Download DISCRETE SUBGROUPS OF VECTOR SPACES AND LATTICES

DISCRETE SUBGROUPS OF VECTOR SPACES AND LATTICES.
IAN KIMING
The following lemma gives a more precise result than Lemma 2.9.2 of [1]. A
lattice in a real vector space U is a subgroup of the form Zu1 + . . . + Zun where
u1 , . . . , un is a basis of U (so forget the erroneous definition of ‘lattice’ in section
2.12 of [1]).
Lemma 1. Let V be a real vector space of dimension d. Let M be a discrete
subgroup of V (i.e., a subgroup that is discrete as a subset of V ).
Then M is a lattice in the subspace of V that it generates as vector space. In
particular, M is a free abelian group of rank ≤ d.
Proof. Let U be the subspace of V that is generated by M (i.e., U is the subspace
consisting of all R-linear combinations of elements of M ). Let n := dim U so
that n ≤ d. Now, M must contain a basis of U as vector space; suppose that
u1 , . . . , un ∈ M is a basis of U . It is easily seen that M is discrete as a subset of
U.
Consider then the subgroup M0 := Zu1 + . . . + Zun of M . Then M0 ∼
= Zn as
the ui are R-linearly independent, and hence in particular Z-linearly independent.
Since U is generated over R by the ui we see that any element u ∈ U can be written:
(∗)
with λi ∈ R , 0 ≤ λi < 1 , m0 ∈ M0 .
u = λ 1 u 1 + . . . λ n u n + m0 ,
As M ⊆ U we see in particular that any element of M can be written in the
form (∗). We deduce that any coset of M/M0 has a representative m of form
(])
m = λ1 u1 + . . . λ n un ,
with λi ∈ R , 0 ≤ λi < 1 .
But if m ∈ M has shape (]) then
X
X
kmk ≤
|λi |kui k ≤
kui k .
i
i
But then m is in the intersection
M ∩ {u ∈ U | kuk ≤
X
kui k}
i
which is a finite set because M is discrete in U . We conclude that M/M0 is finite.
Thus M is finitely generated (as abelian group): For M is clearly generated by
M0 and a system of representatives for the cosets M/M0 , — and the latter is a
finite system. Clearly, as a subgroup of a real vector space M is also torsionfree.
Now, by Theorem A.4.4 of [1] we have that M is free of some rank t, and more
precisely, there exists a Z-basis m1 , . . . , mt of M and natural numbers e1 , . . . , es ,
s ≤ t, such that e1 m1 , . . . , es ms is a Z-basis of M0 . Then:
M/M0 ∼
= Zt−s × Z/Ze1 × . . . × Z/Zes .
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IAN KIMING
Since M/M0 is finite we must then have t = s. Since M0 is free of rank n we must
have s = n. Thus, t = s = n ≤ d.
Now, since both (u1 , . . . , un ) and (e1 m1 , . . . , en mn ) are Z-bases of M0 we have
that the latter arises from the first via an invertible base change matrix with coefficients in Z. So e1 m1 , . . . , en mn must be R-linearly independent because the
u1 , . . . , un are R-linearly independent (because they are a basis of U as real vector
space). It follows that also m1 , . . . , mn are R-linearly independent. So, M is a
lattice in U .
Not entirely uninteresting remarks to the beginning of section 2.13 of [1]: Setup
as in the beginning of 2.13. We are looking at the map: φ : K → Rn given by:
φ(α)
= (α(1) , . . . , α(n) )
:= (g1 α, . . . , gr1 α, Re(gr1 +1 α), . . . , Re(gr α), Im(gr1 +1 α), . . . , Im(gr α)) .
In the beginning of Theorem 2.13.1 it is said that φ maps a complete module m
to a lattice in Rn . Now, it is clear that φ is an injective homomorphism of abelian
groups. As m is a free abelian group of rank n we have that φ(m) is also a free
abelian group of rank n. With the erroneous definition of the notion of a lattice in
2.12, one would then have that m is a ‘lattice’ in Rn . However, we obviously have
to use the correct definition of lattice. Let us have a look:
With the above lemma it is actually fairly easy to see directly that φ(m) is a
lattice in some subspace of Rn : For according to the lemma, all we have to show
is that φ(m) is a discrete subgroup of Rn . That is a consequence of the following
statement: For any c ∈ R+ there are only finitely many m ∈ m such that |gi m| ≤ c
for i = 1, . . . , n. This statement is proved in exactly the same way as Lemma 2.9.1
is proved: Take a basis µ1 , . . . , µn of m and work with the complementary basis
µ̂1 , . . . , µ̂n .
So, φ(m) is a lattice in some subspace U of Rn . If the dimension of U is t then
by definition of a lattice φ(m) is then free of rank t (as abelian group). But φ is an
isomorphism m ∼
= φ(m) as abelian groups, so φ(m) is free of rank n. Thus t = n,
and the subspace generated by φ(m) is all of Rn , i.e., φ(m) is a lattice in Rn .
So that would be a possible proof of the fact that φ(m) is a lattice in Rn . But
actually, there is a much easier proof, namely by the proof of 2.13.1 itself: We
consider a basis µ1 , . . . , µn of m; then φ(m) is generated as an abelian group by the
(1)
(n)
vectors (µi , . . . , µi ), i = 1, . . . , n. But the proof of 2.13.1 shows that these n
vectors are linearly independent over R as the determinant
(j)
det(µi )
is shown to be non-zero. By definition then, φ(m) is a lattice in Rn .
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]