conclude that the support of some other function intersects the same plane at 0 only, and we are done by Case II above.
It is worth summarizing here what this result means for spectral sets.For a spectral set A ⊂ (Z p ) 3 , perform the avearaging procedure leading to the complementary 4tuple in Corollary 2.5.If any of the functions f , h, f , h does not have the dispersive property described in Proposition 3.2 then A necessarily tiles (Z p ) 3 .If all appearing functions have the dispersive property, they must have a representation where the lines L i are situated such that they do not cover any plane fully, and do not leave any plane empty.Furthermore, d i > 0, m < 0 (the latter is true because m ≥ 0 would imply the Fourier transform being positive on the planes L ⊥ i ).

Lemma 2 . 4
Let G = (Z p ) d , and 1 A * h 1 = 1 G be a weak pd-tiling of G by a set A, and define the ray-type function f as in Lemma 2.3.For any k = 1, . . ., p − 1, let h k (x) = h 1 (kx), and let h = 1 p−1 p−1 k=1 h k .Then, for each k, f * h k = 1 G is a functional pd-tiling, the average h is a ray-type function, and f * h is also a functional pd-tiling such that supp f ∩ supp h = {0}.
(b) If supp h ∩ S = {0}, then the same argument as in (a) shows that h has a decomposition h = i di L i , and therefore x∈G h(x) h(0) = p.Considering h(0) = 1, this implies x∈G h(x) = p, and hence |A| = p 2 by 4. Also, supp f = G, so there exists a line L such that A − A ∩ L = {0}, and hence A ⊕ L = G is a tiling.(c) If supp f ∩ S = {0}, then the same argument as in (a) shows that f has a decomposition f = i d i L i , and therefore t∈ G f (t)f (0) = p.Considering that f (0) = x∈G f (x) and t∈ G f (t) = p 2 f (0), we obtain x∈G f (x) = p 2 , that is |A| = p 2 by(2).We can finish the argument as in (b): as supp f = G, there exists a line L such that A − A ∩ L = {0}, and hence A ⊕ L = G is a tiling.(d)If supp h ∩ S = {0}, then the same argument as in (a) shows that h has a decomposition h = i d i L i , and therefore t∈ G h(t) h(0)