% vim: tw=50
% 13/02/2023 11AM

\subsubsection*{The Second Fundamental Form}

Let's try to measure how much $\Sigma \subset
\RR^3$ deviates from its own tangent planes.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/c5780c7cab8f11ed.png}
\end{center}
Let's take $\sigma : V \to U \subset \Sigma$.
Using Taylor's theorem
\begin{align*}
    \sigma(u + h, v + l)
    &= \sigma(u, v) + h\sigma_u(u, v) +
    l\sigma_v(u, v) + \half(h^2\sigma_{uu}(u, v)
    + 2hl \sigma_{uv}(u, v) + l^2 \sigma_{vv} (u,
    v)) + O(h^3, l^3)
\end{align*}
where $(h, l)$ are small enough so that $(u, v)$
and $(u + h, v + l) \in V$. Let's take projection
in the normal direction:
\begin{align*}
    \langle n, \sigma(u + h), v + l) - \sigma(u,
    v) \rangle
    &= \half (\langle n, \sigma_{uu} \rangle h^2 +
    2\langle n, \sigma_{ub} \rangle hl + \langle
    n, \sigma_{vv} \rangle l^2) + O(h^3, l^3)
\end{align*}

\begin{flashcard}[second-fundamental-form]
\begin{definition*}
    The \emph{second fundamental form} of $\Sigma
    \subset \RR^3$ in the parametrisation $\sigma$
    is the quadratic form:
    \[ L\dd u^2 + 2M \dd u \dd v + N \dd v^2 \]
    where
    \begin{align*}
        L
        &= \cloze{\langle n, \sigma_{uu} \rangle} \\
        M
        &= \cloze{\langle n, \sigma_{uv} \rangle} \\
        N
        &= \cloze{\langle n, \sigma_{vv} \rangle}
    \end{align*}
    and \cloze{$n = \frac{\sigma_u \times
    \sigma_v}{\|\sigma_u \times \sigma_v\|}$.}
\end{definition*}
\end{flashcard}

\begin{lemma}
    Let $V$ be connected and $\sigma : V \to U
    \subset \Sigma$ such that second FFF vanishes
    identically. Then $U$ lies in an affine plane
    in $\RR^3$.
\end{lemma}

\begin{proof}
    Recall $\langle n, \sigma_u \rangle = \langle
    n, \sigma_v \rangle = 0$. Hence
    \begin{align*}
        \langle n_u, \sigma_u \rangle + \langle n,
        \sigma_{uu} \rangle
        &= 0 \\
        \langle n_v, \sigma_v \rangle + \langle n,
        \sigma_{vv} \rangle
        &= 0 \\
        \langle n_v, \sigma_u \rangle + \langle n,
        \sigma_{uv} \rangle
        &= 0
    \end{align*}
    \begin{align*}
        L
        &= \langle n, \sigma{uu} \rangle
        = -\langle n_u, \sigma_u \rangle \\
        M
        &= \langle n, \sigma_{uv} \rangle
        = -\langle n_v, \sigma_u \rangle
        = -\langle n_u, \sigma_v \rangle \\
        N
        &= \langle n, \sigma_{vv} \rangle
        = -\langle n_v, \sigma_v \rangle
    \end{align*}
    So if second FFF vanishes then $n_u$ is
    orthogonal to $\sigma_u$ and $\sigma_v$. Also
    $\langle n, n \rangle = 1$ implies
    (differentiate with respect to $u$) $\langle
    n, n_u \rangle = 0$. Hence $n_u$ is orthogonal
    to $\{n, \sigma_u \sigma_v\}$, hence $n_u
    \equiv 0$. Similarly $n_u \equiv 0$. So $n$ is
    \emph{constant} ($V$ connected and mean value
    inequality). This implies that $\langle
    \sigma, n \rangle = \text{constant}$ (since
    $\langle \sigma, n_u \rangle = \langle
    \sigma_u, n \rangle = \langle \sigma, n_v
    \rangle = \langle \sigma_v, n \rangle = 0$)
    and $U$ is contained in a plane.
\end{proof}

\begin{remark*}
    Recall that FFF in the parametrisation
    $\sigma$ has
    \[ (D\sigma)^\top D\sigma =
    \begin{pmatrix}
        E & F \\
        F & G
    \end{pmatrix}
    =
    \begin{pmatrix}
        \sigma_u \cdot \sigma_v & \sigma_u \cdot
        \sigma_v \\
        \sigma_v \cdot \sigma_u & \sigma_v \cdot
        \sigma_v
    \end{pmatrix}
    \]
    Analogously the second FFF:
    \[ -(Dn)^\top D\sigma =
    \begin{pmatrix}
        L & M \\
        M & N
    \end{pmatrix}
    =
    \begin{pmatrix}
        n_u \cdot \sigma_u & n_u \cdot \sigma_v \\
        n_v \cdot \sigma_u & n_v \cdot \sigma_v
    \end{pmatrix}
    \]
    (using the alternative exrpessions for $L$,
    $M$ and $N$ in the previous proof). So if
    $\sigma : V \to U \subset \Sigma$,
    $\tilde{\sigma} : \tilde{V} \to U \subset
    \Sigma$ are 2 parametrisations with transition
    map
    \[ \varphi : \tilde{V} \to V, \qquad \varphi =
    \sigma^{-1} \circ \tilde{\sigma} \]
    ($\tilde{\sigma} = \sigma \varphi$) then
    \[ n_{\tilde{\sigma}}(\tilde{u}, \tilde{v}) =
    \pm n_\sigma (\varphi(\tilde{u}, \tilde{v})) \]
    (use the $-$ sign if $\det(D\varphi) < 0$).
    Recall the discusssion on orientabioilty Lemma
    1.10).
    \begin{align*}
        \begin{pmatrix}
            \tilde{L} & \tilde{M} \\
            \tilde{M} & \tilde{N}
        \end{pmatrix}
        &= -(Dn_{\tilde{\sigma}})^\top
        D\tilde{\sigma} \\
        &= \pm (D\varphi)^\top
        \begin{pmatrix}
            L & M \\
            M & N
        \end{pmatrix}
        D\varphi
    \end{align*}
    (use $-$ sign if $\varphi$ is orientation
    preserving).
\end{remark*}

\begin{hiddenflashcard}[second-fundamental-form-matrix-expression]
\[
    \begin{pmatrix}
        L & M \\
        M & N
    \end{pmatrix}
    =
    \cloze{-(Dn)^\top D\sigma}
\]
\end{hiddenflashcard}

\begin{example*}
    The cylinder has
    \[ \sigma(u, v) = (a\cos u, \sin u, v) \]
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/bfe3eaacab9311ed.png}
    \end{center}
    Note $\sigma_{uv} = \sigma_{vv} = 0$ hence $M
    = N = 0$. Check that second FFF is
    \[
    \begin{pmatrix}
        -a & 0 \\
        0 & 0
    \end{pmatrix}
    \]
    ($-adu^2$).
\end{example*}

\begin{flashcard}[Gauss-map]
\begin{definition*}[The gauss map]
    Let \cloze{$\Sigma \subset \RR^3$ be a smooth
    \emph{oriented} surface}. The \emph{Gauss map}
    is
    \[ \cloze{n : \Sigma \to S^2 = \{x \in \RR^3 : |x| =
    1\}} \]
    is the map $p \mapsto n(p)$, \cloze{the unit normal
    vector at $p$ (well-defined by the orientation
    of $\Sigma$).}
\end{definition*}
\end{flashcard}

\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/1638c332ab9411ed.png}
\end{center}

\begin{lemma}
    The Gauss map $n : \Sigma \to S^2$ is
    \emph{smooth}.
\end{lemma}

\begin{proof}
    Smoothness can be checked locally. If $\sigma
    : V \to U \subset \Sigma$ is allowable and
    compatible with the orientation, then at
    $\sigma(u, v) = p \in \Sigma$
    \[ n(\sigma(u, v)) = \frac{\sigma_u \times
    \sigma_v}{\|\sigma_u \times \sigma_v\|} \]
    (smooth since $\sigma$ is).
    $n \circ \sigma : V \to S^2 \subseteq \RR^3$.
\end{proof}

\begin{center}
    \includegraphics[width=0.8\linewidth]
    {images/8d31d23aab9411ed.png}
\end{center}

\begin{note*}
    $T_p\Sigma = T_{n(p)} S^2$. Thus we can view
    \[ Dn|_p : T_p \Sigma \to T_{n(p)} S^2 =
    T_p\Sigma \]
    (recall Q9, Example Sheet 1).
\end{note*}

\noindent
We can also view $Dn|p$ acting on tangent vectors
in terms of curves
\[ \gamma : (-\eps, \eps) \to \Sigma, \qquad
\gamma(1) = p, \quad \gamma'(0) = v \]
\[ Dn|_p(v) = Dn|_p(\gamma'(0)) = (n \circ
\gamma)'(0) \]
(by chain rule).