% vim: tw=50
% 06/03/2023 11AM

\begin{proof}
    Let $a \in \RR_+ \cap D$ and $\tau$ a smooth
    path from $0$ to $a$, say
    \[ \tau(t) = (u(t), v(t)), \qquad t \in [0, 1] \]
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/be7add96bc1011ed.png}
    \end{center}
    \begin{align*}
        L(\tau)
        &= \int_0^1 \frac{2|\dot{\tau}(t)|}{1 -
        |\tau(w)|^2} \dd t \\
        &= \int_0^1 \frac{2\sqrt{\dot{u}^2 +
        \dot{v}^2}}{1 - u^2 - v^2} \dd t \\
        &\ge \int_0^1 \frac{2|\dot{u}(t)| \dd t}{1
        - u^2} \\
        &\ge \int_0^1 \frac{2\dot{u}}{1 - u^2} \dd
        t \\
        &= \int_0^a \frac{2\dd s}{1 - s^2} \\
        &= 2 \tanh^{-1}(a)
    \end{align*}
    with equality if and only if $\dot{v} = 0$ and
    $\dot{u} \ge 0$, i.e. $v = 0$ and $u$
    monotonic. So the arc of diameter is globally
    length minimising and if parametrised by
    arc-length it becomes a geodesic with length
    \[ \boxed{d_{g_{\text{hyp}}}(0, a) =
    2\tanh^{-1}(a)} \tag{\dag} \]
    Now $0, a \in \RR_+ \cap D$ are joined by a
    unique geodesic (any geodesic going through $0$
    must be a diameter). An element of $\Mob(D)$
    can be used to send any $p, q \in D$ to $0, a
    \in \RR_+ \cap D$ (check, but see below).
    Since isometries map geodesics to geodesics,
    every $p, q \in D$ is joined by a unique
    geodesic. And M\"obius maps send circles to
    circles and preserve angles, and hence
    orthogonality to $\partial D$. This implies
    our description of geodesics.
\end{proof}

\begin{flashcard}[disc-model-dhyp]
\begin{corollary}
    If $p, q \in D$, then
    \[ d_{\text{hyp}}(p, q) = \cloze{2\tanh^{-1} \left|
    \frac{p - q}{1 - \ol{p}q} \right|} \]
\end{corollary}
\end{flashcard}

\begin{proof}
    $Q(z) = \frac{z - p}{1 - \ol{p}z}$ maps $p$ to
    $0$. Pick $\theta$ such that $e^{i\theta} Q(q)
    \in \RR_+ \cap D$. For $T = e^{i\theta} Q$,
    \begin{align*}
        d_{\text{hyp}}(p, q)
        &= d_{\text{hyp}}(T(p), T(q)) \\
        &= d_{\text{hyp}} (0, T(q)) \\
        &= 2\tanh^{-1} \left| \frac{p - q}{1 -
        \ol{p}q} \right|
        &&\text{(using (\dag))} \qedhere
    \end{align*}
\end{proof}

\begin{flashcard}[half-plane-model-riemannian-metric]
\begin{definition*}
    The hyperbolic upper half plane (also called
    Poincar\'e upper half-plane) ($\mathcal{H}$,
    $g_{\text{hyp}}$) is the set
    \[ \mathcal{H} = \{z \in \CC : \Im z > 0\} \]
    with abstract Riemannian metric
    \[ \cloze{\frac{\dd z^2 + \dd y^2}{y^2}} \]
    (or $\cloze{\frac{|\dd z|^2}{(\Im(z))^2}}$).
\end{definition*}
\end{flashcard}

\begin{lemma}
    The Poincar\'e disc $(D, g_{\text{hyp}})$ and
    the Poincar\'e upper-half plane $(\mathcal{H},
    g_{\text{hyp}})$ are \emph{isometric}.
\end{lemma}

\begin{proof}
    We have maps: $\mathcal{H} \to D$, $w \mapsto
    \frac{w - i}{w + i}$ and $D \to \mathcal{H}$,
    $z \mapsto i \frac{i + z}{i - z}$, which are
    inverse diffeomorphisms (easy to check). If $w
    \in \mathcal{H}$, $T(w) = \frac{w - i}{w + i}
    = z \in D$, then $T'(w) = \frac{2i}{(w +
    i)^2}$, then
    \begin{align*}
        \frac{|\dd z|}{1 - |z|^2}
        &= \frac{|\dd (Tw)|}{1 - |Tw|^2} \\
        &= \frac{|T'(w)| |\dd w|}{1 - |Tw|^2} \\
        &= \frac{2|\dd w|}{|w + i|^2 \left( 1 -
        \left| \frac{w - i}{w + i} \right|^2
        \right)} \\
        &= \frac{|\dd w|}{w\Im(w)} \qedhere
    \end{align*}
\end{proof}

\begin{hiddenflashcard}[isometry-H-to-D]
Isometry from half plane model to disk model? \\
\[ \cloze{z \mapsto \frac{z - i}{z + i}} \]
\end{hiddenflashcard}

\begin{hiddenflashcard}[isometry-D-to-H]
Isometry from disk model to half plane model? \\
\[ \cloze{z \mapsto i \frac{i + z}{i - z}} \]
\end{hiddenflashcard}

\begin{corollary}
    In $(\mathcal{H}, g_{\text{hyp}})$ every pair
    of points is joined by a unique geodesic and
    the geodesics are vertical straight lines and
    semicircles centred on $\RR$.
\end{corollary}

\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/e2b8e5cebc1311ed.png}
\end{center}

\begin{proof}
    Our isometry $\mathcal{H} \stackrel{T}{\to} D$
    is a M\"obius map sending $\RR \cup
    \{\infty\}$ to $\partial D$ and M\"obius maps
    preserve circles and orthogonality.
\end{proof}

\subsubsection*{Remarks}

\begin{enumerate}[(1)]
    \item When we discussed surfaces in $\RR^3$
        with constant Gauss curvature, we saw that
        if something had $\kappa = -1$, its FFF in
        Fermi coordinates was
        \[ \dd u^2 + \cosh^2 u \dd v^2 \]
        and after a change of variables we got
        \[ \frac{\dd v^2 + \dd w^2}{w^2} \]
        (\hyperref[lecture17]{Lecture 17}).
        Thus $\mathcal{H}$ (and hence $D$) has
        Gauss curvature $-1$.
    \item Suppose we looked for a metric $d : D
        \times D \to \RR_{\ge 0}$ on $D$ with
        properties:
        \begin{enumerate}[(a)]
            \item $\Mob(D)$-invariant:
                \[ d(Tx, Ty) = d(x, y) \qquad
                \forall x, y \in D, \forall T \in
                \Mob(D) \]
            \item $\RR_+ \cap D$ is
                ``length-minimising''.
        \end{enumerate}
        $\Mob(D)$-invarianve means that $d$ is
        completely determined by $d(0, a)$ for $a
        \in \RR_+ \cap D$. Let us call this
        $p(a)$. If $\RR_+ \cap D$ is
        ``length-minimising'', then if $0 < a < b
        < 1$,
        \[ \ub{d(0, a)}_{p(a)} + \ub{d(a,
        b)}_{p\left( \frac{b - a}{1 - ab} \right)}
        = \ub{d(0, b)}_{p(b)} \]
        If we furthermore suppose $p$ is
        differentiable and differentiate with
        respect to $b$ and set $a = b$ we get
        \[ p'(a) = \frac{p'(0)}{1 - a^2} \]
        hence $p(a) = C \tanh^{-1}(a)$ for some
        constant $C$. So up to scaling, the length
        metric associated with $g_{\text{hyp}}$ is
        the only metric with these nice
        properties. The scale is chosen so that
        $\kappa \equiv -1$.
\end{enumerate}

\myskip
What about the \emph{full} isometry group of $D(,
g_{\text{hyp}})$ on $(\mathcal{H},
g_{\text{hyp}})$? The result is that we need to
add ``reflections'' in hyperbolic lines. These are
called \emph{inversions}.

\begin{flashcard}[inverse-point-defn]
\begin{definition*}
    Let $\Gamma \subset \hat{\CC} = \CC \cup
    \{\infty\}$ be a circle or a line. We say that
    points $z, z' \in \hat{\CC}$ are
    \emph{inverse} for $\Gamma$ if \cloze{every circle
    through $z$ and orthogonal to $\Gamma$ also
    passes through $z'$.}
\end{definition*}
\end{flashcard}

\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/c136f4b6bc1511ed.png}
\end{center}

\begin{lemma}
    For every circle $\Gamma \subset \hat{\CC}$
    there is a unique inverse point with respect
    to $\Gamma$ for $z$.
\end{lemma}
