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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 $\widehat{AB} = \widehat{XY}$.
\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 $\widehat{AX} > 
\widehat{A'X'}$ (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 $n\widehat{AB} = m \widehat{BC}$ for some $m,n \in \mathbb{Z}$, then 
$$\frac{
\widehat{AB}}{
\widehat{A'B'}} = \frac{\frac{m}{n} 
\widehat{BC}}{\frac{m}{n} 
\widehat{B'C'}} 
= \frac{
\widehat{BC}}{
\widehat{B'C'}}$$
\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{expformfigure}
\includegraphics[trim=4in 6in 1in 1in,clip,width=4truein,height=4truein]{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
$$\frac{
\widehat{AB}}{
\widehat{A'B'}} = f(x), \frac{
\widehat{A'B'}}{
\widehat{A''B''}} = f(y), 
\frac{
\widehat{AB}}{
\widehat{A''B''}} = f(x + y)$$
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 
$$
\widehat{AB} = 
\widehat{A'B'} e^{x/k}$$
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}

\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 $\overline{AB}\perp m$.
\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}


\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[trim=3.5in 5in 1in 1in,clip,width=4truein,height=4truein]{PlayfairAnalog.pdf}
\caption{Figure for Theorem~\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 $T$.  
\end{thm}
\vfill\eject
\noindent
{\bf Definition.}  The set of all points corresponding to 
a given point $P$ with respect to the family of all lines parallel to 
a given line is called a \emph{horosphere}, $\mathcal{H}_P$.

\vskip 10pt
To visualize this:
\begin{itemize}
\item Let $\ell$ be any one of the lines in the family
\item Let $T \supset \ell$ be a plane
\item Then $\mathcal{H}_P \cap T$ is a \emph{horocycle} in the plane $T$
\item That horocycle contains the point $P$ if $P$ is on $\ell$, but not otherwise
\end{itemize}

In fact, a horosphere can be thought of as a sort of ``surface of revolution''
generated by a horocycle in one of the planes containing $\ell$.  

One of Bolyai and Lobachevsky's main ideas: \emph{We can ``do geometry'' 
on the horosphere} if we take \emph{horocycles as the analogs of lines}.  For
instance, it's not hard to see that any two points $P,P'$ on the horosphere
determine coplanar parallel lines $P \in \ell, P'\in \ell'$, hence a horocycle
in the plane containing $\ell,\ell'$ passing through $P,P'$.  This and Theorem~\ref{PlayfairAnalog}
leads to a most surprising and important result:

\begin{thm}
The geometry of points and horocycles on a horosphere is \emph{Euclidean}.
\end{thm}

(That is, Postulates I-V all hold if we replace ``line'' by ``horocycle!'')
\vfill\eject
\noindent
{\it Hyperbolic trig functions}
\vskip 10pt
Let 
$$\cosh(t) = \frac{e^t + e^{-t}}{2}, \quad \sinh(t) = \frac{e^t - e^{-t}}{2}$$
This gives a family of functions parallel to the usual trig functions according
to the following definitions
$$\tanh(t) = \frac{\sinh(t)}{\cosh(t)}, \quad \coth(t) = \frac{\cosh(t)}{\sinh(t)}$$
As you'll see on Problem Set 3, these have many properties similar to the 
usual trig functions, except they correspond to properties of \emph{hyperbolas}
rather than \emph{circles}.  Most important hyperbolic identity:
$$\cosh^2(t) - \sinh^2(t) = 1$$
for all $t$ (see where the hyperbolas are entering?)

Two basic results:

\begin{lem}
\label{BLLemma}
Let $\overrightarrow{PQ}$ and $\overrightarrow{XX'}$ be parallel in the direction of $Q$ and $X'$
with $\overline{XP} \perp \overline{PQ}$ with $PX = u$.  If we also draw in the horocycle
arc through $P$ (for the pencil of parallels including the two given lines), and call 
$s = \widehat{PB}$ the distance along that arc from $P$ to where it intersects $\overline{XX'}$ at
$B$.  Finally let $v = BX$.  Then the following relations hold:
$$e^{v/k} = \cosh(u/k),$$
and
$$s = \sigma \tanh(u/k)$$
where $\sigma$ is a constant that will be determined in the course of the proof.
\end{lem}

We will assume these for now and come back to their proof later.  We will next use them
to deduce the Bolyai-Lobachevsky theorem, Theorem~\ref{BLT}.
\vfill\eject
\noindent
\begin{figure}
\begin{centering}
\label{BLProofFig}
\includegraphics[trim=4in 8in 0in 0in,clip,width=5truein,height=4truein]{BLProof.pdf}
\caption{Figure for proof of Theorem~\ref{BLT}}
\end{centering}
\end{figure}

\noindent
\begin{itemize}
\item Let $\Delta ABC$ be a right triangle with right angle at $C$, and 
\item In three-dimensional space let $\overleftrightarrow{AA'}$ be the 
perpendicular line to the plane of the triangle.
\item Let $\overleftrightarrow{BB'}$ and $\overline{CC'}$ be parallels to 
$\overleftrightarrow{AA'}$ in direction of $A'$.  
\item These lines intersect the horosphere $\mathcal{H}_P$ at $B_1,C_1$.
\item Hence we get a triangle (with horocycle sides) $\Delta AB_1C_1$
in the (Euclidean) geometry on $\mathcal{H}_P$.  
\end{itemize}

\vfill\eject
\noindent
\begin{figure}
\begin{centering}
\includegraphics[trim=4in 8in 0in 0in,clip,width=5truein,height=4truein]{BLProof.pdf}
\caption{Figure for proof of Theorem~\ref{BLT}}
\end{centering}
\end{figure}

\begin{itemize}
\item Write $s_1 = 
\widehat{B_1C_1}$, $s_2 = 
\widehat{AC_1}$, $s_3 = 
\widehat{AB_1}$ and 
$a = BC$, $b = AC$, $c = AB$.  
\item By Lemma~\ref{BLLemma}, we have 
$$s_2 = \sigma \tanh(b/k), \qquad s_3 = \sigma \tanh(c/k)$$
and combining with Lemma~\ref{expform}
$$s_1 = \sigma \tanh(a/k) e^{-CC_1/k}.$$
\item But also by Lemma~\ref{BLLemma}, 
$$e^{CC_1/k} = \cosh(b/k),$$ so 
$$s_1 = \sigma \frac{\tanh(a/k)}{\cosh(b/k)}.$$
\item In the Euclidean triangle $\Delta AB_1C_1$ on the horosphere, we have 
$$s_2 = s_3 \cos(\angle CAB)$$
so 
\begin{equation}
\label{onestar}
\cos(\angle CAB) = \frac{\tanh(b/k)}{\tanh(c/k)}
\end{equation}
Similarly, 
$$\cos(\angle ABC) = \frac{\tanh(a/k)}{\tanh(c/k)}$$
and then 
\begin{equation}
\label{twostar}
\sin(\angle CAB) = \frac{\tanh(a/k)}{\cosh(b/k)\tanh(c/k)}
\end{equation}
\end{itemize}

\begin{itemize}
\item Let $a \to \infty$.  Then $c \to \infty$ as well and $\tanh(c/k) \to 1$
\item At the same time, in the right triangle $\angle ABC$, we have $\angle CAB \to \Pi(b)$
(the angle of parallelism)
\item From (\ref{onestar}) we get
$$\cos(\Pi(b)) = \tanh(b/k)$$
while from (\ref{twostar}) we get
$$\sin(\Pi(b)) = \frac{1}{\cosh(b/k)}$$
\item Now, use the trig identity 
$$\tan(\theta/2) = \frac{1 - \cos(\theta)}{\sin(\theta)}$$
and we deduce
$$\tan(\Pi(b)/2) = \frac{1 - \tanh(b/k)}{1/\cosh(b/k)} = e^{-b/k}.$$
This is what we wanted to prove for the Bolyai-Lobachevsky Theorem(!)
\end{itemize}

A byproduct of this set-up:  The geometry of the horosphere is Euclidean
so in fact we have the relation $s_3^2 = s_1^2 + s_2^2$, which yields:
$$\cosh^2(c/k) = \cosh^2(a/k)\cosh^2(b/k)$$
which is the analog of the Pythagorean Theorem for hyperbolic triangles(!!)


\end{document}
