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Minkowski convex-body theorem, strict form

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $n \geq 1$ , let $Λ \subseteq R^{n}$ be a full lattice with $covol (Λ) > 0$ (Full Euclidean lattice and covolume), and let $C \subseteq R^{n}$ be Lebesgue measurable, convex and centrally symmetric (A convex subset of $R^{m}$ contains every line segment between two of its points). If

$λ_{n} (C) > 2^{n} covol (Λ),$

then $C$ contains a nonzero point of $Λ$ .

Facts & Assumptions

Given: The Axiom of Choice, a full lattice $Λ$ with $covol (Λ) > 0$ , and a Lebesgue measurable convex centrally symmetric set $C$ with $λ_{n} (C) > 2^{n} covol (Λ)$ .

[A1]

The Axiom of Choice gives the Axiom of Countable Choice (AC implies DC implies countable choice), the choice hypothesis of the scaling fact [F2], invoked in step 1.1; Blichfeldt's principle [F1] is applied under the Axiom of Choice assumed in the statement, and no further choice is used.

[F1]

Blichfeldt's principle: for a Lebesgue measurable $S \subseteq R^{n}$ with $λ_{n} (S) > covol (Λ)$ there are distinct $x, y \in S$ with $x - y \in Λ$ (Blichfeldt lattice-point principle).

[F2]

For a nonzero real $c$ , a set $E$ is Lebesgue measurable if and only if $c E$ is, and then $λ_{n} (c E) = ∣ c ∣^{n} λ_{n} (E)$ (For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it).

[F3]

$C$ convex means $(1 - t) x + t y \in C$ for all $x, y \in C$ and $t \in [0, 1]$ ; central symmetry means $- C = C$ , so $- y \in C$ whenever $y \in C$ (A convex subset of $R^{m}$ contains every line segment between two of its points).

Proof

1.1F2A1given

Put $C^{'} : = \frac{1}{2} C = {x / 2 : x \in C}$ . By [F2] with $c = 1 / 2$ , $C^{'}$ is Lebesgue measurable with $λ_{n} (C^{'}) = 2^{- n} λ_{n} (C) > covol (Λ)$ , the Countable Choice hypothesis of [F2] being supplied by [A1].

1.2F3algebra

$C^{'}$ is convex and centrally symmetric: for $u, v \in C^{'}$ write $u = x / 2$ , $v = y / 2$ with $x, y \in C$ ; then $(1 - t) u + t v = ((1 - t) x + t y) / 2 \in C^{'}$ by convexity of $C$ , and $- u = (- x) / 2 \in C^{'}$ by symmetry of $C$ .

2.1F1step 1.1

By [F1] applied to the measurable set $C^{'}$ of step 1.1 there are distinct $u, v \in C^{'}$ with $u - v \in Λ$ .

3.1F3step 2.1

The difference $u - v$ is nonzero because $u \neq v$ , and it lies in $C$ : $2 u \in C$ and $2 v \in C$ by definition of $C^{'}$ , so $- 2 v \in C$ by central symmetry, and convexity of $C$ gives $u - v = \frac{1}{2} (2 u) + \frac{1}{2} (- 2 v) \in C$ .

4.1step 2.1step 3.1∎

Thus $u - v$ is a nonzero point of $Λ$ lying in $C$ , as required.

Remarks

The factor $2^{n}$ is optimal for centrally symmetric convex bodies: for the open cube $C = (- 1, 1)^{n}$ and $Λ = Z^{n}$ one has $λ_{n} (C) = 2^{n} = 2^{n} covol (Λ)$ while $C \cap Z^{n} = {0}$ , so the strict inequality cannot be weakened to $\geq$ . The equality case for compact bodies is treated in the next item, where the strict form is applied to the dilates $(1 + 1 / m) C$ .

Depends on

Used by

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Sources

J. S. Milne, Algebraic Number Theory v3.08 (standard reference, not scraped)
Brian Conrad and Aaron Landesman, Math 154 Algebraic Number Theory (standard reference, not scraped)