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:

$\displaystyle \R^n = \coprod \limits_{\vec x \mathop \in L} \left({D + \vec x}\right)$

where:

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


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:

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

where:

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


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:

$\displaystyle \frac 1 2 S \cap \left({D + \vec x}\right) = \left({\left({\frac 1 2 S - \vec x}\right) \cap D}\right) + \vec x$


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

\((2):\quad\) \(\displaystyle \mu \left({\frac 1 2 S \cap \left({D + \vec x}\right)}\right)\) \(=\) \(\displaystyle \mu \left({\left({\left({\frac 1 2 S - \vec x}\right) \cap D}\right) + \vec x}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \mu \left({\left({\frac 1 2 S - \vec x}\right) \cap D}\right)\)


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

That is:

$\displaystyle \forall \vec x, \vec y \in L: \left({\frac 1 2 S - \vec x}\right) \cap \left({\frac 1 2 S - \vec y}\right) \ne \varnothing \iff \vec x \ne \vec y$

Then:

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

which is a contradiction.

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


This means:

$\displaystyle \exists \vec x, \vec y \in L: \vec x \ne \vec y, \left({\frac 1 2 S - \vec x}\right) \cap \left({\frac 1 2 S - \vec y}\right) \ne \varnothing$

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

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

and therefore:

$\displaystyle \frac 1 2 \left({\vec {p_1} - \vec {p_2} }\right) = \vec x - \vec y \in L$

Since $S$ is convex, we have that:

$\dfrac 1 2 \left({\vec {p_1} - \vec {p_2} }\right) \in S$

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

$\dfrac 1 2 \left({\vec {p_1} - \vec {p_2} }\right) \ne \vec 0$

Hence the result.

$\blacksquare$


Source of Name

This entry was named for Hermann Minkowski.