We can derive a proof of the minkowski convex body theorem starting from Blichfeldt’s theorem.

§ Blichfeldt's theorem

This theorem allows us to prove that a set of large-enough-size in any lattice will have two points such that their difference lies in the lattice. Formally, we have:

  1. A lattice L(B)≡{Bx:x∈Zn}L(B) \equiv \{ Bx : x \in \mathbb Z^n \} for some basis B∈RnB \in \mathbb R^n. The lattice LL is spanned by integer linear combinations of rows of BB.
  2. A body S⊆RnS \subseteq R^n which need not be convex! , which has volume greater than det⁡(B)\det(B). Recall that for a lattice L(B)L(B), the volume of a fundamental unit / fundamental parallelopiped is det(B)det(B).

Blichfeldt's theorem tells us that there exists two points x1,x2∈Sx_1, x_2 \in S such that x1−x2∈Lx_1 - x_2 \in L.

§ Proof

The idea is to:

  1. Chop up sections of SS across all translates of the fundamental parallelopiped that have non-empty intersections with SS back to the origin. This makes all of them overlap with the fundamental parallelopiped with the origin.
  2. Since SS has volume great that det⁡(B)\det(B), but the fundamental paralellopiped only has volume det⁡(B)\det(B), points from two different parallelograms must overlap.
  3. "Undo" the translation to find two points which are of the form x1=l1+δx_1 = l_1 + \delta, x2=l2+δx_2 = l_2 + \delta. they must have the same δ\delta since they overlapped when they were laid on the fundamental paralellopiped. Also notice that l1≠l2l_1 \neq l_2since they came from two different parallograms on the plane!
  4. Notice that x1−x2=l1−l2∈L≠0x_1 - x_2 = l_1 - l_2\in L \neq 0, since we already argued that l1≠l2l_1 \neq l_2. This gives us what we want.

§ Minkowskis' Convex body Theorem from Blichfeldt's theorem

Consider a convex set S⊆RnS \subseteq \mathbb R^n that is symmetric about the origin with volume greater than 2ndet(B)2^n det(B).

Create a new set TT which is S∗0.5S * 0.5. Formally:

T≡S/2={(x1/2,x2,…,xn/2):(x1,x2,…,xn)∈S}T \equiv S/2 = \{ (x_1/2, x_2, \dots, x_n/2) : (x_1, x_2, \dots, x_n) \in S \}

We now see that Vol(T)>det(B)Vol(T) > det(B) to invoke Blichfeldt's theorem. Formally:

Vol(T)=1/2nVol(S)>1/2n(2ndet(B))=det(B)Vol(T) = 1/2^n Vol(S) > 1/2^n (2^n det(B)) = det(B)

We can apply Blichfeldt's theorem to get our hands on two points x1,x2∈Tx_1, x_2 \in T such that x1−x2∈Lx_1 - x_2 \in L.

x1∈T⇒2x1∈S (S=2T)x2∈T⇒2x2∈S (S=2T)2x2∈S⇒−2x2∈S (S is symmetric about origin)12(2x1)+12(−2x2)∈S (S is convex)x1−x2∈S (Simplification)nonzero lattice point ∈S \begin{aligned} &x_1 \in T \Rightarrow 2x_1 \in S ~(S = 2T) \\ &x_2 \in T \Rightarrow 2x_2 \in S ~(S = 2T) \\ &2x_2 \in S \Rightarrow -2x_2 \in S~\text{($S$ is symmetric about origin)} \\ &\frac{1}{2}(2x_1) + \frac{1}{2} (-2x_2) \in S~\text{($S$ is convex)}\\ &x_1 - x_2 \in S~\text{(Simplification)}\\ &\text{nonzero lattice point}~\in S \\ \end{aligned}

§ References