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Let $\| \cdot \|$ denote the maximum norm in Euclidean spaces.

Consider the set $D_{m,n}$ of $m \times n $ real matrices satisfying that the system of inequalities

$$\|Aq-p\|^m < \frac{1}{T}, \|q\|^n < T$$

have a nontrivial integer solutions for all large enough $T$. $p\in \mathbb Z^m, q \in \mathbb Z^n$

By Dirichlet approximation, $D_{1,1}=\mathbb R$. I wonder what happens if $(m,n) \ne (1,1)$. Does $D_{m,n}$ necessarily have full measure? I am not sure if there is an elementary proof or ergodic theory/dynamics (things like Dani's correspondence) is needed.

Source of question: the last paragraph of

https://arxiv.org/pdf/1709.04082.pdf#page=2#

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  • $\begingroup$ What do your superscripts represent in $\|Aq - p\|^m$ and $\|q\|^n$? Are they exponents? If so, is the norm just understood to be distance in the proper Euclidean space(s) (i.e. $\mathbb{R}^m$ and $\mathbb{R}^n$ respectively?) Or was the superscript just meant to be a subscript denoting the proper Euclidean norm? $\endgroup$ Commented Sep 8, 2021 at 16:07
  • $\begingroup$ @RonniePavlov Sorry for my late reply. As I indicated in the first line of question $\|\|$ here denotes the maximum norm in ANY Euclidean space. And yes, the superscript indices $m,n$ mean the exponentials. $\endgroup$
    – No One
    Commented Sep 13, 2021 at 18:33
  • $\begingroup$ arxiv.org/abs/2111.07115 Theorem 1.1 clarifies why $D_{m,n}$ is all matrices. $\endgroup$
    – Calamardo
    Commented Nov 20, 2021 at 13:01
  • $\begingroup$ @MondaleJr. No. The union in that theorem is from $0<c<1$ which exclude the case when $c=1$ (which is what I am asking)... There is a critical difference I believe, in view of Minkowski' convex body theorem $\endgroup$
    – No One
    Commented Nov 20, 2021 at 17:18
  • $\begingroup$ There is no union in the theorem (Theorem 1.1). $\endgroup$
    – Calamardo
    Commented Nov 21, 2021 at 2:38

1 Answer 1

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I think that $D_{m,n}$ is just the set of all $m \times n$ matrices $A$ for all $m,n$. The proof is basically the same as that of Dirichlet approximation, i.e., Pigeonhole Principle.

For any $m, n$, any $A$, and any $k$, define $T = (2k)^{mn}$. Then, take the set $S$ of vectors $q \in \mathbb{Z}^n$ with all entries in $[-2^{m-1} k^m, 2^{m-1} k^m)$. Then $|S| = (2^m k^m)^n = (2k)^{mn}$.

Now, consider all vectors $Aq$, $q \in S$, as elements of $[0,1)^m$ (by taking coordinates mod 1). If we partition $[0,1)^m$ into half-open subcubes of side length $(2k)^{-n}$, then the number of such cubes is $(2k)^{mn} = |S|$. We have two cases; either some $Aq'$ and $Aq''$ lie in the same such subcube for $q' \neq q'' \in S$, or every such subcube contains exactly one $Aq$ for $q \in S$.

In the first case, there exists $p \in \mathbb{Z}^m$ so that $A(q' - q'') - p$ has all entries with absolute value less than $(2k)^{-n}$. In the second case, there exists $p \in \mathbb{Z}^m$ so that $Aq - p$ has all entries with absolute value less than $(2k)^{-n}$. If we're in the first case, we redefine $q = q' - q''$, and note that in either case, $q$ has all entries with absolute value less than $(2k)^m$ (either because $q \in S$ or because $q$ is the difference of vectors in $S$.)

Now, $\|q\|^n < (2k)^{mn} = T$, and $\|Aq - p\|^m < (2k)^{-mn} = T^{-1}$, and since $k$ was arbitrary, we get such a solution for infinitely many $T$.

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  • $\begingroup$ please see people.brandeis.edu/~kleinboc/Pub/kwadcompjournal.pdf#page=2# at the bottom (last paragraph). The authors claim that for $m=n=1$, this is the whole space. But otherwise it is a set of full measure $\endgroup$
    – No One
    Commented Sep 14, 2021 at 17:04
  • $\begingroup$ The point is that we need to show "for all large enough $T$'s" not just for infinitely many $T$'s (which was the Dirichlet's theorem). $\endgroup$
    – No One
    Commented Sep 14, 2021 at 17:25
  • $\begingroup$ Ah, thanks. I don't know how I read "all large enough" as "infinitely many." I guess the same proof should give you a version of your main inequality for all large enough $T$ with a multiplicative factor of $(1 + \epsilon)$ (i.e. the second quantity is less than $(1+\epsilon)T$), but according to the paper you sent, these multiplicative factors change the problem quite drastically. So I don't have any knowledge of how to get this stronger result. Sorry for the unhelpful answer! $\endgroup$ Commented Sep 14, 2021 at 21:29
  • $\begingroup$ No problem, any help or clarification is appreciated! $\endgroup$
    – No One
    Commented Sep 14, 2021 at 22:21

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