Strictly positive definite kernels on a product of circles

Abstract

We supply a Fourier characterization for the real, continuous, isotropic and strictly positive definite kernels on a product of circles. In other words, if $S^1$ is the unit circle in $\mathbb {R}^2$, $\cdot$ is the usual inner product of $\mathbb {R}^2$ and f is a real continuous function on $[-1,1]^2$, we determine necessary and sufficient conditions in order that $f(x\cdot y,z \cdot w)$ be a strictly positive definite kernel on $S^1 \times S^1$.

1 Introduction

Positive definite functions and kernels have a long history in mathematics, entering as an important tool in harmonic analysis and other areas as well. In the spherical setting, they can be traced back to the remarkable paper of Schoenberg published in 1942 [19], where a characterization for the continuous, isotropic and positive definite kernels on a single sphere was obtained. This characterization is far-reaching, having applications in approximation theory, spatial statistics, geomathematics, discrete geometry, etc. We mention [3, 4, 6, 16] and references therein for some applications of positive definite functions and kernels on spheres.

In this paper, we will be concerned with positive definite kernels on a product of circles. As so, we will recast the basic concepts and results from Schoenberg’s work that applies to circles, up to the point we can state what the main contribution in the present paper is.

We will write $S^1$ to denote the unit circle in $\mathbb {R}^{2}$. Continuity of a kernel K on $S^1$ will be attached to the usual geodesic distance on $S^1$ and that will be extended to the product $S^1 \times S^1$ in the usual way. The isotropy or radiality of a kernel K on $S^1$ refers to the existence of a function $K_r$ on $[-1,1]$ so that

$$\begin{aligned} K(x,y)=K_r(x \cdot y), \quad x,y \in S^1, \end{aligned}$$

in which $\cdot$ is the usual inner product of $\mathbb {R}^2$. For a kernel K on $S^1 \times S^1$, isotropy corresponds to the property

$$\begin{aligned} K((x,z),(y,w))=K_r(x \cdot y, z\cdot w), \quad x,y,z,w \in S^1, \end{aligned}$$

in which the function $K_r$ has now domain $[-1,1]^2$. In both cases, we will call $K_r$ the isotropic part of the kernel K. Finally, the positive definiteness of a real kernel K on an infinite set X refers to the validity of the inequality

$$\begin{aligned} \sum _{\mu ,\nu =1}^n c_\mu c_\nu K(x_\mu , x_\nu ) \ge 0, \end{aligned}$$

whenever n is a positive integer, $x_1, x_2, \ldots , x_n$ are distinct points on X and the $c_\mu$ are real scalars. The strict positive definiteness of K demands both, its positive definiteness and that the inequalities above be strict whenever at least one of the $c_\mu$ is nonzero. We will apply these definitions to the cases in which either $X=S^1$ or $X=S^1 \times S^1$.

According to a result of Schoenberg in [19], a real, continuous and isotropic kernel K on $S^1$ is positive definite if, and only if, the isotropic part $K_r$ of K has the form

$$\begin{aligned} K_r(t)=\sum _{k=0}^\infty a_k P_k(t), \quad t \in [-1,1], \end{aligned}$$

in which all the $a_k$ are nonnegative, $P_k$ is the Tchebyshev polynomial (of first kind) of degree k (see [23]), and $\sum _{k=0}^\infty a_k P_k(1)<\infty$. In the nineties, due to the appearance of the so-called radial basis interpolation on spheres, many attempts were made in order to deduce a similar characterization for the strict case [11–15, 18, 20, 22] but that only appeared in [15] (see also [2]): a kernel having Schoenberg’s representation described above is strictly positive definite on $S^1$ if, and only if, the set $\{k:a_{|k|}>0\}$ intersects every full arithmetic progression in $\mathbb {Z}$. Both results described above extends to the complex setting, that is, to the case in which $S^1$ is replaced with the unit circle in $\mathbb {C}$, the positive definite kernel is allowed to assume complex values and the scalars $c_\mu$ are now complex numbers. This extension is also discussed in [15]. It is worth to mention [21] where the very same problem was discussed.

Moving to $S^1 \times S^1$, a theorem proved in [8] includes a characterization for the positive definiteness of a real, continuous and isotropic kernel K on $S^1 \times S^1$ as those having an isotropic part in the form

$$\begin{aligned} K_r(t,s)=\sum _{k,l=0}^\infty a_{k,l} P_k(t)P_l(s), \quad t,s \in [-1,1], \end{aligned}$$

(1.1)

where all the coefficients $a_{k,l}$ are nonnegative and $\sum _{k,l=0}^\infty a_{k,l} P_k(1)P_l(1)<\infty$. This characterization can also be deduced from abstract versions of a classical result of S. Bochner on positive definiteness, a typical example being Theorem 4.11 in [1]. For a function $K_r$ representable as in (1.1), we will write

$$\begin{aligned} J_K:=\big \{(k,l)\in \mathbb {Z}_+^2: a_{k,l}>0\big \}. \end{aligned}$$

The results in this paper will converge to the characterization for strict positive definiteness on $S^1 \times S^1$ described in Theorem 1.1 below. The result is supplementary to one of the main results proved in [7], where a similar problem was considered and solved for the product of higher dimensional spheres.

Theorem 1.1

Let K be a real, continuous, isotropic and positive definite kernel on $S^1 \times S^1$. The following assertions are equivalent.

• (i) K is strictly positive definite;

• (ii) The set $\{(k,l): (|k|,|l|)\in J_K\}$ intersects all the translations of each subgroup of $\mathbb {Z}^2$ having the form $\{(pa, qb): q,p\in \mathbb {Z}\}$, $a,b>0$;

• (iii) The set $\{(k,l): (|k|,|l|)\in J_K\}$ intersects all the translations of each subgroup of $\mathbb {Z}^2$ having the form $(a,b)\mathbb {Z}+(0,d)\mathbb {Z}$, $a,d>0$.

If this is the case, the set $\{(k,l): (|k|,|l|)\in J_K\}$ intersects all the translations of each lattice of $\mathbb {Z}^2$ infinitely many times.

At this time, we have found no practical problems where the characterization described in Theorem 1.1 could enter in a decisive manner. Strict positive definiteness on a product of manifolds is a quite new subject and we are sure that potential applications will appear. In particular, we expect the result to have applicability in approximation theory, probability theory, stochastic processes and code theory.

The paper proceeds as follows. In Sect. 2, we present a list of technical results that will be needed in the presentation of the proof to Theorem . The proof itself appears in Sect. 3.

2 Preliminary results

This section contains several technical results to be used in the proof of the main theorem described in Sect. 1. They are presented following the order they will be required in the proof of Theorem 1.1.

In the first proposition, we explore a little bit deeper the concept of strict positive definiteness on $S^1 \times S^1$. The outcome implies an obvious equivalence for that concept.

It is an easy matter to verify that the strict positive definiteness of the kernel K with isotropic part given by (1.1) depends upon $J_K$ only and not on the actual values of the Fourier coefficients $a_{k,l}$. For distinct points $(x_1,w_1),(x_2,w_2), \ldots ,(x_n,w_n)$ on $S^1 \times S^1$, we will write $A=(A_{\mu \nu })$, in which

$$\begin{aligned} A_{\mu \nu }=K_r(x_\mu \cdot x_\nu , w_\mu \cdot w_\nu ), \quad \mu ,\nu =1,2,\ldots ,n. \end{aligned}$$

We will also represent the points above in polar form:

$$\begin{aligned} x_\mu =(\cos \theta _\mu , \sin \theta _\mu ), \quad w_\mu =(\cos \phi _\mu , \sin \phi _\mu ), \quad \theta _\mu ,\phi _\mu {\in } [0,2\pi ),\quad \mu =1,2,\ldots ,n. \end{aligned}$$

Proposition 2.1

Let K be a nonzero, real, continuous, isotropic and positive definite kernel on $S^1 \times S^1$. For distinct points $(x_1,w_1),(x_2,w_2), \ldots ,(x_n,w_n)$ on $S^1 \times S^1$ and a column vector $c=(c_\mu )$ in $\mathbb {R}^n$, the following statements are equivalent:

• (i) $c^{t}Ac=0$;

• (ii) The double equality

$$\begin{aligned} \sum _{\mu =1}^{n}c_{\mu }e^{ik\theta _{\mu }}e^{il\phi _{\mu }}=\sum _{\mu =1}^{n}c_{\mu }e^{ik\theta _{\mu }}e^{-il\phi _{\mu }}=0 \end{aligned}$$

holds for all $(k,l)\in J_K$.

Proof

The normalization one decides to adopt for the Tchebyshev polynomials is of no importance in this paper. So, we will write

$$\begin{aligned} P_k(x_\mu \cdot x_\nu )=\frac{2}{k} \cos k(\theta _\mu -\theta _\nu ), \quad k>0, \quad x_\mu \in S^1, \quad \mu =1,2,\ldots ,n, \end{aligned}$$

while $P_0(x_\mu \cdot x_\nu )=1$, $\mu =1,2,\ldots ,n$. The equality $c^{t}Ac=0$ is equivalent to

$$\begin{aligned} \sum _{\mu ,\nu =1}^{n} c_\mu c_\nu P_k(x_\mu \cdot x_\nu )P_l(w_\mu \cdot w_\nu )=0, \quad (k,l) \in J_K. \end{aligned}$$

Introducing the polar representation for the points and arranging, the equality appearing in the formula above becomes

$$\begin{aligned}&\left| \sum _{\mu =1}^n c_\mu \cos k\theta _{\mu } \cos l\phi _{\mu }\right| ^2+\left| \sum _{\mu =1}^n c_\mu \cos k\theta _{\mu } \sin l\phi _{\mu }\right| ^2+\left| \sum _{\mu =1}^n c_\mu \sin k\theta _{\mu } \cos l\phi _{\mu }\right| ^2\\&\qquad + \left| \sum _{\mu =1}^n c_\mu \sin k\theta _{\mu } \sin l\phi _{\mu }\right| ^2=0. \end{aligned}$$

Thus, $c^{t}Ac=0$ is equivalent to

$$\begin{aligned} \sum _{\mu =1}^n c_\mu \cos k\theta _{\mu } \cos l\phi _{\mu }=\sum _{\mu =1}^n c_\mu \cos k\theta _{\mu } \sin l\phi _{\mu }=0 \end{aligned}$$

and

$$\begin{aligned} \sum _{\mu =1}^n c_\mu \sin k\theta _{\mu } \cos l\phi _{\mu }=\sum _{\mu =1}^n c_\mu \sin k\theta _{\mu } \sin l\phi _{\mu }=0 \end{aligned}$$

for $(k,j) \in J_K$. An obvious manipulation of these equations taking into account the fact that $c_\mu$ are real numbers leads to the double equality in (ii). $\square$

From now on, we will deal with subgroups of $\mathbb {Z}^2$ and their translations. We will need a classification for the nontrivial subgroups of $\mathbb {Z}^2$. Let S be such a subgroup and let$\{(1,0),(0,1)\}$ be the canonical basis of $\mathbb {Z}^2$. Write $p_1$ to denote the canonical projection of $\mathbb {Z}^2$ onto its first component. If $p_1(S)=0$, then S is a subgroup of $(0,1)\mathbb {Z}$. Otherwise, $p_1(S)$ is a nontrivial subgroup of $(1,0)\mathbb {Z}$, say, $(a,0)\mathbb {Z}$, with $a>0$, and we can pick $y \in \mathbb {Z}^2$ so that $p_1(y)=a$. Now, if $x \in S$, then $p_1(x)\in a \mathbb {Z}$, that is, $x-\alpha y \in (0,1)\mathbb {Z}$, for some $\alpha \in \mathbb {Z}$. In other words, $S=y\mathbb {Z}\oplus (S \cap (0,1)\mathbb {Z})$. If $S \cap (0,1)\mathbb {Z}=0$, then $S=y\mathbb {Z}$. Otherwise, $S=y\mathbb {Z}\oplus b\mathbb {Z}$, in which $b\mathbb {Z}$ is a subgroup of $S \cap (0,1)\mathbb {Z}$. The outcome of this brief discussion is this one.

Lemma 2.2

A nontrivial subgroup of $\mathbb {Z}^2$ belongs to one of the following categories:

• (i) $(0,b)\mathbb {Z}:=\{(0,pb): p \in \mathbb {Z}\}$, $b>0$;

• (ii) $(a,b)\mathbb {Z}:=\{(pa, pb): p \in \mathbb {Z}\}$, $a>0$;

• (iii) $(a,b)\mathbb {Z}+(0,d)\mathbb {Z}:=\{(pa,pb+qd): p,q \in \mathbb {Z}\}$, $a,d>0$.

We advise the reader that there are different ways to describe the subgroups of $\mathbb {Z}^2$ (for instance, the one presented in [17] is slightly different and quite elegant). The subgroups that fit into Lemma 2.2-(iii) will be called lattices. The set of lattices of $\mathbb {Z}^2$ encompasses all the subgroups of rank 2. If $ad=1$, then a lattice becomes the whole $\mathbb {Z}^2$, otherwise it is a proper subgroup of $\mathbb {Z}^2$. The lattices having the form

$$\begin{aligned} (a\mathbb {Z},b\mathbb {Z}):=\{(pa, qb): q,p\in \mathbb {Z}\},\quad a,b>0, \end{aligned}$$

will be called rectangular lattices of $\mathbb {Z}^2$. By translates of subgroups of $\mathbb {Z}^2$, we will mean sets of the form $(j,j')+S$, in which $(j,j')$ is a fixed element of $\mathbb {Z}^2$ and S is a subgroup of $\mathbb {Z}^2$.

Lemma 2.3 below provides a decomposition of a lattice through translations of rectangular lattices.

Lemma 2.3

The lattice $L=(a,b)\mathbb {Z}+(0,d)\mathbb {Z}$, $a,d>0$, can be decomposed in the form

$$\begin{aligned} L=\bigcup _{(j,j')\in A}\left[ (j,j')+(ad\mathbb {Z}, ad\mathbb {Z})\right] , \end{aligned}$$

in which $A=L\cap \{(\alpha ,\beta ) \in \mathbb {Z}^2: 0\le \alpha ,\beta < ad\}$.

Proof

For $p,q \in \mathbb {Z}$, we can certainly write

$$\begin{aligned} (pa,pb+qd)=(j+\alpha ad, j'+ \beta ad), \end{aligned}$$

in which $j,j'\in \{0,1,\ldots ,ad-1\}$. Since

$$\begin{aligned} (j,j')=(p-\alpha d)(a,b)+(\alpha b +q-\beta a)(0,d), \end{aligned}$$

it is clear that $(j,j') \in L$. These arguments show that L is a subset of the union quoted in the statement of the lemma. As for the reverse inclusion, first observe that if $\alpha ,\beta \in \mathbb {Z}$, we have that

$$\begin{aligned} (\alpha ad,\beta ad)=d\alpha (a, b) +(a\beta -b\alpha )(0,d) \in L. \end{aligned}$$

Since L is a subgroup of $\mathbb {Z}^2$, $(j,j')+(\alpha ad,\beta ad) \in L$ whenever $(j,j')\in A$. $\square$

Next, we recall an elementary bi-dimensional version of the Skolem-Mahler-Lech Theorem due to Laurent [9, 10]. The original Skolem-Mahler-Lech Theorem is discussed in details in [5]. This very same bi-dimensional version was used in [17] in order to characterize certain strictly positive definite kernels on complex Hilbert spaces.

Theorem 2.4

Let $\{(x_1,w_1), (x_2,w_2), \ldots , (x_n,w_n)\}$ be a subset of $(\mathbb {C}{\setminus }\{0\})^2$. For n complex numbers $c_1,c_2, \ldots , c_n$, define a double sequence $\{b_{k,l}:k,l \in \mathbb {Z}\}$ through the formula

$$\begin{aligned} b_{k,l}:=\sum _{\mu =1}^{n}c_\mu x_\mu ^k w_\mu ^l, \quad k,l\in \mathbb {Z}. \end{aligned}$$

Then, the set $\{(k,l): b_{k,l}=0\}$ is the union of a finite number of translates of subgroups of $\mathbb {Z}^2$.

The technical lemma below adds to Theorem 2.4 when the points are distinct and belong to $\Omega _2 \times \Omega _2$, in which $\Omega _2$ is the unit circle in $\mathbb {C}$.

Lemma 2.5

Let $(x_1,w_1), (x_2,w_2), \ldots , (x_n,w_n)$ be distinct points in $\Omega _2 \times \Omega _2$. For complex numbers $c_1,c_2, \ldots , c_n$, define

$$\begin{aligned} b_{k,l}=\sum _{\mu =1}^{n}c_\mu x_\mu ^k w_\mu ^l, \quad k,l \in \mathbb {Z}. \end{aligned}$$

If $\{(k,l): b_{k,l}=0\}=\mathbb {Z}^2$, then all the $c_\mu$ are zero.

Proof

We will write the components of the points in polar form $x_\mu =e^{i\theta _\mu }$, $w_\mu =e^{i\phi _\mu }$, $\mu =1,2,\ldots ,n$, and will assume, as we can, that the n points $(\theta _1,\phi _1), (\theta _2, \phi _2), \ldots ,$ $(\theta _n, \phi _n)$ are distinct in $[0,2\pi )^2$. Choose $\alpha ,\beta \in \mathbb {Z}$ in such a way that all the elements in the set

$$\begin{aligned} \left\{ \alpha \frac{\theta _\mu -\theta _\nu }{2\pi }+\beta \frac{\phi _\mu -\phi _\nu }{2\pi }: \mu ,\nu =1,2,\ldots ,n; \mu \ne \nu \right\} \end{aligned}$$

are nonzero. Next, pick $\gamma \in \mathbb {Z}_+$ arbitrarily large so that

$$\begin{aligned} \left\{ \frac{\alpha }{\gamma }\frac{\theta _\mu -\theta _\nu }{2\pi }+\frac{\beta }{\gamma } \frac{\phi _\mu -\phi _\nu }{2\pi }: \mu ,\nu =1,2,\ldots ,n; \mu \ne \nu \right\} \subset (-1,1){\setminus } \{0\}. \end{aligned}$$

For each pair $(\mu ,\nu )$, $\mu \ne \nu$, for which

$$\begin{aligned} \frac{\alpha }{\gamma } \frac{\theta _\mu -\theta _\nu }{2\pi }+\frac{\beta }{\gamma } \frac{\phi _\mu -\phi _\nu }{2\pi } \in \mathbb {Q}, \end{aligned}$$

let $p_{\mu \nu }$ be a positive integer $>\gamma$ satisfying

$$\begin{aligned} p_{\mu \nu }\left( \frac{\alpha }{\gamma } \frac{\theta _\mu -\theta _\nu }{2\pi }+\frac{\beta }{\gamma } \frac{\phi _\mu -\phi _\nu }{2\pi } \right) \in \mathbb {Z}. \end{aligned}$$

Finally, select an integer q so that q is greater then all the $p_{\mu \nu }$ and each set $\{q,p_{\mu \nu }\}$ is coprime. If $\{(k,l): b_{k,l}=0\}=\mathbb {Z}^2$, then we may infer that

$$\begin{aligned} \sum _{\mu =1}^n c_\mu e^{i\theta _\mu k}e^{i\phi _\mu l}=0, \ (k,l) =(0,0),(\alpha q,\beta q),\ldots , ((n-1)\alpha q,(n-1)\beta q). \end{aligned}$$

The matrix of the system above has $\mu \nu$-entries given by

$$\begin{aligned} \left[ e^{i(\alpha \theta _\mu +\beta \phi _\mu )q}\right] ^{\nu },\quad \mu ,\nu =1,2,\ldots ,n, \end{aligned}$$

and, consequently, it is a Vandermonde matrix. So, the proof of the lemma will be complete as long as we show that the n points $e^{i(\alpha \theta _\mu +\beta \phi _\mu )q}$, $\mu =1,2,\ldots , n$, are distinct. But, for $\mu \ne \nu$,

$$\begin{aligned} e^{i(\alpha \theta _\mu +\beta \phi _\mu )q}= e^{i(\alpha \theta _\nu +\beta \phi _\nu )q} \end{aligned}$$

if, and only if,

$$\begin{aligned} q \left( \alpha \frac{\theta _\mu -\theta _\nu }{2\pi } +\beta \frac{\phi _\mu -\phi _\nu }{2\pi }\right) \in \mathbb {Z}. \end{aligned}$$

If all the numbers

$$\begin{aligned} \left( \alpha \frac{\theta _\mu -\theta _\nu }{2\pi } +\beta \frac{\phi _\mu -\phi _\nu }{2\pi }\right) ,\quad \mu \ne \nu , \end{aligned}$$

are irrational, we are done. Otherwise, there would be integers j and $j'$ such that

$$\begin{aligned} \frac{j}{\gamma q}=\frac{ j'}{p_{\mu \nu }}\in (-1,1){\setminus } \{0\}, \end{aligned}$$

for some pair $(\mu ,\nu )$, $\mu \ne \nu$. Since $\{q,p_{\mu \nu }\}$ is coprime, then $p_{\mu \nu }$ would divide $\gamma$, contradicting our choice of $p_{\mu \nu }$. $\square$

The next result reveals that if a proper subset A of $\mathbb {Z}^2$ is a finite union of translates of subgroups of $\mathbb {Z}^2$, then there exists a rectangular lattice H of $\mathbb {Z}^2$ and $(j,j') \in \mathbb {Z}^2$ so that $[(j,j')+H]\cap A=\emptyset$.

Lemma 2.6

Let A be a proper subset of $\mathbb {Z}^2$. If A is a finite union of translates of subgroups of $\mathbb {Z}^2$ and $(j,j')\in \mathbb {Z}^2{\setminus } A$, then there exists a rectangular lattice H of $\mathbb {Z}^2$ such that $(j,j')+H \subset \mathbb {Z}^2{\setminus } A$.

Proof

If A is a finite union of translates of subgroups of $\mathbb {Z}^2$, we can write

$$\begin{aligned} A=F \cup [(j_1,j_1') +G_1] \cup [(j_2,j_2') +G_2] \cup \ldots \cup [(j_r,j_r') +G_r] \end{aligned}$$

in which F is a finite (possibly empty) subset of $\mathbb {Z}^2$, $(j_1,j_1'), (j_2,j_2'), \ldots , (j_r,j_r')\in \mathbb {Z}^2$ and $G_1, G_2, \ldots , G_r$ are nontrivial subgroups of $\mathbb {Z}^2$. It suffices to prove the lemma in the case in which $F=\emptyset$. Indeed, if a solution $(j,j')+H$ is available for that case, we can pick a convenient subgroup $H_1$ of H so that $(j,j')+H_1$ avoids all the elements of F. So, assume that $F=\emptyset$ and fix $(j,j') \in \mathbb {Z}^2 {\setminus } A$. We can assume all the $G_i$ have rank 2. Indeed, if $G_i$ has rank 1 for some i, we can pick $(\alpha ,\beta ) \in \mathbb {Z}^2$ such that

$$\begin{aligned} (\alpha , \beta )\mathbb {Z} \cap [(j_i-j,j_i'-j')+G_i]=\emptyset . \end{aligned}$$

Hence,

$$\begin{aligned}{}[(j,j')+(\alpha , \beta )\mathbb {Z}] \cap [(j_i,j_i')+G_i]=\emptyset , \end{aligned}$$

and, therefore,

$$\begin{aligned} (j,j') \not \in (j_i,j_i')+(\alpha , \beta )\mathbb {Z} +G_i. \end{aligned}$$

In particular, $(\alpha , \beta )\mathbb {Z} +G_i$ is a subgroup of rank 2 and we can replace $(j_i,j_i)+G_i$ with $(j_i,j_i')+(\alpha , \beta )\mathbb {Z} +G_i$ in the union decomposition for A keeping $(j,j')$ in $\mathbb {Z}^2{\setminus } A$. If all the $G_i$ have rank 2, the proof of the lemma proceeds as follows. Let $m_i$ be the index of $G_i$ in $\mathbb {Z}^2$, $i=1,2,\ldots ,r$, and pick a common multiple m of all the $m_i$. The subgroup $(m\mathbb {Z},m\mathbb {Z})$ is a rectangular lattice and, by the definition of index of a subgroup, it follows that

$$\begin{aligned} (m\mathbb {Z},m\mathbb {Z}) \subset G_i \quad i=1,2,\ldots ,r. \end{aligned}$$

In particular,

$$\begin{aligned}{}[(j,j')+(m\mathbb {Z},m\mathbb {Z})]\cap G_i=\emptyset , \quad i=1,2,\ldots ,r, \end{aligned}$$

and, consequently, $(j,j')+(m\mathbb {Z},m\mathbb {Z}) \subset \mathbb {Z}^2{\setminus } A$. $\square$

We conclude the section with a technical result on sets that intersect all the translations of each lattice in $\mathbb {Z}^2$.

Lemma 2.7

If A is a subset of $\mathbb {Z}^2$ that intersects all the translations of each lattice in $\mathbb {Z}^2$, then each intersection is an infinite set.

Proof

Let A be a subset of $\mathbb {Z}^2$ that intersects all the translations of each lattice in $\mathbb {Z}^2$. Let $L=(j,j')+(a,b)\mathbb {Z}+(0,d)\mathbb {Z}$, $a,d>0$, and assume that $A\cap L$ is finite. Write

$$\begin{aligned} (j+p_1a,j'+p_1b+q_1d), (j+p_2a,j'+p_2b+q_2d), \ldots , (j+p_na,j'+p_nb+q_nd), \end{aligned}$$

to denote the elements in the intersection and define

$$\begin{aligned} p:=\max \{|p_1|,|p_2|,\ldots ,|p_n|\} \quad \text{ and } \quad q:=\max \{|q_1|,|q_2|,\ldots ,|q_n|\}. \end{aligned}$$

We will reach a contradiction, analyzing four different cases.

Case 1. $p=q=0$: The intersection contains just one element, $(j,j')$. We now look at the translation

$$\begin{aligned} L':=(j+2a,j'+2b)+(3a,3b)\mathbb {Z} +(0,d)\mathbb {Z}\subset L \end{aligned}$$

of the sublattice $(3a,3b)\mathbb {Z} +(0,d)\mathbb {Z}$ of $(a,b)\mathbb {Z}+(0,d)\mathbb {Z}$. If $(j,j')\in L'$, then

$$\begin{aligned} \left\{ \begin{array}{l} j+2a+3ar = j \\ j'+2b+3br+ds = j' \end{array} \right. \end{aligned}$$

for some $r,s\in \mathbb {Z}$. But, since $a(3r+2)\ne 0$, $r\in \mathbb {Z}$, this is impossible. In particular, $A\cap L'=\emptyset$, a contradiction to our basic assumption.

Case 2. $p=0$ and $q>0$: Here we consider the sublattice $(a,b)\mathbb {Z} +(0,{2(2q+1)}d)\mathbb {Z}$ of $(a,b)\mathbb {Z}+(0,d)\mathbb {Z}$ and look at its translation

$$\begin{aligned} L'':=(j,j'+2qd) + (a,b)\mathbb {Z} +(0,2(2q+1)d)\mathbb {Z} \subset L. \end{aligned}$$

If $(j+ra,j'+2qd+rb+2s(2q+1)d)=(j,j'+q_\mu d)$ for some $\mu \in \{1,2,\ldots ,n\}$ and $r,s \in \mathbb {Z}$, then

$$\begin{aligned} \left\{ \begin{array}{l} ra=0\\ 2qd+rb+2s(2q+1)d=q_\mu d \end{array} \right. \end{aligned}$$

and, consequently, $2q+2s(2q+1)=q_\mu$. However, due to the definition of q, no integer s can satisfy the previous equality. Thus, $L''\cap A=\emptyset$, another contradiction.

Case 3. $p>0$ and $q=0$: Its is similar to the previous case.

Case 4. $p,q>0$: Here we consider the sublattice $(2(2p+1)a,2(2p+1)b)\mathbb {Z} +(0,qd)\mathbb {Z}$ of $(a,b)\mathbb {Z}+(0,d)\mathbb {Z}$ and its translation

$$\begin{aligned} L''':= (j+2pa,j'+2pb)+(2(2p+1)a,2(2p+1)b)\mathbb {Z} +(0,qd)\mathbb {Z} \subset L. \end{aligned}$$

If

$$\begin{aligned}&(j+2pa+2r(2p+1)a, j'+2pb+2r(2p+1)b + sqd)\\&\quad = (j+p_\mu a,j'+p_\mu b+q_\mu d) \end{aligned}$$

for some $\mu \in \{1,2,\ldots ,n\}$ and $r,s \in \mathbb {Z}$, we will have that $2p+2r(2p+1)=p_\mu$. As in Case 2, we can deduce that $L'''\cap A=\emptyset$, a contradiction to our initial assumption on A. $\square$

3 The proof of Theorem 1.1

This section contains a proof for the main theorem announced in the introduction.

Proof

$(i)\Rightarrow (ii)$ Assume K is strictly positive definite and write $S=(a\mathbb {Z}, b\mathbb {Z})$ with $a,b>0$. We will show that

$$\begin{aligned} \{(k,l): (|k|,|l|)\in J_K\} \end{aligned}$$

intersects $(j,j')+S$, whenever $j\in \{0,1,\ldots , a-1\}$ and $j'\in \{0,1,\ldots , b-1\}$. There is nothing to prove if $a=b=1$. In the other cases, we will assume that

$$\begin{aligned} \{(k,l): (|k|,|l|)\in J_K\} \cap (j+a\mathbb {Z}, j'+b\mathbb {Z})=\emptyset , \end{aligned}$$

and will reach a contradiction. In the case in which $a=1$ and $b\ge 2$, the assumption on $\{(k,l): (|k|,|l|)\in J_K\}$ implies that $l-j',-l-j' \not \in b\mathbb {Z}$, whenever $(k,l) \in J_K$. In particular,

$$\begin{aligned} \sum _{\mu =1}^b\left( e^{i2\pi \mu /b}\right) ^{l-j'}=\sum _{\mu =1}^b \left( e^{i2\pi \mu /b}\right) ^{-l-j'}=0, \end{aligned}$$

and, consequently,

$$\begin{aligned} \sum _{\mu =1}^b \left[ \text{ Re } \left( e^{i2\pi \mu j'/b}\right) \right] \left( e^{i2\pi \mu /b}\right) ^l=0, \quad (k,l) \in J_K. \end{aligned}$$

The real scalars $c_\mu := \text{ Re } (e^{i2\pi \mu j'/b})$, $\mu =1,2,\ldots ,b$, are not all zero and the points

$$\begin{aligned} (x_\mu ,w_\mu )=(1, e^{i2\pi \mu /b}), \quad \mu =1,2,\ldots ,b, \end{aligned}$$

are distinct in $S^1\times S^1$. Thus, under the light of Proposition 2.1, we have a contradiction with the strict positive definiteness of K. The case in which $a\ge 2$ and $b=1$ is similar. To conclude the proof, we now assume $a,b\ge 2$ and adapt the procedure employed in the first case. If $(k,l) \not \in (j+a\mathbb {Z}, j'+b\mathbb {Z})$, then either $k-j \not \in a\mathbb {Z}$ or $l-j' \not \in b\mathbb {Z}$. Hence, we may conclude that

$$\begin{aligned} \sum _{\mu =1}^a \left( e^{i2\pi \mu /a}\right) ^{k-j} \sum _{\nu =1}^b\left( e^{i2\pi \nu /b}\right) ^{l-j'}=0, \end{aligned}$$

that is,

$$\begin{aligned} \sum _{\mu =1}^a \sum _{\nu =1}^b e^{-i2\pi \mu j /a}e^{-i2\pi \nu j'/b} \left( e^{i2\pi \mu /a}\right) ^k \left( e^{i2\pi \nu /b}\right) ^l=0. \end{aligned}$$

Repeating the argument with the assumption $(-k,-l) \not \in (j+a\mathbb {Z}, j'+b\mathbb {Z}),$ we conclude that

$$\begin{aligned} \sum _{\mu =1}^a \sum _{\nu =1}^b e^{i2\pi \mu j /a}e^{i2\pi \nu j'/b} \left( e^{i2\pi \mu /a}\right) ^k \left( e^{i2\pi \nu /b}\right) ^l=0. \end{aligned}$$

Thus, since (k, l) is arbitrary,

$$\begin{aligned} \sum _{\mu =1}^a \sum _{\nu =1}^b \left[ \text{ Re } \left( e^{i2\pi \mu j /a}e^{i2\pi \nu j'/b}\right) \right] \left( e^{i2\pi \mu /a}\right) ^k \left( e^{i2\pi \nu /b}\right) ^l=0, \quad (k,l) \in J_K. \end{aligned}$$

By an analogous procedure, now taking into account that

$$\begin{aligned} (-k,l),(k,-l) \not \in (j+a\mathbb {Z}, j'+b\mathbb {Z}), \end{aligned}$$

the conclusion is

$$\begin{aligned} \sum _{\mu =1}^a \sum _{\nu =1}^b \left[ \text{ Re }\left( e^{i2\pi \mu j /a}e^{i2\pi \nu j'/b}\right) \right] \left( e^{i2\pi \mu /a}\right) ^k \left( e^{-i2\pi \nu /b}\right) ^l=0, \quad (k,l) \in J_K. \end{aligned}$$

Therefore, since the numbers $\text{ Re } \left[ e^{i2\pi \mu j /a}e^{i2\pi \nu j'/b}\right]$ are not all zero and the ab points

$$\begin{aligned} (x_\mu ,w_\nu )=(e^{i2\pi \mu /a}, e^{i2\pi \nu /b}), \quad \mu =1,2,\ldots ,a,\quad \nu =1,2,\ldots , b, \end{aligned}$$

are distinct in $S^1 \times S^1$, we have reached a contradiction once again.

$(ii) \Leftrightarrow (iii)$ One implication is a consequence of Lemma 2.3. The other one is obvious.

$(ii) \Rightarrow (i)$ Let us assume that $\{(k,l): (|k|,|l|)\in J_K\}$ intersects all the translations of each rectangular lattice of $\mathbb {Z}^2$. For a fixed $n\ge 2$, n distinct points $(\theta _1,\phi _1), (\theta _2, \phi _2), \ldots ,$ $(\theta _n, \phi _n)$ in $[0,2\pi )^2$ and real numbers $c_1, c_2, \ldots , c_n$, not all zero, we intend to show that either $\sum _{\mu =1}^n c_\mu e^{i\theta _\mu k}e^{-i\phi _\mu l} \ne 0$ or $\sum _{\mu =1}^n c_\mu e^{i\theta _\mu k}e^{i\phi _\mu l}\ne 0$, for some $(k,l) \in J_K$. A help of Proposition 2.1 will lead to the strict positive definiteness of K. In order to achieve the conclusion mentioned above, define

$$\begin{aligned} b_{k,l}=\sum _{\mu =1}^n c_\mu e^{i\theta _\mu k}e^{i\phi _\mu l}, \quad k,l\in \mathbb {Z}. \end{aligned}$$

On one hand, Lemma 2.5 and the fact that at least one $c_\mu$ is nonzero imply that $\{(k,l): b_{k,l}=0\}\ne \mathbb {Z}^2$. Theorem 2.4 asserts that $\{(k,l): b_{k,l}=0\}$ is the union of a finite number of translations of subgroups of $\mathbb {Z}^2$ while Lemma 2.6 guarantees the existence of a rectangular lattice of $\mathbb {Z}^2$, a translation of which belongs to $\mathbb {Z}^2 {\setminus } \{(k,l): b_{k,l}=0\}$. Thus, due to our assumption on $\{(k,l) : (|k|,|l|) \in J_K\}$, we immediately have that

$$\begin{aligned} \{(k,l) : (|k|,|l|) \in J_K\}\not \subset \{(k,l): b_{k,l}=0\}. \end{aligned}$$

Therefore, there must exist at least one pair (k, l) in $\{(k,l): (|k|,|l|)\in J_K\}$ for which

$$\begin{aligned} \sum _{\mu =1}^n c_\mu e^{i\theta _\mu k}e^{i\phi _\mu l}\ne 0. \end{aligned}$$

Since the $c_\mu$ are real, the result follows.

The final statement in the proof of Theorem 1.1 is a consequence of Lemma 2.7. $\quad \square$

The results demonstrated in this paper can be adapted to hold for positive definiteness on the complex circle $\Omega _2$. In that case, we replace $S^1$ with $\Omega _2$, we allow the kernels to assume complex values and the scalars $c_\mu$ in the definition of positive definiteness can be complex numbers (the quadratic form in the definition of positive definiteness is Hermitian). We will sketch what these results are and refer the interested reader to [8, 15] where the necessary adaptations for the proofs can be prospected from.

Let $K: \Omega _2 \times \Omega _2 \rightarrow \mathbb {C}$ be a continuous kernel and assume that

$$\begin{aligned} K((x,z),(y,w))=K_r(x\cdot y, z\cdot w),\quad x,y,z,w\in \Omega _2, \end{aligned}$$

for some function $K_r : \Omega _2 \times \Omega _2 \rightarrow \mathbb {C}$, in which $\cdot$ is now the usual inner product of $\mathbb {C}$. It is positive definite if, and only if, the function $K_r$ is of the form

$$\begin{aligned} K_r(z,w)=\sum _{k,l\in \mathbb {Z}}a_{k,l}z^k w^l, \quad z,w \in \Omega _2, \end{aligned}$$

in which $a_{k,l} \ge 0$, $k,l \in \mathbb {Z}$ and $\sum _{k,l\in \mathbb {Z}}a_{k,l}<\infty$. Taking the above representation for granted, we can define $I_K:=\{(k,l): a_{k,l}>0\}$. For distinct points $(x_1,w_1),(x_2,w_2), \ldots ,$ $(x_n,w_n)$ on $\Omega _2 \times \Omega _2$ and a column vector c in $\mathbb {C}^n$, the quadratic form $\overline{c}^{t}Ac=0$ corresponds to

$$\begin{aligned} \sum _{\mu =1}^{n}c_{\mu }e^{ik\theta _{\mu }}e^{il\phi _{\mu }}=0, \quad (k,l)\in I_K, \end{aligned}$$

in which $\theta _\mu$ and $\phi _\mu$ are the arguments in the polar representation of $x_\mu$ and $w_\mu$ respectively. In particular, this reveals that the proofs we have developed in Sects. 2 and 3 simplify in the present complex setting. In particular, a continuous and positive definite kernel K on $\Omega _2 \times \Omega _2$ as described above is strictly positive definite if, and only if, $I_K$ intersects all the translations of each rectangular lattice of $\mathbb {Z}^2$.

It is worth to mention that, after some period of research, we have not found yet a characterization for the real, continuous, isotropic and strictly positive definite kernels on $S^1 \times S^m$. An elegant characterization seems to demand a mix of arguments, some presented here and others developed in [7].