% vim: tw=50
% 06/02/2023 11AM

\begin{proof}
    $\sigma : V \to U \subset \Sigma$, $\sigma(0)
    = p$, $\tilde{\sigma} : \tilde{V} \to
    \tilde{U} \subset \Sigma$, $\tilde{\sigma}(0)
    = p$. Transition map $\sigma^{-1} \circ
    \tilde{\sigma}$ implies:
    \[ \tilde{\sigma} = \sigma \circ (\sigma^{-1}
    \circ \tilde{\sigma}) \]
    So $D(\sigma^{-1} \circ \tilde{\sigma})|_0$
    isan isomorphism. Chain rule implies
    \[ \Im(D\tilde{\sigma}|_0) = \Im(D\sigma|_0) \]
\end{proof}

\begin{definition*}
    $\Sigma \subset \RR^3$. The normal direction
    at $p$ is $(T_p \Sigma)^\perp$ (Euclidean
    orthogonal ocmplement to $T_p\Sigma$). At each
    $p \in \Sigma$ we have two unit normal
    vectors.
\end{definition*}

\begin{flashcard}[two-sided-surface-in-R3]
\begin{definition*}
    A smooth surface \cloze{in $\RR^3$} is
    \emph{two-sided} \cloze{if it admits a
    \fcemph{continuous} global choice of
    \fcemph{unit} normal vectors.}
\end{definition*}
\end{flashcard}

\begin{lemma}
    A smooth surface in $\RR^3$ is orientable with
    its abstract smooth surface structure if and
    only if it is two-sided.
\end{lemma}

\begin{proof}
    Let $\sigma : V \to U \subset \Sigma$ be
    allowable. Define the \emph{positive} unit
    normal with respect to $\sigma$ at $p$ as the
    unique $n_\sigma(p)$ such that $\{\sigma_u,
    \sigma_v, n_\sigma(p)\}$ and $\{e_1, e_2,
    e_3\}$ induce the same orientation in $\RR^3$
    (ie they are related by a choice of basis
    matrix with positive determinant).
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/8a93721aa61011ed.png}
    \end{center}
    Explicitly:
    \[ n_\sigma(p) = \frac{\sigma_n \times
    \sigma_v}{\|\sigma_u \times \sigma_v\|} \]
    Let $\tilde{\sigma}$ be another allowable
    parametrisation.
    \[ \tilde{\sigma} : \tilde{V} \to \tilde{U}
    \subset \Sigma \]
    and suppose $\Sigma$ is orientable as an
    abstract smooth surface with $\sigma$ and
    $\tilde{\sigma}$ belonging to the same
    oriented atlas.
    \[ \sigma = \tilde{\sigma} \circ \varphi
    \qquad \varphi = \tilde{\sigma}^{-1} \circ
    \sigma \]
    and
    \[ D\varphi|_0 =
    \begin{pmatrix}
        \alpha & \beta \\
        \gamma & \delta
    \end{pmatrix}
    \]
    Chain rule:
    \begin{align*}
        \sigma_u
        &= \alpha
        \tilde{\sigma}_{\tilde{w}} + \gamma
        \tilde{\sigma}_{\tilde{v}}
        \sigma_v
        &= \beta
        \tilde{\sigma}_{\tilde{w}} + \delta
        \tilde{\sigma}_{\tilde{v}}
    \end{align*}
    and
    \[ \sigma_u \times \sigma_v =
    \ub{\det(D\varphi|_0)}_{> 0} \tilde{\sigma}_{\tilde{u}}
    \times \tilde{\sigma}_{\tilde{v}} \tag{$*$} \]
    So the positive unit normal at $p$ was
    intrinsic and does not depend on the
    parametrisation. Since
    \[ \frac{\sigma_u \times \sigma_v}{\|\sigma_u
    \times \sigma_v\|} \]
    is continuous, $\Sigma$ is 2-sided.

    \myskip
    Conversely, if $\Sigma$ is 2-sided we have a
    global choice of $n$, so we can consider the
    subatlas of the smooth atlas such that we have
    a chart $(U, \varphi) = \varphi^{-1} = \sigma$
    and $\{\sigma_u, \sigma_v, u\}$ is an
    orientable basis of $\RR^3$. ($*$) shows that
    transition maps between such charts are
    orientation preserving. Hence $\Sigma$ is
    orientable.
\end{proof}

\begin{hiddenflashcard}[orientable-iff-two-sided]
\begin{lemma*}
    A smoth surface in $\RR^3$ is orientable with
    its abstract smooth surface structure if and
    only if it is two-sided.
\end{lemma*}

\begin{proof}
    \begin{enumerate}
        \item[$(\Rightarrow)$] \cloze{Define
            \[ n_\sigma (u, v) =
            \frac{\sigma_u \times \sigma_v}
            {\|\sigma_u \times \sigma_v\|} \]
            and show that it is independent of the
            choice of parametrisation (when
            working within a particular oriented
            atlas.}
        \item[$(\Leftarrow)$] \cloze{If $\Sigma$ is
            two-sided, then we have a global
            choice of $n$, and we construct the
            atlas such that each chart $(U,
            \varphi)$ always has $\{\sigma_u, \sigma_v,
            n\}$ being the same orientation.}
    \end{enumerate}
\end{proof}
\end{hiddenflashcard}

\begin{remark*}
    Given $\gamma : (-\eps, \eps) \to \RR^3$
    smooth with $\Im(\gamma) \subset \Sigma$ and
    $\gamma(0) = p$
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/89c97fe4a61211ed.png}
    \end{center}
    \[ \gamma(t) = \sigma(u(t), v(t)) \]
    \[ \gamma'(0) = D\sigma|_0 (u'(0), v'(0)) \in
    T_p \Sigma \]
    This gives that
    \begin{align*}
        T_p\Sigma
        &= \{\gamma'(0) : \text{with
        $\gamma$ as above}\} \\
        &= \text{``tangent vector to curves in
        $\Sigma$''}
    \end{align*}
\end{remark*}

\subsubsection*{Examples}

\begin{enumerate}[(1)]
    \item $S^2 \subset \RR^3$
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/da97c6eca61211ed.png}
        \end{center}
        Take any $\gamma : (-\eps, \eps) \to S^2$,
        $\gamma(0) = p$, $|\gamma(t)|^2 = 1$.
        Differentiation at $t = 0$
        \begin{align*}
            \langle \gamma'(0), p \rangle
            &= 0 \\
            \implies (T_p S^2)^\perp
            &= \RR p \\
            \implies n(p)
            &= p
        \end{align*}
        $S^2$ is 2-sided.
    \item The M\"obius band, start with unit
        circle in the $xy$-plane.
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/88947456a61411ed.png}
        \end{center}
        and take an open interval of length $1$.
        Rotate this line in the $cz$-plane s we
        move around the circle such that it has
        rotate by $\frac{\theta}{2}$ after moving
        an angle $\theta$ in the circle (see
        picture). After a full turn the segment
        returns to its original position by iwth
        end points inverted. We can describe the
        surface with
        \[ \sigma(t, \theta) = \left( \left( 1 -
        t\sin \frac{\theta}{2} \right) \cos\theta,
        \left( 1 - t\sin \frac{\theta}{2} \right) \sin
        \theta , t \cos \frac{\theta}{2} \right)  \]
        where $(t, \theta)$ belongs to
        \begin{align*}
            V_1
            &= \left\{ t \in \left( -\half,
            \half \right), \theta \in (0, 2\pi)
            \right\} \\
            &\text{or} \\
            V_2
            &= \left\{ t \in \left( -\half,
            \half \right), \theta \in (-\pi, \pi)
            \right\} \\
        \end{align*}
        Check: if $\sigma_i$ is $\sigma$ on $V_i$
        then $\sigma_i$ is allowable. A
        computation shows
        \[ \sigma_t \times \sigma_\theta(0,
        \theta) = \left(-\cos\theta \cos
        \frac{\theta}{2}, -\sin\theta \cos
        \frac{\theta}{2}, -\sin
        \frac{\theta}{2}\right) \defeq n_\theta \]
        As $\theta \to 0^+$, $h_\theta \to (-1, 0,
        0)$. As $\theta \to 2\pi^-$, $n_\theta \to
        (1, 0, 0)$. Hence the M\"obius band is
        \emph{not} 2-sided.
\end{enumerate}

\newpage
\section{Surfaces in 3-space}

$\gamma : (a, b) \to \RR^3$ smooth, the length of
$\gamma$ is
\begin{flashcard}[length-of-gamma-in-R3]
\prompt{What is the length of $\gamma : (a, b) \to
\RR^3$?}
\[ L(\gamma) = \cloze{\int_a^b \|\gamma'(t)\| \dd
t} \]
\end{flashcard}
If $s : (A, B) \to (a, b)$ is monotone increasing
and we let $\tau(t) = \gamma(s(t))$, then
\begin{align*}
    L(\tau)
    &= \int_A^B \|\tau'(t)\| \dd t \\
    &= \int_A^B \|\gamma'(s(t))\| |\ub{s'(t)}_{\ge
    0}| \dd t \\
    &= \int_a^b \|\gamma'(s)\| \dd s \\
    &= L(\gamma)
\end{align*}

\begin{lemma}
    If $\gamma : (a, b) \to \RR^3$ and $\gamma'(t)
    \neq 0$ for all $t$, then $\gamma$ can be
    paramaterised by \emph{arc-length}, i.e. a
    parameter such that $\|\gamma'(s)\| = 1$ for
    all $s$.
\end{lemma}