# Minkowski's Theorem (Tuesday, November 08)

:::{.remark}
Recall: $\Omega \subseteq \RR^N$ a convex subset, $\Lambda$ a lattice, $\covol \lambda \da \abs{\det{\vector v_i}}$ where $\Lambda = \gens{\vector v_i}_\ZZ$.
Check that this is well-defined independent of choice of generating set.
This is a **covolume** since it is the actual volume of a fundamental domain for $\Lambda \actson \RR^N$, i.e. a subset $Y \subseteq X$ such that $\forall x\in X\exists y\in Y$ such that $G.y=x$.
Suppose $X\in G\dash\Top$, we can then try to find a particularly nice $Y$, but it may be useful to relax the definition of $Y$.
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:::{.example title="?"}
$\ZZ^n\actson \RR^n$ by translations, and we can take $Y = [0, 1)^N$ for a fundamental domain, but the closure of $Y$ is compact and thus somewhat nicer.
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:::{.definition title="Tiling"}
A family of subset $\ts{Y_i}_{i\in I}$ of $X\in \Top$ is a **tiling** of $X$ if 

1. $\union_{i\in I} \cl_X(Y_i) = X$, and
2. $\forall i\neq j, \interior{Y_i} \intersect \interior{Y_j} = \emptyset$.

For $X\in G\dash\Top$, $Y$ is a **fundamental region** iff $\ts{gY}_{g\in G}$ is a tiling of $X$.
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:::{.example title="?"}
$[0, 1]^N$ is a tiling of $\RR^N$.
There is a quotient map $q: \RR^N\surjects \dcosetl{\ZZ^N}{\RR^N}$ which restricts to $\tilde q: [0, 1]^N \surjects \dcosetl{\ZZ^N}{\RR^N}$.
:::

:::{.remark}
There is some theorem of the following form: if $(X, \mu)$ is a Borel-measurable $G\dash$space (with some further hypotheses),

1. There exists a fundamental region $Y$, and
2. Any two fundamental regions have the same measure
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:::{.remark}
Choosing a basis $\Lambda = \gens{v_1,\cdots, v_N}_\ZZ$, the fundamental parallelepiped $P_V \da \ts{\sum_{i=1}^N \lambda_i v_i \st 0\leq \lambda_i \leq 1}$ is a fundamental region for $\Lambda\actson \RR^N$.
Note $\Vol(P_V) = \covol \Lambda$, which follows from elementary properties of determinants, and $P_V = M_V \actson [0,1]^N$.
More generally if $\Omega \subseteq \RR^N$ is bounded and measurable and $M\in \GL_N(RR)$, then $\Vol(M\Omega) = \abs{ \det{M} } \Vol(\Omega)$.
Note also that the measure of $\Lambda$ is zero since it's countable, so this is a *covolume* since it bears an inverse relationship to the volume since it measures the space in between lattice points.
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:::{.exercise title="?"}
Let \( \Lambda_1 \subseteq \Lambda_2 \subseteq \RR^N \) be lattices and show 
\[
\covol \Lambda_1 = [\Lambda_2: \Lambda_1] \cdot \covol \Lambda_2
.\]

> Hint: SNF, or use the fundamental region for the smaller lattice to build a fundamental region for the bigger one.
> E.g. $\covol (2\ZZ)^N = 2^N = 1\cdot [\ZZ^N : (2\ZZ)^N]$.

:::

:::{.definition title="Convex bodies"}
A subset $\Omega \subseteq \RR^N$ is a **convex body** iff it's nonempty, convex, symmetric about zero, and bounded.
One sometimes requires nonempty interior.
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:::{.exercise title="?"}
Let $\Omega$ be a convex subset of $\RR^N$ and show TFAE:

- $\Omega$ is flat, i.e. contained in a hyperplane,
- $\interior{\Omega} = \emptyset$,
- $\Vol \Omega = 0$.

:::

:::{.remark}
Let \( \Omega \) be a convex body in $\RR^N$ and \( \Lambda \) a lattice.
Note \( \Omega \intersect \Lambda \ni 0 \), are there any nonzero lattice points?
I.e. what is \( \Omega \intersect (\Lambda\smz)\)?

This intersection can be empty if $\vol \Omega$ is small compared to $\covol \Lambda$; is the intersection guaranteed to be nonempty if it is comparatively large instead?
:::

:::{.slogan}
Use linear changes of variables to either simplify $\Omega$ or $\Lambda$.
:::

:::{.remark}
On *linear equivariance*: note that the ratio
\[
\Vol(M \Omega)\over \covol(M \Lambda) = 
{\abs{\det M} \over \abs{\det M} } {\Vol( \Omega) \over \covol (\Lambda)} =
{\Vol( \Omega) \over \covol (\Lambda)} 
.\]
Here *equivariant* means the group action changes things in a predictable way, and here we see that the ratio of two linear equivariants is a linear *invariant*.
So without loss of generality, take \( \Lambda = \ZZ^N \) -- does there exist a $V_N \in (0, \infty)$ such that if $\Vol( \Omega) > V_N$, the intersection \( \Omega \intersect \ZZ^N \) is nonempty?
:::

:::{.example title="?"}
Let \( \Omega= (-1, 1)^N \), this is no nonzero lattice points and has volume $2^N$, so $V_N \geq 2^N$ if it exists.
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:::{.theorem title="Minkowski's convex body theorem"}
If $\Omega$ is a convex body and $\Lambda$ a lattice, if $\Vol( \Omega)\over \covol \Lambda > 2^N$ then \( \Omega \intersect (\Lambda\smz) \neq \emptyset \).
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:::{.remark}
Dilation: for $\alpha\in (0, \infty)$, one can dilate \( \Lambda \) to \( \alpha \Lambda \) to get $\Vol( \alpha \Lambda) = \alpha^n \Vol( \Lambda)$.
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:::{.exercise title="?"}
Show that if $\Omega$ is closed and $\Vol \Omega= 2^N \cdot \covol \Lambda$ then \( \Omega \intersect (\Lambda\smz) \neq \emptyset \). 

> Hint: check that $(1+\eps) \Omega$ has a lattice point for every $\eps$ and use that $\Omega$ is closed.

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## Proof of Minkowski's Convex Body Theorem

:::{.remark}
**Step 1**: Blichfeldt's lemma.
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:::{.definition title="Packable subsets"}
A subset $\Omega \subseteq \RR^N$ is **packable** if $\ts{x + \Omega}_{x\in \ZZ^N}$ are pairwise disjoint.

If $\Omega$ is not packable, there is some $x\neq y\in \ZZ^N$ and $z_1, z_2\in \Lambda$ with $x+z_1 = y+z_2$, which happens iff $\exists z_1\neq z_2\in \Omega$ such that $z_1 -z_2\in \ZZ^N$.
:::

:::{.lemma title="Blichfeldt's Lemma"}
If $\Omega \subseteq \RR^N$ is packable and measurable, then $\Vol( \Omega)\leq 1$.
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:::{.proof title="?"}
For $x = \tv{x_1,\cdots, x_n} \in \ZZ^N$, define $\Omega_x \da \Omega \intersect \prod_{i=1}^N [x_i, x_{i+1})$.

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This allows writing \( \Omega = \disjoint_{x\in \ZZ^N} \Omega_x \), and $\Vol \Omega = \sum_{x\in \ZZ^N} \Vol( \Omega_x)$.
If $\Omega$ is packable, since $\ts{x + \Omega_x}_{x\in \ZZ^N}$ are pairwise disjoint, we have
\[
\Vol( \Disjoint_{x\in \ZZ^N} -x + \Omega_x ) = \sum_{x\in \ZZ^N} \Vol( x + \Omega_x) = \sum_{x\in \ZZ^N} \Omega_x = \Vol( \Omega)
.\]

Note that every $-x + \Omega_x$ is in the unit cube, thus so is their union, so the total volume is at most one.
I.e. $-x + \Omega_x \subseteq [0, 1]^N\, \forall x\in \ZZ^N\implies \Union_{x\in \ZZ^N} (-x + \Omega_x) \subseteq [0, 1]^N$, and taking volumes yields an inequality.
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:::{.proof title="of Minkowski"}
**Step 2**: It suffices to consider $\Lambda = \ZZ^N$ and $\Vol( \Omega) > 2^N$, and further using $M = \diag(1/2, \cdots, 1/2)$, we can reduce to $\Lambda = \qty{{1\over 2} \ZZ}^N$ where $\Vol \Omega > 2^N \cdot 2^{-N} = 1$.

We'll show that if $\Omega$ is a convex body with $\Vol \Omega> 1$ then there exists an $x$ in the intersection $(\qty{ {1\over 2}\ZZ }^N \intersect \Omega)\smz$.
Apply Blichfeldt's lemma: $\Omega$ is not packable, so there exist $z_1\neq z_2$ with $z_1 - z_2 \in \ZZ^N$.
This occurs iff $(1/2)(z_1-z_2) \in \qty{{1\over 2}\ZZ}^N$, and $z_2\in \Omega \iff -z_2\in \Omega$, so ${1\over 2}z_1 - {1\over 2}(-z_2) \in \Omega$, but $z_1 - z_2\neq 0$ and we're done.
This proves Minkowski's theorem.
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:::{.remark}
Up next: for $K$ a number field, $\deg K = n$, we'll define an additive embedding $\sigma: K \injects \RR^N$ where for $\Lambda \subseteq K$ any $\ZZ\dash$lattice, the image $\sigma( \Lambda)$ is again a lattice in $\RR^N$ and $\covol(\sigma( \Lambda)) \approx \delta( \Lambda)$, so the covolume is well-controlled by the discriminant.
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