1.3.3. Lattices and norms [025Y]
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1.3.3. Lattices and norms
From now on and until the end of the subsection , we assume that is non-trivial. Let be an -submodule of . We say that is a lattice of if and
for some norm of . Note that the condition does not depend on the choice of the norm since all norms on are equivalent. For a lattice of , we define to be
Note that forms a norm of . Moreover, for a norm of ,
is a lattice of .
Proposition 1.14.
Let be a lattice of . We assume that, as an -module, admits a free basis . Then is an orthonormal basis of with respect to .
Proof.
For and ,
so that . β
Let us consider the following lemmas.
Lemma 1.15.
A subgroup of is either discrete or dense in .
Proof.
Clearly we may assume that , so that . We set . If , then . Indeed, for , let be an integer such that . Thus , and hence . Therefore, is discrete.
Next we assume that . Then there is a sequence in such that for all and . If we set , then and . For an open interval of (), we choose and an integer such that and . Then we have and
so that . Thus is dense. β
Lemma 1.16.
Let be a norm of and . Then
Moreover, and for all with .
Proof.
The first assertion is obvious because, for , if and only if .
For , let with . Then , that is, , and hence .
Finally we consider the second inequality, that is, for . Clearly we may assume that . As , there is with . By the first assertion, we can choose such that . If , then
Thus . This is a contradiction, so that . Therefore,
as required. β
Proposition 1.17.
We assume that is discrete. Then we have the following:
- (1)
Every lattice of is a finitely generated -module.
- (2)
If we set for a norm of of , then .
Proof.
(1) Let be an orthogonal basis of with respect to (cf. PropositionΒ 1.3). As is discrete, there is with . If we set for , then forms an orthonormal basis of with respect to . Therefore,
Thus we have (1) because is noetherian.
(2) follows from LemmaΒ 1.16. β
Proposition 1.18.
We assume that is not discrete. If we set for a norm of of , then .
Proof.
Proposition 1.19.
We assume that the absolute value is not discrete. Let be a norm of and . For any , there is a sub-lattice of such that is finitely generated over and .