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% 27/02/2024 10AM

Recall: $\sigma_1, \ldots, \sigma_r$ are the
embeddings $K \to \CC$ with real image, $\tau_1,
\ol{\tau_1}, \ldots, \tau_s, \ol{\tau_s}$ are the
other embeddings, $d = r + 2s$. We defined
\begin{align*}
  \Sigma : K
  &\to \RR^d \\
  \Sigma(\alpha)
  &= (\sigma_1(\alpha), \ldots, \sigma_r(\alpha),
  \Re(\tau_1(\alpha)), \Im(\tau_1(\alpha)),
  \ldots, \Re(\tau_s(\alpha)),
  \Im(\tau_s(\alpha)))
\end{align*}
$\Sigma(\O_K) \subset \RR^d$ is a \gls{latt}, i.e. it
is an additive subgroup of $\RR^d$ generated by
$d$ linearly independent elements.
\[ \coVol(\Sigma(\O_K)) = 2^{-s} \sqrt{\nfdisc(K)}
.\]
Let $I \subset \O_K$ be an ideal, then $\Sigma(I)
\subset \Sigma(\O_K)$ is a sub\gls{latt}, and
\[ \coVol(\Sigma(I)) = 2^{-s} \sqrt{\disc(I)} =
2^{-s} \N(I) \sqrt{\nfdisc(K)} .\]
(where $\disc(I)$ is the discriminant of a
generating tuple).

$\mathcal{N} : \RR^d \to \RR$,
\[ \mathcal{N}(x_1, \ldots, x_d) = \prod_{j = 1}^r
|x_j| \prod_{j = 1}^s (|x_{r + j}|^2 + |x_{r + j +
1}|^2) .\]
Note $\alpha \in K$, $\mathcal{N}(\Sigma(\alpha))
= |N(\alpha)|$. Need to prove that the lattive
$\Sigma(\O_K)$ contains a non-zero element in the
region:
\[ \{x \in \RR^d : \mathcal{N}(x) \le \N(I)
\sqrt{\nfdisc(K)}\} \]
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/bbbc85e288014802.png}
\end{center}

\subsubsection*{Geometry of numbers}

Convex means that if $x, y \in S$ and $a \in (0,
1)$, then $ax + (1 - a)y \in S$.

Symmetric to $0$ means that if $x \in S$, then $-x
\in S$.

\begin{flashcard}[minkowski-thm-lemma]
\begin{lemma*}
  Let $\Lambda \subset \RR^d$ be a \gls{latt}, and
  let $S \subset \RR^d$ be a Borel set with
  $\Vol(S) > \coVol(\Lambda)$, \cloze{then there exists $x \neq
  y$ in $S$ such that $x - y \in \Lambda$.}
\end{lemma*}

\begin{proof}
  \cloze{Let $F$ be a \gls{fundom} for $\Lambda$. Note
  that $\RR^d$ is the disjoint union of
  \[ \{F + a : a \in \Lambda\} .\]
  Define: $S(a) = (S \cap (F + a)) - a$ for $a \in
  \Lambda$. Observe that $S(a) \subset F$.
  \[ \Vol(S) = \sum_{a \in \Lambda} \Vol(S \cap
  (F + a)) = \sum_{a \in \Lambda}  \] % TODO
  % TODO
  Then $\exists a \neq b \in \Lambda$ and $x \in
  S(a) \cap S(b)$. Then $x + a \neq x + b \in S$,
  and $(x + a) - (x + b) = a - b \in \Lambda$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[minkowski-theorem]
\begin{theorem*}[Minkowski's theorem]
  \label{mink_thm}
  \cloze{Let $\Lambda \in \RR^d$ be a \gls{latt}, and let $S
  \subset \RR^d$ be convex and symmetric to $0$.
  Suppose $\coVol(S) > 2^d \coVol(\Lambda)$. Then
  $\exists x \in \Lambda \cap S$ such that $x \neq
  0$.}
\end{theorem*}

\begin{proof}
  \cloze{We apply the lemma for the set
  \[ \half S = \left\{ \half x : x \in S \right\}
  .\]
  Then $\Vol \left( \half S \right) = 2^{-d}
  \Vol(S)$. We get $x \neq y \in \half S$ such
  that $x - y \in \Lambda$. THen $2x, -2y \in S$,
  and by symmetry, $x - y = \half (2x) + \half
  (-2y) \in S$ by convexity.}
\end{proof}
\end{flashcard}

\begin{example*}[non-example]
  $\Lambda = \ZZ^d$, $S = (-1, 1)^d$, $\coVol(S) =
  2^d = 2^d \coVol(\Lambda)$, $S \cap \Lambda =
  \{0\}$.
\end{example*}

\vspace{-1em}
Is $S$ is closed in addition, then $>$ can be
replaced by $\ge$.

\begin{flashcard}[mink-bound-1-proof]
\begin{proof}[Proof of \nameref{mink_bound_1}]
  Consider $S = [-Y, Y]^d$ for some $Y \in \RR$.
  Then $\Vol(S) = 2^d Y^d$, and $|\mathcal{N}(x)|
  \le 2^s Y^d$ for $x \in S$. \nameref{mink_thm}
  gives $S \cap \Lambda \neq \{0\}$ if $\Vol(S) >
  2^s \coVol(\Lambda)$.

  % TODO
\end{proof}
\end{flashcard}
