# proof of Minkowski’s theorem

###### Theorem 1.

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

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

 $\mathbb{R}^{n}=\coprod_{x\in\mathcal{L}}(D+x)$

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

 $\frac{1}{2}\mathfrak{K}=\coprod_{x\in\mathcal{L}}\left(\frac{1}{2}\mathfrak{K}% \cap(D+x)\right).$

Now, note that

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

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

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

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

 $2^{-n}\mu(\mathfrak{K})=\mu\left(\frac{1}{2}\mathfrak{K}\right)=\mu\left(% \coprod_{x\in\mathcal{L}}\left(\frac{1}{2}\mathfrak{K}\cap(D+x)\right)\right)=% \sum_{x\in\mathcal{L}}\mu\left(\left(\frac{1}{2}\mathfrak{K}-x\right)\cap D% \right)\leq\mu(D)=\Delta$

which is a contradiction. Thus there must exist $x\neq y\in\mathcal{L}$ and $c_{1},c_{2}\in\mathfrak{K}$ such that

 $\frac{1}{2}c_{1}-x=\frac{1}{2}c_{2}-y.$

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

###### Corollary 1.

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

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

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 $x_{1},x_{2},\ldots,x_{n},\ldots$ with $x_{i}\in\Lambda\cap C_{i}-\{0\}$. But $\Lambda$ is discrete, so there must be a subsequence constant at a nonzero element

 $x\in\Lambda\bigcap\left(\bigcap_{i=1}^{\infty}C_{i}-\{0\}\right)=\Lambda\cap% \overline{C}-\{0\}.$

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

Title proof of Minkowski’s theorem ProofOfMinkowskisTheorem 2013-03-22 17:53:41 2013-03-22 17:53:41 rm50 (10146) rm50 (10146) 5 rm50 (10146) Proof msc 11H06