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:
A function is said to be reduction-inducing if
there exist real numbers such that, for and , we have
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
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.
For , we have . For any , if , we have and . In fact, .
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.
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:
Let . Then there exists a minimum of with for all .
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
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 .
For , consider the set
By Minkowski's Lattice Point Theorem, we have for
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
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.