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\begin{document}
\centerline{MATH 392 -- Geometry Through History}
\centerline{Bolyai-Lobachevsky Theorem and Hyperbolic Pythagorean Theorem}
\centerline{February 22, 24, 26}
\vskip 10pt
\noindent
{\it Recap}

\vskip 10pt
In the geometry of the hyperbolic plane, the following notions are keys to 
the general properties:

\begin{itemize}
\item Given a line $\ell$, the collection of \emph{all lines parallel to $\ell$}
is called ${\bf P}_\ell$.  
\item Given a pencil ${\bf P}$ of lines (of any one of the three types -- but we'll
be concerned mostly with the parallel pencils), we say two points 
$X$ and $Y$ \emph{correspond with respect to ${\bf P}$} if the line of the pencil that 
passes through the midpoint of the line segment $\overline{XY}$ meets that segment
at a right angle (or equivalently, the lines of the pencil through $X$ and through $Y$
make equal angles with the segment $\overline{XY}$).
\item {\it In the case ${\bf P} = {\bf P}_{\ell}$ is the pencil of lines 
parallel to some line $\ell$}, the set of all points corresponding to a given point $X$
is called a {\it horocycle} through $X$.   In Euclidean geometry, horocycles are lines;
in hyperbolic geometry, horocycles are \emph{not} lines, but they can be thought of as
``circles with center at infinity.'')
\item Given a pencil of parallel lines, there is a horocycle
passing through every point in the plane.  Two horocycles are \emph{concentric}
of they come from the same pencil of parallel lines.
\end{itemize}

\noindent
Our next goal is to prove a result known as the Bolyai-Lobachevsky theorem:

\begin{thm}[Bolyai-Lobachevsky]  
\label{BLT}
Let $\Pi(x)$ denote the angle of parallelism 
of a segment of length $x$.  Then 
$$\tan(\Pi(x)/2) = e^{-x/k}$$
for some constant $k$.
\end{thm}

Along the way, we will take a ``detour'' into \emph{three-dimensional} hyperbolic space
and see a result analogous to the Pythagorean theorem that holds for triangles
in the hyperbolic plane.
\vfill\eject

\noindent
{\it Some preliminary results}

\begin{prop}
\label{horarcs}  
\begin{enumerate}
\item[(1)] Let $A,B$, $X,Y$ be two pairs of points on the same horocycle.
If the line segments $\overline{AB}$ and $\overline{XY}$ satisfy $AB = XY$,
then the horocycle arcs satisfy $\arc{AB} = \arc{XY}$.
\item[(2)] Let $A,A'$ be on one line of a parallel pencil, while $X,X'$ lie
on another line of the same parallel pencil.  If $A,X$ lie on the same horocycle
and similarly for $A',X'$ then $AA' = XX'$.
\item[(3)] In the situation of the previous part, if $\overrightarrow{AA'}$ and 
$\overrightarrow{XX'}$
are in the direction of parallelism, then $\arc{AA'} > \arc{XX'}$ (that is, horocycle
arcs intercepted by the lines of a pencil \emph{decrease} in the direction of parallelism). 
\end{enumerate}
\end{prop}

\noindent
For proofs, see Lemma 4.8 in McCleary.


\begin{prop}
\label{expform}
The ratio of the lengths of arcs of two concentric horocycles intercepted by two lines
of a parallel pencil is expressible in terms of an exponential function of the 
distance between the arcs (along any one of the lines -- see Proposition~\ref{horarcs}, part (2)).
\end{prop}

\begin{itemize}
\item First suppose $A,B,C$ are on one horocycle and $A',B',C'$ are on 
a concentric horocycle.
\item If $n \arc{AB} = m \arc{BC}$ for some $m,n \in \mathbb{Z}$, then 
$$\frac{\arc{AB}}{\arc{A'B'}} = \frac{\frac{m}{n} \arc{BC}}{\frac{m}{n} \arc{B'C'}} 
= \frac{\arc{BC}}{\arc{B'C'}}$$
\item So, the ratio depends only on the distance between the horocycles.
\item If the ratio is irrational, approximate by rational numbers and take a limit.
\end{itemize}
\vfill\eject

\begin{figure}
\begin{centering}
\label{expformfig}
\includegraphics[width=4truein,height=3truein]{expformfig.pdf}
\caption{Figure for Proposition~\ref{expform}}
\end{centering}
\end{figure}

\begin{itemize}
\item Write $x = AA'$ and $y = A'A''$ along the line.
\item Then we have
$$\frac{\arc{AB}}{\arc{A'B'}} = f(x), \frac{\arc{A'B'}}{\arc{A''B''}} = f(y), 
\frac{\arc{AB}}{\arc{A''B''}} = f(x + y)$$
and 
$$f(x+y) = f(x)f(y)$$
\item (Assuming $f$ differentiable), we have, using the previous bullet:
$$f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h} = f(x) \lim_{h\to 0} \frac{f(h) - 1}{h} = f(x)\cdot \frac{1}{k}$$
for some $k$.
\item Then standard facts from calculus show
$$f(x) = e^{x/k}$$
so 
$$\arc{AB} = \arc{A'B'} e^{x/k}$$
Note, this is increasing in the \emph{opposite of the direction of parallelism}.
\end{itemize}

\vfill\eject
\noindent
Next, following Bolyai and Lobachevsky, we will make a detour into \emph{three-dimensional}
space.  Assume Postulates I-IV, and Saccheri's HAA hold in all planes.  
\vskip 10pt
\noindent
{\bf Definition.}  Two lines $\ell_1,\ell_2$ in space are \emph{parallel} if they lie
in the same plane and are parallel in that plane. 
\vskip 10pt
\noindent
Note that there are also \emph{skew} pairs of lines, and coplanar pairs of lines
that do not meet, but are not parallel.  Since the definition of parallel is 
hyperbolic one(!)

First there are 
some results that should look very familiar, since they are the same as in Euclidean 
three-dimensional space.  You will be asked to provide proofs on Problem Set 3, due
Monday, February 29.

\begin{lem}
\label{parlines}
Let $\ell_1,\ell_2$ be parallel lines in space and $\ell_1 \subset T_1$, $\ell_2 \subset T_2$, 
where $T_1,T_2$ are distinct planes.  If $T_1$ and $T_2$ meet, the line $m = T_1\cap T_2$
is parallel to both $\ell_1$ and $\ell_2$.
\end{lem}

\vfill\eject

\noindent
{\bf Definition.}  We say a line $\ell$ is \emph{perpendicular to a plane} $T$ if 
$\ell$ meets $T$ at a point $P$ and every line through $P$ in $T$ meets $\ell$
at a right angle.


\begin{lem}
\label{perpline}
The following are equivalent for a line $\ell$ intersecting a plane $T$ at a point $P$.
\begin{enumerate}
\item[(1)] $\ell \perp T$
\item[(2)] There exist lines $m_1\ne m_2$ through $P$ in $T$ such that $\ell$ meets
$m_1$ and $m_2$ at right angles. 
\end{enumerate}
\end{lem}

\begin{cor}
\label{lineperpplane}
Given a line $\ell$ and a point $P$, there is a unique plane containing $P$
and perpendicular to $\ell$.
\end{cor}

\begin{lem}
\label{perpllineprop}
If $\ell \perp T$ with $\ell \cap T = P$, let $m$ be any line in $T$ and $A \in m$
such that $\overline{AP}\perp m$.  If $B$ is any point on $\ell$, then $\overline{AB}\perm m$.
\end{lem}

\begin{cor}
Given a plane $T$ and a point $P$ in $T$, there is a unique line through $P$ that 
is perpendicular to $T$.  
\end{cor}

\vfill\eject

\noindent
{\bf Definition.}  Two planes $T,T'$ are perpendicular if there is a line $\ell \subset T$
that is perpendicular to $T'$ (or equivalently, there is a line $\ell' \subset T'$ that
is perpendicular to $T$.

\begin{lem} 
\label{perpplanes}
\begin{enumerate}
\item[(1)]  If $T$ and $T'$ are perpendicular planes, and $\ell$ is a line on $T$, then 
$\ell \perp T$ is equivalent to $\ell \perp m$ for $m = T\cap T'$.
\item[(2)]  If $\ell_1 \perp T$ and $\ell_2 \perp T$, then $\ell_1,\ell_2$
are coplanar and non-intersecting (but not necessarily parallel -- this is a
difference in the hyperbolic case!)
\end{enumerate}
\end{lem}

\begin{lem}
\label{perpplanethroughline}
\begin{enumerate}
\item[(1)] Given a line $\ell$ and a plane $T$ not containing $\ell$, there is a unique plane
$T'$ containing $\ell$ that is perpendicular to $T$.  
\item[(2)] Given any two distinct planes
$T_1,T_2$, there exist planes $T$ that are perpendicular to both.
\end{enumerate}
\end{lem}

\noindent
{\bf Definitions.}  
\begin{enumerate}
\item[(a)]  Let $T'$ be the unique plane containing a given line $\ell$ and perpendicular
to a plane $T$ (note that the assumption this plane is unique rules out the case where 
$\ell \perp T$).  Then $m = T\cap T'$ is called the \emph{perpendicular projection of
$\ell$ onto $T$}.  
\item[(b)] $\ell \subset T'$ is \emph{parallel to $T$} if $\ell$ and the perpendicular
projection $m$ into $T$ are parallel lines in $T'$
\item[(c)] \emph{Two planes are parallel} if the intersections with
some common perpendicular plane are parallel lines.
\end{enumerate}

\vfill\eject

\begin{figure}
\begin{centering}
\label{expformfig}
\includegraphics[width=4truein,height=3truein]{PlayfairAnalog.pdf}
\caption{Figure for Proposition~\ref{PlayfairAnalog}}
\end{centering}
\end{figure}

\noindent
Next, we have a \emph{very important} analog of the Playfair Postulate in
Euclidean plane geometry:

\begin{thm}
\label{PlayfairAnalog}
Through a line $\ell$ parallel to a plane $T$, there exists a unique
plane that contains $\ell$ and is parallel to $T$.  

\vfill\eject


\end{document}
