% vim: tw=50
% 27/02/2023 11AM

\noindent
The local existence of geodesics gives rise to
parametrisations with very \emph{nice properties}.

\myskip
Fix $p \in \Sigma$. Consider a geodesic arc
$\gamma$ going through $p$ and parametrised by
arc-length. For $t$ small enough, let $\gamma_t$
be the unique geodesic such that
\begin{itemize}
    \item $\gamma_t(0) = \gamma(t)$
    \item $\gamma_t'(0)$ is orthogonal to
        $\gamma'(t)$ and has unit length.
\end{itemize}
Define $\sigma(u, v) = \gamma_v(u)$ for $u \in
(-\eps, \eps)$, $v \in (-\eps, \eps)$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/8d822c06b69011ed.png}
\end{center}

\begin{lemma}
    For $\eps$ and $\delta$ sufficiently small,
    $\sigma$ defines an allowable parametrisation
    of an open set in $\Sigma$.
\end{lemma}

\begin{proof}
    Smoothness follows from Corollary 3.4
    (smoothness of geodesics with respect to
    initial conditions). At $(0, 0)$, $\sigma_u$
    and $\sigma_v$ are orthogonal and have norm 1
    by construction. Thus
    \[ D\sigma|_0 : \RR^2 \to T_p\Sigma \]
    is a linear isomorphism. Now applying the
    inverse function theorem as in Example Sheet 1
    Question 9, we deduce that $\sigma$ is a local
    diffeomorphism at $(0, 0)$ and hence for
    $\eps, \delta$ small enough it is an allowable
    parametrisation.
\end{proof}

\begin{flashcard}[Fermi-coordinates]
\begin{proposition}
    Any smooth surface $\Sigma$ in $\RR^3$ admits
    a local parametrisation for which the FFF is
    of the form
    \cloze{
    \[ \dd u^2 + G \dd v^2 \]
    i.e. $E = 1$, $F = 0$.}
\end{proposition}
\end{flashcard}

\begin{proof}
    Consider $\sigma(u, v) = \delta_v(u)$ as
    above. If we fix $v_0$, the curve $u \to
    \gamma_{v_0}(u)$ is a geodesic parametrised by
    arc-length. So $E = \langle \sigma_u, \sigma_u
    \rangle = 1$. Also one of the geodesic
    equations is
    \[ \dfrac{}{t} (F\dot{u} + G \dot{v}) = \half
    (E_v \dot{u}^2 + 2F_v \dot{u} \dot{v} + G_v
    \dot{v}^2) \]
    and $v = v_0$, $u(t) = t$. $\dfrac{}{t}(F) =
    0$ i.e. $F_u = 0$. So $F$ is independent of
    $u$. But when $u = 0$, then by construction of
    $\gamma_v$ as orthogonal to $\gamma$ at
    $\gamma(v)$, we see that $F = 0$ everywhere.
\end{proof}

\subsubsection*{Remarks}

\begin{enumerate}[(1)]
    \item Somtimes these coordinates are called
        \emph{Fermi coordinates}.
    \item Careful:
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/6871ce88b69211ed.png}
        \end{center}
        $u$-fixed, $v$ vary is typically
        \emph{not} a geodesic.
    \item In these coordinates we also have:
        \begin{enumerate}[(a)]
            \item $G(0, v) = 1$,
            \item $G_u(0, v) = 0$
        \end{enumerate}
        (a) holds because $\sigma_v$ has length
        $1$ at $u = 0$. To see (b), we use that $u
        = 0$, $v = t$ is a geodesic and
        \[ \dfrac{}{t}(E\dot{u} + F \dot{v}) =
        \half (E_u \dot{u}^2 + 2F_u \dot{u}
        \dot{v} + G_u \dot{v}^2) \]
        because $0 = \half G_u (0, t)$.
    \item One can show that if $E = 1$ and $F =
        0$, then the Gauss curvature is given by
        \begin{flashcard}[Gauss-curvature-in-Fermi-coordinates]
        \prompt{Gauss curvature in Fermi
        coordinates?}
        \[ \boxed{\kappa =
        \cloze{-\frac{(\sqrt{G})_{uu}}{\sqrt{G}}}}
        \tag{\dag} \]
        \prompt{\cloze{Recall Fermi coordinates
        are when $E = 1$ and $F = 0$.}}
        \end{flashcard}
        (cf Theorem 8.1 in Wilson's book). This
        proves theorema egregium! The computation
        that proves (\dag) is not too hard, but
        beyond the scope of this course. But we'll
        use (\dag) in our next discussion.
\end{enumerate}

\subsubsection*{Constant Gaussian Curvature}

First a general remark / Exercise: If $\Sigma
\subset \RR^3$ and $f : \RR^3 \to \RR^3$ is
dilation by $\lambda \neq 0$:
\[ f(x, y, z) = \lambda(x, y, z) \]
then $\kappa_{f(\Sigma)} = \frac{1}{\lambda^2}
\kappa_\Sigma$. To check this note that the
coefficients $E, F, G$ rescale by $\lambda^2$ and
$L, M, N$ by $\lambda$.

\myskip
QUestion: What do constant curvature surfaces look
like? By dilations it sufficies to understand
surfaces of constant curvature $1, -1, 0$.

\begin{flashcard}[Constant-curvature-surfaces]
\begin{proposition}
    $\Sigma \subset \RR^3$ smooth surface.
    \begin{enumerate}[(a)]
        \item If $\kappa_\Sigma \equiv 0$, \cloze{then
            $\Sigma$ is locally isometric to
            $(\RR^2, \dd u^2 + \dd v^2)$.}
        \item If $\kappa_\Sigma \equiv 1$, \cloze{then
            $\Sigma$ is locally isometric to
            $(S^2, \dd u^2 + \cos^2(u) \dd v^2)$.}
    \end{enumerate}
\end{proposition}

\prompt{
\begin{proof}
\cloze{
    Consider a parametrisation with $E = 1$, $F =
    0$ and $G$ such that
    \[ \boxed{G(0, v) = 1 \quad \text{and} \quad G_u(0,
    v) = 0} \]
    Recall that
    \[ \boxed{kappa =
    -\frac{(\sqrt{G})_{uu}}{\sqrt{G}}}
    .\]
    So we get a differential equation for
    $(\sqrt{G})_{uu}$. Solve this using initial
    conditions to work out the FFF.
}
\end{proof}
}
\end{flashcard}

\begin{proof}
    We know that $\Sigma$ admits a parametrisation
    with $E = 1$, $F = 0$ and $G(0, v) = 1$ and
    $G_u(0, v) = 0$. Also
    \[ \kappa = -\frac{(\sqrt{G})_{uu}}{\sqrt{G}} \]
    If $\kappa \equiv 0$, we get $(\sqrt{G})_{uu}
    = 0$, hence $\sqrt{G} = A(v) u + B(v)$.
    Conditions on $G$ give $B = 1$ and $A = 0$.
    Then the FFF is $\dd u^2 + \dd v^2$.

    \myskip
    If $\kappa = 1$, $(\sqrt{G})_{uu} + \sqrt{G} =
    0$, so
    \[ \sqrt{G} = A(v) \sin u + B(v) \cos u \]
    conditions on $G$ gives $A = 0$ and $B \equiv
    1$, so FFF is $\dd u^2 + \cos^2 u \dd v^2$. In
    the parametrisation
    \[ \sigma(u, v) = (\cos u \cos v, \cos u \sin
    v, \sin u) \]
    of $S^2$ this is the FFF.
\end{proof}

\begin{remark*}
    We can certainly do the same for $\kappa =
    -1$, and we'll get FFF $\dd u^2 + \cosh^2 (u)
    \dd v^2$. A surface of revolution with FFF
    $\dd u^2 + \cosh^2(u) \dd v^2$ is given by
    rotating
    \[ \eta(u) = \left(\cosh u, 0, \int_0^h
    \sqrt{1 - \sinh^2 x} \dd x \right) \]
    This has $f'^2 + g'^2 = 1$ and hence $\kappa =
    -f'' / f$ (Question 5 on Example Sheet 2).
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/5c5d851cb69511ed.png}
    \end{center}
    Or\ldots we forget about $\RR^3$ and think in
    more abstract terms.
    The change of variables $v = e^v \tanh u$, $w
    = e^v \sech u$ turns $\dd u^2 + \cosh^2 (u)
    \dd v^2$ into
    \[ \frac{\dd v^2 + \dd w^2}{w^2} \]
    which is ``the standard presentation'' of the
    hyperbolic plane.
\end{remark*}
