# How to Obtain Reduction Maps for n-dimensional Infrastructures.

So far, we have seen how -dimensional infrastructures can be defined. In the case of one-dimensional infrastructures, we saw that there is a (more or less) obvious way how to define a reduction map, which does not extend to the -dimensional case. We next want to motivate how a reduction map can be defined given and , using additional information which might be easier to obtain.

First, introduce on a lexicographic order as follows: for and , define

Note that this choice is rather random and can easily be replaced by other choices.

Assume that is a lattice, a finite set and injective. Consider the projection , and . Defining a function is the same as defining a function which is invariant under , i.e. satisfies for all ; in that case, we can set . Note that the condition translates to .

Hence, we have a discrete set which is invariant under translation by , and we want to define a function satisfying .

Both of the two sections which follow describe one way to obtain such and . The way describes in the second section fits perfectly for all totally real number fields : think of as the image of the ring of integers under all embeddings , i.e.

The first section resembles more the general global field situation. The set will consist of a finite set of ideals with bounded norms. The degree map will be the logarithm of the norm, and the 's correspond to the degrees of the infinite places.

#### Constructing a Reduction Map.

In this section, we describe a way to construct a reduction map , given .

The main idea in the following is that if we want to define for , to consider the area

and look at all elements . By adding additional (numeric) information to every of these elements, one obtains an order (by comparing the additional information) which hopefully has a largest element, or a finite set of largest elements. From these largest elements, one chooses the largest one with respect to the lexicographic order as .

To make this “additional information” more precise, we consider special functions which should behave in a good way:

Definition.

A function is said to be reduction-inducing if

1. there exist real numbers such that, for and , we have

and

2. for every , we have

Note that by this definition, there exist such that

for all . Moreover, note that these functions with fixed correspond to functions by

for .

Let be a reduction-inducing function. For and , consider

Note that since for , and , we see that is finite for every choice of . If , we have , and as we get for . Hence, exists. Then, define .

Let be a constant such that for all , we have

Note that since is reduction-inducing, a maximal such exists.

Lemma.

For , we have . For any , if , we have and . In fact, .

Proof.

For the first statement, it suffices to show . But note that if , we would have and hence , a contradiction.

For the second statement, note that . Moreover, for , whence we get . This shows the inequality on . Now clearly , whence . Now , whence . As .

#### Using Minima of Lattices.

In this section, we describe how to obtain and from an -dimensional lattice . We require that for every , we either have for for all . More precisely, consider the map , . We assume that there exists a constant with for all .

In fact, one can replace by any discrete subset with some additional properties which give similar results as Minkowski's Lattice Point Theorem.

Definition.

A minimum of is an element such that for all with for all , we either have or for all . Denote the set of all minima by .

First, we will show that such minima exist:

Lemma.

Let . Then there exists a minimum of with for all .

Proof.

This follows from the fact that is discrete. For , define

As is discrete, is always finite.

In particular, is finite. Assume that is not a minimum (in which case we could choose ). Then there exists some with for some . In that case, . Now either is a minimum, in which case we choose , or it is not. In that case, we can repeat the procedure with instead of . As the size of these sets decreases every step and the sets are finite but non-empty, we eventually must find some which is a minimum.

Define on by

and consider the map

First, if, and only if, . Let

Lemma.

Let . Then, there exists some with and . In particular, .

Here, is the determinant of the lattice , i.e. the volume of one fundamental parallelepiped of .

Proof.

For , consider the set

By Minkowski's Lattice Point Theorem, we have for

i.e.

Since is closed and discrete, a limit argument shows that this also holds for . By the previous lemma, there exists a minimum of which lies in ; let . Now for as , whence .

Now , whence . Now , whence . Therefore, we get

i.e. .

Define ; this is a discrete subgroup of . We assume that is a lattice in , i.e. contains a basis of . We can define and such that , if , is the projection. To get an -dimensional infrastructure, we are left to define .

For that, we proceed as in the proof of the second lemma in this section. For , consider

Let be minimal with , and let . Then satisfies the properties in the statement of the lemma, i.e. lies near to itself. Moreover, one quickly checks that for all .

Hence, we obtain an -dimensional infrastructure.