Towards resolving Keller’s cube tiling conjecture in dimension seven
Towards resolving Keller’s cube tiling conjecture in dimension seven
复制标题
解决凯勒七维立方体平铺猜想
作者：
A. Kisielewicz
一个cube tiling是一族两两不相交的cube [0，1）d + T = {[0，1）d + t：t ∈ T}，使得cube t∈T（[0，1）d + t）= cube d。两个立方体[0，1）d + t，[0，1）d + s称为孪生对，如果|tj−sj| i ∈ [d] = {1，n，d}，ti = si，i ∈ [d] n {j}.在1930年，凯勒证实，在每一个立方体瓷砖的阿卡德有一个双胞胎对。设L（T，x，i）是向量t ∈ T的所有第i个坐标ti的集合，使得（[0，1] d + t）且ti ≤ xi.设r−（T）=minx∈ Rdmax 1 ≤i≤d| L（T，x，i）|$r^-（T）=min_{xin mathbb{R}^d} max_{1leq ileq d}| L（T，x，i）|$和r+（T）=maxx∈ Rdmax 1 ≤i≤d| L（T，x，i）|$r^ +（T）=max_{xin mathbb{R}^d} max_{1leq ileq d}| L（T，x，i）|$.在2019年之前，已知凯勒猜想对d ≤ 6维为真，对所有d = 8维为假。此外，在7维中，如果r−（T）≤ 2或r+（T）= 5，则已知为真。本文解决了r+（T）= 4的情况。在2019年底，当论文仍在审查中时，Brakensiek等人解决了r+（T）∈ {3，4，6}的情况，从而证明了7维的凯勒猜想。
Abstract A cube tiling of ℝd is a family of pairwise disjoint cubes [0, 1)d + T = {[0, 1)d + t: t ∈ T} such that ∪t∈T([0, 1)d + t) = ℝd. Two cubes [0, 1)d + t, [0, 1)d + s are called a twin pair if |tj−sj| = 1 for some j ∈ [d] = {1, ⋅, d} and ti = si for every i ∈ [d]∖{j}. In 1930, Keller conjectured that in every cube tiling of ℝd there is a twin pair. For x ∈ ℝd and i ∈ [d], let L(T, x, i) be the set of all ith coordinates ti of vectors t ∈ T such that ([0, 1)d + t)∩([0, 1]d + x)≠∅ and ti ≤ xi. Let r−(T)=minx∈Rdmax1≤i≤d|L(T,x,i)|$r^-(T)=min_{xin mathbb{R}^d} max_{1leq ileq d}|L(T,x,i)|$ and r+(T)=maxx∈Rdmax1≤i≤d|L(T,x,i)|$r^ + (T)=max_{xin mathbb{R}^d} max_{1leq ileq d}|L(T,x,i)|$. Before 2019 it was known that Keller’s conjecture is true for dimensions d ≤ 6 and false for all dimensions d = 8. Moreover, in dimension 7 it was known to be true if r−(T) ≤ 2 or r+(T) = 5. The present paper resolves the case r+(T) = 4. At the end of 2019, when the paper was still under review, Brakensiek et al. resolved the cases r+(T) ∈ {3, 4, 6}, proving thereby Keller’s conjecture in dimension 7.