Translative packing constant strictly larger than lattice packing constant Simply put, my question is this: what is the smallest dimension, if any,
where we can know for sure that a convex body exists whose translative
packing constant is strictly larger than its lattice packing constant?
Some relevant facts I know:


*

*In two dimensions, the translative packing constant is always
equal to the lattice packing constant for convex bodies. For 
non-convex bodies, already in two dimensions there are
counterexamples. Bezdek and Kuperberg give a good exposition of this.

*In ten dimensions, the best packing of spheres seems to
be non-lattice. In lower dimensions, the best lattice packing
seems to be also the best packing. The best known lattice packing
in dimensions above 8 is not actually known to be best (except in
24 dimensions). Therefore, even for spheres, there is no dimension
where lattices are proved to be suboptimal. See this MO question for more on that.

*Convex bodies that tile by translation can also tile
by lattice translations. So the example cannot be a tiling
body. This is famous result due to Venkov, Alexandrov, and McMullen, and Gruber's book on Convex and Discrete Geometry gives a nice treatment.

*Except for tiling bodies, which are ruled out be the previous point, very few bodies have known translative packing constants. However, all we need to have an example is a lower bound for the translative packing constant and an upper bound for the lattice packing constant. There are known methods to compute the lattice packing constant for polytopes, and in 3D one such method has been implemented by Betke and Henk.
So, it seems that in dimension 10, there are convex bodies,
namely spheres, that pack better by translation than
by lattice translation. However, this is not rigorously known.
If true, this example can probably be extended to higher
dimensions by forming cylinders. However, it seems to me
that if we allow nonspherical bodies, the dimension
where translative packing starts to beat out lattice
packing should be siginificantly lower. Is there an example? In dimensions low enough (e.g. 3D), candidates can be checked computationally and rigorously established if they check out.
 A: This is not an answer, but rather a documentation of a failed attempt to obtain an answer.
One problem for which we know there is a significant difference between unrestricted sets of translations and lattices is in tilings with equal-size cubes. Every lattice tiling with n-cubes has a pair of cubes sharing an entire (n-1)-dimensional face (Hajós). On the other hand, in dimensions $n\ge8$, there are translational tilings with n-cubes in which no two cubes share an entire (n-1)-dimensional face ($n\ge10$ due to Lagarias and Shor, improve to 8 by Mackey, n=7 case open, see Keller's conjecture). Moreover, a stronger property holds in these tilings: the center of each (n-1)-face does not lie in the relative interior of any other (n-1)-face. I call such a tiling a center-to-boundary tiling.
Now consider a slightly stellated n-cube: the convex hull of a unit n-cube and a point for each face lying a height $\epsilon$ above the center of the face. It is easy to perturb a center-to-boundary tiling of n-cubes to a packing with mean volume $(1+\epsilon)^n$. I wanted to show that since lattice tilings have shared faces along at least one direction, and a lattice packing of slightly stellated cubes will be near a lattice tiling, that the mean volume (i.e. lattice determinant) of the packing will have to be $\ge 1+(n+1)\epsilon + o(\epsilon)$, but this turns out to be false. Here is an example of a family of lattices that pack $\epsilon$-stellated 5-cubes, but have determinant $1+(47/8)\epsilon+O(\epsilon^2)$: $L=A\mathbb{Z}^5$, where
$$A=\left(\begin{array}{ccccc}
1+2\epsilon&-\tfrac12&0&0&0\\
0&1+\epsilon&-\tfrac12&\tfrac12&0\\
0&0&1+\epsilon&0&-\tfrac14\\
0&0&0&1+\epsilon&\tfrac12\\
0&0&-\tfrac12\epsilon&0&1+\epsilon
\end{array}\right)\text.$$
Now, still $47/8>5$, but this example does not bode well for the program. For large $n$, I believe similar constructions are possible that achieve determinants $\le 1+n\epsilon+o(\epsilon)$. It might still be that this is impossible for $n=8$, but if so, it could be delicate to show.
