Defect of an octahedron in a rational lattice

TL;DR

This paper establishes a new upper bound on the defect of an octahedron in rational lattices, correcting previous inaccuracies and providing a rigorous proof for the bound involving lattice modifications.

Contribution

It presents a correct proof for an upper bound on the defect of an octahedron in rational lattices, resolving prior errors in the literature.

Findings

Derived an explicit upper bound involving n, m, and logarithmic factors.

Corrected and clarified previous incorrect proofs in the literature.

Provided a rigorous mathematical proof for the defect bound.

Abstract

Consider an arbitrary $n$-dimensional lattice $\Lambda$ such that $\mathbb{Z}^n \subset \Lambda \subset \mathbb{Q}^n$. Such lattices are called {\it rational} and can always be obtained by adding $m \le n$ rational vectors to $\mathbb{Z}^n$. {\it Defect } $d({\cal E},\Lambda)$ of the standard basis $ {\cal E}$ of ${\mathbb Z}^n$ ($n$ unit vectors going in the directions of the coordinate axes) is defined as the smallest integer $d$ such that certain $ (n-d) $ vectors from $ {\cal E} $ together with some $d$ vectors from the lattice $\Lambda$ form a basis of $\Lambda$. Let $||...||$ be $L^1$-norm on $\mathbb{Q}^n$. Suppose that for each non-integer $x \in \Lambda$ inequality $||x|| > 1$ holds. Then the unit octahedron $O^n = \left\{{ x} \in \mathbb{R}^n: ||x|| \leqslant 1\right\}$ is called admissible with respect to $\Lambda$ and $d({\cal E},\Lambda)$ is also called defect of the octahedron $O^n$ with respect to $\cal{E}$ and is denoted as $d(O^n_{{\cal E}}, \Lambda)$. Let $ d_n^m = \max_{\Lambda \in {\cal A}_m} d(O^n_{{\cal E}},\Lambda), $ where $ {\cal A}_m $ is the set of all {\it rational} lattices that can be obtained by adding $m$ rational vectors to $\mathbb{Z}^n$: $ \Lambda = \left \langle {\mathbb Z}^n, { a}_1, \dots, { a}_m \right \rangle_{{\mathbb Z}}, { a}_1, \dots, { a}_m \in {\mathbb Q}^n. $ In this article we show that there exists an absolute positive constant $ C $ such that for any $m < n $ $$ d_n^m \leq C \frac{n \ln (m+1)}{\ln \frac{n}{m}} \left(\ln\ln \left(\frac{n}{m}\right)^m \right)^2 $$ This bound was also claimed in $[1],[2]$, however the proof was incorrect. In this article along with giving correct proof we highlight substantial inaccuracies in those articles.