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Article #115

Re: A Conjecture About Packings of Balls

#115
From: Greg Kuperberg
Date: Sun, 08 Nov 1992 23:52
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In article <9211060833.AA05802@wisdom.weizmann.ac.il> you write:
>For a finite ball packing P in R^3 [with balls round but not
>necessarily congruent], let
>a(P)=2(number of tangencies)/(number of balls)
...
>Conjecture: A = sup a(P) = 12.
...
>I think I can improve the naive argument above to get something
>better than A<=24, but I don't expect to reach A=12.

Your expectation is true and your conjecture is false.  :-)

A regular dodecahedron in flat Euclidean space has angles slightly less
than 120 degrees.  On the surface of the 3-sphere, a very large
dodecahedron has angles close to 180 degrees, while a very small one is
close to flat.  Therefore there is one of intermediate size with angles
exactly 120 degrees and as it happens 120 of them tile the 3-sphere.
Inscribing a sphere in each one, we get a packing of the 3-sphere by
120 spherical caps of a certain size, each kissing twelve others.
I'll call this the hyperdodecahedral packing.

Stereographic projection preserves spheres, so we can project
this packing onto Euclidean space, sending some point not contained
in a sphere to the point at infinity.  Thus we get a finite cluster
D with a(D) = 12 exactly.  However, this is trouble, because the sup
of a(D) is not a maximum.  You can always take two clusters of the same
type and make them touch at a single point, increasing a(D) slightly.
In this case you get a(D+D) = 12+1/120.

If you are interested in the actual value of A, then here is my best
lower bound:   Take the above cluster D.  If you project it from a
point that lies symmetrically between four spheres, you get a regular
tetrahedron of four big spheres with 116 little spheres on the inside.
I'll call the interior arrangement "hyperdodecahedral grout".  You can
put hyperdodecahedral grout inside any other sphere packing with
tetrahedral crevices, including the dodecahedral arrangement itself,
which yields a better kind of grout.  If you do this once you get a
packing with average kissing number 7152/581 > 12.309, and if you
iterate, the supremum is (I think) 3486/283 > 12.318.

Interestingly, a lot of simpler constructions, like taking the
cannonball packing and placing a sphere in each octahedral crevice,
yield a(P) = 12 on the nose.  Perhaps there is one with a(P) = 13 or 12
1/2 or something like that.

This construction is vaguely similar to the example due to A. Bezdek
and W. Kuperberg of packings of long, flat ellipsoids whose density
exceeds the conjectured packing density of spheres in three
dimensions.  (I should finally mention that with a comment of J.  M.
Wills and another comment of mine, you get a sequence of ellipsoid
packings which, in the limit, have a density 2.4% greater than that of
the cannonball packing.)
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From: greg@math.uchicago.edu (Greg Kuperberg)
Subject: Re: A Conjecture About Packings of Balls
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Date: Sun, 8 Nov 1992 23:52:32 GMT
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