% vim: tw=50
% 17/02/2023 11AM

\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/9ee2617aaeb411ed.png}
\end{center}
All points lie on one side of the tangent plane.
\[ \sigma(u, v) = (u, v, u^3 - 3v^2u) \]
at $p = \sigma(0, 0)$, $\kappa(p) = 0$, but
locally $\Sigma$ meets both sides of its tangent
plane. Monkey saddle: see picture.

\begin{flashcard}[compact-surface-has-elliptic-point]
\begin{proposition}
    Let $\Sigma$ be a compact surface in $\RR^3$.
    Then $\Sigma$ \cloze{has an \fcemph{elliptic
    point}\prompt{ and $\Sigma$ is \fcemph{two-sided}}.}
\end{proposition}
\end{flashcard}

\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/0531638aaeb611ed.png}
\end{center}

\begin{proof}
    $\Sigma$ compect implies $\Sigma \subset
    \ol{B(0, R)}$ for some $R$ large enough.
    Decrease $R$ to the minimal such value. Up to
    applying a rotation and translation we may
    assume that the point of contact is on the $z$
    axis. Locally near $p$ view $\Sigma$ as the
    graph of a smooth function $f$ such that
    \[ f - \sqrt{\kappa^2 - u^2 - v^2} \le 0 \]
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/269900a0aeb611ed.png}
    \end{center}
    $f : V \to \RR$, $V$ open in $\RR^2$. Since
    $f$ has a local maximum at $(0, 0)$ $\implies$
    $f_v = f_0 = 0$ at $(0, 0)$, $f(0, 0) = p$.
    \[ F(u, v) = f(u, v) - \sqrt{R^2 - u^2 - v^2}
    \le 0 \tag{$*$} \]
    An easy computation shows: $F_u = F_v = 0$ at
    $(0, 0)$,
    \begin{align*}
        F_{uu}
        &= f_{uu} + \frac{1}{R}
        &&(\text{at $(0, 0)$})
        F_{uv}
        &= f_{uv}
        &&(\text{at $(0, 0)$})
        F_{vv}
        &= f_{vv} + \frac{1}{R}
        &&(\text{at $(0, 0)$})
    \end{align*}
    ($*$) $\implies$ (Taylor expansion and use
    that $0$ is a local maximum):
    \[ \implies \left( f_{uu} + \frac{1}{R}
    \right) h^2 + 2f_{uv} hl + \left( f_{vv} +
    \frac{1}{R} \right) l^2 \le 0 \]
    for $(h, l)$ small enough.
    \[ \implies f_{uu} h^2 + 2f_{uv} hl + f_{vv}
    l^2 \le -\frac{1}{R} (h^2 + l^2) \]
    \[ \implies 
    \begin{pmatrix}
        L & M \\
        M & N
    \end{pmatrix}
    =
    \begin{pmatrix}
        f_{uu} & f_{uv} \\
        f_{uv} & f_{vv}
    \end{pmatrix}
    \]
    Is negative definite at $(0, 0)$. At $(0, 0$,
    $E = G = 1$, $F = 0$, hence $\kappa(p) > 0$.
\end{proof}

\subsubsection*{Gauss Curvature and Area}

\begin{theorem}
    $\Sigma \subset \RR^3$, $p \in \Sigma$,
    $\kappa(p) \neq 0$. Let $U \subset \Sigma$ be
    a small open neighbourhood of $p$ and consider
    a sequence $p \in A_i \subset U \subset
    \Sigma$ ($A_i$ open) such that $A_i$ ``shrink
    to $p$'' in the sense that $\forall \eps > 0$,
    $A_i \subset B(p, \eps) \subset \RR^3$ for all
    $i$ large enough. Then
    \[ |\kappa(p)| = \lim_{i \to \infty}
    \frac{\Area_{S^2}(n(A_i))}{\Area_\Sigma(A_i)} \]
    \begin{center}
        \includegraphics[width=0.2\linewidth]
        {images/e8652316aeb711ed.png}
    \end{center}
    i.e. the Gaussian curvature is an
    $i$-finiteseimal measure of how much the Gauss
    map $n$ \emph{distorts areas}.
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/112b8a06aeb811ed.png}
    \end{center}
\end{theorem}

\begin{hiddenflashcard}[Gauss-curvature-area]
\begin{theorem}[Gauss curvature and area]
    Let $\Sigma$ be a smooth surface in $\RR^3$.
    \cloze{Let $p \in \Sigma$ with $\kappa(p)$
    \fcemph{non-zero}. Let $U \subset \Sigma$ be a small open
    neighbourhood of $p$ and consider a sequence
    $p \in A_i \subset U \subset \Sigma$ such that
    $\forall \eps > 0$, $A_i \subset B(p, \eps)
    \subset \RR^3$ for all $i$ large enough. Then
    \[ |\kappa(p)| = \lim_{i \to \infty}
    \frac{\Area_{S^2}(n(A_i))}{\Area_\Sigma(A_i)} \]
    }
\end{theorem}

\begin{proof}
\cloze{
    .
}
\end{proof}
\end{hiddenflashcard}

\begin{proof}
    $\kappa$ is all ``local'', so take $\sigma : V
    \to U \subset \Sigma$ with $\sigma(0) = p$ and
    let $V_i = \sigma^{-1}(A_i) \subset V$ open.
    Since $A_i$ shrinks to $p$,
    \[ \bigcap_{i \ge 1} V_i = \{(0, 0)\} \]
    \begin{align*}
        \Area_\Sigma(A_i)
        &= \int_{V_i} \sqrt{EG - F^2} \dd u \dd v
        \\
        &= \int_{V_i} \|\sigma_u \times \sigma_v\|
        \dd u \dd v
    \end{align*}
    Now $n \circ \sigma : V \to S^2 \subset
    \RR^2$. $Dn|_p \circ D\sigma|_0$ has rank 2
    since $\kappa(p) \neq 0$. This $n \circ
    \sigma$ defines an allowable parametrisation
    in an open neighbourhood of $n(p) \subset S^2$
    by the inverse function theorem.
    \[ \Area_{S^2}(n(A_i)) = \int_{V_i} \|n_u
    \times n_v\| \dd u \dd v \]
    (Same source as ($*$) but for $S^2$!)
    \[ \|n_u \times n_v\| = \|Dn(\sigma_u) \times
    Dn(\sigma_v)\| \]
    Recall from last lecture
    \begin{align*}
        Dn(\sigma_u)
        &= a_{11}\sigma_u + a_{21}\sigma_v \\
        Dn(\sigma_v)
        &= a_{12}\sigma_u + a_{22}\sigma_v \\
        \implies Dn(\sigma_u) \times Dn(\sigma_v)
        &= (a_{11}\sigma_u + a_{21}\sigma_v)
        \times (a_{12}\sigma_u + a_{22}\sigma_v)
        \\
        &= \ub{(a_{11}a_{22} -
        a_{12}a_{21})}_{\det(Dn) = \kappa(p)}
        \sigma_u \times \sigma_v \\
        \implies \Area_{S^2}(n(A_i))
        &= \int_{V_i} \|n_u \times n_v\| \\
        &= \int_{V_i} |\det Dn| \|\sigma_u \times
        \sigma_v\| \dd u \dd v \\
        &= \int_{V_i} |\kappa(u, v)|\|\sigma_u
        \times \sigma_v\| \dd u \dd v
    \end{align*}
    Since $\kappa$ is continuous, given $\eps >
    0$, $\exists \delta > 0$ such that $|\kappa(u,
    v) - \kappa(0, 0)| < \eps$ for all $(u, v) \in
    B((0, 0), \delta) \subset V$. So if $i \ge
    i_0$ we have
    \[ |\kappa(p)| - \eps \le |\kappa(u, v)| \le
    |\kappa(p)| + \eps \qquad \forall (u, v) \in
    V_i \]
    \begin{align*}
        \implies(|\kappa(p)| - \eps) \int_{V_i}
        \|\sigma_u \times \sigma_v\| \dd u \dd v
        \\
        &\le \int_{V_i} |\kappa(u, v)| \|\sigma_u
        \times \sigma_v\| \dd u \dd v \\
        &\le (|\kappa(p)| + \eps) \ub{\int_{V_i}
        \|\sigma_u \times \sigma_v\| \dd u \dd
        v}_{\Area_\Sigma(A_i)} \\
        \implies |\kappa(p)| - \eps
        &\le
        \frac{\Area_{S^2}(n(A_i))}{\Area_\Sigma(A_i)}
        \\
        &\le |\kappa(p)| + \eps
    \end{align*}
    for all $i \ge i_0$.
\end{proof}
