7
$\begingroup$

Let $A:\mathbb{Z}^n\to \mathbb{Z}^n$ be non-singular. Consider a box $B=[0,N_1]\times [0,N_2] \times \dotsc \times [0,N_n]$. Let $p_1,\dotsc,p_n$ be primes (distinct, if you wish) and let $L = p_1\mathbb{Z} \times \dotsb \times p_n \mathbb{Z}$. Can one give a good bound on $A^{-1} L \cap B$? For instance, might $$|A^{-1} L \cap B| \ll \det(A) \prod_{1\leq i\leq n} \left(\frac{N_i}{p_{\pi(i)}} + 1\right)$$ hold for some permutation $\pi$ of $\{1,2,\dotsc,n\}$?

Assume all $p_i$ considerably larger than $n$ if needed. Also assume the matrix entries of $A$ to be bounded, if needed.

Note: I am not looking for an estimate with error term of size $\prod_{1\leq j\leq n-1} N_j$, say. That would be easy.

$\endgroup$
5
  • $\begingroup$ what are $\mathbb{Z}_i$ for different $i$? $\endgroup$ Jul 17, 2019 at 18:28
  • $\begingroup$ Sorry, meant $\mathbb{Z}$ $\endgroup$ Jul 17, 2019 at 18:39
  • $\begingroup$ If all $N_i$ are the same ($=N$) and the entries of $A$ are of absolute value $\leq c$, then it's easy to get a fairly good upper bound ($\leq \prod_i (c n N/p_i+1)$. The interesting case is that of $N_i$ not all of the same size. $\endgroup$ Jul 17, 2019 at 18:58
  • $\begingroup$ (Of course one can always chop a box into cubes, but I am hoping for something better ) $\endgroup$ Jul 17, 2019 at 20:01
  • $\begingroup$ The Community Bot bumps this question every $120$ days. $\endgroup$ Dec 4, 2021 at 22:59

1 Answer 1

0
$\begingroup$

Probably not an answer, but too much for a comment:

Actually you can represent $L$ as a transformation of $\mathbb Z^n$ via $L=\Lambda \mathbb Z^n$, where $\Lambda=(\lambda_{ij})$ is the diagonal matrix such that $\lambda_{ii}=pi$. In the same way you can represent $N$ as $ME_n$ where $M=(m)_{ij}$ is the diagonal matrix with $m_{ii}=(N_i)$ and the unit cube $E_n=[0,1]^n \subset \mathbb R^n$ in the real vector space $\mathbb R^n$. Then we get: \begin{equation} |A^{-1}L\cap N|=|\mathbb Z^n \cap (\Lambda^{-1}AM)E_n| \end{equation} This suggests that $\det(\Lambda^{-1}AM)$ is useful, here. But this would be true only, if we would consider semi-open complete intervals in $\mathbb R^n$. So we must add another $1$ to each dimension on the $\mathbb Z^n$. This leads to the formula: \begin{equation} |\mathbb Z^n \cap (\Lambda^{-1}AM)E_n| \leq \det (\Lambda^{-1}AM +I) \end{equation}

Maybe, some improvement can be achieved using the singular value decomposition \begin{equation} \Lambda^{-1}AM = U \Sigma V^T. \end{equation} This would give you the intersection of a rotated orthogonal grid $U^T\mathbb Z^n$ with some rotated parallelepiped $\Sigma V^TE_n$: \begin{equation} |\mathbb Z^n \cap (\Lambda^{-1}AM)E_n|= |\mathbb Z^n \cap U\Sigma V^TE_n| = |U^T\mathbb Z^n \cap \Sigma V^TE_n| \end{equation}

$\endgroup$
1
  • $\begingroup$ so $det(A)∏_{i≤n}(N_i/p_{π(i)}+1)$ iff $1 ≪ \det A$ for some permutation $π$. $\endgroup$ Jul 18, 2019 at 14:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.