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For any set $S\subseteq \mathbb{Z}\times\mathbb{Z}= \mathbb{Z}^2$ and $a\in \mathbb{Z}^2$, we set $a+S = \{a+s: s\in S\}$, where $+$ is the componentwise addition in $\mathbb{Z}^2$. Moreover, for any collection of subsets ${\frak S}\subseteq {\cal P}(\mathbb{Z}^2)$ we let $a + {\frak S} = \{a+S: S\in{\frak S}\}$.

We say that a partition ${\frak T}$ of $\mathbb{Z}^2$ is a mono-tiling if for any $T_0\neq T_1\in {\frak T}$ there is $a\in\mathbb{Z}^2$ such that $T_0 = a + T_1$. We call the mono-tiling ${\frak T}$ aperiodic if for all $y\in \mathbb{Z}^2$ we have ${\frak T}\neq y + {\frak T}$.

Does $\mathbb{Z}^2$ have an aperiodic mono-tiling?

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  • $\begingroup$ No rotation? And just the existence of mono-tiling, not whether a tile admits only mono-tilings? $\endgroup$ Commented May 26, 2023 at 9:15
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    $\begingroup$ For a long time \cal A (or whatever symbol in place of A) shows in MO LaTeX as a small square. This is very annoying. Can one fix this \cal malfunctioning? $\endgroup$ Commented May 26, 2023 at 9:29
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    $\begingroup$ @WlodAA: see meta.mathoverflow.net/questions/5508/… $\endgroup$ Commented May 26, 2023 at 13:10
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    $\begingroup$ Usually an aperiodic monotile refers to one whose associated subshift (of all tilings) has no periodic points. The title could also be changed to "monotiling". $\endgroup$ Commented May 26, 2023 at 13:31

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Yes. A $2$-by-$2$ square $\{0,1\}^2$ can tile $\mathbb{Z}^2$ with just one period. So $\{0,2\}^2$ can tile $2\mathbb{Z}^2 \leq \mathbb{Z}^2$ with just one period. Break other periods in the other cosets.

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  • $\begingroup$ I was about to post the same thing, but I feel that 2-by-2 square actually can not tile $\mathbb{Z}^2$ with only one period -- you can shift by the vector $(2, 0)$ (which you probably think of), but also by $(2, 1)$ (or some variant of it). But now that I think of it, it can be fixed by considering $\{0, 1, 2\}\times \{0, 1, 2\}$ and use some non-repeating sequence to make your argument actually work. $\endgroup$ Commented May 26, 2023 at 13:31
  • $\begingroup$ Yes, that's what I mean. The SFT is basically a union of a vertical and a horizontal one-dimensional binary full shift (with a bit of overlap). $\endgroup$ Commented May 26, 2023 at 13:34
  • $\begingroup$ If this was math.SE I'd have explained :) $\endgroup$ Commented May 26, 2023 at 13:36
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    $\begingroup$ Ah, I'm stupid and your example actually work, I can't draw pictures apparently (although a little explanation might be better). $\endgroup$ Commented May 26, 2023 at 13:37
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    $\begingroup$ If an MO answer doesn't cause any headaches, what are we even doing. I'll add something later $\endgroup$ Commented May 26, 2023 at 13:38

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