Felix' Math Place » infrastructure 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 Rigorous Arithmetic in the Arakelov Divisor Class Group of a Number Field. http://math.fontein.de/2010/07/27/rigorous-arithmetic-in-the-arakelov-divisor-class-group-of-a-number-field/ http://math.fontein.de/2010/07/27/rigorous-arithmetic-in-the-arakelov-divisor-class-group-of-a-number-field/#comments Tue, 27 Jul 2010 09:50:37 +0000 Felix Fontein http://math.fontein.de/?p=778 This year at the IX. Algorithmic Number Theory Symphosium, held in Nancy, I had a poster in the poster session. You can see it here (click to see a larger version):

You can also get a PDF version here (9.1 MB).
The poster discusses how to effectively compute in the Arakelov divisor class group \Pic^0(K) of a number field K, which is assumed to be totally real in the current implementation described in the poster, but the same method works as long as there is at least one real embedding of K. In case K is totally imaginary, the only thing which gets more complicated is doing comparisms. The arithmetic uses f-representations as the main tool, i.e. it allows to compute in the infrastructure of K.

]]> http://math.fontein.de/2010/07/27/rigorous-arithmetic-in-the-arakelov-divisor-class-group-of-a-number-field/feed/ 1 Infrastructures and Global Fields. http://math.fontein.de/infrastructures/ http://math.fontein.de/infrastructures/#comments Thu, 23 Jul 2009 05:59:40 +0000 Felix Fontein http://math.fontein.de/?page_id=259 The following posts give an introduction to infrastructures and how to obtain these from global fields:

  1. The Discrete Logarithm Problem and Generalizations.
  2. One-dimensional Infrastructures.
  3. Interpreting One-dimensional Infrastructures as Groups: f-Representations.
  4. n-dimensional Infrastructures.
  5. How to Obtain Reduction Maps for n-dimensional Infrastructures.
  6. Obtaining Infrastructures from Global Fields.

See also my article on infrastructures at Wikipedia.

]]> http://math.fontein.de/infrastructures/feed/ 1 Obtaining Infrastructures from Global Fields. http://math.fontein.de/2009/07/21/obtaining-infrastructures-from-global-fields/ http://math.fontein.de/2009/07/21/obtaining-infrastructures-from-global-fields/#comments Tue, 21 Jul 2009 09:39:48 +0000 Felix Fontein http://math.fontein.de/?p=196 Basics on Global Fields.

Let K be a global field, i.e. an algebraic number field or an algebraic function field with a finite constant field. In the first case, let k^* be the roots of unity and k = k^* \cup \{ 0 \}. In the latter case, let k be the exact field of constants.

Let S = \{ \frakp_1, \dots, \frakp_{n+1} \} be the set of infinite places of K. If K is a number field, the elements of S correspond to embeddings of K into \C up to complex conjugation. Define q := \exp(1), and for \frakp \in S let \sigma : K \to \C be a corresponding embedding. Then define \nu_\frakp(f) := -\log \abs{\sigma(f)} for f \in K^* and \deg \frakp := 1 if \sigma(K) \subseteq \R, or \deg \frakp := 2 otherwise, and define \G_\frakp := \R. If K is a function field, let q := \abs{k}, i.e. k = \F_q; in this case, there exists an element x \in K \setminus k whose poles are exactly the elements of S, i.e. are the places of K lying above the infinite place of k(x). In all cases, S is finite and non-empty.

For a non-archimedean place \frakp of K, let \calO_\frakp be the valuation ring and \frakm_\frakp its maximal idea, and denote the discrete valuation by \nu_\frakp. Then set \deg \frakp := \log_q \abs{\calO_\frakp / \frakm_\frakp} and \abs{f}_\frakp := q^{-\nu_\frakp(f) \deg \frakp}, f \in K. Define \G_\frakp := \Z. In the number field case, let \G := \R, and otherwise \G := \Z.

Denote the set of places of K by \calP_K. The divisor group of K is \Div(K) := \coprod_{\frakp \in \calP} \G_\frakp, and for D = \sum_{\frakp \in \calP_K} n_\frakp \frakp define \deg D := \sum_{\frakp \in \calP_K} n_\frakp \deg \frakp. This is a homomorphism \deg : \Div(K) \to \G; denote its kernel by \Div^0(K). For f \in K^*, (f) := \sum_{\frakp \in \calP_K} \nu_\frakp(f) \frakp \in \Div^0(K) is a principal divisor; let the group of all these be denoted by \Princ(K). Then \Pic(K) := \Div(K) / \Princ(K) is the divisor class group of K and \Pic^0(K) := \Div^0(K) / \Princ(K) its degree zero part.

The support of a divisor D = \sum_{\frakp \in \calP_K} n_\frakp \frakp is the set \support(D) = \{ \frakp \in \calP_K \mid n_\frakp \neq 0 \}. Consider the subgroups \Div_{fin}(K) :={} & \{ D \in \Div(K) \mid \support(D) \cap S = \emptyset \} \\ \text{and} \qquad \Div_\infty(K) :={} & \{ D \in \Div(K) \mid \support(D) \subseteq S \}; then \Div(K) = \Div_{fin}(K) \oplus \Div_\infty(K). Moreover, let \Div_\infty^0(K) := \Div^0(K) \cap \Div_\infty(K). The set \calO := \calO_S := \{ f \in K \mid \nu_\frakp(f) \ge 0 \text{ for all } \frakp \in S \} is a Dedekind domain, whose maixmal ideals correspond to the places in \calP_K \setminus S. Moreover, the fractional ideal group \Id(\calO_S) is isomorphic to \Div_{fin}(K) by \divisor(\fraka) = \sum_{\frakp \not\in S} n_\frakp \frakp, in case \fraka = \prod_{\frakp \not\in S} (\frakm_\frakp \cap \calO_S)^{-n_\frakp}; the inverse is given by the restriction of Formula does not parse: \ideal : \Div(K) \to \Id(\calO_S), \sum n_\frakp \frakp \mapsto \prod_{\frakp \not\in S} (\frakm_\frakp \cap \calO_S)^{-n_\frakp} to \Div_{fin}(K). The group of fractional principal ideals \PId(\calO_S) equals Formula does not parse: \ideal(\Princ(K)). The quotient \Id(\calO_S) / \PId(\calO_S) is the ideal class group \Pic(\calO_S) of \calO_S. Putting all these things together, we get the following diagram with exact rows and columns: \displaystyle \xymatrix{ & 0 \ar[d] & 0 \ar[d] & 0 \ar[d] & \\ 0 \ar[r] & \calO_S^* / k^* \ar[r] \ar[d] & \Div^0_\infty(K) \ar[r] \ar[d] & T \ar[r] \ar[d] & 0 \\ 0 \ar[r] & K^* / k^* \ar[r] \ar[d] & \Div^0(K) \ar[r] \ar[d] & \Pic^0(K) \ar[r] \ar[d] & 0 \\ 0 \ar[r] & K^* / \calO_S^* \ar[r] \ar[d] & \Id(\calO_S) \ar[r] \ar[d] & \Pic(\calO_S) \ar[r] \ar[d] & 0 \\ & 0 & H \ar@{=}[r] \ar[d] & H \ar[d] & \\ & & 0 & 0 & } Here, T and H are essentially defined by the diagram, i.e. are the kernels and cokernels of the respective maps. In the number field case, H = 0, and in the function field case, H \cong (\deg \frakp \mid \frakp \in \calP_K) / (\deg \frakp \mid \frakp \in S).

A Geometry of Numbers in Global Fields.

Let \fraka \in \Id(\calO_S) and t_1, \dots, t_{n+1} \in \G. Define \displaystyle B(\fraka, (t_1, \dots, t_{n+1})) := \{ f \in \fraka \mid \forall i : \abs{f}_{\frakp_i} \le q^{t_i \deg \frakp_i} \}. If D := \divisor(\fraka) + \sum_{i=1}^{n+1} t_i \frakp_i \in \Div(K) and L(D) is the Riemann-Roch space of D, then L(D) = B(\fraka, (t_1, \dots, t_{n+1})). In particular, the set is finite and invariant under multiplication by elements of k; in case K is a function field, L(D) is a finite-dimensional k-vector space, whose dimension is described by the Riemann-Roch theorem. In the number field case, we can make statements on L(D) with Minkowski’s Lattice Point Theorem.

Consider the map \displaystyle \Psi : K^* \to \G^n, \quad f \mapsto (-\nu_{\frakp_1}(f), \dots, -\nu_{\frakp_n}(f)). Then \Lambda := \Psi(\calO^*) \cong \Z^n is a lattice by Dirichlet’s Unit Theorem, and \ker \Psi|_{\calO^*} = k^*. We get \calO^* \cong k^* \times \Z^n, and n is called the unit rank of \calO_S. This \Lambda will be the lattice for our n-dimensional infrastructure.

Reduced Ideals.

The elements of X will be principal reduced fractional ideals, modulo an equivalence relation. We begin by defining minima, which are similar to the ones introduced in the previous post for lattices.

Definition.
Let \fraka \in \Id(\calO_S) and \mu \in \fraka \setminus \{ 0 \}. We say that \mu is a minimum of \fraka if every f \in \fraka \setminus \{ 0 \} with \abs{f}_{\frakp_i} \le \abs{\mu}_{\frakp_i} for all i satisfies \abs{f}_{\frakp_i} = \abs{\mu}_{\frakp_i} for all i. Denote the set of all minima of \fraka by \calC(\fraka).

Using them, we can define reduced ideals:

Definition.
An ideal \fraka \in \Id(\calO_S) is said to be reduced if 1 \in \fraka is a minimum. Write \Red_S(K) for the set of all reduced ideals of \calO_S. For \frakb \in \Id(\calO_S) let \Red_S(\frakb) := \{ \fraka \in \Red_S(K) \mid \exists f \in K^* : f \fraka = \frakb \}.

The equivalence relation we need is defined by \displaystyle \fraka \sim_S \fraka' :\Leftrightarrow \exists f \in K^* : \fraka = f \fraka' \wedge \forall \frakp \in S : \abs{f}_\frakp = 1 for \fraka, \fraka' \in \Id(\calO_S). We then get the following results:

Theorem.
  1. We have that \Red_S(K) is a finite set.
  2. In case \deg \frakp = 1 for some \frakp \in S, we get \fraka \sim_S \fraka' for \fraka, \fraka' \in \Red(K) if, and only if, \fraka = \fraka'.
  3. We have that \calO^* acts on \calC(\fraka) by multiplication.
  4. The map \displaystyle \calC(\fraka) / \calO^* \to \Red(\fraka), \quad \mu \calO^* \mapsto \frac{1}{\mu} \fraka is a bijection.
  5. If \fraka \in \Red(K) and \frakb \in \Id(\calO) satisfies \fraka \sim_S \frakb, then \frakb \in \Red(\fraka).
□

The proofs of these and the following results or hints to the proofs can be found here. We next construct the map d:

Theorem (Infrastructure, Part I).
Fix an ideal \fraka \in \Id(\calO). Define X_\fraka := \Red(\fraka)/_{\sim_S} and define \displaystyle d_\fraka : X \to \G^n / \Lambda, \quad [\tfrac{1}{\mu} \fraka]_{\sim_S} \mapsto \Psi(\mu) + \Lambda. Then d_\fraka is well-defined and injective.
□

For a, a' \in K^*, write \displaystyle a \sim_S a' :\Longleftrightarrow \forall \frakp \in S : \abs{a}_\frakp = \abs{a'}_\frakp. Define \hat{X} := \calC(\fraka)/_{\sim_S} and \displaystyle \hat{d} : \hat{X} \to \G^n, \quad [\mu]_\sim \mapsto \Psi(\mu). Then (\hat{X}, \hat{d}) is the unrolled version of (X, d): if \pi : \G^n \to \G^n / \Lambda, x \mapsto x + \Lambda is the projection, and \psi : \hat{X} \to X, [\mu]_\sim \mapsto [\frac{1}{\mu} \fraka]_\sim, then the following diagram commutes: \displaystyle \xymatrix{ \hat{X} \ar[d]_{\psi} \ar[r]^{\hat{d}} & \G^n \ar[d]^{\pi} \\ X \ar[r]_{d} & \G^n/\Lambda } In particular, \hat{d}(\hat{X}) is the set \hat{X} from the previous post.

The Reduction Map, f-Representations, and the Infrastructure.

We proceed by defining f-representations, as giving these is equivalent to give a reduction map. Fix an ideal \fraka \in \Id(\calO_S).

First, define for f, f' \in K^* \displaystyle f \le_S f' :\Longleftrightarrow (\abs{f}_{\frakp_{n+1}}, \abs{f}_{\frakp_1}, \dots, \abs{f}_{\frakp_n}) \le_{\ell ex} (\abs{f'}_{\frakp_{n+1}}, \abs{f'}_{\frakp_1}, \dots, \abs{f'}_{\frakp_n}), where \le_{\ell ex} is the lexicographic order on \R^{n+1}.

Definition.
A tuple ([\frakb]_{\sim_S}, (t_1, \dots, t_n)) \in \Red_S(\fraka)/_{\sim_S} \times \G^n is said to be an f-representation if 1 is a smallest element of \displaystyle B(\frakb, (t_1, \dots, t_n, 0)) \setminus \{ 0 \} with respect to \le. Denote the set of all f-representations by \fRep(\fraka).

One quickly sees that this is well-defined. We have two auxilliary results:

Lemma (Uniqueness).
Let A = ([\frakb]_{\sim_S}, (t_1, \dots, t_n)) \in \fRep(\fraka) and f \in K^* such that \displaystyle B = ([\tfrac{1}{f} \frakb]_{\sim_S}, (t_1 + \nu_{\frakp_1}(f), \dots, t_n + \nu_{\frakp_n}(f))) \in \fRep(\fraka). Then \abs{f}_\frakp = 1 for all \frakp \in S, i.e. A = B.
□
Lemma (Reduction).
Let v = (v_1, \dots, v_n) \in \G^n. Then there exists a smallest \ell \in \G such that B_\ell := B(\fraka, (v_1, \dots, v_n, \ell)) \setminus \{ 0 \} \neq \emptyset. If \mu is minimal with respect to \le in that B_\ell, then \displaystyle ([\tfrac{1}{\mu} \fraka]_{\sim_S}, (v_1 + \nu_{\frakp_1}(\mu), \dots, v_n + \nu_{\frakp_n}(\mu))) \in \fRep(\fraka) and \Phi(\mu) + (v_1 + \nu_{\frakp_1}(\mu), \dots, v_n + \nu_{\frakp_n}(\mu)) + \Lambda = v + \Lambda.
□

From that, we get the following result:

Theorem (Infrastructure, Part II).
Let \fraka \in \Id(\calO). Then the map \Phi :{} & \fRep(\fraka) \to \G^n / \Lambda \\ & ([\tfrac{1}{\mu} \fraka]_{\sim_S}, (t_1, \dots, t_n)) \mapsto \Psi(\mu) + (t_1, \dots, t_n) + \Lambda is a bijection.
□

This allows to equip \fRep(\fraka) with a group operation. We will see that the group operation of \fRep(\calO_S) can be described in a very explicit form. This extends to a broader interpretation of the infrastructure, whence we will do this in the next section.

Before ending this section, we want to state a result which shows that f-representations are small.

Theorem.
Let ([\frakb]_{\sim_S}, (t_1, \dots, t_n)) \in \fRep(\fraka), then \divisor(\frakb) \ge 0, t_i \ge 0 for all i and \displaystyle \deg \divisor(\fraka) + \sum_{i=1}^n t_i \deg \frakp_i \le \kappa, where \displaystyle \kappa := \begin{cases} g + \deg \frakp_{n+1} - 1 & \text{if } K \text{ is a function field} \\ s \log \tfrac{2}{\pi} + \tfrac{1}{2} \log \abs{\Delta} & \text{if } K \text{ is a number field;} \end{cases} here, g is the genus of K in case K is a function field, and in case K is a number field, s denotes the number of places of degree two and \Delta is the discriminant of \calO_S.
□

Therefore, f-representations are small.

The Infrastructure and the Divisor Class Group.

Assume for a moment that \deg \frakp_{n+1} = 1, or that K is a number field. Then we have a short exact sequence \displaystyle \xymatrix{ 0 \ar[r] & T \ar[r] & \Pic^0(K) \ar[r] & \Pic(\calO_S) \ar[r] & 0, } and T \cong \G^n / \Lambda \cong \fRep(\fraka). This means that the divisor class group \Pic^0(K) is covered by copies of \G^n/\Lambda, where the copies are indexed by the elements of the divisor class group. If \fraka and \fraka' are in the same ideal class, X_\fraka and X_{\fraka'} differ by a translation, i.e. they give essentially the same infrastructure; in fact, \fRep(\fraka) = \fRep(\fraka'). Hence, one could get the idea to cover \Pic^0(K) by \fRep(\fraka), where \fraka ranges over the distinct ideal classes, i.e. by \fRep(K) := \bigcup_{\fraka \in \Id(\calO_S)} \fRep(\fraka). It turns out that this is indeed the case, and the arithmetic on \fRep(\calO_S) and \Pic^0(K) turn out to be the same under the bijection we get.

In case K is a function field and \deg \frakp_{n+1} > 1, we have T \not\cong \G^n / \Lambda in general (this is the case if, and only if, \deg \frakp_{n+1} = \gcd(\deg \frakp_1, \dots, \frakp_n, \frakp_{n+1})), and \Pic^0(K) \to \Pic(\calO_S) does not needs to be surjective. It would be nice to change the above sequence to \displaystyle \xymatrix{ 0 \ar[r] & \G^n/\Lambda \ar[r] & \Pic^0(K) \ar[r] & \Pic(\calO_S) \ar[r] & 0 } in any case, but this is not possible with \Pic^0(K) as it is; we have to replace it by something bigger. It turns out that the right replacement is \Pic(K) / \ggen{[\frakp_{n+1}]}, which is canonically isomorphic to \Pic^0(K) in case \deg \frakp_{n+1} = \gcd(\deg \frakp \mid \frakp \in \calP_K). We then get the diagram \displaystyle \xymatrix{ 0 \ar[r] & T \ar[r] \ar@{^(->}[d] & \Pic^0(K) \ar@{^(->}[d] \ar[r] & \Pic(\calO_S) \ar@{=}[d] & \\ 0 \ar[r] & \G^n/\Lambda \ar[r] & \Pic(K) / \ggen{[\frakp_{n+1}]} \ar[r] & \Pic(\calO_K) \ar[r] & 0 } with exact rows.

The complete result is stated in the following theorem:

Theorem (Infrastructure, Part III).
  1. Let K be a number field. Then the map \Phi :{} & \fRep(K) \to \Pic^0(K), \\ & ([\frakb]_{\sim_S}, (t_1, \dots, t_n)) \mapsto \biggl[ \divisor(\frakb) + \sum_{i=1}^n t_i \frakp_i - \frac{\dots}{\deg \frakp_{n+1}} \frakp_{n+1} \biggr], where \dots equals \deg \divisor(\frakb) + \sum_{i=1}^n t_i \deg \frakp_i, is a bijection.
  2. Let K be a function field. Then the map \Phi :{} & \fRep(K) \to \Pic(K) / \ggen{[\frakp_{n+1}]}, \\ & ([\frakb]_{\sim_S}, (t_1, \dots, t_n)) \mapsto \biggl[ \divisor(\frakb) + \sum_{i=1}^n t_i \frakp_i \biggr] + \ggen{[\frakp_{n+1}]} is a bijection.
Moreover, \Phi|_{\fRep(\calO_S)} is a group homomorphism, where the group structure on \fRep(\calO_S) is the one induced by the bijection \fRep(\calO_S) \to \G^n/\Lambda.
□

Finally, we explicitly describe the group operation induced by this bijection on \fRep(K) without using the bijection itself.

Theorem.
Let \Phi be the bijection from the previous theorem, and let A = ([\fraka]_{\sim_S}, (t_1, \dots, t_n)), A' = ([\fraka']_{\sim_S}, (t'_1, \dots, t'_n)) \in \fRep(K).
  1. Set B_\ell := B(\fraka \fraka', (t_1 + t'_1, \dots, t_n + t'_n, \ell)) \setminus \{ 0 \} for \ell \in \G. There exists a minimal \ell with B_\ell \neq \emptyset, and if \mu is a smallest element of B_\ell with respect to \le, we get \displaystyle B := ([\tfrac{1}{\mu} \fraka \fraka']_{\sim_S}, (t_i + t'_i + \nu_{\frakp_i}(\mu))_{i=1,\dots,n}) \in \fRep(K) with \Phi(A) + \Phi(A') = \Phi(B).
  2. Set B_\ell := B(\fraka^{-1}, (-t_1, \dots, -t_n, \ell)) \setminus \{ 0 \} for \ell \in \G. There exists a minimal \ell with B_\ell \neq \emptyset, and if \mu is a smallest element of B_\ell with respect to \le, we get \displaystyle C := ([\tfrac{1}{\mu} \fraka^{-1}]_{\sim_S}, (-t_i + \nu_{\frakp_i}(\mu))_{i=1,\dots,n}) \in \fRep(K) with -\Phi(A) = \Phi(C).

This shows that the n-dimensional infrastructure we defined has a very close connection to the arithmetic of the divisor class group. This connection was first shown for real hyperelliptic curves by H.-G. Rück and S. Paulus, “Real and Imaginary Quadratic Representations of Hyperelliptic Function Fields”. The first relation between the infrastructure of number fields and the Arakelov divisor class group was described by R. Schoof in his paper Computing Arakelov class groups.

What about… Baby Steps?

As I mentioned, there is no known construction for baby steps in general n-dimensional infrastructures, but there exists a construction for infrastructures obtained from global fields. I want to describe this construction here.

For i \in \{ 1, \dots, n + 1 \} and \fraka \in \Red(K), consider \displaystyle B_\ell := \biggl\{ f \in \fraka \;\biggm|\begin{matrix} \abs{f}_{\frakp_j} \le 1 \text{ for all } j \neq i, \\ \exists j' : \abs{f}_{\frakp_{j'}} < 1, \; \abs{f}_{\frakp_i} \le \ell \end{matrix} \biggr\} \setminus \{ 0 \} for \ell > 0. There exists a minimal \ell such that B_\ell \neq \emptyset. In case \deg \frakp_i = 1, B_\ell contains exactly one k^*-orbit, which gives a unique element \mu \in B_\ell. Otherwise, one has to add an order (lexicographic order as \le above) to chose an element. In any case, define \bs_i([\fraka]_{\sim_S}) := [\frac{1}{\mu} \fraka]_{\sim_S}; then this gives a function \Red(K) \to \Red(K) resp. \Red(\frakb) \to \Red(\frakb) for any \frakb \in \Id(\calO). Opposed to the one-dimensional case, this function neither has to be injective nor surjective, as examples below will show.

We begin with a “small” example: the infrastructure (X_{\calO_S}, d_{\calO_S}) of the function field defined by y^3 = x^6 + x^5 + x^4 + 4 x^2 over \F_7. The red arrows show \bs_1, the blue arrows \bs_2 and the green arrows \bs_3. The small black circles denote usual minima, the larger black circles denote elements of \Lambda, and the shaded areas denote translates of an fundamental parallelepiped of \Lambda:

\setlength{\unitlength}{12cm}\psset{unit=12cm}\begin{picture}(1,1)(0,0)\put(0,0){\white\line(1,0){0.001}}\put(1,1){\white\line(-1,0){0.001}}\put(0,0){\begin{pspicture}(0,0)(1,1) \begin{psclip}{\psframe[linestyle=none](0,0)(1,1)} \pspolygon[linewidth=0pt,fillstyle=solid,fillcolor=lightgray,linecolor=white](0.712334, 1.783083)(0.342881, 1.536781)(0.589183, 0.921026)(0.958636, 1.167328) \pspolygon[linewidth=0pt,fillstyle=solid,fillcolor=lightgray,linecolor=white](-0.026571, 1.290479)(-0.396024, 1.044177)(-0.149722, 0.428423)(0.219731, 0.674724) \pspolygon[linewidth=0pt,fillstyle=solid,fillcolor=lightgray,linecolor=white](0.589183, 0.921026)(0.219731, 0.674724)(0.466032, 0.058970)(0.835485, 0.305272) \pspolygon[linewidth=0pt,fillstyle=solid,fillcolor=lightgray,linecolor=white](-0.149722, 0.428423)(-0.519175, 0.182121)(-0.272873, -0.433634)(0.096580, -0.187332) \pspolygon[linewidth=0pt,fillstyle=solid,fillcolor=lightgray,linecolor=white](1.204938, 0.551573)(0.835485, 0.305272)(1.081787, -0.310483)(1.451240, -0.064181) \pspolygon[linewidth=0pt,fillstyle=solid,fillcolor=lightgray,linecolor=white](0.466032, 0.058970)(0.096580, -0.187332)(0.342881, -0.803087)(0.712334, -0.556785) \psline[linewidth=0.333pt,linecolor=darkgray](0.557594, 1.000000)(0.589183, 0.921026) \psline[linewidth=0.333pt,linecolor=darkgray](0.089620, 1.000000)(0.219731, 0.674724) \psline[linewidth=0.333pt,linecolor=darkgray](0.707644, 1.000000)(0.589183, 0.921026) \psline[linewidth=0.333pt,linecolor=darkgray](0.589183, 0.921026)(0.835485, 0.305272) \psline[linewidth=0.333pt,linecolor=darkgray](0.589183, 0.921026)(0.219731, 0.674724) \psline[linewidth=0.333pt,linecolor=darkgray](0.219731, 0.674724)(0.466032, 0.058970) \psline[linewidth=0.333pt,linecolor=darkgray](0.219731, 0.674724)(0.000000, 0.528237) \psline[linewidth=0.333pt,linecolor=darkgray](0.000000, 0.054117)(0.021647, 0.000000) \psline[linewidth=0.333pt,linecolor=darkgray](1.000000, 0.414948)(0.835485, 0.305272) \psline[linewidth=0.333pt,linecolor=darkgray](0.835485, 0.305272)(0.957594, 0.000000) \psline[linewidth=0.333pt,linecolor=darkgray](0.835485, 0.305272)(0.466032, 0.058970) \psline[linewidth=0.333pt,linecolor=darkgray](0.466032, 0.058970)(0.489620, 0.000000) \psline[linewidth=0.333pt,linecolor=darkgray](0.466032, 0.058970)(0.377578, 0.000000) \pscircle(0.466032, 0.921026){0.01} \pscircle(0.096580, 0.674724){0.01} \pscircle(0.096580, 0.921026){0.01} \pscircle(0.712334, 0.674724){0.01} \pscircle(0.712334, 0.797875){0.01} \pscircle(0.958636, 0.674724){0.01} \pscircle(0.835485, 0.797875){0.01} \pscircle(0.342881, 0.428423){0.01} \pscircle(0.342881, 0.551573){0.01} \pscircle(0.589183, 0.428423){0.01} \pscircle(0.466032, 0.551573){0.01} \pscircle(0.712334, 0.305272){0.01} \pscircle(0.589183, 0.797875){0.01} \pscircle(0.589183, 0.921026){0.01} \pscircle(0.589183, 0.921026){0.01} \pscircle(0.589183, 0.921026){0.013} \pscircle(0.589183, 0.921026){0.016} \pscircle(0.712334, 0.551573){0.01} \pscircle(0.219731, 0.182121){0.01} \pscircle(0.096580, 0.305272){0.01} \pscircle(0.342881, 0.058970){0.01} \pscircle(0.219731, 0.551573){0.01} \pscircle(0.219731, 0.674724){0.01} \pscircle(0.219731, 0.674724){0.01} \pscircle(0.219731, 0.674724){0.013} \pscircle(0.219731, 0.674724){0.016} \pscircle(0.342881, 0.305272){0.01} \pscircle(0.958636, 0.058970){0.01} \pscircle(0.958636, 0.182121){0.01} \pscircle(0.835485, 0.182121){0.01} \pscircle(0.835485, 0.305272){0.01} \pscircle(0.835485, 0.305272){0.01} \pscircle(0.835485, 0.305272){0.013} \pscircle(0.835485, 0.305272){0.016} \pscircle(0.466032, 0.058970){0.01} \pscircle(0.466032, 0.058970){0.01} \pscircle(0.466032, 0.058970){0.013} \pscircle(0.466032, 0.058970){0.016} \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.466032, 1.290479)(0.546080, 1.111910)(0.589183, 0.921026) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.096580, 1.044177)(0.176628, 0.865608)(0.219731, 0.674724) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.342881, 1.044177)(0.472190, 0.994917)(0.589183, 0.921026) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.466032, 0.921026)(0.527608, 0.927184)(0.589183, 0.921026) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.466032, 1.167328)(0.539923, 1.050335)(0.589183, 0.921026) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(-0.026571, 0.797875)(0.102737, 0.748615)(0.219731, 0.674724) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.096580, 0.674724)(0.158155, 0.680882)(0.219731, 0.674724) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.096580, 0.921026)(0.170470, 0.804033)(0.219731, 0.674724) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.712334, 0.674724)(0.792382, 0.496156)(0.835485, 0.305272) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.712334, 0.797875)(0.773910, 0.804033)(0.835485, 0.797875) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.958636, 0.674724)(1.087944, 0.625464)(1.204938, 0.551573) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.835485, 0.797875)(0.903218, 0.742457)(0.958636, 0.674724) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.342881, 0.428423)(0.422930, 0.249854)(0.466032, 0.058970) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.342881, 0.551573)(0.404457, 0.557731)(0.466032, 0.551573) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.589183, 0.428423)(0.718492, 0.379162)(0.835485, 0.305272) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.466032, 0.551573)(0.533765, 0.496156)(0.589183, 0.428423) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.712334, 0.305272)(0.773910, 0.311429)(0.835485, 0.305272) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.589183, 0.797875)(0.650759, 0.804033)(0.712334, 0.797875) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.589183, 0.921026)(0.835485, 0.945656)(1.081787, 0.921026) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.712334, 0.551573)(0.786225, 0.434580)(0.835485, 0.305272) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(-0.026571, 0.182121)(0.053477, 0.003552)(0.096580, -0.187332) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(-0.026571, 0.305272)(0.035004, 0.311429)(0.096580, 0.305272) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.219731, 0.182121)(0.349039, 0.132860)(0.466032, 0.058970) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.096580, 0.305272)(0.164313, 0.249854)(0.219731, 0.182121) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.342881, 0.058970)(0.404457, 0.065127)(0.466032, 0.058970) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.219731, 0.551573)(0.281306, 0.557731)(0.342881, 0.551573) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.219731, 0.674724)(0.466032, 0.699355)(0.712334, 0.674724) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.342881, 0.305272)(0.416772, 0.188278)(0.466032, 0.058970) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(-0.149722, 0.428423)(0.096580, 0.453053)(0.342881, 0.428423) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(-0.026571, 0.058970)(0.047319, -0.058024)(0.096580, -0.187332) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.958636, 0.058970)(1.038684, -0.119599)(1.081787, -0.310483) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.958636, 0.182121)(1.020211, 0.188278)(1.081787, 0.182121) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.835485, 0.182121)(0.897061, 0.188278)(0.958636, 0.182121) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.835485, 0.305272)(1.081787, 0.329902)(1.328089, 0.305272) \pscurve[arrowsize=1.5pt 3,linecolor=darkred]{->}(0.466032, 0.058970)(0.712334, 0.083600)(0.958636, 0.058970) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.466032, 0.921026)(0.398299, 0.976444)(0.342881, 1.044177) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.096580, 0.674724)(0.028847, 0.730142)(-0.026571, 0.797875) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.096580, 0.921026)(-0.155880, 0.957971)(-0.396024, 1.044177) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.712334, 0.674724)(0.644601, 0.730142)(0.589183, 0.797875) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.712334, 0.797875)(0.644601, 0.853293)(0.589183, 0.921026) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.958636, 0.674724)(0.761595, 0.779403)(0.589183, 0.921026) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.835485, 0.797875)(0.706177, 0.847136)(0.589183, 0.921026) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(1.081787, 0.551573)(1.014054, 0.606991)(0.958636, 0.674724) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(1.081787, 0.797875)(0.829328, 0.834821)(0.589183, 0.921026) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.342881, 0.428423)(0.275148, 0.483840)(0.219731, 0.551573) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.342881, 0.551573)(0.275148, 0.606991)(0.219731, 0.674724) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.589183, 0.428423)(0.392142, 0.533101)(0.219731, 0.674724) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.466032, 0.551573)(0.336724, 0.600834)(0.219731, 0.674724) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.712334, 0.305272)(0.644601, 0.360690)(0.589183, 0.428423) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.589183, 0.797875)(0.521450, 0.853293)(0.466032, 0.921026) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.589183, 0.921026)(0.502978, 1.161170)(0.466032, 1.413630) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.712334, 0.551573)(0.459875, 0.588519)(0.219731, 0.674724) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.219731, 0.182121)(0.022689, 0.286799)(-0.149722, 0.428423) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.096580, 0.305272)(-0.032729, 0.354532)(-0.149722, 0.428423) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.342881, 0.058970)(0.275148, 0.114388)(0.219731, 0.182121) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.219731, 0.551573)(0.151998, 0.606991)(0.096580, 0.674724) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.219731, 0.674724)(0.133525, 0.914869)(0.096580, 1.167328) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.342881, 0.305272)(0.090422, 0.342217)(-0.149722, 0.428423) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.958636, 0.058970)(0.890903, 0.114388)(0.835485, 0.182121) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.958636, 0.182121)(0.890903, 0.237539)(0.835485, 0.305272) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(1.204938, 0.058970)(1.007896, 0.163648)(0.835485, 0.305272) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(1.081787, 0.182121)(0.952478, 0.231381)(0.835485, 0.305272) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(1.328089, 0.182121)(1.075629, 0.219066)(0.835485, 0.305272) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.589183, -0.064181)(0.521450, -0.008763)(0.466032, 0.058970) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.835485, -0.187332)(0.638444, -0.082654)(0.466032, 0.058970) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.712334, -0.064181)(0.583026, -0.014921)(0.466032, 0.058970) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.835485, 0.182121)(0.767752, 0.237539)(0.712334, 0.305272) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.835485, 0.305272)(0.749279, 0.545416)(0.712334, 0.797875) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.958636, -0.064181)(0.706177, -0.027236)(0.466032, 0.058970) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.466032, -0.064181)(0.398299, -0.008763)(0.342881, 0.058970) \pscurve[arrowsize=1.5pt 3,linecolor=darkgreen]{->}(0.466032, 0.058970)(0.379827, 0.299114)(0.342881, 0.551573) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.712334, 1.290479)(0.669231, 1.099595)(0.589183, 0.921026) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.589183, 1.413630)(0.613813, 1.167328)(0.589183, 0.921026) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.835485, 1.167328)(0.724649, 1.031862)(0.589183, 0.921026) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.096580, 1.044177)(0.102737, 0.982602)(0.096580, 0.921026) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.342881, 1.044177)(0.299779, 0.853293)(0.219731, 0.674724) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.219731, 1.167328)(0.244361, 0.921026)(0.219731, 0.674724) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.466032, 0.921026)(0.355197, 0.785560)(0.219731, 0.674724) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.096580, 0.674724)(-0.014256, 0.539258)(-0.149722, 0.428423) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.096580, 0.921026)(0.041162, 0.853293)(-0.026571, 0.797875) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.712334, 0.674724)(0.718492, 0.613149)(0.712334, 0.551573) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.712334, 0.797875)(0.718492, 0.736300)(0.712334, 0.674724) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.958636, 0.674724)(0.915533, 0.483840)(0.835485, 0.305272) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.835485, 0.797875)(0.860115, 0.551573)(0.835485, 0.305272) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(1.081787, 0.551573)(0.970951, 0.416107)(0.835485, 0.305272) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.958636, 1.044177)(0.780067, 0.964129)(0.589183, 0.921026) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(1.081787, 0.797875)(1.026369, 0.730142)(0.958636, 0.674724) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.342881, 0.428423)(0.349039, 0.366847)(0.342881, 0.305272) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.342881, 0.551573)(0.349039, 0.489998)(0.342881, 0.428423) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.589183, 0.428423)(0.546080, 0.237539)(0.466032, 0.058970) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.466032, 0.551573)(0.490663, 0.305272)(0.466032, 0.058970) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.712334, 0.305272)(0.601498, 0.169806)(0.466032, 0.058970) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.589183, 0.797875)(0.410614, 0.717827)(0.219731, 0.674724) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.589183, 0.921026)(0.595341, 0.859451)(0.589183, 0.797875) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.712334, 0.551573)(0.656916, 0.483840)(0.589183, 0.428423) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.219731, 0.182121)(0.176628, -0.008763)(0.096580, -0.187332) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.096580, 0.305272)(0.121210, 0.058970)(0.096580, -0.187332) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.342881, 0.058970)(0.232046, -0.076496)(0.096580, -0.187332) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.219731, 0.551573)(0.041162, 0.471525)(-0.149722, 0.428423) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.219731, 0.674724)(0.225888, 0.613149)(0.219731, 0.551573) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.342881, 0.305272)(0.287464, 0.237539)(0.219731, 0.182121) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.958636, 0.058970)(0.964794, -0.002606)(0.958636, -0.064181) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.958636, 0.182121)(0.964794, 0.120545)(0.958636, 0.058970) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(1.204938, 0.428423)(1.026369, 0.348374)(0.835485, 0.305272) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.835485, 0.182121)(0.656916, 0.102073)(0.466032, 0.058970) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.835485, 0.305272)(0.841643, 0.243696)(0.835485, 0.182121) \pscurve[arrowsize=1.5pt 3,linecolor=darkblue]{->}(0.466032, 0.058970)(0.472190, -0.002606)(0.466032, -0.064181) \end{psclip} \end{pspicture}}\end{picture}

Unfortunately, the second example is too large for WordPress.

]]> http://math.fontein.de/2009/07/21/obtaining-infrastructures-from-global-fields/feed/ 0 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.

□

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.

□

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}.

□

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 Interpreting One-dimensional Infrastructures as Groups: f-Representations. http://math.fontein.de/2009/07/20/interpreting-one-dimensional-infrastructures-as-groups-f-representations/ http://math.fontein.de/2009/07/20/interpreting-one-dimensional-infrastructures-as-groups-f-representations/#comments Mon, 20 Jul 2009 03:46:16 +0000 Felix Fontein http://math.fontein.de/?p=109 Let (X, d) be a one-dimensional infrastructure of circumference R > 0. We have seen that we obtain two operations \bs : X \to X and \gs : X \times X \to X together with a reduction map red : \R/R\Z \to X, and we have \gs(x, x') = red(d(x) + d(x')) for x, x' \in X. This gives us a binary operation which is in general not associative. For several reasons, it would be interesting to embed the infrastructure into a group. An obvious choice for such a group would be \R/R\Z, as we can identify X as a subset of \R/R\Z by identifying it with d(X). Such an interpretation of the infrastructure as part of a “circular group” has first been considered by Hendrik Lenstra in his 1980 paper “On the computation of regulators and class numbers of quadratic fields”.

The idea is to consider pairs (x, f) \in X \times \R; the map d : X \times \R \to \R/R\Z, (x, f) \mapsto d(x) + f is obviously surjective, but far from being injective. Hence, it would be a good idea to restrict X \times \R to a subset which makes this map both injective and surjective. We want this subset to contain X \times \{ 0 \}; note that d|_{X \times \{ 0 \}} : X \times \{ 0 \} \to \R/R\Z is injective and essentially equals d : X \to \R/R\Z by the identification X \leftrightarrow X \times \{ 0 \}, x \mapsto (x, 0). We obtain the following definition:

Definition.
A set of f-representations of (X, d) is a set \fRep \subseteq X \times \R such that d|_{\fRep} : \fRep \to \R/R\Z is bijective and such that X \times \{ 0 \} \subseteq \fRep.

But how to obtain such a subset? In fact, we already have all ingredients ready, as the following proposition shows, by relating f-representations to reduction maps:

Theorem.
Let (X, d) be a one-dimensional infrastructure. Let A be the set of reduction maps and B be the set of f-representations for (X, d). The map \displaystyle \Phi : A \to B, \quad red \mapsto \{ (x, f) \in X \times \R \mid red(d(x) + f) = x \} is a bijection, with its inverse given by \displaystyle \Psi : B \to A, \quad \fRep \mapsto \pi_1 \circ (d|_{\fRep})^{-1}, where \pi_1 : X \times \R \to X, (x, f) \mapsto x is the projection onto the first component.
□

Now fix one corresponding choice of red and \fRep, and let \psi : \fRep \to \R/R\Z, (x, f) \mapsto d(x, f) = d(x) + f be the associated bijection. Then \displaystyle \gs(x, x') = red(d(x) + d(x')) = \pi_1(\psi^{-1}(\psi(x, 0) + \psi(x', 0))) for all x, x' \in X with \pi_1 : (x, f) \mapsto x being the projection. Moreover, using the bijection \psi, \fRep becomes a group which we will write additively by (x, f) + (x', f') := \psi^{-1}(\psi(x, f) + \psi(x', f')), which extends the giant steps.

Finally, let red be the map we originally defined for infrastructures, i.e. red(r) = d^{-1}(r - \min\{ f \in \R \mid f \ge 0, \; r - f \in d(X) \}) for r \in \R/R\Z. In that case, one can compute the group operation on \fRep without having to evalute \psi^{-1} or d, with only computing baby and giant steps and relative distances, i.e. d(\bs(x)) - d(x) \ge 0 resp. d(x) + d(x') - d(\gs(x, x')) \ge 0 for x, x' \in X:

Theorem.
Let D_{\min} := \min\{ d(\bs(x)) - d(x) \mid x \in X \} and D_{\max} := \max\{ d(\bs(x)) - d(x) \mid x \in X \}, and let (x, f), (x', f') \in \fRep. Then (x, f) + (x', f') can be computed with one giant step computation and at most \ceil{\frac{3 D_{\max}}{D_{\min}}} baby step computations as follows:
  1. Compute x'' := \gs(x, x') and f'' := f + f' + ( d(x) + d(x') - d(\gs(x, x')) ).
  2. Compute x''' := \bs(x'') and f''' := f'' - ( d(x''') - d(x'') ).
  3. If f''' \ge 0, set (x'', f'') := (x''', f''') and go to step 2.
  4. Return the pair (x'', f'').
□

In the case of infrastructures obtained from global fields, we are also able to describe the group operation in \fRep without having to evaluate \psi^{-1} or d. In fact, evaluating \psi^{-1} or d is essentially the discrete logarithm problem, for which so far no general polynomial methods for solving it exist in the case of global fields.

]]> http://math.fontein.de/2009/07/20/interpreting-one-dimensional-infrastructures-as-groups-f-representations/feed/ 0 One-dimensional Infrastructures. http://math.fontein.de/2009/07/20/one-dimensional-infrastructures/ http://math.fontein.de/2009/07/20/one-dimensional-infrastructures/#comments Mon, 20 Jul 2009 03:45:16 +0000 Felix Fontein http://math.fontein.de/?p=100 One-dimensional infrastructures first appeared in the 1970′s in Daniel Shanks‘ work on real quadratic number fields \Q(\sqrt{D}), D > 1 a squarefree integer, when he tried to fasten regulator computations. The previous algorithms used continued fraction expansion to obtain the regulator in \calO(D^{1/2 + \varepsilon}) binary operation, \varepsilon > 0 arbitrary. Shanks found out that one can obtain a multiplication like operation, which he dubbed giant steps, as opposed to the baby steps taken by one step in the continued fraction expansion. He described a baby step-giant step method to compute the regulator in \calO(D^{1/4 + \varepsilon}) binary operations, requiring \calO(D^{1/4 + \varepsilon}) bytes of storage. His methods were analysed, written up more clearly and extended by various people, including Hendrik Lenstra, Hugh Williams, Johannes Buchmann, Rene Schoof, and many others. Extensions of the method to function fields exist as well, most notably due to the work of Andreas Stein and Renate Scheidler.

I begin with giving an abstract definition of a one-dimensional infrastructure.

Definition (One-dimensional infrastructure).
Let R > 0 be a real number. A one-dimensional infrastructure of circumference R is a pair (X, d), where X \neq \emptyset is a finite set and d : X \to \R/R\Z is an injective map.

If you interpret \R/R\Z as a circle of circumference R (think of it as folding up the real line, such that two numbers whose difference is an integer multiple of R are identified), a one-dimensional infrastructure can be seen as a circle with a finite number of dots on it. The map d gives the distance between 0 \in \R/R\Z and some element x \in X on the circle, whence d is called the distance map.

Now one can define two operations on a one-dimensional infrastructure. Due to Shanks’ nomenclature, these are called baby steps and giant steps. To define a baby step, let x \in X. Then consider the set F_x := \{ f \in \R \mid f > 0, \; d(x) + f \in d(X) \}. It is non-empty as R \in F_x and bounded from below. Moreover, it is discrete as X is finite; therefore, f := \min F_x exists and d(x) + f \in d(X), say d(x) + f = d(x') with x' \in X. In that case, we define \bs(x) := x'. This gives a bijective map \bs : X \to X which, in case \abs{X} > 1, has no fixed points. If \R/R\Z is interpreted as a circle and X identified with d(X), \bs will send each point to the “next one” in positive direction on the circle.

To define giant steps, let x, x' \in X. For that, note that \R/R\Z is naturally a group, whence we can add d(x) and d(x'). Now d(x) + d(x') \in \R/R\Z, but in general d(x) + d(x') \not\in d(X). But we can use a similar trick as in the baby step case: we jump back to the previous point of d(X). For that, define F_{x,x'} := \{ f \in \R \mid f \ge 0, \; d(x) + d(x') - f \in d(X) \}. It is bounded from above, non-empty and discrete, whence f := \max F_{x,x'} exists with d(x) + d(x') - f' \in d(X), say d(x) + d(x') - f = d(y) for y \in X; then we define \gs(x, x') := y. This gives a binary operation \gs : X \times X \to X which is in general not associative.

But even though, we have \displaystyle d(\gs(x, x')) \approx d(x) + d(x') in general, assuming that D := \max\{ d(\bs(x)) - d(x) \mid x \in X \} is small (here, we identify d(\bs(x)) - d(x) \in \R/R\Z with its smallest non-negative representant). More precisely, we have d(\gs(x, x')) + f = d(x) + d(x') with 0 \le f < D, whence the giant step operation is “almost” associative.

Finite Cyclic Groups as One-dimensional Infrastructures.

Let G = \ggen{g} be a finite cyclic group of order R. For h \in G, one can write h = g^n with n \in \Z; note that n = \log_g h \in \Z/R\Z is the discrete logarithm of h with respect to g. Hence, we get the isomorphism G \cong \Z/R\Z induced by \log_g : G \to \Z/R\Z. As \Z/R\Z \subseteq \R/R\Z, we get the injective map d : G \to \R/R\Z, h \mapsto \log_g h, turning (G, d) into a one-dimensional infrastructure.

Let h, h' \in G; then we get \bs(h) = g h and \gs(h, h') = h h', i.e. baby steps are multiplications by the generator g and the giant steps equals the group operation. In particular, this provides an example for giant steps being associative.

Therefore, one-dimensional infrastructures can be seen as generalizations of finite cyclic groups.

Remarks.

Finally, we want to sketch some ideas, which will allow generalizing infrastructures to higher dimensions. For that, let (X, d) be a one-dimensional infrastructure. First, define the map red : \R/R\Z \to X as follows. For r \in \R/R\Z, define F_r := \{ f \in \R \mid f \ge 0, \; r - f \in d(X) \}. Again, F_r is non-empty, bounded from below and discrete, whence f := \min F_r exists and r - f \in d(X), say r - f = d(x) for some x \in X. Define red(r) := x. Then red satisfies red \circ d = \id_X and \gs(x, x') = red(d(x) + d(x')) for all x, x' \in X.

If red' : \R/R\Z \to X would be any other map satisfying red' \circ d = \id_X, one would obtain another giant step function \gs' : X \times X \to X, (x, x') \mapsto red'(d(x) + d(x')). In case X comes from a finite cyclic group, as above, \gs' would again be the group operation. If this is not the case, \gs and \gs' could be two distinct binary operations on X. If red' satisfies d(red'(r)) \approx r for all r \in \R/R\Z, we would also have \displaystyle d(\gs'(x, x')) \approx d(x) + d(x') \text{ for all } x, x' \in X.

This shows that our choice of red is rather random; we could also define red(r) = d^{-1}(d(x) + f) with f = \min \{ f \in \R \mid f \ge 0, \; r + f \in d(X) \}, or chose f such that \abs{f} = \min\{ \abs{f} \mid r + f \in d(X) \}, with some additional condition to rule out ties. Any other arbitrary choice of red is also possible, as long as red \circ d = \id_X is satisfied. We will later see that our definition of red is exactly the one we obtain in a canonical way if we obtain infrastructures from global fields of unit rank one. We call such maps reduction maps.

]]> http://math.fontein.de/2009/07/20/one-dimensional-infrastructures/feed/ 0