关于$\mathbb{R}^n$的子空间的逼近指数
On the approximation exponents for subspaces of $\mathbb{R}^n$
摘要：
本文将经典的丢番图逼近理论推广到由W. M.施密特在1967年。设$A$和$B$是$\mathbb{R}^n$的两个子空间，分别具有$d$和$e$以及$d+e\leqslant n$。$A$和$B$之间的接近度由$t=\min(d,e)$正则角$0\leqslant θ_1\leqslant \cdots\leqslant θ_t\leqslant π/2$测量;我们设置$ψ_j(A,B)=\sinθ_j$。如果$B$是一个有理子空间，那么它的复杂性由它的高度$H(B)=\mathrm{covol}(B\cap\mathbb{Z}^n)$来度量。我们用$μ_n(A\vert e)_j$表示定义为$β>0$集合的上界（可能等于$+\infty$）的逼近指数，使得不等式$ψ_j(A,B)\leqslant H(B)^{-β}$对维数为$e$的无穷多个有理子空间$B$成立。我们感兴趣的最小值$\mathringμ_n(d\vert e)_j$采取$μ_n(A\vert e)_j$时$A$范围通过一组子空间的尺寸$d$的$\mathbb{R}^n$，使所有合理的子空间$B$的尺寸$e$有$\dim (A\cap B)<j$。我们发现，$\mathringμ_4(2\vert 2)_1=3$，$\mathringμ_5(3\vert 2)_1\le 6$和$\mathringμ_{2d}(d\vert \ell)_1\leqslant 2d^2/(2d-\ell)$。我们还证明了在一般情况下，这意味着$\mathringμ_n(d\vert d)_d\xrightarrow[n\to+\infty]{} 1/d$的下界。
英文摘要：
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 $0\leqslant θ_1\leqslant \cdots\leqslant θ_t\leqslant π/2$; we set $ψ_j(A,B)=\sinθ_j$. 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 $μ_n(A\vert e)_j$ the exponent of approximation defined as the upper bound (possibly equal to $+\infty$) of the set of $β>0$ such that the inequality $ψ_j(A,B)\leqslant H(B)^{-β}$ holds for infinitely many rational subspaces $B$ of dimension $e$. We are interested in the minimal value $\mathringμ_n(d\vert e)_j$ taken by $μ_n(A\vert e)_j$ 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 $\mathringμ_4(2\vert 2)_1=3$, $\mathringμ_5(3\vert 2)_1\le 6$ and $\mathringμ_{2d}(d\vert \ell)_1\leqslant 2d^2/(2d-\ell)$. We also prove a lower bound in the general case, which implies that $\mathringμ_n(d\vert d)_d\xrightarrow[n\to+\infty]{} 1/d$.
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