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$\kappa(p) = \det(Dn|_p)$ ($\kappa = \frac{LN -
M^2}{EG - F^2}$) The Gauss urvature is constraint
by 2 \emph{amazing theorems}.

\myskip
The first is called ``theorema egregium''
(remarkable theorem)

\begin{flashcard}[theorema-egregium]
\begin{theorem*}[Theorema Egregium]
    \cloze{The Guass curvature of a smooth surface in
    $\RR^3$ is an \emph{isometry invariant} i.e.
    if $f : \Sigma_1 \to \Sigma_2$ is an isometry,
    then
    \[ \kappa_1(p) = \kappa_2(f(p)) \qquad \forall
    p \in \Sigma_1 \]
    In fact, $\kappa$ can be computed exclusively
    in terms of $I_p$ even though it was defined
    using $I_p$ and $\II_p$.}
\end{theorem*}
\end{flashcard}

\noindent
The second result is a \emph{global result}:

\begin{flashcard}[gauss-bonnet]
\begin{theorem*}[Gauss-Bonnet Theorem]
    \cloze{If $\Sigma$ is a \fcemph{compact} smooth surface in
    $\RR^3$, then
    \[ \int_\Sigma \kappa \dd A_\Sigma = 2\pi
    \chi(\Sigma) \]
    }
\end{theorem*}
\end{flashcard}

This is an amazing result because the left side of
this equation is a quantity involving a lot of
complicated geometric notions, but the right hand
side is purely a topological property!

\myskip
The proofs of these theorems will be in Part II
Differential Geometry.

\newpage
\section{Geodesics}

Recall, if $\gamma : [a, b] \to \RR^3$ is smooth
then
\[ \length(\gamma) = L(\gamma) = \int_a^b
|\gamma'(0)| \dd t \]

\begin{flashcard}[energy]
\begin{definition*}
    The \emph{energy} of $\gamma$ is
    \[ E(\gamma) \defeq \cloze{\int_a^b |\gamma'(t)|^2
    \dd t} \]
\end{definition*}
\end{flashcard}

\noindent
Think
\[ \Omega_{pq} = \{\text{all smooth curves $\gamma
: [a, b] \to \RR^3$, $\gamma(a) = p$, $\gamma(b) =
q$}\} \]
\begin{center}
    \includegraphics[width=0.3\linewidth]
    {images/2bd0db70b11211ed.png}
\end{center}
$E : \Omega_{pq} \to \RR$. In fact what we really
want is given $\Sigma \subset \RR^3$, $\gamma :
[a, b] \to \Sigma$ ``Variational Principles.''

\begin{flashcard}[one-parameter-variation]
\begin{definition*}
    Let $\gamma : [a, b] \to \Sigma \subset \RR^3$
    be smooth. A one-parameter variation \cloze{(with
    fixed end points) of $\gamma$ is a smooth map
    \[ \Gamma : (-\eps, \eps) \times [a, b] \to
    \Sigma \]
    such that $\gamma_s \defeq \Gamma(s,
    \bullet)$, then
    \begin{enumerate}[(a)]
        \item $\gamma_0(t) = \gamma(t)$ for all
            $t$,
        \item $\gamma_s(a)$, $\gamma_s(b)$ are
            independent of $s$.
    \end{enumerate}
    }
    \fcscrap{
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/86c4b754b11211ed.png}
    \end{center}
    }
\end{definition*}
\end{flashcard}

\begin{flashcard}[geodesic]
\begin{definition*}
    A smooth curve $\gamma : [a, b] \to \Sigma$ is
    a \emph{geodesic} \cloze{if for every
    \fcemph{variation}
    $\gamma_s$ of $\gamma$ with \fcemph{fixed end
    points}
    we have:
    \[ \left. \dfrac{}{s} \right|_{s = 0} E(\gamma_s) =
    0 \]
    i.e. $\gamma$ is a ``critical point'' of the
    energy functional on curves from $\gamma(a)$
    to $\gamma(b)$.}
\end{definition*}
\end{flashcard}

\noindent
Suppose $\gamma$ has image contained in the image
of a parametrisation $\sigma$ and we write
\[ \gamma_s(t) = \sigma(u(s, t), v(s, t)) \]
FFF is $E \dd u^2 + 2F \dd u\dd v + G \dd v^2$.
\[ R \defeq E \dot{u}^2 + 2F \dot{u} \dot{v} +
G\dot{v}^2 \]
($E$, $F$, $G$ are functions of $u(s, t)$, $v(s,
t)$)
where $\dot{u} = \pfrac{u}{t}$, $\dot{v} =
\pfrac{v}{t}$. $R$ depends on $s$
\[ E(\gamma_s) = \int_a^b R \dd t \]
\begin{align*}
    \pfrac{R}{s}
    &= (E_u \dot{u}^2 + 2F_u \dot{u} \dot{v} + G_u
    \dot{v}^2) \pfrac{u}{s} + (E_v \dot{u}^2 +
    2F_v \dot{u} \dot{v} + G_v \dot{v}^2)
    \pfrac{v}{s} \\
    &~~~+ 2(E\dot{u} + F\dot{v})
    \pfrac{\dot{u}}{s} + 2(F\dot{u} + G\dot{v})
    \pfrac{\dot{v}}{s}
\end{align*}
so $\dfrac{}{s} E(\gamma_s) = \int_a^b
\pfrac{R}{s} \dd t$

\begin{note*}
    $\pfrac{\dot{u}}{s} = \frac{\partial^2
    u}{\partial s \partial t}$,
    $\pfrac{\dot{v}}{s} = \frac{\partial^2
    v}{\partial s \partial t}$, and we can
    integrate by parts and note $\pfrac{u}{s}$,
    $\pfrac{v}{s}$ vanish at the end points of $a$
    and $b$. We get
    \[ \left. \dfrac{}{s} \right|_{s = 0}
    E(\gamma_s) = \int_a^b \left[ A \pfrac{u}{s} +
    B \pfrac{v}{s} \right] \dd t \tag{$*$} \]
    where
    \[ A = E_u \dot{u}^2 + 2F_u \dot{u} \dot{v} +
    G_u \dot{v}^2 - 2 \dfrac{}{t} (E\dot{u} + F
    \dot{v}) \]
    \[ B = E_v \dot{u}^2 + 2F_v \dot{u} \dot{v} +
    G_v \dot{v}^2 - 2 \dfrac{}{t}(F\dot{u} + G
    \dot{v}) \]
    Note that we have absolute freedom for
    choosing the ``rotational vector field''
    \[ W(t) = \left( \pfrac{u}{s}(0, t),
    \pfrac{v}{s}(0,t) \right) \]
    \begin{center}
        \includegraphics[width=0.4\linewidth]
        {images/0395f6b0b11511ed.png}
    \end{center}
    Going back to ($*$) we see that $\gamma$ is a
    geodesic if and only if $A = B = 0$. That is
    $\gamma$ is a geodesic if and only if
    $\gamma(t) = \sigma(u(t), v(t))$, then we have
    the \emph{geodesic equations}
    \begin{flashcard}[geodesic-equations]
    \prompt{Geodesic equations?}
    \[ \cloze{\dfrac{}{t} (E\dot{u} + F\dot{v}) = \half
    (E_u \dot{u}^2 + 2F_u \dot{u} \dot{v} + G_u
    \dot{v}^2)} \]
    \[ \cloze{\dfrac{}{t} (F\dot{u} + G\dot{v}) = \half
    (E_v \dot{u}^2 + 2F_v \dot{u} \dot{v} + G_v
    \dot{v}^2)} \]
    \end{flashcard}
\end{note*}

\subsubsection*{Remarks}

\begin{enumerate}[(1)]
    \item If $W(t)$ with $W(a) = W(b) = 0$ then
        \[ \gamma_s(t) = \sigma((u(t), v(t)) +
        sw(t)) \]
        for $s$ small enough is a variation of
        $\gamma$ with fixed and points and
        vertical field $W$.
    \item Recall Q10 in Example Sheet 4 of IA
        Analysis:
        \[ \int_a^b f(x)g(x) \dd x = 0 \]
        for all $g : [a, b] \to \RR$, $g(a) =
        g(b)$, then we had to prove that $f \equiv
        0$.
\end{enumerate}