The continuous periodic tiling conjecture

Equivalent formulations 2

Other statements of this same problem, merged from separate entries. Each is equivalent to the statement above — proving any one settles them all.

1. The continuous periodic tiling conjecture

   Let $\Sigma$ be a bounded measurable subset of Euclidean space $\mathbb{R}^d$. A discrete set $\Lambda\subseteq\mathbb{R}^d$ gives a measurable tiling when the translates partition $\mathbb{R}^d$ up to null sets; call the tiling periodic when $\Lambda$ is a finite union of cosets of a lattice in $\mathbb{R}^d$. Continuous periodic tiling conjecture. If $\Sigma$ tiles $\mathbb{R}^d$ by translations, then $\Sigma$ periodically tiles $\mathbb{R}^d$ by translations. The conjecture is known in the one-dimensional settings $G=\mathbb{Z}$ and $G=\mathbb{R}$, but the higher-dimensional problem is not resolved here.

   source: Rachel Greenfeld and Terence Tao, “A counterexample to the periodic tiling conjecture (announcement)”, arXiv:2209.08451 (2022).

2. The continuous periodic tiling conjecture

   Let $\Omega\subset\mathbb{R}^d$ be a bounded measurable set of positive measure. A continuous translational tiling is a set $T\subset\mathbb{R}^d$ such that

   $$\Omega\oplus T=_{\mathrm{a.e.}}\mathbb{R}^d.$$

   A tiling set is periodic if it is invariant under translations by some lattice $\Lambda\subset\mathbb{R}^d$.

   Continuous periodic tiling conjecture. The tiling equation $\Omega\oplus X=_{\mathrm{a.e.}}\mathbb{R}^d$ is not aperiodic. Equivalently, if $\operatorname{Tile}(\Omega;\mathbb{R}^d)$ is nonempty, then it contains a periodic set $T$.

   This is proposed as the continuous analogue of the discrete periodic tiling conjecture. The source does not state whether it is resolved.

   source: Rachel Greenfeld, “Translational tilings: structured or wild?”, arXiv:2509.25576 (2025).