Gruber P. Convex and Discrete Geometry

Proof. It is sufficient to consider the case where L = Z d .
Right-hand inequality: Let
Ci = λi
2
C for i = 1,...,d.
23 Successive Minima 377
It follows from the definition of successive minima that there are d linearly independent
points l1,...,ld ∈ Z d , such that
li ∈ λiC for i = 1,...,d.
After assigning to {l1,...,ld} a basis of Z d by Theorem 21.3 (i) and then transforming
this basis into the standard basis of Z d by an integer unimodular d × d matrix
according to Theorem 21.1, we may assume that
Next, let
li ∈ Ei = � x = (x1,...,xi , 0,...,0) ∈ E d� for i = 1,...,d.
Ldk = � l = (u1,...,ud) ∈ Z d :|ui| ≤k � , Lik = Ei ∩ Ldk for k = 1, 2,...
Since C is bounded, there is a constant α>0, which depends only on C, such that
(2) V (Cd + Ldk) ≤ (2k + 1 + α) d .
By the definition of λ1 we have, for C1 = λ1
2 C, (int C1 + l) ∩ (int C1 + m) =∅for
l, m ∈ Z d , l �= m. Thus
(3) V (C1 + Ldk) = (2k + 1) d V (C1) = (2k + 1) d λd 1
V (C).
2d The main step of the proof is to show the following estimate:
� �
λi+1
d−i
(4) V (Ci+1 + Ldk) ≥ V (Ci + Ldk) for i = 1,...,d − 1.
λi
If λi+1 = λi, the inequality (4) is trivial. We may thus assume that λi+1 >λi. Then
the following statement holds:
(5) (int Ci+1 + l) ∩ (int Ci+1 + m) =∅
for l = (u1,...,ud), m = (v1,...,vd) ∈ Z d ,
where (ui+1,...,ud) �= (vi+1,...,vd).
Otherwise the i + 1 linearly independent lattice points l1,...,li, l − m would be
contained in the interior of λi+1C = Ci+1 − Ci+1, in contradiction to the definition
of λi+1. Proposition (5) implies that
Let
(6) V (Ci + Ldk) = (2k + 1) d−i V (Ci + Lik),
V (Ci+1 + Ldk) = (2k + 1) d−i V (Ci+1 + Lik).
E ⊥ i = � x = (0,...,0, xi+1,...,xd) ∈ E d�