The Coin Exchange Problem and the Structure of Cube Tilings

It is shown that if [0, 1)d + t, t ∈ T , is a unit cube tiling of Rd, then for every x ∈ T , y ∈ Rd, and every positive integer m the number |T ∩ (x+Zd)∩ ([0,m)d+y)| is divisible by m. Furthermore, by a result of Coppersmith and Steinberger on cyclotomic arrays it is proven that for every finite discrete box D = D1×· · ·×Dd ⊆ x+Zd of size m1×· · ·×md the number |D∩T | is a linear combination of m1, . . . ,md with non-negative integer coefficients. Several consequences are collected. A generalization is presented. An interest in cube tilings of R originated from the following question raised by Hermann Minkowski [23]: Characterize lattices Λ ⊂ R such that [0, 1) + λ, λ ∈ Λ, is a cube tiling. Minkowski conjectured that such a lattice Λ is of the form AZ, where A is a lower triangular matrix with ones on the main diagonal. By a simple inductive argument, it is equivalent to showing that Λ contains an element of the standard basis. Geometrically, it means that the tiling [0, 1)+λ, λ ∈ Λ, contains a column. This inspired Ott-Heinrich Keller [11] to consider the problem of the existence of columns in arbitrary cube tilings. In [12] he conjectured that starting from dimension 7 there are cube tilings without columns. This has been confirmed in dimension 10 by Jeffrey Lagarias and Peter Shor [18] and in dimension 8 by John Mackey [22]. There are quite a few other papers stemming from Keller’s problem [2, 3, 5, 6, 8, 13, 14, 19, 21, 24, 28]. A new stimulus came from Fuglede’s conjecture [7]. Several papers appeared at almost the same time where the set determining a cube tiling is characterized as follows: [0, 1)+t, t ∈ T , is a cube tiling of R if and only if the system of functions exp(2π i〈t, x〉), t ∈ T , is an orthonormal basis of L([0, 1]) ([9, 10, 15, 17]). A reader who seeks an exposition concerning cube tilings is advised to consult [16, 26, 29]. Let us suppose that [0, 1) + t, t ∈ T , is a cube tiling and that x ∈ T . The simplest unit cube tiling to which the cube [0, 1) +x belongs is [0, 1) +u, u ∈ x+Z. We discuss the electronic journal of combinatorics 19(2) (2012), #P26 1 here in what way these two tilings are intertwined. To be more specific, we show that there are certain number-theoretic characteristics of the intersection (x+ Z) ∩ T . A standard block in R is a set of the form X = X1× · · ·×Xd, where Xi ∈ {[0, 1),R}. The translate X + x of a standard block X ⊆ R on x ∈ R is called a block in R. A block which is a translate of [0, 1) is said to be a unit cube. If F = F1×· · ·×Fd is a block in R, then the set NF = {i : Fi = R} is called the cosupport of F . For future reference, we define the set ZdNF by the equation ZdNF = {(k1, . . . , kd) ∈ Z d : ki = 0,whenever i 6∈ NF}. If F is a block, then its position vector v = v(F ) ∈ R is defined as follows vi = { xi, if Fi = [0, 1) + xi, 0, if Fi = R. Clearly, F − v is a standard block. A family F of disjoint blocks contained in R is a block tiling of R if ⋃ F = R. If F consists of unit cubes only, then we refer to F as a cube tiling. Suppose that there is a unit cube J which belongs to a block tiling F . Then the family of cubes {(J + k) : k ∈ Z}∩F is called a simple component of F (containing J). Theorem 1 Suppose that a unit cube J belongs to a block tiling F . Let C be a simple component of F containing J . Then there is a block tiling G such that (1) v(G)i ∈ v(J)i + Z, for G ∈ G and i 6∈ NG; (2) J ∈ G and the simple component of G containing J equals C . In particular, C consists of all unit cubes belonging to G . Proof. We may assume that J = [0, 1). Let us fix i ∈ {1, . . . , d} and α ∈ (0, 1). Let F = {F ∈ F : v(F )i ∈ α+Z}. If F is non-empty, then since F is a tiling, the family S := {F ∈ F : v(F )i = α} is non-empty. For F ∈ S , let F̄ be a block defined so that F̄i = R and F̄j = Fj whenever j 6= i. Let Gα = {F̄ : F ∈ S }. Again by the fact that F is a tiling, we deduce that ⋃ F = ⋃ Gα. If F is empty, then we let Gα to be empty. Observe now that the family F (i) := (F \ ⋃ α∈(0,1) F ) ∪ ⋃ α∈(0,1) Gα is a block tiling which has the following properties: the simple components of F and F (i) containing J coincide; if F ∈ F , then v(F )i ∈ Z. Let us define inductively a sequence of block tilings (F [i] : i = 1, . . . , d): F [1] = F , F [2] = (F ), . . . , F [d] = (F [d−1])(d). the electronic journal of combinatorics 19(2) (2012), #P26 2 It is clear that the simple components of F and F [d] containing J coincide and v(F ) ∈ Z whenever F ∈ F . Therefore, it suffices to declare G = F . We say that a set T ⊂ R determines a cube tiling F of R if T = v(F ) = {v(F ) : F ∈ F}. Let T determine a cube tiling F and t ∈ T . Then the set (t+ Z) ∩ T consists of all position vectors of a certain simple component of F . As an immediate consequence we have Theorem 2 Given a set T determining a cube tiling of R. Let x ∈ T and let D ⊂ x+Z be a finite discrete box with sidelengths equal to m; that is, D = D1 × · · · × Dd and m = |D1| = · · · = |Dd|. Then |T ∩D| is divisible by m. Proof. Let F be the cube tiling determined by T . Let C be the simple component of F which contains J = [0, 1) + x. Let G be the block tiling of R as described in Theorem 1. Since for each G ∈ G we have G = [0, 1) + v(G) + ZdNG , it follows that the set P = {(v(G) + ZdNG) ∩D : G ∈ G } \ {∅} is a partition of D. Observe that NG is non-empty if G ∈ G \ C . Thus, HG := (v(G) + ZdNG) ∩D is empty or it is a discrete box such that Hi = Di, for i ∈ NG. Now, it follows that |HG| is divisible by m. Therefore, the number |T ∩D| = m − ∑