The dual spectral-set conjecture for spectra and tiling sets

About 27 years old · traced to 1999

Let $L\subset\mathbb{R}^{d}$ . A spectrum is a set $L$ for which there exists a set $\Omega$ such that $(\Omega,L)$ is a spectral pair, and a tiling set is a set $L$ for which there exists a set $\Omega'$ such that $(\Omega',L)$ is a tiling pair. The dual spectral-set conjecture. $L$ is a spectrum if and only if $L$ is a tiling set, i.e., there exists a set $\Omega$ so that $(\Omega,L)$ is a spectral pair if and only if there exists a set $\Omega'$ so that $(\Omega',L)$ is a tiling pair. This is formulated as the dual counterpart of Fuglede's spectral-set conjecture; its general validity remains open.

References

Primary source

Palle E. T. Jorgensen and Steen Pedersen, “Spectral pairs in Cartesian coordinates”, arXiv:math/9912131 (2001).

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Progress summary

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Open

The equivalence is known in a narrow one-dimensional setting, but no general proof or counterexample has been found.

The conjecture asks whether the same set can serve as the frequency set for a spectral pair and as the translation set for a tiling pair. It is presented as the dual counterpart of Fuglede’s conjecture; a 2010 survey records it explicitly as open.

Known results

In one dimension, for a union of two translated lattices, spectrality is equivalent to being a tiling set; the lattices need not be disjoint.
For disjoint lattices, a weaker version holds for possibly unbounded sets of finite positive measure.
The result is restricted to dimension one and leaves the analogous question for $d>1$ open.

Current status (as of August 2026): The conjecture is proved only for the recorded one-dimensional two-lattice class; its general validity, including higher dimensions, remains open, with no publicly reported proof, counterexample, or verification.

Sources

No solutions have been posted yet.