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\begin{document}

\title{A Introductory Course on Volume Conjecture}
\author{Kim Hyuk}
\date{\today}
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\section{A brief introduction to Volume conjecture}
\section{Linear Fractional Transformation and 2-dimensional hyperbolic geometry}
\subsection{Linear Fractional Transformation (LFT)}
\subsection{Geometry}
\subsection{Cross Ratio}
\subsection{Poincar\'{e} Upper Half Plane and Disk}
\subsection{2D Hyperbolic Geometry}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Special Polygons}







\subsubsection{Lambert Quadrilateral}

.\\
Fig 16.\\ ($\phi<\frac{\pi}{2}$)
.\\
Then the following rules hold.\begin{align}
&\cos\phi=\sinh a\sinh b\\
&\displaystyle{\sin\phi =\frac{\cosh a}{\cosh c}\left(=\frac{\sinh a\cosh b}{\sinh c}\right)}\\
&\tanh c=\tanh a \cosh b
\end{align}

(Note that $a$ and $b$ determine the quadrilateral!)
\begin{proof}
Fig 17.\\
.\begin{align}\cos\phi&=-\cos\gamma\cos\delta+\sin\gamma\sin\delta\cosh f &\text{ (by Cosine Rule II)}\nonumber\\
&=-\sin\alpha\sin\beta+\cos\beta\cos\alpha\cosh f&\text{ ($\alpha+\delta=\beta+\gamma=\frac{\pi}{2}$)}\nonumber
\end{align}
Also, $\cos\beta=\sin\alpha\cosh b$ (Right Triangle) and $\cos\alpha=\sin\beta\cosh a$.\\
$\therefore$ The equation above becomes

\begin{align}\sin\alpha\sin\beta(\cosh a\cosh b\cosh f-1)&=
\sin\alpha\sin\beta(\cosh^2f-1)\nonumber\\=\sin\alpha\sin\beta\sinh^2f&=\sinh
a\sinh b\text{ (by Sine Rule)}\nonumber
\end{align}
\\
\textit{2nd Proof for 2.5.10}.\\
Fig 18.\\
.\begin{align}\cos^2\frac{\phi}{2}&=[e^{-i\theta},e^{i\theta},ie^{-\tau i},ie^{\tau i}]\nonumber\\&=\frac{1}{2}(\cot\theta\cot\tau+1)\nonumber
\end{align}
$\therefore$ $\sinh a\sinh b=\cot\theta\cot\tau=2\cos^2\frac{\phi}{2}-1=\cos\phi$\\\\
See Fig 17.

\textit{Proof for (2.5.11) and (2.5.12)}.\\

\begin{align}
\sin\phi =\frac{\sin\gamma}{\sinh c}\sinh f=\frac{\sinh f}{\sinh c}\cos\beta=\frac{\sinh a\cosh b}{\sinh c}\nonumber\\
\text{The last equality holds since $\cos\beta=\cosh
b\sin\alpha=\cosh b\frac{\sinh a}{\sinh f}$ from the identities
for right triangle.}\nonumber
\end{align}

From $\sin^2\phi+\cos^2\phi=1$,
\begin{align}1&=\frac{\sinh^2a\cosh^2b}{\sinh^2c}+\sinh^2a\sinh^2b\nonumber\\
&=\sinh^2a(\frac{\cosh^2b}{\sinh^2c}+\cosh^2b)-\sinh^2a\nonumber\\
&=\sinh^2a\cosh^2b(\frac{1+\sinh^2c}{\sinh^2c})-\sinh^2a\nonumber\\
&=\frac{\sinh^2a\cosh^2b\cosh^2c}{\sinh^2c}-\sinh^2a\nonumber
\end{align}
Hence
$\displaystyle{\cosh^2a=\frac{\sinh^2a\cosh^2b\cosh^2c}{\sinh^2c}}$.
and it follows that $\tanh c=\tanh a\cdot \cosh b$
\end{proof}


Saccheri Quadrilateral\\
Fig 19.\\
If $\phi=0$, then $\sinh a\sinh b=1$\\\\
\textit{Remark}.\\
Fig 20.\\
.\\ $\sin\phi=\displaystyle{\frac{\cos(-ia)}{\cosh c}}$ from
(2.5.11). We observe that these rules also follows from the
corresponding rules of a right triangle with Formal Substitution
as in the picture (using "extended model").
\end{document}
