The Minkowski-Hlawka theorem (as stated by Gruber in his lovely book Convex and Discrete Geometry) says that if $S$ is a Jordan measurable set in $\mathbb{R}^n$ with volume < 1. Then there is a lattice $L \subset \mathbb{R}^n$ with determinant $1$ which contains no point of $S$, with the possible exception of the origin.

In the special case where the set $S$ is a star-shaped body (with respect to the origin), this inequality can be written as $$ 1 < \text{vol}(S)/\Delta(S) , $$ where $\Delta(S)$ is the critical determinant of $S$ (i.e., the infimum of the determinants of all lattices that intersect $S$ only at the origin).

Question. Does the the linear-invariant functional $S \mapsto \text{vol}(S)/\Delta(S)$ attain its greatest lower bound when defined on (1) the space of star-shaped (compact) bodies and (2) the space of convex bodies that contain the origin as an interior point.

I'm really just interested in the question of existence of minima and specially interested in the case of convex bodies. In any case, even in the "easy" case of $0$-symmetric convex bodies in $\mathbb{R}^2$, where the existence of minima is obvious, the minimum value is still just a conjecture (the Reinhardt conjecture).

  • $\begingroup$ related: mathoverflow.net/questions/125531/… $\endgroup$ – Yoav Kallus May 16 '13 at 20:35
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    $\begingroup$ Also worth noting, the even easier case: the minimum of $S\mapsto\operatorname{vol}(S)/\Delta(S-S)$ is actually known among two-dimensional convex bodies (the triangle). $\endgroup$ – Yoav Kallus May 16 '13 at 20:50
  • $\begingroup$ Looking again at Sergei's answer to the other question, it seems like what you're asking for here should fail: let T=S∪(−S), then among all S′ such that T=S′∪(−S') (and therefore Δ(S)=Δ(S′)) we minimize the volume by letting S′ be the intersection of T with a half-space through the origin. Since this minimum is never achieved so long as the origin is in the interior, the minimum of vol(S)/Δ(S) is never achieved. Am I missing something? $\endgroup$ – Yoav Kallus May 16 '13 at 21:08
  • $\begingroup$ I guess T doesn't have to be bisectible into two convex regions, but, still, I don't see a good reason why the minimum should be achieved. $\endgroup$ – Yoav Kallus May 16 '13 at 22:16
  • $\begingroup$ @Yoav: Thanks for your input. I don't see why the min should be achieved either and I'm starting to suspect it is not. What puzzles me is that no one seems to mention this. $\endgroup$ – alvarezpaiva May 17 '13 at 7:17

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