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On the probability of n equidistant points in high-dimensional lattices

Stefan Gerdjikov, Martin Minchev, Mladen Savov

Abstract

Consider $n$ $d$-dimensional vectors with iid entries from a lattice distribution $X$. We show that the probability that all distances between them are equal is asymptotically \[ C_n\cdot\frac{1}{d^{(m-1)/2}} \quad \text{for} \quad d \to \infty \quad \text{and} \quad m = \binom{n}{2}, \] with an explicit constant in terms of the first 4 moments of $X$. Moreover, we generalise this result to encompass all finitely supported $X$, as well as under different distances. Our method relies on the relatively rarely used multidimensional local limit theorem and an analysis of the lattice on $\mathbb{Z}^{\binom{n}{2}}$ spanned by the image of the \emph{overlapping} map \[ H : \{0,1\}^n \to \{0,1\}^{\binom{n}{2}}, \quad (v_1, \dots, v_n) \mapsto \Bigl( \mathbf{1}_{\{v_i \neq v_j\}} \Bigr)_{1 \le i < j \le n}. \]

On the probability of n equidistant points in high-dimensional lattices

Abstract

Consider -dimensional vectors with iid entries from a lattice distribution . We show that the probability that all distances between them are equal is asymptotically with an explicit constant in terms of the first 4 moments of . Moreover, we generalise this result to encompass all finitely supported , as well as under different distances. Our method relies on the relatively rarely used multidimensional local limit theorem and an analysis of the lattice on spanned by the image of the \emph{overlapping} map

Paper Structure

This paper contains 8 sections, 14 theorems, 123 equations, 4 figures.

Table of Contents

  1. Introduction
  2. The lattice spanned by the overlapping map H
  3. Main results
  4. The binary case
  5. The lattice case
  6. The general case
  7. Further understanding of the involved lattices
  8. Appendix

Key Result

Theorem 2.1

Let $H$ be the map from def: overlap or alternatively defined for a set $I\subset [1:n]$ by Then the vectors form a basis of the lattice induced by $\mathtt{Im} H$, that is $\mathscr{L}\!\! \begin{tikzpicture}[baseline=(X.base)]\node[inner sep=0, xslant=0.3] (X) {$at$};\end{tikzpicture} H := \mathscr{L}\!\! \begin{tikzpicture}[baseline=(X.base)]\node[inner sep=0, xslant=0.3] (X) {$at$};\end{ti

Figures (4)

  • Figure 1: The vectors $H([1:1]), H([1:2]), H([1:n-2])$ and $H([1:n-1])$ with empty cells representing $0$s. We recall that we put a one in $H(I)$ for every $(i,j)$ such that $i \in I, j\notin I$ or vice versa.
  • Figure 2: Illustrating the relation $H(\{i\}) + H(\{j\}) - H(\{i,j\}) = 2 e_{i, j}$. On the left: $H(\{i\})$ is shown in green and $H(\{j\})$ in red; on the right: $H(\{i,j\})$.
  • Figure 3: Illustrating the relation $H([i+1:j]) = H([1:j]) - H([1:i]) + \sum_{k = 1}^i \sum_{\ell=i+1}^j 2e_ {k,\ell}$. On the left: $H([1:i])$ is shown in green and $H([1:j])$ in red; on the right: $H([i+1:j])$.
  • Figure 4: The vectors $H^\ast_e([1:1]), H^\ast_e([1:2]), \dots H^\ast_e([1:n-2])$ and $H^\ast_e([1:n-1])$, with the embedded cell shaded in grey. Only the first vector in Figure \ref{['fig: H123']} have a non-zero $(1,2)$th coordinate so $H^\ast_e$ differs from $H$ only for it.

Theorems & Definitions (32)

  • Theorem 2.1
  • proof
  • Theorem 3.1
  • proof
  • Lemma 3.2
  • proof : Proof of Lemma \ref{['lem: H^ast']}
  • Theorem 3.3
  • Remark 3.4
  • Remark 3.5
  • proof : Proof of Theorem
  • ...and 22 more