arXiv:1402.4965v1 [math.MG] 20 Feb 2014 neogtdothdo,o rnae caern(Figure rh a truncated prism, a hexagonal a or parallelotope, octahedron, a elongated be an can only and be can .Introduction 1. ercoecnee tteoii of origin the at centered one metric aallga racnrlysmerchxgn(iue1); (Figure symmetric centrally a or n nyif only and hr qaiyi tandpeieyfrteelpe.I 19 In ellipses. the for precisely θ attained is equality where China. [email protected] R. P. University, 100871, Peking Beijing Sciences, Mathematical of School Zong Chuanming and Xue Fei Coverings Lattice on Note A n let and iigof tiling aallhdo.I te od,teei utbelattic suitable a is there words, other In parallelohedron. θ eoetefml falcnrlysmerchxgn conta symmetric centrally all of family the denote θ etal ymti hexagon symmetric centrally n o ovnec,let convenience, For l l l dmninlsmlxwt ntegs n13,Krcnr[7 Kerschner 1939, In edges. unit with simplex -dimensional ( ( ( Let lelohedron Let If Abstract. n18,ES eoo 3 icvrdta,in that, discovered [3] Fedorov E.S. 1885, In etal ymti ovxbody convex symmetric centrally K C B ) 2 ) K 2 = ) ≤ ≤ K θ saltiecvrn of covering lattice a is Λ + t 2 E θ 3 ( π/ K eoean denote l / n ( π/ od o l w-iesoa ovxdmisadteequal the and domains convex two-dimensional all for holds 2 K K . eoetedniyo h hnettasaiecvrn of covering translative thinnest the of density the denote ) √ √ eoetedniyo h hnetltiecvrn of covering lattice thinnest the of density the denote ) P 7hlsfraltodmninlcnrlysmercconve symmetric centrally two-dimensional all for holds 27 satinl.Frmr bu oeig,w ee o[] 4 a [4] [1], to refer we coverings, about more For . a is Whenever 27 that . parallelogram n14 n 90 .FjsTt 5 n 6 rvdthat proved [6] and [5] Toth Fejes L. 1950, and 1946 In n B P dmninlcne oyadlet and body convex -dimensional n satln of tiling a is Λ + eoethe denote n ≥ H θ ,teei atc covering lattice a is there 3, l uhthat such ( K E min = ) E n C n dmninlui aladlet and ball unit -dimensional iue1 Figure npriua,let particular, In . 2 uhthat such tcnb aiysonthat shown easily be can it , H E 1 ∈H H n . etal ymti hexagon symmetric centrally vol( vol( satln of tiling a is Λ + K H C E ) ) . 2 osntcnanayparal- any contain not does aallhdo sete a either is parallelohedron a P C uhthat such Λ e 0 ´r 2 rvdthat proved F´ary [2] 50, eoean denote eoeacnrlysym- centrally a denote in ndin ined C 2). mi , ombic E E of Λ + 3 2 proved ] hrfr,let Therefore, . parallelohedron a K T n ehave we , K n -dimensional E domains, x t od if holds ity P eoethe denote E otisa contains n E θ d[8]. nd n sa is Λ + t ya by θ n ( t B by ( by C 2 = ) = ) K K H . 2

parallelotope elongated 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

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