Packing, tiling, orthogonality and completeness

Let $\mathrm{\Omega }\subseteq {\mathbf{R}}^d$ be an open set of measure 1. An open set $D \subseteq {\bf R}^d$iscalleda\mathrm{‘}\mathrm{‘}tightorthogonalpackingregio{n}^{\mathrm{\prime }\mathrm{\prime }}for$\Omega$if$D-D$ does not intersect the zeros of the Fourier Transform of the indicator function of $\mathrm{\Omega }$ and $D$ has measure 1. Suppose that $\mathrm{\Lambda }$ is a discrete subset of ${\mathbf{R}}^d$. The main contribution of this paper is a new way of proving the following result (proved by different methods by Lagarias, Reeds and Wang and, in the case of $\mathrm{\Omega }$ being the cube, by Iosevich and Pedersen: $D$ tiles ${\mathbf{R}}^d$ when translated at the locations $\mathrm{\Lambda }$ if and only if the set of exponentials $E_\Lambda = {\exp 2\pi i \lambda\cdot x: \lambda\in\Lambda}$isanorthonormalbasisfor$L^2(\Omega)$\mathrm{.}(When$\Omega$istheunitcubein${\bf R}^d$ then it is a tight orthogonal packing region of itself.) In our approach orthogonality of $E_{\mathrm{\Lambda }}$ is viewed as a statement about ``packing’’ ${\mathbf{R}}^d$ with translates of a certain nonnegative function and, additionally, we have completeness of $E_{\mathrm{\Lambda }}$ in $L^2(\mathrm{\Omega })$ if and only if the above-mentioned packing is in fact a tiling. We then formulate the tiling condition in Fourier Analytic language and use this to prove our result.