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Given a Barlow packing of $\mathbb{R}^n$ by balls with at most a finite number of different radii, the centers of the balls will form a Delone set in $\mathbb{R}^n.$

For a highest density sphere packing, or at least a Barlow packing of highest density among Barlow packings, must the corresponding Delone set be a Meyer set? A Patterson set? In the cases of dimensions 2 and 3 where the optimal packing using a single radius is known, the sets can be chosen to be lattices.

I looked for papers exploring this connection and could only find this one https://www-fourier.ujf-grenoble.fr/PUBLIS/publications/REF_678.pdf, which is good, but doesn't address the big picture questions above.

EDIT: As the comments indicate, we should restrict to Barlow packings. In this case the Delone sets always appear to be of finite local complexity.

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  • $\begingroup$ I am particularly interested in whether the topology and dynamics of tiling spaces cold be used to prove results about sphere packings. Would the tiling spaces associated to optimal sphere packings need to have extra symmetry or could we give restrictions on the dimensions of their cohomology? $\endgroup$
    – mkreisel
    Commented Dec 1, 2012 at 13:47
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    $\begingroup$ Regarding "in the cases of dimensions 2 and 3 where the optimal packing is known, the sets end up being lattices," I assume you're not talking about a finite number of different radii any more but about a single radius. $\endgroup$
    – Yoav Kallus
    Commented Dec 14, 2012 at 18:54
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    $\begingroup$ Also keep in mind that a sphere packing in three dimensions, even if it achieves the highest density and is saturated (in the sense that there is no space to add an extra ball without moving the others) can fail to have finite local complexity. You probably need a stronger definition of saturation. $\endgroup$
    – Yoav Kallus
    Commented Dec 14, 2012 at 22:35
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    $\begingroup$ I believe that's right: you can have pockets of arbitrary (saturated) arrangements as long as the total volume of those pockets have zero density. However, under the most stringent definitions of saturation, you can have only Barlow stackings, and those, I believe have finite local complexity. For ideas about what it means for a packing to be saturated, see Kuperberg "Notions of Denseness" (arXiv:math/9908003). $\endgroup$
    – Yoav Kallus
    Commented Dec 14, 2012 at 23:07
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    $\begingroup$ To be clear, "Barlow packing" means a stacking of close-packed triangular layers in ways other than h.c.p. or f.c.c. It does not, to my knowledge, means packings that are saturated under the strongest notions of saturation. $\endgroup$
    – Yoav Kallus
    Commented Dec 15, 2012 at 0:21

1 Answer 1

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If you consider the Barlow lattices, these are obtained by taking planar hexagonal packings of spheres, and nesting them together in layers. For each layer, there are two possible ways of placing the layer above it: alt text
(source: cnx.org)

Let's normalize the Barlow lattices to have one common layer (the A-layer). The heights of the (centers of spheres of the) other layers will be multiples of the height of a tetrahedron, so are the same for each Barlow lattice. The centers of the vertices of the next layer above have two possibilities, one with projection given by the B-circles, and one given by the C-circles (these are symmetric under a transformation preserving A-circles, so let's assume the next layer is the B-circles). Then the next layer can have projection either the C-circles or the A-circles (the diagram corresponds to the face-centered cubic lattice, so the pattern there is ABCABC...). Thus, we see that for any Barlow lattice D, the differences $D-D$ will be a subset of the superposition of 3 face-centered cubic lattices which have the A-,B-,C-layers at each level, and thus will be a discrete set. So it will be a Meyer set (I'm going by the definition of a Meyer set as a discrete set $M$ such that $M-M$ is also Delone).

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  • $\begingroup$ Thanks! It seems like this question hasn't yet been fully explored... $\endgroup$
    – mkreisel
    Commented Dec 20, 2012 at 15:21

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