Worked solution: Smith–Myers–Kaplan–Goodman-Strauss's aperiodic monotiles: the hat and the Spectre (2023)

Step 3 of 8: First proof: coupling the hat tiling to two incompatible periodic tilings

In plain words

Suppose, for contradiction, that some tiling by the hat did repeat periodically, like wallpaper. Shrinking every short edge of the hats down to nothing turns that hypothetical periodic hat-tiling directly into a periodic tiling by chevrons; shrinking every long edge instead turns it into a periodic tiling by comets. Both derived tilings would have to repeat with exactly the same underlying pattern of repetition as the original, just rescaled.

But chevrons and comets have different areas, in a ratio that is not a perfect square, so no rescaling of one triangular grid of repetition can ever match the other exactly — like trying to tile a floor with two grids of squares whose spacings are in the ratio $\sqrt{2/3}$ , which never lines up no matter how you shift or rotate it. This impossible mismatch is the contradiction that rules out any periodic hat-tiling.

g: \mathcal{T}_4 \to \mathcal{T}_8 \ \text{would need scale factor } \sqrt{2/3}, \ \text{impossible on the triangular lattice}

Detailed analysis

Assume for contradiction that there is a strongly periodic tiling $\mathcal{T}$ by the hat. Contracting every edge of length $1$ to length $0$ turns $\mathcal{T}$ into a strongly periodic tiling $\mathcal{T}_4$ by chevrons ( $\mathrm{Tile}(0,\sqrt{3})$ , area $3\sqrt{3}$ ); contracting instead every edge of length $\sqrt{3}$ to $0$ gives a strongly periodic tiling $\mathcal{T}_8$ by comets ( $\mathrm{Tile}(1,0)$ , area $2\sqrt{3}$ ) (Smith, Myers, Kaplan & Goodman-Strauss 2023, §3). Because $\mathcal{T}_4$ and $\mathcal{T}_8$ are combinatorially derived from the same tiling $\mathcal{T}$ , there is an affine map $g$ carrying every translation symmetry of $\mathcal{T}_4$ to the corresponding translation symmetry of $\mathcal{T}_8$ , scaling areas by the fixed ratio $\tfrac{2\sqrt{3}}{3\sqrt{3}} = \tfrac{2}{3}$ .

The technical heart of the argument shows $g$ must in fact be a similarity (a shape-preserving rescaling), by tracking how $g$ acts on "worms" — infinite chains of rhombi obtained by further subdividing the chevron tiling — and showing the combinatorics forces $g$ to scale uniformly in every direction. But a similarity scaling areas by $\tfrac{2}{3}$ must scale lengths by $\sqrt{2/3}$ , and no similarity with that particular scale factor can map one set of translation symmetries of the triangular lattice onto another, because $\sqrt{2/3}$ is irrational in a way incompatible with the lattice's own geometry (Smith, Myers, Kaplan & Goodman-Strauss 2023, §3, via a lemma on lattices in the plane).

This contradiction shows no strongly periodic tiling $\mathcal{T}$ by the hat can exist, and since a planar tile with a weakly periodic tiling always has a strongly periodic one too, the hat has no periodic tiling of either kind. Notably, this half of the proof needs no computer at all — it is a classical geometric argument. What it does not show is that the hat tiles the plane in the first place; that existence question is answered independently in the next steps.

Terms in this step

Strongly / weakly periodic tiling: A tiling is weakly periodic if some translation maps it to itself; it is strongly periodic if translations mapping it to itself already tile the whole plane on their own (a discrete lattice of them). For nice tiles, having one kind implies having the other.

Knowledge used in this step

Geometric transformations