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

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    $\begingroup$ 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

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 $C_n$.


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