Structure Results for Multiple Tilings in 3D

Abstract

We study multiple tilings of 3-dimensional Euclidean space by a convex body. In a multiple tiling, a convex body $P$ is translated with a discrete multiset $\Lambda$ in such a way that each point of ${\mathbb {R}}^d$ gets covered exactly $k$ times, except perhaps the translated copies of the boundary of $P$. It is known that all possible multiple tilers in ${\mathbb {R}}^3$ are zonotopes. In ${\mathbb {R}}^2$ it was known by the work of Kolountzakis (Discrete Comput Geom 23(4):537–553, 2000) that, unless $P$ is a parallelogram, the multiset of translation vectors $\Lambda$ must be a finite union of translated lattices (also known as quasi periodic sets). In that work (Kolountzakis, Discrete Comput Geom 23(4):537–553, 2000) the author asked whether the same quasi-periodic structure on the translation vectors would be true in ${\mathbb {R}}^3$. Here we prove that this conclusion is indeed true for ${\mathbb {R}}^3$. Namely, we show that if $P$ is a convex multiple tiler in ${\mathbb {R}}^3$, with a discrete multiset $\Lambda$ of translation vectors, then $\Lambda$ has to be a finite union of translated lattices, unless $P$ belongs to a special class of zonotopes. This exceptional class consists of two-flat zonotopes $P$, defined by the Minkowski sum of two 2-dimensional symmetric polygons in ${\mathbb {R}}^3$, one of which may degenerate into a single line segment. It turns out that rational two-flat zonotopes admit a multiple tiling with an aperiodic (nonquasi-periodic) set of translation vectors $\Lambda$. We note that it may be quite difficult to offer a visualization of these 3-dimensional non-quasi-periodic tilings, and that we discovered them by using Fourier methods.

1 Introduction

The study of multiple tilings of Euclidean space began in 1936, when the famous Minkowski facet-to-facet conjecture [18] for classical tilings was extended to the setting of $k$-tilings with the unit cube, by Furtwängler [4]. Minkowski’s facet-to-facet conjecture states that for any lattice tiling of ${\mathbb {R}}^d$ by translations of the unit cube, there exist at least two translated cubes that share a facet (face of co-dimension 1). This conjecture was strengthened by Keller [10] who conjectured the same conclusion for any cube tiling, not just lattice tilings. It was also strengthened in a different direction by Furtwängler [4] who, again, conjectured the same conclusion for any multiple lattice tiling.

To define a multiple tiling, suppose we translate a convex body $P$ with a discrete multiset $\Lambda$, in such a way that each point of ${\mathbb {R}}^d$ gets covered exactly $k$ times, except perhaps the translated copies of the boundary of $P$. We then call such a body a $k$-tiler, and such an action has been given the following names in the literature: a k-tiling, a tiling at level k, a tiling with multiplicity k, and sometimes simply a multiple tiling. We may use any of these synonyms here, and we immediately point out, for polytopes $P$, a trivial but useful algebraic equivalence for a tiling at level $k$:

$$\begin{aligned} \sum _{\lambda \in \Lambda } \mathbf{1}_{P+\lambda }(x) = k, \end{aligned}$$

(1)

for almost all $x \in {\mathbb {R}}^d$, where $\mathbf{1}_{P}$ is the indicator function of the polytope $P$.

Furtwängler’s conjecture was disproved by Hajós [8] for dimension larger than 3 and for $k \ge 9$ while Furtwängler himself [4] proved it for dimension at most 3. Hajós [9] also proved Minkowski’s conjecture in all dimensions. The ideas of Furtwängler were subsequently extended (but still restricted to cubes) by the important work of Perron [19], Robinson [20], Szabó [24], Gordon [5] and Lagarias and Shor [14]. These authors showed that for some levels $k$ and dimensions $d$ and under the lattice assumption as well as not, a facet-to-facet conclusion for $k$-tilings is true in ${\mathbb {R}}^d$, while for most values of $k$ and $d$ it is false.

There is a vast literature on the study of coverings of Euclidean space by a convex body, and an equally vast body of work on classical tilings by translations of one convex body, which must necessarily be a polytope (see for example [3, 7]). On the one hand, when we consider a $k$-tiling polytope $P$, we obtain an exact covering of ${\mathbb {R}}^d$, in the sense that almost every point of ${\mathbb {R}}^d$ gets covered exactly $k$ times. On the other hand, the family of $k$-tilers is much larger than the family of $1$-tilers. Hence the study of $k$-tilings lies somewhere between coverings and $1$-tilings.

It was known to Bolle [2] that in ${\mathbb {R}}^2$, every $k$-tiling convex polytope has to be a centrally symmetric polygon, and using combinatorial methods Bolle [2] gave a characterization for all polygons in ${\mathbb {R}}^2$ that admit a $k$-tiling with a lattice $\Lambda$ of translation vectors. Kolountzakis [13] proved that if a convex polygon $P$ tiles ${\mathbb {R}}^2$ multiply with any discrete multiset $\Lambda$, then $\Lambda$ must be a finite union of two-dimensional lattices. The ingredients of Kolountzakis’ proof include the idempotent theorem for the Fourier transform of a measure. Roughly speaking, the idempotent theorem of Meyer [17] tells us that if the square of the Fourier transform of a measure is itself, then the support of the measure is contained in a finite union of lattices. To put our main result into its proper context, we record here the precise result of Kolountzakis. A multiple tiling is called quasi-periodic if its multiset of discrete translation vectors $\Lambda$ is a finite union of translated lattices, not necessarily all of the same dimension.

Theorem

(Kolountzakis, 2002 [11]) Suppose that $K$ is a symmetric convex polygon which is not a parallelogram. Then $K$ admits only quasi-periodic multiple tilings if any.

Here we extend this result to ${\mathbb {R}}^3$, and we also find a fascinating class of polytopes analogous to the parallelogram of the theorem above. To describe this class, we first recall the definition of a zonotope, which is the Minkowski sum of a finite number of line segments. In other words, a zonotope equals a translate of $[-\mathbf {v}_1, \mathbf {v}_1] + \cdots + [-\mathbf {v}_N, \mathbf {v}_N]$, for some positive integer $N$ and vectors $\mathbf {v}_1,\ldots ,\mathbf {v}_N \in {\mathbb {R}}^d$. A zonotope may equivalently be defined as the projection of some $l$-dimensional cube. A third equivalent condition is that for a $d$-dimensional zonotope, all of its $k$-dimensional faces are centrally symmetric, for $1 \le k \le d$. For example, the zonotopes in ${\mathbb {R}}^2$ are the centrally symmetric polygons.

We shall say that a polytope $P\subseteq {\mathbb {R}}^3$ is a two-flat zonotope if $P$ is the Minkowski sum of $n+m$ line segments which lie in the union of two different two-dimensional subspaces $H_1$ and $H_2$. In other words, $H_1$ contains $n$ of the segments and $H_2$ contains $m$ of the segments (if one of the segments belongs to both $H_1$ and $H_2$ we list it twice, once for each plane). Equivalently, $P$ may be thought of as the Minkowski sum of two $2$-dimensional symmetric polygons one of which may degenerate into a single line segment.

Recently, a structure theorem for convex $k$-tilers in ${\mathbb {R}}^d$ was found, and is as follows.

Theorem

(Gravin, Robins, Shiryaev 2011 [6]) If a convex polytope $k$-tiles ${\mathbb {R}}^d$ by translations, then it is centrally symmetric and its facets are centrally symmetric.

In the present context of ${\mathbb {R}}^3$, it follows immediately from the latter theorem that a $k$-tiler $P \subset {\mathbb {R}}^3$ is necessarily a zonotope. In this paper we extend the result of Kolountzakis [11] from ${\mathbb {R}}^2$ to ${\mathbb {R}}^3$, providing a structure theorem for multiple tilings by polytopes in three dimensions.

Main Theorem

Suppose a polytope $P\; k$-tiles ${\mathbb {R}}^3$ with a discrete multiset $\Lambda$, and suppose that $P$ is not a two-flat zonotope. Then $\Lambda$ is a finite union of translated lattices.

It turns out that if $P$ is a rational two-flat zonotope, then $P$ admits a $k$-tiling with a non-quasi-periodic set of translation vectors $\Lambda$, as we show in our Corollary 7.1. For some of the classical study of $1$-tilings, and their interesting connections to zonotopes, the reader may refer to the work of [15, 16, 22, 25], and [1]. Here we find it very useful to use the intuitive language of distributions [21, 23] in order to think—and indeed discover—facts about $k$-tilings. To that end we introduce the distribution (which is locally a measure)

$$\begin{aligned} \delta _\Lambda := \sum _{\lambda \in \Lambda } \delta _\lambda , \end{aligned}$$

(2)

where $\delta _\lambda$ is the Dirac delta function at $\lambda \in {\mathbb {R}}^d$. To develop some intuition, we may check formally that

$$\begin{aligned} \delta _\Lambda * \mathbf{1}_P = \sum _{\lambda \in \Lambda } \delta _\lambda * \mathbf{1}_P = \sum _{\lambda \in \Lambda } \mathbf{1}_{P+\lambda }, \end{aligned}$$

so that from the first definition (1) of $k$-tiling, we see that a polytope $P$ is a $k$-tiler if and only if

$$\begin{aligned} \delta _\Lambda * \mathbf{1}_P=k. \end{aligned}$$

(3)

The paper is modularized into short sections that highlight each step separately, and the organization runs as follows. In Sect. 3 we compute the Fourier transform of any $4$-legged frame of a polytope, and show that its zeros form a certain countable union of hyperplanes. In Sect. 4 we find a sufficient condition, which we call the intersection property, for the Fourier transform of ${\delta _\Lambda }$ to have a discrete support. Then we show that if $P$ is a $k$-tiler, and if the intersection property holds for all $4$-legged frames of $P$, then $\mathrm{supp\,}\widehat{\delta _\Lambda }$ (the support of the distribution ${{\widehat{\delta _\Lambda }}}$, the Fourier Transform of the distribution $\delta _\Lambda$) is a discrete set in ${\mathbb {R}}^3$, of bounded density.

In Sect. 5, we prove that the intersection property implies the quasi-periodicity of $\Lambda$.

In Sect. 6 we discover a fascinating family of $k$-tilers, associated to a non-discrete $\mathrm{supp\,}{\widehat{\delta _\Lambda }}$. We prove that if $P$ tiles ${\mathbb {R}}^3$ with multiplicity, by translations with a discrete multiset $\Lambda$, and the intersection property fails to hold, then $P$ must be a two-flat zonotope.

The proof of the Main Theorem is also given in Sect. 6; this proof is quite short since it just strings together all of the results of the previous sections. In the final section, we show that each rational two-flat zonotope admits a very peculiar non-quasi-periodic $k$-tiling. We note that it may be quite difficult to offer a visualization of these $3$-dimensional non-quasi-periodic tilings, and that we discovered them by using Fourier methods.

2 Preliminaries

Suppose the polytope $P$ tiles multiply with the translates $\Lambda \subseteq {\mathbb {R}}^d$. We will need to understand some basic facts about how the $\Lambda$ points are distributed, for example in Theorem 5.1 below.

Definition 2.1

(Uniform density) A multiset $\Lambda \subseteq {\mathbb {R}}^d$ has asymptotic density $\rho$ if

$$\begin{aligned} \lim _{R\rightarrow \infty } {\#(\Lambda \cap B_R(x)) \over {|{B_R(x)}|}} \rightarrow \rho \end{aligned}$$

uniformly in $x\in {\mathbb {R}}^d$. In this case we write $\rho = \mathrm{dens\,}\Lambda$. Another (weaker) notion is that of bounded density—we say that $\Lambda$ has (uniformly) bounded density if ${{\#(\Lambda \cap B_R(x))} /{{|{B_R(x)}|}}} \le M$ for $x\in {\mathbb {R}}^d$, and $R>1$. We say then that $\Lambda$ has density (uniformly) bounded by $M$.

Since we intend to speak of the Fourier Transform of $\delta _\Lambda$ it is important to us that if $\Lambda$ has bounded density then $\delta _\Lambda$ is a tempered distribution [21, 23], and, therefore, its Fourier Transform $\widehat{\delta _\Lambda }$ is well defined. And, it is almost obvious by comparing volumes that, if a polytope $P$ $k$-tiles ${\mathbb {R}}^d$ with translates $\Lambda$ then $\Lambda$ has density $k/{|{P}|}$.

For any symmetric polytope $P$, and any face $F \subset P$, we define $F^{-}$ to be the face of $P$ symmetric to $F$ with respect to $P$’s center of symmetry. We call $F^-$ the opposite face of $F$.

Throughout the paper, we use the notation ${\mathbf {x}}^\perp$ to denote the perpendicular subspace, of codimension $1$, to the vector ${\mathbf {x}}$. We also use the standard convention of boldfacing all vectors, to differentiate between $\mathbf {v}$ and $v$, for example. We furthermore use the convention that $[\mathbf {e}]$ denotes the $1$-dimensional line segment from $0$ to the endpoint of the vector $\mathbf {e}$. Whenever it is clear from context, we will also write $[e]$ to denote the same line segment—for example, in the case that $e$ denotes an edge of a polytope.

We let ${\mathsf {Z}}(f)$ be the zero set of the function $f$.

Definition 2.2

[$4$-legged-frame of a polytope]

• (a) Suppose $P\in {\mathbb {R}}^3$ is a zonotope (symmetric polytope with symmetric facets). A collection of four (one-dimensional) edges of $P$ is called a $4$ -legged-frame if whenever $e$ is one of the edges then there exist two vectors ${\varvec{\tau }}_1$ and ${\varvec{\tau }}_2$ such that the four edges are

  $$\begin{aligned}{}[e],\ [e]+{\varvec{\tau }}_1, [e]+{\varvec{\tau }}_2 \text { and } [e]+{\varvec{\tau }}_1+{\varvec{\tau }}_2, \end{aligned}$$

  and such that the edges $[e]$ and $[e]+{\varvec{\tau }}_1$ belong to the same face of $P$ and the edges $[e]+{\varvec{\tau }}_2$ and $[e]+{\varvec{\tau }}_1+{\varvec{\tau }}_2$ belong to the opposite face.

• (b) For a set of four legs as above the leg measure is the measure supported on the legs and is equal to arc-length on the two legs $[e]$ and $[e]+{\varvec{\tau }}_1+{\varvec{\tau }}_2$ and minus arc-length on the two legs $[e]+{\varvec{\tau }}_1$ and $[e]+{\varvec{\tau }}_2$. We denote this measure by $\mu _{e,\tau _1,\tau _2}$. The leg measure is defined up to sign.

Remark 2.1

A set of four legs of a symmetric polytope with symmetric faces is determined uniquely if we know two opposite edges on a face (these edges are the first two legs). The other two legs are then the corresponding opposite edges on the opposite face (Fig. 1).

Fig. 1

Four legs of a convex polytope and leg-measure. The colored segments are edges of the polytope. Translating an edge by $\pm \tau _1$ gives you the opposite edge on a face. Translating by $\pm \tau _2$ takes you to the opposite face (Color figure online)

3 The Fourier Transform of a $4$-Legged Frame

Lemma 3.1

Suppose $\mathbf {e}, {\varvec{\tau }}_1, {\varvec{\tau }}_2 \in {\mathbb {R}}^3$ are linearly independent and consider the measure

$\mu =\mu _{e,\tau _1,\tau _2}$ $($see Definition 2.2). Then the zero-set of the Fourier Transform $\hat{\mu }$, is

$$\begin{aligned} {\mathsf {Z}}(\hat{\mu }) = {\mathbb {H}}_{_{-0}}(\mathbf {e}) \cup {\mathbb {H}}({\varvec{\tau }}_1) \cup {\mathbb {H}}({\varvec{\tau }}_2), \end{aligned}$$

(4)

where, if ${\mathbf {x}}$ is a non-zero vector and $\displaystyle {{\mathbf {x}}}^* = \tfrac{{{\mathbf {x}}}}{{|{{{\mathbf {x}}}}|}^2}$ is its geometric inverse,

$$\begin{aligned} {\mathbb {H}}({\mathbf {x}}) = {\mathbb {Z}}{\mathbf {x}}^* + {\mathbf {x}}^\perp \end{aligned}$$

(5)

and

$$\begin{aligned} {\mathbb {H}}_{_{-0}}({\mathbf {x}}) = \big ({\mathbb {Z}}\setminus \{0\}\big ){\mathbf {x}}^* + {\mathbf {x}}^\perp . \end{aligned}$$

(6)

Here ${\mathbf {x}}^\perp$ is the hyperplane orthogonal to the vector ${\mathbf {x}}$, so that ${\mathbb {H}}({\mathbf {x}}) = {\mathbb {Z}}{\mathbf {x}}^* + {\mathbf {x}}^\perp$ is a collection of parallel hyperplanes, orthogonal to ${\mathbf {x}}$ spaced by $1/{|{{\mathbf {x}}}|}$.

Proof

Translating a measure does not alter the zero-set of its FT so we may translate $\mu$ so that $0$ is the midpoint of the first line segment, which now runs from $-\mathbf {e}/2$ to $\mathbf {e}/2$. Denoting by $\nu$ the arc-length measure on this line segment and writing $\alpha =\delta _0-\delta _{\tau _1}$ and $\beta =\delta _0-\delta _{\tau _2}$ we obtain $\mu$ as a convolution:

$$\begin{aligned} \mu = \nu *\alpha *\beta . \end{aligned}$$

Taking the FT we get that

$$\begin{aligned} {\mathsf {Z}}(\hat{\mu }) = {\mathsf {Z}}(\hat{\nu }) \cup {\mathsf {Z}}(\hat{\alpha }) \cup {\mathsf {Z}}(\hat{\beta }). \end{aligned}$$

Based on the calculation of the FT of the indicator function of $[-\tfrac{1}{2}, \tfrac{1}{2}]$

$$\begin{aligned} \int \limits _{-1/2}^{1/2} \mathrm{e}^{-2\pi i \xi x}\,\mathrm{d}x = \frac{\sin \pi \xi }{\pi \xi } \end{aligned}$$

we conclude that

$$\begin{aligned} \widehat{\nu }(\mathbf {u}) = {|{\mathbf {e}}|}\frac{\sin \pi \langle \mathbf {u}, \mathbf {e}\rangle }{\pi \langle \mathbf {u}, \mathbf {e}\rangle }. \end{aligned}$$

One also immediately obtains the formulas

$$\begin{aligned} \widehat{\alpha }(\mathbf {u}) = 2i \mathrm{e}^{-\pi i \langle {\varvec{\tau }}_1, \mathbf {u}\rangle } \sin {\pi \langle {\varvec{\tau }}_1, \mathbf {u}\rangle } \end{aligned}$$

and

$$\begin{aligned} \widehat{\beta }(\mathbf {u}) = 2i \mathrm{e}^{-\pi i \langle {\varvec{\tau }}_2, \mathbf {u}\rangle } \sin {\pi \langle {\varvec{\tau }}_2, \mathbf {u}\rangle }. \end{aligned}$$

Since $\widehat{\nu },\; \widehat{\alpha }$ and $\widehat{\beta }$ vanish precisely on ${\mathbb {H}}_{_{-0}}(\mathbf {e}),\; {\mathbb {H}}({\varvec{\tau }}_1)$ and ${\mathbb {H}}({\varvec{\tau }}_2)$ respectively, the proof of Lemma 3.1 is complete. $\square$

4 A Sufficient Condition for $\widehat{\delta _\Lambda }$ to have Discrete Support

Theorem 4.1

Suppose $P$ is a symmetric polytope in ${\mathbb {R}}^3$ with symmetric faces and $\Lambda$ is a multiset of points in ${\mathbb {R}}^3$ such that $P$ tiles at level $k$, a positive integer, when translated at the locations $\lambda \in \Lambda$. Then we have

$$\begin{aligned} \mathrm{supp\,}\widehat{\delta _\Lambda } \subseteq \{0\} \cup \bigcap _{e, \tau _1, \tau _2} \big ( {\mathbb {H}}_{_{-0}}(\mathbf {e}) \cup {\mathbb {H}}({\varvec{\tau }}_1) \cup {\mathbb {H}}({\varvec{\tau }}_2) \big ), \end{aligned}$$

(7)

where $\delta _\Lambda$ is the measure corresponding to $\Lambda$ defined in (2) and the intersection in (7) is taken over all $4$-legged frames $(e, \tau _1, \tau _2)$ of $P$.

Proof

We know from [6] (see Lemmas $3.1$ and $3.2$ in [6]) that if $P$ tiles with $\Lambda$ and $\mu$ is a leg measure on $P$ then $\mu$ also tiles with $\Lambda$, at level 0. In other words $\mu *\delta _\Lambda = 0$. Since $P$ tiles when translated by $\Lambda$ it follows that ${|{\Lambda \cap [-R,R]^3}|} = O(R^3)$, hence $\delta _\Lambda$ is a tempered distribution and we may take its FT which gives us $\widehat{\mu }\widehat{\delta _\Lambda }=0$. This implies (see the details in [12, Sect. 1.2])

$$\begin{aligned} \mathrm{supp\,}\widehat{\delta _\Lambda } \subseteq \{0\} \cup {\mathsf {Z}}(\hat{\mu }). \end{aligned}$$

But the measure $\mu$ is exactly the one described in Lemma 3.1 and since this must be true for all sets of four legs of $P$ we conclude (7). $\square$

Corollary 4.1

Suppose $P$ is a $k$-tiler with a discrete multiset $\Lambda$, in ${\mathbb {R}}^3$. Let the following intersection property hold:

$$\begin{aligned} \bigcap _{e, \tau _1. \tau _2} \big ( \mathbf {e}^\perp \cup {\varvec{\tau }}_1^\perp \cup {\varvec{\tau }}_2^\perp \big ) = \{0\}, \end{aligned}$$

(8)

where the intersection above is taken over all sets of $4$-legged frames of $P$.

Then $\mathrm{supp\,}\widehat{\delta _\Lambda }$ is a discrete set in ${\mathbb {R}}^3$, of bounded density.

Proof

The sets which are being intersected in (7) are all unions of planes. For this set to be non-discrete it must be the case that it contains an entire line of direction, say $\mathbf {u}\in {\mathbb {R}}^3\setminus \{0\}$.

This in turn implies that there is a selection $X_\ell$ of $\mathbf {e}$, ${\varvec{\tau }}_1$ or ${\varvec{\tau }}_2$ for each set $\ell$ of four legs such that $\mathbf {u}\in X_\ell ^\perp$. This contradicts condition (8).

Having established that the intersection in the right hand side of (7) is a discrete point set we observe that the larger set

$$\begin{aligned} \bigcap _{\mathbf {e}, \tau _1, \tau _2} {\mathbb {H}}(\mathbf {e}) \cup {\mathbb {H}}({\varvec{\tau }}_1) \cup {\mathbb {H}}({\varvec{\tau }}_2) \end{aligned}$$

(9)

is a finite union of discrete groups, each of them of the form

$$\begin{aligned} \bigcap _{\ell } {\mathbb {H}}_{\ell }, \end{aligned}$$

where $\ell$ runs through all possible sets of four legs of $P$ and for each $\ell =\{e,\tau _1,\tau _2\}$ the set ${\mathbb {H}}_{\ell }$ is one of ${\mathbb {H}}(\mathbf {e}), {\mathbb {H}}({\varvec{\tau }}_1), {\mathbb {H}}({\varvec{\tau }}_2)$. Since each discrete group has bounded density so has the set (9) and $\mathrm{supp\,}\widehat{\delta _\Lambda }$ as its subset. $\square$

5 The Intersection Property Implies Quasi-periodicity of $\Lambda$

In this section we show how the discreteness of $\mathrm{supp\,}{\widehat{\delta _\Lambda }}$ implies a rather rigid structure for $\Lambda$. Below we quote the result from [11], where the multidimensional statements had been proved in general, despite the fact that the final conclusions in [11] are given only for dimension $2$.

Theorem 5.1

(Kolountzakis, 2002) Suppose that for the multiset $\Lambda \subseteq {\mathbb {R}}^d$

• (1) $\Lambda$ has uniformly bounded density;

• (2) $S:= \mathrm{supp\,}\widehat{\delta _{\Lambda }}$ is discrete;

• (3) $|S \cap B_R(0)| \le C\cdot R^d,$ for some positive constant $C$.

Then $\Lambda$ is a finite union of translated $d$-dimensional lattices.

Next, we verify the conditions of the theorem above, for our $3$-dimensional $k$-tilers with a multiset $\Lambda$.

Claim 5.1

Suppose that convex polytope $P$ $k$-tiles ${\mathbb {R}}^{d}$ with $\Lambda$ and the intersection property (8) of Corollary 4.1 is true. Then $\Lambda$ is a finite union of translated $d$-dimensional lattices.

Proof

We just need to verify conditions (1), (2) and (3) of Theorem 5.1.

Hypothesis (1) simply follows from the fact that in each sufficiently large ball $B_{R}(x)$ of ${\mathbb {R}}^{d}$ every point is covered exactly $k$ times by the translations of $P$ with the set $\Lambda \cap B_{R'}(x)$, where $R'=R+\mathrm{diam\,}P$.

Hypotheses (2) and (3) follow from Corollary 4.1. $\square$

6 $k$-Tilers Associated to a Non-discrete $\mathrm{supp\,}{\widehat{\delta _\Lambda }}$

In this section we study the convex polytopes that admit exceptional multiple tilings, in the sense that the multiset of translations $\Lambda$ is not a finite union of $3$-dimensional lattices. A class of these exceptions is easily provided by prisms (Minkowski sums of a symmetric polygon with a line segment, not in the polygons plane), for which one can lift a $2$-dimensional $k$-tiling up into the third dimension by separately $1$-tiling the tube above each projection.

By Claim 5.1 for such a tiling, the intersection property (8) cannot be true. Therefore, there exists a line (in fact a $1$-dimensional subspace) $l \subseteq {\mathbb {R}}^{3}$ such that

$$\begin{aligned} l \subseteq \bigcap _{\mathbf {e}, \tau _1. \tau _2} \big ( \mathbf {e}^\perp \cup {\varvec{\tau }}_1^\perp \cup {\varvec{\tau }}_2^\perp \big ) . \end{aligned}$$

(10)

It was already shown in [6] that a multiple tiler in ${\mathbb {R}}^3$ must be a zonotope, i.e. a Minkowski sum of line segments. Here we will show that given the non-discreteness of $\mathrm{supp\,}{\widehat{\delta _\Lambda }}$, we can deduce that a zonotope is a Minkowski sum of two $2$-dimensional symmetric polygons. And in Sect. 7 we provide examples of such exceptional tilings, under a mild commensurability condition for such zonotopes.

Theorem 6.1

Suppose a polytope $P$ tiles ${\mathbb {R}}^3$ with multiplicity by translations over a multiset $\Lambda$ and condition (10) holds. Then $P$ is a two-flat zonotope.

Proof

We let $L$ be a plane orthogonal to $l$ and supporting $P$; (10) is then equivalent to

$$\begin{aligned} \forall \mathbf {e}, {\varvec{\tau }}_1, {\varvec{\tau }}_2, \text { either } \mathbf {e}\parallel L, \text { or } {\varvec{\tau }}_1 \parallel L, \text { or } {\varvec{\tau }}_2 \parallel L. \end{aligned}$$

(11)

Let $F = L\cap P$. The dimension of the face $F$ can be $0$, $1$ or $2$. Consider any facet $G$ of $P$ that has at least one common vertex with $F$, and let $e$ be an edge of $G$ that shares exactly one vertex $v$ with $F$ (so $G \ne F$). Consider the $4$-legged frame determined by $G$ and $e$ with ${\varvec{\tau }}_1,{\varvec{\tau }}_2$ being the corresponding translation vectors. Since $v$ is in $L$, by (11) one of the three vertices $v+\mathbf {e}$, $v+{\varvec{\tau }}_1$ and $v+{\varvec{\tau }}_2$ lies in $L$, and therefore lies in $F$ as well. By our choice of $\mathbf {e},\; v+\mathbf {e}$ is a vertex of $G$ but not of $F$. Thus either $v+{\varvec{\tau }}_1\in F$, or $v+{\varvec{\tau }}_2\in F$:

• (1) If $v+{\varvec{\tau }}_1\in F$, then ${\varvec{\tau }}_1 \in G \cap F$, so we see that ${\varvec{\tau }}_1$ is an edge of $G$. Hence $G$ is a parallelogram.

• (2) If $v+{\varvec{\tau }}_2\in F$, then $F$ connects $G$ with its opposite face $G^-$.

See Fig. 2.

Fig. 2

The two possibilities for the facet $G$ with respect to $F$. (1) $v+{\varvec{\tau }}_1\in F.$ (2) $v+{\varvec{\tau }}_2\in F$

It is our goal to find a facet $G$ which satisfies property (2). If, to the contrary, every facet adjacent to $F$ satisfies property (1), then each facet adjacent to $F$ is a parallelogram sharing an edge with $F$. It follows that exactly three edges meet at every vertex of $F$, and all edges of these parallelograms that are not edges of $F$ or parallel to $F$, are parallel to each other (see Fig. 3). Now since $F$ is centrally symmetric, consider two opposite parallel edges $e^+$ and $e^-$ of $F$, and corresponding parallelogram facets $G$ and $G^-$. The facets $G$ and $G^-$ are parallel and therefore opposite in $P$, therefore $G$ enjoys property (2).

Fig. 3

This is the case that each facet adjacent to $F$ is a parallelogram sharing an edge with $F$, giving us a prism

Now that we have found a facet $G$ such that $F$ connects $G$ with $G^-$, we also note that since $P$ is centrally symmetric, $G$ also connects $F$ and $F^-$. We will now show that $P=F+G$, under the minor assumption that $F$ and $G$ do not share an edge. The case that $F$ and $G$ do in fact share an edge may be handled in exactly the same manner, so without loss of generality we assume throughout the rest of the proof that $F$ and $G$ share none of their edges. Since $P$ is a zonotope, $P=F+G+H$ for some polytope $H$. To arrive at a contradiction, we assume to the contrary that $H$ is not a single point, and let $h_0$ be any edge of $H$. Let $F=[f_1]+\dots +[f_k]$, $G=[g_1]+\dots +[g_\ell ]$, and $H=[h_0]+[h_1]+\dots +[h_m],$ where $k\ge 0$, $\ell \ge 2$, and $m\ge 0.$ We may assume that all line segments $f_i, g_i, h_i$ have the origin as their midpoint and thus the center of $P$ is also at the origin. We further consider a normal vector $\mathbf {f}_{\perp }$ to the face $F$ of $P$. When $F$ is a $2$-dimensional face of $P$, $\mathbf {f}_{\perp }$ cannot be orthogonal to any line segment $h_i\in H$ and $g_i\in G$. If $F$ is $0$ or $1$ dimensional face of $P$ (see Fig. 4), we have an infinite collection of perpendicular vectors to $F$ and we may choose $\mathbf {f}_{\perp }$ to be not orthogonal to any line segment $h_i\in H$ and $g_i\in G$. For each edge $g_i$ (resp. $h_i$) we define $\mathbf {g}_i^+$ (resp. $\mathbf {h}_i^+$) to be the vector from the origin to the endpoint of $g_i$ (resp. $h_i$) such that $\langle \mathbf {g}_i^+ , \mathbf {f}_{\perp } \rangle >0$ (resp. $\langle \mathbf {h}_i^+ , \mathbf {f}_{\perp } \rangle >0$). In the same way we define $\mathbf {g}_i^-$ (resp. $\mathbf {h}_i^-$) s.t. $\langle \mathbf {g}_i^- , \mathbf {f}_{\perp } \rangle <0$ (resp. $\langle \mathbf {h}_i^- , \mathbf {f}_{\perp } \rangle <0$). Now one may easily see that the location of the face $F$ in ${\mathbb {R}}^3$ is given by $[f_1]+\dots +[f_k]+\mathbf {g}_1^{+}+\dots +\mathbf {g}_\ell ^{+} + \mathbf {h}_0^{+}+\dots +\mathbf {h}_m^{+}$ as a set of extremal points of the linear functional corresponding to $\mathbf {f}_{\perp }$. Similarly the location of the face $F^-$ in ${\mathbb {R}}^3$ is given by $[f_1]+\dots +[f_k]+\mathbf {g}_1^{-}+\dots +\mathbf {g}_\ell ^{-} + \mathbf {h}_0^{-}+\dots +\mathbf {h}_m^{-}.$ Therefore the distance between $F$ and $F^-$ is

$$\begin{aligned} \Big \langle \mathbf {f}_{\perp }, \sum _{i=1}^{\ell }\mathbf {g}_i^{+} + \sum _{i=0}^{m}\mathbf {h}_i^{+} -\sum _{i=1}^{\ell }\mathbf {g}_i^{-} - \sum _{i=0}^{m}\mathbf {h}_i^{-} \Big \rangle > \Big \langle \mathbf {f}_{\perp }, \sum _{i=1}^{\ell }\big (\mathbf {g}_i^{+} -\mathbf {g}_i^{-}\big ) \Big \rangle . \end{aligned}$$

Fig. 4

Here $F$ is a lower-dimensional face of $P$, namely an edge of $P$, and we see how we can get from the face $F$ to the face $F^-$ by walking along the vectors $\mathbf {g}_i^+ - \mathbf {g}_i^-$. Here $\mathbf {f}_\perp$ is chosen to be a vector orthogonal to $F$ and not orthogonal to any of the line segments $h_j$

On the other hand, since $G$ connects $F$ and $F^-$ we have $F=F^-+\sum _{i\in I}\big (\mathbf {g}_i^+-\mathbf {g}_i^-\big ),$ for a set $I$ of edges in $G$. The latter implies that the distance between $F$ and $F^-$ is not more than

$$\begin{aligned} \Big \langle \mathbf {f}_{\perp }, \sum _{i=1}^{\ell }\big (\mathbf {g}_i^{+} -\mathbf {g}_i^{-}\big ) \Big \rangle , \end{aligned}$$

a contradiction. $\square$

Remark 6.1

One of $F$ or $G$ can be $1$-dimensional, in which case $P$ becomes a $3$-dimensional prism.

Main Theorem

Suppose a polytope $P$ $k$-tiles ${\mathbb {R}}^3$ with a multiset $\Lambda$, and suppose that $P$ is not a two-flat zonotope. Then $\Lambda$ is a finite union of translated lattices.

Proof

If $P$ is not a two-flat zonotope, then due to Theorem 6.1, condition (10) is violated. Therefore, the intersection property (8) in Corollary 4.1 holds. Claim 5.1 now concludes the proof. $\square$

7 Many Two-Flat Zonotopes have Weird Tiling Sets

In this section we prove that, under a mild commensurability condition, two-flat zonotopes admit tilings which are not quasi-periodic (“weird”).

Theorem 7.1

Suppose $P$ is a two-flat zonotope in ${\mathbb {R}}^3$ which is the Minkowski sum of the segments

$$\begin{aligned}{}[\mathbf {v}_1], \dots , [\mathbf {v}_n], \quad [\mathbf {w}_1], \dots , [\mathbf {w}_m], \end{aligned}$$

where $\mathbf {v}_1, \ldots , \mathbf {v}_n \in H_1$ and $\mathbf {w}_1,\ldots ,\mathbf {w}_m \in H_2$ and $H_1,\; H_2$ are two different two dimensional subspaces. Suppose also that the additive group generated by $\mathbf {v}_1,\ldots ,\mathbf {v}_n,\mathbf {w}_1,\ldots ,\mathbf {w}_m$ is discrete and that the $\mathbf {v}_j$ span $H_1$.

Then $P$ admits a tiling by translations at a set $\Lambda \subseteq {\mathbb {R}}^3$ which is not a finite union of translated lattices.

Corollary 7.1

If $P \subseteq {\mathbb {R}}^3$ is a two-flat rational zonotope then $P$ admits tilings by sets which are not finite unions of translated lattices.

Proof of Theorem 7.1

We begin the analysis by noting that $P$ can be paved by parallelepipeds, whose sides are among the vectors $\mathbf {v}_j$ and $\mathbf {w}_j$ (proof is by induction on the number of line segments whose Minkowski sum is the zonotope). Therefore we can write its indicator function as a finite sum of indicator functions of parallelepipeds.

$$\begin{aligned} {\mathbf{1}}_P(x) = \sum _{j=1}^M {\mathbf{1}}_{B_j}(x),\quad \text{ for } \text{ a.e. } x, \end{aligned}$$

where each $B_j$ is a parallelepiped, whose three sides are equal to some of the $\mathbf {v}_j$ and $\mathbf {w}_j$.

Suppose now that the parallelepiped $B$ is centered at the origin and has sides parallel to the three linearly-independent vectors $\mathbf {a}, \mathbf {b}, \mathbf {c}$. We can write the indicator function of $B$ as a convolution

$$\begin{aligned} \mathbf{1}_{B} = \frac{{|{\det (\mathbf {a},\mathbf {b},\mathbf {c})}|}}{{|{\mathbf {a}}|}\cdot {|{\mathbf {b}}|}\cdot {|{\mathbf {c}}|}} \mu _{\mathbf {a}}*\mu _{\mathbf {b}}*\mu _{\mathbf {c}}, \end{aligned}$$

where $\mu _{\mathbf {a}}$ is the measure that equals arc-length on the line segment from $-\mathbf {a}/2$ to $\mathbf {a}/2$, and $\mu _{\mathbf {b}}, \mu _{\mathbf {c}}$ are similarly defined. Since (see Sect. 3)

$$\begin{aligned} \widehat{\mu _{\mathbf {a}}}({\varvec{\xi }}) = {|{\mathbf {a}}|} \frac{\sin {\pi {\langle {\varvec{\xi }}, \mathbf {a} \rangle }}}{\pi {\langle {\varvec{\xi }}, \mathbf {a} \rangle }} \end{aligned}$$

and similarly for $\widehat{\mu _{\mathbf {b}}}$, $\widehat{\mu _{\mathbf {c}}}$, we obtain the formula

$$\begin{aligned} \widehat{\mathbf{1}_B}({\varvec{\xi }}) = {|{\det (\mathbf {a},\mathbf {b},\mathbf {c})}|} \frac{\sin {\pi {\langle {\varvec{\xi }}, \mathbf {a} \rangle }}}{\pi {\langle {\varvec{\xi }}, \mathbf {a} \rangle }} \cdot \frac{\sin {\pi {\langle {\varvec{\xi }}, \mathbf {b} \rangle }}}{\pi {\langle {\varvec{\xi }}, \mathbf {b} \rangle }} \cdot \frac{\sin {\pi {\langle {\varvec{\xi }}, \mathbf {c} \rangle }}}{\pi {\langle {\varvec{\xi }}, \mathbf {c} \rangle }}. \end{aligned}$$

(12)

Each parallelepiped $B_j$ in the decomposition of $P$ is a translate of a parallelepiped of the type $B$, above, with some of the vectors $\mathbf {v}_j$, $\mathbf {w}_j$ in place of $\mathbf {a}, \mathbf {b}, \mathbf {c}$. Hence the Fourier Transform $\widehat{\mathbf{1}_{B_j}}$ has the same zeros as the Fourier Transform of its centered translate and these are

$$\begin{aligned} {{\mathsf Z}\big ({\widehat{{\mathbf{1}_{B_j}}}}\big )} = \big (({\mathbb {Z}}')\mathbf {a}^*+\mathbf {a}^\perp \big ) \cup \big (({\mathbb {Z}}')\mathbf {b}^*+\mathbf {b}^\perp \big ) \cup \big (({\mathbb {Z}}')\mathbf {c}^*+\mathbf {c}^\perp \big ), \end{aligned}$$

where ${\mathbb {Z}}'={\mathbb {Z}}\setminus \{0\}$ and again $\mathbf {a}^*=\mathbf {a}/{|{\mathbf {a}}|}^2$ is the geometric inverse of $\mathbf {a}$, etc. Write now

$$\begin{aligned} G = \langle \mathbf {v}_1, \ldots , \mathbf {v}_n \rangle \end{aligned}$$

for the additive subgroup (lattice) of $H_1$ generated by the $\mathbf {v}_j$’s and $G^* \subseteq H_1$ for its dual lattice $in\; H_1$

$$\begin{aligned} G^* = \{\mathbf {u}\in H_1: \forall \mathbf {g}\in G \quad {\langle \mathbf {u}, \mathbf {g} \rangle } \in {\mathbb {Z}}\}. \end{aligned}$$

(13)

We claim now that for each $j$

$$\begin{aligned} H_1^\perp + \big ( G^*\setminus (\mathbf {v}_1^\perp \cup \cdots \cup \mathbf {v}_n^\perp ) \big ) \ \subseteq \ {{\mathsf Z}\big ({\widehat{{\mathbf{1}_{B_j}}}}\big )}. \end{aligned}$$

(14)

This follows since at least one side of $B_j$ is equal to a vector $\mathbf {v}_j$ which makes the corresponding factor in (12) vanish on any element of $G^*$ which is not orthogonal to $\mathbf {v}_j$. And since that factor in (12) is constant along $H_1^\perp$ we obtain the claim. Since (14) holds for all $j$ we obtain

$$\begin{aligned} H_1^\perp + \big ( G^*\setminus (\mathbf {v}_1^\perp \cup \cdots \cup \mathbf {v}_n^\perp ) \big ) \ \subseteq \ {{\mathsf Z}\big ({\widehat{{\mathbf{1}_{P}}}}\big )}. \end{aligned}$$

(15)

Pick now any non-zero $c_1, c_2, \ldots , c_n \in {\mathbb {R}}$. We claim that the measure

$$\begin{aligned} \tau := \mathbf{1}_P *\delta _G * \big [\delta _0 - \delta _{c_1 \mathbf {v}_1}\big ] * \dots * \big [\delta _0 - \delta _{c_n \mathbf {v}_n}\big ] = 0, \end{aligned}$$

(16)

where $\delta _G = \sum _{\mathbf {g}\in G} \delta _{\mathbf {g}}$. For this it is enough to show that the Fourier Transform of the above measure

$$\begin{aligned} \widehat{\tau }({\varvec{\xi }}) = \widehat{\mathbf{1}_P}({\varvec{\xi }}) (1-e^{2\pi i c_1{\langle \mathbf {v}_1, {\varvec{\xi }} \rangle }})\cdots (1-e^{2\pi i c_n{\langle \mathbf {v}_n, {\varvec{\xi }} \rangle }}) \widehat{\delta _G} \end{aligned}$$

is identically $0$. By the Poisson Summation Formula (the Fourier Transform is taken in the sense of distributions)

$$\begin{aligned} \widehat{\delta _L} = \frac{1}{\mathrm{vol\,}{L}} \delta _{L^*}, \end{aligned}$$

(17)

for each lattice $L=A{\mathbb {Z}}^d$ in ${\mathbb {R}}^d$ and dual lattice $L^*=A^{-\top }{\mathbb {R}}^d$ (where $A \in GL(d,{\mathbb {R}})$), it follows that $\widehat{\delta _G}$ is a measure with support on the lines orthogonal to $H_1$ that go through the points of $G^*$:

$$\begin{aligned} \mathrm{supp\,}{\widehat{\delta _G}} = G^* + H_1^\perp . \end{aligned}$$

By (15) the function $\widehat{\mathbf{1}_P}({\varvec{\xi }})$ kills $\widehat{\delta _G}$ except at the lines of the form $\mathbf {g}^*+H_1^\perp$ with $\mathbf {g}^* \in G^*$ is orthogonal to some $\mathbf {v}_j$. But at these lines one of the factors

$$\begin{aligned} (1-e^{2\pi i c_1{\langle \mathbf {v}_1, {\varvec{\xi }} \rangle }})\cdots (1-e^{2\pi i c_n{\langle \mathbf {v}_n, {\varvec{\xi }} \rangle }}) \end{aligned}$$

vanishes, so indeed $\widehat{\tau }$ is zero. Now rewrite the measure $(\delta _0-\delta _{c_1\mathbf {v}_1})\cdots (\delta _0-\delta _{c_n\mathbf {v}_n})$ in the form

$$\begin{aligned} \sum _{k=1}^N \delta _{\mathbf {u}_k^+} - \sum _{k=1}^N \delta _{\mathbf {u}_k^-}\quad (\text{ where } N=2^{n-1}). \end{aligned}$$

Equivalently, we can rewrite (16) as the equality

$$\begin{aligned} \mathbf{1}_P * \delta _G * \sum _{k=1}^N \delta _{\mathbf {u}_k^+} = \mathbf{1}_P * \delta _G * \sum _{k=1}^N \delta _{\mathbf {u}_k^-}. \end{aligned}$$

(18)

Define the multisets

$$\begin{aligned} S = G+\{\mathbf {u}_1^+,\ldots ,\mathbf {u}_N^+\} \quad \text{ and } T = G+\{\mathbf {u}_1^-,\ldots ,\mathbf {u}_N^-\} \end{aligned}$$

(19)

whose ground sets are the supports of the discrete measures

$$\begin{aligned} \delta _G * \sum _{k=1}^N \delta _{\mathbf {u}_k^+} \text{ and } \delta _G * \sum _{k=1}^N \delta _{\mathbf {u}_k^-}, \end{aligned}$$

and their multiplicities at each point are those described by these measures.

In what follows we exploit (18) to give an example of a multiple tiling by $P$ with a discrete set $\Lambda$, which by no means can be expressed as a finite union of translated lattices.

We notice first that since $P$ is a zonotope decomposing into parallelepipeds of sides among the vectors $\mathbf {v}_j, \mathbf {w}_j$, it $k$-tiles ${\mathbb {R}}^3$, for some $k$, with the lattice

$$\begin{aligned} \Gamma = \langle \mathbf {v}_1, \ldots , \mathbf {v}_n, \mathbf {w}_1, \ldots , \mathbf {w}_m \rangle \end{aligned}$$

generated by the $\mathbf {v}_j, \mathbf {w}_j$. The reason is that each of the parallelepipeds $B_j$ tiles with a subgroup of $\Gamma$ (the group generated by its side vectors) and therefore it tiles multiply with $\Gamma$ itself. Clearly, $P$ also $(Nk)$-tiles ${\mathbb {R}}^3$ by the union of $N$ translations of the lattice $\Gamma$ by the vectors $\mathbf {u}_1^+,\dots ,\mathbf {u}_N^+$.

Let $\{\gamma _j:\ j\in {\mathbb {Z}}\}$ be a complete set of coset representatives of $G$ in $\Gamma$. Define the set of translates

$$\begin{aligned} \Lambda = \bigcup _{j \in {\mathbb {Z}}} ( E_j + \gamma _j ), \end{aligned}$$

where for each $j\in {\mathbb {Z}}$ we choose $E_j = S$ or $E_j = T$ arbitrarily. We claim that for any such choice of the $E_j$ the $\Lambda$-translates of $P$ form a $(Nk)$-tiling of ${\mathbb {R}}^3$. Indeed the claim is true if all $E_j=S$ as it is a restatement of the fact that $P$ $(Nk)$-tiles with $\Gamma +\{\mathbf {u}_1^+,\ldots ,\mathbf {u}_N^+\}$, which itself follows from the fact that $P$ $k$-tiles with $\Gamma$. Observe now that if we change any single $E_j$ from $S$ to $T$ we are adding the quantity

$$\begin{aligned} \mathbf{1}_P * \delta _G * \sum _{i=1}^N\delta _{\mathbf {u}_i^-} * \delta _{\gamma _j} - \mathbf{1}_P * \delta _G * \sum _{i=1}^N\delta _{\mathbf {u}_i^+} * \delta _{\gamma _j} \end{aligned}$$

(20)

to the constant function

$$\begin{aligned} \mathbf{1}_P*\delta _{\Lambda }, \end{aligned}$$

which therefore remains the same since (20) is identically $0$ by (18). We conclude that we have a $(Nk)$-tiling no matter how the $E_j$ are chosen (one has to make the remark here that in any given bounded region of space the fact that $P+\Lambda$ is a $(Nk)$-tiling or not is affected by finitely many choices for the $E_j$).

Choose now all $E_j=S$ with the exception of $E_0=T$. We claim that the corresponding set $\Lambda$ is not a finite union of translated fully-dimensional lattices. Indeed, by the Poisson Summation Formula (17) we have that, if

$$\begin{aligned} \Lambda ' = \bigcup _{j\in {\mathbb {Z}}} \big ( S+\gamma _j \big ) = \Gamma + \{\mathbf {u}_1^+,\dots ,\mathbf {u}_N^+\} \end{aligned}$$

then $\widehat{\delta _{\Lambda '}}$ is a discrete measure in ${\mathbb {R}}^3$ and this should also be true for $\widehat{\delta _\Lambda }$ if $\Lambda$ too were a finite union of translated lattices. Thus the difference

$$\begin{aligned} \widehat{\delta _{\Lambda '}} - \widehat{\delta _\Lambda } \end{aligned}$$

would also be a discrete measure. But

$$\begin{aligned} \delta _{\Lambda '}-\delta _{\Lambda }&= \delta _{S+\gamma _0} - \delta _{T+\gamma _0}\\&= \delta _{\gamma _0} * \delta _G * \sum _{i=1}^N \big ( \delta _{\mathbf {u}_i^+}-\delta _{\mathbf {u}_i^-} \big ) \\&= \delta _{\gamma _0} * \delta _G * (\delta _0-\delta _{c_1\mathbf {v}_1})*\cdots * (\delta _0 - \delta _{c_n\mathbf {v}_n}). \end{aligned}$$

So

$$\begin{aligned} \widehat{\delta _{\Lambda '}} - \widehat{\delta _{\Lambda }} = e^{2\pi i {\langle \gamma _0, {\varvec{\xi }} \rangle }} \prod _{j=1}^n (1-e^{2\pi i c_j{\langle \mathbf {v}_j, {\varvec{\xi }} \rangle }}) \widehat{\delta _G}. \end{aligned}$$

(21)

Recall now that the support of the measure $\widehat{\delta _G}$ are all straight lines orthogonal to $H_1$ passing through a point of $G^*$, the dual lattice to $G$ in $H_1$. The factors in the right hand side of (21) vanish at the set

$$\begin{aligned} \bigcup _{j=1}^n \Big ( {\mathbb {Z}}\frac{{\mathbf {v}}_j^*}{c_j} + {\mathbf {v}}_j^\perp \Big ). \end{aligned}$$

(22)

Each set in this union consists of a series of planes normal to $\mathbf {v}_j$ and spaced by a length of $\big (c_j {|{\mathbf {v}_j}|}\big )^{-1}$. Each of the straight lines that make up the support of $\widehat{\delta _G}$ is parallel to each such plane and, therefore, each such line is either entirely contained in (22) or is disjoint from it. It follows that, since the right-hand side of (21) is not identically zero, its support contains at least one straight line of the direction $H_1^\perp$ and is not a discrete set, as we had to show. $\square$

Remark 7.1

One may easily extend the previous construction of $\Lambda$ to the examples that cannot be expressed as a linear combination of finitely many possibly lower-dimensional lattices.

Proof

In the previous construction of $\Lambda$ we could let $E_j$ to be either $S$, or $T$ for each $j$ and still get a legitimate $(Nk)$-tiling of ${\mathbb {R}}^3$. In general we could have a big family (of cardinality $2^{{\mathbb {Z}}}$) of possible $Nk$-tilings of ${\mathbb {R}}^3$. We call a tiling weird if it is not quasi-periodic. In what follows we show that our big family has a weird member $\Lambda ^{\dag }$. In our construction we will need the following claim.$\square$

Claim 7.1

The set of integers ${\mathbb {Z}}$ can be colored with two colors in such a way that every arithmetic progression has infinitely many numbers of each color.

Proof

There are countably many arithmetic progressions in ${\mathbb {Z}}$. We enumerate them all denoting $A_i$ the $i$th progression in the enumeration, such that any progression appears infinitely many times. We begin to color ${\mathbb {Z}}$ step by step in such a way, that on the $n$th step all progressions $A_i$ for $i$ from $1$ to $n$ have numbers of both colors. At the $n$th step we find two numbers of $A_n$ that are not yet colored, and color them differently. It is always possible to do so, because at step $n$ only finitely many numbers of ${\mathbb {Z}}$ are already colored, and $A_n$ has infinitely many numbers. With such a coloring every arithmetic progression would contain infinitely many integers of each of the colors. $\square$

We can pick our complete set of the coset representatives $\{\gamma _j:\ j\in {\mathbb {Z}}\}$ so that it contains $\gamma _1\cdot {\mathbb {Z}}$ as a subset.

In order to construct $\Lambda ^{\dag }$, we consider a coloring of $\gamma _1\cdot {\mathbb {Z}}$ with two colors (red and black) so that every arithmetic progression there has infinitely many red and infinitely many black members. We let $E_j=S$ if corresponding coset representative $\gamma _j\notin \gamma _1\cdot {\mathbb {Z}}$. For coset representatives in $\gamma _1\cdot {\mathbb {Z}}$, if the point $\gamma _1\cdot j$ is red we choose $E_j=S$, if the point $\gamma _1\cdot j$ is black we choose $E_j=T$ in $\Lambda ^{\dag }$.

We further notice that due to the freedom to choose $c_k$’s in the definition of $\mathbf {u}^{-}_{k}$’s and $\mathbf {u}^{+}_{k}$’s, one can pick $c_k$’s so that multisets $S$ and $T$ have different multiplicities at $0$. Indeed, we may pick $c_k$’s so that for any set of indexes $I\subset [n]$ the corresponding linear combination $\sum _{k\in I} c_k\cdot \mathbf {v}_k\notin G$. Then $G+\sum _{k=1}^{N}\mathbf {u}^{+}_{k}$ has $0$ at multiplicity $1$, while $G+\sum _{k=1}^{N}\mathbf {u}^{-}_{k}$ has $0$ at multiplicity $0$. Furthermore, at each point $\gamma _1\cdot \ell$ of $\gamma _1\cdot {\mathbb {Z}}$ the multisets $S+\ell \cdot \gamma _1$ and $T+\ell \cdot \gamma _1$ have different multiplicities as well. Thus we get irregular behavior of $\Lambda ^{\dag }$ on the line ${\mathbb {Z}}\cdot \gamma _1$. In particular, $\Lambda ^{\dag }$ simultaneously contains and misses infinitely many members of each infinite coset of ${\mathbb {Z}}\cdot \gamma _1$.

In other words, we construct such $\Lambda ^{\dag }$, that $\Lambda ^{\dag }$ simultaneously hits and misses infinitely many points of any $1$-dimensional sublattice of ${\mathbb {Z}}\cdot \gamma _1$ (note that (i) we can pick sufficiently small $c_i$’s in the definition of $\mathbf {u}^{+}_{i}$’s and $\mathbf {u}^{-}_{i}$’s so that the sets $S$ and $T$ are disjoint; (ii) that ${\mathbb {Z}}\cdot \gamma _1$ intersects each $S+\gamma _1\cdot j$ exactly at one point).

Now if we assume that $\Lambda ^{\dag }$ may be expressed as a finite linear combination of translated lattices $\delta _{_{\Lambda ^{\dag }}}=q_1\cdot \delta _{_{\Lambda _1}} + \cdots +q_m\cdot \delta _{_{\Lambda _m}}$ (to simplify notations, we will write $q_1\cdot \Lambda _1 + \cdots +q_m\cdot \Lambda _m$ instead of $\delta _{_{\Lambda ^{\dag }}}$), then

$$\begin{aligned} \Lambda ^{\dag }\cap {\mathbb {Z}}\cdot \gamma _1 = \sum _{i=1}^m q_i\cdot \big (\Lambda _i\cap {\mathbb {Z}}\cdot \gamma _1\big ). \end{aligned}$$

Each $\Lambda _i\cap {\mathbb {Z}}\cdot \gamma _1$ is a coset of ${\mathbb {Z}}\cdot \gamma _1$. Therefore, $\Lambda _i\cap {\mathbb {Z}}\cdot \gamma _1$ is either empty, or is a single point, or is an arithmetic progression in ${\mathbb {Z}}\cdot \gamma _1$ with the common difference $d_i$. We denote the set of all the indices of the latter $\Lambda _i$’s by $M\subset \{1,\dots ,m\}$. We further consider an arithmetic progression $A$ of ${\mathbb {Z}}\cdot \gamma _1$ with the common difference $\prod _{i\in M} d_i$. We notice that for any $i\in M$ either $A\cap \Lambda _i=A$, or $A\cap \Lambda _i=\emptyset$. Since $A\subset {\mathbb {Z}}\cdot \gamma _1$, we have

$$\begin{aligned} \Lambda ^{\dag }\cap A = \sum _{i=1}^m q_i\cdot \big (\Lambda _i\cap A\big )=\sum _{i\notin M}q_i\cdot \big (\Lambda _i\cap A\big )+ A\cdot \sum _{\begin{array}{c} i\in M:\\ \Lambda _i\cap A\ne \emptyset \end{array}}q_i. \end{aligned}$$

According to the definition of $M$ the set $\sum _{i\notin M}q_i\cdot \delta _{_{\Lambda _i\cap A}}$ has finite support. Since $A$ is an arithmetic progression in ${\mathbb {Z}}\cdot \gamma _1$ and due to our construction of $\Lambda ^{\dag }$, the support of

$$\begin{aligned}\delta _{_{\Lambda ^{\dag }\cap A}} - \delta _{_A}\cdot \sum _{\begin{array}{c} i\in M:\\ \Lambda _i\cap A\ne \emptyset \end{array}}q_i \end{aligned}$$

cannot be finite, a contradiction. $\square$