On the Minkowski distances and products of sum sets

TL;DR

This paper investigates the properties of Minkowski distances and sum-product estimates in the Minkowski plane, establishing lower bounds on the number of distinct distances among points and deriving near-optimal sum-product inequalities.

Contribution

It extends the Elekes/Sharir/Guth/Katz approach to the Minkowski plane, overcoming obstacles posed by null intervals to prove new bounds on distances and sum-product estimates.

Findings

At least rac{N}{\u221a{ }N} distinct distances among N points in f^{1,1}

A near-optimal sum-product estimate for finite sets in f

Identification and discounting of null intervals to adapt incidence bounds

Abstract

Given two points $p,q$ in the real plane, the signed area of the rectangle with the diagonal $[pq]$ equals the square of the Minkowski distance between the points $p,q$. We prove that $N>1$ points in the Minkowski plane $\R^{1,1}$ generate $\Omega(\frac{N}{\log{N}})$ distinct distances, or all the distances are zero. The proof follows the lines of the Elekes/Sharir/Guth/Katz approach to the Erd\H os distance problem, analysing the 3D incidence problem, arising by considering the action of the Minkowski isometry group $ISO^*(1,1)$. The signature of the metric creates an obstacle to applying the Guth/Katz incidence theorem to the 3D problem at hand, since one may encounter a high count of congruent line intervals, lying on null lines, or "light cones", all these intervals having zero Minkowski length. In terms of the Guth/Katz theorem, its condition of the non-existence of "rich planes" generally gets violated. It turns out, however, that one can efficiently identify and discount incidences, corresponding to null intervals and devise a counting strategy, where the rich planes condition happens to be just ample enough for the strategy to succeed. As a corollary we establish the following near-optimal sum-product type estimate for finite sets $A,B\subset \R$, with more than one element: $$|(A\pm{B})\cdot{(A\pm{B})}|\gg{\frac{|A||B|}{\log{|A|}+\log{|B|}}}.$$