A Note on Lattice Coverings Fei Xue and Chuanming Zong School of Mathematical Sciences, Peking University, Beijing 100871, P. R. China. [email protected] Abstract. Whenever n 3, there is a lattice covering C +Λ of En by a centrally symmetric convex≥ body C such that C does not contain any paral- lelohedron P that P + Λ is a tiling of En. 1. Introduction Let K denote an n-dimensional convex body and let C denote a centrally sym- metric one centered at the origin of En. In particular, let P denote an n-dimensional parallelohedron. In other words, there is a suitable lattice Λ such that P + Λ is a tiling of En. In 1885, E.S. Fedorov [3] discovered that, in E2 a parallelohedron is either a parallelogram or a centrally symmetric hexagon (Figure 1); in E3 a parallelohedron can be and only can be a parallelotope, a hexagonal prism, a rhombic dodecahedron, an elongated octahedron, or a truncated octahedron (Figure 2). parallelogram centrally symmetric hexagon Figure 1 arXiv:1402.4965v1 [math.MG] 20 Feb 2014 Let θt(K) denote the density of the thinnest translative covering of En by K and let θl(K) denote the density of the thinnest lattice covering of En by K. For convenience, let Bn denote the n-dimensional unit ball and let T n denote the n-dimensional simplex with unit edges. In 1939, Kerschner [7] proved θt(B2) = θl(B2) = 2π/√27. In 1946 and 1950, L. Fejes Toth [5] and [6] proved that θt(C)= θl(C) 2π/√27 holds for all two-dimensional centrally symmetric convex domains, where≤ equality is attained precisely for the ellipses. In 1950, F´ary [2] proved that θl(K) 3/2 holds for all two-dimensional convex domains and the equality holds if and only≤ if K is a triangle. For more about coverings, we refer to [1], [4] and [8]. If K + Λ is a lattice covering of E2, it can be easily shown that K contains a centrally symmetric hexagon H such that H + Λ is a tiling of E2. Therefore, let denote the family of all centrally symmetric hexagons contained in K, we have H l vol(K) θ (K)= min . H∈H vol(H) 1 2 parallelotope hexagonal prism elongated octahedron rhombic dodecahedron truncated octahedron Figure 2 This provides a practice method to determine the value of θl(K), in particular when K is a polygon. Then, it is natural to raise the following problem in higher dimensions (see [9]): Problem 1. Whenever K + Λ is a lattice covering of En, n 3, is there always a parallelohedron P satisfying both P K and P + Λ is a tiling≥ of En? ⊆ This note presents a counterexample to this problem. 2. A Counterexample to Problem 1 π π For convenience, we write α = cos 3 , β = sin 3 and take γ to be a small positive number. We note that (1, 0), (α, β), ( α, β), ( 1, 0), ( α, β) and (α, β) are the vertices of a regular hexagon. Let C denote− a three-dimensional− − − centrally− symmetric convex polytope as shown in Figure 3 with twelve vertices v1 = (1, 0, 1+ γ), v2 = (α, β, 1 γ), v3 = ( α, β, 1+ γ), v4 = ( 1, 0, 1 γ), v5 = ( α, β, 1+ γ), v6 = − − − − − − (α, β, 1 γ), v7 = (1, 0, 1 + γ), v8 = (α, β, 1 γ), v9 = ( α, β, 1 + γ), − − − − − − − v10 = ( 1, 0, 1 γ), v11 = ( α, β, 1+ γ) and v12 = (α, β, 1 γ), and let Λ − − − − − − − − − to be the lattice with a basis a1 = (1+α, β, 0), a2 = (1+α, β, 0) and a3 = (0, 0, 2). In fact, C can be obtained from an hexagonal prism of height− 2(1 + γ) by cutting off six tetrahedra, all of them are congruent to each others. It can be easily verified that vi = v6+i + a3 holds for all i = 1, 2,..., 6 and C + Λ is a lattice covering of En. If C contains a parallelohedron P such that P + Λ is a tiling of E3, then P must contain all the 3 v2 v3 v1 v4 v6 v5 v8 v9 v7 v10 v12 v11 Figure 3 twelve vertices v1, v2, . ., v12 of C and therefore P = C. However, C is apparently not a parallelohedron. Thus, C + Λ is a counterexample to Problem 1 in E3. If K is a counterexample to Problem 1 in En−1, defining K′ to be the cylinder over K, one can easily show that K′ will be a counterexample to Problem 1 in En. Therefore, we have proved the following result by explicit examples: Theorem 1. Whenever n 3, there is a lattice covering C + Λ of En by a centrally symmetric convex body C such≥ that C does not contain any parallelohedron P that P + Λ is a tiling of En. Acknowledgements. This work is supported by 973 Programs 2013CB834201 and 2011CB302401, the National Science Foundation of China (No.11071003), and the Chang Jiang Scholars Program of China. References 1. P. Brass, W. Moser and J. Pach, Research Problems in Discrete Geometry, Springer-Verlag, New York, 2005. 2. I. F´ary, Sur la densit´edes r´eseaux de domaines convexes, Bull. Soc. Math. France 178 (1950), 152-161. 3. E.S. Fedorov, Elements of the study of figures, Zap. Mineral. Imper. S. Petersburgskogo Obsˇcˇ, 21(2) (1885), 1-279. 4. G. Fejes T´oth and W. Kuperberg, Packing and covering with convex sets, Handbook of Convex Geometry (P.M. Gruber and J.M. Wills, eds.), North-Holland, Amsterdam 1993, 799–860. 5. L. Fejes Toth, Eine Bemerkung ¨uber die Bedeckung der Eben durch Eibereiche mit Mittelpunkt, Acta Sci. Math. Szeged 11 (1946), 93-95. 6. L. Fejes Toth, Some packing and covering theorems, Acta Sci. Math. Szeged 12 (1950), 62-67. 7. R. Kerschner, The number of circles covering a set, Amer. J. Math. 61 (1939), 665-671. 8. J. Pach and P.K. Agarwal, Combinatorial Geometry, John Wiley & Sons, 1995. 9. C. Zong, Minkowski bisectors, Minkowski cells, and lattice coverings, arXiv:1402.3395..