Gruber P. Convex and Discrete Geometry

446 Geometry of Numbers
In this section upper and lower estimates for the number of neighbours in lattice
packings are given.
Upper Estimates for the Number of Neighbours
The following are classical estimates due to Minkowski [743].
Theorem 30.2. Let C be a proper convex body and L a lattice in E d such that {C +l :
l ∈ L} is a packing. Then C has at most 3 d − 1 neighbours. If C is strictly convex, it
has at most 2 d+1 − 2 neighbours.
Proof. Let D = 1 2 (C − C). Since we have the following,
{C + l : l ∈ L} packing ⇔{D + l : l ∈ L} packing,
C + l neighbour of C ⇔ D + l neighbour of D,
C strictly convex ⇔ D strictly convex,
it is sufficient to prove the result for D instead of C. Since
{D + l : l ∈ L} packing ⇔ L is admissible for 2D,
D + l neighbour of D ⇔ l ∈ bd 2D,
it is sufficient to show the following:
(1) the lattice L, which is admissible for 2D, has at most 1 2 (3d − 1) pairs of
points ±l on bd 2D. Here 1 2 (3d − 1) can be replaced by 2 d − 1if2D is
strictly convex.
First, let C and thus 2D be strictly convex and let {b1,...,bd} be a basis of L.
If ±l =±(u1b1 +···+udbd) ∈ L ∩ bd 2D, then we cannot have ui ≡ 0mod
2fori = 1,...,d; otherwise all ui are even and 1 2l ∈ (L ∩ int 2D) \{o}, which
contradicts the admissibility of L for 2D. If±l =±(u1b1 +···+udbd), ±m =
±(v1b1 +···+vdbd) ∈ (L ∩ bd 2D) where ±l �= ±m, then we cannot have ui ≡ vi
mod 2 for i = 1,...,d; otherwise 1 2 (l − m) ∈ L \{o} and by the strict convexity
of 2D we have 1 2 (l − m) ∈ int 2D, which again contradicts the admissibility of L
for 2D. Since there are 2d − 1 residue classes for (u1,...,ud) modulo 2, excluding
(0,...,0), there can be at most 2d − 1 different pairs of points ±l ∈ L ∩ bd 2D, as
required.
If, second, C and thus 2D is not strictly convex, similar arguments with congruences
modulo 3 lead to the bound 1 2 (3d − 1). ⊓⊔
Remark. Helmut Groemer [406] pointed out that the following simple geometric
argument shows that, in any packing of translates of a convex body C, the number of
neighbours of a fixed translate is at most 3 d − 1: It is sufficient to prove this for the
difference body D instead of C. Since D is symmetric in o and convex, a translate
of D is a neighbour of D if and only if it is contained in 3D. Considering volumes,
we see that in 3D there is space for at most 3 d non-overlapping translates of D,
including D.