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<m^2+n^2<(\sqrt{2p})^2=2p 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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