Minkowski's convex body theorem

Let $𝐿 \leq ℤ ℝ 𝑛$ be a complete lattice and $𝑆 \subset ℝ 𝑛$ be a bounded convex subset symmetric about the origin. If

v o l (𝑆) > 2 𝑛 c o v o l (𝐿),

then $𝑆$ contains a nonzero element of $𝐿$ . #m/thm/geo Moreover, the constant $2 𝑛$ cannot be any smaller. If $𝑆$ is compact, then the same conclusion holds for the weaker hypothesis¹

v o l (𝑆) \geq 2 𝑛 c o v o l (𝐿) .

Proof of first part

For the first part, suppose $v o l (𝑆) > 2 𝑛 c o v o l (𝐿)$ and consider the sublattice $2 𝐿 \leq ℤ 𝐿$ . By Covolume of a classical lattice, we have $c o v o l (𝐿') = 2 𝑛 c o v o l (𝐿)$ .

Let $𝐹' \subset ℝ 𝑛$ be a measurable fundamental domain for $𝐿'$ , and consider the map $𝑇 : ℝ 𝑛 \to 𝐹'$ induced by the projection $ℝ 𝑛 \to ℝ 𝑛 / 𝐿'$ . Since

v o l (𝑆) > v o l (𝐹') \geq v o l (𝑇 (𝑆))

by the hypothesis, the Measure theoretic pigeonhole principle implies that $𝑇 ↾ 𝑆$ is not injective, i.e. there exist distinct $𝑥', 𝑦' \in 𝑆$ such that $𝑇 (𝑥') = 𝑇 (𝑦')$ , whence $0 \neq 𝑃' : = 𝑥' - 𝑦' \in 𝐿'$ . Let $𝑃 = 12 𝑃' \in 𝐿$ . By symmetry of $𝑆$ , $- 𝑦' \in 𝑆$ and by convexity

𝑃 = 12 𝑥' + 12 (- 𝑦') \in 𝑆

so $𝑃$ is the required nonzero element.

Sharpness

It is already evident for $𝐿 = ℤ 𝑛 \leq ℤ ℝ 𝑛$ and $𝑆 = s p a n (- 1, 1) {ˆ 𝐞 𝑖} 𝑛 𝑖 = 1$ that the constant $2 𝑛$ cannot be made smaller.

#state/develop | #lang/en | #SemBr

2022. Algebraic number theory course notes, ¶3.6, p. 62 ↩

Minkowski's convex body theorem

Footnotes