2 typo

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

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