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\newpage
\section{Further Topics}

\subsubsection*{Gauss Bonnet theorem revisited}

Recall:
\begin{enumerate}[(i]
    \item In a spherical triangle $T$ with
        internal angles $\alpha, \beta, \gamma$ we
        saw in Example Sheet 2 that
        \[ \Area_{S^2}(T) = \alpha + \beta +
        \gamma - \pi \]
        while a hyperbolic triangle has area
        \[ \Area_{\HH}(T) = \pi - \alpha - \beta -
        \gamma \]
    \item We also saw that for compact surfaces
        $\Sigma \subset \RR^3$,
        \[ \int_\Sigma \kappa \dd A = 2\pi
        \chi(\Sigma) \]
\end{enumerate}

\begin{flashcard}[local-gauss-bonnet]
\begin{theorem}[Local Gauss-Bonnet]
    Let $\Sigma$ be an abstract smooth surface
    with Riemannian metric $g$. \cloze{Take a
    \emph{geodesic polygon} $R$ on $\Sigma$, i.e.
    it is \fcemph{homeomorphic to a disc} and its boundary
    is decomposed into \fcemph{finitely many geodesic
    arcs}. Then
    \[ \int_R \kappa \dd A = \sum_{i = 1}^n
    \alpha_i - (n - 2)\pi \]
    where $n = \text{\# of arcs}$, and $\alpha_i$
    are the internal angles of the polygon.}
\end{theorem}
\end{flashcard}

\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/e0e3b894c32211ed.png}
\end{center}

\begin{theorem}[Global Gauss-Bonnet]
    If $\Sigma$ is a compact smooth surface with
    Riemannian metric $g$, then
    \[ \int_\Sigma \kappa \dd A = 2\pi
    \chi(\Sigma) \]
\end{theorem}

\begin{remark*}
    \begin{enumerate}[(i)]
        \item $\kappa$ and $\dd A$ can be defined
            just using $g$.
        \item For our hyperbolic surfaces observed
            by identifying the edges of a regular
            $4g$-gon with angles $\frac{\pi}{2g}$
            \begin{align*}
                \int_\Sigma 1 \dd A
                &= \Area(\operatorname{Polygon})
                \\
                &= (4g - 2)\pi - \sum_{n = 1}^{4g}
                \frac{\pi}{2g} \\
                &= (4g - 4)\pi \\
                &= 2(2g - 2)\pi
            \end{align*}
            and we had $\chi(\Sigma_g) = 2 - 2g$
            and $\kappa \equiv -1$, so this agrees
            with the Global Gauss-Bonnet.
        \item If $\Sigma$ is a flat surface so
            $\kappa = 0$ and $\gamma$ is a closed
            geodesic, i.e. $\gamma : \RR \to
            \Sigma$ and $\exists T > 0$ such that
            $\gamma(t + T) = \gamma(t)$ for all $t
            \in \RR$, then $\gamma$ cannot bound
            a disc:
            \begin{center}
                \includegraphics[width=0.4\linewidth]
                {images/b7f9de02c32411ed.png}
            \end{center}
            Indeed, if we had such a $\gamma$,
            then
            \begin{center}
                \includegraphics[width=0.4\linewidth]
                {images/c66d1418c32411ed.png}
            \end{center}
            is a geodesic polygon with internal
            angles $\pi$ and local Gauss-Bonnet
            gives:
            \[ 0 = \int_R \kappa \dd A = \sum_{i =
            1}^n \alpha_i - (n - 2)\pi = 2\pi \]
            (since $n = 2$), so such a polygon
            would violate Local Gauss-Bonnet.
            However, in a non-flat metric, we
            clearly \emph{can} have such a
            geodesic:
            \begin{center}
                \includegraphics[width=0.6\linewidth]
                {images/ddb982b4c32411ed.png}
            \end{center}
    \end{enumerate}
\end{remark*}

\subsubsection*{Flat metrics}

Question: Can we understand \emph{all} flat
metrics on $T^2$?

\myskip
The key to get a flat metric on $T^2$ was an atlas
where transition functions were
\emph{isometries}.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/4e1eb646c32511ed.png}
\end{center}
So \emph{any} parallelogram $Q \subset \RR^2$
delivers a flat metric $g_Q$ on $T^2$.
\[ (T^2, g_Q) = \RR^2 / \ZZ v_1 \oplus \ZZ v_2 \]
Observation:
\[ \Area_{g_Q}(T^2) = \Area_{\text{Euclidean}}(Q) \]
Since $Q_1$ and $Q_2$ could have different ares,
the metrics $g_{Q_1}$ and $g_{Q_2}$ are not
\emph{isometric}.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/b3940544c32511ed.png}
\end{center}
We'll consider flat metrics up to dilations:
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/c02c6c00c32611ed.png}
\end{center}
Under this assumption, given $Q$, we can put
vertices at $0 \in \RR^2$, $1 \in \RR^2$ and
$\tau \in \mathcal{H}$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/fa98bf6ac32611ed.png}
\end{center}
This defines a map
\[ \mathcal{H} \to \{\text{flat metrics on $T^2$}\}
/ \text{Dilations} \]
But \emph{diffeomorphisms} acct on the set of
\emph{flat metrics}. Given $g$ and $g : T^2 \to
T^2$ diffeomorphism, we can ``pull-back'' $g$ by
$f$:
\[ Df |_p : T_p \Sigma \to T_{f(p)} \Sigma \]
$f^* g$ is given by
\[ \langle u, w \rangle_{f^* g} = \langle
Df|_p(v), Df|_p(w) \rangle_g \]
$v, w \in T_p\Sigma$. $(\Sigma, f^* g)
\stackrel{f}{\longrightarrow} (\Sigma, g)$
isometry?

\myskip
Now $\SL(2, \ZZ)$ acts on $T^2$ by
diffeomorphisms: it acts on $\RR^2$ and preserves
the lattice $\ZZ^2$
\[
\begin{pmatrix}
    2 & 1 \\
    1 & 1
\end{pmatrix}
: \RR^2 \to \RR^2 \]
$\RR^2 / \ZZ^2 \to \RR^2 / \ZZ^2$. $\SL(2, \ZZ)$
also acts on $\mathcal{H}$ by M\"obius maps (as
isometries of $g_{\text{hyp}}$!)

\begin{theorem}
    The map
    \[ \mathcal{H} \to \{\text{flat metrics on
    $T^2$}\} / \text{Dilations} \]
    descends to a map
    \[ \mathcal{H} / \SL(2, \ZZ) \to \{\text{flat
    metrics on $T^2$}\} / \text{Dilations and
    Diffomorphisms$^+$} \]
    which is a \fbox{bijection}. We say that
    $\mathcal{H} / \SL(2, \ZZ)$ is the
    \emph{moduli space of flat metrics of $T^2$}.
\end{theorem}
