Bounded common fundamental domains for two lattices

TL;DR

The paper proves that any two full-rank lattices L and M in R^d with equal volume admit a bounded measurable set E that tiles R^d under translations by both L and M. The authors combine cut-and-project/model-set techniques, bounded distance equivalence, and equidecomposability to construct E, treating first the dense-sum case and then the general case via a structured lattice decomposition into L1⊕L2 and M1⊕M2, assembling E from a bounded tile in the dense case plus finite combinatorial ingredients. The result also yields a compactly supported Weyl–Heisenberg window for Gabor bases when the product of the determinants is unity, improving localization properties of such bases. Overall, the work solves a longstanding question on the existence of bounded common fundamental domains for two equal-volume lattices and broadens the toolbox for time-frequency tiling problems. Its methods blend geometric, combinatorial, and aperiodic-tiling concepts with tiling-equivalence results.

Abstract

We prove that for any two lattices $L, M \subseteq \mathbb{R}^d$ of the same volume there exists a measurable, bounded, common fundamental domain of them. In other words, there exists a bounded measurable set $E \subseteq \mathbb{R}^d$ such that $E$ tiles $\mathbb{R}^d$ when translated by $L$ or by $M$. In fact, the set $E$ can be taken to be a finite union of polytopes. A consequence of this is that the indicator function of $E$ forms a Weyl--Heisenberg (Gabor) orthogonal basis of $L^2(\mathbb{R}^d)$ when translated by $L$ and modulated by $M^*$, the dual lattice of $M$.

Theorems & Definitions (10)

• Theorem 1
• Theorem 2
• Theorem 3
• Theorem 4
• Definition 1
• Theorem 5
• proof : Proof of Theorem \ref{['th:dense']}
• Lemma 1
• proof
• Lemma 2