Lattice Tilings by Cubes: Whole, Notched and Extended

We discuss some problems of lattice tiling via Harmonic Analysis methods. We consider lattice tilings of R by the unit cube in relation to the Minkowski Conjecture (now a theorem of Hajós) and give a new equivalent form of Hajós’s theorem. We also consider “notched cubes” (a cube from which a reactangle has been removed from one of the corners) and show that they admit lattice tilings. This has also been been proved by S. Stein by a direct geometric method. Finally, we exhibit a new class of simple shapes that admit lattice tilings, the “extended cubes”, which are unions of two axis-aligned rectangles that share a vertex and have intersection of odd codimension. In our approach we consider the Fourier Transform of the indicator function of the tile and try to exhibit a lattice of appropriate volume in its zero-set. 1991 Mathematics Subject Classification. Primary 52C22; Secondary 42. the electronic journal of combinatorics 5 (1998), #R14 2 §0. Introduction 0.1 Results. We obtain some results about translational tilings of R with some simple classes of of polyhedra as tiles (cubes as well as “notched” and “extended” cubes–see §2 for a definition of the latter shapes). The approach we use is to study the zero-set of the Fourier Transform (FT) of the indicator function of the tile. If that set contains a lattice except 0 then the set tiles R when translated at the locations of the dual lattice. This means that the translated copies of the tile cover (almost) every point in R a constant number of times–see Theorem 2. In §1 we use our harmonic analysis approach to derive a new equivalent form of the Minkowski conjecture (every lattice tiling of R with the unit cube contains two cubes which share an entire (d−1)-dimensional face) which was proved by Hajós [Haj] in 1941. This new form of the Minkowski conjecture (Theorem 6) is an elementary number-theoretic statement that involves no inequalities and could conceivably lead to a new, elementary proof of the conjecture. In §2 we prove that certain classes of polyhedra tile R if translated by an appropriate lattice. The notched cube (see Figure 1) has already been shown by Stein [St] to tile R by a lattice (Conlan [Con] has done this in some cases). Stein’s method was a direct geometric one. We give a new proof that the notched cube is a tile using our approach. That is, we find lattices in the zero-set of the FT of the indicator function of the notched cube, which is a very explicit function (see (11)). We find all the tilings discovered by Stein, which, by a deeper theorem of Schmerl [Sch], is the complete list of possible translational tilings (lattice or not) of the notched cube. However, our approach for the notched cube leads us to the discovery of a whole class of simple tiles of R (the “extended cubes”–see Figure 2), for which we know of no geometric proof of the fact that they tile. These tiles consist of two axis-aligned rectangles which share a vertex and have intersection of odd codimension, and the lengths of their sides can be completely arbitrary. The tiling lattices for each of these tiles are very simple to describe. Furthermore, the proof that the notched cube tiles essentially proves that the extended cubes tile as well, as the FT of the two indicator functions (that of the notched cube and of that of the extended cube) have the same form and differ only at the values of some parameters. 0.2 Translational tiling in R. Let f ∈ L(R) and A ⊂ R be a discrete point set. We say that (the tile) f tiles R with (the tile set) A and with weight w if for almost all (Lebesgue) x ∈ R we have ∑ a∈A f(x− a) = w, (1) where the series above converges absolutely. If f is the indicator function of a (measurable) set T ⊂ R then we also say that T (the tile) tiles R with weight w, which then has to be a nonnegative integer. When w = 1 we sometimes write R = T ⊕ A. (2) the electronic journal of combinatorics 5 (1998), #R14 3 We restrict our attention to tile sets A of bounded density. That is, we demand that # ( A ∩ (x+ [0, 1]) ) ≤ C, (3) for all x ∈ R and for some constant C, a requirement which is automatically fulfilled whenever f ≥ 0. 0.3 A spectral condition for tiling. In [KLa] a necessary and sufficient condition was given for (1) to hold. It was proved for dimension d = 1 only. Here we state it for arbitrary d. We ommit the proof as it is identical to the one-dimensional case. For a tempered distribution μ we denote by μ̂ its Fourier Transform (see for example [Str]). Theorem 1 Assume that f ∈ L(R) has Fourier Transform f̂ ∈ C(R) and that the discrete set A ⊂ R is of bounded density. Write