% vim: tw=50
% 08/03/2023 10AM

\begin{proof}
    Recall M\"obius maps send circles (in
    $\hat{\CC}$) to circles and preserve cycles.
    So if $z$ and $z'$ are inverse for $\Gamma$
    and $T \in \Mob$, then $Tz$ and $Tz'$ are
    inverse for $T(\Gamma)$. If $\Gamma = \RR \cup
    \{\infty\}$, then $J(z) = \ol{z}$ gives
    inverse points; this map satisfies the
    requirements and is unique such.

    \myskip
    Now if $\Gamma$ is any circle, there exists $T
    \in \Mob$ such that $T(\RR \cup \{\infty\}) =
    \Gamma$. Define inversion in $\Gamma$ by
    $J_\Gamma = TJT^{-1}$. This works (one should
    also check that this is unique, but this is
    left as an exercise).
\end{proof}

\begin{definition*}
    The map $z \to J_\Gamma(z)$ sending $z$ to the
    unique inverse point $z'$ for $z$ with respect
    to $\Gamma$ is called \emph{inversion} in
    $\Gamma$. (This fixes all points in $\Gamma$
    and exchanges the two complementary regions).
\end{definition*}

\subsubsection*{Examples}

\begin{enumerate}[(1)]
    \item If $\Gamma$ is a straight line (circle
        in $\hat{\CC}$ through $\infty \in
        \hat{\CC}$), then $J_\Gamma$ is reflection
        in $\Gamma$.
    \item If $S^1 = \{z : |z| = 1\}$ then
        $J_{S^1}(z) = \frac{1}{\ol{z}}$ (see
        Example Sheet 4).
\end{enumerate}

\begin{remark*}
    A composition of \emph{two} inversions is a
    M\"obius map. Let $C(z) = \ol{z}$ be inversion
    in $ \RR \cup \{\infty\}$. So if $\Gamma
    \subset \hat{\CC}$, $J_\Gamma = TCT^{-1}$
    where $T(\RR \cup \{\infty\}) = \Gamma$. Now
    given circles $\Gamma_1$ and $\Gamma_2$ and
    $T_i(\RR \cup \{\infty\}) = \Gamma_i$,
    \begin{align*}
        J_{\Gamma_1} \circ J_{\Gamma_2}
        &= (J_{\Gamma_1} \circ C) \circ (C \circ
        J_{\Gamma_2}) \\
        &= (C \circ J_{\Gamma_1})^{-1} \circ (C
        \circ J_{\Gamma_2})
    \end{align*}
    and also note
    \[ C \circ J_\Gamma = C \circ T \circ C \circ
    T^{-1} \]
    We know $T^{-1} \in \Mob$ so want to prove $C
    \circ T \circ C \in \Mob$. Let
    \[ T(z) = \frac{az + b}{cz + d} \]
    Then
    \[ C \circ T \circ C(z) = \frac{\ol{a}z +
    \ol{b}}{\ol{c} z + \ol{d}} \in \Mob \]
    Using this, we have constructed a M\"obius map
    $J_{\Gamma_1} \circ J_{\Gamma_2}$ which maps
    $\Gamma_1$ to $\Gamma_2$.
\end{remark*}

\begin{lemma}
    An orientation preserving isometry of
    $(\HH, g_{\text{hyp}})$ is an element
    of $\Mob(\HH)$, where $\HH = D$ or $\HH
    \mathcal{H}$. The \emph{full} isometry group
    is generated by inversions in hyperbolic lines
    (circles perpendicular to $\partial \HH$).
\end{lemma}

\begin{proof}
    Suffices to prove this in either model. In
    $D$, inversion in $\RR \cap D$, i.e.
    conjugation, preserves $g_{\text{hyp}} =
    \frac{4|\dd z|}{(1 - |z|^2)^2}$. Now $\Mob(D)$
    acts transitively on geodesics, and it acts by
    isometries, so all inversions in hyperbolic
    lines are isometries. Have we got them all?
    Now suppose $\alpha \in \operatorname{Isom}(D,
    g_{\text{hyp}})$. Define $a = \alpha(0) \in D$
    and using $T(z) =  \frac{z - a}{1 - \ol{a}z}$
    we see that $T \circ \alpha(0) = 0$. Now
    $\exists$ a rotation $R \in \Mob(D)$ such that
    $R \circ T \circ \alpha$ sends $D \cap \RR_+$
    to itself (fixed set wise).
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/6adfb4eabda511ed.png}
    \end{center}
    \[ D(R \circ J \circ \alpha)|_0 =
    \begin{pmatrix}
        1 \\
        0
    \end{pmatrix}
    =
    \begin{pmatrix}
        1 \\
        0
    \end{pmatrix}
    \]
    Composing with conjugation $C$ if necessary,
    there exists $A \in \operatorname{Isom}(D, g_{\text{hyp}})$
    of the required form such that $A \circ
    \alpha$ fixes $\RR \cap D$ pointwise and fixes
    $i\RR \cap D$ pointwise, so
    \[ D(A \circ \alpha)|_0 = \id \]
    Now implies $A \circ \alpha = \id$. Hence
    $\alpha$ has the required form. In Example
    Sheet 4, you'll prove that every M\"obius map
    is a product of inversions. If $\alpha$
    preserved orientation and fixed $\RR \cap D$,
    it necessarily fixes $i \RR \cap D$ and so in
    fact $\alpha = (R \circ T)^{-1} \in \Mob$.
\end{proof}

\begin{remark*}
    In the $\mathcal{H}$ model,
    \begin{align*}
        \Mob(\mathcal{H})
        &= \PSL_2(\RR) \\
        &= \left\{ z \mapsto \frac{az + b}{cz + d}
        :
        \begin{pmatrix}
            a & b \\
            c & d
        \end{pmatrix}
        \in \SL_2(\RR) \right\}
    \end{align*}
    (Note $\RR$ rather than $\CC$!) and
    $d_{\text{hyp}}(p, q) = 2\tanh^{-1} \left|
    \frac{p - q}{p - \ol{q}} \right|$.
\end{remark*}

\begin{hiddenflashcard}[dhyp-half-plane-model]
\prompt{Distance in half plane model? \\}
\[ d_{\text{hyp}}(p, q) = \cloze{2\tanh^{-1} \left|
\frac{p - q}{p - \ol{q}} \right|} \]
\end{hiddenflashcard}

\begin{hiddenflashcard}[isometry-group-of-H-model]
\[ \Mob(\mathcal{H}) = \cloze{\left\{z \mapsto
\frac{az + b}{cz + d} : ab - cd = 1 \right\}} \]
\end{hiddenflashcard}

\begin{flashcard}[classification-of-MobH]
\begin{definition*}
    Let $\alpha \in \Isom^+(\HH) = \Mob(\HH)$.
    Suppose $\alpha \neq \id$. We say that
    $\alpha$ is:
    \begin{itemize}
        \item \cloze{\emph{elliptic} if $\alpha$ fixes $p \in
            \HH$.
            \fcscrap{
            \begin{center}
                \includegraphics[width=0.3\linewidth]
                {images/f9b11852bda611ed.png}
            \end{center}
            }
            If $p = 0 \in D$ this is a
            rotation.}
        \item \cloze{\emph{parabolic} if $\alpha$ fixes a
            unique point of $\partial \HH$
            \fcscrap{
            \begin{center}
                \includegraphics[width=0.3\linewidth]
                {images/08d0ce86bda711ed.png}
            \end{center}
            }
            If $p = \infty \in \partial
            \mathcal{H}$ then $\alpha(z) =
            \alpha(z + t)$.}
        \item \cloze{\emph{hyperbolic} if $\alpha$ fixes
            2 points of $\partial \HH$.}
            \fcscrap{
            \begin{center}
                \includegraphics[width=0.4\linewidth]
                {images/21b583ecbda711ed.png}
            \end{center}
            }
    \end{itemize}
\end{definition*}
\end{flashcard}

\noindent
Exercise: All elements of $\Mob(\HH)$ fall into
one of $3$ cases.

\begin{flashcard}[hyperbolic-pair-of-lines-classification]
\begin{definition*}
    Let $l, l'$ be hyperbolic lines. We say that
    $l, l'$ are
    \begin{itemize}
        \item \cloze{\emph{parallel} if they emet in
            $\partial \HH$ but not in $\HH$.}
        \item \cloze{\emph{ultraparallel} if they don't
            meet in $\HH \cup \partial \HH$.}
        \item \cloze{\emph{intersecting} if they meet in
            $\HH$.}
    \end{itemize}
    \fcscrap{
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/8624b352bda711ed.png}
    \end{center}
    }
\end{definition*}
\end{flashcard}

\noindent
The parallel postulate of Euclid fails!

\subsubsection*{Hyperbolic Triangles}

\begin{flashcard}[hyperbolic-triangle-defn]
\begin{definition*}
    A \emph{hyperbolic triangle} is \cloze{the region
    bounded by $3$ hyperbolic lines, not two of
    which are ultraparallel. Vertices lying at
    infinity ($\partial \HH$) are called
    \emph{ideal} vertices.

    \myskip
    An \emph{ideal triangle} has all its vertices
    ideal.}
\end{definition*}
\end{flashcard}

\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/64a4ffc4bda811ed.png}
\end{center}
Next lecture we'll study triangles in more detail
and compute their area.
