% vim: tw=50
% 15/02/2023 11AM

\noindent
Recall FFF $I_p : T_p \Sigma \times T_p \Sigma \to
\RR$, $I_p(v, w) = \langle v, w \rangle_{\RR^3}$.

\begin{flashcard}[proof-FFF-self-adjoint]
\begin{lemma}
    $Dn|_p : T_p \Sigma \to T_p\Sigma$ is
    self-adjoint with respect to $I_p$, i.e.
    \[ I_p(Dn|_p (v), w) = I_p(v, Dn|_p(w)) \]
\end{lemma}

\prompt{
\begin{proof}
\cloze{
    Note that $\{\sigma_u, \sigma_v\}$ is a basis
    of $T_p\Sigma$, so just check that $\langle
    Dn|_p(\sigma_u), \sigma_v \rangle = \langle
    \sigma_u, Dn|_p(\sigma_v) \rangle$. (Note that
    these expressions are the same as $\langle
    n_u, \sigma_v \rangle = \langle \sigma_u, n_v
    \rangle$). Do this by considering the
    derivative of $\langle n, \sigma_v \rangle$
    and similarly for $u$.
}
\end{proof}
}
\end{flashcard}

\begin{proof}
    Take $\sigma$ a parametrisation with
    $\sigma(0) = p$. Then $\{\sigma_u, \sigma_v\}$
    is a basis of $T_p \Sigma$. To prove
    self-adjoint if suffices to check that
    \[ \langle n_u, \sigma_v \rangle =
    \langle \sigma_u, n_v \rangle \]
    (because $n_u = Dn|_p(\sigma_u)$ and
    $n_v = Dn|_p(\sigma_v)$). Note that
    \[ \langle n, \sigma_u \rangle = \langle n,
    \sigma_v \rangle = 0 \]
    Differentiate the first term with respect to
    $v$:
    \[ \langle n_v, \sigma_u \rangle + \langle n,
    \sigma_{uv} \rangle = 0 \]
    Similarly differentiate the second:
    \[ \langle n_u, \sigma_v \rangle + \langle n,
    \sigma_{uv} \rangle = 0 \]
    So $\langle n_u, \sigma_v \rangle = \langle
    n_v, \sigma_u \rangle$ as desired.
\end{proof}

\myskip
(Recall $M = -\langle n_v, \sigma_u \rangle =
-\langle n_u, \sigma_v \rangle$).

\myskip
Let's try to find the matrix of $Dn|_p$ in the
basis $\{\sigma_u, \sigma_v\}$.
\begin{align*}
    n_u
    &= Dn|_p(\sigma_u)
    = a_{11} \sigma_u + a_{21} \sigma_v \\
    n_v
    &= Dn|_p(\sigma_v)
    = a_{12} \sigma_u + a_{22} \sigma_v
\end{align*}
Taking products of the above with $\sigma_u$ and
$\sigma_v$ (check!):
\[ -
\ub{
\begin{pmatrix}
    L & M \\
    M & N
\end{pmatrix}
}_{Q}
=
\ub{
\begin{pmatrix}
    E & F \\
    F & G
\end{pmatrix}
}_{P}
\ub{
\begin{pmatrix}
    a_{11} & a_{12} \\
    a_{21} & a_{22}
\end{pmatrix}
}_{A} \]
\[ Q = -PA = A^\top P \]
If $V = D\sigma|_0(\hat{v})$, $w =
D\sigma|_0(\hat{w})$, then
\begin{align*}
    -\hat{v}^\top
    \begin{pmatrix}
        L & M \\
        M & N
    \end{pmatrix}
    \hat{w}
    &=
    -v^\top
    \begin{pmatrix}
        E & F \\
        F & G
    \end{pmatrix}
    \begin{pmatrix}
        a_{11} & a_{12} \\
        a_{21} & a_{22}
    \end{pmatrix}
    \hat{w} \\
    &= I_p(v, -Dn|_p(w)) \\
    &= I_p(-Dn|_p(v), w)
\end{align*}
Then the second FF has an intrinsic form given by
the \emph{symmetric} bilinear form
\[ \II : T_p\Sigma \times T_p\Sigma \to \RR \]
given by
\[ \II(v, w) = I_p(-Dn_p(v), w) \]

\begin{flashcard}[Gauss-curvature]
\begin{definition*}
    Let $\Sigma \subset \RR^3$ smooth surface. The
    \emph{Gauss curvature} \cloze{$\kappa : \Sigma \to
    \RR$ of $\Sigma$ is the function
    \[ p \mapsto \det(Dn|_p) \]}
\end{definition*}
\end{flashcard}

\begin{remark*}
    This is always \emph{well-defined} even if
    $\Sigma$ is not oriented. We can always choose
    a local expression for $n$. If we replace it
    by $-n$, the determinant will \emph{not}
    change (this is because $\det(A) = \det(-A)$
    if $A$ is $2 \times 2$).
\end{remark*}

\noindent
Computing $\kappa$: If we pick $\sigma$ using
(\dag) we see that taking $\det$:
\[ LN - M^2 = (EG - F^2) \kappa \]
hence
\begin{flashcard}[curvature-from-FFF-and-second-FF]
\[ \kappa \fcscrap{= \det(A)} = \cloze{\frac{LN - M^2}
{EG - F^2}} \]
\end{flashcard}

\begin{example*}
    Cylinder we saw last time. We computed second
    fundamental form is
    \[ \sigma(u, v) = (a\cos u, a\sin u, v) \]
    \[
    \begin{pmatrix}
        -a & 0 \\
        0 & 0
    \end{pmatrix}
    \implies \kappa(p) = 0 \]
    for all $p$.
    \begin{center}
        \includegraphics[width=0.3\linewidth]
        {images/1d76a19aad2511ed.png}
    \end{center}
    $n : \Sigma \to \text{Equator} \subset S^2$.
    So if $\gamma : (-\eps, \eps) \to \Sigma$ is a
    vertical curve, then
    \[ Dn|_p(\gamma'(0)) = (n \circ \gamma)'(0) \]
    \[ \implies \det(Dn|_p) = 0 \]
\end{example*}

\begin{flashcard}[flat-surface-in-R3]
\begin{definition*}
    $\Sigma$ is said to be \emph{flat} if \cloze{$\kappa
    \equiv 0$ on $\Sigma$.}
\end{definition*}
\end{flashcard}

\begin{example*}
    If $\Sigma$ is a graph of a smooth function
    $f$, then it is easy to check that
    \begin{align*}
        E
        &= 1 + f_u^2 \\
        G
        &= 1 + f_v^2 \\
        F
        &= f_u f_v \\
        EG - F^2
        &= 1 + f_u^2 + f_v^2 \\
        L
        &= \frac{f_{uu}}{\sqrt{EG - F^2}} \\
        M
        &= \frac{f_{uv}}{\sqrt{EG - F^2}} \\
        N
        &= \frac{f_{vv}}{\sqrt{EG - F^2}} \\
        \kappa
        &= \frac{f_{uu} f_{vv} - f_{uv}^2}{(1 +
        f_u^2 + f_v^2)^2}
    \end{align*}
    So depends on the \emph{Hessian} of $f$.
\end{example*}

\begin{flashcard}[elliptic-hyperbolic-parabolic]
\begin{definition*}
    $\Sigma \subset \RR^3$, $p \in \Sigma$. We say
    that $p$ is:\prompt{ (classifying curvatures)}
    \begin{itemize}
        \item \cloze{\emph{elliptic} if $\kappa(p)
            > 0$}
        \item \cloze{\emph{hyperbolic} if
            $\kappa(p) < 0$}
        \item \cloze{\emph{parabolic} if
            $\kappa(p) = 0$}
    \end{itemize}
\end{definition*}
\end{flashcard}

\subsubsection*{Graphs}

\begin{enumerate}[(1)]
    \item $f(u, v) = \frac{u^2 + v^2}{2}$ at $(0,
        0)$, $\kappa(0, 0, 0) = 1$.
    \item $f(u, v) = \frac{(u^2 - v^2)}{2}$,
        $\kappa(0, 0, 0) = -1$
\end{enumerate}

\begin{flashcard}[planes-elliptic-hyperbolic]
\prompt{Surface lying on sides of tangent plane}
\begin{lemma}
    \begin{enumerate}[(a)]
        \item \cloze{In a sufficiently small
            neighbourhood of an elliptic point
            $p$, $\Sigma$ lies entirely on one
            side of the affine tangent plane $p +
            T_p\Sigma$.}
        \item \cloze{In a sufficiently small
            neighbourhood of a hyperbolic point,
            $\Sigma$ meets both sides of its
            affine tangent plane.}
            \fcscrap{
            \begin{center}
                \includegraphics[width=0.6\linewidth]
                {images/dbb61a5ead2611ed.png}
            \end{center}
            }
    \end{enumerate}
\end{lemma}

\prompt{
\begin{proof}
\cloze{
    Recall that if $w = n\sigma_n + l\sigma_v$
    then $\half \II_p(w, w)$ measures the
    signed distance from $\sigma(h, l)$ to $p +
    T_p\Sigma$ ($\sigma(0) = p$), and gives
    \[ \half(Lh^2 + 2Mhl + Nl^2) + O(h^3, l^3) \]
    If $p$ eliptic, then this quadratic form has
    eigenvalues of the same sign so is either
    always positive or always negative for small
    $h, l$, so $\Sigma$ locally lies on one side
    only. If $p$ hyperbolic, then the quadratic
    form has eigenvalues of opposite sign, so
    for any neighbourhood, the quadratic form is
    negative in places and positive in others, so
    $\Sigma$ always takes values on each side.
}
\end{proof}
}
\end{flashcard}

\begin{proof}
    Take a parametrisation $\sigma$ near $p$.
    \[ \kappa = \frac{LN - M^2}{EG - F^2} \]
    and $EG - F^2 > 0$. Recall also that if
    \[ w = n\sigma_n + l\sigma_v \in T_p\Sigma ,\]
    then $\half \II_p(w, w)$ measured the signed
    distance from $\sigma(h, l)$ to $p +
    T_p\Sigma$ ($\sigma(0) = p$) measured via the
    inner product with positive normal:
    \[ \half (Lh^2 + 2M hl + Nl^2) + O(h^3, l^3) \]
    $p$ elliptic implies
    \[
    \begin{pmatrix}
        L & M \\
        M & N
    \end{pmatrix}
    \]
    has eigenvalues of the same sign, so positive
    or negative definite at $p$, so in a
    neighbourhood of $p$, $\kappa$ is signed
    distance only has one sign locally.

    \myskip
    If $p$ is hyperbolic, then $\II_p$ is
    indefinite, so $\Sigma$ meets both sides of $p
    + T_p\Sigma$.
\end{proof}

\begin{remark*}
    If $p$ is parabolic we can't conclude either.
\end{remark*}