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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 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.

[edit] 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.^[1]

[edit] Example

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.

[edit] Proof

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 $f (p 1) = f (p 2)$ for some pair of points $p 1, p 2$ in S. Moreover, we know from the definition of f that $p 2 = p 1 + (2 i,2 j)$ for some integers i and j, where i and j are not both zero.

Then since S is symmetric about the origin, $- p 1$ is also a point in S. Since S is convex, the line segment between $- p 1$ and $p 2$ lies entirely in S, and in particular the midpoint of that segment lies in S. In other words,

$\frac{1}{2}\left(-p_1 + p_2\right) = \frac{1}{2}\left(-p_1 + p_1 + (2i, 2j)\right) = (i, j)$

lies in S. (i,j) is a lattice point, and is not the origin since i and j are not both zero, and so we have found the point we're looking for.

[edit] Applications

A corollary of this theorem is the fact that every class in the ideal class group of a number field K contains an integral ideal of norm not exceeding a certain bound, depending on K, called Minkowski's bound.

[edit] See also

[edit] Notes

^ 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.

[edit] References

Hazewinkel, Michiel, ed. (2001), "Minkowski theorem", Encyclopaedia of Mathematics, Kluwer Academic Publishers, ISBN 978-1556080104, http://eom.springer.de/m/m064090.htm
Minkowski's theorem on PlanetMath
Stevenhagen, Peter. Number Rings.

[0] 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.

[1]

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