On the approximation exponents for subspaces of $\mathbb{R}^n$

TL;DR

This paper extends Diophantine approximation theory to subspaces of Euclidean space, analyzing approximation exponents related to angles and heights of rational subspaces, and establishing exact and asymptotic bounds for these exponents.

Contribution

It introduces new bounds and exact values for approximation exponents of subspaces, generalizing classical Diophantine approximation results to higher-dimensional subspace settings.

Findings

Exact value of minimal approximation exponent for certain subspace configurations.

Upper bounds for approximation exponents in higher dimensions.

Asymptotic behavior of exponents as dimension tends to infinity.

Abstract

This paper follows the generalisation of the classical theory of Diophantine approximation to subspaces of $\mathbb{R}^n$ established by W. M. Schmidt in 1967. Let $A$ and $B$ be two subspaces of $\mathbb{R}^n$ of respective dimensions $d$ and $e$ with $d+e\leqslant n$. The proximity between $A$ and $B$ is measured by $t=\min(d,e)$ canonical angles $0\leqslant \theta_1\leqslant \cdots\leqslant \theta_t\leqslant \pi/2$; we set $\psi_j(A,B)=\sin\theta_j$. If $B$ is a rational subspace, his complexity is measured by its height $H(B)=\mathrm{covol}(B\cap\mathbb{Z}^n)$. We denote by $\mu_n(A\vert e)_j$ the exponent of approximation defined as the upper bound (possibly equal to $+\infty$) of the set of $\beta>0$ such that the inequality $\psi_j(A,B)\leqslant H(B)^{-\beta}$ holds for infinitely many rational subspaces $B$ of dimension $e$. We are interested in the minimal value $\mathring{\mu}_n(d\vert e)_j$ taken by $\mu_n(A\vert e)_j$ when $A$ ranges through the set of subspaces of dimension $d$ of $\mathbb{R}^n$ such that for all rational subspaces $B$ of dimension $e$ one has $\dim (A\cap B)<j$. We show that $\mathring{\mu}_4(2\vert 2)_1=3$, $\mathring{\mu}_5(3\vert 2)_1\le 6$ and $\mathring{\mu}_{2d}(d\vert \ell)_1\leqslant 2d^2/(2d-\ell)$. We also prove a lower bound in the general case, which implies that $\mathring{\mu}_n(d\vert d)_d\xrightarrow[n\to+\infty]{} 1/d$.