Tiling R 5 by Crosses

An n-dimensional cross comprises 2n + 1 unit cubes: the center cube and reflections in all its faces. It is well known that there is a tiling of R n by crosses for all n. AlBdaiwi and the first author proved that if 2n + 1 is not a prime then there are 2 ℵ 0 non-congruent regular (= face-to-face) tilings of R n by crosses, while there is a unique tiling of R n by crosses for n = 2, 3. They conjectured that this is always the case if 2n + 1 is a prime. To support the conjecture we prove in this paper that also for R 5 there is a unique regular, and no non-regular, tiling by crosses. So there is a unique tiling of R 3 by crosses, there are 2 ℵ 0 tilings of R 4 , but for R 5 there is again only one tiling by crosses. We guess that this result goes against our intuition that suggests " the higher the dimension of the space, the more freedom we get ". Tilings of R n by unit cubes go back to 1907 when Minkowski conjectured [16] that each lattice tiling of R n by unit cubes contains twins, a pair of cubes sharing a complete n − 1 dimensional face. This conjecture was proved by Hajós [6] in 1942. In 1930, when Minkowski's conjecture was still open, Keller [12] suggested that the lattice condition in the conjecture is redundant, that the nature of the problem is purely geometric, and not algebraic as assumed by Minkowski. Thus he conjectured that each tiling of R n by unit cubes contains twins. It is trivial to see that each tiling of R 2 by unit cubes contains twins, and it is also easy to verify it for R 3. However, a proof that each tiling of R n , 4 ≤ n ≤ 6, contains twins takes in aggregate 80 pages, see [15]. There was no progress on Keller's conjecture for more than 50 years. Only in 1992 Lagarias and Shor [13] constructed a tiling of R n , n ≥ 10, by unit cubes with no twins. First they found such a 1