Felix' Math Place » n-dimensional http://math.fontein.de Focussed on, but not limited to Computational Number Theory Sat, 30 Jul 2011 12:35:49 +0000 en hourly 1 http://wordpress.org/?v=3.1 How to Obtain Reduction Maps for n-dimensional Infrastructures. http://math.fontein.de/2009/07/21/how-to-obtain-reduction-maps-for-n-dimensional-infrastructures/ http://math.fontein.de/2009/07/21/how-to-obtain-reduction-maps-for-n-dimensional-infrastructures/#comments Tue, 21 Jul 2009 05:43:54 +0000 Felix Fontein http://math.fontein.de/?p=211 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 \displaystyle 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. \displaystyle \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 \displaystyle 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:

Definition.
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 \displaystyle \deg x + \sum_{i=1}^n b_i \lambda_i = \deg (x + \lambda); and
  2. for every x = (x_1, \dots, x_n) \in \hat{X}, we have Formula does not parse: \displaystyle 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 \displaystyle 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 \displaystyle \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 \displaystyle 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 \displaystyle 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.

Lemma.
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.
Proof.

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.

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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.

Definition.
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:

Lemma.
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.
Proof.

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

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Define \sim on \R^{n+1} by \displaystyle (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 \displaystyle \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 \displaystyle \hat{X} := \Phi(\min \Gamma) = \{ \Phi(\mu) \mid \mu \text{ minimum of } \Gamma \}.

Lemma.
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.

Proof.

For \ell > 0, consider the set \displaystyle 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 \displaystyle 2^n \prod_{i=1}^n \exp(t_i) \cdot 2 \ell = \mathrm{vol}(B_\ell) > 2^{n+1} \abs{\det \Gamma}, i.e. \displaystyle \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 \displaystyle \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}.

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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 \displaystyle 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.

]]> http://math.fontein.de/2009/07/21/how-to-obtain-reduction-maps-for-n-dimensional-infrastructures/feed/ 0 n-dimensional Infrastructures. http://math.fontein.de/2009/07/20/n-dimensional-infrastructures/ http://math.fontein.de/2009/07/20/n-dimensional-infrastructures/#comments Mon, 20 Jul 2009 08:40:46 +0000 Felix Fontein http://math.fontein.de/?p=195 For one-dimensional infrastructures, we have a circle \R/R\Z together with a finite, non-empty set X and an injective map d : X \to \R/R\Z. Now \R/R\Z = \R^n / \Lambda, where n = 1 and \Lambda is the one-dimensional lattice \Lambda = R \Z. Hence, one could say that an n-dimensional infrastructure is a torus \R^n/\Lambda together with X and d : X \to \R^n/\Lambda as above. From the discussion in the remarks of this post we see that we need some kind of reduction map to define giant steps (and also f-representations) in the one-dimensional case, even though there a pretty canonical reduction map is given. In the case of \R^n/\Lambda, we do not have something similar to a given positive direction. Moreover, definiting the “nearest” element of a finite subset of \R^n/\Lambda to some t \in \R^n/\Lambda is even more complicated and offers more choices which appear more or less obvious. Only the selection of different norms on \R^n lead to several possible definitions of such a map. Hence, we should require such a map in the definition:

Definition.
Let \Lambda \subseteq \R^n be a lattice. Then, an n-dimensional infrastructure is a non-empty finite set X together with an injective map d : X \to \R^n / \Lambda and another map red : \R^n/\Lambda \to X satisfying red \circ d = \id_X.

Again, as in the one-dimensional case, one can define giant steps: \displaystyle \gs(x, x') := red(d(x) + d(x')), \quad x, x' \in X. Moreover, one gets the same relation between reduction maps and f-representations, whence we define \displaystyle \fRep := \fRep(X, d, red) := \{ (x, f) \in X \times \R^n \mid red(d(x) + f) = x \}. Then the map \displaystyle \Psi : \fRep(X, d, red) \to \R^n/\Lambda, \quad (x, f) \mapsto d(x) + f is a bijection, and we can use this bijection to equip \fRep(X, d, red) with a group law by \displaystyle (x, f) + (x', f') = \Psi^{-1}(\Psi(x, f) + \Psi(x', f')), \quad (x, f), (x', f') \in \fRep.

Discrete Infrastructure.

We say that an n-dimensional infrastructure (X, d, red) with lattice \Lambda is discrete if \Lambda \subseteq \Z^n, d(X) \subseteq \Z^n/\Lambda and if red does not depends on fractions. To make the last part more precise, define \displaystyle floor : \R^n \to \Z^n, \quad (x_1, \dots, x_n) \mapsto (\floor{x_1}, \dots, \floor{x_n}); if \Lambda \subseteq \Z^n, floor induces a map \R^n/\Lambda \to \Z^n/\Lambda. Now, that red does not depends on fractions simply means that red factors through floor, i.e. that we can write red = red' \circ floor with red' : \Z^n/\Lambda \to X. Moreover, if in the following we specify discrete infrastructures, we often just define red for values in \Z^n/\Lambda. In that case, for elements v \in \R^n/\Lambda \setminus \Z^n/\Lambda, define red(v) := red(floor(v)). In case (X, d, red) is discrete, consider the subset \displaystyle \fRep_{disc} := \fRep_{disc}(X, d, red) := \{ (x, f) \in \fRep \mid f \in \Z^n \}. Then \displaystyle \Psi|_{\fRep_{disc}} : \fRep_{disc} \to \Z^n/\Lambda is an isomorphism.

Finite Abelian Groups as Infrastructures.

Let G be a finite abelian group, generated by g_1, \dots, g_n. Consider the relation lattice \Lambda \subseteq \Z^n for g_1, \dots, g_n, defined by \displaystyle (v_1, \dots, v_n) \in \Lambda \Leftrightarrow \prod_{i=1}^n g_i^{v_i} = 1. Then \Lambda is the kernel of \Z^n \to G, (v_1, \dots, v_n) \mapsto \prod_{i=1}^n g_i^{v_i}, and \displaystyle \varphi : \Z^n/\Lambda \to G, \quad (v_1, \dots, v_n) + \Lambda \mapsto \prod_{i=1}^n g_i^{v_i} is a group isomorphism. Define X := G, d := \varphi^{-1} and red := \varphi (or, more precisely, red := \varphi \circ floor); then (X, d, red) is an n-dimensional infrastructure. Moreover, for g, g' \in G, we have \gs(g, g') = g g', whence \gs equals the group operation of G. Hence, every finite abelian group can be seen in a natural way as an infrastructure. Moreover, this shows that d can be thought of as an analogue to the discrete logarithm map, and red is an analogue of the power map n \mapsto g^n. In particular, we obtained the goal described in the first post of this series: we found a generalization of the discrete logarithm problem to a non-associative algebraic structure. In the next post, I will how such infrastructures can be obtained from global fields; this gives a rich source of examples for n-dimensional infrastructures.

What about baby steps?

Note that in the above discussion, I simply ignored baby steps. In the one-dimensional case, \R has a canonical direction (namely the positive one) and so has \R/R\Z, whence saying “go to the next element” makes sense. Opposed to that, in \R^n, there are infinitely many directions, no one better than another. Even if we fix a direction, “go to the next element in that direction” seems to not really make sense. So far, I have not seen any definition of baby steps in this case which works for all n-dimensional infrastructures. Note that in the case of infrastructures obtained from global fields, one has some kind of canonical baby steps (even though there are still some choices left). In fact, there are n + 1 of them. To define them, though, one needs more information than just X, d and red: one needs information about a (n + 1)-st dimension, both for constructing the reduction map and for defining baby steps. ]]> http://math.fontein.de/2009/07/20/n-dimensional-infrastructures/feed/ 0