## Minkowski’s Lattice Point Theorem

We present a proof of a famous theorem of Hermann Minkowski which spawned a beautiful branch of number theory known as the Geometry of Numbers. This theorem gives a condition on the volume of a centrally symmetric convex body in $n$-dimensional Euclidean space to contain at least one lattice point except the trivial point $\Vec{0}$.

Definitions and Notations

Definition 1 In an $n$-dimensional vector space $V$ over the field $\mathbb{R}$, a lattice is a free abelian subgroup of $V$ of the following form

$\Gamma=v_1\mathbb{Z}+v_2\mathbb{Z}+\cdots+v_m\mathbb{Z}$

where $v_1,v_2,\ldots,v_m$ are linearly independent vectors in $V$. Then $\Gamma$ is a free abelian group of rank $m$. A lattice is said to be complete if $m=n$. Assuming $\Gamma$ to be complete, we define the fundamental mesh in $\Gamma$ with respect the the $v_i$ to be

$\displaystyle \Phi=\{x_1v_1+x_2v_2+\cdots+x_nv_n:x_i\in\mathbb{R},x_i\in[0,1),1\leq i\leq n\}$

The volume of the lattice $\Gamma$ is defined to be the volume of the fundamental mesh $\Phi$ denoted as $\mathrm{Vol}(\Phi)$, i.e. $\mathrm{Vol}(\Gamma)=\mathrm{Vol}(\Phi)$. There are a few equivalent expressions for this volume. One expression convenient for computations is given by the following relation

$\mathrm{Vol}(\Phi)=\sqrt{\det[\langle v_i,v_j\rangle]}$

where $\langle \cdot,\cdot\rangle$ is the usual inner-product (or a symmetric positive definite bilinear form) defined on the vector space $V$.

Definition 2 (Convex and Centrally Symmetric Body)

A subset $X$ of $V$ is said to be centrally symmetric if for any point $x\in X$ we have $-x\in X$, and the subset is convex if for any two distinct points $x,y$ in $X$ the line segment $\{tx+(1-t)y:t\in[0,1]\}$ is contained in $X$. For example, a ball is convex in $\mathbb{R}^3$ but a solid torus is not.

Minkowski’s Lattice Point Theorem

Theorem: Let $V$ be an $n$-dimensional Euclidean vector space ($\mathbb{R}$-vector space) and $\Gamma$ be a complete lattice of $V$. Let $X$ be a centrally symmetric convex subset of $V$ such that

$\mathrm{Vol}(X)>2^n\mathrm{Vol}(\Gamma)$

Then $X$ contains at least one point other than $\vec{0}$ in $\Gamma$.

Proof: Consider the dilation $\frac{1}{2}X=\left\{\frac{1}{2}x:x\in X\right\}$. For any $\gamma\in \Gamma$, consider the translated sets $\frac{1}{2}X+\gamma$. We will show that there exist $\gamma_1,\gamma_2 \in \Gamma$ such that $\gamma_1\neq\gamma_2$ and

$\left(\frac{1}{2}X+\gamma_1\right)\cap\left(\frac{1}{2}X+\gamma_2\right)\neq\emptyset$

For the sake of contradiction, assume the sets $\frac{1}{2}X+\gamma,\gamma\in\Gamma$ are pairwise disjoint. Then the intersections $\Phi\cap\left(\frac{1}{2}X+\gamma\right),\gamma\in\Gamma$ are also disjoint for a fundamental mesh $\Phi$. This gives us the following inequality involving volumes

$\mathrm{Vol}(\Phi)\geq\sum_{\gamma\in \Gamma}\mathrm{Vol}\left(\Phi\cap(\frac{1}{2}X+\gamma)\right)$

We know that translation preserves volumes, so we consider the following translations

$\Phi\cap\left(\frac{1}{2}X+\gamma\right)\longrightarrow(\Phi-\gamma)\cap\frac{1}{2}X$

Hence for each $\gamma\in \Gamma,$ we have

$\mathrm{Vol}\left(\Phi\cap(\frac{1}{2}X+\gamma)\right)=\mathrm{Vol}\left((\Phi-\gamma)\cap\frac{1}{2}X\right)$

We claim that the $\Phi-\gamma$ cover $V$ as $\gamma$ varies over $\Gamma$. Let $x\in V$. Since $\{v_1,v_2,\ldots,v_n\}$ is a set of $n$ linearly independent vectors in $V$ and $\mathrm{dim}(V)=n$, it follows that $\{v_1,v_2,\ldots,v_n\}$ is a basis of $V$. Then $\exists$ $\lambda_1,\lambda_2,\ldots,\lambda_n\in\mathbb{R}$ such that $x=\sum_{i=1}^{n}\lambda_iv_i$. We know that every real number $r$ can be written as $r=\lfloor r\rfloor+\{r\}$ where $\lfloor r\rfloor\in\mathbb{Z}$ and $\{r\}\in[0,1)$. Then

$x=\sum_{i=1}^{n}\lfloor\lambda_i\rfloor v_i+\sum_{i=1}^{n}\{\lambda_i\}v_i$

Note that $\sum_{i=1}^{n}\lfloor\lambda_i\rfloor v_i\in\Gamma$ and $\sum_{i=1}^{n}{\lambda_i}v_i\in\Phi$. Taking $\gamma'=-\sum_{i=1}^{n}\lfloor\lambda_i\rfloor v_i,$ we observe that $x\in\Phi-\gamma'$. Hence, $V\subset\bigcup_{\gamma\in\Gamma}(\Phi-\gamma)$. Since $(\Phi-\gamma)\subset V$ for all $\gamma\in\Gamma,$ we have $\bigcup_{\gamma\in\Gamma}(\Phi-\gamma)\subset V$. Hence $\bigcup_{\gamma\in\Gamma}(\Phi-\gamma)=V$, and so the sets $(\Phi-\gamma)\cap\frac{1}{2}X$ cover $\frac{1}{2}X$. Therefore, we finally have

$\mathrm{Vol}(\Phi)\geq\sum_{\gamma\in\Gamma}\mathrm{Vol}\left((\Phi-\gamma)\cap\frac{1}{2}X\right)=\mathrm{Vol}\left(\frac{1}{2}X\right)=\frac{1}{2^n}\mathrm{Vol}(X)$

which is a contradiction to our initial assumption since $\mathrm{Vol}(\Gamma)=\mathrm{Vol}(\Phi)$. Hence, we can choose $\gamma_1,\gamma_2\in\Gamma, \gamma_1\neq\gamma_2$ such that

$\left(\frac{1}{2}X+\gamma_1\right)\cap\left(\frac{1}{2}X+\gamma_2\right)\neq\emptyset$

Therefore, there exist $x_1,x_2\in X, x_1 \neq x_2$, such that

$\frac{1}{2}x_1+\gamma_1=\frac{1}{2}x_2+\gamma_2$

Since $X$ is centrally symmetric and convex, $-x_2\in X$ and thus $\gamma_0=\gamma_1-\gamma_2=\frac{1}{2}x_1-\frac{1}{2}x_2\in X$ Therefore $\gamma_0\neq\Vec{0}$ and $\gamma_0\in\Gamma\cap X$. This completes the proof of the theorem.

Remark

Since $\gamma_0\neq\Vec{0}$ we also have $-\gamma_0\in\Gamma\cap X$ and $-\gamma_0\neq\gamma_0$. This means $X$ contains at least two distinct lattice points. What else can you find hidden in the proof?

A Useful Corollary

Taking $V=\mathbb{R}^n$ and $\Gamma=\mathbb{Z}^n$, we observe that any convex centrally symmetric body in $\mathbb{R}^n$ of volume strictly bigger that $2^n$ contains at least one point with integer coordinates other than $\Vec{0}$.

Minkowski’s Theorem in Action

Now we present a number theoretic application of Minkowski’s theorem. The following result was proved by Axel Thue using the pigeonhole principle. We give a proof using Minkowski’s lattice point theorem.

Theorem: Primes of the form $4k+1$ can be expressed as a sum of two squares.

Proof. Let $p$ be a prime of the form $4k+1$. Then $-1$ is a quadratic residue modulo $p$ or, equivalently, there exists $a\in\mathbb{Z}$ such that $a^2+1\equiv0\pmod{p}$. Consider the two vectors $v_1=(p,0)$ and $v_2=(a,1)$ in $\mathbb{R}^2$. Let $\alpha v_1+\beta v_2=(0,0)$ for some $\alpha,\beta\in\mathbb{R}$. This gives us $p\alpha+a\beta=0,\beta=0$ and hence $\alpha=\beta=0$. Therefore $v_1,v_2$ are linearly independent. Then $\Gamma=v_1\mathbb{Z}+v_2\mathbb{Z}$ is a complete lattice in $\mathbb{R}^2$ with $\mathrm{Vol}(\Gamma)=p$.

Let $(x,y)\in\Gamma$. There exist $A,B\in\mathbb{Z}$ such that $(x,y)=Av_1+Bv_2$. This implies $x=Ap+Ba,y=B$. Hence,

$x^2+y^2=(Ap)^2+2ABpa+(Ba)^2+B^2\equiv B^2(a^2+1)\equiv0\pmod{p}$

Consider the open disc $D$ of radius $\sqrt{2p}$ centered at the origin $(0,0)$. We have, $\mathrm{Vol}(D)=\mathrm{area}(D)=\pi(\sqrt{2p})^2=2\pi p>4p=2^2\mathrm{Vol}(\Gamma)$

Thus, $D$ is convex and centrally symmetric. By Minkowski’s theorem, there exists a lattice point apart from the origin in $D$. Let this point be $(m,n)$. Then $0 and $p\mid (m^2+n^2)$ and hence $m^2+n^2=p$. We are done!

References

[1] Andreescu, T. and Dospinescu, G., 2008. Problems from the Book.

[2] Neukirch, J., 2013. Algebraic number theory (Vol. 322). Springer Science & Business Media.

[3] Cassels, J.W.S., 2012. An introduction to the geometry of numbers. Springer Science & Business Media.

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