Gruber P. Convex and Discrete Geometry

21.4 Polar Lattices
21 Lattices 365
For each lattice L in E d , there exists a sort of dual lattice, called the polar lattice of
L, which is a useful tool in various contexts. Polar lattices seem to have appeared
first in the work of Bravais in crystallography, see the discussion by Rigault [838].
There is a sort of weak duality between a convex body and a lattice, on the one hand,
and the polar body and the polar lattice, on the other hand. For an example of this
duality, see Theorem 23.2.
In this section, we introduce the notion of the polar lattice of a given lattice and
show how their bases are related.
The Polar Lattice of a Given Lattice
Our aim here is to show the following simple result, where B −T is the transposed of
the inverse of the d × d matrix B.
Theorem 21.5. Let L be a lattice in E d . Then
L ∗ = � m ∈ E d : l · m ∈ Z for each l ∈ L �
is a lattice, called the dual or polar lattice of L. If {b1,...,bd} is a basis of L and
B the (non-singular) matrix with columns b1,...,bd, then the columns of the matrix
B ∗ = B −T form a basis {b ∗ 1 ,...,b∗ d } of L∗ ,thedual or polar basis of the given basis.
Proof. Let {b1,...,bd} be a basis of L. Then b ∗ 1 ,...,b∗ d
To see that
(1) bi · b ∗ k =
� 1fori = k,
0fori �= k,
note that, for the d × d matrix (bi · b ∗ k ),wehave,
(bi · b ∗ k ) = BT B −T = (B −1 B) T = I T = I,
where I is the d × d unit matrix. Since b ∗ 1 ,...,b∗ d
sufficient, for the proof of the theorem, to show that
(2) L ∗ = � v1b ∗ 1 +···+vdb ∗ d : vi ∈ Z � .
are linearly independent.
are linearly independent, it is
First, let m ∈ L ∗ . Since b ∗ 1 ,...,b∗ d are linearly independent, m = w1b ∗ 1 +···+wdb ∗ d
for suitable wi ∈ R. Then
wi = bi · (w1b ∗ 1 +···+wdb ∗ d ) = bi · m ∈ Z
by (1) and the definition of L ∗ . Hence m ∈ � v1b ∗ 1 +···+vdb ∗ d : vi ∈ Z � . Second,
let m ∈ � v1b ∗ 1 +···+vdb ∗ d : vi ∈ Z � ,saym = v1b ∗ 1 +···+vdb ∗ d , where vi ∈ Z.
Then we have
l · m = (u1b1 +···+udbd) · (v1b ∗ 1 +···+vdb ∗ d ) = u1v1 +···+udvd ∈ Z,
for any l = u1b1 +···+udbd ∈ L by (1), and thus m ∈ L ∗ . This concludes proof
of (2) and thus of the theorem. ⊓⊔