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 $$|A^{-1} L \cap B| \ll \det(A) \prod_{i\leq n} (N_i/p_{\pi(i)} + 1)$$ 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 $\prod_{j\leq n-1} N_j$, say. That would be easy.

  • $\begingroup$ what are $\mathbb{Z}_i$ for different $i$? $\endgroup$ – Fedor Petrov Jul 17 '19 at 18:28
  • $\begingroup$ Sorry, meant $\mathbb{Z}$ $\endgroup$ – H A Helfgott Jul 17 '19 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. $\endgroup$ – H A Helfgott Jul 17 '19 at 18:58
  • $\begingroup$ (Of course one can always chop a box into cubes, but I am hoping for something better ) $\endgroup$ – H A Helfgott Jul 17 '19 at 20:01

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}

  • $\begingroup$ so $det(A)∏_{i≤n}(N_i/p_{π(i)}+1)$ iff $1 ≪ \det A$ for some permutation $π$. $\endgroup$ – Tobias Schlemmer Jul 18 '19 at 14:41

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