14
$\begingroup$

In Lemma 2 of [1], Heath-Brown proves the following (I state a simplified version of a more general result):

Let $\Lambda \subset \mathbb{Z}^2$ be a lattice of determinant $d(\Lambda)$. Then $$\# \{ (x_1,x_2) \in \Lambda: \max_i |x_i| \leq B, \gcd(x_1,x_2) = 1\} \leq 16\left (\frac{B^2}{d(\Lambda)} + 1\right).$$

My question is whether this generalises to arbitrary dimensions.

Does an analogous result hold for lattices in $\Lambda \subset \mathbb{Z}^n$? Namely, is $$\# \{ (x_1,\dots, x_n) \in \Lambda: \max_i |x_i| \leq B, \gcd(x_1,\dots,x_n) = 1\} \leq C_n\left (\frac{B^n}{d(\Lambda)} + 1 \right)$$ for some constant $C_n$?

If it helps, I'm primarily interested in the case $n=3$.

Obviously I'm aware of standard lattice point counting techniques, but these usually give an error term of the shape $O(\text{boundary of the region/first successive minima})$, and I don't know how to control this in my case. So I'm just looking for uniform upper bounds where this term doesn't appear.

[1] Heath-Brown - Diophantine approximation with Square-free numbers

$\endgroup$
  • 1
    $\begingroup$ The proof is definitely specialized to $n=2$ case, as HB defines a basis for the lattice by choosing the shortest (lin. ind.) vectors, which is a special phenomenon taking place only in low dimensions. Note that the LLL algorithm will only give you a ''good'' basis but not necessarily the shortest vectors. $\endgroup$ – Asaf Apr 20 '18 at 14:32
  • $\begingroup$ @Asaf: Yes this was why it was not obvious to me how to generalise the proof to higher dimensions. $\endgroup$ – Daniel Loughran Apr 20 '18 at 14:46
  • $\begingroup$ Do you care if the $C_n$ is explicit, or is $\ll_n$ enough for your purposes? $\endgroup$ – Greg Martin Apr 20 '18 at 17:11
  • $\begingroup$ I don't need an explicit $C_n$; $\ll_n$ is fine. $\endgroup$ – Daniel Loughran Apr 20 '18 at 18:17
14
$\begingroup$

No, there is no result in this form because in dimension 3 or higher it is allowed to have some non-first minima relatively small even when the first minimum is very small.

For example, for any $N>0 $ consider the lattice $$ \Lambda_N = \frac 1 N \mathbb Z \times \mathbb Z \times \mathbb Z^{n-3}\times N\mathbb Z$$ with unit determinant. Then the set $$ S_N = \left\{\left(\frac k N,1 , 0,\ldots, 0\right):\ -N\leq k\leq N\right\}$$ is a set of $2N+1$ primitive vectors of $\Lambda_N$ contained in the unit hypercube.

Now, to answer your question, take $B=N$ prime and the lattices $N\Lambda_N\subset \mathbb Z^n$ to see that there is no such constant $C_n$.

$\endgroup$

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.