# Satz von Minkowski Minkowski's theorem hide This article has multiple issues. Bitte helfen Sie mit, es zu verbessern oder diskutieren Sie diese Probleme auf der Diskussionsseite. (Erfahren Sie, wie und wann Sie diese Vorlagennachrichten entfernen können) Dieser Artikel enthält eine Liste von Referenzen, weiterführende Lektüre oder externe Links, Die Quellen bleiben jedoch unklar, da Inline-Zitate fehlen. (Februar 2017) This article cites its sources but does not provide page references. (September 2010) A set in ℝ2 satisfying the hypotheses of Minkowski's theorem.

In Mathematik, Minkowski's theorem is the statement that every convex set in {Anzeigestil mathbb {R} ^{n}} which is symmetric with respect to the origin and which has volume greater than {Anzeigestil 2 ^{n}} contains a non-zero integer point (meaning a point in {Anzeigestil mathbb {Z} ^{n}} that is not the origin). The theorem was proved by Hermann Minkowski in 1889 and became the foundation of the branch of number theory called the geometry of numbers. It can be extended from the integers to any lattice {Anzeigestil L} and to any symmetric convex set with volume greater than {Anzeigestil 2 ^{n},d(L)} , wo {Anzeigestil d(L)} denotes the covolume of the lattice (the absolute value of the determinant of any of its bases).

Inhalt 1 Formulierung 2 Beispiel 3 Nachweisen 4 Anwendungen 4.1 Bounding the shortest vector 4.2 Applications to number theory 4.2.1 Primes that are sums of two squares 4.2.2 Lagrange's four-square theorem 4.2.3 Dirichlet's theorem on simultaneous rational approximation 4.2.4 Algebraic number theory 5 Complexity theory 6 Siehe auch 7 Weiterlesen 8 Externe Links 9 References Formulation Suppose that L is a lattice of determinant d(L) in the n-dimensional real vector space ℝn and S is a convex subset of ℝn that is symmetric with respect to the origin, meaning that if x is in S then −x is also in S. Minkowski's theorem states that if the volume of S is strictly greater than 2n d(L), then S must contain at least one lattice point other than the origin. (Since the set S is symmetric, it would then contain at least three lattice points: the origin 0 and a pair of points ± x, where x ∈ L 0.) Example The simplest example of a lattice is the integer lattice ℤn of all points with integer coefficients; its determinant is 1. Für n = 2, the theorem claims that a convex figure in the Euclidean plane symmetric about the origin and with area greater than 4 encloses at least one lattice point in addition to the origin. The area bound is sharp: if S is the interior of the square with vertices (±1, ±1) then S is symmetric and convex, and has area 4, but the only lattice point it contains is the origin. This example, showing that the bound of the theorem is sharp, generalizes to hypercubes in every dimension n.

Proof The following argument proves Minkowski's theorem for the specific case of L = ℤ2.

Proof of the {textstyle mathbb {Z} ^{2}} Fall: Consider the map {Anzeigestil f:Sto mathbb {R} ^{2}/2L,Quad (x,j)mapsto (x{in gewisser Weise {2}},j{in gewisser Weise {2}})} Intuitiv, this map cuts the plane into 2 durch 2 squares, then stacks the squares on top of each other. Clearly f (S) has area less than or equal to 4, because this set lies within a 2 durch 2 square. Assume for a contradiction that f could be injective, which means the pieces of S cut out by the squares stack up in a non-overlapping way. Because f is locally area-preserving, this non-overlapping property would make it area-preserving for all of S, so the area of f (S) would be the same as that of S, which is greater than 4. That is not the case, so the assumption must be false: f is not injective, meaning that there exist at least two distinct points p1, p2 in S that are mapped by f to the same point: f (p1) = f (p2).

Because of the way f was defined, the only way that f (p1) can equal f (p2) is for p2 to equal p1 + (2ich, 2j) for some integers i and j, not both zero. Das ist, the coordinates of the two points differ by two even integers. Since S is symmetric about the origin, −p1 is also a point in S. Since S is convex, the line segment between −p1 and p2 lies entirely in S, and in particular the midpoint of that segment lies in S. Mit anderen Worten, {Anzeigestil {tfrac {1}{2}}links(-p_{1}+p_{2}Rechts)={tfrac {1}{2}}links(-p_{1}+p_{1}+(2ich,2j)Rechts)=(ich,j)} is a point in S. But this point (ich, j) is an integer point, and is not the origin since i and j are not both zero. Deswegen, S contains a nonzero integer point.

Bemerkungen: The argument above proves the theorem that any set of volume {textstyle >!das(L)} contains two distinct points that differ by a lattice vector. This is a special case of Blichfeldt's theorem. The argument above highlights that the term {textstyle 2^{n}das(L)} is the covolume of the lattice {textstyle 2L} . To obtain a proof for general lattices, it suffices to prove Minkowski's theorem only for {textstyle mathbb {Z} ^{n}} ; this is because every full-rank lattice can be written as {textstyle Bmathbb {Z} ^{n}} for some linear transformation {textstyle B} , and the properties of being convex and symmetric about the origin are preserved by linear transformations, while the covolume of {textstyle Bmathbb {Z} ^{n}} ist {textstyle |!das(B)|} and volume of a body scales by exactly {textstyle {frac {1}{das(B)}}} under an application of {textstyle B^{-1}} . Applications Bounding the shortest vector Minkowski's theorem gives an upper bound for the length of the shortest nonzero vector. This result has applications in lattice cryptography and number theory.

Satz (Minkowski's bound on the shortest vector): Lassen {textstyle L} be a lattice. Then there is a {textstyle xin Lsetminus {0}} mit {textstyle |x|_{unendlich }leq left|das(L)Rechts|^{1/n}} . Im Speziellen, by the standard comparison between {textstyle l_{2}} und {textstyle l_{unendlich }} norms, {textstyle |x|_{2}leq {quadrat {n}},links|das(L)Rechts|^{1/n}} .

Proof Let {textstyle l=min{|x|_{unendlich }:xin Lsetminus {0}}} , und einstellen {textstyle C={j:|j|_{unendlich }2^{n}|d(L)|} , dann {textstyle C} contains a non-zero lattice point, which is a contradiction. Daher {textstyle lleq |d(L)|^{1/n}} . Q.E.D.

Bemerkungen: The constant in the {textstyle L^{2}} bound can be improved, for instance by taking the open ball of radius {textstyle

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