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Minkowski second theorem on successive minima

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $C \subseteq R^{n}$ be compact, convex, centrally symmetric (A convex subset of $R^{m}$ contains every line segment between two of its points) with nonempty interior, and let $Λ \subseteq R^{n}$ be a full lattice with $covol (Λ) > 0$ (Full Euclidean lattice and covolume). Let $λ_{1} \leq \dots \leq λ_{n}$ be the successive minima of $C$ with respect to $Λ$ (Successive minima of a convex body). Then

$\frac{2^{n}}{n!} covol (Λ) \leq (\prod_{i = 1}^{n} λ_{i}) vol (C) \leq 2^{n} covol (Λ) .$

Facts & Assumptions

Given: A compact convex centrally symmetric $C \subseteq R^{n}$ with nonempty interior, a full lattice $Λ$ with $covol (Λ) > 0$ , and the successive minima $λ_{1} \leq \dots \leq λ_{n}$ of $C$ with respect to $Λ$ .

[A1]

The Axiom of Choice implies the Axiom of Countable Choice (AC implies DC implies countable choice), which supplies the hypotheses of the linear-change-of-variables fact [F6] invoked in steps 1.3 and 4.1, and of the Lebesgue-measure and Tonelli facts [F7] invoked in steps 2.1 and 3.1; no other selection is made.

[F1]

The successive minima are defined by $λ_{i} = \inf {t > 0 : \dim span (t C \cap Λ) \geq i}$ , and for $C$ compact with nonempty interior and $Λ$ full one has $0 < λ_{1} \leq \dots \leq λ_{n} < \infty$ ; also $λ_{0} : = 0$ by convention (Successive minima of a convex body).

[F2]

There exist linearly independent $a_{1}, \dots, a_{n} \in Λ$ with $a_{i} \in λ_{i} C$ for every $i$ (Attained successive minima and adapted flag).

[F3]

With $U : = int (C)$ the centroid map $Φ : U \to R^{n}$ of Successive-minima volume deformation and collision avoidance is Borel measurable, has $vol (Φ (U)) = (\prod_{i} λ_{i}) vol (C)$ , and no two distinct points of $Φ (U)$ differ by an element of $2 Λ$ (Successive-minima volume deformation and collision avoidance).

[F4]

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

[F5]

If $b_{1}, \dots, b_{n}$ is a $Z$ -basis of a full lattice $Λ$ , then $covol (Λ) = ∣ \det (b_{1}, \dots, b_{n}) ∣$ ; consequently $covol (2 Λ) = ∣ \det (2 b_{1}, \dots, 2 b_{n}) ∣ = 2^{n} covol (Λ)$ (Full Euclidean lattice and covolume, For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ , The determinant of a triangular matrix is the product of its diagonal entries).

[F6]

Assume the Axiom of Countable Choice. For a linear $T : R^{n} \to R^{n}$ with matrix $A$ , if $\det A \neq 0$ then $T [E]$ is measurable and $λ_{n} (T [E]) = ∣ \det A ∣ λ_{n} (E)$ for every Lebesgue measurable $E$ ; if $\det A = 0$ , then $T [E]$ is measurable and null for every $E \subseteq R^{n}$ (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not).

[F7]

Under the Axiom of Countable Choice, product Lebesgue measure agrees with Euclidean Lebesgue measure on Borel sets, and Tonelli's theorem permits iterated integration; thus the volume of a Borel subset of $R^{n}$ can be computed by its coordinate integrals (for $n = 1$ , use the one-dimensional integral directly) (On Borel subsets of R^{m+n}, the product lambda_m times lambda_n agrees with lambda_{m+n}, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product). Also under Countable Choice, $λ_{n}$ is a complete measure and is therefore additive over finite unions of pairwise disjoint measurable sets (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume).

[F8]

A convex set contains every convex combination of finitely many of its points, and central symmetry means $- C = C$ (A convex subset of $R^{m}$ contains every line segment between two of its points).

[F9]

For $A \in M_{n} (Z)$ with $\det A \neq 0$ the index $[Z^{n} : A Z^{n}]$ equals $∣ \det A ∣$ and is a positive integer; for real square matrices $\det (A B) = \det (A) \det (B)$ and $\det (2 I_{n}) = 2^{n}$ (The index of a full-rank subgroup of $Z^{n}$ is the absolute determinant of a generating matrix, For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ , The determinant of a triangular matrix is the product of its diagonal entries).

Proof

1.1F1given

The minima satisfy $0 < λ_{1} \leq \dots \leq λ_{n} < \infty$ , so every $λ_{i}$ is a positive finite real number.

1.2given

Let $Δ : = {y \in R^{n} : \sum_{i = 1}^{n} ∣ y_{i} ∣ \leq 1}$ and $S : = {y \in R^{n} : y_{i} \geq 0, \sum_{i} y_{i} \leq 1}$ .

1.3F6A1algebra

For each sign vector $ε = (ε_{1}, \dots, ε_{n}) \in {\pm 1}^{n}$ let $ε S : = {(ε_{1} y_{1}, \dots, ε_{n} y_{n}) : y \in S}$ . These $2^{n}$ measurable sets cover $Δ$ . If $ε \neq ε^{'}$ , choose $i$ with $ε_{i} \neq ε_{i}^{'}$ ; then $ε S \cap ε^{'} S$ lies in the coordinate hyperplane $H_{i} = {y : y_{i} = 0}$ . The projection onto $H_{i}$ is singular and has image $H_{i}$ , so [F6] and [A1] give $λ_{n} (H_{i}) = 0$ . Thus the sign pieces overlap only on null sets. Each sign map is an invertible diagonal linear map with determinant of absolute value $1$ , so [F6] and [A1] give $λ_{n} (ε S) = λ_{n} (S)$ .

1.4F3F5

For the upper bound, [F3] gives $vol (Φ (U)) = (\prod_{i} λ_{i}) vol (C)$ for the Borel set $Φ (U)$ , and $covol (2 Λ) = 2^{n} covol (Λ)$ by [F5].

2.1F7step 1.2

By Tonelli's theorem [F7] applied to the indicator of the simplex, $λ_{n} (S) = \int_{0}^{1} \int_{0}^{1 - x_{1}} \dots \int_{0}^{1 - x_{1} - \dots - x_{n - 1}} 1 d x_{n} \dots d x_{1} = \frac{1}{n!}$ .

2.2step 1.2algebra

$Δ = conv {\pm e_{1}, \dots, \pm e_{n}}$ : every $y$ with $\sum_{i} ∣ y_{i} ∣ \leq 1$ is $\sum_{i} y_{i} e_{i}$ , a convex combination of the vectors $\pm e_{i}$ after moving negative coefficients and adding the origin, and conversely every convex combination of the $\pm e_{i}$ satisfies the inequality.

2.3F2step 1.1

Choose linearly independent $a_{1}, \dots, a_{n} \in Λ$ with $a_{i} \in λ_{i} C$ by [F2], and let $B$ be the matrix with columns $a_{1} / λ_{1}, \dots, a_{n} / λ_{n}$ ; by step 1.1 the columns are well defined, and they are linearly independent, so $B$ is invertible and $P : = B [Δ] = {B y : \sum_{i} ∣ y_{i} ∣ \leq 1}$ is measurable.

2.4F3F4F5step 1.4

If $vol (Φ (U)) > covol (2 Λ)$ then [F4] applied to the lattice $2 Λ$ produces distinct $x, y \in Φ (U)$ with $x - y \in 2 Λ$ , contradicting the collision-free clause of [F3]; hence $vol (Φ (U)) \leq covol (2 Λ) = 2^{n} covol (Λ)$ .

3.1F7step 1.3step 2.1

Disjointifying the finite cover in step 1.3 changes each piece only by a null set, so finite additivity and steps 1.3 and 2.1 give $vol (Δ) = \sum_{ε} λ_{n} (ε S) = 2^{n} / n!$ .

3.2F8step 2.2step 2.3

Each $a_{i} / λ_{i}$ lies in $C$ , and $- a_{i} / λ_{i}$ lies in $C$ by central symmetry; hence every convex combination of the $2 n$ points $\pm a_{i} / λ_{i}$ lies in $C$ by [F8]. Since $B [Δ] = {\sum_{i} y_{i} (a_{i} / λ_{i}) : \sum_{i} ∣ y_{i} ∣ \leq 1} = conv {\pm a_{i} / λ_{i}}$ by the same convex-combination identity as step 2.2, we have $P \subseteq C$ .

3.3F5step 2.3

Let $b_{1}, \dots, b_{n}$ be a $Z$ -basis of $Λ$ and $M = (b_{1} \dots b_{n})$ , so $covol (Λ) = ∣ \det M ∣$ ; each $a_{i} \in Λ$ has $a_{i} = M c_{i}$ with a unique $c_{i} \in Z^{n}$ , and $A = M D$ for $D = (c_{1} \dots c_{n}) \in M_{n} (Z)$ .

4.1F6F9A1step 3.1step 2.3step 3.2

Therefore $vol (C) \geq vol (P) = ∣ \det B ∣ vol (Δ) = 2^{n} ∣ \det B ∣ / n!$ by [F6], whose Countable Choice hypothesis is supplied by [A1], and steps 3.1 and 3.2; writing $A = (a_{1} \dots a_{n})$ we have $B = A diag (1 / λ_{1}, \dots, 1 / λ_{n})$ , so $\det B = \det A / \prod_{i} λ_{i}$ by [F9], and multiplying the volume inequality by $\prod_{i} λ_{i} > 0$ gives $(\prod_{i} λ_{i}) vol (C) \geq 2^{n} ∣ \det A ∣ / n!$ .

4.2F9step 2.3step 3.3

Since $A$ is invertible and $\det A = \det M \det D$ by [F9], also $\det D \neq 0$ ; hence $∣ \det D ∣ = [Z^{n} : D Z^{n}]$ is a positive integer by [F9], in particular at least $1$ .

5.1F5step 4.1step 3.3step 4.2

It follows that $∣ \det A ∣ = ∣ \det M ∣ ∣ \det D ∣ \geq ∣ \det M ∣ = covol (Λ)$ , and combining with step 4.1 gives $(\prod_{i} λ_{i}) vol (C) \geq 2^{n} ∣ \det A ∣ / n! \geq (2^{n} / n!) covol (Λ)$ .

6.1step 5.1step 2.4∎

Steps 5.1 and 2.4 combine into $(2^{n} / n!) covol (Λ) \leq (\prod_{i} λ_{i}) vol (C) \leq 2^{n} covol (Λ)$ .

Remarks

The two bounds have different shapes. The lower bound is geometric: the adapted vectors $a_{i} \in λ_{i} C$ turn the cross-polytope of side data into a subset of $C$ , and the determinant comparison against a lattice basis produces the index factor $∣ \det D ∣ \geq 1$ . The upper bound is measure theoretic: the centroid deformation of [F3] has volume $(\prod_{i} λ_{i}) vol (C)$ and avoids collisions modulo $2 Λ$ , so Blichfeldt's principle bounds that volume by $covol (2 Λ)$ . For $Λ = Z^{n}$ , the cube $C = [- 1, 1]^{n}$ attains the upper bound: all $λ_{i} = 1$ and $vol (C) = 2^{n}$ . The cross-polytope $C = {x : \sum_{i} ∣ x_{i} ∣ \leq 1}$ attains the lower bound: again all $λ_{i} = 1$ , since $C$ contains the standard basis vectors and no $t C$ with $t < 1$ contains a nonzero lattice point, while $vol (C) = 2^{n} / n!$ by step 3.1. The cube attains both bounds only when $n = 1$ .

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Sources

Ben Green, Additive Combinatorics, Lecture 3 §3.7 (standard reference, not scraped)
Martin Henk, Successive Minima and Lattice Points (standard reference, not scraped)