proof of Minkowski’s theorem

Theorem 1.

Let $\mathrm{L}$ be an arbitrary lattice in ${\mathrm{R}}^n$ and let $\mathrm{\Delta }$ be the area of a fundamental parallelepiped. Any convex region $\mathrm{K}$ symmetrical about the origin with $\mu \mathit{}\mathrm{(}\mathrm{K}\mathrm{)}\mathrm{>}{\mathrm{2}}^n\mathit{}\mathrm{\Delta }$ contains a point of the lattice $\mathrm{L}$ other than the origin.

Proof. Let $D$ be any fundamental parallelepiped. Then obviously

$${ℝ}^n=\coprod _{x\in ℒ}(D+x)$$

(where $\coprod$ means disjoint union) and thus

$$\frac{1}{2}𝔎=\coprod _{x\in ℒ}\left(\frac{1}{2}𝔎\cap (D+x)\right).$$

Now, note that

$$\frac{1}{2}𝔎\cap (D+x)=\left(\left(\frac{1}{2}𝔎-x\right)\cap D\right)-x$$

(draw a picture!) and thus, since measure is preserved by translation,

$$\mu \left(\frac{1}{2}𝔎\cap (D+x)\right)=\mu \left(\left(\frac{1}{2}𝔎-x\right)\cap D\right)$$

so that if all the $\frac{1}{2}𝔎-x$ are disjoint, we have

$$2^{-n}\mu (𝔎)=\mu \left(\frac{1}{2}𝔎\right)=\mu \left(\coprod _{x\in ℒ}\left(\frac{1}{2}𝔎\cap (D+x)\right)\right)=\sum _{x\in ℒ}\mu \left(\left(\frac{1}{2}𝔎-x\right)\cap D\right)\le \mu (D)=\mathrm{\Delta }$$

which is a contradiction. Thus there must exist $x\ne y\in ℒ$ and $c_1,c_2\in 𝔎$ such that

$$\frac{1}{2}{c}_1-x=\frac{1}{2}{c}_2-y.$$

Thus $x-y=\frac{1}{2}(c_2-c_1)\in 𝔎$ since $𝔎$ is convex and centrally symmetric, and certainly $x-y\in ℒ$, so we have found a nonzero element of $𝔎\cap \mathrm{\Lambda }$.

Corollary 1.

Let $\mathrm{L}$ be an arbitrary lattice in ${\mathrm{R}}^n$ and let $\mathrm{\Delta }$ be the area of a fundamental parallelepiped. Any compact convex region $\mathrm{K}$ symmetrical about the origin with $\mu \mathit{}\mathrm{(}\mathrm{K}\mathrm{)}\mathrm{\ge }{\mathrm{2}}^n\mathit{}\mathrm{\Delta }$ contains a point of the lattice $\mathrm{L}$ other than the origin.

Note that this corollary requires that $𝔎$ be compact in addition to being convex and centrally symmetric, but slightly relaxes the volume condition on $𝔎$.

Proof. Apply the previous case to $C_n=\left(1+\frac{1}{n}\right)C$, i.e. dilate $C$. This gives a sequence of points $x_1,x_2,\mathrm{\dots },x_n,\mathrm{\dots }$ with $x_i\in \mathrm{\Lambda }\cap C_i-\{0\}$. But $\mathrm{\Lambda }$ is discrete, so there must be a subsequence constant at a nonzero element

$$x\in \mathrm{\Lambda }\bigcap \left(\bigcap _{i=1}^{\mathrm{\infty }}{C}_i-\{0\}\right)=\mathrm{\Lambda }\cap \overline{C}-\{0\}.$$

Since $C$ is compact and thus closed, $x\in C$.

Title	proof of Minkowski’s theorem
Canonical name	ProofOfMinkowskisTheorem
Date of creation	2013-03-22 17:53:41
Last modified on	2013-03-22 17:53:41
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	5
Author	rm50 (10146)
Entry type	Proof
Classification	msc 11H06