% vim: tw=50
% 10/02/2023 11AM

$\Sigma \ub \RR^3$, $p \in \Sigma$. Take allowable
parametrisations:
\begin{align*}
    \sigma
    &: V \to U \subset \Sigma, \qquad \sigma(0) =
    p \\
    \tilde{\sigma}
    &: \tilde{V} \to U \subset \Sigma, \qquad
    \tilde{\sigma}(0) = p
\end{align*}
Transition map $f : \tilde{\sigma}^{-1} \circ
\sigma : V \to \tilde{V}$. We have FFFs:
\[
\begin{pmatrix}
    E & F \\
    F & G
\end{pmatrix}
\qquad \text{for $\sigma$} \]
\[
\begin{pmatrix}
    \tilde{E} & \tilde{F} \\
    \tilde{F} & \tilde{G}
\end{pmatrix}
\qquad \text{for $\tilde{\sigma}$} \]

\begin{flashcard}[change-of-FFF-formula]
\begin{lemma}
    If $f = \tilde{\sigma}^{-1} \circ \sigma$ is a
    \fcemph{transition map}, then
    \[
    \begin{pmatrix}
        \cloze{E} & \cloze{F} \\
        \cloze{F} & \cloze{G}
    \end{pmatrix}
    = \cloze{(Df)^\top}
    \begin{pmatrix}
        \cloze{\tilde{E}} & \cloze{\tilde{F}} \\
        \cloze{\tilde{F}} & \cloze{\tilde{G}}
    \end{pmatrix}
    \cloze{Df} \]
    \fcscrap{($^\top$ is transpose)}
\end{lemma}

\prompt{
\begin{proof}
    \cloze{
    \begin{align*}
        \begin{pmatrix}
            E & F \\
            F & G
        \end{pmatrix}
        &= \cdots \\
        &= (D\sigma)^\top (D\sigma) \\
        &= (D\tilde{\sigma} Df)^\top (D\tilde{\sigma}
        Df) \\
        &= (Df)^\top (D\tilde{\sigma})^\top
        D\tilde{\sigma} Df \\
        &= (Df)^\top
        \begin{pmatrix}
            \tilde{E} & \tilde{F} \\
            \tilde{F} & \tilde{G}
        \end{pmatrix}
    \end{align*}
    }
\end{proof}
}
\end{flashcard}

\begin{proof}
    \begin{align*}
        \begin{pmatrix}
            E & F \\
            F & G
        \end{pmatrix}
        &=
        \begin{pmatrix}
            \sigma_u \cdot \sigma_u & \sigma_u
            \cdot \sigma_v \\
            \sigma_v \cdot \sigma_u & \sigma_v
            \cdot \sigma_v
        \end{pmatrix}
        \\ 
        &=
        \begin{pmatrix}
            \sigma_u & \sigma_u \\
            \sigma_v & \sigma_v
        \end{pmatrix}
        \begin{pmatrix}
            \sigma_u & \sigma_v \\
            \sigma_u & \sigma_v
        \end{pmatrix}
        \\
        &= (D\sigma)^\top (D\sigma) \\
        &= (D\tilde{\sigma} Df)^\top
        (D\tilde{\sigma} Df)
        &&\text{using$\sigma = \tilde{\sigma}
        \circ f$} \\
        &= (Df)^\top (D\tilde{\sigma})^\top
        D\tilde{\sigma} Df \\
        &= (Df)^\top
        \begin{pmatrix}
            \tilde{E} & \tilde{F} \\
            \tilde{F} & \tilde{G}
        \end{pmatrix}
        Df \qedhere
    \end{align*}
\end{proof}

\subsubsection*{FFF and Angles}

\begin{center}
    \includegraphics[width=0.3\linewidth]
    {images/f18af1e2a93411ed.png}
\end{center}
$v, w \in \RR^3$, $v \cdot w = |v| |w|
\cos\theta$. If $v, w \in T_p\Sigma$, $\cos\theta
= \frac{v \cdot w}{|v| |w|}$ ($*$).
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/2588bccca93511ed.png}
\end{center}
\begin{align*}
    w
    &= D\sigma|_0 (w_0) \\
    v
    &= D\sigma|_0 (v_0) \\
    v \cdot w
    &= v_0^\top
    \begin{pmatrix}
        E & F \\
        F & G
    \end{pmatrix}
    w_0
\end{align*}
So using ($*$) we can compute angles using the FFF
of $\sigma$.

\begin{flashcard}[conformal-FFF]
\begin{lemma}
    $\sigma$ is \emph{conformal} (preserves
    angles) \cloze{exactly when $E = G$ and $F =
    0$.}
\end{lemma}

\prompt{
\begin{proof}
\cloze{
    Let $\alpha(t) = (u(t), v(t))$,
    $\tilde{\alpha}(t) = (\tilde{u}(t),
    \tilde{v}(t))$ be curves with non-zero
    derivative that intersect for $t
    = 0$.
    Use $\cos(\theta) = \frac{v \cdot w}{|v|
    |w|}$.
    We calculate:
    \begin{align*}
        w
        &= D\sigma|_0 (w_0) \\
        &= (E\dot{u}^2 + 2F\dot{u}\dot{v} +
        G\dot{v}^2)^{1/2} \\
        v
        &= D\sigma|_0 (v_0) \\
        &= (E\dot{\tilde{u}}^2 +
        2F\dot{\tilde{u}}\dot{\tilde{v}} +
        G\dot{\tilde{v}}^2)^{1/2} \\
        v \cdot w
        &= v_0^\top
        \begin{pmatrix}
            E & F \\
            F & G
        \end{pmatrix}
        w_0 \\
        &= (E\dot{u}\dot{\tilde{u}} +
        F(\dot{u}\dot{\tilde{v}} +
        \dot{\tilde{u}}\dot{v}) +
        G\dot{v}\dot{\tilde{v}})
    \end{align*}
    For the forward direction, consider $\alpha(t)
    = (t, 0)$, $\tilde{\alpha}(t) = (0, t)$. Then
    the angle is $\frac{\pi}{2}$, so $\cos\theta =
    0$ then we deduce $F = 0$. Similarly use
    $\alpha(t) = (t, t)$, $\tilde{\alpha}(t) = (t,
    -t)$ to deduce $E = G$.

    \myskip
    If $E = G$ and $F = 0$ then
    \[ \cos \theta = \frac{\dot{u} \dot{\tilde{u}}
    + \dot{v} \dot{\tilde{v}}}{(\dot{u}^2 +
    \dot{v}^2)^{1/2}(\dot{\tilde{u}} +
    \dot{\tilde{v}}^2)^{1/2}} \]
    hence conformal.
}
\end{proof}
}
\end{flashcard}

\begin{proof}
    Consider curves
    \begin{align*}
        \alpha(t)
        &= (u(t), v(t)) \\
        \tilde{\alpha}(t)
        &= (\tilde{u}(t), \tilde{v}(t))
    \end{align*}
    $\alpha(0) = \tilde{\alpha}(0) \in V$. The
    curves $\sigma \circ \alpha$ and $\sigma \circ
    \tilde{\alpha}$ meet at $p$ with angle
    $\theta$ given by
    \[ \cos \theta = \frac{E
    \dot{u}\dot{\tilde{u}} + F( \dot{u}
    \dot{\tilde{v}} + \dot{\tilde{u}} \dot{v}) + G
    \dot{v} \dot{\tilde{v}}}{(E \dot{u}^2 + 2F
    \dot{u} \dot{v} + G
    \dot{v}^2)^{1/2}(E\dot{\tilde{u}}^2 + 2F
    \dot{\tilde{u}} \dot{\tilde{v}} + G
    \dot{\tilde{v}}^2)^{1/2}}
    \tag{\dag} \]
    If $\sigma$ is conformal and $\alpha(t) = (t,
    0)$, $\tilde{\alpha}(t) = (0, 1)$ meeting at
    angle $\frac{\pi}{2}$ in $V$, we get using
    (\dag)
    \[ 0 = F \]
    Similarly using $\alpha(t) = (t, t)$ and
    $\tilde{\alpha}(t) = (t, -t)$ we get $E = G$.

    \myskip
    Conversely, if $\sigma$ is such that $E = G$
    and $F = 0$ then with respect to the FFF is
    just
    \[ \rho(\dd u^2 + \dd v^2) ,\]
    where $\rho = E = G : V \to \RR$. From (\dag)
    we see that
    \[ \cos\theta = \frac{\dot{u} \dot{\tilde{u}}
    + \dot{v} \dot{\tilde{v}}}{(\dot{u}^2 +
    \dot{v}^2)^{1/2} (\dot{\tilde{u}}^2 +
    \dot{\tilde{v}}^2)^{1/2}} \]
    i.e. angles do \emph{not} change.
\end{proof}

\subsubsection*{Areas}

Recall that the area of a parallelogram spanned by
vectors $v$ and $w$ is
\[ |v \times w| = (|v|^2 |w|^2 - (v \cdot
w)^2)^{1/2} \]
\begin{center}
    \includegraphics[width=0.3\linewidth]
    {images/31b97bcea93711ed.png}
\end{center}
Suppose we have $\sigma : V \to U \subset \Sigma$,
$\sigma(0) = p$ and consider $\sigma_u, \sigma_v
\in T_p\Sigma$
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/c268970ea93711ed.png}
\end{center}
They span a parallelogram in $T_p\Sigma$ of area:
\[ (|\sigma_u|^2 |\sigma_v|^2 - (\sigma_u \cdot
\sigma_v)^2)^{1/2} = \sqrt{EG - F^2} \]

\begin{flashcard}[area-formula-defn]
\begin{definition*}
    $\Area(U) = \cloze{\int_V \sqrt{EG - F^2} \dd u \dd
    v}$.
\end{definition*}
\end{flashcard}

\begin{note*}
    Suppose $\sigma : V \to U$, $\tilde{\sigma} :
    \tilde{V} \to U$ allowable parametrisations.
    $\tilde{\sigma} = \sigma \circ \varphi$,
    $\varphi = \sigma^{-1} \circ \tilde{\sigma}$,
    $\varphi : \tilde{V} \to V$ transition map. By
    Lemma 2.3:
    \[
    \begin{pmatrix}
        \tilde{E} & \tilde{F} \\
        \tilde{F} & \tilde{G}
    \end{pmatrix}
    = (D\varphi)^\top
    \begin{pmatrix}
        E & F \\
        F & G
    \end{pmatrix}
    D\varphi \]
    Hence (taking determinants)
    \[ \sqrt{\tilde{E}\tilde{G} - \tilde{F}^2} =
    |\det(D\varphi)| \sqrt{EG - F^2} \]
    Note the change of variables formula for
    integration shows that
    \[ \int_{\tilde{V}} \sqrt{\tilde{E} \tilde{G}
    - \tilde{F}^2} \dd \tilde{u} \dd \tilde{v} =
    \int_V \sqrt{EG - F^2} \dd u \dd v \]
    so $\Area(U)$ is \emph{instrinsic} and
    well-defined.
\end{note*}

\begin{example*}
    Consider the graph
    \[ \Sigma = \{(u, v, f(u, v)) : (u, v) \subset
    \RR^2\} \]
    with $f : \RR^2 \to \RR$ smooth.
    \begin{align*}
        \sigma(u, v)
        &= (u, v, f(u, v)) \\
        \sigma_u
        &= (1, 0, f_u) \\
        \sigma_v
        &= (0, 1, f_v) \\
        EG - F^2
        &= 1 + f_u^2 + f_v^2
    \end{align*}
    If $U_R \subset \Sigma$ is $\sigma(B(0, R))$
    then
    \begin{align*}
        \Area(U_R)
        &= \int_{B(0, R)} \sqrt{1 + f_u^2 + f_v^2}
        \dd u \dd v \\
        &\ge \pi R^2
    \end{align*}
    \begin{center}
        \includegraphics[width=0.3\linewidth]
        {images/f83ca82aa9eb11ed.png}
    \end{center}
    With equality only when $f_u = f_v = 0$ in
    $B(0, R)$, i.e. only when $U_R$ is a subset of
    a $z = \text{const}$ plane. So projection from
    $\Sigma$ to $\RR_{xy}^2$ is \emph{not} area
    preserving unless $\Sigma$ is a plane parallel
    to $\RR_{xy}^2$.
\end{example*}

\begin{example*}
    Contrast (Archimedes).
    \begin{center}
        \includegraphics[width=0.3\linewidth]
        {images/5f5b5cf4a9ec11ed.png}
    \end{center}
    The horizontal radial projection (with centre
    the $z$-axis) from $S^2$ to the cylinder is
    \emph{area-preserving} (see Example Sheet 2).
\end{example*}


