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Both directions of Fuglede's conjecture fail in dimension two

This paper resolves the long-standing open problem in dimension two by constructing explicit counterexamples in a finite Abelian group and lifting them to $\mathbb{R}^2$ , thereby proving that both directions of Fuglede's conjecture fail.

Original authors: Tao Zhang

Published 2026-07-20

📖 4 min read🧠 Deep dive

CC BY 4.0

Original authors: Tao Zhang

Original paper licensed under CC BY 4.0 (http://creativecommons.org/licenses/by/4.0/). ✨ This is an AI-generated explanation of the paper below. It is not written or endorsed by the authors. For technical accuracy, refer to the original paper. Read full disclaimer

Imagine you are a master architect trying to cover a floor. You have a specific shape of tile, and you want to know if you can cover the entire infinite floor using only that shape, sliding it around without any overlaps or gaps. This is the art of tiling. Now, imagine you are a musician trying to play a song inside a room. You want to know if you can pick a set of musical notes (frequencies) that, when played together, perfectly describe every possible sound inside that room without any redundancy. This is the art of being spectral.

For decades, mathematicians wondered if these two very different skills were actually the same thing. Could a shape that is perfect for tiling a floor always be perfect for creating a musical spectrum, and vice versa? This idea, known as Fuglede's conjecture, linked the geometry of shapes with the physics of sound waves. It was a beautiful theory that held true for simple, one-dimensional lines and for many nice, round, or boxy shapes. However, as mathematicians looked at more complex, jagged, or high-dimensional spaces, the theory began to crack. They found shapes that could tile but couldn't sing, and shapes that could sing but couldn't tile. But a stubborn mystery remained: did this failure happen in the flat, two-dimensional world we live in—the plane? For a long time, no one could prove it either way.

This paper by Tao Zhang answers that question with a definitive "yes." The author proves that in two dimensions, the connection between tiling and music is broken. The paper constructs two very specific, somewhat bizarre shapes. The first shape is a perfect tiler—it can cover a grid without gaps—but it is "tone-deaf," meaning it cannot be described by a clean set of musical notes. The second shape is a perfect "singer," capable of generating a full musical spectrum, but it is a terrible tiler that cannot cover a grid without overlapping or leaving holes.

To understand how this was done, think of the mathematician's strategy as a game of "zooming out." First, the author built these tricky shapes inside a small, finite world made of a grid of numbers (specifically a group called $\mathbb{Z}_{60} \times \mathbb{Z}_{12}$ ). In this tiny, finite universe, the author showed that the "tiling" shape had three different ways to be completed into a full grid, but none of those completions shared a common musical key. Conversely, the "singing" shape had three different musical keys, but none of them could help it tile the grid.

The magic happens in the next step. The author used a mathematical "transference principle" to lift these small, finite shapes up into the real, infinite two-dimensional plane. Imagine taking a small, pixelated stamp and pressing it onto a giant sheet of graph paper, repeating it over and over to create a massive, jagged mosaic. The paper proves that if the small stamp fails to be a perfect singer or a perfect tiler in the finite world, the giant mosaic will fail in the exact same way in the real world.

The result is a pair of explicit counterexamples. The first is a bounded shape made of 60 points in the finite group, which lifts to a massive shape made of 564,540 unit squares that tiles the plane but has no spectrum. The second is a shape made of 60 points in the finite group, which lifts to a massive shape made of 564,540 unit squares that has a spectrum but cannot tile the plane. By lifting these to the real plane, the author creates these massive shapes to settle the debate: in two dimensions, being able to tile a floor and being able to sing a song are two completely different talents. A shape can have one, the other, both, or neither, and there is no rule that forces them to go together. The conjecture that they were equivalent is false.

Technical Summary: Failure of Fuglede's Conjecture in Dimension Two

Problem Statement
Fuglede's conjecture, proposed in 1974, posits that a measurable set $\Omega \subset \mathbb{R}^d$ of positive finite measure is spectral (admits an orthogonal basis of exponential functions) if and only if it is a translational tile (can tile $\mathbb{R}^d$ by translations). While the conjecture holds for convex domains and under various geometric or arithmetic constraints, it is known to fail in both directions for all dimensions $d \geq 3$ . The status of the conjecture in dimensions one and two remained an open problem. This paper addresses the unresolved two-dimensional case.

Methodology
The authors employ a strategy of constructing counterexamples in finite Abelian groups and transferring them to Euclidean space via a finite-to-infinite transference principle. The core methodology involves:

Finite Group Construction: The authors work within the rank-two finite Abelian group $G = \mathbb{Z}_{60} \times \mathbb{Z}_{12}$ $G = Z_{60} \times Z_{12}$ . They construct two specific 60-point subsets of this group:
- A set $A_-$ that tiles $G$ but possesses no spectrum.
- A set $A_+$ that is spectral but does not tile $G$ .
Layer-Amplification: To achieve these properties, the authors utilize a "five-layer" amplification technique. They start with a "seed" set $E \subset \mathbb{Z}_{12}^2$ $E \subset Z_{12}^{2}$ (a 12-point graph).
- For the non-spectral tile, they identify three distinct tiling complements of $E$ that share no common spectrum. These are arranged in five layers (using an auxiliary factor $\mathbb{Z}_5$ ) to create a set where the incompatibility of the complements prevents the existence of any global spectrum.
- For the non-tiling spectral set, they identify three distinct spectra of $E$ that share no common tiling complement. A parallel five-layer construction ensures the resulting set is spectral but cannot tile.
Isomorphism and Transfer: The constructions are performed in $H \times \mathbb{Z}_5$ (where $H = \mathbb{Z}_{12}^2$ ). The Chinese Remainder Theorem provides an isomorphism $\Phi: H \times \mathbb{Z}_5 \to \mathbb{Z}_{60} \times \mathbb{Z}_{12}$ , preserving the group structure without increasing the rank.
Finite-to-Infinite Transference: Using a refined version of the transference principle (Proposition 2.1), the finite counterexamples are lifted to bounded subsets of $\mathbb{R}^2$ . Specifically, for sufficiently large $k$ , the set $A[k] = A + \{ (n_1 j_1, n_2 j_2) : 0 \leq j_1, j_2 < k \}$ retains the spectral or tiling properties of the original finite set $A$ . These discrete sets are then expanded into unions of unit squares in $\mathbb{R}^2$ .

Key Contributions and Results
The paper establishes the following main theorem:

Theorem 1.1: There exists a bounded translational tile in $\mathbb{R}^2$ that is not spectral, and there exists a bounded spectral set in $\mathbb{R}^2$ that does not tile by translations.

The specific results include:

Explicit Construction of a Non-Spectral Tile: The authors construct a set $A_- \subset \mathbb{Z}_{60} \times \mathbb{Z}_{12}$ which tiles the group but has no spectrum. Through transference, this yields a bounded translational tile in $\mathbb{R}^2$ (a finite union of unit squares) that is not spectral.
Explicit Construction of a Non-Tiling Spectral Set: The authors construct a set $A_+ \subset \mathbb{Z}_{60} \times \mathbb{Z}_{12}$ which is spectral but does not tile the group. Through transference, this yields a bounded spectral set in $\mathbb{R}^2$ (specifically, a disjoint union of 564,540 half-open unit squares) that does not tile by translations.
Resolution of the Rank-Two Barrier: The paper demonstrates that Fuglede's conjecture fails even in groups with only two cyclic factors ( $\mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2}$ ), overcoming a restrictive setting where the conjecture was previously known to hold in many specific finite Abelian families.

Significance
The paper claims to resolve the two-dimensional case of Fuglede's conjecture, proving that both implications of the conjecture fail in $\mathbb{R}^2$ . This completes the picture for unrestricted measurable sets, showing that the equivalence between spectrality and tiling fails in every dimension $d \geq 2$ . The work extends the known negative results from $d \geq 3$ down to the plane, utilizing a novel construction in a rank-two finite Abelian group to bypass previous positive results in low-rank settings. The authors note that their constructions are explicit and rely on the interplay between tiling complements and spectra in finite groups, lifted to the continuous domain.

Technical Summary: Failure of Fuglede's Conjecture in Dimension Two

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