通过十字形平铺$R^{5}$

摘要：

一个$n$ -维交叉包括$2n+1$单位立方体：中心立方体和在其所有面的反射.众所周知，有一个平铺的$R^{n}$的交叉为所有$n.$ AlBdaiwi和第一作者证明，如果$2n+1$不是一个素数，然后有$2^{\aleph_{0}}$ \非全等定期（=面对面）平铺的$R^{n}$的交叉，而有一个独特的平铺的$R^{n}$的交叉为$n=2,3$。他们指出，如果$2n+1$是素数，情况总是如此。为了支持这一猜想，我们证明在本文中，也为$R^{5}$有一个独特的定期，没有不定期，平铺的十字架。因此，$R^{3}$通过杂交有唯一的平铺，$R^{4},$有$2^{\aleph_{0}}$平铺，但对于$R^{5}$，仍然只有一个通过杂交的平铺。我们猜测这个结果与我们的直觉相反，即“空间的维度越高，我们获得的自由越多”。

英文摘要：

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 $2^{\aleph_{0}}$ \ 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 $2^{\aleph_{0}}$ 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".

机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。