Squaring the circle: embedding $S^1$ in $\ell_1$
Let $(S^1,\delta)$ be the unit circle endowed with the arc length metric. This paper concerns the question of which subsets of $(S^1,\delta)$ 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 $(X, \delta)$ in $\ell_{1}$. As a result, no subset of $S^{1}$ with Lebesgue measure greater than $\pi$ 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 $\ell_1$.