TITLE: Polar body of a convex body that avoids a lattice
QUESTION [32 upvotes]: Let $K \subset {\bf R}^d$ be a symmetric convex body (an open bounded convex neighbourhood of the origin with $K = -K$) with the property that $K + {\bf Z}^d \neq {\bf R}^d$, i.e. the projection of $K$ to the standard torus ${\bf R}^d/{\bf Z}^d$ is not surjective, or equivalently $K$ is disjoint from some coset $x + {\bf Z}^d$ of the standard lattice.  My question is: what does this say about the polar body
$$K^* := \{ \xi \in {\bf R}^d: \xi \cdot x < 1 \hbox{ for all } x \in K \}?$$
Intuitively, the property $K + {\bf Z}^d \neq {\bf R}^d$ is a "smallness" condition on K, and is thus a "largeness" condition on $K^*$.
If $K^*$ contains a non-trivial element $n$ of $2 {\bf Z}^d$, then $K$ is contained in the strip $\{ x: |n \cdot x| < 1/2 \}$, and will thus avoid the coset $x+{\bf Z}^d$ whenever $x \cdot n = 1/2$.  So this is a sufficient condition for $K + {\bf Z}^d \neq {\bf R}^d$, but it is not necessary.  Indeed, if one takes $K$ to be the octahedron 
$$K := \{ (x_1,\ldots,x_d) \in {\bf R}^d: |x_1|+\ldots+|x_d| < d/2 \}$$
then $K$ avoids $(1/2,\ldots,1/2)+{\bf Z}^d$, but the dual body
$$ K^* = \{ (\xi_1,\ldots,\xi_d) \in {\bf R}^d: |\xi_1|,\ldots,|\xi_d| < 2/d \}$$
is quite far from reaching a non-trivial element of $2 {\bf Z}^d$.
On the other hand, by using the theory of Mahler bases or Fourier analysis one can show that if $K + {\bf Z}^d \neq {\bf R}^d$, then $K^*$ must contain a non-trivial element of $\varepsilon_d {\bf Z}^d$ for some $\varepsilon_d > 0$ depending only on $d$.  However the bounds I can get here are exponentially poor in $d$.
Based on the octahedron example (which intuitively seems to be the "biggest" convex set that still avoids a coset of ${\bf Z}^d$), one might tentatively conjecture that if $K + {\bf Z}^d \neq {\bf R}^d$, then the closure of $K^*$ contains a non-trivial element of $\frac{2}{d} {\bf Z}^d$.  I do not know how to prove or disprove this conjecture (though I think the $d=2$ case might be worked out by ad hoc methods, and the $d=1$ case is trivial), so I am posing it here as a question.

REPLY [3 votes]: This is not really an answer, but a comment on a natural variation of the original problem: assume a convex body in $\mathbb{R}^n$ that is not necessarily centrally symmetric contains only one lattice point as an interior point. What can be said about the dual of $K$ defined with respect to the unique lattice point in its interior?
In general, you can say that the volume of the dual body cannot be too small. In this paper Balacheff, Tzanev, and I conjecture that the volume of the dual cannot be less than $(n+1)/n!$. We can prove this in dimension two under the following equivalent guise:
Theorem. The area of a convex body in the plane that intersects every
integer line $mx + ny = 1$ is at least $3/2$. Moreover, equality holds only
for the triangle with vertices $(1, 0)$, $(0, 1)$, $(−1,−1)$ and its images under
(integer) unimodular transformations.
In higher dimensions, we rely (heavily!) on Greg Kuperberg's lower bound for the volume product to prove.
Theorem. The volume of a convex body in $\mathbb{R}^n$ that intersects every
integer hyperplane $m_1x_1 + · · · + m_nx_n = 1$ is at least $(\pi/8)^n(n + 1)/n!$.
But the fun part of the paper is that is also shows that these inequalities are (in the case of the first theorem) or should be extended to inequalities about Finsler metrics or even contact manifolds.