% vim: tw=50
% 08/02/2023 11AM

\noindent
$\Sigma \subset \RR^3$, $\sigma : V \to U \subset
\Sigma$ allowable. If $\gamma : (a, b) \to U$ is
smooth, write
\[ \boxed{\gamma(t) = \sigma(u(t), v(t))} \]
with $(u, v) : (a, b) \to V$ smooth. Then
\[ \gamma'(t) = \sigma_u u'(t) + \sigma_v v'(t) \]
then
\begin{align*}
    \|\gamma'(t)\|^2
    &= E(u'(t))^2 + 2Fu'(t) v'(t) + G(v'(T))^2
\end{align*}
where
\begin{align*}
    E
    &= \langle \sigma_u, \sigma_u \rangle
    = \|\sigma_u\|^2 \\
    F
    &= \langle \sigma_u, \sigma_v \rangle
    = \langle \sigma_v, \sigma_u \rangle \\
    G
    &= \langle \sigma_v, \sigma_v \rangle
    = \|\sigma_v\|^2
\end{align*}
(smooth functions on $V$). They depend only on
$\sigma$ and \emph{not} $\gamma$.

\begin{definition*}
    The \emph{first fundamental form} (FFF) in the
    parametrisation $\sigma$ is the expression
    \[ E \dd u^2 + 2F \dd u \dd v + G \dd v^2 \]
    The notation is designed to remind you that
    \[ L(\gamma) = \int_a^b \sqrt{Eu'^2 + 2F u'v'
    + G v'^2} \dd t ,\]
    where $\gamma(t) = \sigma(u(t), v(t))$.
\end{definition*}

\begin{remark*}
    The FFF is  sometimes defined as the quadratic
    form in $T_p \Sigma$ given by the restriction
    of the standard inner product in $\RR^3$.
    \[ I_p(w) = |w|^2 = \langle w, w
    \rangle_{\RR^3}, w \in T_p\Sigma \]
    After picking $\sigma$, $\sigma(0) = p$ and
    after writing $w = D\sigma|_0 (u', v')$ we
    have
    \[ I_p(w) = Eu'^2 + 2F u'v' + Gv'^2 \]
    This is an example of a \emph{Riemannian
    metric}.
\end{remark*}

\subsubsection*{Examples}

\begin{enumerate}[(1)]
    \item The $xy$-plane $\RR^3$:
        \[ \sigma(u, v) = (u, v, 0) \]
        so
        \[ \sigma_u = (1, 0, 0), \qquad \sigma_v =
        (0, 1, 0)\]
        and FFF is $\dd u^2 + \dd v^2$, $E = G =
        1$, $F = 0$. In polar coordinates
        \[ \sigma(r, \theta) = (r\cos\theta,
        r\sin\theta, 0) \]
        $r \in (0, \infty$, $(0, 2\pi)$, we have
        \[ \sigma_r = (\cos\theta, \sin\theta, 0) \]
        \[ \sigma_\theta = (-r\sin\theta,
        r\cos\theta, 0) \]
        and FFF is $\dd r^2 + r^2 \dd \theta^2$,
        $E = 1$, $F = 0$, $G = r^2$.
        \[
        \begin{pmatrix}
            E & F \\
            F & G
        \end{pmatrix}
        \]
\end{enumerate}

\begin{flashcard}[isometric-surfaces-in-R3]
\begin{definition*}
    $\Sigma, \Sigma' \subset \RR^3$ smooth
    surfaces. We say that $\Sigma$ and $\Sigma'$
    are \emph{isometric} if \cloze{there exists
    $f : \Sigma
    \to \Sigma'$ \fcemph{diffeomorphic} such that for every
    \fcemph{smooth} curve $\gamma : (a, b) \to \Sigma$,
    \[ L_\Sigma(\gamma) = L_{\Sigma'}(f \circ
    \gamma) \]}
\end{definition*}
\end{flashcard}

\begin{example*}
    If $\Sigma' = f(\Sigma)$ where $f : \RR^3 \to
    \RR^3$ is a ``rigid motion'' i.e. $f(x) = Ax +
    b$, $A \in O(3)$, $b \in \RR^3$ (so $f$
    preserves $\langle \bullet, \bullet
    \rangle_{\RR^3}$). $f : \Sigma \to \Sigma'$
    \emph{isometry} because
    \begin{align*}
        |(f \circ \gamma)'(t)|
        &= |A\gamma'(t)| \\
        &= |\gamma'(t)|
    \end{align*}
    hence lengths of curves are preserved. Often
    we're interested in local statements.
\end{example*}

\begin{flashcard}[locally-isometric-surfaces-in-R3]
\begin{definition*}
    $\Sigma, \Sigma'$ are \emph{locally isometric}
    (near points $p \in \Sigma$ and $p' \in
    \Sigma'$) \cloze{if there exists open neighbourhoods
    $p \in U \subset \Sigma$ and $p' \in U'
    \subset \Sigma'$ which are isometric.}
\end{definition*}
\end{flashcard}

\begin{lemma}
    $\Sigma, \Sigma' \subset \RR^3$ are locally
    isometric near $p \in \Sigma$ and $p' \in
    \Sigma'$ if and only if there exists allowable
    parametrisation
    \begin{align*}
        \sigma
        &: V \to U \subset \Sigma \\
        \sigma'
        &: V \to U' \subset \Sigma'
    \end{align*}
    for which the FFFs are equal in $V$ ($E = E'$,
    $F = F'$, $G = G'$).
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/13b4f54ca7a511ed.png}
    \end{center}
\end{lemma}

\begin{hiddenflashcard}[locally-isometric-surfaces-in-R3-iff]
    $\Sigma$ and $\Sigma'$ are locally isometric
    near $p \in \Sigma$ and $p' \in \Sigma'$ if
    and only if \cloze{there exists allowable
    parametrisations $\sigma : V \to U \subset
    \Sigma$ and $\sigma' : V \to U' \subset
    \Sigma'$ for which the FFFs are equal in $V$
    ($E = E'$, $F = F'$, $G = G'$).}
\end{hiddenflashcard}

\begin{hiddenflashcard}[locally-isometric-surfaces-in-R3-iff-proof]
\begin{lemma*}
    $\Sigma, \Sigma' \subset \RR^3$ are locally
    isometric near $p \in \Sigma$ and $p' \in
    \Sigma'$ if and only if there exists allowable
    parametrisation
    \begin{align*}
        \sigma
        &: V \to U \subset \Sigma \\
        \sigma'
        &: V \to U' \subset \Sigma'
    \end{align*}
    for which the FFFs are equal in $V$ ($E = E'$,
    $F = F'$, $G = G'$).
\end{lemma*}

\begin{proof}
    \begin{enumerate}
        \item[$(\Rightarrow)$] \cloze{Show that
            \[ \left| \dfrac{}{t} \ub{\sigma'
            \circ \sigma^{-1}}_{f} \circ \gamma \right|^2
            = \left| \dfrac{}{t} \gamma(t)
            \right|^2 \]
            hence $f$ is an isometry, so they're
            locally isometric.
            }
        \item[$(\Leftarrow)$] \cloze{Determine $E$
            by considering
            $\dfrac{}{\eps}L(\gamma_\eps)$ where
            $\gamma_\eps : [0, \eps] \to \Sigma$,
            $t \mapsto \sigma(t, 0)$. Similarly,
            use $\chi_\eps, \lambda_\eps: [0,
            \eps]$ with $\chi_\eps(t) = \sigma(0,
            t)$ and $\lambda_\eps(t) = \sigma(t,
            t)$.}
    \end{enumerate}
\end{proof}
\end{hiddenflashcard}

\begin{proof}
    We know (by definition) that the FFF of
    $\sigma$ determines the lengths of all curves
    on $\sigma(V) = U$. If we have $\sigma$ and
    $\sigma'$ as in the lemma, then
    \[ \sigma' \circ \sigma^{-1} : U \to U' \]
    is an isometry since
    \[ \sigma^{-1}(\gamma(t)) = (u(t), v(t)) \]
    \begin{align*}
        \left| \dfrac{}{t} \ub{\sigma' \circ
        \sigma^{-1}}_{f} \circ \gamma \right|^2
        &= \left| \dfrac{}{t} \sigma'(u(t), v(t))
        \right|^2 \\
        &= E' \dot{u}^2 + 2F' \dot{u} \dot{v} + G'
        \dot{v}^2 \\
        &= E \dot{u}^2 + 2F \dot{u} \dot{v} + G
        \dot{v}^2 \\
        &= \left| \dfrac{}{t} \gamma(t) \right|^2
        \implies l(\sigma' \circ \sigma^{-1} \circ
        \gamma)
        &= L(\gamma)
    \end{align*}
    For the converse, we'll show first that the
    lengths of the curves in $U$ determine the FFF
    of $\sigma$.
    \[ \sigma : (0, f) \to U \subset \Sigma,
    \sigma(0) = p \]
    $\gamma_\eps : [0, \eps] \to U$, $t \mapsto
    \sigma(t, 0)$. Then
    \begin{align*}
        \dfrac{}{\eps} L(\gamma_\eps)
        &= \dfrac{}{\eps} \int_0^\eps \sqrt{E(t,
        0)} \dd t \\
        &= \sqrt{E(\eps, 0)}
    \end{align*}
    So length of curve determine $E(0, 0)$.
    Similarly, $\chi_\eps : [0, \eps] \to U$, $t
    \mapsto \sigma(0, t)$ determine $G(0, 0)$.
    Finally considering $\lambda_\eps : [0, \eps]
    \to U$, $t \mapsto \sigma(t, t)$ we get
    $\sqrt{(E + 2F + G)(0, 0)}$, so knowing $E$
    and $G$ we get $F$. So if $f : U \to U'$ is a
    local isometry take any allowable
    parametrization $\sigma' : V \to U'$, then
    $\sigma = f^{-1} \circ \sigma'$ is such that the
    FFFs of $\sigma$ and $\sigma'$ agree.
\end{proof}

\begin{example*}[Cone]
    $u > 0$, $v \in (0, 2\pi)$.
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/240c0488a7a711ed.png}
    \end{center}
    \[ \sigma(u, v) = (au \cos v, au \sin v, u) \]
    parametrizes the complement of the line of the
    cone. FFF:
    \[ (1 + a^2) \dd u^2 + a^2 u^2 \dd v^2 \]
    If we cut open the cone and unfold it we get a
    plane such that
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/3d457754a7a711ed.png}
    \end{center}
    $\theta_0 = \frac{2\pi a}{\sqrt{1 + a^2}}$.
    Parametrize the plane sector by
    \[ \sigma(r, \theta) = \left(\sqrt{1 + a^2}
    r\cos \left( \frac{a\theta}{\sqrt{1 + a^2}}\right),
    \sqrt{1 + a^2} r \sin \left(
    \frac{a\theta}{\sqrt{1 + a^2}} \right) , 0 \right) \]
    $r > 0$, $\theta \in (0, 2\pi)$
    Check: FFF: $(1 + a^2) \dd r^2 + r^2 a^2 \dd
    \theta^2$, $V = (0, \infty) \times (0, 2\pi)$.
    So the cone is locally isometric to the plane!
\end{example*}

\begin{remark*}
    The cone can't be \emph{globally isometric} to
    the plane, since they are not even
    homeomorphic
\end{remark*}