Gruber P. Convex and Discrete Geometry

Thus
V (C) ≥ V (O) = 2d
� �
� l1
� det ,...,
d ! λ1
ld
��
��
λd
= 2d | det(l1,...,ld)| 2
≥
d ! λ1 ···λd
d
d ! λ1 ···λd
Polar Lattices and Polar Bodies
23 Successive Minima 379
. ⊓⊔
Given a convex body C in E d with o ∈ int C, its polar body C ∗ is defined by:
C ∗ ={y : x · y ≤ 1 for all x ∈ C},
compare Sect. 9.1. The following theorem is due to Mahler [679]:
Theorem 23.2. Let C be an o-symmetric convex body and L a lattice in E d and let
λi = λi(C, L), λ ∗ j = λ j(C ∗ , L ∗ ) for i, j = 1,...,d. Then
(7) 1 ≤ λd−k+1λ ∗ k ≤
4 d
V (C)V (C ∗ ) ≤ (d !)2 for k = 1,...,d.
Proof. We need the following inequality of Mahler, see Theorem 9.6:
(8) V (C)V (C ∗ ) ≥ 4d
.
(d !) 2
The definition of polar lattice implies that
(9) d(L) d(L ∗ ) = 1.
For the proof of the left-hand inequality, choose linearly independent points
l1,...,ld ∈ L and m1,...,md ∈ L ∗ such that
li ∈ λi bd C, m j ∈ λ ∗ j bd C∗ for i, j = 1,...,d.
Then ± 1 λi li ∈ bd C and ± 1
λ∗ m j ∈ bd C
j
∗ and the definition of C∗ implies that
± li
λi
· m j
λ ∗ j
≤ 1orλiλ ∗ j ≥±li · m j.
Taking into account the definition of L ∗ , it follows that
(10) λiλ ∗ j ≥ 1orli · m j = 0fori, j = 1,...,d.
Let k ∈{1,...,d}. Since m1,...,mk are linearly independent, the set {x : x · m1 =
··· = x · mk = 0} is a subspace of E d of dimension d − k. Thus, at least one of the
d −k +1 linearly independent points l1,...,ld−k+1 is not contained in this subspace.