## Minkowski's Convex Body Theorem

Suppose $S$ is a convex set in an "n-dimensional" vector space in a lattice $L$ which is symmetric about the origin. Then, Minkowski's Theorem states that if the volume of the set $V(S) > 2^nd(L)$, then $S$ must contain at least one lattice point other than the origin.

Before diving deeper, let's first clearly understand some terms used in the theorem..

### Convex

A set $C$ is said to be convex in $\mathbb{R}^n$ if for all points $a,b \in \mathbb{C}$, the segment joining the points $a$ and $b$ is completely contained in $C$. \newline

An example for a set in $\mathbb{R^2}$ is as follows.

A set $C$ is said to be symmetric about the origin if for every point $x \in C$, $-x \in C$. In other words, the reflection of every point in $C$ across the origin must also be in $C$. For example

### Lattice

A lattice is an array of points that differ in equal intervals in any dimension. The most widely used lattice is the well-known "integer lattice ($\mathbb{Z}^2$)". For example, non-integer lattices, are hexagonal and parallelogram lattices in the Euclidean Plane.

### Determinant of a lattice

The determinant of a lattice L is the volume of the smallest area enclosed by the lattice point. This is represented by $d(L)$. For example, the determinant of the lattices in the above will just be the area of "1 hexagon" and "1 parallelogram formed by 4 adjacent lattice points" respectively.

Now, we understand what the theorem says completely. Let's move on to building the foundation for our proof.

### Blichfeldt's Theorem

If R is a bounded set in $\mathbb{R}^2$ with area greater than 1, then R contains two distinct points $(x_1,y_1)$ and $(x_2,y_2)$ such that the point ${(x_2-x_1,y_2-y_1)}$ is an integer point in $\mathbb{R}^2$

Proof: Let $S = \{(x,y) | 0 \le x < 1 \text{and} 0 \le y <1\}$. For every point $a \in \mathbb{Z}^2$, define $S_a = S + a$ be the translation of $S$ along the line segment with endpoints $(0,0)$ and $a$. Note that $a$ will be the only integer point in $S_a$  Since, $R$ is bounded, we know that there will be a finite number of points $z$ such that $R_z = S_z \cap R$ is non empty. Let $R_z-z$ be the translation of $R_z$ back to $S$ along the line segment with endpoints $(0,0)$ and $z$. Since the translations are plane isometry, the area of the sets during translation will be preserved, thus, $A(R_z-z) = A(R_z)$. We have,
$$\sum_{z\in \mathbb{Z}^2}A(R_z-z) = \sum_{z \in \mathbb{Z}^2}A(R_z) = A(R) > 1$$
This means that that all the subsets $R_z$ are stacked one upon another after the translation. Since, the sum of area is greater than 1, we can say that, after the translation, there will be at least integer points $z_1$ and $z_2$ such that the areas of $(R_{z_1}-z_1)$ and $(R_{z_2}-z_2)$ overlap. That is, $(R_{z_1} - z_1) \cap (R_{z_2} - z_2) \neq \phi$. Define $r \in (R_{z_1} - z_1) \cap (R_{z_2} - z_2)$. Let $r+z_1 = r_{z_1} \in R_{z_1}$ and $r+z_2 = r_{z_2} \in R_{z_2}$ by definition. We have,
$$r_{z_1} - r_{z_2} = (r+z_1) - (r+z_2) = z_1-z_2 \in \mathbb{Z}$$
Thus, there are two distinct points in R $r_{z_1}$ and $r_{z_2}$, whose difference in co-ordinates is an integer point.

Remark: Note that this theorem is also true for "non-integer lattices", say $L$. The only difference would be that $R$ should have an area greater than $d(L)$. The proof will be similar to this.

This is an extremely helpful theorem which almost finishes the proof of "Minkowski's Theorem". Now, let's move on to prove a simpler version of "Minkowski's Convex Body Theorem".

### Simpler version

Let $R$ be a convex region in $\mathbb{R}^2$ that is symmetric about the origin and has an area greater than $2^2d(\mathbb{Z}^2) = 4$. Then $R$ contains an integer point other than the origin.

Proof: Define the set $R'= \left\{\frac{1}{2}x | x \in R\right\}$. Since $R'$ is just $R$ scaled down by a factor of $0.5$, it is also convex and symmetric about the origin. Moreover, $A(R') = \frac{1}{4} A(R) > 1$. By Blichfeldt's Theorem, there exists two distinct points $m,n \in R'$ such that $m-n$ is an integer point. Note that $2m, 2n \in R$. Since, $R$ is symmetric about the origin, $-2n \in R$. We know that $R$ is convex. Thus, every point on the line segment with endpoints $2m$ and $-2n$ lies inside $R$. Therefore, their midpoint also lies in $R$. We have
$$\frac{2m + (-2n)}{2} = m-n$$ should be in $R$. But since, $m-n$ is an integer point and $m$ and $n$ are distinct, the statement is proved.

Note that the symmetric condition yield that along with $m-n$, the point $-(m-n)$ must also lie in $R$. Thus, there are at least 3 integer points in $R$.

### Minkowski's Theorem for Arbitrary Lattices

For, arbitrary lattices, Minkowski's Theorem can be proved with the help of "Pick's Theorem". Although this will not be discussed in this blog, you can refer to its wonderful proof here

### Applications of Minkowski's Theorem

The two squares theorem: For a prime $p \equiv 1 \pmod 4$, we can always find some $a,b \in \mathbb{Z}$, such that $p = a^2 + b^2$

Proof: By, Fermat's Christmas Theorem, $(-1)$ is a quadratic residue modulo $p$. Thus, there exists some $b$ such that $q^2 \equiv -1 \pmod p$. Let $z_1 = (1,q)$ and $z_2 = (0,p)$. Define a lattice $L$ such that $z_1$, $z_2$, $(0,0)$ and $(z_1-z_2)$ are all lattice points. Note that these four points form a parallelogram with area $p$. Thus, $d(L) = p$. \newline

Now consider a set $C = \left\{(x,y) | x^2 + y^2 < 2p\right\}$ in $\mathbb{R}^2$. We have,
$$V(C) = \pi\sqrt{2p}^2 = 2\pi p > 4p = 2^2d(L)$$
By "Minkwoski's Theorem", we know that $C$ contains some point $(a,b) \in L/\{0\}$. Since, $(a,b)$ can be written as a linear combination of $z_1$ and $z_2$, let $(a,b) = az_1 + bz_2 = (a, aq + bp) \in \mathbb{R}^2$, which implies
$$a^2 + b^2 = a^2 + (aq + bp)^2 \equiv (q^2 + 1)a^2 \equiv 0 \pmod p$$
Since, $0 < a^2 + b^2 < 2p$, we must have $a^2 + b^2 = p$. Hence, proved.

Post by Akshat Pandey.

About the guest blogger: Hi, I am Akshat Pandey, an 11th grader. I am an olympiad math freak who also enjoys playing the piano, badminton and table tennis.

Post modified and set up by Sunaina Pati. This post is part of Akshat's lecture notes for Sophie fellowship's WeMP.

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