Multiple lattice tiles and Riesz bases of exponentials
Multiple lattice tiles and Riesz bases of exponentials
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多个格子瓦片和指数的 Riesz 底
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发表时间：
2013
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通讯作者：
M. N. Kolountzakis
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作者：
M. N. Kolountzakis
设$OmeasubseteqRR^d$是一个有界可测集合，$Lambda子集q RR^d$是格。还假设当在位置$Lambda$转换时，$Omega$Tiles在级别$k$上倍增。这意味着$Lambda$-翻译的$Omega$几乎覆盖了$Rr^d$$恰好$k$次的每一点。这里我们证明了存在一个指数集$exp(2pi I Tcdox)$，$tin T$，其中$T$是$RR^d$的某个可数子集，它构成了$L^2(欧米茄)$的Riesz基。这一结果是Grepstad和Lev最近用准晶理论的方法在额外假设$Omega$有测度0的边界下证明的。我们的方法更基本，几乎完全基于线性代数。频率集$T$被证明是对偶格$Lambda^*$的移位副本的有限并。它可以只知道$Lambda$和$k$，并且对于与$Lambda$相乘的所有$Omega$都是相同的。
Suppose $OmegasubseteqRR^d$ is a bounded and measurable set and $Lambda subseteq RR^d$ is a lattice. Suppose also that $Omega$ tiles multiply, at level $k$, when translated at the locations $Lambda$. This means that the $Lambda$-translates of $Omega$ cover almost every point of $RR^d$ exactly $k$ times. We show here that there is a set of exponentials $exp(2pi i tcdot x)$, $tin T$, where $T$ is some countable subset of $RR^d$, which forms a Riesz basis of $L^2(Omega)$. This result was recently proved by Grepstad and Lev under the extra assumption that $Omega$ has boundary of measure 0, using methods from the theory of quasicrystals. Our approach is rather more elementary and is based almost entirely on linear algebra. The set of frequencies $T$ turns out to be a finite union of shifted copies of the dual lattice $Lambda^*$. It can be chosen knowing only $Lambda$ and $k$ and is the same for all $Omega$ that tile multiply with $Lambda$.