Balancing Sets of Vectors

TL;DR

This paper investigates the minimal size of vector sets that balance inner products over roots of unity and explores a combinatorial problem related to subset intersections, providing new bounds for prime-related cases.

Contribution

It introduces bounds on the minimal number of vectors needed for balancing sets and extends a combinatorial problem to prime cases, offering new lower bounds.

Findings

Lower bound for $K(n,p)$ is $n(p-1)$ using algebraic methods.

Established that for primes $p>3$, the minimal number $m(p)$ is at least $p$.

Connected algebraic and combinatorial problems to derive these bounds.

Abstract

Let $n$ be an arbitrary integer, let $p$ be a prime factor of $n$. Denote by $\omega_1$ the $p^{th}$ primitive unity root, $\omega_1:=e^{\frac{2\pi i}{p}}$. Define $\omega_i:=\omega_1^i$ for $0\leq i\leq p-1$ and $B:=\{1,\omega_1,...,\omega_{p-1}\}^n$. Denote by $K(n,p)$ the minimum $k$ for which there exist vectors $v_1,...,v_k\in B$ such that for any vector $w\in B$, there is an $i$, $1\leq i\leq k$, such that $v_i\cdot w=0$, where $v\cdot w$ is the usual scalar product of $v$ and $w$. Gr\"obner basis methods and linear algebra proof gives the lower bound $K(n,p)\geq n(p-1)$. Galvin posed the following problem: Let $m=m(n)$ denote the minimal integer such that there exists subsets $A_1,...,A_m$ of $\{1,...,4n\}$ with $|A_i|=2n$ for each $1\leq i\leq n$, such that for any subset $B\subseteq [4n]$ with $2n$ elements there is at least one $i$, $1\leq i\leq m$, with $A_i\cap B$ having $n$ elements. We obtain here the result $m(p)\geq p$ in the case of $p>3$ primes.