# Minkowski's theorem

A set in R2 satisfying the hypotheses of Minkowski's theorem.

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.

## 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]

## 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.

## 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, which means the pieces of S cut out by the squares stack up in a non-overlapping way. Since 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 f is not injective, and $f(p_1) = f(p_2)$ for some pair of points $p_1, p_2$ in S. Moreover, we know from the definition of f that $p_2 = p_1 + (2i, 2j)$ 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.

## Applications

An application of this theorem is the result 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: the finiteness of the class number of an algebraic number field follows immediately.

Minkowski's theorem is also useful to prove Lagrange's four-square theorem, which states that every natural number can be written as the sum of the squares of four natural numbers.

## Notes

1. ^ 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.