[Paper Review] Packing near the tiling density and exponential bases for product domains

Mihail N. Kolountzakis|arXiv (Cornell University)|Jun 8, 2016

Mathematical Analysis and Transform Methods11 references3 citations

TL;DR

This paper establishes that if a product domain $ A \times \Omega \subseteq \mathbb{R}^{d+n} $ is spectral, then both $ A $ and $ \Omega $ are spectral, extending previous results beyond intervals to unions of two intervals. By proving that high-density packings of $ |\widehat{\chi_\Omega}|^2 $ imply tiling, the authors show that spectrality of $ A \times \Omega $ forces both factors to be spectral, even when $ A $ is a union of two disjoint intervals.

ABSTRACT

A set $Ω$ in a locally compact abelian group is called spectral if $L^2(Ω)$ has an orthogonal basis of group characters. An important problem, connected with the so-called Spectral Set Conjecture (saying that $Ω$ is spectral if and only if a collection of translates of $Ω$ can partition the group), is the question of whether the spectrality of a product set $Ω= A imes B$, in a product group, implies the spectrality of the factors $A$ and $B$. Recently Greenfeld and Lev proved that if $I$ is an interval and $Ω\subseteq {\mathbb R}^d$ then the spectrality of $I imes Ω$ implies the spectrality of $Ω$. We give a different proof of this fact by first proving a result about packings of high density implying the existence of tilings by translates of a function. This allows us to improve the result to a wider collection of product sets than those dealt with by Greenfeld and Lev. For instance when $A$ is a union of two intervals in ${\mathbb R}$ then we show that the spectrality of $A imes Ω$ implies the spectrality of both $A$ and $Ω$.

Motivation & Objective

• To resolve the spectral factor problem for product domains, particularly when one factor is a union of two intervals.
• To extend Greenfeld and Lev's result on intervals to a broader class of product sets.
• To establish a general principle linking high-density packings of $ |\widehat{\chi_\Omega}|^2 $ to the existence of tilings.
• To prove that spectrality of $ A \times \Omega $ implies spectrality of both $ A $ and $ \Omega $, even when $ A $ is a union of two intervals.
• To provide a new proof technique based on packing density and tiling equivalence, independent of previous approaches.

Proposed method

• Introduces a general principle: if a function $ f = |\widehat{\chi_\Omega}|^2 $ packs with density arbitrarily close to $ |\Omega|^2 $, then it tiles.
• Uses the equivalence between orthogonality and packing (via Bessel’s inequality) and completeness and equality (via Parseval’s identity) in $ L^2(\Omega) $.
• Applies a density-based tiling criterion: if the packing density of $ f $ exceeds a threshold related to its $ L^2 $-norm, then $ f $ tiles.
• Applies this to the function $ f = |\widehat{\chi_A}|^2 $ for $ A = I \cup J $, a union of two intervals, and analyzes its zero set using Fourier analysis.
• Uses the fact that if $ D - D $ avoids the zero set of $ \widehat{\chi_A} $, then $ A \times \Omega $ spectral implies $ \Omega $ spectral.
• Employs a duality argument: if $ D $ is chosen so that $ D - D $ avoids the zero set of $ \widehat{\chi_A} $, and $ |D| \geq 1 $, then $ A \times \Omega $ spectral implies $ \Omega $ spectral.

Experimental results

Research questions

• RQ1 Does the spectrality of a product set $ A \times \Omega $ imply the spectrality of both $ A $ and $ \Omega $, even when $ A $ is a union of two intervals?
• RQ2 Can high-density packings of $ |\widehat{\chi_\Omega}|^2 $ guarantee the existence of a tiling by translates of $ |\widehat{\chi_\Omega}|^2 $?
• RQ3 What conditions on the zero set of $ \widehat{\chi_A} $ for $ A = I \cup J $ ensure that $ A \times \Omega $ spectral implies $ \Omega $ spectral?
• RQ4 Is the spectral factor problem decoupled for product domains when $ A $ is not a single interval but a union of two intervals?
• RQ5 Can the Fuglede conjecture be extended to product domains where one factor is a union of two intervals?

Key findings

• If $ A = I \cup J $ is a union of two disjoint intervals with $ |I| + |J| = 1 $, and $ A \times \Omega \subseteq \mathbb{R}^{1+n} $ is spectral, then both $ A $ and $ \Omega $ are spectral.
• When $ |I| \neq |J| $, the Fourier transform $ \widehat{\chi_A} $ has no zeros in $ (-1,1) $, which allows the use of a density-based tiling argument.
• When $ |I| = |J| = 1/2 $, the zero set of $ \widehat{\chi_A} $ is explicitly characterized as $ 2\mathbb{Z} \setminus \{0\} \cup (2\mathbb{Z}+1)\Delta $, where $ \Delta = 1/(2|m_1 - m_2|) $, enabling the construction of a suitable $ D $ with $ |D| = 1 $ and $ D - D $ avoiding the zero set.
• The construction of such a $ D $ with $ |D| = 1 $ and $ D - D $ disjoint from the zero set of $ \widehat{\chi_A} $ implies that $ \Omega $ is spectral via Theorem 2(a).
• The result generalizes Greenfeld and Lev’s theorem: spectrality of $ [0,1] \times \Omega $ implies spectrality of $ \Omega $, and extends it to $ A = I \cup J $, a union of two intervals.
• The paper proves that $ [0,1]^d \times B \subseteq \mathbb{R}^{d+n} $ spectral implies $ B \subseteq \mathbb{R}^n $ is spectral, via induction and the main theorem.