仅用两块瓦片或一块非交换瓦片的不可判定平移铺砌
Undecidable translational tilings with only two tiles, or one nonabelian tile
AI总结：
本文构造了仅用两块瓦片或一块非交换瓦片在ZFC中不可判定的平移铺砌实例，证明了其非周期性，并揭示了少量瓦片编码任意铺砌问题的能力。
摘要：
我们构造了一个有限交换群 $G_0$ 的群 $G = \mathbb{Z}^2 \times G_0$ 的例子，以及 $G_0$ 的子集 $E$ 和 $G$ 的两个有限子集 $F_1,F_2$，使得 $\mathbb{Z}^2\times E$ 是否可以通过 $F_1,F_2$ 的平移进行铺砌在 ZFC 中是不可判定的。特别地，这意味着该铺砌问题是非周期的，即（在 ZFC 的标准全域中）存在使用瓦片 $F_1,F_2$ 对 $E$ 的平移铺砌，但不存在周期性铺砌。此前，此类非周期或不可判定的平移铺砌仅针对十一块或更多瓦片的集合（主要在 $\mathbb{Z}^2$ 中）进行过构造。对于足够大的 $d$，类似的构造也适用于 $G = \mathbb{Z}^d$。如果允许群 $G_0$ 是非交换的，该构造的一个变体可以仅用一块瓦片 $F$ 产生不可判定的平移铺砌。该证明过程首先观察到，单个铺砌方程能够编码任意铺砌方程系统，而一旦有两块或更多瓦片，该系统又能编码某些特定泛函方程的任意系统。特别地，可以使用两块瓦片来编码任意数量瓦片的铺砌问题。
英文摘要：
We construct an example of a group $G = \mathbb{Z}^2 \times G_0$ for a finite abelian group $G_0$, a subset $E$ of $G_0$, and two finite subsets $F_1,F_2$ of $G$, such that it is undecidable in ZFC whether $\mathbb{Z}^2\times E$ can be tiled by translations of $F_1,F_2$. In particular, this implies that this tiling problem is aperiodic, in the sense that (in the standard universe of ZFC) there exist translational tilings of $E$ by the tiles $F_1,F_2$, but no periodic tilings. Previously, such aperiodic or undecidable translational tilings were only constructed for sets of eleven or more tiles (mostly in $\mathbb{Z}^2$). A similar construction also applies for $G = \mathbb{Z}^d$ for sufficiently large $d$. If one allows the group $G_0$ to be non-abelian, a variant of the construction produces an undecidable translational tiling with only one tile $F$. The argument proceeds by first observing that a single tiling equation is able to encode an arbitrary system of tiling equations, which in turn can encode an arbitrary system of certain functional equations once one has two or more tiles. In particular, one can use two tiles to encode tiling problems for an arbitrary number of tiles.
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。
