The Andoni--Krauthgamer--Razenshteyn characterization of sketchable norms fails for sketchable metrics

TL;DR

This paper demonstrates that the AKR characterization of sketchable norms does not extend to all metric spaces, providing counterexamples of large metric spaces that are sketchable but lack bounded embeddings into -psilon spaces.

Contribution

The authors construct large metric spaces that are sketchable yet cannot be embedded into -psilon spaces with bounded distortion, disproving the extension of AKR's characterization beyond normed spaces.

Findings

Existence of arbitrarily large sketchable metric spaces with unbounded embedding distortion.

Counterexamples show AKR's characterization fails for general metric spaces.

The result clarifies the limitations of geometric embeddings in sketching algorithms.

Abstract

Andoni, Krauthgamer and Razenshteyn (AKR) proved (STOC 2015) that a finite-dimensional normed space $(X,\|\cdot\|_X)$ admits a $O(1)$ sketching algorithm (namely, with $O(1)$ sketch size and $O(1)$ approximation) if and only if for every $\varepsilon\in (0,1)$ there exist $\alpha\geqslant 1$ and an embedding $f:X\to \ell_{1-\varepsilon}$ such that $\|x-y\|_X\leqslant \|f(x)-f(y)\|_{1-\varepsilon}\leqslant \alpha \|x-y\|_X$ for all $x,y\in X$. The "if part" of this theorem follows from a sketching algorithm of Indyk (FOCS 2000). The contribution of AKR is therefore to demonstrate that the mere availability of a sketching algorithm implies the existence of the aforementioned geometric realization. Indyk's algorithm shows that the "if part" of the AKR characterization holds true for any metric space whatsoever, i.e., the existence of an embedding as above implies sketchability even when $X$ is not a normed space. Due to this, a natural question that AKR posed was whether the assumption that the underlying space is a normed space is needed for their characterization of sketchability. We resolve this question by proving that for arbitrarily large $n\in \mathbb{N}$ there is an $n$-point metric space $(M(n),d_{M(n)})$ which is $O(1)$-sketchable yet for every $\varepsilon\in (0,\frac12)$, if $\alpha(n)\geqslant 1$ and $f_n:M(n)\to \ell_{1-\varepsilon}$ are such that $d_{M(n)}(x,y)\leqslant \|f_n(x)-f_n(y)\|_{1-\varepsilon}\leqslant \alpha(n) d_{M(n)}(x,y)$ for all $x,y\in M(n)$, then necessarily $\lim_{n\to \infty} \alpha(n)= \infty$.