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

\begin{proposition}
    $d_g$ is a metric.
\end{proposition}

\begin{proof}
    We only show that $d_g(p, q)$ for $p \neq q$
    for $p \neq q$. (See remarks from last
    lecture). Since
    \[ g =
    \begin{pmatrix}
        E(p) & F(p) \\
        F(p) & G(p)
    \end{pmatrix}
    \]
    is positive definite, there is $\eps$
    sufficiently small such that
    \[
    \begin{pmatrix}
        E(p) - \eps^2 & F(p) \\
        F(p) & G(p) - \eps^2
    \end{pmatrix}
    \]
    is also positive definite. Moreover, the
    matrix
    \[
    \begin{pmatrix}
        E(p') - \eps^2 & F(p') \\
        F(p') & G(p') - \eps^2
    \end{pmatrix}
    \]
    remains positive definite $\forall p' \in B(p,
    \delta) \subset V$. (Euclidean ball). Thus, for
    any $p' \in B(p, \delta)$ and $v = (v_1, v_2)
    \in \RR^2$ we have
    \begin{align*}
        \|v\|_{p'}^2
        &= E(p') v_1^2 + 2F(p') v_1 v_2 + G(p')
        v_2^2 \\
        &\ge \eps^2(v_1^2 + v_2^2)
    \end{align*}
    Hence if $\gamma$ is a curve in $B(p, \delta)$
    then we have
    \[ L_g(\gamma) \ge \eps L_{\text{Euclidean}}
    (\gamma) \]
    Given $p \neq q$, let $\gamma : [a, b] \to V$
    be any curve connecting $p$ to $q$. If
    $\gamma$ is not contained in $B(p, \delta)$
    then there exists $t_0 \in [a, b]$ such that
    $\gamma|_{[a, t_0]}$ is in $B(p, \delta)$, but
    $\gamma(t_0)$ is on the boundary of the ball.
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/515e54d8b9b611ed.png}
    \end{center}
    Thus $L_g(\gamma) \ge L_g(\gamma|_{[a, t_0]})
    \ge \eps \delta$. If however $\gamma$ is
    contained in $B(p, \delta)$, then
    \[ L_g(\gamma) \ge \eps d_{\text{Euclidean}}
    (p, q) \]
    Taking inf over all such $\gamma$ we get
    \[ d_g(p, q) \ge \eps \min\{\delta,
    d_{\text{Euclidean}}(p, q)\} > 0 \qedhere \]
\end{proof}

\begin{hiddenflashcard}[proof-d-g-metric]
Proof that $d_g$ is a metric: \\
\cloze{
Everything except proving $d_g(p, q) > 0$ is
fairly straightforward. To prove $> 0$, pick
$\eps$ small enough so that
\[
\begin{pmatrix}
    E(p) - \eps^2 & F(p) \\
    F(p) & G(p) - \eps^2
\end{pmatrix}
\]
is positive definite at $p$, pick neighbourhood
$B(p, \delta)$ such that it is positive definite
in the ball. Then within the ball,
\[ L_g(\gamma) \ge \eps
L_{\text{Euclidean}}(\gamma) \]
for any $\gamma$ entirely inside the ball. If $p
\neq q$, then pick $\gamma$ to be any path between
$p$ and $q$. Either $\gamma$ is always in the
ball, in which case length $\gamma$ is at least
$\eps$ times euclidean, which is $> 0$, or
otherwise $\gamma$ is not always inside the ball,
in which case it reaches the boundary, so length
at least $\eps\delta$. So $d_g(p, q) > 0$.
}
\end{hiddenflashcard}

\begin{remark*}
    $d_g$ gives the same topology that $V \subset
    \RR^2$ inherits from $\RR^2$ (can check this
    as an exercise).
\end{remark*}

\subsection{Hyperbolic Geometry}

\begin{definition*}
    We define an abstract Riemannian metric on the
    disc
    \[ D = B(0, 1) = \{z \in \CC : |z| < 1\} \]
    by
    \[ g_{\text{hyp}} = \frac{4(\dd u^2 + \dd
    v^2)}{(1 - u^2 - v^2)^2} = \frac{4|\dd z|^2}{(1
    - |z|^2)^2} .\]
    In other words,
    \[ E = G = \frac{4}{(1 - u^2 - v^2)^2} \qquad
    F = 0 \]
\end{definition*}

\begin{hiddenflashcard}[disk-model-riemannian-metric]
Riemannian metric for the disk model? \\
\[ g_{\text{hyp}} = \cloze{\frac{4(\dd u^2 + \dd
v^2)}{(1 - u^2 - v^2)^2} = \frac{4|\dd z|^2}{(1 -
|z|^2)^2}} \]
or in other words
\[ E = G = \cloze{\frac{4}{(1 - u^2 - v^2)^2}}
\qquad F = \cloze{0} \]
\end{hiddenflashcard}

\noindent
Recall the M\"obius group
\[ \Mob = \left\{ z \mapsto
\frac{az + b}{cz + d} :
\begin{pmatrix}
    a & b \\
    c & d 
\end{pmatrix}
\in \GL(2, \CC) \right\} \]
acts on $\CC \cup \{\infty\}$.

\begin{lemma}
    \[ \Mob(D) = \{T \in \Mob : T(D) = D\} =
    \left\{ z \mapsto e^{i\theta} \frac{z - a}{1 -
    \ol{a}z} : |a| < 1 \right\} \]
\end{lemma}

\begin{hiddenflashcard}[mobius-group-disk]
\[ \Mob(D) = \cloze{\left\{ z \mapsto e^{i\theta}
\frac{z - a}{1 - \ol{a}z} : |a| < 1 \right\}} \]
\end{hiddenflashcard}

\begin{proof}
    First we note:
    \begin{align*}
        \left| \frac{z - a}{1 - \ol{a}z} \right|
        = 1
        &\iff (z - a)(\ol{z} - \ol{a}) = (1 -
        \ol{a}z)(1 - a\ol{z}) \\
        &\iff z\ol{z} - \cancel{a\ol{z}} -
        \cancel{\ol{a}z} + a\ol{a} = 1 -
        \cancel{a\ol{z}} - \cancel{\ol{a}z} +
        a\ol{a} z\ol{z} \\
        &\iff |z|^2(1 - |a|^2) = 1 - |a|^2 \\
        &\iff |z| = 1
    \end{align*}
    So $z \mapsto e^{i\theta} \frac{z - a}{1 -
    \ol{a}z}$ preserves $|z| = 1$ and maps $a$ to
    $0$, thus it belongs to $\Mob(D)$. To show
    that they are all of this form, pick $T \in
    \Mob(D)$. If $a = T^{-1}(0)$ and $Q(z) =
    \frac{z - a}{1 - \ol{a}z} \in \Mob(D)$, then
    $TQ^{-1}(0) = 0$ and preserves $|z| = 1$ and
    hence (\textcolor{red}{check!}) it must be of
    the form $z \mapsto e^{i\theta} z$.
\end{proof}

\begin{lemma}
    The Riemannian metric $g_{\text{hyp}}$ is
    invariant under $\Mob(D)$, i.e. it acts by
    hyperbolic isometries.
\end{lemma}

\begin{proof}
    $\Mob(D)$ is generated by $z \mapsto
    e^{i\theta} z$ and $z \mapsto \frac{z - a}{1 -
    \ol{a}z}$, $|a| < 1$. The first (rotation)
    clearly preserves $g_{\text{hyp}} =
    \frac{4|\dd z|^2}{(1 - |z|^2)^2}$. For the
    second type, let
    \[ w = \frac{z - a}{1 - \ol{a}z} ,\]
    so
    \begin{align*}
        \dd w
        &= \frac{\dd z}{1 - \ol{a}z} +  \frac{(z -
        a)}{(1 - \ol{a}z)^2} \ol{a} \dd z \\
        &= \frac{\dd z(1 - |a|^2)}{(1 -
        \ol{a}z)^2} \\
        \frac{|\dd w|}{1 - |w|^2}
        &= \frac{|\dd z|(1 - |a|^2)}{|1 -
        \ol{a}z|^2 \left( 1 - \left| \frac{\ -
        a}{1 - \ol{a}z} \right|^2 \right)} \\
        &= \frac{|az| (1 - |a|^2)}{|1 - \ol{a}z|^2
        - |z - a|^2}
        &&(\text{check}) \\
        &= \frac{|\dd z|}{1 - |z|^2} \qedhere
    \end{align*}
\end{proof}

\myskip
``Another view''
\[ g_{\text{hyp}} = \lambda \id \]
$\lambda(z) = \frac{4}{(1 - |z|^2)^2}$, $f(z) =
\frac{z - a}{1 - \ol{a}z}$, $|a| < 1$. To check
isometry:
\[ (Df|_z)^\top
\ub{(g_{\text{hyp}})_{f(z)}}_{\lambda(f(z)) \id}
Df|_z = \lambda(z) \id \]
i.e.
\[ \lambda(f(z)) (Df|_z)^\top Df|_z = \lambda (z)
\id \]
\[ \lambda(f(z)) |f'(z)|^2 = \lambda(z) \]
and this is checked as previously!

\begin{lemma}
    \begin{enumerate}[(i)]
        \item Every pair of points in $(D,
            \text{hyp})$ is joined by a unique
            geodesic (up to reparametrisation).
        \item The geodesics are diameters of the
            discs and circular arcs orthogonal to
            $\partial D$.
    \end{enumerate}
\end{lemma}

\begin{center}
    \includegraphics[width=0.3\linewidth]
    {images/92b63244b9ba11ed.png}
\end{center}
Geodesics in the hyperbolic disc. The whole
geodesics are called hyperbolic lines (defined on
all $\RR$).
