Let $\Lambda$ be a lattice in $\mathbb{R}^n$ and let $|\mathbf{x}|$ denote the L2 norm. There is a fairly standard argument involving successive minima to obtain the estimate on $N(R)$ which is the number of points $\mathbf{x}$ on $\Lambda$ satisfying $|\mathbf{x}| \leq R$.

Let $R_1, ..., R_n$ be positive real numbers. I was wondering if there was a way to estimate the number of points $\mathbf{x}$ on $\Lambda$ satisfying $|x_i| \leq R_i$ (for each $1\leq i \leq n$)?

PS I was thinking maybe one can apply a linear transformation to make the counting for a box with equals sides...

I am just not really sure what I can possibly do if I wanted such an estimate. Any comments or suggestions would be appreciated. Thank you very much!

  • $\begingroup$ All norms on $\mathbb{R}^n$ are equivalent, which means there are constants $c,C$ such that $c\|x\|_1 \leq \|x\|_2 \leq C\|x\|_1$ $\endgroup$ – Marcel Bischoff Jul 1 '17 at 23:07
  • 1
    $\begingroup$ This is a standard application of Davenport's lemma, even though this particular result is much older. If you need more than what Davenport's lemma can give you, then likely more information is needed. $\endgroup$ – Stanley Yao Xiao Jul 1 '17 at 23:50
  • 1
    $\begingroup$ @StanleyYaoXiao What is Davenport's lemma and what does it imply in this situation? $\endgroup$ – Johnny T. Jul 2 '17 at 0:31

Let $\mathcal{S}\subset\mathbb{R}^n$ be a convex compact set lying in the closed ball of radius $R$ centered at the origin. Then $$\left|\#(\mathcal{S}\cap\Lambda)-\frac{\mathrm{vol}(\mathcal{S})}{\det(\Lambda)}\right|\ll_n 1+\max_{1\leq m\leq n-1}\frac{R^m}{\lambda_1\dots\lambda_m},$$ where $\lambda_1\leq\dots\leq\lambda_n$ are the successive minima of $\Lambda$. In particular, the above bound applies for the box $\mathcal{S}=[-R_1,R_1]\times\dots\times[-R_n,R_n]$, upon choosing $R:=R_1+\dots+R_n$.

The above result is essentially Lemma 1 in Schmidt - Northcott's theorem on heights. II. The quadratic case (Acta Arith. 70 (1995), 343-375.), and it is based on Davenport's theorem. Schmidt's right hand side is slightly different, but the above version also follows by combining (2.1), (2.4), and the display after (2.4).

P.S. See also my answer to your earlier question.


This is not trivial (and Davenport's lemma, as in the comments, does not give great bounds). For more, see W. Schmidt's Diophantine Approximation (Chapter IV).

  • $\begingroup$ Would you happen to know what is the bound you get by Davenport's lemma? $\endgroup$ – Johnny T. Jul 2 '17 at 0:40
  • $\begingroup$ @JohnnyT. Look at the link, and all will be revealed - it would take too long to type. $\endgroup$ – Igor Rivin Jul 2 '17 at 1:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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