Translational tilings without full periodicity

Source and verification

This article explains family 155 of the October 6, 2026 OpenAI collection. Research conclusions are attributed to its manuscripts. The review covers statements, introductions, and selected proof mechanisms; it is not an independent verification of the complete proofs.

🏷️ Translational tilings and periods

Let $T$ be a finite subset of the integer lattice ${\mathbb{Z}}^3$. A translation set $A\subseteq {\mathbb{Z}}^3$ tiles the lattice by $T$ when every lattice point has exactly one representation $a+t$, with $a\in A$ and $t\in T$. We write

$$A\oplus T={\mathbb{Z}}^3.$$

This is a covering requirement and a disjointness requirement simultaneously. Rotations and reflections of $T$ are not allowed. A tiling is fully periodic if $A$ is invariant under three linearly independent lattice translations, equivalently under a subgroup of finite index in ${\mathbb{Z}}^3$. Periodicity is a property of the translation set, not just of the shape.

The distinction between one nonperiodic tiling and a tile that permits no fully periodic tiling is essential. A tile might admit both periodic and nonperiodic arrangements. The assertion below excludes every fully periodic arrangement of the chosen tile.

Manuscript claim

There is a finite $T\subset {\mathbb{Z}}^3$ admitting a translational tiling but no fully periodic translational tiling. Its unit-cube thickening $\Omega =T+[0,1{]}^3$ also tiles ${\mathbb{R}}^3$ up to null sets, but admits no tiling whose translation set is invariant under a full-rank lattice, even when arbitrary real translation vectors are allowed. (OpenAI, 2026)

“No fully periodic tiling” does not assert the absence of every individual nonzero period. The manuscript also identifies three as the least lattice dimension for this phenomenon: in dimensions one and two, a finite tile that tiles admits some fully periodic tiling. The Euclidean thickening statement should not be replaced by a claim about all measurable planar tiles.

🏷️ A finite-state source of nonperiodicity

A useful way to understand the construction is to start with a constrained array rather than a geometric solid. Let $I$ be a finite set of column indices, and consider an array $W(n,m)$, with $n\in I$ and $m\in \mathbb{Z}$, taking values in a finite alphabet. For each integer slope $d$ and intercept $e$, inspect the finite word

$$(W(n,dn+e){)}_{n\in I}.$$

The manuscript supplies a finite-word constraint system with two properties: an admissible array exists with every column nonconstant, and no admissible array satisfying those nonconstancy conditions can have a vertical period. This is the combinatorial source of the eventual obstruction.

The hard issue is preserving the implication in both directions while converting these arrays into tilings. It would not suffice to encode one convenient array: a different tiling of the final tile might evade the constraints and become periodic. Every tiling must decode into an admissible array with the required nonconstancy.

🏷️ Encoding constraints by disjoint translates

The first encoding stage uses several tiling equations with a common unknown translation set. A finite cyclic coordinate records symbols and local tests. Different tests impose compatibility of the encoded words, activation of relevant symbols, and consistency around cycles. Two additional seed channels ensure that decoded columns cannot be constant.

There is a subtle asymmetry between construction and decoding. The explicitly constructed model may have one active symbol at a location, while the equations can permit several active symbols in an arbitrary solution. The decoding argument must therefore make its constraints valid for every permitted choice of an active symbol. Simply assuming the arbitrary solution has the tidy form of the model would leave a gap.

The finite coordinate systems have pairwise coprime orders. Their product can consequently be represented by one cyclic coordinate. A further stacking construction, using a fresh prime, combines the simultaneous equations into a single translational tiling problem. At this intermediate point the ambient space has two infinite integer coordinates and one finite cyclic coordinate.

The final lattice construction replaces that finite coordinate by geometric structure in a third infinite coordinate. If a fully periodic tiling of the resulting finite tile existed, its decoded array would inherit a forbidden period. Existence, on the other hand, follows by encoding the explicitly supplied admissible array. These two directions explain why the proof needs both a model and universal decoding tests.

🏷️ Why thickening needs an additional argument

For any lattice tiling $A\oplus T={\mathbb{Z}}^3$, replacing each point of $T$ by a unit cube immediately gives a tiling of ${\mathbb{R}}^3$ outside cube boundaries. This proves Euclidean existence. It does not establish the Euclidean periodicity obstruction: a putative periodic tiling could use translation vectors outside ${\mathbb{Z}}^3$.

The manuscript uses a rigidity mechanism involving shapes with a large common portion and displaced marked cubes. Under a hypothetical periodic tiling, these marked pieces constrain relative translations, recover the relevant lattice structure, and force the invariance needed for decoding. Merely observing that the tile has integer vertices would not justify this conclusion.

The graduate-level lesson is an exactness issue. Passing from a discrete model to a geometric realization must preserve all solutions relevant to the forbidden behavior. Here the unrestricted real translation set is precisely where that preservation requires a new argument.

🏷️ Key steps and proof outlines

A full-difference partition prevents duplicate contributions

The stacking step has a short probabilistic construction. Given $s$ simultaneous tiling equations, choose a sufficiently large fresh odd prime $q$ and independently color each element of the cyclic coordinate with one of $s$ colors. For a nonzero difference $d$, the pairs $(0,d),(2d,3d),\dots$ are disjoint and all have difference $d$. The probability that none has both endpoints in a specified color is at most $(1-s^{-2}{)}^{\lfloor q/2\rfloor }$.

A union bound over colors and nonzero differences gives failure probability at most $s(q-1)(1-s^{-2}{)}^{\lfloor q/2\rfloor }<1$ for large $q$. Hence there is a partition $E_1,\dots ,E_s$ with $E_{\nu }-E_{\nu }$ equal to the whole cyclic coordinate for every $\nu$. Operationally, any two translates of the same $E_{\nu }$ intersect. This intersection property will force uniqueness in an arbitrary stacked tiling. (OpenAI, 2026)

Fiber counting recovers every original equation

Suppose the finite tiles $F_1,\dots ,F_s$ in an ambient discrete space $H$ must share one translation set $A$. Form the stacked tile $F^{\mathrm{st}}={\bigcup }_{\nu }(F_{\nu }\times E_{\nu })$. A common solution immediately gives the stacked solution $A\times \{0\}$: the finite-coordinate value chooses the unique color, then the corresponding original equation supplies the unique representation in $H$.

Conversely, take any exact stacked tiling with translation set $B$. Fix a horizontal point $g$. A contribution of color $\nu$ covers some translated vertical set $t+E_{\nu }$. There cannot be two such contributions of the same color, because they would overlap. Let their number be $n_{\nu }(g)\in \{0,1\}$. Exact coverage of the vertical fiber gives

$$\sum _{\nu =1}^s{n}_{\nu }(g)|E_{\nu }|=q=\sum _{\nu =1}^s|E_{\nu }|.$$

Every summand size is positive, so every $n_{\nu }(g)$ equals one. Projection of $B$ onto $H$ is also injective: two translations above the same horizontal point, together with any element of $F_1$, would create two color-one contributions. Thus the projection is an ordinary set, and the fiber count says it solves every original tiling equation exactly.

This proof uses both covering and disjointness. Coverage alone permits extra contributions; disjointness alone permits missing colors. It is the combination that makes every solution decode, not merely the explicitly constructed model solution.

Periodicity is preserved in the required direction

If $B$ has a finite-index period subgroup, its projection is a period subgroup of the decoded translation set. It still has finite index because projecting the finite quotient surjects onto the quotient by the projected subgroup. Therefore a fully periodic stacked tiling would yield a fully periodic common solution, contradicting the earlier array constraint.

The fresh prime is chosen coprime to the previous finite-coordinate order. Their product can then be represented by one cyclic coordinate, which is necessary for the following realization in three infinite lattice coordinates. An arbitrary product of finite coordinates would not automatically have the same cyclic description.

Activation and Euclidean rigidity close two further loopholes

Before stacking, the activation tests must force every admissible decoded choice to obey the word rules. A constructed tiling may activate one symbol at a site, while another legal tiling may activate several. The universal decoding argument therefore applies to all choices among those symbols. The two seed channels separately prevent constant decoded columns, so a vertical period cannot evade the source array’s nonperiodicity theorem by degenerating to a constant column.

After replacing the finite coordinate with a third lattice direction, the source’s marked-cube realization must preserve exact tilings and transfer a full-rank period back to the encoded system. Thickening then proves Euclidean existence immediately, but not Euclidean nonperiodicity. The additional rigidity argument allows arbitrary real translation vectors and extracts the needed lattice constraints from a large common shape and displaced marked cubes. Integer vertices of the tile alone would not force integer translation differences.

Thus the proof has three distinct universal-decoding interfaces: simultaneous equations from a stacked tile, word constraints from every simultaneous solution, and discrete period information from a hypothetical periodic Euclidean tiling. The full-difference fiber argument above makes the first interface explicit; the marked-cube and activation theorems supply the other two in the manuscript.

See Also

• Plane coloring also transfers a discrete obstruction into Euclidean geometry while carefully retaining the quantifier over arbitrary configurations.
• Irrational triangle billiards concerns a different form of nonperiodicity, defined through trajectories rather than translation sets.

🏷️ Manuscripts and verification scope

• A translational tile with no fully periodic tiling in dimension three — pinned manuscript and companion source.

The family scope page describes selected formalization coverage. The page selects the lattice counterexample, its Euclidean unit-cube thickening with arbitrary real translations, and lattice dimensional minimality. No Lean build, axiom audit, or manuscript-to-formal-statement equivalence check was performed for these notes.

OpenAI collection provides prerequisites and the reading sequence; coverage register records completed and pending families.

🐻  OpenAI 2026. A translational tile with no fully periodic tiling in dimension three.