Let $\mathbf{p}$ be a primitive point in the lattice $\mathbb{Z}^J$ and denote the $J-1$ dimensional vector space $V = \mathbf{p}^{\perp} \subseteq \mathbb{R}^J$. Let $\Lambda' = \mathbb{Z}^J \cap V $, which is a lattice of rank $J-1$.

What I am interested in is : does there always exist an integral basis $\{ \omega_1, ..., \omega_{J-1} \}$ of $\Lambda'$ such that for each $j$, $ |\omega_j | \leq |\mathbf{p}|$?

The statement is definitely true when $J=2$. I was curious for $J>2$. I thought it is maybe true, but I haven't been able to show it yet (or disprove it). I would appreciate any hint/solution! (I apologize in advance if this question happens to be not at the mathoverflow level.)

  • $\begingroup$ Which norm are you using on $\mathbb R^J$? $\endgroup$ – Anthony Quas Jun 1 '14 at 22:47
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    $\begingroup$ You want an estimate for the so-called last successive minimum of the lattice. This is provided by Minkowski's second theorem and in your case, it gives an upper bound $c_J|p|$ for some constant $c_J$ depending on the dimension only. I don't think you can take the constant to be $1$ but I don't have a counterexample offhand. $\endgroup$ – Felipe Voloch Jun 1 '14 at 23:01
  • $\begingroup$ @AnthonyQuas I am using the $l^2$-norm here $\endgroup$ – SJY Jun 2 '14 at 15:51
  • $\begingroup$ @FelipeVoloch Thank you for this comment! The only thing I was confused was that I have only seen Minkowski's theorems for lattice that are full rank. Here my lattice $\Lambda'$ does not have a full rank. Does the theorem still work the same way? $\endgroup$ – SJY Jun 2 '14 at 15:54
  • $\begingroup$ @SJY You have to apply the theorem to the lattice inside $V$. $\endgroup$ – Felipe Voloch Jun 2 '14 at 15:59

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