Minkowski's Theorem

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Theorem

Let $L$ be a lattice in $\R^n$.

Let $d$ be the covolume of $L$.

Let $\mu$ be a translation invariant measure on $\R^n$

Let $S$ be a convex subset of $\R^n$ that is symmetric about the origin, i.e. such that:

$\forall p \in S : -p \in S$

Let the volume of $S$ be greater than $2^n d$.


Then $S$ contains a non-zero point of $L$.


Proof

Let $D$ be any fundamental parallelepiped.

Then by definition:

$\ds \R^n = \coprod \limits_{\vec x \mathop \in L} \paren {D + \vec x}$

where:

$A + \vec x := \set {\vec a + \vec x : \vec a \in A}$


By Intersection with Subset is Subset:

$\dfrac 1 2 S \cap \R^n = \dfrac 1 2 S \iff \dfrac 1 2 S \subseteq \R^n$

Hence by Intersection Distributes over Union:

$(1): \quad \ds \frac 1 2 S = \coprod \limits_{\vec x \mathop \in L} \paren {\frac 1 2 S \cap \paren {D + \vec x} }$

where:

$\dfrac 1 2 S := \set {\dfrac 1 2 \vec s: \vec s \in S}$




Consider the intersection of $D + \vec x$ and $\dfrac 1 2 S$.

This is obtained by adding $\vec x$ to every point in $D$ then taking those points that are also in $S$.

However, this is the same as subtracting $\vec x$ from all elements of $\dfrac 1 2 S$, taking the elements that are also in $D$ and adding $\vec x$ to restore them to their original position.

Thus:

$\ds \frac 1 2 S \cap \paren {D + \vec x} = \paren {\paren {\frac 1 2 S - \vec x} \cap D} + \vec x$


Since, by hypothesis, $\mu$ is translation invariant:

\(\text {(2)}: \quad\) \(\ds \map \mu {\frac 1 2 S \cap \paren {D + \vec x} }\) \(=\) \(\ds \map \mu {\paren {\paren {\frac 1 2 S - \vec x} \cap D} + \vec x}\)
\(\ds \) \(=\) \(\ds \map \mu {\paren {\frac 1 2 S - \vec x} \cap D}\)


Aiming for a contradiction, suppose the sets $\ds \set {\frac 1 2 S - \vec x: \vec x \in L}$ are pairwise disjoint.

That is:

$\ds \forall \vec x, \vec y \in L: \paren {\frac 1 2 S - \vec x} \cap \paren {\frac 1 2 S - \vec y} \ne \O \iff \vec x \ne \vec y$

Then:

\(\ds \map \mu S\) \(=\) \(\ds 2^n \map \mu {\frac 1 2 S}\) Dilation of Lebesgue-Measurable Set is Lebesgue-Measurable
\(\ds \) \(=\) \(\ds 2^n \map \mu {\coprod \limits_{\vec x \mathop \in L} \paren {\frac 1 2 S \cap \paren {D + \vec x} } }\) from $(1)$ above
\(\ds \) \(=\) \(\ds 2^n \map \mu {\coprod \limits_{\vec x \mathop \in L} \paren {\paren {\frac 1 2 S - \vec x} \cap D} }\) from $(2)$ above
\(\ds \) \(=\) \(\ds 2^n \sum \limits_{\vec x \mathop \in L} \map \mu {\paren {\frac 1 2 S - \vec x} \cap D}\) Definition of Measure
\(\ds \) \(\le\) \(\ds 2^n \map \mu D\) Measure is Countably Subadditive
\(\ds \) \(=\) \(\ds 2^n d\)

which is a contradiction.

So $\ds \set {\frac 1 2 S - \vec x: \vec x \in L}$ are not pairwise disjoint.


This means:

$\ds \exists \vec x, \vec y \in L: \vec x \ne \vec y, \paren {\frac 1 2 S - \vec x} \cap \paren {\frac 1 2 S - \vec y} \ne \O$

Therefore there exist $\vec {p_1}, \vec {p_2} \in L$ such that $\vec {p_1} \ne \vec {p_2}$ and:

$\ds \frac 1 2 \vec {p_1} - \vec x = \frac 1 2 \vec {p_2} - \vec y$

and therefore:

$\ds \frac 1 2 \paren {\vec {p_1} - \vec {p_2} } = \vec x - \vec y \in L$

Since $S$ is convex, we have that:

$\dfrac 1 2 \paren {\vec {p_1} - \vec {p_2} } \in S$

As $\vec {p_1} \ne \vec {p_2}$ by definition:

$\dfrac 1 2 \paren {\vec {p_1} - \vec {p_2} } \ne \vec 0$

Hence the result.

$\blacksquare$


Source of Name

This entry was named for Hermann Minkowski.