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So far, we have seen how n-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 n-dimensional case. We next want to motivate how a reduction map can be defined given X and d, using additional information which might be easier to obtain.

First, introduce on \R^n a lexicographic order as follows: for a = (a_1, \dots, a_n) and b = (b_1, \dots, b_n), define

a \le b :\Longleftrightarrow \exists i \in \{ 1, \dots, n \} : a_i \le b_i \wedge \forall j < i : a_i = b_i.

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

Assume that \Lambda \subseteq \R^n is a lattice, X \neq \emptyset a finite set and d : X \to \R^n / \Lambda injective. Consider the projection \pi : \R^n \to \R^n/\Lambda, x \mapsto x + \Lambda and \hat{X} := \pi^{-1}(d(X)). Defining a function \psi : \R^n / \Lambda \to X is the same as defining a function \varphi : \R^n \to \hat{X} which is invariant under \Lambda, i.e. satisfies \varphi(t + \lambda) = \varphi(t) + \lambda for all \lambda \in \Lambda; in that case, we can set \psi(t + \Lambda) := d^{-1}(\varphi(t) + \Lambda). Note that the condition \psi \circ d = \id_X translates to \varphi|_{\hat{X}} = \id_{\hat{X}}.

Hence, we have a discrete set \hat{X} \subseteq \R^n which is invariant under translation by \Lambda, and we want to define a function \varphi : \R^n \to \hat{X} satisfying \varphi|_{\hat{X}} = \id_{\hat{X}}.

Both of the two sections which follow describe one way to obtain such \hat{X} and \varphi. The way describes in the second section fits perfectly for all totally real number fields K: think of \Gamma as the image of the ring of integers \calO_K under all embeddings \sigma_1, \dots, \sigma_{n+1} : K \to \R, i.e.

\Gamma = \{ (\sigma_1(x), \dots, \sigma_{n+1}(x)) \mid x \in \calO_K \}.

The first section resembles more the general global field situation. The set X will consist of a finite set of ideals with bounded norms. The degree map will be the logarithm of the norm, and the b_i'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 \varphi, given \hat{X}.

The main idea in the following is that if we want to define \varphi(t) for t = (t_1, \dots, t_n) \in \R^n, to consider the area

B_t := \{ (x_1, \dots, x_n) \in \R^n \mid \forall i : x_i \le t_i \}

and look at all elements \hat{X} \cap B_t. 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 \le as \varphi(t).

To make this “additional information” more precise, we consider special functions \deg : \hat{X} \to \R which should behave in a good way:


A function \deg : \hat{X} \to \R is said to be reduction-inducing if

  1. there exist real numbers b_1, \dots, b_n > 0 such that, for x \in \hat{X} and \lambda = (\lambda_1, \dots, \lambda_n) \in \Lambda, we have

    \deg x + \sum_{i=1}^n b_i \lambda_i = \deg (x + \lambda);


  2. for every x = (x_1, \dots, x_n) \in \hat{X}, we have

    B_x := \{ x' = (x_1', \dots, x_n') \in \hat{X} \mid x_i' \le x_i, \; \deg x' > \deg x \} = \emptyset.

Note that by this definition, there exist a, A \in \R such that

a \le \deg (x_1, \dots, x_n) - \sum_{i=1}^n x_i b_i \le A

for all x = (x_1, \dots, x_n) \in \hat{X}. Moreover, note that these functions with a, A, b_1, \dots, b_n fixed correspond to functions \deg' : X \to [a, A] by

\deg (x_1, \dots, x_n) = \deg' d^{-1}((x_1, \dots, x_n) + \Lambda) + \sum_{i=1}^n x_i b_i

for x = (x_1, \dots, x_n) \in \hat{X}.

Let \deg : \hat{X} \to \R be a reduction-inducing function. For t = (t_1, \dots, t_n) \in \R^n and \ell \in \R, consider

B_{t,\ell} := \{ x \in \hat{X} \cap B_t \mid \deg x \ge \ell \}.

Note that since \deg x \le A + \sum_{i=1}^n x_i b_i for x = (x_1, \dots, x_n) \in \hat{X} \cap B_t, and x_i \le t_i, we see that B_{t,\ell} is finite for every choice of \ell. If A + \sum_{i=1}^n t_i b_i < \ell, we have B_{t,\ell} = \emptyset, and as X \neq \emptyset we get \abs{B_{t,\ell}} \to \infty for \ell \to -\infty. Hence, \ell(t) := \max\{ \ell' \mid B_{t,\ell'} \neq \emptyset \} exists. Then, define \varphi(t) := \max_{\le} B_{t,\ell(t)}.

Let C \in \R be a constant such that for all t = (t_1, \dots, t_n) \in \R^n, we have

B_{t,\ell} \neq \emptyset \quad \text{for } \ell := \sum_{i=1}^n t_i b_i + C.

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


For x \in \hat{X}, we have \varphi(x) = x. For any t = (t_1, \dots, t_n) \in \R^n, if x = (x_1, \dots, x_n) = \varphi(t), we have 0 \le t_i - x_i \le \frac{A - C}{b_i} and C \le \ell(t) - \sum_{i=1}^n t_i b_i \le A. In fact, \sum_{i=1}^n (t_i - x_i) b_i \le A - C.


For the first statement, it suffices to show x \in B_{x,\ell(x)}. But note that if x \not\in B_{x,\ell(x)}, we would have \ell(x) > \deg x and hence B_{x,\ell(x)} \subseteq B_x, a contradiction.

For the second statement, note that \ell(t) = \deg (x_1, \dots, x_n) \le \sum_{i=1}^n x_i b_i + A. Moreover, B_{t,\ell} \neq \emptyset for \ell = \sum_{i=1}^n t_i b_i + C, whence we get \ell(t) - \sum_{i=1}^n t_i b_i \ge \ell - \sum_{i=1}^n t_i b_i = C. This shows the inequality on \ell(t). Now clearly x_i \le t_i, whence 0 \le t_i - x_i. Now A + \sum_{i=1}^n x_i b_i \ge \deg (x_1, \dots, x_n) = \ell(t) \ge C + \sum_{i=1}^n t_i b_i, whence \sum_{i=1}^n (t_i - x_i) b_i \le A - C. As t_i - x_i \ge 0.

Using Minima of Lattices.

In this section, we describe how to obtain \hat{X} and \varphi from an (n + 1)-dimensional lattice \Gamma \subseteq \R^{n+1}. We require that for every t = (t_1, \dots, t_{n+1}) \in \Gamma, we either have t = 0 for t_i \neq 0 for all i. More precisely, consider the map N : \R^{n+1} \to \R, (x_1, \dots, x_{n+1}) \mapsto \prod_{i=1}^n x_i. We assume that there exists a constant c > 0 with N(x) \ge c for all x \in \Gamma \setminus \{ 0 \}.

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


A minimum of \Gamma is an element \mu = (\mu_1, \dots, \mu_{n+1}) \in \Gamma \setminus \{ 0 \} such that for all z = (z_1, \dots, z_{n+1}) \in \Gamma with \abs{z_i} \le \abs{\mu_i} for all i, we either have z = 0 or \abs{z_i} = \abs{\mu_i} for all i. Denote the set of all minima by \min \Gamma.

First, we will show that such minima exist:


Let t = (t_1, \dots, t_{n+1}) \in \Gamma \setminus \{ 0 \}. Then there exists a minimum \mu = (\mu_1, \dots, \mu_{n+1}) of \Gamma with \abs{\mu_i} \le \abs{t_i} for all i.


This follows from the fact that \Gamma is discrete. For s = (s_1, \dots, s_{n+1}), define

B_s := \{ (x_1, \dots, x_{n+1} \in \Gamma \setminus \{ 0 \} \mid \abs{x_i} \le \abs{s_i} \text{ for all } i \}.

As \Gamma is discrete, B_s is always finite.

In particular, B_t is finite. Assume that t is not a minimum (in which case we could choose \mu = t). Then there exists some s = (s_1, \dots, s_{n+1}) \in B_t with \abs{s_i} < \abs{t_i} for some i. In that case, s \in B_s \subsetneqq B_t. Now either s is a minimum, in which case we choose \mu = s, or it is not. In that case, we can repeat the procedure with B_s instead of B_t. As the size of these sets decreases every step and the sets are finite but non-empty, we eventually must find some s \in B_t which is a minimum.

Define \sim on \R^{n+1} by

(s_1, \dots, s_{n+1}) \sim (t_1, \dots, t_{n+1}) :\Longleftrightarrow \forall i : \abs{s_i} = \abs{t_i},

and consider the map

\Phi : \Gamma \setminus \{ 0 \} \to \R^n, \quad (t_1, \dots, t_{n+1}) = (\log \abs{t_1}, \dots, \log \abs{t_n}).

First, \Phi(a) = \Phi(b) if, and only if, a \sim b. Let

\hat{X} := \Phi(\min \Gamma) = \{ \Phi(\mu) \mid \mu \text{ minimum of } \Gamma \}.

Let t = (t_1, \dots, t_n) \in \R^n. Then, there exists some \mu = (\mu_1, \dots, \mu_n) \in \hat{X} with 0 \le t_i - x_i and \sum_{i=1}^n (t_i - x_i) \le \log \abs{\det \Gamma}. In particular, t_i - x_i \le \log \abs{\det \Gamma}.

Here, \det{\Gamma} is the determinant of the lattice \Gamma, i.e. the volume of one fundamental parallelepiped of \Gamma.


For \ell > 0, consider the set

B_\ell := \{ (x_1, \dots, x_{n+1}) \in \R^{n+1} \mid \abs{x_i} \le \exp(t_i), \; \abs{x_{n+1}} \le \ell \}.

By Minkowski's Lattice Point Theorem, we have B_\ell \cap \Gamma \neq 0 for

2^n \prod_{i=1}^n \exp(t_i) \cdot 2 \ell = \mathrm{vol}(B_\ell) > 2^{n+1} \abs{\det \Gamma},


\ell > \abs{\det \Gamma} \exp\biggl( -\sum_{i=1}^n t_i \biggr).

Since B_\ell is closed and \Gamma discrete, a limit argument shows that this also holds for \ell = \abs{\det \Gamma} \exp\bigl( -\sum_{i=1}^n t_i \bigr). By the previous lemma, there exists a minimum s = (s_1, \dots, s_{n+1}) of \Gamma which lies in B_\ell; let \mu := (\mu_1, \dots, \mu_n) := \Phi(s). Now \mu_i = \log \abs{s_i} \le \log \exp(t_i) = t_i for 1 \le i \le n as s \in B_\ell, whence 0 \le t_i - \mu_i.

Now N(s) \ge c, whence \sum_{i=1}^n \mu_i \ge \log c - \log \abs{s_{n+1}}. Now \abs{s_{n+1}} \le \ell, whence -\log \abs{s_{n+1}} \ge -\log \ell \ge -\log \abs{\det \Gamma} + \sum_{i=1}^n t_i. Therefore, we get

\sum_{i=1}^n \mu_i \ge -\log \abs{\det \Gamma} + \sum_{i=1}^n t_i,

i.e. \sum_{i=1}^n (t_i - \mu_i) \le \log \abs{\det \Gamma}.

Define \Lambda := \{ x \in \R^n \mid \forall \mu \in \hat{X} : x + \mu \in \hat{X} \}; this is a discrete subgroup of \R^n. We assume that \Lambda is a lattice in \R^n, i.e. contains a basis of \R^n. We can define X and d : X \to \R^n/\Lambda such that \hat{X} = \pi^{-1}(d(X)), if \pi : \R^n \to \R^n/\Lambda, t \mapsto t + \Lambda is the projection. To get an n-dimensional infrastructure, we are left to define \varphi.

For that, we proceed as in the proof of the second lemma in this section. For t = (t_1, \dots, t_n) \in \R^n, consider

B_\ell := \biggl\{ \Psi(x) \;\biggm| \begin{matrix} x = (x_1, \dots, x_{n+1}) \in \min \Gamma, \\ \abs{x_i} \le \exp(t_i), \; \abs{x_{n+1}} \le \ell \end{matrix} \biggr\}.

Let \ell > 0 be minimal with B_\ell \neq \emptyset, and let \varphi(t) := \max_{\le} B_\ell. Then \varphi(t) satisfies the properties in the statement of the lemma, i.e. lies near to t itself. Moreover, one quickly checks that \varphi(t + \lambda) = \varphi(t) + \lambda for all \lambda \in \Lambda.

Hence, we obtain an n-dimensional infrastructure.


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