MINKOWSKI’S LATTICE POINT THEOREM.
IAN KIMING
There are a couple of minor points in section 2.12 of [1] the can be a little
improved.
First, the definition of a lattice in Rn is that a lattice is a subgroup of the additive
group of Rn of form Ze1 +. . .+Zen where e1 , . . . , en are linearly independent vectors
in Rn . (Sometimes one also defines ‘non-full lattices’ which are subgroups of form
Ze1 + . . . + Zem where m < n and e1 , . . . , em are linearly independent, but we don’t
need that). One then refers to e1 , . . . , en as a basis of the lattice.
The definition
in the book is not correct: For instance, in R2 the subgroup
√
Z(1, 0)+Z( 2, 0) would be a lattice
√according to the definition of the book, because
it is free of rank 2 √
(as (1, 0) and ( 2, 0) are Z-linearly independent); however, the
vectors (1, 0) and ( 2, 0) are linearly dependent as vectors in R2 , so this subgroup
is not what one calls a lattice.
Some people prefer to define the fundamental parallelepiped of a lattice G with
basis e1 , . . . , en as the set:
X
G0 := {
λi ei | 0 ≤ λi < 0}
i
the reason being that G0 is then obviously a system of representatives for the cosets
in Rn /G.
For the proof of Minkowski’s lattice point theorem below it is however immaterial
whether one uses this definition or that of the book.
I feel that a part of the proof of Minkowski’s lattice point theorem given on page
56 of [1] can be formulated a bit more clearly. I will give 2 versions of the theorem
with full details for convenience.
Theorem 1 (Minkowski’s lattice point theorem.). Let G be a lattice in Rn with
volume v of a fundamental parallelepiped.
Let M be a bounded, convex, and centrally symmetric subset of Rn of finite
volume vol(M ). If vol(M ) > 2n · v then M contains a point of G different from 0.
The same conclusion holds under the weaker assumption that vol(M ) ≥ 2n · v if
we additionally assume that M is closed.
Proof. Notice that 0 ∈ M : For as vol(M ) > 0, the set M is not empty. If P ∈ M
then −P ∈ M because M is symmetric; then 0 = 21 · P + 12 · (−P ) ∈ M because M
is convex.
Let us start with the assumption that M is a bounded, convex, and centrally
symmetric with vol(M ) > 2n · v.
Let e1 , . . . , en be a basis of G, and let G0 be the associated fundamental parallelepiped. Then 2G := {2x | x ∈ G} is a lattice with basis 2e1 , . . . , 2en and
1
2
IAN KIMING
associated fundamental parallelepiped 2G0 . So we clearly have:
[
(2x + 2G0 ) = Rn .
x∈G
Put Mx := M ∩ (2x + 2G0 ) for x ∈ G. As M is bounded, only finitely many Mx
are non-empty. Clearly, M = ∪x∈G Mx , so we may deduce:
vol(2G0 )
2n vol(G0 ) = 2n · v
X
X
< vol(M ) ≤
vol(Mx ) =
vol((2x + 2G0 ) ∩ M )
=
x∈G
=
X
x∈G
vol((2G0 ) ∩ (−2x + M )) .
x∈G
Formally, the sums over x ∈ G are infinite, but in each case only finitely many
of the terms are 6= 0.
Now, if any two subsets (2G0 ) ∩ (−2x + M ) and (2G0 ) ∩ (−2y + M ) have empty
intersection whenever x 6= y we would deduce
X
vol((2G0 ) ∩ (−2x + M )) ≤ vol(2G0 ) ,
x∈G
contradicting the above inequality. Therefore there exist x, y ∈ G, x 6= y, such that
(2G0 ) ∩ (−2x + M ) and (2G0 ) ∩ (−2y + M ) intersect non-trivially. Let z be a point
in the intersection. We can then write:
z = −2x + m1 = −2y + m2 ,
with m1 , m2 ∈ M . As M is centrally symmetric we have −m2 ∈ M ; as M is convex
this implies x − y = 21 m1 + (1 − 12 )(−m2 ) ∈ M . But G 3 x − y 6= 0, so we are done
in this case.
Let us then assume only that vol(M ) ≥ 2n · vol(G0 ) but additionally that M is
closed.
Let k ∈ N. The set (1 + k1 )M (= {(1 + k1 )m | m ∈ M }) is easily seen to be
bounded, centrally symmetric, and convex. Also,
1
1
1
vol((1 + )M ) = (1 + )n · vol(M ) ≥ (1 + )n 2n · vol(G0 ) > 2n · vol(G0 ) .
k
k
k
By what we have already proved we may thus conclude that for any k ∈ N, the
set (1 + k1 )M contains some lattice point xk 6= 0. Now, if k ≥ l then (1 + k1 )M is
contained in (1 + 1l )M : For if m ∈ M then
1+
1+
as 0 ≤
1
1+ k
1+ 1l
1
k
1
l
· m + (1 −
1+
1+
1
k
1
l
)·0∈M
≤ 1; but then (1 + k1 ) · m = (1 + 1l ) · m0 for some m0 ∈ M .
Thus, all points xk are in particular in 2M . As 2M is a bounded set there
can only be finitely many distinct points among the xk . In particular, there is
a constant subsequence of the sequence (xk ). The constant x of such a constant
subsequence is a lattice point which is in (1 + k1 )M for infinitely many k. So, x is
in fact in (1 + k1 )M for all k. Also, since xk 6= 0 for all k, and since x equals xk for
some k, we have x 6= 0.
MINKOWSKI’S LATTICE POINT THEOREM.
3
Now, (1 + k1 )−1 x, k ∈ N, is a sequence of points in M . This sequence clearly
converges to x. As M is closed we have x ∈ M , and hence the desired conclusion.
References
[1] H. Koch: ‘Number Theory. Algebraic Numbers and Functions’. Graduate Studies in Mathematics 24, AMS 2000.
Department of Mathematics, University of Copenhagen, Universitetsparken 5, DK2100 Copenhagen Ø, Denmark.
E-mail address: [email protected]