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Minkowski's theorem states that if $K\subseteq\mathbb{Z}^n$ is a convex compact set, $K=-K$, and $\mathrm{volume}(K)\geq 2^n$, then $K$ contains a nonzero integral vector.

Can this bound be improved in the special case when $K$ is the image of the unit $L_p$ ball under a (full-dimensional) linear transformation, for some fixed $p$? I am particularly interested the case $p=1$.

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    $\begingroup$ This paper by Elkies, Odlyzko, and Rush gives a bound in the other direction, namely a lower bound for the packing density of $L_p$ balls: dx.doi.org/10.1007/BF01232282 . Hopefully Noam will stop by and tell us whether there is an easy upper bound. $\endgroup$
    – Yoav Kallus
    May 29, 2013 at 0:11
  • $\begingroup$ This is interesting, thanks for the reference Yoav! $\endgroup$
    – Marcel Celaya
    May 29, 2013 at 19:37
  • $\begingroup$ Notice that your question is equivalent if we consider arbitrary (full-rank) determinant-one lattices and stick with unit balls. From this, we see that the bound can't be improved much because, for a random determinant-one lattice (under the Haar measure) the expected number of non-zero lattice points in any measurable set is exactly the volume of the set. (See math.uga.edu/~pete/Siegel45.pdf.) Therefore, for any body with volume strictly less than one, there exists a determinant-one lattice with no non-zero points inside the body. $\endgroup$
    – Noah Stephens-Davidowitz
    Jun 13, 2017 at 6:30
  • $\begingroup$ (I.e. Minkowski is tight up to a scaling by a factor of 2 for any symmetric convex body $K$.) $\endgroup$
    – Noah Stephens-Davidowitz
    Jun 13, 2017 at 6:30

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