A linear programming approach to Fuglede’s conjecture in $\mathbb {Z}_p^3$

Abstract

We present an approach to Fuglede’s conjecture in $\mathbb {Z}_p^3$ using linear programming bounds, obtaining the following partial result: if $A\subseteq \mathbb {Z}_p^3$ with $p^2-p\sqrt{p}+\sqrt{p}<|A|<p^2$, then A is not spectral.

1 Introduction

A fundamental question in harmonic and functional analysis predicts that a geometric property of a measurable set (tiling) coincides with an analytic one (spectrality). This was posed initially by Fuglede [18] in the setting of finite dimensional Euclidean spaces, following a question of Segal in the commutativity of certain partial differential operators.

Conjecture 1.1

Let $\Omega \subseteq \mathbb {R}^d$ be a bounded measurable set with $m(\Omega )>0$. Then, $\Omega$ tiles $\mathbb {R}^d$ by translations (i. e. almost every $x\in \mathbb {R}^d$ can be written uniquely as $\omega +t$, where $\omega \in \Omega$ and t belongs to a fixed set of translations T) if and only if $L^2(\Omega )$ accepts an orthogonal basis of complex exponential functions, $e^{2\pi i{\left\langle {\lambda ,x}\right\rangle }}$, for $\lambda \in \Lambda$ ($\Lambda$ is called the spectrum of $\Omega$).

While Conjecture 1.1 has been largely disproved, namely for $d\ge 3$ [16, 23, 36], there is a great recent interest in the setting of finite Abelian groups. Besides, even Fuglede himself hinted at different settings for Conjecture 1.1.

Let G be a finite Abelian group. We say that a subset $A\subseteq G$ tiles G by translations (or just that A is a tile) if there is a subset $T\subseteq G$ such that each element of G can be expressed uniquely as $a+t$, with $a\in A$, $t\in T$. This fact will be denoted as $A\oplus T=G$, and T is called the tiling complement of A. In general, we will denote by $A+T$ the sumset ${\left\{ {a+t:a\in A, t\in T}\right\} }$; if every element of $A+T$ has a unique representation as a sum of an element of A and an element of T, we emphasize this by writing $A\oplus T$.

We may also define the following inner product on the space of complex functions defined on $A\subseteq G$:

$$\begin{aligned} {\left\langle {f,g}\right\rangle }_A=\sum _{x\in A}f(x)\overline{g(x)}. \end{aligned}$$

$\widehat{G}$ denotes as usual the dual group of G, i. e. the group of multiplicative characters $\chi :G\rightarrow S^1$. We call $B\subseteq \widehat{G}$ a spectrum of A if $|A|=|B|$ and ${\left\langle {\chi ,\psi }\right\rangle }_A=0$, $\forall \chi \ne \psi \in B$; in other words, the characters of B, when restricted to A, form an orthogonal basis in $\mathbb {C}^A$. In this case, A is called spectral and (A, B) a spectral pair.

There is a connection between the discrete and continuous setting of Fuglede’s conjecture, as was first suggested by Łaba [25], who connected this question to the tiling results by Coven & Meyerowitz in $\mathbb {Z}$ [8], and Tao, in his disproof of Fuglede’s conjecture in $\mathbb {R}^d$, $d\ge 5$ [36]. The principal argument in Tao’s paper is the following:

a counterexample in G with d generators can be lifted to a counterexample in $\mathbb {R}^d$

and this holds for both directions, namely the spectral to tiling and tiling to spectral directions, denoted henceforth as S-T and T-S, respectively. Tao found a spectral subset in $\mathbb {Z}_3^5$ which has 6 elements [36]; such a subset cannot be a tile, as the cardinality of a tile divides the order of the group.

Counterexamples in $\mathbb {Z}_8^3$ [23] and $\mathbb {Z}_{24}^3$ [16] were given shortly thereafter, which led to counterexamples in $\mathbb {R}^3$ in both directions; since S-T and T-S are hereditary properties, these counterexamples show that Fuglede’s conjecture fails in $\mathbb {R}^d$, $d\ge 3$, in both directions.

The original conjecture is still open in $\mathbb {R}$ and $\mathbb {R}^2$. The counterexamples mentioned above, raised naturally the following question.¹:

Question 1

For which finite Abelian groups G do the properties of tiling and spectrality coincide?

In the last few years, there is a wealth of results on Fuglede’s conjecture in the discrete setting (there are of course the notable results [14, 15, 29] where this conjecture has been confirmed in $\mathbb {Q}_p$ and for all convex domains in $\mathbb {R}^d$). So far, all the results are positive in cyclic groups [21, 26,27,28, 30, 34, 39] or groups with two generators [12, 38]. A counterexample in a group of the form $\mathbb {Z}_N^2$ would also be lifted to a counterexample in $\mathbb {R}^2$, whereas the confirmation of Fuglede’s conjecture in every cyclic group would be an enormous step to confirming this conjecture in $\mathbb {R}$ [11] (the only missing ingredient would then be the rationality of spectrum [4, 24]).

For groups with at least three generators, it is natural to search for counterexamples, as we know that Fuglede’s conjecture fails in $\mathbb {R}^3$. Using the fundamental theorem on the factorization of finite Abelian groups, a counterexample in $\mathbb {Z}_p^d$ for all primes p can be lifted to a counterexample to a finite Abelian group of $\ge d$ generators. This endeavor has so far been very successful in the S-T direction [17]:

Theorem 1.2

S-T fails in $\mathbb {Z}_p^4$ for every odd prime p, and also in $\mathbb {Z}_2^{10}$.

This implies that S-T fails in every Abelian group with $\ge 10$ generators, as well as every group with $\ge 4$ generators having odd order. For the T-S direction there no other counterexamples known besides $\mathbb {Z}_{24}^3$ (and every group that contains a copy thereof), which led to the following conjecture [35]:

Conjecture 1.3

T-S holds in every finite Abelian p-group.

Summarizing, the state of the art regarding groups of the form $\mathbb {Z}_p^d$ is the following:

• (1) Fuglede’s conjecture is true if $d\le 2$ [19].

• (2) For $d\ge 4$ and p odd prime, S-T fails [17, 32].

• (3) S-T fails in $\mathbb {Z}_2^{10}$ [17].

• (4) Fuglede’s conjecture holds in $\mathbb {Z}_2^6$ [17].

• (5) T-S holds in $\mathbb {Z}_p^3$ [1].

• (6) S-T holds in $\mathbb {Z}_p^3$, and $p\le 7$ [13].

For $p=2$, the question is still open for $\mathbb {Z}_2^d$, $d= 7, 8, 9$, and the T-S direction for all $d\ge 7$, whereas for odd p the S-T direction is unknown only for $d=3$ (and the T-S direction for all $d\ge 4$).

Here, we will approach the S-T direction in $\mathbb {Z}_p^3$ using linear programming bounds. This method provides estimates for the size of sets B whose set of differences $B-B$ avoids a certain "forbidden set". The method of linear programming bounds has seen widespread applications in coding theory [10, 20], combinatorial designs [9], discrete geometry, most notably in the sphere packing problem by Viazovska [7, 37], who used the version by Cohn and Elkies [6], which is also known as "Delsarte’s method" [10].

This method involves the notion of a positive definite function, which is not new in this conjecture [22, 29]. The novelty of the present approach is the notion of a blocking set in the finite projective plane, whose size estimates lead to the main result:

Theorem 1.4

Let $A\subseteq \mathbb {Z}_p^3$ be a set satisfying $p^2-p\sqrt{p}+\sqrt{p}<|A|<p^2$. Then A is not spectral.

It has been shown [13] that a spectral subset $A\subseteq \mathbb {Z}_p^3$ which is neither a singleton nor the entire group, must have pk elements, where $1\le k\le p$. If $k=1$ or p, then A is also a tile, so we may restrict our attention to the range $1<k<p$. In [13] it was also shown that if $k=p-2$ or $p-1$, then A cannot be spectral, so Theorem 1.4 extends this result.

The paper is organized as follows: in Sect. 2 we summarize all the known facts about spectral subsets and tiles in $\mathbb {Z}_p^3$; we also include the T-S proof in $\mathbb {Z}_p^3$ to make this article as self-contained as possible. In Sect. 3 we introduce the first main tool for the proof of Theorem 1.4, whereas in Sect. 4 we introduce the second, which involves the geometric structure of the finite projective plane. Finally, in Sect. 5 we are ready to prove Theorem 1.4, and in Sect. 6 we discuss several possible extensions of the approach presented herein.

2 Spectral sets and tiles in $\mathbb {Z}_p^3$

All results and arguments (or generalizations thereof) in this section have already appeared in [1, 13, 19]. As usual, the Fourier transform of $f:G\rightarrow \mathbb {C}$ is denoted by $\hat{f}$ or $\textbf{F}f$ and satisfies

$$\begin{aligned} \hat{f}(\xi )=\sum _{x\in G}f(x)\xi (-x)={\left\langle {f,\xi }\right\rangle }, \end{aligned}$$

for every $\xi \in \widehat{G}$. With this definition, the Fourier inversion formula is

$$\begin{aligned} f(x)=\frac{1}{|G|}\sum _{\xi \in \widehat{G}}\hat{f}(\xi )\xi (x), \end{aligned}$$

(2.1)

and the Plancherel-Parseval formula becomes

$$\begin{aligned} |G|\sum _{x\in G}|f(x)|^2=\sum _{\xi \in \widehat{G}}|\hat{f}(\xi )|^2. \end{aligned}$$

The convolution $f*g$ is defined as

$$\begin{aligned} f*g(x)=\sum _{y\in G}f(y)g(x-y), \end{aligned}$$

and the Fourier transform converts convolution to pointwise product, i. e.

$$\begin{aligned} \widehat{f*g}=\hat{f}\cdot \hat{g}. \end{aligned}$$

The operator $\textbf{U}=\frac{1}{\sqrt{|G|}}\textbf{F}$ is unitary, and $\textbf{U}^2$ preserves all even functions, that is, $\hat{\hat{f}}=|G|f$ for every even function $f:G\rightarrow \mathbb {C}$. Also, if we denote by $f_-$ the function $f_-(x)=f(-x)$, then $|\widehat{f*f_-}|=|\hat{f}|^2$.

It is a fundamental fact that G is isomorphic to $\widehat{G}$, albeit non-canonically. For $G=\mathbb {Z}_p^3$, we fix such an isomorphism $x\mapsto \xi _x$ for each $x=(x_1,x_2,x_3)\in \mathbb {Z}_p^3$, where

$$\begin{aligned} \xi _x(y)=\zeta _p^{{\left\langle {x,y}\right\rangle }}. \end{aligned}$$

Here, $\zeta _p=e^{2\pi i/p}$ and

$$\begin{aligned} {\left\langle {x,y}\right\rangle }=x_1y_1+x_2y_2+x_3y_3, \;\;\; \text { for }y=(y_1,y_2,y_3)\in \mathbb {Z}_p^3. \end{aligned}$$

Under this identification, we write $\hat{f}(x)$ instead of $\hat{f}(\xi _x)$, and each character $\xi _x$ with $x\ne 0$ is an epimorphism from $\mathbb {Z}_p^3$ to the multiplicative group of the p-th roots of unity; the kernel of $\xi _x$ is precisely the set of all y that are "orthogonal" to x, that is, ${\left\langle {x,y}\right\rangle }=0$. When $\mathbb {Z}_p^3$ is viewed as a 3-dimensional space over the finite field $\mathbb {Z}_p$, the kernel is then a plane orthogonal to x, which we denote by $x^\perp$. A fact that challenges our intuition from the Euclidean space $\mathbb {R}^3$ is that we may have ${\left\langle {x,x}\right\rangle }=0$, or equivalently, we may have $V+V^\perp \ne \mathbb {Z}_p^3$ for subspaces $V\subseteq \mathbb {Z}_p^3$, but this won’t be an obstacle to our investigations.

We may also identify the spectrum of a spectral set $A\subseteq \mathbb {Z}_p^3$ with a subset of $\mathbb {Z}_p^3$ itself. So, $B\subseteq \mathbb {Z}_p^3$ is a spectrum of A if ${\left\langle {\xi _b,\xi _{b'}}\right\rangle }_A=0$, for every pair $b, b'$ of distinct elements in B. Hence,

$$\begin{aligned} 0={\left\langle {\xi _b,\xi _{b'}}\right\rangle }_A=\sum _{a\in A}\xi _b(a)\overline{\xi _{b'}(a)}=\sum _{x\in A}\textbf{1}_A(x)\xi _{b-b'}(x)=\widehat{\textbf{1}}_A(b'-b), \end{aligned}$$

where the function $\textbf{1}_A$ denotes the indicator function on the set A, yielding a basic fact on spectra.

Proposition 2.1

A set of pairwise orthogonal characters, restricted on A and indexed by $B\subseteq \mathbb {Z}_p^3$, satisfies

$$\begin{aligned} B-B:={\left\{ {b-b':b,b'\in B}\right\} }\subseteq Z(\widehat{\textbf{1}}_A)\cup {\left\{ {0}\right\} }, \end{aligned}$$

where $Z(f)={\left\{ {x\in \mathbb {Z}_p^3:f(x)=0}\right\} }$ is the zero set of the function f. In addition, $|A|=|B|$ holds, if and only if B is a spectrum of A.

Next we will study the main properties of spectral sets, especially the nontrivial ones, i. e. those that are neither singletons nor the entire group. First of all, any nontrivial spectral set $A\subseteq \mathbb {Z}_p^3$ must satisfy $Z(\widehat{\textbf{1}}_A)\ne \varnothing$. Therefore, the zero set $Z(\widehat{\textbf{1}}_A)$ is a union of punctured lines, i. e. one-dimensional subspaces of $\mathbb {Z}_p^3$ with the origin missing. Indeed, if $x\in Z(\widehat{\textbf{1}}_A)$ and $\lambda \in \mathbb {Z}_p^*$ is arbitrary, then

$$\begin{aligned} \widehat{\textbf{1}}_A(\lambda x)=\sum _{a\in A}\zeta _p^{\left\langle {\lambda x,-a}\right\rangle }=\sum _{a\in A}\sigma (\zeta _p^{\left\langle {x,-a}\right\rangle })=\sigma {\left( {\sum _{a\in A}\zeta _p^{\left\langle {x,-a}\right\rangle }}\right) }=\sigma (\widehat{\textbf{1}}_A(x))=0, \end{aligned}$$

where $\sigma \in \text {Gal}(\mathbb {Q}(\zeta _p)/\mathbb {Q})$ which satisfies $\sigma (\zeta _p)=\zeta _p^\lambda$. We usually denote by L and H lines and planes, and by $L^*$, $H^*$ their punctured counterparts, when L and H contain the origin.

The structure of the vanishing sums of p-th roots of unity² is the next crucial tool towards a first characterization of spectral subsets of $\mathbb {Z}_p^3$.

Proposition 2.2

If $a_0,\dotsc ,a_{p-1}$ are integers such that

$$\begin{aligned} a_0+a_1\zeta _p+\cdots +a_{p-1}\zeta _p^{p-1}=0, \end{aligned}$$

(2.2)

then $a_0=a_1=\cdots =a_{p-1}$.

Proof

Taking all possible Galois automorphisms on both sides of (2.2), we get

$$\begin{aligned} a_0+a_1\zeta _p^\lambda +\cdots +a_{p-1}\zeta _p^{\lambda (p-1)}=0, \;\;\; \forall 1\le \lambda \le p-1, \end{aligned}$$

or equivalently, the vector $\textbf{a}=(a_0,\dotsc ,a_{p-1})$ is orthogonal to each of the $p-1$ vectors $(1,\zeta _p^\lambda ,\dotsc ,\zeta _p^{\lambda (p-1)})$. Therefore, it must be parallel to the all-1 vector, $\textbf{1}=(1,1,\dotsc ,1)$, proving the desired fact. $\square$

A simple consequence of Proposition 2.2 is the fact that $x\in Z(\widehat{\textbf{1}}_A)$ if and only if A is equidistributed with respect to the p planes parallel to $x^\perp$. Indeed, if we denote

$$\begin{aligned} H_j={\left\{ {y\in \mathbb {Z}_p^3:{\left\langle {x,y}\right\rangle }=j}\right\} }, \end{aligned}$$

we have

$$\begin{aligned} 0=\widehat{\textbf{1}}_A(x)=\sum _{a\in A}\zeta _p^{\left\langle {x,a}\right\rangle }=\sum _{j=0}^{p-1}|A\cap H_j|\zeta _p^j, \end{aligned}$$

hence $|A\cap H_0|=|A\cap H_1|=\cdots =|A\cap H_{p-1}|$.

Theorem 2.3

Let $A\subseteq \mathbb {Z}_p^3$ be a nontrivial spectral set. Then $|A|=pk$, with $1\le k\le p$. Moreover,

• (1) if $|A|=p$ or $p^2$, then A is a tile.

• (2) if $1<k<p$ (assuming that such a spectral set exists), then $Z(\widehat{\textbf{1}}_A)$ intersects every punctured plane, but contains none.

Proof

Since we have shown that $\widehat{\textbf{1}}_A$ vanishes at some $x\in \mathbb {Z}_p^3$, A must be equidistributed with respect to the p planes parallel to $x^\perp$. This shows that $p\mid |A|$.

If $|A|>p^2$, then the same must hold for a spectrum B. Let $x\ne 0$ be arbitrary, and consider the intersections of B with the planes parallel to $x^\perp$; at least one of them must contain at least $p+1$ elements. But then, $B-B$ intersects all punctured lines in $x^\perp$. Indeed, let $b_0,\dotsc ,b_p\in B$ such that ${\left\langle {b_j,x}\right\rangle }$ is constant for $0\le j\le p$. Then, the $p+1$ elements of $B-b_0$, $0, b_1-b_0,\dotsc ,b_p-b_0$, are all in $x^\perp$; therefore, for every line $L\subseteq x^\perp$, there is a parallel one $L'$ in $x^\perp$ that contains at least two elements of $B-b_0$. The difference of these two elements belongs to $(B-B)\cap L^*$. Since x is arbitrary, and $L\subseteq x^\perp$ is arbitrary as well, the above argument shows that $B-B$ intersects every punctured line, hence Proposition 2.1 yields that $\widehat{\textbf{1}}_A$ vanishes on every nonzero element of $\mathbb {Z}_p^3$, and then (2.1) shows that $\textbf{1}_A(x)=\frac{|A|}{p^3}$ for all $x\in \mathbb {Z}_p^3$, which can only hold if $A=\mathbb {Z}_p^3$, contradicting the fact that A is nontrivial. Thus, $|A|=pk$, with $1\le k\le p$.

Suppose now that $k=1$; since $\widehat{\textbf{1}}_A$ vanishes at some $x\in \mathbb {Z}_p^3$, A must be equidistributed with respect to the planes parallel to $H=x^\perp$. which shows that every such plane must contain exactly one element of A. It is then obvious that $A\oplus H=\mathbb {Z}_p^3$, and A is a tile.

Next, suppose that $k=p$. If $A-A$ intersects every punctured line, then by Proposition 2.1 with the roles of A and B reversed, $\widehat{\textbf{1}}_B$ would vanish everywhere except for the origin, which would yield again $B=\mathbb {Z}_p^3$ as above, a contradiction. So, there is a line L through the origin such that $(A-A)\cap L^*=\varnothing$, hence $A+L=A\oplus L$. Since $|A|=p^2$ and $|L|=p$, we will have $A\oplus L=\mathbb {Z}_p^3$, showing that A is a tile.

Finally, assume that there exists a spectral set A with $1<k<p$. If there existed a plane H through the origin such that $Z(\widehat{\textbf{1}}_A)\cap H^*=\varnothing$, then $(B-B)\cap H={\left\{ {0}\right\} }$ for a spectrum B of A, by Proposition 2.1. But then no two elements of B would be congruent modulo the subgroup H, which would imply $|B|\le |\mathbb {Z}_p^3/H|=p<|A|$, a contradiction. Thus, $Z(\widehat{\textbf{1}}_A)$ intersects every punctured plane.

If there existed a plane H through the origin such that $H^*\subseteq Z(\widehat{\textbf{1}}_A)$, the plane itself would satisfy $H-H=H\subseteq Z(\widehat{\textbf{1}}_A)\cup {\left\{ {0}\right\} }$, which would imply that the $|H|=p^2$ characters $\xi _x$, for $x\in H$, would be pairwise orthogonal, when restricted to A. This is impossible, as the dimension of complex functions on A has cardinality $|A|<p^2$. Thus, $Z(\widehat{\textbf{1}}_A)$ contains no punctured plane, completing the proof. $\square$

We end this Section with the proof that T-S holds in $\mathbb {Z}_p^3$.

Theorem 2.4

Every tile $A\subseteq \mathbb {Z}_p^3$ is spectral.

Proof

Let T be a tiling complement of A. $A\oplus T=\mathbb {Z}_p^3$ written in functional notation is $\textbf{1}_A*\textbf{1}_T=\textbf{1}_{\mathbb {Z}_p^3}$. Applying the Fourier transform on both sides we get $\widehat{\textbf{1}}_A\widehat{\textbf{1}}_T=p^3\textbf{1}_0$, hence

$$\begin{aligned} {\text {supp}}(\widehat{\textbf{1}}_A)\cap {\text {supp}}(\widehat{\textbf{1}}_T)={\left\{ {0}\right\} }. \end{aligned}$$

(2.3)

We have already seen that both supports are unions of punctured lines, along with the origin. If ${\text {supp}}(\widehat{\textbf{1}}_A)=\mathbb {Z}_p^3$, then ${\text {supp}}(\widehat{\textbf{1}}_T)={\left\{ {0}\right\} }$, which implies that $T=\mathbb {Z}_p^3$ and A a singleton, and vice versa, if ${\text {supp}}(\widehat{\textbf{1}}_T)=\mathbb {Z}_p^3$, we get $A=\mathbb {Z}_p^3$. So, if we assume that A is nontrivial, so that $|A|=p$ or $p^2$, we get that both $\widehat{\textbf{1}}_A$ and $\widehat{\textbf{1}}_T$ must vanish somewhere.

Suppose first that $|A|=p$. Consider a punctured line $L^*\subseteq Z(\widehat{\textbf{1}}_A)$. Then, the line L is a spectrum of A by Proposition 2.1, as $|A|=|L|$ and $L-L=L\subseteq Z(\widehat{\textbf{1}}_A)\cup {\left\{ {0}\right\} }$.

Finally, suppose that $|A|=p^2$, hence $|T|=p$. Consider a line L through two points of T; now let a plane H through the origin that is orthogonal to the direction of L. For any $x\in H^*$, we must have $\widehat{\textbf{1}}_T(x)\ne 0$ by Proposition 2.2, since T is not equidistributed with respect to the planes parallel to $x^\perp$ (the one containing L has at least 2 elements of T). Therefore, (2.3) yields $H^*\subseteq Z(\widehat{\textbf{1}}_A)$, and H is a spectrum of A by Proposition 2.1, since $H-H=H\subseteq Z(\widehat{\textbf{1}}_A)\cup {\left\{ {0}\right\} }$ and $|H|=|A|$. $\square$

3 Delsarte’s method and balanced functions

Regardless whether $A\subseteq \mathbb {Z}_p^3$ is spectral or not, we may ask how many pairwise orthogonal characters exist, when they are restricted to A. Of course, there are at most $|A|$ many such characters, with equality if and only if A is spectral. In general, Proposition 2.1 shows that if we have a set of characters, indexed by $B\subseteq \mathbb {Z}_p^3$, such that ${\left\langle {\xi _b,\xi _{b'}}\right\rangle }_A=0$ for every $b,b'\in B$ with $b\ne b'$, then the difference set $B-B$ must avoid ${\text {supp}}(\widehat{\textbf{1}}_A)\setminus {\left\{ {0}\right\} }$, which may be viewed as a "forbidden set".

Definition 3.1

Let $E\subseteq G$, such that $0\in E=-E$. We call $h:G\rightarrow \mathbb {R}$ a witness function with respect to E, if it satisfies the following properties:

• (1) h is an even function, such that $h(x)\le 0$, $\forall x\in G\setminus E$.

• (2) $\hat{h}(\xi )\ge 0$, $\forall \xi \in \hat{G}$, and $\hat{h}(0)>0$.

The following theorem, which is also known as Delsarte’s method, is the main tool for estimating the maximal size of $|B|$, where A accepts pairwise orthogonal characters, restricted on A and indexed by B. The proof is adapted from [31], in the setting of finite additive groups.

Theorem 3.2

With notation as above, let $B\subseteq G$ be such that $B-B\subseteq {\left\{ {0}\right\} }\cup (G\setminus E)$, and let h be a witness function for E. Then,

$$\begin{aligned} |B|\le |G|\frac{h(0)}{\hat{h}(0)}. \end{aligned}$$

Proof

It holds

$$\begin{aligned} \hat{\textbf{1}}_B(\xi )=\sum _{b\in B}\xi (-b). \end{aligned}$$

Put

$$\begin{aligned} S=\sum _{\xi \in \hat{G}}|\hat{\textbf{1}}_B(\xi )|^2\hat{h}(\xi ). \end{aligned}$$

Since $\hat{h}$ is nonnegative, we obtain the estimate

$$\begin{aligned} S\ge |\hat{\textbf{1}}_B(0)|^2\hat{h}(0)=|B|^2\hat{h}(0). \end{aligned}$$

On the other hand,

$$\begin{aligned} S&=\sum _{\xi \in \hat{G}}\hat{\textbf{1}}_B(\xi )\overline{\hat{\textbf{1}}_B(\xi )}\hat{h}(\xi )\\&=\sum _{\xi \in \hat{G}}{\left( {\sum _{b\in B}\xi (-b)\sum _{b'\in B}\xi (b')}\right) }\hat{h}(\xi )\\&=\sum _{\xi \in \hat{G}}\sum _{b,b'\in B}\xi (b'-b)\hat{h}(\xi )\\&=|G|\sum _{b,b'\in B}h(b'-b)\\&\le |G||B|h(0), \end{aligned}$$

using the Fourier inversion, the definition of a witness function and the hypothesis $B-B\subseteq {\left\{ {0}\right\} }\cup (G\setminus E)$. Combining the two estimates above we obtain

$$\begin{aligned} |B|^2\hat{h}(0)\le S\le |B||G|h(0), \end{aligned}$$

thus,

$$\begin{aligned} |B|\le |G|\frac{h(0)}{\hat{h}(0)}, \end{aligned}$$

as desired. $\square$

Our goal is to show that there can be no spectral set, as described in Theorem 2.3(2), by finding a witness function h with respect to $E={\text {supp}}(\widehat{\textbf{1}}_A)$ that satisfies

$$\begin{aligned} \frac{h(0)}{\hat{h}(0)}<\frac{|A|}{p^3}=\frac{k}{p^2}. \end{aligned}$$

We note that the function $h_A=|\widehat{\textbf{1}}_A|^2$ attains equality, as

$$\begin{aligned} \widehat{|\widehat{\textbf{1}}_A|^2}=\widehat{\widehat{\textbf{1}_A*\textbf{1}_{-A}}}=p^3\textbf{1}_A*\textbf{1}_{-A} \end{aligned}$$

is a nonnegative function, and

$$\begin{aligned} \frac{h_A(0)}{\hat{h}_A(0)}=\frac{|A|^2}{p^3|A|}=\frac{|A|}{p^3}, \end{aligned}$$

proving the trivial fact from linear algebra, that the number of pairwise orthogonal functions on A are at most $|A|$. The last equality holds true also for the functions $h_{\lambda A}$, $1\le \lambda \le p-1$, as well as for their average,

$$\begin{aligned} h=\sum _{\lambda =1}^{p-1}h_{\lambda A}. \end{aligned}$$

The latter is constant on every punctured line, as

$$\begin{aligned} h(x)={\text {Tr}}_{\mathbb {Q}(\zeta _p)/\mathbb {Q}}|\widehat{\textbf{1}}_A(x)|^2, \;\;\; \forall x\in \mathbb {Z}_p^3, \end{aligned}$$

where ${\text {Tr}}_{\mathbb {Q}(\zeta _p)/\mathbb {Q}}$ denotes the trace of the Galois extension $\mathbb {Q}(\zeta _p)/\mathbb {Q}$. Moreover, h takes only integer values. A function that is constant on every punctured line in $\mathbb {Z}_p^3$ will be called balanced (equivalently, it is a homogeneous function of degree 0). The following lemma shows us that it suffices to restrict our attention to balanced functions, in search for a witness function with desirable conditions, since the "forbidden" set ${\text {supp}}(\widehat{\textbf{1}}_A)$ is a union of lines through the origin.

Lemma 3.3

Suppose that h is a witness function with respect to the forbidden set $E\subseteq \mathbb {Z}_p^3$, which is a union of lines through the origin. Then, there is a balanced witness function f, also with respect to E, which satisfies $h(0)=f(0)$ and $\hat{h}(0)=\hat{f}(0)$.

Proof

Consider the function $f:\mathbb {Z}_p^3\rightarrow \mathbb {R}$ which is defined by

$$\begin{aligned} f(x)=\frac{1}{p-1}\sum _{\lambda =1}^{p-1}h(\lambda x), \;\;\; \forall x\in \mathbb {Z}_p^3. \end{aligned}$$

f is obviously balanced, as h is, and for every $x\notin E$ we have $\lambda x\notin E$ for $1\le \lambda \le p-1$, as E is a union of lines. Therefore, $f(x)\le 0$, $\forall x\notin E$. We also have $\hat{f}\ge 0$, as

$$\begin{aligned} \hat{f}(y)&=\frac{1}{p-1}\sum _{\lambda =1}^{p-1}\sum _{x\in \mathbb {Z}_p^3}h(\lambda x)\zeta _p^{\left\langle {y,-x}\right\rangle }\\&= \frac{1}{p-1}\sum _{\lambda =1}^{p-1}\sum _{x\in \mathbb {Z}_p^3}h(x)\zeta _p^{\left\langle {\lambda ^{-1}y,-x}\right\rangle }\\&=\frac{1}{p-1}\sum _{\lambda =1}^{p-1}\hat{h}(\lambda ^{-1}y)\ge 0. \end{aligned}$$

Putting $y=0$ above we get $\hat{f}(0)=\hat{h}(0)$, and obviously satisfies $f(0)=h(0)$, completing the proof. $\square$

Fourier transforms of balanced functions are also balanced; indeed,

$$\begin{aligned}{} & {} \widehat{\textbf{1}}_0=\textbf{1}_{\mathbb {Z}_p^3}, \;\;\; \widehat{\textbf{1}}_L=p\textbf{1}_{L^\perp }, \;\;\; \widehat{\textbf{1}}_{L^*}=p\textbf{1}_{L^\perp }-\textbf{1}_{\mathbb {Z}_p^3}, \widehat{\textbf{1}}_E=p^2\textbf{1}_{E^\perp }, \;\;\; \nonumber \\{} & {} \widehat{\textbf{1}}_{E^*}=p^2\textbf{1}_{E^\perp }-\textbf{1}_{\mathbb {Z}_p^3}, \;\;\; \widehat{\textbf{1}}_{\mathbb {Z}_p^3}=p^3\textbf{1}_0, \end{aligned}$$

(3.1)

where L is a line through the origin.

4 Fourier analysis on the finite projective plane

Now that we have restricted on balanced functions, we may work with functions on the finite projective plane of order p, denoted by $\textbf{P}\mathbb {F}_p^2$, along with the point O, which corresponds to the origin of $\mathbb {Z}_p^3$. A balanced function $f:\mathbb {Z}_p^3\rightarrow \mathbb {C}$ corresponds to the function $\widetilde{f}:\textbf{P}\mathbb {F}_p^2\cup {\left\{ {O}\right\} }\rightarrow \mathbb {C}$, which is defined by

$$\begin{aligned} \widetilde{f}([x_1:x_2:x_3])=f(x_1,x_2,x_3), \;\;\; \forall x_1, x_2, x_3\in \mathbb {Z}_p. \end{aligned}$$

$\widetilde{f}$ is well-defined since f is balanced, and we note that we extend the notation on projective coordinates, by putting $O=[0:0:0]$. Since the Fourier transforms of balanced functions are also balanced, we define the Fourier transform for functions defined on $\textbf{P}\mathbb {F}_p^2\cup {\left\{ {O}\right\} }$ so that it satisfies $\widetilde{\hat{f}}=\widehat{\widetilde{f}}$. Similarly, for a set $D\subseteq \mathbb {Z}_p^3$, that is a union of lines through the origin, we denote by $\widetilde{D}$ the subset of $\textbf{P}\mathbb {F}_p^2\cup {\left\{ {O}\right\} }$ satisfying

$$\begin{aligned} \widetilde{D}={\left\{ {[x_1:x_2:x_3]:(x_1,x_2,x_3)\in D}\right\} }. \end{aligned}$$

If $P=[x_1:x_2:x_3]\in \textbf{P}\mathbb {F}_p^2$, we denote by $P^\perp$ the set

$$\begin{aligned} P^\perp ={\left\{ {[y_1:y_2:y_3]\ne O:x_1y_1+x_2y_2+x_3y_3=0}\right\} }. \end{aligned}$$

The indicator function of a subset $K\subseteq \textbf{P}\mathbb {F}_p^2\cup {\left\{ {O}\right\} }$ will be denoted by $\delta _K$, to make it clearer that we work on $\textbf{P}\mathbb {F}_p^2\cup {\left\{ {O}\right\} }$ and not $\mathbb {Z}_p^3$. If K consists of a single point P, then we write $\delta _P$ instead of $\delta _{\left\{ {P}\right\} }$. Lastly, the constant function on $\textbf{P}\mathbb {F}_p^2\cup {\left\{ {O}\right\} }$ which equals 1, is denoted simply by $\textbf{1}$. The equations (3.1) are then transformed to

$$\begin{aligned} \widehat{\delta }_O=\textbf{1}, \;\;\; \widehat{\delta }_P=p\delta _{P^\perp }+p\delta _O-\textbf{1}. \end{aligned}$$

(4.1)

When we pass from $\mathbb {Z}_p^3\setminus {\left\{ {0}\right\} }$ to $\textbf{P}\mathbb {F}_p^2$, punctured lines collapse to points and punctured planes collapse to lines; moreover, coplanar punctured lines correspond to collinear points on the projective plane.

Now consider a subset $A\subseteq \mathbb {Z}_p^3$ having pk elements, where $1<k<p$. As we have seen from Theorem 2.3(2), if A is spectral, then $Z(\widehat{\textbf{1}}_A)$ intersects every punctured plane, but contains none. Therefore, the set $Z=\widetilde{Z(\widehat{\textbf{1}}_A)}\subseteq \textbf{P}\mathbb {F}_p^2$ intersects every projective line, but contains none.

Definition 4.1

A set $S\subseteq \textbf{P}\mathbb {F}_p^2$ is called a blocking set if it intersects every projective line, but contains none. S is called a minimal blocking set, if $S\setminus {\left\{ {P}\right\} }$ is not a blocking set, for every $P\in S$.

So the set Z is a blocking set, and so is the complement of every blocking set, by definition. We denote by S the complement of Z in $\textbf{P}\mathbb {F}_p^2$, or equivalently,

$$\begin{aligned} S=\widetilde{{\text {supp}}(\widehat{\textbf{1}}_A)}\setminus {\left\{ {O}\right\} }. \end{aligned}$$

(4.2)

The following bounds on the size of blocking sets were given by [3, 5].

Theorem 4.2

Let $S\subseteq \textbf{P}\mathbb {F}_p^2$ be a blocking set. Then,

$$\begin{aligned} \frac{3}{2}(p+1)\le |S|\le p^2-\frac{1}{2}p-\frac{1}{2}. \end{aligned}$$

If S is a minimal blocking set, then

$$\begin{aligned} |S|<p\sqrt{p}+1. \end{aligned}$$

Our ultimate goal is to prove the S-T direction in $\mathbb {Z}_p^3$. If counterexamples exist, they are described by Theorem 2.3(2). In view of Lemma 3.3, it suffices to show that there is a balanced witness function h with respect to ${\text {supp}}(\widehat{\textbf{1}}_A)$, such that

$$\begin{aligned} \frac{h(0)}{\hat{h}(0)}<\frac{k}{p^2}, \end{aligned}$$

where $|A|=pk$, and $Z(\widehat{\textbf{1}}_A)$ intersects every punctured plane, but contains none. Equivalently, it suffices to find a function $\widetilde{h}:\textbf{P}\mathbb {F}_p^2\cup {\left\{ {O}\right\} }\rightarrow \mathbb {R}$, such that $\widetilde{h}(P)\le 0$, $\forall P\in Z$, and $\widehat{\widetilde{h}}\ge 0$, such that

$$\begin{aligned} \frac{\widetilde{h}(O)}{\widehat{\widetilde{h}}(O)}<\frac{k}{p^2}. \end{aligned}$$

We will actually restrict to functions $\widetilde{h}$ that vanish on Z, so that $\widetilde{h}\ge 0$ as well. Such functions (i. e. functions f such that $f, \hat{f}\ge 0$) are called positive definite.

5 Proof of Theorem 1.4

Let S be the set defined as in (4.2), and let $S'\subseteq S$ be a minimal blocking set.³ We define

$$\begin{aligned} \widetilde{h}=\delta _{S'}+(|S'|-p)\delta _O. \end{aligned}$$

$\widetilde{h}$ obviously vanishes on Z, so to show that it is indeed a witness function with respect to S, we have to show that $\widetilde{h}$ is positive definite. Since $\widetilde{h}\ge 0$, we only have to compute the Fourier transform using (4.1),

$$\begin{aligned} \widehat{\widetilde{h}}=\sum _{P\in S'}p\delta _{P^\perp }+p|S'|\delta _O-|S'|\textbf{1}+|S'|\textbf{1}-p\textbf{1}=p{\left( {\sum _{P\in S'}\delta _{P^\perp }+|S'|\delta _O-\textbf{1}}\right) }. \end{aligned}$$

The fact that $\widehat{\widetilde{h}}\ge 0$ follows from the fact that $S'$ is a blocking set; indeed, as for an arbitrary $Q\in \textbf{P}\mathbb {F}_p^2$ it holds

$$\begin{aligned} \widehat{\widetilde{h}}(Q)=p{\left( {\sum _{P\in S'}\delta _{P^\perp }(Q)-1}\right) }=p{\left( {\sum _{P\in S'}\delta _{Q^\perp }(P)-1}\right) }=p(|S'\cap Q^\perp |-1)\ge 0. \end{aligned}$$

Finally, $\widetilde{h}(O)=|S'|-p>0$, $\widehat{\widetilde{h}}(O)=p(|S'|-1)>0$, and

$$\begin{aligned} \frac{\widetilde{h}(O)}{\widehat{\widetilde{h}}(O)}=\frac{|S'|-p}{p(|S'|-1)}. \end{aligned}$$

When is the latter less than $k/p^2$? Putting $k=p-m$, this is equivalent to

$$\begin{aligned} |S'|-1<\frac{p(p-1)}{m}, \end{aligned}$$

which is satisfied if

$$\begin{aligned} m<\frac{p-1}{\sqrt{p}}, \end{aligned}$$

as

$$\begin{aligned} \frac{p-1}{\sqrt{p}}=\frac{p(p-1)}{p\sqrt{p}}<\frac{p(p-1)}{|S'|-1}, \end{aligned}$$

by virtue of Theorem 4.2. This shows that if

$$\begin{aligned} p^2-p\sqrt{p}+\sqrt{p}<p(p-m)=|A|<p^2, \end{aligned}$$

then A cannot be spectral, completing the proof of Theorem 1.4.

6 Conclusion

Delsarte’s method provided a partial result for Fuglede’s conjecture in $\mathbb {Z}_p^3$. This can be improved further, using Lemma 3 from [13]; translated to the terminology of this paper, this shows that either a spectral set $A\subseteq \mathbb {Z}_p^3$ with $|A|=pk$, $1<k<p$ (if it exists!), is contained in k parallel planes, or $\widetilde{h}$ is supported on at least 3 points on every projective line. In the latter case, the set S is a 3-fold blocking set, and if $S'\subseteq S$ is a minimal 3-fold blocking set, we define

$$\begin{aligned} \widetilde{h}=\delta _{S'}+(|S'|-3p)\delta _O. \end{aligned}$$

Working exactly as in Sect. 5, we obtain that A cannot be spectral if

$$\begin{aligned} p{\left( {p-\frac{3(p-1)}{\sqrt{3p-5}+1}}\right) }<|A|<p^2, \end{aligned}$$

(6.1)

using the upper bound on the size of a 3-fold blocking set in $\textbf{P}\mathbb {F}_p^2$, given by [2]

$$\begin{aligned} \frac{1}{2}p\sqrt{12p-20}+p. \end{aligned}$$

This method has limitations; even if S were a t-fold blocking set, and $S'\subseteq S$ a minimal subset with the same property, then the witness function

$$\begin{aligned} \widetilde{h}=\delta _{S'}+(|S'|-tp)\delta _O \end{aligned}$$

(6.2)

would improve (6.1) using [2, Theorem  1.1], but not to the point where sets of cardinality $|A|=2p$ are included, since $t\le p-2$. We note that we cannot have $t\ge p-1$, since that would imply that no three points of Z are collinear; if that were true, then considering all lines through a single point of Z would give the upper bound $|Z|\le p+2$, which violates Theorem 4.2, since Z is also a blocking set.

If we are in the former case, where S is not a 3-fold blocking set, then the best we can do with this approach is to examine whether S contains minimal blocking sets of small cardinality. However, even the lower bound described by Theorem 4.2, would improve Theorem 1.4 to

$$\begin{aligned} p\cdot \frac{p^2+5p}{3p+1}<|A|<p^2, \end{aligned}$$

at best case scenario, which does not include $|A|=kp$ for roughly p/3 values of k.

So, how can we approach uniformly the cases where $|A|=kp$, with k small? The limitations of the method described so far is that we consider only witness functions of the form (6.2), which seems to tackle the cases where $|A|$ is large. A cleverer selection of a witness function supported on $S\cup {\left\{ {O}\right\} }$, perhaps in a probabilistic fashion, might give the answer to this question.