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%\title{Hyperbolic Geometry Lecture Note 2}
%\author{Hyuk Kim}
%\affil{Seoul National University}

%\maketitle

\theoremstyle{definition}   \newtheorem{problem}{Problem}
\theoremstyle{definition}   \newtheorem{proposition}{Proposition}[subsection]
\theoremstyle{definition}   \newtheorem{lemma}{Lemma}[subsection]
\theoremstyle{definition}   \newtheorem{corollary}{Corollary}[subsection]
\theoremstyle{definition}   \newtheorem{definition}{Definition}[subsection]
\theoremstyle{definition}   \newtheorem{theorem}{Theorem}[subsection]
\theoremstyle{definition}   \newtheorem{remark}{Remark}[subsection]
\theoremstyle{definition}   \newtheorem*{fact}{Fact}
\theoremstyle{definition}   \newtheorem*{introduction}{Introduction}
\theoremstyle{definition}   \newtheorem*{note}{Note}
\theoremstyle{definition}   \newtheorem*{claim}{Claim}


\section{A brief introduction to Volume conjecture}

\section{Linear Fractional Transformation and 2-dimensional hyperbolic geometry}
\section{Inversive geometry and hyperbolic geometry}

\subsection{Inversion(or reflection) and M$\ddot{\text{o}}$bius transformation}
\subsection{M$\ddot{\text{o}}$bius transformations as conformal maps}
\subsection{M$\ddot{\text{o}}$bius transformation as a cross-ratio preserving maps}
\subsection{M$\ddot{\text{o}}$bius transformation as a sphere preserving map}


\subsection{Poincare extension}

\begin{definition}
    Define $i:M(\widehat{\mathbb{R}^{n}})\rightarrow M(\widehat{\mathbb{R}^{n+1}})$ by $\sigma=J_{S}\mapsto \tilde{\sigma}=J_{\tilde{S}}$ and $i:\phi=\sigma_{1}\circ \cdots \circ \sigma_{k}\mapsto \tilde{\phi}=\tilde{\sigma_{1}}\circ \cdots \circ \tilde{\sigma_{k}}$, where $\tilde{S}$ is the sphere in $\mathbb{R}^{n+1}$ for which $\tilde{S}\cap \widehat{\mathbb{R}^{n}}=S$.
\end{definition}

Check if $i$ is uniquely well-defined and 1-1 :

Suppose $\tilde{\phi}_{1},\tilde{\phi}_{2}$ are two extensions of
$\phi\in M(\widehat{\mathbb{R}^{n}})$ $\Longrightarrow$
$\tilde{\phi}_{1}\circ\tilde{\phi}_{2}^{-1}=id$
 on $\mathbb{R}^{n}$and preserves $\mathbb{H}^{n+1}$ $\Longrightarrow$
$\tilde{\phi_{1}}\circ\tilde{\phi_{2}}^{-1}=id$ by Proposition
3.5.2.

\begin{theorem}
    ${}$
\begin{description}
    \item{i)} $\phi\in M(\mathbb{H}^{n+1}):=\{\phi\in M(\widehat{\mathbb{R}^{n}})\,|\, \text{$\phi$ is an automorphism on $\mathbb{H}^{n+1}$}\}$ $\Longrightarrow$ $\phi|_{\widehat{\mathbb{R}^{n}}}\in M(\widehat{\mathbb{R}^{n}})$
    \item{ii)} $i(M(\widehat{\mathbb{R}^{n}})) = M(\mathbb{H}^{n+1})$
    \item{iii)} $\phi\in M(\mathbb{H}^{n+1})$ $\Leftrightarrow$ $\phi=J_{S_{1}}\circ \cdots \circ J_{S_{k}}, S_{i}$ $\bot$ $\widehat{\mathbb{R}^{n}}$
\end{description}
\end{theorem}

\begin{proof}
${}$
\begin{description}
    \item{i)} $\phi\in M(\mathbb{H}^{n+1})\Longrightarrow \phi| :\widehat{\mathbb{R}^{n}}=\partial \mathbb{H}^{n+1} \circlearrowleft$.
    $\phi|$ preserves cross-ratio since $\phi$ does. $\Longrightarrow \phi|\in M(\widehat{\mathbb{R}^{n}})$

    \item{ii)} $(\subset)$ : clear. $(\supset)$ : $\forall \phi\in M(\mathbb{H}^{n+1})$, consider $\phi|$.
    Then $\tilde{\phi|}\circ \phi^{-1}=id$ on $\widehat{\mathbb{R}^{n}}$ and $\mathbb{H}^{n+1}\circlearrowleft$
    $\Longrightarrow\tilde{\phi|}\circ \phi^{-1}=id$ $\Longrightarrow$ $\phi=\tilde{\phi|}=i(\phi|)$ .
\end{description}

    \item{iii)} $(\Longleftarrow)$ : clear. $(\Longrightarrow)$ : clear from ii).
\end{proof}

Note that $\forall \phi\in M(\mathbb{H}^{n+1})$, $\phi|\in
M(\widehat{\mathbb{R}^{n}})$ as in the proof of i) and
$\phi=\tilde{\phi|}$ by the proof of ii). Therefore if $\phi\in
M(\widehat{\mathbb{R}^{n}})$ is a similarity, then $\tilde{\phi}$
is the unique similarity on $\mathbb{H}^{n+1}$ whose restriction
is $\phi$.

Now consider the ball model. Recall $\eta =
J_{\widehat{\mathbb{R}^{n}}}\circ J_{S(e_{n+1}, \sqrt{2})} :
\mathbb{B}^{n+1}\rightarrow \mathbb{H}^{n+1}$,
$S^{n}=\partial\mathbb{B}^{n+1}\rightarrow \partial
\mathbb{H}^{n+1}=\widehat{\mathbb{R}^{n}}$. Then

\begin{equation*}
    M(\mathbb{B}^{n+1})=\eta^{-1}\circ M(\mathbb{H}^{n+1})\circ \eta.
\end{equation*}


\begin{proposition}
    Let $\phi\in M(\mathbb{B}^{n+1})$. Then the followings are
    equivalent.
\begin{description}
    \item{i)} $\phi(\infty)=\infty$
    \item{ii)} $\phi(0)=0$
    \item{iii)} $\phi\in O(n+1)$
\end{description}
\end{proposition}

\begin{proof}
    i) $\Longleftrightarrow$ ii) since $\phi$ preserves the inversion $J_{S(0,1)}$.
    If i) holds, then ii) also holds and $\phi$ is a similarity:$x\mapsto \lambda Ax$, where $A\in O(n)$.
    Now $\phi:\mathbb{B}^{n+1}\circlearrowleft$ $\Longrightarrow$ $|\lambda|=1$ and hence iii) follows. Now iii)$\Longrightarrow$ii) is clear.
\end{proof}


\subsection{Hyperbolic metric}

\subsubsection{$\mathbb{B}^{n}$ case}
Let $\phi\in \mathcal{M}(\mathbb{B}^{n})$ and $x^{*}=\sigma_{1}(x)=\frac{x}{|x|^{2}}$. Note that
\begin{equation*}
    |x^{*}-u^{*}|^{2}=\sum_{i=1}^{n}\left( \frac{x_{i}}{|x|^{2}}-\frac{u_{i}}{|u|^{2}}\right)=\sum_{i=1}^{n}\frac{|x|^{2}-2x_{i}u_{i}+|u^{2}|}{|x|^{2}|u^{2}|}=\left( \frac{|x-u|}{|x||u|} \right)^{2}.
\end{equation*}
This yields
\begin{equation*}
    [x,x^{*},u,u^{*}] = \frac{|x-u||x^{*}-u^{*}|}{|x-\frac{x}{|x|^{2}}||u-\frac{u}{|u|^{2}}|} = \frac{|x-u|^{2}}{(1-|x|^{2})(1-|u|^{2})}.
\end{equation*}
Put $u=x+dx$, $y=\phi x$ and since the M$\ddot{\text{o}}$bius transformation $\phi$ preserves cross ratio, we conclude
\begin{equation*}
    \frac{2|dy|}{1-|y|^{2}} = \frac{2|dx|}{1-|x|^{2}}.
\end{equation*}
In other words, the Poincare matric is invariant under M$\ddot{\text{o}}$bius transformations.

\subsubsection{$\mathbb{H}^{n+1}$ case}
The inversive point is given as $x^{*}=(x_{1},\cdots,x_{n-1},-x_{n})$ for any $x\in \mathbb{H}^{n+1}$. Then
\begin{equation*}
[x,x^{*},u,u^{*}] = \frac{|x-u||x^{*}-u^{*}|}{|x-x^{*}||u-u^{*}|} = \frac{|x-u|^{2}}{4x_{n+1}u_{n+1}},
\end{equation*}
and by letting $u=x+dx$, we see that
\begin{equation*}
    \frac{|dx|^{2}}{4x_{n+1}^{2}}
\end{equation*}
is an invariant metric and $\frac{|dx|}{x_{n+1}}$ is called the Poincare metric.

\subsubsection{Canonical embedding}
Fig3.1

\begin{proposition}
    $M(\mathbb{H}^{n+1}) = Isom
    (\mathbb{H}^{n+1})$
\end{proposition}

\begin{proof}

    \item{($\subset$)} : clear.\\
    \item{($\supset$)} : It suffices to show that $M(\mathbb{B}^{n+1})= Isom(\mathbb{B}^{n+1})$. $M(\mathbb{B}^{n+1})\subset Isom(\mathbb{B}^{n+1})$
    and is already "full", $i.e.$, transitive and isotropy group $=O(n+1)$. Indeed $g\in Isom(M)$, $M$ connected Riemannian,
    such that $g(x)=x$ and $dg(x)=id$, then $g=id$: Note $g=id$ on a neighborhood of $x$(since it fixes radial
    geodesics), and
\begin{eqnarray*}
    A&=&\{ x\in M \,|\, \text{$g(x)=x$ and $dg(x)=id$} \}\\
&\Rightarrow & \text{$A$ is open and closed}\\
&\Rightarrow & A=M.
\end{eqnarray*}

\end{proof}

\subsection{Isometry types}

\subsubsection{$\mathbb{B}^{n+1}$ case}

$\phi \in M(\mathbb{B}^{n+1})\Longrightarrow \phi :
\overline{\mathbb{B}^{n+1}}\circlearrowleft$ $\Longrightarrow$
$\phi$ has a fixed point in $\overline{\mathbb{B}^{n+1}}$ by
Brouwer fixed point theorem.

\begin{claim}
$\phi$ has more than two fixed points on $S^{n}=\partial
\mathbb{B}^{n+1}$ $\Longrightarrow$ $\phi$ has a fixed point in
$\mathbb{B}^{n+1}$.
\end{claim}

\begin{proof}
    Work on $\mathbb{H}^{n+1}$ and suppose $\#|Fix|\ge 3$ on $\widehat{\mathbb{R}^{n}}=\partial \mathbb{H}^{n+1}$.
    We may assume $\phi(\infty)=\infty$, $\phi(0)=0$, $\phi(e_{1})=e_{1}$.\\

$\Longrightarrow$ $\phi(x)=Ax$, $A\in O(n+1)$

$\Longrightarrow$ $\phi(x)$ fixes $x_{n+1}$ axis since it is
perpendicular to $\mathbb{R}^{n}$, $i.e.$, fixes points in
$\mathbb{H}^{n+1}$
\end{proof}

Therefore we have the following trichotomy for a conjugacy class
of $\phi$ :
\begin{description}
    \item{(1)} $\phi$ fixes a point in $\mathbb{B}^{n+1}$ : elliptic
    \item{(2)} $\phi$ fixes exactly one point on $\partial \mathbb{B}^{n+1} = S^{n}$ : parabolic
    \item{(3)} $\phi$ fixes exactly two points on $\partial \mathbb{B}^{n+1} = S^{n}$ : loxodormic or hyperbolic
\end{description}

\subsubsection{$\mathbb{H}^{2}$ and $\mathbb{H}^{3}$ case}
\begin{equation*}
g(z) = \frac{az+b}{cz+d},\quad A:=\begin{bmatrix} a & b \\ c & d\end{bmatrix}
\end{equation*}
Fixed point : $z=g(z)$ $\Longrightarrow$ $cz^{2}+(d-a)z-b=0$ $\Longrightarrow$ $D=(d-a)^{2}+4bc = (a+d)^{2}-4(ad-bc)=$ tr$^{2}(A)-4\det(A)$.
Then we define
\begin{equation*}
    \text{tr}(g):=\frac{(\text{tr}A)^{2}}{\det(A)},
\end{equation*}
which is an invariant of a projective transformation. Note that we
have the following trichotomy for $G=PSL_{2}(\mathbb{R})$
\begin{description}
    \item{(1)} $g$ is elliptic $\Longleftrightarrow$ $D<0$ $\Longleftrightarrow$ tr$^{2}<4$
    \item{(2)} $g$ is parabolic $\Longleftrightarrow$ $D=0$ $\Longleftrightarrow$ tr$^{2}=4$
    \item{(3)} $g$ is hyperbolic $\Longleftrightarrow$ $D>0$ $\Longleftrightarrow$ tr$^{2}>4$
\end{description}


\begin{proposition}
    Let $G=PSL_{2}(\mathbb{C})$ or $PSL_{2}(\mathbb{R})$. Then $\forall f,g\in G$ we have $f\sim g$($i.e.,$ $f$ is conjugate to $g$)
    $\Longleftrightarrow$ tr$^{2}(f)$=tr$^{2}(g)$
\end{proposition}

\begin{proof}
\item{$\Longrightarrow)$} : Clear.\\
\item{$\Longleftarrow)$} :
\begin{description}
    \item{(1)} tr$^{2}(g)=4$ $\Longrightarrow$ there exists a unique fixed point, say $\infty$.

        $\Longrightarrow$ $g(z)=az+b$ and $a=1$ ($\exists$ another fixed point otherwise)

        $\Longrightarrow$ $g\sim f:z\mapsto z+1$ ($\because$ $f=h^{-1}\circ g\circ h$ with $h(z)=bz$)

    \item{(2)} tr$^{2}(g)\ne 4$ $\Longrightarrow$ there are two fixed points, say $0,\infty$

        $\Longrightarrow$ $g\sim f(z)=az$ ($a\ne 1$ $a\ne 0$) $\Longrightarrow$ tr$^{2}=a+\frac{1}{a}+2$

            \begin{description}
    \item{(i)} $|a|=1$ $\Longrightarrow$ $g$ is elliptic
    \item{(ii)} $|a|\ne 1$ $\Longrightarrow$ $g$ is loxodromic (hyperbolic if $a$ is real)
            \end{description}

\end{description}
Notice that, tr$^{2}$ determines $a,\frac{1}{a}$ and $g(z)=az\sim f(z)=\frac{1}{a}z$ via $h(z)=-1/z$, and this proves the proposition.
\end{proof}


In the above proof, we notice
\begin{equation*}
(2) i)\Longrightarrow
\text{tr}^{2}g=a+\frac{1}{a}+2=a+\overline{a}+2=2\cos \theta+2\in
[0,4).
\end{equation*}
Conversely, if tr$^{2}g\in [0,4)$, then by the above dichotomy
$g\sim f(z) = az$ with $a=e^{i\theta}$, $i.e.$, elliptic. Hence we
have the following map,
\begin{equation*}
    \text{tr}^{2}:(G\setminus \{id\}) / \sim \quad \longrightarrow \quad \mathbb{C}
\end{equation*}
such that $\text{tr}^{2}$(elliptic)$=[0,4)$,
tr$^{2}$(parabolic)$=4$, and loxodromic otherwise.




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