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化圆为方:将$S^1$嵌入$\ell_1$

Squaring the circle: embedding $S^1$ in $\ell_1$

Ian Doust, Anthony Weston

arXiv 2608.10422首次发表:更新:

AI 中文总结

本文研究单位圆子集的$\ell_1$等距嵌入问题,给出了可嵌入与不可嵌入的测度判据,并由此得到$L_1[0,1]$不可嵌入$\ell_1$、度量图可嵌入$\ell_1$当且仅当为树等推论。

AI 中文摘要

设$(S^1,δ)$为配备弧长度量的单位圆。本文研究$(S^1,δ)$的哪些子集可以等距嵌入到序列空间$\ell_1$中。众所周知,$S^1$的每个有限子集都允许这样的嵌入,但无限子集(甚至包括整个圆周)的情况可能因文献中相互矛盾的术语而变得模糊。值得注意的是,圆周避开了等距可嵌入性的经典障碍,因为它既是负型的也是超度量的。\n本文证明,若$X \subseteq S^{1}$是闭集,且$X \cap X^{\ast}$的勒贝格测度为正(其中$X^{\ast}$是$X$的对径集),则$(X, δ)$不可能等距嵌入$\ell_{1}$。由此可得,勒贝格测度大于$\pi$的$S^{1}$子集都不能等距嵌入$\ell_1$。反之,我们证明了任何与任意半圆的交集测度为零的$S^1$闭子集都可以等距嵌入$\ell_1$。这些结果有若干推论:它们表明经典巴拿赫空间$L_1[0, 1]$(仅作为度量空间考虑)不能等距嵌入$\ell_{1}$;其次,它们给出了一个简单证明,即度量图$(M, d)$等距嵌入$\ell_1$当且仅当它是树。与$S^{1}$不同,我们证明了一些相关的度量空间(如柱面和平坦环面)包含不能等距嵌入$\ell_1$的有限子集。

英文摘要

Let $(S^1,δ)$ be the unit circle endowed with the arc length metric. This paper concerns the question of which subsets of $(S^1,δ)$ can be embedded isometrically into the sequence space $\ell_1$. It is well known that every finite subset of $S^1$ admits such an embedding, but the situation for infinite subsets, even including the whole circle, has perhaps been obscured by conflicting terminology in the literature. It is worth noting that the circle avoids the classical obstructions to isometric embeddability, since it is both of negative type and hypermetric. In this paper we show that if $X \subseteq S^{1}$ is a closed set and the Lebesgue measure of $X \cap X^{\ast}$ is positive, where $X^{\ast}$ is the antipodal set of $X$, then it is impossible to isometrically embed $(X, δ)$ in $\ell_{1}$. As a result, no subset of $S^{1}$ with Lebesgue measure greater than $π$ can be isometrically embedded in $\ell_1$. Conversely, we show that any closed subset of $S^1$ whose intersection with any half-circle has measure zero can be isometrically embedded in $\ell_1$. These results have several consequences. They imply that the classical Banach space $L_1[0, 1]$, considered purely as a metric space, does not isometrically embed in $\ell_{1}$. Secondly, they yield a simple proof that a metric graph $(M, d)$ embeds isometrically in $\ell_1$ if and only if it is a tree. In contrast to $S^{1}$, we show that some related metric spaces, such as the cylinder and the flat torus, contain finite subsets that cannot be isometrically embedded in $\ell_1$.

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