klein.tex

\chapter{Projective model}\label{chap:klein}

The \textit{projective model} is another model of the hyperbolic plane discovered by Eugenio Beltrami; it is often called the {}\emph{Klein model}.
The projective and conformal models are saying exactly the same thing but in two different languages.
Some problems  are easier in projective model, and others in the conformal model.
Therefore, it is worth knowing both. 

\section{Special bijection on the h-plane}
\label{sec:special-bijection}

Consider the conformal disc model.
Assume its absolute is a unit circle $\Omega$ centered at~$O$.
Choose a coordinate system $(x,y)$ on the plane with the origin at $O$, 
so the circle $\Omega$ is described by the equation $x^2+y^2=1$.


\label{pic:stereographic_projection-klein}
\begin{wrapfigure}{o}{48mm}
\centering
\vskip-3mm
\includegraphics{mppics/pic-256}
\caption*{Plane thru $P$, $O$, and $S$.}
\vskip-3mm
\end{wrapfigure}

Let us think that our plane is the coordinate $(x,y)$-plane in the Euclidean space; denote it by $\Pi$.
Let $\Sigma$ be the unit sphere centered at $O$;
it is described by the equation 
$x^2+y^2+z^2=1$.
Set $S\z=(0,0,-1)$ and $N\z=(0,0,1)$; 
these are the south and north poles of~$\Sigma$.

Consider the stereographic projection $\Pi\to\Sigma$ from $S$;
given a point $P\in\Pi$ denote by $P'$ its image in $\Sigma$.
Note that the  h-plane is mapped to the {}\emph{northern hemisphere},
that is, to the set of points $(x,y,z)$ in $\Sigma$ described by the inequality~$z>0$.

For a point $P'\in \Sigma$ consider its footpoint $\hat P$
on $\Pi$;
this is the closest point to~$P'$.

The composition of these two maps $P\z\leftrightarrow P'\z\leftrightarrow\hat P$
gives a bijection from the h-plane to itself.
Furthermore, $P=\hat P$
 if and only if  $P\in \Omega$ or $P=O$.

%\begin{thm}{Exercise}\label{ex:P-->hat-P}
%Suppose that $P\leftrightarrow \hat P$ is the bijection described above.
%Assume that $P$ is a point of the h-plane distinct from the center of the absolute and $Q$ is its inverse across the absolute.
%Show that the inversion of $\hat P$ across the absolute is the midpoint of $[PQ]$. 
%\end{thm}

\begin{thm}{Lemma}\label{lem:P-hat-chord}
Let $P\leftrightarrow\hat P$ be the bijection of the h-plane described above.
Suppose $(PQ)_h$ is an h-line with the ideal points $A$ and~$B$.
Then $\hat P,\hat Q\in[AB]$.

Moreover, 
$$\frac{A\hat Q\cdot B\hat P}{\hat QB\cdot \hat PA}
=
\left(\frac{AQ\cdot BP}{QB\cdot PA}\right)^2.
\eqlbl{eq:lem:P-hat-chord}$$
In particular, if $A,P,Q,B$ appear in the same order, then
$$PQ_h=\tfrac12\cdot\ln\frac{A\hat Q\cdot B\hat P}{\hat QB\cdot \hat PA}.$$
\end{thm}

\parit{Proof.}
Consider the stereographic projection $\Pi\to \Sigma$ from the south pole~$S$.
It fixes $A$ and $B$;
denote by $P'$ and $Q'$ the images of $P$ and~$Q$;

According to Theorem~\ref{thm:inversion-3d}\textit{\ref{thm:inversion-3d:cross-ratio}},
$$\frac{AQ\cdot BP}{QB\cdot PA}=\frac{AQ'\cdot BP'}{Q'B\cdot P'A}.\eqlbl{eq:(AB;PQ)=(AB;P'Q')}$$

By Theorem~\ref{thm:inversion-3d}\textit{\ref{thm:inversion-3d:angle}}, 
each circline in $\Pi$ that is perpendicular to $\Omega$ 
is mapped to a circle in $\Sigma$ that is still perpendicular to~$\Omega$.
It follows that the stereographic projection sends $(PQ)_h$ to the intersection of the northern hemisphere of $\Sigma$ with a plane perpendicular to~$\Pi$.

Denote this plane by $\Lambda$;
it contains the points $A$, $B$, $P'$, $\hat P$ and the circle $\Gamma=\Sigma\cap\Lambda$.
(It also contains $Q'$ and $\hat Q$ but we will not use these points for a while.)

{

\begin{wrapfigure}{r}{30mm}
\vskip-0mm
\centering
\includegraphics{mppics/pic-258}
\caption*{$\Lambda$}
\end{wrapfigure}

Note that 
\begin{itemize}
\item 
$A,B,P'\in\Gamma$,
\item $[AB]$ is a diameter of $\Gamma$,
\item $(AB)=\Pi\cap\Lambda$,
\item $\hat P\in [AB]$
\item $(P'\hat P)\perp (AB)$.
\end{itemize}



Since $[AB]$ is the diameter of $\Gamma$, 
by Corollary~\ref{cor:right-angle-diameter},
the angle $AP'B$ is right. 
Hence $\triangle A\hat PP'\z\sim \triangle AP'B\z\sim \triangle P'\hat PB$.
In particular
$$\frac{AP'}{BP'}=\frac{A\hat P}{P'\hat P}=\frac{P'\hat P}{B\hat P}.$$
Therefore
$$\frac{A\hat P}{B\hat P}=\left(\frac{AP'}{BP'}\right)^2
\quad\text{and similarly}\quad
\frac{A\hat Q}{B\hat Q}=\left(\frac{AQ'}{BQ'}\right)^2.
\eqlbl{eq:AP/BP}$$
Finally, 
\ref{eq:(AB;PQ)=(AB;P'Q')}+\ref{eq:AP/BP} imply \ref{eq:lem:P-hat-chord}.

}

The last statement follows from \ref{eq:lem:P-hat-chord} and the definition of the h-distance.
Indeed,
\begin{align*}
PQ_h&\df\ln\frac{A Q\cdot B P}{QB\cdot PA}=
\\
&=\ln\left(\frac{A \hat Q\cdot B \hat P}{\hat QB\cdot \hat PA}\right)^{\frac12}=
\\
&=\tfrac12\cdot\ln\frac{A \hat Q\cdot B \hat P}{\hat QB\cdot \hat PA}.
\end{align*}
\qedsf

{

\begin{wrapfigure}[8]{r}{36mm}
\centering
\vskip-8mm
\includegraphics{mppics/pic-260}
\end{wrapfigure}

\begin{thm}{Exercise}\label{ex:hex}
Let $\Gamma_1$, $\Gamma_2$, and $\Gamma_3$ 
be three circles perpendicular to a circle~$\Omega$.
Let $[A_1B_1]$, $[A_2B_2]$, and $[A_3B_3]$ denote
the common chords of $\Omega$ and $\Gamma_1$, $\Gamma_2$, $\Gamma_3$ respectively.
Show that the chords $[A_1B_1]$, $[A_2B_2]$, and $[A_3B_3]$ intersect at one point inside $\Omega$ if and only if $\Gamma_1$, $\Gamma_2$, and $\Gamma_3$ intersect at two points.
\end{thm}

\begin{thm}{Exercise}\label{ex:P<->hatP}
Show that $2\cdot OP_h=O\hat P_h$, where $O$ is the center of absolute, and $P\leftrightarrow \hat P$ is the bijection of the h-plane described above.
\end{thm}


}

\section{Projective model}
\label{sec:proj-model}

The following picture illustrates the bijection $P\mapsto \hat P$ of the h-plane described in the previous section --- 
\begin{figure}[!ht]
\centering
\includegraphics{mppics/pic-262}
\end{figure}
if you take the picture on the left and apply the map $P\z\mapsto \hat P$,
you get the picture on the right.
The pictures are {}\emph{conformal} and \index{projective!model}\emph{projective models} of the hyperbolic plane respectively.
The map $P\mapsto \hat P$ serves as a \textit{translation} between the models.

In the projective model, things look different;
some become simpler, and others become more complex.

\parbf{Lines.}
In the projective model, the h-lines are represented as chords of the absolute, more precisely, chords without their endpoints.

This observation can be used to transfer statements about lines and points from the Euclidean plane to the h-plane.
As an example let us state a version of Pappus' theorem for h-plane.

\begin{thm}{Hyperbolic Pappus' theorem}\label{thm:pappus-h}\index{Pappus' theorem}
Assume that two triples of h-points $A$, $B$, $C$,
and $A'$, $B'$, $C'$ in the h-plane are h-collinear.
Suppose that the h-points $X$, $Y$, and $Z$ are defined by
\begin{align*}
X&=(BC')_h\cap(B'C)_h,
&
Y&=(CA')_h\cap(C'A)_h,
&
Z&=(AB')_h\cap(A'B)_h.
\end{align*}
Then the points $X$, $Y$, $Z$ are h-collinear.
\end{thm}

In the projective model, this statement follows immediately from the original Pappus' theorem \ref{thm:pappus}.
The same can be done for Desargues' theorem \ref{thm:desargues}.
The same argument shows that the construction of a tangent line with a ruler only described in Exercise~\ref{ex:tangent ruler} works in the h-plane as well.
These statements are not at all evident in the conformal model.

\parbf{Circles and equidistants.}
In the projective model, the h-circles and equidistants are ellipses and their open arcs.
This follows since the stereographic projection sends circles on the plane to circles on the unit sphere and the footpoint projection of the circle back to the plane is an ellipse.
(One may define an \index{ellipse}\emph{ellipse} as a footpoint projection of a circle.)



\parbf{Distance.}
Consider a pair of h-points $P$ and $Q$.
Let $A$ and $B$ be the ideal points of the h-line in the projective model;
that is, $A$ and $B$ are the intersections of the Euclidean line $(PQ)$ with the absolute.

\begin{wrapfigure}{o}{36mm}
\vskip-2mm
\centering
\includegraphics{mppics/pic-264}
\vskip-2mm
\end{wrapfigure}

Then by Lemma~\ref{lem:P-hat-chord},
$$PQ_h=\tfrac12\cdot\ln\frac{AQ\cdot BP}{QB\cdot PA},\eqlbl{eq:proj-h-dist}$$
assuming the points $A, P, Q, B$ appear on the line in the same order.

\parbf{Angles.}
The hyperbolic angle measure in the projective model is very different from Euclidean, making it hard to determine from the picture. 
\label{klein-angles}
For example, all the intersecting h-lines on the picture 
are perpendicular.

To find the angle measure,
one may apply a motion of the h-plane that moves 
the vertex of the angle to the center of the absolute;
once done, the hyperbolic and Euclidean angles have the same measure.
In particular, if $O$ is the center of the absolute, then 
$$\measuredangle_hAOB=\measuredangle AOB.$$

\begin{thm}{Observation}\label{obs:h-p-perp}
If $O$ is the center of the absolute, then 
\[(OA)\z\perp (AB)\qquad\Longleftrightarrow\qquad(OA_h)\perp (AB)_h.\]

\end{thm}

\parit{Proof.}
The Euclidean reflection across $(OA)$ induces the h-reflection across $(OA)_h$.
Let $B'$ be the reflection of $B$.
Observe that $(OA)\perp (AB)$ $\Longleftrightarrow$ $A\in(BB')$ $\Longleftrightarrow$  $A\in(BB')_h$ $\Longleftrightarrow$  $(OA_h)\perp (AB)_h$.
\qeds

\parbf{Motions.}
The motions of the h-plane in the conformal and projective models are relevant to inversive transformations and projective transformations in the same way.
Namely: 
\begin{itemize}
\item Inversive transformations that preserve the h-plane describe motions of the h-plane in the conformal model.
\item Projective transformations that preserve the h-plane describe motions of the h-plane in the projective model.\footnote{The idea described in the solution of Exercise~\ref{ex:cone}.
The sketch of proof of \ref{thm:circle-center-proj} can be used to construct many projective transformations of this type.}
\end{itemize}

The following exercise is a hyperbolic analog of Exercise~\ref{ex:s-medians}. 

\begin{thm}{Exercise}\label{ex:h-median}
Let $P$ and $Q$ be points in the h-plane that lie at the same distance from the center of the absolute.
Observe that in the projective model, the h-midpoint of $[PQ]_h$ coincides with the Euclidean midpoint of $[PQ]_h$.

Conclude that if an h-triangle is inscribed in an h-circle, then its medians meet at one point.

Recall that an h-triangle might also be inscribed in a horocycle or an equidistant.
Think about how to prove the statement in this case.
\end{thm}

\begin{thm}{Exercise}\label{ex:h-altitudes}
Show that the altitudes of a hyperbolic triangle either intersect at one point or are pairwise disjoint. 
\end{thm}


{

\begin{wrapfigure}[4]{r}{34mm}
\vskip-11mm
\centering
\includegraphics{mppics/pic-266}
\end{wrapfigure}

\begin{thm}{Exercise}\label{ex:klein-perp}
Let $\ell$ and $m$ be  h-lines in the projective model.
Let $s$ and $t$ denote the Euclidean lines tangent to the absolute
at the ideal points of $\ell$. 
Show that
if the lines $s$, $t$, and the extension of $m$ meet at one point, then $\ell\perp m$. 
\end{thm}

}

\begin{thm}{Exercise}\label{ex:klein-for-angle-parallelism}
Use the projective model to derive the formula for the angle of parallelism (Proposition~\ref{prop:angle-parallelism}). 
\end{thm}

\begin{thm}{Exercise}\label{ex:klein-inradius}
Use the projective model to find an inradius of an ideal triangle.
\end{thm}

\begin{thm}{Advanced exercise}\label{ex:pyth-h-proj}
Prove the hyperbolic Pythagorean theorem (\ref{thm:pyth-h-poincare}) using the following idea.
\end{thm}

Recall that the hyperbolic Pythagorean theorem\index{Pythagorean theorem} (\ref{thm:pyth-h-poincare}) states that
\[\cosh c=\cosh a\cdot\cosh b,
\eqlbl{eq:hyp-pyth-proj}\]
where $a\z=BC_h$, $b=CA_h$, and $c=AB_h$ and
$\triangle_hACB$ is an h-triangle with a right angle at~$C$.

\begin{wrapfigure}{o}{34mm}
\vskip-8mm
\centering
\includegraphics{mppics/pic-268}
\end{wrapfigure}

\parit{Idea.}
We can assume that $A$ is the center of the absolute.
By \ref{obs:h-p-perp} the Euclidean triangle $ABC$ is right.
Set 
$s=BC$, $t =CA$, $u\z= AB$.
According to the Euclidean Pythagorean theorem (\ref{thm:pyth}), we have
$$u^2=s^2+t^2.\eqlbl{eq:hyp-proj}$$
It remains to express $a$, $b$, and $c$ using $s$, $u$, and $t$ and show that \ref{eq:hyp-proj} implies~\ref{eq:hyp-pyth-proj}.

\section{Bolyai's construction}

Assume we need to construct a line thru $P$ that is asymptotically parallel to the given line $\ell$ in the h-plane.

If $A$ and $B$ are ideal points of $\ell$ in the projective model, 
then we could simply draw the Euclidean line $(PA)$.
However, the ideal points do not lie in the h-plane; therefore there is no way to use them in the construction.

In the following construction we assume that you know a ruler-and-compass construction of the perpendicular line; see Exercise~\ref{ex:construction-perpendicular}.

\begin{thm}{Bolyai's construction}
\begin{enumerate}
\item Drop a perpendicular from $P$ to~$\ell$; denote it by~$m$.
Let $Q$ be the footpoint of $P$ on~$\ell$.
\item Erect a perpendicular from $P$ to~$m$; denote it by~$n$.
\item Mark a point $R$ on $\ell$ distinct from $Q$.
\item Drop a perpendicular from $R$ to~$n$; denote it by~$k$. 
\item Draw the circle $\Gamma$ with center $P$ and the radius $QR$. 
Mark a point of intersection of $\Gamma$ with~$k$; denote it by $T$.
\item The line $(PT)_h$ is asymptotically parallel to~$\ell$.
\end{enumerate}
\end{thm}

\begin{thm}{Exercise}\label{ex:Boyai-in-Euclid}
Explain what happens if one performs the Bolyai construction in the Euclidean plane.
\end{thm}

The following proposition implies that Bolyai's construction works.

\begin{thm}{Proposition}\label{prop:boyai}
Assume $P$, $Q$, $R$, $S$, $T$ are points in the h-plane
such that 
$S\in (RT)_h$,
$(PQ)_h\perp (QR)_h$,
$(PS)_h\perp(PQ)_h$,
$(RT)_h\perp (PS)_h$ and 
$(PT)_h$ and $(QR)_h$ are asymptotically parallel.
Then $QR_h=PT_h$.
\end{thm}


\parit{Proof.}
We will use the projective model.
Without loss of generality, we may assume that $P$ is the center of the absolute.
By~\ref{obs:h-p-perp},
the corresponding Euclidean lines are also perpendicular;
that is, $(PQ)\perp (QR)$, $(PS)\perp(PQ)$, and $(RT)\z\perp (PS)$.

Let $A$ be the common ideal point of $(QR)_h$ and $(PT)_h$.
Let $B$ and $C$ denote the remaining ideal points of $(QR)_h$ and $(PT)_h$ respectively.

The Euclidean lines $(PQ)$, $(TR)$, and $(CB)$ are parallel.
Therefore,  
\[\triangle AQP\sim \triangle ART \sim\triangle ABC.\]

\begin{wrapfigure}{o}{55mm}
\vskip-0mm
\centering
\includegraphics{mppics/pic-270}
\vskip2mm
\end{wrapfigure}

In particular,
\[\frac{AC}{AB}=\frac{AT}{AR}=\frac{AP}{AQ}.\]
It follows that
\[\frac{AT}{AR}=\frac{AP}{AQ}=\frac{BR}{CT}=\frac{BQ}{CP}.\]
In particular, 
\[\frac{AT\cdot CP}{TC\cdot PA}=\frac{AR\cdot BQ}{RB\cdot QA}.\]
Applying the formula for h-distance \ref{eq:proj-h-dist}, we get that $QR_h=PT_h$.
\qeds

\begin{wrapfigure}{r}{35mm}
\vskip-6mm
\centering
\includegraphics{mppics/pic-271}
\vskip2mm
\end{wrapfigure}

\begin{thm}{Advanced exercise}\label{ex:common-perp}
Look at the diagram and
recover the construction of a common perpendicular $n$ to the h-lines $\ell$ and $m$.

Prove that it works.
\end{thm}