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\newpage
\section{Hyperbolic Surfaces}

We start by discussing \emph{abstract Riemannian
metrics}.

\begin{flashcard}[abstract-riemannian-metric]
\begin{definition*}[Riemannian metric]
    Let $V \subset \RR^2$ be an open set. An
    abstract Riemannian metric on $V$ is \cloze{a smooth
    map:
    \[ V \to \{\text{positive definite symmetric
    forms}\} \subset \RR^4 \]
    \[ p \mapsto
    \begin{pmatrix}
        E(p) & F(g) \\
        F(g) & G(p)
    \end{pmatrix}
    = g(p) \]
    $E > 0$, $G > 0$ and $EG - F^2 > 0$.}
\end{definition*}
\end{flashcard}

\noindent
\begin{center}
    \includegraphics[width=0.2\linewidth]
    {images/681d0094b83e11ed.png}
\end{center}
If $v$ is a vector at $p \in V$, then its norm is:
\[ \|v\|_g^2 = v^\top
\begin{pmatrix}
    E(p) & F(p) \\
    F(p) & G(p)
\end{pmatrix}
v \]
and if $\gamma : [a, b] \to V$ is smooth, then its
length
\begin{align*}
    L(\gamma)
    &= \int_a^b \|\dot{\gamma}(t)\|_g \dd t \\
    &= \int_a^b (E\dot{u}^2 + 2F\dot{u}\dot{v} + G
    \dot{v}^2) \dd t
\end{align*}
where $\gamma(t) = (u(t), v(t))$.

\begin{flashcard}[isometric-riemannian-metric]
\begin{definition*}
    Given $(V, g)$, $(\tilde{V}, \tilde{g})$, we
    say that they are \emph{isometric} \cloze{if there
    exists a diffeomorphism $f : V \to \tilde{V}$
    such that
    \[ \| Df|_p(v)\|_{\tilde{g}} = \|v\|_g \qquad
    \forall v \in T_pV = \RR^2, \forall p \in V
    \fcscrap{\tag{$*$}} \]}
    \fcscrap{This is \emph{equivalent} to saying
    that $f$ preserves the length of curves.}
\end{definition*}
\end{flashcard}

\begin{note*}
    $Df_p : T_pV \to T_{f(p)}\tilde{V}$. Let's
    spell out this condition ($*$) using $g$ and
    $\tilde{g}$.
    \begin{align*}
        \|Df|_p(v)\|_{\tilde{g}}^2
        &= (Df|_p v)^\top \tilde{g}_{f(p)} Df|_p v
        \\
        &= v^\top (Df|_p)^\top \tilde{g}_{f(p)}
        Df|_p v \\
        &= \|v\|_g^2 \\
        &= v^\top g v
    \end{align*}
    This holds for all $v$ iff
    \[ \boxed{(Df|_p)^\top \tilde{g}_{f(p)} Df|_p
    = g} \tag{\dag} \]
    Recall that (\dag) is exactly the
    transformation law from Lemma 2.3 (Lecture
    \#9).
\end{note*}

\begin{flashcard}[riemannian-metric-on-a-surface]
\begin{definition*}[Riemannian metric on a surface]
    Let $\Sigma$ be an abstract smooth surface, so
    $\Sigma = \bigcup_{i \in I} U_i$, $U_i \subset
    \Sigma$ open and $\varphi : U_i \to V_i
    \subset \RR^2$ homeomorphism with $V_i$ open
    such that
    \[ \varphi_i \varphi_j^{-1} : \varphi_j(U_i
    \cap U_j) \to \varphi_i(U_i \cap U_j) \]
    is smooth for all $i, j$.
    
    \myskip
    A \emph{Riemannian metric} on $\Sigma$ usually
    denoted by $g$ is \cloze{a choice of Riemannian
    metrics $g_i$ on each $V_i$ which are
    compatible in the following sense:
    \myskip
    For all $i, j$, $\varphi_i
    \varphi_j^{-1}$ is an isometry between
    $\varphi_j(U_i \cap U_j)$ and
    $\varphi_i(U_i \cap U_j)$, i.e. if we
    let $f = \varphi_i \varphi_j^{-1}$,
    then
    \[ (Df|_p)^\top
    \begin{pmatrix}
        E_i & F_i \\
        F_i & G_i
    \end{pmatrix}
    _{f(p)}
    Df|_p =
    \begin{pmatrix}
        E_j & F_j \\
        F_j & G_j
    \end{pmatrix}
    _p \qquad \forall p \in \varphi_j(U_i \cap
    U_j) \]}
\end{definition*}
\end{flashcard}

\subsubsection*{Examples}

\begin{enumerate}[(1)]
    \item Recall the torus $T^2 = \RR^2 / \ZZ^2$
        \begin{center}
            \includegraphics[width=0.3\linewidth]
            {images/d46d5bacb84011ed.png}
        \end{center}
        We exhibited charts where transition
        functions were restriction of
        translations. Equip each $V_i \subset
        \RR^2$ (image of such a chart) with the
        Euclidean metric $\dd u^2 + \dd v^2$, i.e.
        the $V_i \mapsto \id_{2 \times 2}$. If $f$
        is a translation then $Df = \id$ so
        $(Df)^\top I Df = I$ is obvious! So $T^2$
        inherits a global Riemannian metric
        everywhere locally isometric to $\RR^2$
        (hence \emph{flat}). Since geodesics are
        well-defined for abstract Riemannian
        metrics (Energy only depends on $g$!) they
        are also well-defined on $T^2$ and they
        are just projections of straight lines in
        $\RR^2$:
        \begin{center}
            \includegraphics[width=0.6\linewidth]
            {images/e7efc42ab84111ed.png}
        \end{center}
        Exercise: Show that there are infinitely
        many cloesd geodesics and also infinitely
        many non-closed ones (think about lines
        with rational / irrational slope).
        \begin{note*}
            This flat metric on $T^2$ is
            \emph{not} induced by any embedding of
            $T^2$ in $\RR^3$! (For example because
            it would have to have an elliptic
            point).
        \end{note*}
    \item The real projective plane admits a
        Riemannian metric with constant curvature
        $+1$. Inded, in lecture 2 we exhibited an
        atlas for $\mathbb{RP}^2$ with charts of the form
        $(U, \varphi)$ where $U = q\hat{U}$, $q :
        S^2 \to \mathbb{RP}^2$, $\hat{U} \subset
        S^2$ open small enough so that $\hat{U}$
        subset of open hemisphere and $\varphi : U
        \to V \subset \RR^2$ was $\varphi :
        \hat{\varphi} \circ q^{-1} |_U$, where
        $\hat{\varphi} : \hat{U} \to V$ chart of
        $S^2$. Transition maps for this atlas
        were all the identity or induced by the
        antipodal map. But both are isometries of
        the round metric in $S^2$.
    \item Exercise: The Klein bottle has a flat
        Riemannian metric induced from $\RR^2$.
        \begin{center}
            \includegraphics[width=0.3\linewidth]
            {images/95f956eeb84211ed.png}
        \end{center}
\end{enumerate}

\begin{proposition}
    Given a Riemannian metric $g$ on a
    \emph{connected} open set $V \subset \RR^2$,
    define the length metric
    \[ d_g(p, q) = \inf_\gamma L(\gamma) \]
    where $\gamma$ varies over all piece-wise
    smooth paths in $V$ from $p$ to $q$ and
    $L(\gamma)$ is computed using $g$. Then $d_g$
    is a metric in $V$ (in the sence of metric
    spaces!)
    \begin{center}
        \includegraphics[width=0.4\linewidth]
        {images/010e8cf6b84311ed.png}
    \end{center}
\end{proposition}

\subsubsection*{Remarks / Examples}

\begin{enumerate}[(1)]
    \item Given $p, q \in V$, there is always a
        piece-wise smooth path connecting $p$ to
        $q$.
    \item $d_g(p, q) \ge 0$. Also $d_g(p, q) =
        d_g(q, p)$ and $d_g(p, r) \le d_g(p, q) +
        d_g(q, r)$
        \begin{center}
            \includegraphics[width=0.3\linewidth]
            {images/48a4ee98b84311ed.png}
        \end{center}
        The only non-trivial claim is that $d_g(p,
        q)= 0$ implies $p = q$ (proof next
        lecture).
    \item All this works on \emph{any} abstract
        smooth \emph{connected} surface $(\Sigma,
        g)$ equipped with a Riemmanian metric $g$.
\end{enumerate}
