Minkowski's theorem

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In mathematics, Minkowski's theorem is the statement that any convex set in Rn which is symmetric with respect to the origin and with volume greater than 2n d(L) contains a non-zero lattice point. The theorem was proved by Hermann Minkowski in 1889 and became the foundation of the branch of number theory called the geometry of numbers.



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.[1]


The simplest example of a lattice is the set Zn of all points with integer coefficients; its determinant is 1. For n = 2 the theorem claims that a convex figure in the 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, has area 4, but the only lattice point it contains is the origin. This observation generalizes to every dimension n.


The following argument proves Minkowski's theorem for the special case of L=Z2. It can be generalized to arbitrary lattices in arbitrary dimensions.

Consider the map f: S \to \mathbb{R}^2, (x,y) \mapsto (x \bmod 2, y \bmod 2). Intuitively, this map cuts the plane into 2 by 2 squares, then stacks the squares on top of each other. Clearly f(S) has area ≤ 4. Suppose f were injective. Then each of the stacked squares would be non-overlapping, so f would be area-preserving, and the area of f(S) would be greater than 4, since S has area greater than 4. That is not the case, so f(p1) = f(p2) for some pair of points p1,p2 in S. Moreover, we know from the definition of f that p2 = p1 + (2i,2j) for some integers i and j, where i and j are not both zero.

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