Filling space with hypercubes of two sizes – The pythagorean tiling in higher dimensions

Abstract We construct a unilateral lattice tiling of Rn into hypercubes of two differnet side lengths p or q. This generalizes the Pythagorean tiling in R2. We also show that this tiling is unique up to symmetries, which proves a variation of a conjecture by Bölcskei from 2001. For positive integers p and q, this tiling also provides a tiling of (Z/(pn+qn)Z)n.


INTRODUCTION
In 1907, Minkowski [18] conjectured that any tiling of ℝ into -dimensional unit hypercubes whose center points form a lattice has to contain two hypercubes that share a full facet. Keller [10] generalized this conjecture by allowing any tiling of ℝ into -dimensional unit hypercubes. Perron [19] proved Keller's conjecture for ⩽ 6 in 1940 and Hajos [8] proved Minkowski's original conjecture in 1942.
In 1992, Lagarias and Shor [12] proved Keller's conjecture to be false, in particular they showed it is false for ⩾ 10. The gaps were filled by Mackey [14] for dimensions 8 and 9 and Brakensiek et. al. [4] for dimension 7, showing that 7 is the smallest dimension where Keller's conjecture is true. For additional information, see the survey on unit cubes by Zong [24].
In ℝ 2 , there are some related results.

F I G U R E 1 The Pythagorean tiling
• ℝ 2 can be tiled into squares of pairwise distinct integer side lengths in 2 ℵ 0 ways. For ⩾ 3, this is not possible not even if we only require neighboring cubes to have different sizes [5]. • Sprague [22] gave the first example of a tiling of a square into squares of pairwise different integer side lengths (see also [6] and [7]). • Meir and Moser [17] asked whether a square of area ∑ ∞ =1 1∕ 2 = 2 ∕6 can be tiled by the squares of side length 1∕ for ∈ ℤ ⩾1 . The problem remains open but Januszewski and Zielonka [9] showed that for 1∕2 < ⩽ 2∕3, the square of area ∑ ∞ =1 1∕ 2 can be tiled by the squares of side length 1∕ for ∈ ℤ ⩾1 and Tao [23] showed that for every 2∕3 < < 1 there exists an 0 ∈ ℕ such that the square of area ∑ ∞ = 0 1∕ 2 can be tiled by the squares of side length 1∕ for ∈ ℤ ⩾ 0 .
In order to state the main result, we need the following definitions: A tiling is a partition into closed sets without holes, called tiles, whose pairwise intersection is of Lebesgue measure zero. We further call sets of Lebesgue measure zero null sets. For tilings of ℝ into hypercubes, we call a tiling unilateral if no two hypercubes of the same size share a full facet and equitransitive if any two hypercubes of the same size can be mapped to each other by an isomorphism that keeps the tiling unchanged. Lattice tilings are tilings where the center points of all hypercubes of the same size form a lattice and are therefore also equitransitive.
In 2001, Bölcskei [3] proved that Roger's filling ( Figure 2) is the only unilateral and equitransitive tiling of ℝ 3 into cubes of two sizes and conjectured this to be true for any higher dimension.
We prove the following variation of the conjecture, where we consider lattice tiling instead of equitransitive tilings: Theorem 1. Let , ∈ ℝ + with < and ∈ ℕ ⩾2 . There exists exactly one unilateral lattice tiling of ℝ into hypercubes of side length or up to symmetries. Remark 1. Since lattice tilings are equitransitive, in regards to existence Theorem 1 is stronger than Conjecture 1 and in regards to uniqueness it is the other way around.

EXISTENCE
We split Theorem 1 into two parts: The first is Theorem 2 and the other is dealt with in § 4.
Remark 2. After a preprint of this article appeared on arXiv, the author was informed by Mihalis Kolountzakis that Theorem 2 follows from [11,Theorem 8].
Let be the following × matrix: Theorem 2 follows directly from the following Lemma. Proof of Claim 2. Consider the following matrix 1 : Let 1 , .., be the columns of and 1 , … , the columns of 1 . We obtain 1 in the following way. To show that the tiling is unilateral, the followimg Lemma is sufficient.

SYMMETRIES AND PERIODICITY
The group of symmetries of the tiling has to keep the cube invariant, so it is a supgroup of the hyperoctahedral group which has order 2 !. Again we work with the matrix formulation of the problem and write the group operations as matrix multiplications from the left. Here the hyperoctahedral group can be written in the following way: , ±1} × |every row and column contains exactly one ±1 entry} and we find the biggest supgroup of ′ which fixes ℤ . In other words, ∈ ′ is in the stabilizer of the tiling ⟺ ( ) = has an integer solution '⇒': Let ∶= ± such that ≠ and ≠ . Then In the first two cases, − is not in ℤ because it is a vector connecting two inner points of [0, ) . Since − 2 is also connecting two inner points of [0, ) , − ∉ ℤ and therefore ∉ ℤ . Since every vector is of the form ± or the negation of it for all ∈ [1, ] there exists ∈ [1, ] such that = ± . Now suppose , = ±1, then = ± and therefore On the periodicity of the tiling along the coordinate axis, we have the following result: Remark 4. For = 2 and = 1, the tiling gives a lower bound on the maximal number of hypercubes of side length two that can be packed in (ℤ∕(2 + 1)ℤ) , which coincides with the optimal solution (see [1]). Finding such packings is used to determine the Shannon capacity of odd cycles (see [13, 16 21]). □

UNIQUENESS
We now show that every unilateral lattice tiling of ℝ into hypercubes of two sizes is equivalent to the tiling described in the previous sections. It is known that such a tiling does not exist if we only use cubes of one size (see Hajós [8]) and so we can describe the tiling by means of two cubes of different size, translated by ℤ for an × matrix of full rank. . is touching and ′′ in full opposite sides of distance so it can only by filled by small cubes and since small cubes cannot touch, by at most one small cube (see Figure 5).