LattE background

Minkowski's Theorem

Theorem 1   Let $ \zeta$ be a convex set in $ {\mathbb{R}}^2$, with $ \zeta$ centrally symmetric and $ area(\zeta)>4$. Then $ \zeta$ must contain a lattice point besides $ (0,0)$.

Definition 2   A set is convex whenever $ x,y\in\zeta$ implies $ \overline{xy}\in\zeta$.

Definition 3   An object $ \zeta$ is centrally symmetric if whenever $ (0,0)\in\zeta$ and $ P\in\zeta$ implies $ \overline{0P}\in\zeta$ and $ -\overline{0P}\in\zeta$.

1. Take $ \frac{1}{2}\zeta=\zeta'$

\includegraphics[width=2 in]{blob.eps}

2. Assign a copy of $ \zeta'$ to each lattice point in an $ n\times n$ grid.

\includegraphics[width=3 in]{nxn.eps}

3. From the above figure we can see $ area(\zeta')>1$, thus the area covered by the set s is $ (n+1)^2A'$.

4. In fact, all objects cover more (go over a bit). All $ (n+1)^2$ objects are contained in a square of side $ (n+2s)$, where $ s$ is the maximum distance from the center $ (0,0)$ to any point of $ \zeta'$.

\includegraphics[width=2.75 in]{n2s.eps}

5. Now we want to show the inequality

$\displaystyle (n+1)^2area(\zeta')>(n+2s)^2

to be true. CLAIM:
  $\displaystyle (n+1)^2(area(\zeta'))-(n+2s)^2$ $\displaystyle =$ $\displaystyle (area(\zeta')-1)(n^2+2n)(area(\zeta)-2s)$
      $\displaystyle {} +area(\zeta')-4s^2$
    $\displaystyle >$ $\displaystyle 0.$

Since $ area(\zeta)>1$ implies the leading coefficient is positive for large $ n$, if we choose $ n$ large then we're done! Conclusion, the blobs will overlap for some pair of blobs, but by translating $ \zeta'$ and another blob $ \zeta''$ cross too! (All is being repeated).

6. Let $ (a'',b'')$ be in the intersection of $ \zeta', \zeta''$. Then

$\displaystyle (a'',b'')=(a',b')+(p,q),$   where$\displaystyle (a',b')\in\zeta'.

Note that $ (-a',-b')\in\zeta'$ as well.
\includegraphics[width=2 in]{overlap.eps}
Consider the midpoint of $ (-a',-b')$ and $ (a'',b'')$, then $ (\frac{p}{2},\frac{q}{2})\in\zeta'\Rightarrow(p,q)\in\zeta$.


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