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\begin{document}

\titlehead{Notes from a course
  at the Nesin Mathematics Village,
  July 15--21, 2019,
given jointly with Ay\c se Berkman}
%\title{Geometry of Bolyai and Lobachevski}
\title{Hyperbolic Geometry}
\author{David Pierce}
\publishers{Matematik B\"ol\"um\"u\\
Mimar Sinan G\"uzel Sanatlar \"Universitesi\\
\url{mat.msgsu.edu.tr/~dpierce/}\\
\url{polytropy.com}}
\date{August 5, 2019}
\maketitle

\tableofcontents
\listoffigures

\addchap{Introduction}

In \textbf{Chapter \ref{ch:an},}
we start with the \textbf{Poincar\'e half-plane,}
which we shall denote by
\begin{equation*}
  \Hp{\R}.
\end{equation*}
This is an example of a \textbf{hyperbolic plane,}
namely a set of points with designated lines and circles---%
\textbf{hyperbolic lines} and \textbf{hyperbolic circles}---%
having certian properties.
In $\Hp{\R}$,
the points are those of $\R^2$ (or $\C$)
with positive $y$-coordinate or (imaginary part).
A segment of a hyperbolic line
is the path of the shortest journey between the endpoints,
if speed is proportional to $y$-coordinate.
We can compute the time of the journey with an integral.
A hyperbolic circle is the set of points,
reaching any of which takes the same time.

The hyperbolic lines
turn out to be semicircles and vertical rays;
the hyperbolic circles, ordinary circles.
We can define these in the upper half-plane
associated with any \textbf{Euclidean field,}
namely an ordered field $K$ whose every positive element is a square.
Thus we obtain the hyperbolic plane that we call
\begin{equation*}
  \Hp K.
\end{equation*}
Here a hyperbolic segment may not have a length that is a number;
but we can still tell when two segments are equal.
An \textbf{isometry} of the plane takes every segment to an equal segment.
We work out what the isometries are.

In \textbf{Chapter \ref{ch:Hilbert},}
we show that every model of Hilbert's axioms for a hyperbolic plane
is isomorphic to a Poincar\'e half-plane over some Euclidean field.
This work is based on Hilbert's
``New Development of Bolyai-Lobachevskian Geometry,''
which,
having been translated from
\emph{Mathematische Annalen} \textbf{57,}
is Appendix III
in the 1971 English edition of Hilbert's \emph{Foundations of Geometry}
\cite{Hilbert-10}.
In Hilbert's words, we show
\begin{quote}
  that it is possible to
develop Bolyai-Lobachevskian geometry in the plane exclusively with
the plane axioms without the use of the continuity axioms.
\end{quote}
Our specific aim in Chapter \ref{ch:Hilbert} is what
Hartshorne \cite[p.\ 388]{MR1761093} calls
\begin{quote}
  Hilbert's tour de force,
  the creation of an abstract field
  out of the geometry of a hyperbolic plane.
\end{quote}
We set out just enough axioms for this.
Thus we differ from Borsuk and Szmielew in
\emph{Foundations of Geometry} \cite{Borsuk-Szmielew}
(one of the ``additional references''
in a note added to the end of Hilbert's paper); for they
\begin{quote}
  develop Euclidean and Bolyai-Lobachevskian 
geometry on the basis of an axiom system due, in principle, to Hilbert. It 
should be noted at once, however, that the authors develop these  
geometries, in principle, as far as necessary to be able to prove them  
categorical, i.e., to show that the Cartesian space known from analytic  
geometry is up to isomorphism the only model of Euclidean geometry, and 
Klein space (constructed with the help of notions known from the analytic 
geometry of projective space) is up to isomorphism the only model of 
Bolyai-Lobachevskian geometry. 
\end{quote}
The field interpreted in these models will be the field of real numbers.
However, in the most general sense,
any field (or even skew-field) yields a ``Cartesian space'';
any Euclidean field
yields a ``Klein space.''

In a footnote on the passage quoted above,
Hilbert refers to similar work in plane elliptic geometry,
by Dehn and Hessenberg,
then in absolute geometry
(``even without any assumption
about intersecting or nonintersecting lines''),
by Hjelmslev,
in \emph{Mathematische Annalen} \textbf{60, 61,} and
\textbf{64.}


Like the Ancients (as Netz \cite{MR1683176} points out),
dispensing with verbal descriptions,
we may let diagrams alone define certain points.

In the past, I have taught hyperbolic geometry axiomatically,
as in Chapter \ref{ch:Hilbert},
but through Lobachevski \cite{Lobachevski}.
I have then introduced the Poincar\'e half-plane as a model,
and used this model for drawing diagrams;
but students have had trouble thinking of curved lines as straight.
Chapter \ref{ch:an}
is an attempt to solve this problem.

The general problem of presenting hyperbolic geometry may be insoluble,
if only because ill defined.
Different courses will have different emphases for different purposes.
The present course concerns the logical interactions
between geometry and algebra.
We use analysis at the beginning,
only to justify thinking of certain curved lines as straight.
The notes are still a fairly rough draft.

\chapter{Field}\label{ch:an}

\section{Lines}\label{sect:lines}

  In the words at the head of Euclid's \emph{Elements}
  \cite{Euclid-Heiberg,MR17:814b,Euclid-homepage,bones,MR1932864},
  ``a \textbf{line} is breadthless length''
  (\gr{Gramm`h m~hkos >aplat'es}).
  Such a line may be curved:
  the circumference of a circle will be an example.
  
  The English word ``line'' is
  the noun corresponding to the adjective ``linen.''
  This means flaxen
  and is related to the Latin \textsc{linvm,} meaning flax.
  A line is a thread or cord, originally flaxen.
  A cord may lie in a heap,
  or be wound around a spool, not straight at all;
  or it may be pulled taut, into a \emph{straight line.}%%%%%
  %Γραμμὴ δὲ μῆκος ἀπλατές
  \footnote{%
    After the definition of a line in the \emph{Elements,}
    there is an obscure definition of a straight line.
    In \emph{The Forgotten Revolution}
    \cite[pp.\ 323--4]{MR2038833},
     Lucio Russo argues plausibly
     that this definition
is only a later addition to the \emph{Elements,}
its origin being a student's crib-sheet.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

  In the Cartesian plane $\R^2$,
the \textbf{straight lines} are the solu\-tion-sets of equations
\begin{equation}\label{eqn:abc}
  ax+by=c,
\end{equation}
where at least one of $a$ and $b$ is not $0$,
but each, along with $c$, is an element of $\R$.
If $A$ and $B$ are distinct points in the plane
and have coordinates $(a_0,a_1)$ and $(b_0,b_1)$ respectively,
then there is a unique straight line passing through them,
and the segment denoted by $AB$ has length given by the rule
\begin{equation}\label{eqn:|AB|}
  \abs{AB}=
  \sqrt{(a_0-b_0)^2+(a_1-b_1)^2}.
\end{equation}
We may also use the notation $AB$
for the whole straight line that contains $A$ and $B$.

We consider now lengths of other lines.
If $(f_0,f_1)$ or $f$
is an injective, continuously differentiable function
into $\R^2$ from an interval $[0,T]$,
then
we can understand $f$ as a \textbf{journey,}
which proceeds
\begin{compactitem}
\item
  from \emph{time} $0$ to time $T$,
  \item
  from \emph{position} $f(0)$ to position $f(T)$.
\end{compactitem}
Suggestively,
we may say that
\begin{compactitem}
\item
  the domain of $f$ is the \textbf{period,} and
\item
  the range of $f$ is the \textbf{path,}
\end{compactitem}
of the journey.
The \textbf{speed} of the journey at each time $t$
is given by
\begin{equation*}
  \abs{f'(t)}
  =\sqrt{f_0{}'(t)^2+f_1{}'(t)^2}.
\end{equation*}
The \textbf{length} of the journey,
or the \textbf{distance} from its beginning to its end,
is ambiguous: it could be the length of the path or the period.
\begin{compactitem}
  \item
    The length of the period is just $T$.
    \item
The length of the path of $f$ is given by
\begin{equation*}%\label{eqn:l}
  L
  =\int_0^T\abs{f'(t)}\dee t
  =\int_0^T\abs{f'}.
\end{equation*}
\end{compactitem}
The path of $f$
may also be the range of a
(still injective and continuously differentiable)
function $\gamma$ on an interval $[a,b]$, where
\begin{align*}
  \gamma(a)&=f(0),&
  \gamma(b)&=f(T).
\end{align*}
With the substitution
\begin{align}\label{eqn:sub}
  \gamma(u)&=f(t),
  &\gamma'(u)\dee u&=f'(t)\dee t,
\end{align}
we obtain the equation
\begin{equation*}%\label{eqn:f'g'}
  \int_0^T\abs{f'}=\int_a^b\abs{\gamma'}.
\end{equation*}
It may be that the speed of $f$
is constrained by the rule
\begin{equation}\label{eqn:f'}
  \abs{f'}=\sigma\circ f,
\end{equation}
where
$\sigma$ is a continuous function into $\R$, defined
on a subset of $\R^2$ that includes the path of $f$
(which is the range of $\gamma$).
We are going to work with a nontrivial example
of such a constraint
in the next section and beyond.

Meanwhile,
rewriting \eqref{eqn:f'} as
\begin{equation*}
  \frac{\abs{f'}}{\sigma\circ f}=1,
\end{equation*}
we obtain
\begin{equation*}
  T=\int_0^T\frac{\abs{f'}}{\sigma\circ f}.
\end{equation*}
As distance is the time integral of speed,
so time is the distance integral of the inverse of speed.
Again using the substitution \eqref{eqn:sub},
we obtain
\begin{equation}\label{eqn:T}
  T=\int_a^b\frac{\abs{\gamma'}}{\sigma\circ\gamma}.
\end{equation}
Thus, knowing only a path from a point $A$ to a point $B$
in the domain of $\sigma$,
we know how long a journey takes on that path
under the constraint \eqref{eqn:f'}.

The constrained journey
from $A$ to $B$ that has the shortest \emph{period}
    may not have a straight path.
    We call it a \textbf{geodesic segment}
    (with respect to $\sigma$).
    The length of that shortest period
    is then the \textbf{geodesic distance} between $A$ and $B$.
    If a point $D$ is at the same distance from a point $C$
    that $B$ is from $A$,
    then the two geodesic segments $AB$ and $CD$ are
    \textbf{equal.}
    An unbounded line whose segments are all geodesic will be a
    \textbf{geodesic line.}

    In another of the definitions
    (\gr{<'oroi}) at the head of Euclid's \emph{Elements,}
    \begin{quote}
      A \textbf{circle} is a plane figure
      bounded by one line,
      % [called the \textbf{circumference}],
      all straight lines extending to which
      from one of the points lying inside the figure
      are equal to one another;
      and the \textbf{center} is that point of the circle.
      //
      \gr{K'uklos >est`i sq~hma >ep'ipedon <up`o mi~as gramm~hs
         %Κύκλος  ἐστὶ   σχῆμα  ἐπίπεδον   ὑπὸ   μιᾶς  γραμμῆς
        perieq'omenon, %[<`h kale~itai perif'ereia],
       %περιεχόμενον  [ἣ   καλεῖται  περιφέρεια],
        pr`os <`hn >af'' <en`os shme'iou t~wn >ent`os
       %πρὸς  ἣν   ἀφ'   ἑνὸς   σημείου  τῶν  ἐντὸς
        to~u sq'hmatos keim'enwn p~asai a<i prosp'iptousai
       %τοῦ  σχήματος  κειμένων  πᾶσαι  αἱ  προσπίπτουσαι
        e>uje~iai %[pr`os t`hn to~u k'uklou perif'ereian]
       %εὐθεῖαι   [πρὸς  τὴν  τοῦ  κύκλου  περιφέρειαν]
        >'isai >all'hlais e>is'in.
      %ἴσαι  ἀλλήλαις   εἰσίν.
      K'entron d`e to~u k'uklou t`o shme~ion kale~itai.}
      %Κέντρον  δὲ  τοῦ  κύκλου  τὸ  σημεῖον  καλεῖται.
    \end{quote}
    Now we have the notion of a \textbf{geodesic circle,}
    whose circumference is the locus of points
    whose geodesic distances from a given point are all the same.
    We may confuse a circle with its circumference.


    Let us define a \textbf{geometry}
    to be an ordered triple of three sets,
    the sets consisting respectively of
    points, straight lines, and circles.
    A standard example then is
    \begin{equation*}
      \Aff{\R},
    \end{equation*}
    whose
    \begin{compactitem}
      \item
    points compose $\R^2$,
      \item
        straight lines are defined
        by equations \eqref{eqn:abc}
        as above,
      \item
        circles are defined
        by equations
        \begin{equation*}
          \abs{z-a}^2=r^2,
        \end{equation*}
        where the variable $z$ ranges over $\C$,
        and also $a\in K(\mi)$
        and $r$ in $K$.
    \end{compactitem}
    We shall make use of this example,
    but also another,
    which we are about to define,
    where the ``straight lines'' will be geodesic lines;
    circles, geodesic circles.

    
\section{Poincar\'e Half-plane}%\label{sect:P-plane}

The points of our new geometry will compose
the upper half-plane,
    \begin{equation*}
      \{(x,y)\in\R^2\colon y>0\},
    \end{equation*}
    which is also $\{z\in\C\colon\Im(z)>0\}$.
    Geodesics will be determined as in \S\ref{sect:lines}
    by one of the equivalent rules
\begin{align*}
  \sigma(x,y)&=y,&
  \sigma(z)&=\Im(z).
\end{align*}
The geometry will be the \textbf{Poincar\'e half-plane,}
which we may denote by
\begin{equation*}
  \Hp{\R}.
\end{equation*}
    It will be an example of a \emph{hyperbolic plane.}
    In such a geometry, to refer to geodesics,
    we may use the adjective \textbf{hyperbolic;}
    however, we intend no allusion
    to the conic section called an hyperbola.
    For example, the length of the geodesic segment
    bounded by $A$ and $B$
    is the \emph{hyperbolic distance} from $A$ to $B$.
    When we do not apply the adjective
    geodesic or hyperbolic,
    we mean a term in the usual sense:
    the Euclidean sense,
    to be considered further in \S\ref{sect:Euc-ax}.
    
    \begin{theorem}\label{thm:geod}
      Between two points of $\Hp{\R}$,
            the geodesic segment is
      \begin{compactenum}[1)]
      \item
        the straight segment joining the points,
        if they are on the same vertical line;
      \item
        otherwise,
        the arc that they bound on a circle
        whose center is on the $x$-axis.
      \end{compactenum}
    \end{theorem}

    \begin{proof}
      \begin{asparaenum}
      \item
        Suppose the points are $(c,a)$ and $(c,b)$.
        Along the straight path
        from one to the other,
        \eqref{eqn:T} takes the form
        \begin{equation*}
          T=\int_a^b\frac{\dee y}y.
        \end{equation*}
        On an arbitrary path $\gamma$,
        we compute
        \begin{equation*}
          \int_{\gamma}\frac{\sqrt{\dee x^2+\dee y^2}}y
          =\int_a^b\frac{\sqrt{(\dee x/\dee y)^2+1}}y\dee y
          \geq T.
        \end{equation*}
      \item
        The points being $A$ and $B$,
        we let the perpendicular bisector of $AB$
        cut the $x$-axis at $O$.
        Letting $I$ be the projection of $A$ on the $x$-axis,
        we define
        \begin{align}\label{eqn:ab}
          \alpha&=IOA,
          &\beta&=IOB,
        \end{align}
        assuming then
        \begin{equation*}%\label{eqn:0abp}
          0<\alpha<\beta<\uppi.
        \end{equation*}
        Using a system of polar coordinates centered at $O$,
        for travel along the circular arc from $A$ to $B$
        centered at $O$,
        if the arc has radius $r$,
        from \eqref{eqn:T} we compute
        \begin{equation*}
          T
          =\int_{\alpha}^{\beta}\frac{r\dee\theta}{r\sin\theta}
          =\int_{\alpha}^{\beta}\frac{\dee\theta}{\sin\theta}.
        \end{equation*}
        This is independent of $r$.
        If we follow an arbitrary path,
        \begin{equation*}
  \int_{\gamma}\frac{\sqrt{\dee r^2+(r\dee\theta)^2}}{r\sin\theta}
  =\int_{\alpha}^{\beta}\frac{\sqrt{(\dee r/r\dee\theta)^2+1}}
          {\sin\theta}\dee\theta
          \geq
          T.
          \qedhere
        \end{equation*}
      \end{asparaenum}
    \end{proof}

    A geodesic circle in $\Hp{\R}$
    will be an ordinary circle,
    but its center will be displaced
    so that its distance from the $x$-axis will be,
    not the arithmetic,
    but the geometric mean
    of the distances of the nearest and farthest points
    of the circle from the axis.
    (Distances from the $x$-axis
    can only be meant in the ordinary, Euclidean sense,
    since the points of the axis
    do not belong to $\Hp{\R}$.)
    
    The following theorem,
    complicated to state in words,
    is illustrated by \figgref{fig:hyp-circ}{fig:hyp-circ-2}.
    \begin{figure}
      \centering
      \begin{pspicture}(0,-4.4)(10,7)
        %\psgrid
        \psset{PointSymbol=none}
        \pstGeonode[PosAngle={90,0,180}]
        (2.2,6.4)A(9.4,0)D(0.6,0){D'}
        \pstProjection[PosAngle=-45]{D'}DA[O]
        \pstMiddleAB[PosAngle=-90]{D'}DX
        \pstArcOAB XD{D'}
        \pstArcnOAB[linestyle=dashed] XD{D'}
        \pstInterLC[PosAngleA=-135,PosAngleB=165] OAXD{B'}B
        \pstTranslation[PointName=none] ADB[b]
        \pstInterLL[PointName=none]bBODx
        \pstTranslation[PointName=none] DBx
        \pstInterLL[PosAngle=-45]{x'}xOAC
        \pstMiddleAB[PointName=none] ACm
        \pstCircleOA[linestyle=dotted] mA
        \ncline[linestyle=dashed]{o-o}{D'}D
        \pstInterCC[PosAngleA=195,PosAngleB=60] XDmA{E'}E
        {\psset{arrows=o-}\ncline XB\ncline XE}
        \ncline BD\ncline DA\ncline{D'}A
        \ncline[linestyle=dashed] O{B'}
        \psline{o-}(O)(A)
        \pstRightAngle AEC
        \pstRightAngle C{E'}A
        \psset{linestyle=dashed}
        \ncline EC\ncline{E'}C
      \end{pspicture}
      \caption{A hyperbolic circle}
      \label{fig:hyp-circ}
    \end{figure}

    \begin{theorem}\label{thm:circ}
      In $\Hp{\R}$, the geodesic circle
      whose circumference passes through a point $E$
      and whose geodesic center is a point $B$
      is the (Euclidean) circle
      \begin{compactitem}
      \item
         whose diameter is $AC$,
       \item
         where $A$ and $C$ are on the same vertical line with $B$,
        \item
          this line cuts the $x$-axis at $O$,
          \item
            $\abs{OA}\cdot\abs{OC}=\abs{OB}^2$,
            \item
            and moreover
      \begin{compactenum}[1)]
      \item
        $A$ is $E$, if this lies on $OB$;
      \item
        otherwise,
        $A$ is where $DE$ cuts $OB$,
        \begin{compactitem}
        \item
          $D$ being where the circle with center $X$
        and passing through $E$
        cuts the $x$-axis,
      \item
        $X$ being where
the perpendicular bisector of $BE$
cuts the $x$-axis.
        \end{compactitem}
      \end{compactenum}
      \end{compactitem}
\end{theorem}
    
    \begin{proof}
In \figgref{fig:hyp-circ}{fig:hyp-circ-2},
    \begin{figure}
      \centering
      \begin{pspicture}(-0.6,-4)(8,10)
%        \psgrid
        \psset{PointSymbol=none}
        \pstGeonode[PosAngle={90,0,180}]
        (3.5,9.4)A(7.4,0)D(0.6,0){D'}
        \pstProjection[PosAngle=-135]{D'}DA[O]
        \pstMiddleAB[PosAngle=-90]{D'}DX
        \pstArcOAB XD{D'}
        \pstArcnOAB[linestyle=dashed] XD{D'}
        \pstInterLC[PosAngleA=-90,PosAngleB=135] OAXD{B'}B
        \pstTranslation[PointName=none] ADB[b]
        \pstInterLL[PointName=none]bBODx
        \pstTranslation[PointName=none] DBx
        \pstInterLL[PosAngle=-135]{x'}xOAC
        \pstMiddleAB[PointName=none] ACm
        \pstCircleOA[linestyle=dotted] mA
        \ncline[linestyle=dashed]{o-o}{D'}D
        \pstInterCC[PosAngleA=200,PosAngleB=0] XDmA{E'}E
        {\psset{arrows=o-}\ncline XB\ncline XE}
        \ncline BD\ncline DA\ncline{D'}A
        \ncline[linestyle=dashed] O{B'}
        \psline{o-}(O)(A)
        \pstRightAngle AEC
        \pstRightAngle C{E'}A
        \psset{linestyle=dashed}
        \ncline EC\ncline{E'}C
      \end{pspicture}
      \caption{Another hyperbolic circle}
      \label{fig:hyp-circ-2}
    \end{figure}
    \begin{compactitem}
    \item
      the altitudes $AO$, $DE'$, and $D'E$
      of triangle $ADD'$ meet
      (as the altitudes of a triangle always do) at $C$;
    \item
      therefore
      the circle with diameter $DD'$
      passes through $E$ and $E'$;
    \item
      the circle also cuts $AO$      
      at $B$ and $B'$.
    \end{compactitem}
    As a result,
    the points $E$ and $E'$ lie on the circle
    with diameter $AC$.
    Together with the points $B$ and $O$,
    \begin{compactenum}[1)]
    \item
      the point $E$ determines $A$, and
    \item
      the point $A$ determines $C$;
    \end{compactenum}
        we show this now.
    \begin{asparaenum}
    \item
      Given $E$ not on $BO$,
      we let angle $BOX$ be right,
      but also require $X$ to be equidistant from $B$ and $E$.
      Letting the circle with center $X$
      passing through $B$ (and therefore $E$)
      cut $OX$ at $D$,
      we let $A$ be the intersection of $DE$ and $BO$.
      \item
      Given $A$ on $BO$,
      we choose $X$ arbitrarily so that angle $BOX$ is right,
      and then we draw the circle with center $X$
      passing through $B$,
      letting the circle cut
      \begin{compactitem}
      \item
        $OX$ at $D$ and $D'$,
      \item
        $AD$ and $AD'$ at $E$ and $E'$ respectively,
      \item
        $OB$ also at $B'$.
      \end{compactitem}
      We let $C$ be the intersection of $AO$ and $DE'$
      (and therefore of $DE'$ as well).
      By similarity of triangles $AEC$ and $AOD$,
      as well as the theorem (found in Euclid)
      on secants of circles,
      and since $\abs{OB'}=\abs{OB}$,
      \begin{equation*}
        \abs{AC}\cdot\abs{AO}
        =\abs{AE}\cdot\abs{AD}
        =\abs{AB}\cdot\abs{AB'}
        =\abs{AO}^2-\abs{OB}^2.
      \end{equation*}
      Thus $C$
      is independent of the choice of $X$.
    \end{asparaenum}

    It is now enough to show that $A$, $C$, and $E$
    are all at the same geodesic distance from $B$.
    \begin{asparaitem}
      \item
      $AB$ and $BC$ are equal as geodesic segments,
      because,
      if $A$ and $B$ have $y$-coordinates
      $a$ and $b$ respectively,
      then, as in the proof of Theorem \ref{thm:geod},
      the geodesic length of $AB$ is given by
              \begin{equation*}
          \int_a^b\frac{\dee y}y
          =\left.\log y\strut\right|_a^b
          =\log b-\log a=\log\frac ba.
              \end{equation*}
              \item
                $AB$ and $BE$ are equal as geodesic segments,
                because,
letting
    \begin{align*}
      \alpha&=OXB,
      &\beta&=OXE,
    \end{align*}
so that
    \begin{align*}
      \frac{\alpha}2&=ODB,
      &\frac{\beta}2&=ODE,
    \end{align*}
    we have the geodesic length of $EB$
    as given by
  \begin{equation*}
          \int_{\alpha}^{\beta}\frac{\dee\theta}{\sin\theta}
=    \log\frac{\tan(\beta/2)}{\tan(\alpha/2)}
=\frac ab,
  \end{equation*}
  where we use
  \begin{equation}\label{eqn:log-tan}
    \int\frac{\dee\theta}{\sin\theta}
    =\log\left|\tan\frac{\theta}2\right|+C,
  \end{equation}
which we can derive as follows.
        \begin{equation*}
          \int_{\alpha}^{\beta}\frac{\dee\theta}{\sin\theta}
     =\int_{\alpha}^{\beta}\frac{\sin\theta\dee\theta}{\sin^2\theta}
     =\int_{\alpha}^{\beta}\frac{\sin\theta\dee\theta}{1-\cos^2\theta}\\
          =-\int_{\cos\alpha}^{\cos\beta}\frac{\dee t}{1-t^2}
        \end{equation*}
        by the substitution
        \begin{align*}
          t&=\cos\theta,&\dee t&=-\sin\theta\dee\theta.
        \end{align*}
        Since
        \begin{equation*}
          \frac1{1-t^2}=\frac1{(1+t)(1-t)}
          =\frac12\left(\frac1{1+t}+\frac1{1-t}\right),
        \end{equation*}
        so that
        \begin{multline*}
          -\int\frac{\dee t}{1-t^2}
          =\frac12\left(\log\abs{1-t}-\log\abs{1+t}\right)+C\\
          =\frac12\log\left|\frac{1-t}{1+t}\right|+C
          =\frac12\log\frac{\abs{1-t^2}}{(1+t)^2}+C,
        \end{multline*}
        we conclude
        \begin{equation*}
          \int\frac{\dee\theta}{\sin\theta}
          =\frac12\log\frac{1-\cos^2\theta}{(1+\cos\theta)^2}+C
          =\log\frac{\abs{\sin\theta}}{1+\cos\theta}+C.
        \end{equation*}
        \begin{figure}
          \centering
          \begin{pspicture}(0,-0.6)(8,4)
            \psset{PointSymbol=none,PointName=none}
            \pstGeonode(0,0)B(4,0)O(8,0)A
            \pstArcOAB OAB
            \pstRotation[RotAngle=65]OA[C]
            \ncline BO\nbput{$1$}
            \ncline OC\ncline BC
            \pstProjection OAC[D]
            \ncline OD\nbput{$\cos\theta$}\ncline DA
            \ncline[linestyle=dotted]CD\naput{$\sin\theta$}
            \pstMarkAngle AOC{$\theta$}
            \pstMarkAngle ABC{$\theta/2$}
          \end{pspicture}
          \caption{Angle and its half}
          \label{fig:tan}
        \end{figure}
        We now obtain \eqref{eqn:log-tan}
        from the identity (read from \figref{fig:tan})
        \begin{equation*}%\label{eqn:tan}
          \tan\frac{\theta}2
          =\frac{\sin\theta}{1+\cos\theta}.\qedhere
        \end{equation*}
    \end{asparaitem}
\end{proof}

    We note in passing that
        some sources may use the formula
    \begin{equation*}
      \tan\frac{\theta}2=\frac{1-\cos\theta}{\sin\theta}
      =\csc\theta-\cot\theta.
    \end{equation*}
    

  \section{Euclidean Fields}%\label{ch:alg}

The field $\R$ of real numbers
is an example of a
\textbf{Euclidean field,}
because it is an ordered field
whose every positive element has a square root.
Thus a Euclidean field
is an ordered field whose ordering is definable by the rule
\begin{equation*}
  x\leq y\liff\Exists zx+z^2=y.
\end{equation*}
Examples of ordered fields are $\Q$ and $\R(t)$,
but these are not Euclidean.
  The latter field consists of ratios
  \begin{equation*}
    \frac{a_0+a_1t+\dots+a_mt^m}{b_0+b_1t+\dots+b_nt^n},
  \end{equation*}
  where $a_i$ and $b_j$ are from $\R$,
  and not all $b_j$ are $0$.
  The ordering is determined by
  \begin{align*}
    t&>0,&1+tx>0
  \end{align*}
  for all $x$ in $\R(t)$.
  This means $t$ is a positive infinitesimal;
  in particular, letting $x$ be $-n$ for some positive integer $n$, we have
  \begin{equation*}
    0<t<\frac1n.
  \end{equation*}
  The presence of infinitesimals
  (and therefore infinite elements,
  such as $1/t$)
  means $\R(t)$ is not \emph{Archimedean.}


Every ordered field $K$
is included in a smallest Euclidean ordered field,
\begin{equation*}
  K\Euc.
\end{equation*}
We obtain it by continually replacing $K$ with
\begin{equation*}
  K[X]/(X^2-a),
\end{equation*}
where $a$ is a positive element of $K$ with no square root.
The field $K\Euc$ is countable if $K$ is.
Since $\R$ is not countable, it is not
$\Q\Euc$.
The latter is the smallest Euclidean field;
it contains $\surd2$, $\sqrt{\surd2+\surd3}$, and so forth.

    \begin{figure}
      \centering
  \psset{PointSymbol=none,PointName=none,unit=1.5cm}
  \begin{pspicture}(-2,0)(2,4)
    \psline[linestyle=dotted](-2,0)(2,0)
    \pstGeonode(0,1)O
    (0,0.25)E(0,4)F
    (!10 tan 0)C
    (!10 30 add tan 0){C1}
    (!10 60 add tan 0){C2}
    (!10 90 add tan 0){C3}
    (!10 120 add tan 0){C4}
    (!10 150 add tan 0){C5}
    \pstMiddleAB EFG
    \pstCircleOA GE
    \pstInterCC GECOBD
    \pstArcnOAB CDB
    \pstInterCC GE{C1}O{B1}{D1}
    \pstArcnOAB {C1}{D1}{B1}
    \pstInterCC GE{C2}O{B2}{D2}
    \pstArcnOAB {C2}{D2}{B2}
    \pstInterCC GE{C3}O{B3}{D3}
    \pstArcnOAB {C3}{D3}{B3}
    \pstInterCC GE{C4}O{B4}{D4}
    \pstArcnOAB {C4}{D4}{B4}
    \pstInterCC GE{C5}O{B5}{D5}
    \pstArcnOAB {C5}{D5}{B5}
      \end{pspicture}
      \caption{Hyperbolic circle and radii}
      \label{fig:radii}
    \end{figure}
If $K$ is a Euclidean field,
we can form the field $K(\mi)$,
where as usual $\mi$ satisfies
  \begin{equation*}
    z^2+1=0.
  \end{equation*}
We turn the upper half-plane over $K$,
\begin{equation*}
  \{(x,y)\in K^2\colon y>0\},
\end{equation*}
which is
\begin{equation*}
  \{z\in K(\mi)\colon\Im(z)>0\},
\end{equation*}
into a hyperbolic plane,
as we did in case $K$ was $\R$;
the result will be a geometry that we can denote by
\begin{equation*}
  \Hp K.
\end{equation*}
However,
since we may no longer have the analytic properties of $\C$,
we have not got an obvious theoretical way
to define geodesic lines.
We just declare that the analogues of Theorem
    \ref{thm:geod} and
    \ref{thm:circ} shall be true.
    By fiat then, the curves
    emanating from the point in \figref{fig:radii}
    are equal radii of a circle.

    If
    $B$ lies between $A$ and $C$ on the $x$-axis in $K^2$,
    then the vertical line though $B$
    actually cuts the circle with diameter $AC$,
    as it would in case $K$ were $\R$.
    There are two solutions to the system of equations
    \begin{align*}
      \left(x-\frac{a+c}2\right)^2+y^2
      &=\left(\frac{a-c}2\right)^2,&
      x&=b
    \end{align*}
    over $K$,
    because positive elements of $K$ have square roots.

    \section{Geometric Axioms}\label{sect:Euc-ax}

    The first four of the thirteen books of Euclid's
    \emph{Elements} are the original exposition of
    \emph{Euclidean plane geometry.}%%%%%
    \footnote{Books \textsc v and \textsc{vi}
      develop a theory of proportion;
      \textsc{vii--ix,} of numbers;
      \textsc{x,} of irrationality;
      \textsc{xi--xii,} of solid geometry.}
    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    The exposition begins with some definitions,
    two of which---for line and circle---%
    we have seen in \S\ref{sect:lines}.
    Here is another Euclidean definition,
    in a fairly literal translation,
    complete with subjunctive mood:
    \begin{quote}%\sloppy
      Whenever a straight line, erected on a straight line,
      make the adjacent angles equal to one another,
      a \textbf{right} angle
      is either of the equal angles,
      and \textbf{perpendicular} is called the former line
      to that on which it stands.
      //
      \gr{<'Otan d`e e>uje~ia >ep'' e>uje~ian staje~isa
         %Ὅταν   δὲ  εὐθεῖα   ἐπ'   εὐθεῖαν   σταθεῖσα
        t`as >efex~hs gwn'ias >'isas >all'hlais poi~h|,
       %τὰς  ἐφεξῆς   γωνίας  ἴσας   ἀλλήλαις   ποιῇ,
        >orj`h >ekat'era t~wn >'iswn gwni~wn >esti,
       %ὀρθὴ   ἑκατέρα   τῶν  ἴσων   γωνιῶν  ἐστι,
        ka`i <h >efesthku~ia e>uje~ia k'ajetos kale~itai,
       %καὶ  ἡ  ἐφεστηκυῖα   εὐθεῖα   κάθετος  καλεῖται,
        >ef'' <`hn >ef'esthken.}
       %ἐφ' ἣν ἐφέστηκεν.
    \end{quote}
    Next are five \emph{postulates}
    (\gr{a>it'hmata}).
    Their precise meaning is controversial;
    we paraphrase them for our purposes:
    \begin{compactenum}
    \item
      Any two points can be joined by
      a segment of a straight line.
      That Euclid intends this segment to be unique is clear
        from his proof of \emph{SAS,}
        discussed presently.
    \item
      Any segment of a straight line can be extended.
    \item
      A circle can be drawn through any point with any center.
    \item
      All right angles are equal.
    \item
      If a straight line crosses two others,
      making the internal angles on the same side
      less than two right angles,
      then the two lines, extended on that side,
      will intersect.
    \end{compactenum}
    There are five additional ``common notions''
    (\gr{koina`i >'ennoiai}):%%%%%
    \footnote{Heiberg \cite{Euclid-Heiberg} lists nine common notions,
      but brackets the four whereby,
      \begin{inparaenum}[(1)]
        \item
      if unequals be added to equals, the wholes are equal;
    \item
      doubles and
    \item
      halves of the same are equal to one another;
    \item
      two straight lines cannot bound a space.
      \end{inparaenum}
      Mourmouras \cite{Euclid-homepage} omits to bracket the last.}
    %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    \begin{compactenum}
    \item
      What are equal to the same are equal to one another.
    \item
      If equals be added to equals, the wholes are equal.
    \item
      If equals be subtracted from equals, the remainders are equal.
    \item
      What are congruent to one another are equal to one another.
    \item
      The whole is greater than the part.
    \end{compactenum}
        Modern mathematicians may ignore any distinction
    between postulates and common notions,
    but call them all axioms.
    They may find in Euclid's axioms
    various inadequacies,
    which Hilbert attempts to remedy in
    \emph{The Foundations of Geometry} \cite{Hilbert-10}.

    A point on a straight line
    divides the line into two opposite \textbf{rays.}
    Hilbert accepts the following as an axiom,
    because his system does not have circles.

    \begin{theorem}\label{thm:I.3}
      By Euclid's first three postulates
      and first three common notions,
      from a given ray,
      a segment equal to any given segment can be cut.
    \end{theorem}

    \begin{proof}
      Let the given segment be $AB$
      in \figref{fig:I.3},
    \begin{figure}
      \centering
      \begin{pspicture}(9,7)
        \psset{PointSymbol=none}
%        \psgrid[subgriddiv=1]
        \pstGeonode[PosAngle={180,-45,-135,-45}]
        (1.5,2.5)A(3,3)B(6,3)C(8,4.5)D
        \pstInterCC[PosAngle=90,PointNameB={}] BCCBE{E'}
        \pstInterLC[PosAngle=20,PointNameA={}] BEBA{F'}F
        \pstInterLC[PosAngle=160,PointNameB={}] CEEFG{G'}
        \pstInterLC[PosAngle=80,PointNameB={}] DCCGH{H'}
        \ncline AB\ncline CD\ncline EB\ncline EC
        \ncline BC
        \pstArcOAB BCE
        \pstArcOAB CEB
        \pstArcOAB BAF
        \pstArcOAB EFG
        \pstArcOAB CGH
        \psset{linestyle=dotted}
        \pstArcnOAB BCE
        \pstArcnOAB CEB
        \pstArcnOAB BAF
        \pstArcnOAB EFG
        \pstArcnOAB CGH
      \end{pspicture}
      \caption{Cutting $AB$ from $CD$}
      \label{fig:I.3}
    \end{figure}
    and let the given ray begin at $C$ and pass through $D$.
    All segments in the following are drawn and extended
    with Postulates 1 and 2.
    \begin{align*}
      BE=BC&\And BC=EC,&&\text{[Postulate 3]}\\
      BE&=EC,&&\text{[Common Notion 1]}\\
      BF&=BA,&&\text{[Postulate 3]}\\
      EG&=EF,&&\text{[Postulate 3]}\\
      CG&=BF,&&\text{[Common Notion 2 or 3]}\\
      CH&=CG.&&\text{[Postulate 3]}\qedhere
    \end{align*}
    \end{proof}

    Likewise,
    Hilbert accepts as an axiom that,
    on either side of a given ray,
    an angle equal to a given angle can be drawn.
    Euclid does the drawing,
    again by means circles,
    now using,
    in justification,
    the Side-Side-Side (SSS) theorem.

    Hilbert accepts as an axiom
    \textbf{Side-Angle-Side,} or \textbf{SAS:}
      If two sides and the included angle in one triangle
      are equal respectively to two sides and the included angle
      in another triangle,
      then the remaining angles are equal respectively.
      (It follows that the remaining sides are equal too.)

    Hilbert uses ``equal'' and ``congruent'' as synonyms
    for a relation whose properties,
    even transitivity, are defined by axioms, as above.
    For Euclid, equality
    \begin{compactitem}
      \item
        is present in the radii of a circle,
        \item
          is \emph{explained} as congruence.
    \end{compactitem}
    Congruent things can be made to coincide.
    Thus Side-Angle-Side is a theorem for Euclid,
    because the equations in the hypothesis
    mean that one triangle can be laid exactly on top of the other.
    
    Being congruent is not the same
    as having the same \emph{measurement,}
    although today we may take the latter to determine equality.
    This is how we proved Theorem \ref{thm:circ}.

    In $\Hp{\R}$,
    two hyperbolic rays emanating from the same point form a
    \textbf{hyperbolic angle,}
    whose size is that of the angle formed by the rays
    tangent to the geodesic rays at their common endpoint,
    as in \figref{fig:angle}.
    \begin{figure}
      \centering
      \psset{PointName=none,PointSymbol=none}
      \begin{pspicture}(7,3)
%        \psgrid
        \pstGeonode(0,0)A(7,0)B(2.5,0)O(5,0)P
        \pstInterCC OAPBCD
        \pstRotation[RotAngle=90]CP[E]
        \pstRotation[RotAngle=-90]CO[F]
        {\psset{linestyle=dashed}\ncline CE\ncline CF}
        \pstMarkAngle ECF{}
        \ncline[linestyle=dotted] AB
        \psset{arrows=o-o}
        \pstArcnOAB OAB\pstArcOAB PBA
      \end{pspicture}
      \caption{Geodesic angle}
      \label{fig:angle}
    \end{figure}
    Note that the latter angle always has positive size,
    if the hyperbolic rays are distinct,
    since such rays are never tangent to one another.
    
    If we are working in $\Hp K$
    for an arbitrary Euclidean field $K$,
    a circle still has tangents,
    namely straight lines,
    the equation of each of which has a single common solution
    with the equation of the circle.
    Thinking of $K^2$ as a vector space,
    understanding a point $X$ of the space as $(x_0,x_1)$,
    defining
    \begin{equation*}
      X\cdot Y=x_0y_0+x_1y_1,
    \end{equation*}
    we define angles $ABC$ and $DEF$ to be \textbf{equal,}
    provided
    \begin{equation*}
      \frac{\abs{(A-B)\cdot(C-B)}}{\abs{A-B}\cdot\abs{C-B}}
=
     \frac{\abs{(D-E)\cdot(F-E}}{\abs{D-E}\cdot\abs{F-E}}.
    \end{equation*}
    The angle $ABC$ is \textbf{right,}
    provided $(A-B)\cdot(C-B)=0$.
We now have the first three of Euclid's postulates
    in our setting, as follows.

    \begin{theorem}\label{thm:abs}
      In $\Hp K$ for an arbitrary Euclidean field $K$,
      \begin{compactenum}[1)]
      \item
        for any two points,
        there is a unique geodesic segment between them;
      \item
        any geodesic segment
        can be extended to a longer one,
      \item
        a geodesic circle can be drawn
        passing through any point
        with any center.
      \item
        all geodesic right angles
        are equal to one another.
      \end{compactenum}
    \end{theorem}

    In $\Hp K$,
    though we can always extend a hyperbolic \emph{segment,}
    a hyperbolic \emph{line} has at least one
    ``bound'' or ``limit'' in $K$.
    Given the semicircle in the upper half-plane
    bounded on the $x$-axis by $(\alpha,0)$ and $(\beta,0)$,
    let us say, as Hilbert does,
    that $\alpha$ and $\beta$ are the \textbf{ends}
    of the corresponding hyperbolic line.
    We may denote the line by
    \begin{equation*}
      \myline{\alpha,\beta}.
    \end{equation*}
    (See also \S\ref{sect:ends}.)
    The ends of the vertical hyperbolic line
    starting from $(\alpha,0)$
    are $\alpha$ and $\infty$, and the line itself is
    \begin{equation*}
      \myline{\alpha,\infty}.
    \end{equation*}
    Thus
    \begin{compactitem}
      \item
        a hyperbolic line has two ends;
        \item
          a hyperbolic segment, two endpoints;
          \item
    a hyperbolic ray, a single end and a single endpoint.
    \end{compactitem}
    Two hyperbolic rays are \textbf{parallel}
    if they have the same end.
    Then a hyperbolic ray is parallel to a hyperbolic line,
    and two hyperbolic lines are parallel, if they share an end.
    Parallel hyperbolic rays do not intersect;
    but some non-intersecting hyperbolic rays are not parallel.

    
   In each part of \figref{fig:horocycle},
    \begin{figure}
        \psset{PointName=none,PointSymbol=none}
\subfloat[]{
      \begin{pspicture}(-2.5,-2.5)(2.5,2.5)
        \pstGeonode(0,0)O(0,-2.5)A(-2.5,-2.5)X(0,2.5)Y(2.5,-2.5)Z
        \ncline[linestyle=dotted]XZ
        \pstCircleOA[linestyle=dashed] OA
        \multido{\i=1+1}{5}{
          \pstGeonode
          (! \i\space 5 1 add div 180 mul 90 sub dup cos 2.5 mul
          exch sin 2.5 mul){B\i}
        \pstMediatorAB[linestyle=none] A{B\i}{m\i}{k\i}
          \pstInterLL{m\i}{k\i}AX{C\i}
          \pstArcnOAB{C\i}A{B\i}}
        \ncline AY
        \multido{\i=1+1}{5}{
          \pstGeonode
          (! \i\space 5 1 add div 180 mul 90 add dup cos 2.5 mul
          exch sin 2.5 mul){B\i}
        \pstMediatorAB[linestyle=none] A{B\i}{m\i}{k\i}
          \pstInterLL{m\i}{k\i}AX{C\i}
          \pstArcOAB{C\i}A{B\i}}
        \psdot[dotstyle=o](A)
\end{pspicture}}
\hfill
\subfloat[]{
  \begin{pspicture}(5,5)
    \pstGeonode(0,0)O(5,0)A(0,2)B(5,2)C(0,5)D
    \multido{\i=1+1}{6}{
      \pstHomO[HomCoef=\i\space 6 div]BC[E\i]
      \pstTranslation BD{E\i}[F\i]
      \psline{->}(E\i)(F\i)}
    \psline{->}(B)(D)
    \ncline[linestyle=dotted]OA
        \ncline[linestyle=dashed]BC
  \end{pspicture}
  }
      \caption{Parallel geodesic rays from horocycles}
      \label{fig:horocycle}
    \end{figure}
    geodesic rays are
    \begin{compactitem}
      \item
        parallel to one another,
        \item
    at right angles to a
    \textbf{horocycle,} which is a line (not geodesic),
    the perpendicular bisectors of whose chords
    all have a common end.
    \end{compactitem}

    In place of Euclid's fifth postulate, we have the following.

    \begin{theorem}\label{thm:paral}
      In $\Hp K$ for any $K$,
        for any hyperbolic line
        and any point not lying on it,
        there are two distinct hyperbolic rays,
        emanating from that point,
        each parallel to the given line,
        but together forming an acute angle,
        within which any hyperbolic ray
        emanating from the same point
        cuts the given line.
    \end{theorem}

    \begin{proof}
      If the geodesic line is $\myline{\alpha,\beta}$,
      and the point is $C$,
      then the two rays are those
      that we may write as $C\alpha$ and $C\beta$.
      See \figref{fig:rays}.
      \begin{figure}
      \centering
      \psset{PointName=none,PointSymbol=none}
      \subfloat[]{%
        \begin{pspicture}(7,6)
%        \psgrid
        \pstGeonode(0,0.6)A(7,0.6)B(3,2)C(4.5,5.4){C'}
        \pstMiddleAB ABM
        \pstMediatorAB[linestyle=none] ACde
          \pstInterLL deABO
        \pstMediatorAB[linestyle=none] BCde
          \pstInterLL deABP
        \pstMediatorAB[linestyle=none] A{C'}de
          \pstInterLL deAB{O'}
        \pstMediatorAB[linestyle=none] B{C'}de
          \pstInterLL deAB{P'}
        \ncline[linestyle=dotted] AB
        \psset{arrows=o-o}
        \pstArcOAB MBA
        \psset{arrows=-o,linestyle=dashed}
        \pstArcOAB OCA\pstArcnOAB PCB
        \pstArcOAB {O'}{C'}A\pstArcnOAB {P'}{C'}B
        \uput[d](A){$\alpha$}
        \uput[d](B){$\beta$}
        \uput[d](C){$C$}
        \uput[u](C'){$C'$}
        \end{pspicture}
        }
      \hfill
      \subfloat[]{%
        \begin{pspicture}(3,6)
          \pstGeonode(0.5,2){C''}(3,0.6)B(3,5.4)F(0.5,0.6)A
          \pstMediatorAB[linestyle=none] B{C''}DE
          \pstInterLL DEABO
          \ncline[arrows=->] BF
          \pstTranslation BFA[G]
          \ncline[linestyle=dashed,arrows=->]{C''}G
          \ncline[linestyle=dotted] AB
        \psset{arrows=-o,linestyle=dashed}
        \pstArcnOAB O{C''}B
        \uput[d](B){$\alpha$}
        \uput[dl]({C''}){${C''}$}
        \uput[u](F){$\infty$}
        \uput[u](G){$\infty$}
        \end{pspicture}
      }
      \caption{Hyperbolic parallel postulate}
      \label{fig:rays}
    \end{figure}
    \end{proof}

    \section{Hyperbolic Automorphisms}\label{sect:hyp-aut}
    
    An \textbf{automorphism} of a geometry
    in the sense of \S\ref{sect:lines}
    is a permutation of the points
    that induces a permutation of the straight lines
    and a permutation of the circles,
    sending centers to centers.
    We seek automorphisms of $\Hp K$.

    \begin{theorem}
      Every automorphism of $\Hp K$
      takes equal segments
      to equal segments.
    \end{theorem}

    \begin{proof}
      An automorphism preserves the diagram
      of \figref{fig:I.3}
      used for the proof of Theorem \ref{thm:I.3}.
    \end{proof}

    For all $z$ in $K(\mi)\cup\{\infty\}$,
    we define
    \begin{equation}\label{eqn:infty}
      \left.\qquad
      \begin{gathered}
    z+\infty=\infty=\infty+z,\\
      z\neq0\implies
    \frac z0=\infty=
    z\cdot\infty=
    \infty\cdot z,\\
    z\neq\infty\implies
    \frac z{\infty}=0.
      \end{gathered}
    \qquad\right\}
    \end{equation}
    We leave $0/0$, $0\cdot\infty$, $\infty\cdot0$, and $\infty/\infty$
    undefined; but we do say
    \begin{equation*}
      \conj{\infty}=\infty.
    \end{equation*}
  We define three permutations of $K(\mi)\cup\{\infty\}$
  by
  \begin{align*}
    \mathrm r(z)&=-\conj z,
    &\upiota(z)&=\frac1{\conj z},
    &\uprho(z)&=-\frac1z.    
  \end{align*}
  They are indeed permutations,
  since each is its own inverse.
  We shall call the first two
  \textbf{reflection} and \textbf{inversion}
  respectively;
  the third,
  \textbf{rotation,}
   for reasons that will become apparent.
   Each of the three permutations is its own inverse
   and is the composite
   of the other two;
   since they have order $2$,
   they commute with one another.
   
  We define also two families of operations on $K(\mi)\cup\{\infty\}$ by
  \begin{align*}
    \mathrm d_a(z)&=az,
    &\uptau_b(z)&=z+b,
  \end{align*}
  where $a$ and $b$ are in $K$, and $a>0$.
  The operations are also permutations, since
  \begin{align*}
    \mathrm d_a{}\inv&=\mathrm d_{a\inv},&
    \uptau_b\inv&=\uptau_{-b}.
  \end{align*}  
  Here $\mathrm d_a$ is \textbf{dilation} by $a$,
  and $\uptau_b$ is \textbf{translation} by $b$.

  Each of the permutations of $K(\mi)\cup\{\infty\}$ that we have defined
  is also a permutation of either of
  \begin{compactitem}
  \item
    the upper half-plane $\{z\in K(\mi)\colon\Im(z)>0\}$,
  \item
    $K\cup\{\infty\}$.
  \end{compactitem}
  Except ``rotation,''
  our terms allude to what the operations do in $\Aff K$.
  In $\Hp K$,
  \begin{compactitem}
%  \item
%    $\uptau_b$ is a sort of ``rotation'' about $\infty$;
  \item
    $\mathrm d_a$ is a translation along $\myline{0,\infty}$;
  \item
    $\mathrm r$
    is reflection about $\myline{0,\infty}$;
  \item
    $\upiota$ is reflection about $\myline{-1,1}$;
  \item
    $\uprho$ is rotation by $\uppi$ about $\mi$.
  \end{compactitem}

  \begin{theorem}\label{thm:aut}
    $\uptau_b$, $\mathrm d_a$, $\mathrm r$, $\upiota$,
    and $\uprho$
    are automorphisms of $\Hp K$.
  \end{theorem}

  \begin{proof}
    Obviously $\uptau_b$, $\mathrm d_a$, and $\mathrm r$
    preserve hyperbolic linearity and circularity.
    It is now enough to show that $\upiota$ does the same,
    since $\uprho=\mathrm r\circ\upiota$.
    We show that $\upiota$
  \begin{compactenum}[1)]
  \item
    fixes $\myline{0,\infty}$,
  \item
    interchanges
    \begin{compactenum}
    \item
      $\myline{\alpha,\infty}$ and
      $\myline{\alpha\inv,0}$, if $\alpha\neq0$;
    \item
      $\myline{\alpha-\rho,\alpha+\rho}$ and
      $\myline{(\alpha-\rho)\inv,(\alpha+\rho)\inv}$,
      if $\alpha\neq\rho$.
    \end{compactenum}
  \end{compactenum}
  To do so, we note that in $\{z\in K(\mi)\colon\Im(z)>0\}$,
  \begin{compactitem}
    \item
   $\myline{\alpha,\infty}$
  is defined by
    $z+\conj z=2\alpha$;
  \item
  $\myline{\alpha-\rho,\alpha+\rho}$, by
    $\abs{z-\alpha}^2=\rho^2$.
  \end{compactitem}
  We show that $\upiota$
  \begin{compactenum}[1)]
  \item
    fixes
      $z+\conj z=0$,
  \item
    interchanges
    \begin{compactenum}
    \item
    $z+\conj z=2\alpha$ and
      $\left|z-\displaystyle\frac1{2\alpha}\right|^2
      =\left(\displaystyle\frac1{2\alpha}\right)^2$,
  \item
      $\abs{z-\alpha}^2=\rho^2$ and
      $\left|z-\displaystyle\frac{\alpha}{\alpha^2-\rho^2}\right|
      =\left(\displaystyle\frac{\rho}{\alpha^2-\rho^2}\right)^2$.
    \end{compactenum}
\end{compactenum}
  We do this by replacing $z$ with $\conj z\inv$ in one equation,
  then deriving the other.
  It is clear that $z+\conj z=0$ remains unchanged.
  In the other cases,
      \begin{compactenum}[a)]
    \item
      the line is sent to the curve given by
      \begin{gather*}
        \conj z\inv+z\inv=2\alpha,\\
        z+z\inv=2\alpha\abs z^2,\\
        \abs z^2-\frac1{2\alpha}z-\frac1{2\alpha}\conj z=0,
      \end{gather*}
      and thus to the indicated circle;
    \item
      the first circle is sent to the curve given by
          \begin{gather*}
      \abs{\conj z\inv-\alpha}^2=\rho^2,\\
      \abs{1-\alpha\conj z}^2=\rho^2\abs z^2,\\
      \left(\alpha^2-\rho^2\right)\abs z^2
      -\alpha\conj z-\alpha z=-1,\\
      \abs z^2-\frac{\alpha\conj z}{\alpha^2-\rho^2}
      -\frac{\alpha z}{\alpha^2-\rho^2}
      =\frac1{\rho^2-\alpha^2},
          \end{gather*}
          and thus to the second circle.
      \end{compactenum}
      So $\upiota$ permutes hyperbolic lines.
      In particular, it permutes the circles
  not passing through $0$
  \emph{and} that have centers on the $x$-axis.
  Therefore it does the same without the last condition.
  Moreover,
  when the circles are in the upper half-plane,
  $\upiota$ also takes hyperbolic centers
  to hyperbolic centers, since
  \begin{equation*}
    ac=b^2\implies
    \frac1a\cdot\frac1c=\left(\frac1b\right)^2.\qedhere
  \end{equation*}
  \end{proof}
  
  By definition,
  \begin{align*}
        \SL K&=
\left\{
    \begin{pmatrix}
      a&b\\c&d
    \end{pmatrix}
    \colon ad-bc=1\right\},
    &I&=
    \begin{pmatrix}
      1&0\\0&1
    \end{pmatrix}.
  \end{align*}
  An \textbf{action} of a group (such as $\SL K$)
  on a structure (such as a geometry)
  is a homomorphism from the group
  into the group of automorphisms of the structure.
  
  \begin{theorem}\label{thm:act}
    $\SL K$ acts on $\Hp K$
    by the rule
  \begin{equation*}
        \begin{pmatrix}
      a&b\\c&d
    \end{pmatrix}z=\frac{az+b}{cz+d}.
  \end{equation*}
  The range of the action
  is generated by $\uprho$ and the $\uptau_b$ and $\mathrm d_a$.
\end{theorem}

  \begin{proof}
    The listed automorphisms
  are the images of
  \begin{align*}
    &\begin{pmatrix}
      0&-1\\1&0
    \end{pmatrix},
    &&\begin{pmatrix}
      1&b\\1&0
    \end{pmatrix},
    &&\begin{pmatrix}
      \surd a&0\\0&\surd a\inv
    \end{pmatrix}
  \end{align*}
  respectively.
  Composition of automorphisms
  will correspond to multiplication of matrices,
  since
  \begin{gather*}
    \begin{pmatrix}
      a&b\\c&d
    \end{pmatrix}
    \begin{pmatrix}
      e&f\\g&h
    \end{pmatrix}
    =
    \begin{pmatrix}
      ae+bg&af+bh\\ce+dg&cf+dh
    \end{pmatrix},\\
      \frac{a\displaystyle\frac{ez+f}{gz+h}+b}
           {c\displaystyle\frac{ez+f}{gz+h}+d}
%           =\frac{a(ez+f)+b(gz+h)}{c(ez+f)+d(gz+h)}\\
           =\frac{(ae+bg)z+af+bh}
           {(ce+dg)z+cf+dh}.
  \end{gather*}
  Finally,
  if $ad-bc=1$, then
  \begin{gather*}
    c\neq0\implies          \frac{az+b}
           {cz+d}
           =\frac ac+\frac{bc-ad}{c(cz+d)}
           =(\uptau_{a/c}\circ\uprho\circ\uptau_{cd}\circ\mathrm d_{c^2})z,\\
           c=0\implies          \frac{az+b}
           {cz+d}
=\frac{az+b}d=\frac bd+\frac ad\cdot z
        =(\uptau_{b/d}\circ\mathrm d_{a/d})z.\qedhere
  \end{gather*}
  \end{proof}

  \begin{lemma}\label{lem:SL-trans}
  The action of $\SL K$ on $\Hp K$
  takes any point to any point.
  \end{lemma}

  \begin{proof}
    $c+d\mi
    =(\uptau_c\circ\mathrm d_{d/b}\circ\uptau_{-a})(a+b\mi)$.
  \end{proof}
  
\begin{theorem}\label{thm:SL}
  The action of $\SL K$ on $\Hp K$
    takes any ray to any ray.
\end{theorem}

\begin{proof}
  By Lemma \ref{lem:SL-trans},
  it is enough to consider rays with endpoint $\mi$.
  Such a ray has the end $\tan\phi$ for some $\phi$,
  and then the normal to the ray at $\mi$
  makes angle $2\phi$ to the $x$-axis,
  as in \figref{fig:i-a}.
    \begin{figure}
    \centering
    \begin{pspicture}(8,5)
%      \psgrid[subgriddiv=1]
      \psset{PointName=none,PointSymbol=none}
      \pstGeonode(0,0.6)C(8,0.6)A(6.5,0.6)O
      \pstMiddleAB CAM
      \pstRotation[RotAngle=90]OC[I]
      \pstInterLC OIMAJ{J'}
      \pstRotation[RotAngle=90]JM[K']
      \pstHomO[HomCoef=1.5]{K'}J[K]
      \pstTranslation KJJ[L]
      \pstMarkAngle OJA{$\phi$}
      \pstMarkAngle OCJ{$\phi$}
      \pstMarkAngle AMJ{$2\phi$}
      \psset{linestyle=dashed}
      \ncline MJ\ncline JA
      \ncline KL\ncline CJ
      \psset{linestyle=dotted}
      \ncline CA
      \psset{linestyle=solid}
      \pstRightAngle KJM
      \pstArcOAB[arrows=o-o] MAC
      \ncline[arrows=o-] OJ
      \uput[d](A){$\tan\phi$}
      \uput[d](C){$-\cot\phi$}
      \uput[d](O){$0$}
%      \uput[d](M){$-\cot(2\phi)$}
      \uput[ur](J){$\mi$}
    \end{pspicture}

    \caption{Hyperbolic line through $\mi$}
    \label{fig:i-a}
  \end{figure}  
  Since
  \begin{equation*}
      \frac{a\mi+b}
           {c\mi+d}
           =\mi
           \iff
           a\mi+b=d\mi-c
           \iff
             d=a\And
             c=-b,
  \end{equation*}
  an element of $\SL K$ that fixes $\mi$ is precisely a matrix
  \begin{equation*}
    \begin{pmatrix}
      \cos\theta&\sin\theta\\-\sin\theta&\cos\theta
    \end{pmatrix}
  \end{equation*}
  for some $\theta$.
  We may denote the resulting operation on $\Hp K$ by
         \begin{equation*}
           \uprho_{2\theta},
         \end{equation*}  
         since,
         applying the matrix to the end $\tan\phi$, we compute
  \begin{multline*}
    \frac{\cos\theta\tan\phi+\sin\theta}
         {-\sin\theta\tan\phi+\cos\theta}
         =
    \frac{\cos\theta\sin\phi+\sin\theta\cos\phi}
         {-\sin\theta\sin\phi+\cos\theta\cos\phi}\\
         =\frac{\sin(\theta+\phi)}{\cos(\theta+\phi)}
         =\tan(\theta+\phi).
  \end{multline*}
  Thus $\uprho_{2\theta}$ effects rotation about $\mi$
  by $2\theta$.
\end{proof}

\begin{porism}
  The kernel of the action of $\SL K$ on $\Hp K$
  is the two-element subgroup generated by
  \begin{equation*}
    \begin{pmatrix}
      -1&0\\0&-1
    \end{pmatrix}.
  \end{equation*}
\end{porism}

We denote by
\begin{equation*}
  \PSL K
\end{equation*}
the quotient of $\SL K$ by the kernel in the porism.
We may consider the quotient
as a group of automorphisms of $\Hp K$.

\begin{theorem}\label{thm:conj}
  $\PSL K$
  is closed under conjugation by $\mathrm r$.
\end{theorem}

\begin{proof}
  By Theorem \ref{thm:act},
  we need only note
  \begin{align*}
    \mathrm r\circ\uptau_b\circ\mathrm r&=\uptau_{-b},
    &\mathrm r\circ\mathrm d_a\circ\mathrm r&=\mathrm d_a,
    &\mathrm r\circ\uprho\circ\mathrm r&=\uprho.\qedhere
  \end{align*}
\end{proof}

We can now understand
the group of automorphisms of $\Hp K$
generated by all of the elements
listed in Theorem \ref{thm:aut}
as the semidirect product
\begin{equation*}
  \PSL K\rtimes\langle\mathrm r\rangle.
\end{equation*}

\begin{theorem}\label{thm:refl}
  For any one point
  and two ends in $\Hp K$,
  the group
  $\PSL K\rtimes\langle\mathrm r\rangle$
  of automorphisms of $\Hp K$
  has an element
  fixing the point and interchanging the ends.
\end{theorem}

\begin{proof}
  By Theorem \ref{thm:SL}, there is at most one such element,
  and we may assume the point is $\mi$.
  Letting the ends be $\tan\theta_0$ and $\tan\theta_1$,
  we have
  \begin{equation*}
    (\uprho_{2(\theta_0+\theta_1)}\circ\mathrm r)\tan\theta_j
    =\tan\theta_{1-j}.\qedhere
  \end{equation*}
\end{proof}

\begin{theorem}\label{thm:seg-pres}
  In $\Hp K$,
  if two segments are equal,
  then an automorphism in
  $\PSL K$
takes one segment to the other.  
\end{theorem}

\begin{proof}
  By Theorem \ref{thm:SL}
  and
  \figref{fig:I.3},
  used for the proof of Theorem \ref{thm:I.3},
  if two segments are equal,
  some automorphism in
  $\PSL K$
  takes one to the other.
\end{proof}

    \begin{theorem}\label{thm:aut-met}
  The automorphisms in
  $\PSL K\rtimes\langle\mathrm r\rangle$
  preserve lengths in $\Hp K$.
    \end{theorem}

    \begin{proof}
      In the proof of Theorem \ref{thm:SL},
        since
  \begin{equation*}
    \tan(\theta+\phi)=\tan(\theta'+\phi)
    \iff\uprho_{2\theta}=\uprho_{2\theta'},
  \end{equation*}
   $\uprho_{2\theta}$
  and is the unique automorphism
  in $\PSL K$
  to take the given ray
  to the ray with the same endpoint $\mi$
  and end $\tan(\theta+\phi)$.
  Hence, when a segment $AB$ and a ray with endpoint $C$ are given,
  a unique element $\sigma$ of $\PSL K$
  sends $A$ to $C$
  and $B$ to a point $D$ of the ray.
  By Theorem \ref{thm:I.3},
  there is a point $D'$ of the ray such that
  $AB=CD'$.
  Then $\sigma$ must be the automorphism,
  which exists by Theorem \ref{thm:seg-pres},
  that takes $B$ to $D'$; so $D'$ is $D$.
    \end{proof}

    \begin{theorem}
  The automorphisms in
  $\PSL K\rtimes\langle\mathrm r\rangle$
  preserve angle measurements in $\Hp K$.
    \end{theorem}

    \begin{proof}
      This being clear
      for the other generators,
we need only show it for $\upiota$.      
  Since
  \begin{equation*}
    \upiota=\mathrm d_a\circ\upiota\circ\mathrm d_a,
  \end{equation*}
  it is enough to consider angles
  whose vertices are on the unit circle.
  More than that,
  if $\rot$ is rotation in $\Aff K$ by $\theta$ about $0$,
  then this preserves angles, and
  \begin{equation*}
    \rot[-\theta]\circ\upiota\circ\rot,
  \end{equation*}
  so
  it is enough to consider angles with vertices at $\mi$.
  Since $\iota=\mathrm r\circ\uprho$,
  it is enough to observe that,
  by the proof of Theorem \ref{thm:SL},
  $\rho$ preserves angles at $\mi$.
    \end{proof}
    
    \begin{theorem}\label{thm:SAS}
      SAS holds in $\Hp K$.
    \end{theorem}

    In $\Hp K$ we can now prove all results
    of so-called \emph{absolute} geometry,
    namely Euclidean geometry without a parallel postulate
    (either Euclid's or our Theorem \ref{thm:paral}).
    For example, we have the following two theorems.
    
    \begin{theorem}\label{thm:SSS}
      SSS holds in $\Hp K$.
    \end{theorem}

    \begin{theorem}\label{thm:col}
      In $\Hp K$,
      three points are collinear
      if each is equidistant from the same two points.
    \end{theorem}
  
    An \textbf{isometry} of a geometry
    that satisfies the absolute axioms
    is a permutation of the points
    that preserves lengths.
    An isometry is automatically an automorphism.
    \begin{compactitem}
      \item
    An isometry of $\Hp K$
    preserves angle measurements,
    by Theorem \ref{thm:SSS}.
  \item
    Every element of $\PSL K\rtimes\langle\mathrm r\rangle$
    is an isometry of $\Hp K$,
    by Theorem \ref{thm:aut-met}.
    \end{compactitem}
    
    
    \begin{theorem}\label{thm:Isom=PSL}
      Every isometry of $\Hp K$ is an automorphism in
      $\PSL K\rtimes\langle\mathrm r\rangle$.
    \end{theorem}

    \begin{proof}
      By Theorem \ref{thm:col},
      only the identical isometry
      fixes each of three non-collinear points.
      Thus every isometry is determined
      by where it sends such points.
      By Theorems \ref{thm:SSS},
      \ref{thm:SL}, \ref{thm:refl}, and \ref{thm:aut-met},
      some automorphism
      in $\PSL K\rtimes\langle\mathrm r\rangle$
      agrees with the isometry at those points,
      and therefore everywhere.
    \end{proof}

    It is not clear whether $\Hp K$ has automorphisms
    other than those that we have found.
    For example, $\Aff K$ has automorphisms,
    such as dilations,
    that are not isometries.

    In $\Hp K$ and $\Aff K$,
    every automorphism preserves right angles,
    because it preserves the diagram in \figref{fig:rgt}.
    \begin{figure}
      \centering
      \begin{pspicture}(6,4)
        \psset{PointSymbol=none}
        \pstGeonode[PosAngle={180,0}](2,2)A(4,2)B
        \pstCircleOA AB
        \pstCircleOA BA
        \pstInterCC[PosAngleA=90,PosAngleB=-90] ABBACD
        \ncline AB\ncline CD
      \end{pspicture}
      \caption{Perpendicular bisector construction}
      \label{fig:rgt}
    \end{figure}


    We shall not know till \S\ref{sect:Bolyai}
    that automorphisms of $\Hp K$ preserve parallelism.


\chapter{Plane}\label{ch:Hilbert}

\section{Axioms}\label{sect:axioms}

Hilbert's axioms for \textbf{hyperbolic geometry}
(strictly, hyperbolic \emph{plane geometry}) are as follows.
(The naming and numeration are his,
not the exact words.)
\begin{compactdesc}
  \item[I.  Incidence.]\mbox{}
\begin{compactdesc}
\item[1, 2.]
  Two points lie on exactly one line.
\item[3.]
  For every line,
  at least two points lie on it,
  but one does not.
\end{compactdesc}
\item[II. Order.]\mbox{}
\begin{compactdesc}
\item[1.]
  If $B$ lies between $A$ and $C$, then
  \begin{compactitem}
  \item
    all three are distinct points of one line, and
  \item
    $B$ lies between $C$ and $A$.
  \end{compactitem}
\item[2.]
  Between any two points lies a third.
\item[3.]
  Of three points of a line,
  no more than one lies between the other two.
\item[4.]
  If $ABC$ is a proper triangle,
  a line crossing $AB$ crosses either $AC$ or $CB$
  (if not $C$ itself).
\end{compactdesc}
\item[III. Congruence.]\mbox{}
\begin{compactdesc}
\item[1.]
  On a given line,
  on either side of a given point (this uses order),
  we can mark off a segment congruent to a given segment.
\item[2.]
  Congruence of segments is transitive.
\item[3.]
  Sums of congruent segments are congruent.
\item[4.]
  A given ray can always serve
  as the side of an angle congruent to a given angle.
\item[5.]
  Side-Angle-Side.
\end{compactdesc}
\item[IV. Parallels.]\mbox{}
  For any line $b$ and point $A$ not on it,
  there are two rays emanating from $A$
  that do not cut $b$,
  but form an acute angle,
  inside of which,
  any ray from $A$ cuts $b$.
\end{compactdesc}
In the Parallel Axiom,
any point of $b$ divides it into two rays,
each \textbf{parallel} to one of the rays emanating from $A$.
According to Hilbert,
that parallelism is symmetric and transitive
``follows immediately.''
Apparently this does not mean the proofs are trivial.
For symmetry, a note refers to ``a method due to Gauss.
Cf.\ Bonola-Liebmann,
\emph{Die nichteuklidische Geometrie} (Leipzig, 1908 and 1921)''
and Bonola \cite[pp.\ 70--1]{Bonola}
(Hilbert himself gives no page references).
Gauss's proof has two cases;
Lobachevski's \cite{Lobachevski}, with one, seems simpler.

\section{Ends}\label{sect:ends}

A ray and the rays parallel to it compose an \textbf{end.}
(This is consistent with the usage initiated in \S\ref{sect:Euc-ax}.)
A point $A$ and an end $\alpha$ determine the ray $A\alpha$.
A line with ends $\alpha$ and $\beta$ is unique;
we shall denote such a line by
\begin{equation*}
  \myline{\alpha,\beta},
\end{equation*}
since the ends of a line have no intrinsic order.
However,
Hilbert calls the line $(\alpha,\beta)$.
We shall construct this line in Lemma \ref{lem:3},
using Lemma \ref{lem:2},
which needs Lemma \ref{lem:1}.
(The lemmas are Hilbert's,
along with Lemmas \ref{lem:4} and \ref{lem:5};
the labelling of theorems as such is our addition.)

\begin{lemma}\label{lem:1}
  Lines making equal alternate angles
  with a third are not parallel in any direction.
\end{lemma}

\begin{proof}
  If parallel in one direction, then in the other, by symmetry
  (which can be spelled out).
\end{proof}

\begin{lemma}\label{lem:2}
  Two lines,
  as $AP$ and $BB'$ in \figref{fig:2},
that neither intersect nor are parallel
have a common perpendicular,
$MN$.
\end{lemma}
\begin{figure}
  \centering
  \psset{unit=7mm}
    \begin{pspicture}(-4,0)(7,9.5)
%      \psgrid[subgriddiv=1]
      \newcommand{\myrat}{2.6}
      \psset{PointSymbol=none}
      \pstGeonode%[PosAngle={-90,-90}]
          [PointName=none]
      (0,0)O(6,0)Z
      \pstHomO[HomCoef=0.6,PointName=none]OZ[D]
      \pstRotation[RotAngle=-90,PointName=none]ZO
      \pstHomO[HomCoef=0.6]Z{O'}[B]
      \pstHomO[HomCoef=\myrat]ZB[B']
      \pstHomO[HomCoef=\myrat,PointName=none]Z{O,D}[{O',D'}]
      \pstInterCC[PointNameB={},PosAngleA=-135] ODZBAa
      \pstInterCC[PointNameB={},PosAngleA=180] ODZ{B'}Pp
      \pstInterLC[PointNameA={},PosAngleB=180] ZAZ{B'}s{A'}
      \pstInterCC[PointNameA={},PosAngleB=-105] OD{O'}{A'}qQ
      \pstInterLC[PointNameA={}] ZBZQrR
      \pstInterLC[PointName=none] ZQZP{Y'}Y
      \pstTranslation[PointName=none] YBQ[X]
      \pstInterLL QXZB{R'}
      \pstInterCC[PointNameB={},PosAngleA=-135] ODZ{R'}{Q'}V
      \pstMiddleAB[PointName=none]ZRm
      \pstRotation[RotAngle=90,PointName=none]{R'}Z[z]
      \pstInterLC[PointName=none]{R'}zmZxy
      \pstInterLC[PointNameA={}] ZBZy{M'}M
      \pstInterCC[PointNameB={},PosAngleA=-135] ODZMN{N'}
      \pstArcOAB[linewidth=2.4pt] O{Q'}P
      \ncline[linewidth=2.4pt]{R'}{B'}
      \pstArcOAB ZBA
      \pstArcOAB Z{B'}P
      \psset{linestyle=dashed}
      \pstArcnOAB{O'}{A'}Q
      \pstArcnOAB ZQR
      \pstArcOAB Z{R'}{Q'}
      \psset{linestyle=dotted}
      \pstArcOAB[linewidth=2.4pt] ZMN
      \psline(-4,0)(7,0)
      \psset{PointName=none}
      \pstMediatorAB[linestyle=none,PointSymbol=none] BDim
      \psset{arrows=-o}
      \pstInterLL imOZx
      \pstArcnOAB O{Q'}D
      \pstArcOAB xBD
      \pstTranslation Bx{B'}[y]
      \pstInterLL{B'}yOZz
      \pstArcOAB z{B'}O
      \pstArcnOAB{O'}QZ
    \end{pspicture}
    \caption{Common perpendicular}
    \label{fig:2}
    \end{figure}

\begin{proof}
  The construction and demonstration is as follows (reading by columns):
  \begin{align*}
    &\begin{gathered}
      AB,PB'\perp BB',\\
      B'P>BA=B'A',\\
      B'A'Q=BAQ',\\
      AQ'=A'Q,
     \end{gathered}
    &&
    \begin{gathered}
      QR\perp BB',\\
      BR'=B'R,\\
      B'A'QR\cong BAQ'R',\\
      Q'R'=QR,
    \end{gathered}
  \end{align*}
  and we let $MN$ join the midpoints of $RR'$ and $QQ'$.
  In the simpler case, where $B'P=BA$, then $MN$ joins the midpoints
  of $B'B$ and $PA$.
  Otherwise, the point $Q$ exists on $AP$
  because the parallel to $A'Q$ from $B'$
  must cut $PA$, because that parallel
  neither cuts nor, by Lemma \ref{lem:1},
  is parallel to the parallel to $PA$ drawn from $B$.
\end{proof}

We shall use now an infinitary version of Side-Angle-Side:
if $AB=A'B'$,
and $BA\alpha$ and $B'A'\alpha'$ are equal as angles,
then so are $AB\alpha$ and $A'B'\alpha'$.

\begin{lemma}\label{lem:3}
  For any distinct ends $\alpha$ and $\beta$
  as in \figref{fig:3},
  the line $\myline{\alpha,\beta}$
  is the common perpendicular $EF$
  of
  the bisectors of angles $CA\alpha$ and $CB\beta$,
  where $OA=OB$.
\end{lemma}
\begin{figure}
  \centering
  \newcommand{\findcenter}[5]{%
    \psset{PointName=none,PointSymbol=none}
    \pstMediatorAB[linestyle=none]{#1}{#2}{#1#2i}{#1#2m}
      \pstInterLL{#1#2i}{#1#2m}{#4}{#5}{#3}}
  \begin{pspicture}(-6,0)(4,5.5)
%    \psgrid
    \psset{PointName=none,PointSymbol=none}
%      \psset{PointSymbol=none}
      \pstGeonode(0,0)o(0,1)c(-6,0)a(4,0)b
    \newcommand{\myrat}{2}
      \pstHomO[HomCoef=\myrat]oc[O]
      \pstHomO[HomCoef=\myrat]oO[d]
      \pstMiddleAB cdm
                {\findcenter aO{caO}ab}
                \pstArcOAB{caO}Oa
                {\findcenter bO{cbO}ab}
                \pstArcnOAB{cbO}Ob
      \pstInterCC{caO}OmcA{A'}
                 {\findcenter Ab{cAb}ab}
                 \ncline[linestyle=dotted] ab
                \pstArcOAB[arrows=o-]{cAb}bA
      \pstInterCC{cbO}Omc{B'}B
                {\findcenter Ba{cBa}ab}
                \pstArcOAB[arrows=-o]{cBa}Ba
                \pstOutBissectBAC[linestyle=none]{caO}A{cAb}{biA}
                \pstInterLL{biA}Aab{cbiA}
                \pstOutBissectBAC[linestyle=none]{cbO}B{cBa}{biB}
                \pstInterLL{biB}Bab{cbiB}
                \pstMiddleAB abM
                \pstInterCC Ma{cbiA}A{E'}E
                \pstArcOAB{cbiA}AE
                \pstInterCC Ma{cbiB}BF{F'}
                \pstArcnOAB{cbiB}BF
                \pstArcOAB[linestyle=dotted] MFE
                \uput[d](a){$\alpha$}
                \uput[d](b){$\beta$}
                \pstInterCC{cBa}B{cAb}AC{C'}
                \uput[u](C){$C$}
                {\findcenter AB{cAB}ab}
                \pstArcOAB[linestyle=dashed]{cAB}BA
                \pstInterCC{cBa}B{cbiA}A{D'}D
                \uput[ur](D){$D$}
   \uput[d](O){$O$}
   \uput[d](A){$A$}
   \uput[d](B){$B$}
   \uput[ul](E){$E$}
   \uput[ur](F){$F$}
  \end{pspicture}
  \caption{Line with given ends}
    \label{fig:3}
  \end{figure}

\begin{proof}
  We shall show
  \begin{compactenum}[1)]
  \item
    the bisectors
    neither intersect nor are parallel;
  \item
    $EF$ lies on $\myline{\alpha,\beta}$.
  \end{compactenum}
  To establish the latter claim,
  we observe (usually by symmetry)
  \begin{gather*}
    \alpha OB\cong\beta OA,\\
    OBC=OAC,\\
    \alpha AE=\alpha BF,\\
    AE=BF,\\
    \alpha AE\cong\alpha BF,
  \end{gather*}
  so the rays from $E$ and $F$ respectively having end $\alpha$
  make the same angle with $EF$,
  and therefore they coincide with $EF$ by Lemma \ref{lem:1}.

  Suppose if possible the angle bisectors meet at a point $M$.
  Then $AM=BM$ and $\alpha AM=\alpha BM$, so
  \begin{equation*}
    \alpha AM\cong\alpha BM,
  \end{equation*}
  and in particular $\alpha MA=\alpha MB$, which is absurd.

  Suppose if possible the angle bisectors have the common end $\mu$.
  If $DB'$, equal to $DA$, is marked off along $DB$, then
  \begin{equation*}
    \alpha AD\cong\mu B'D,
  \end{equation*}
  and so $\mu B'D=\alpha AD=\mu BD$,
  which is absurd by Lemma \ref{lem:1} unless $B'$ coincides with $B$.
  In this case we have another absurdity,
  \begin{equation*}
    DAB=DBA=CBA=CAB.\qedhere
  \end{equation*}
\end{proof}

\section{Euclidean Field}

Now let us denote
\begin{compactitem}
  \item
    some particular end by $\infty$,
  \item
    the set of remaining ends by $K$,
  \item
    a particular element of $K$ by $0$,
  \item
    a particular point on $\myline{0,\infty}$ by $O$,
  \item
    one end of the perpendicular to $\myline{0,\infty}$ at $O$ by $1$.
\end{compactitem}
We shall make $K$ into a Euclidean field,
as suggested by \figref{fig:end-circ}.
\begin{figure}
  \subfloat[Disk model]{%
    \begin{pspicture}(-2.7,-2.5)(2.4,2.5)
%      \psgrid
    \newcommand{\myrad}{2}
    \pscircle[linestyle=dotted](0,0){\myrad}
    \uput[0](\myrad;0){$1$}
    \uput[90](\myrad;90){$\infty$}
    \uput[180](\myrad;180){$-1$}
    \uput[270](\myrad;270){$0$}
    \psset{arrows=o-o}
    \psline(\myrad;0)(\myrad;180)
    \psline(\myrad;90)(\myrad;270)
    \uput[ur](0,0){$O$}
  \end{pspicture}
  }
  \hfill
  \subfloat[Half-plane model]{%
  \begin{pspicture}(-2.8,-0.5)(2.8,4.5)
    \newcommand{\myrad}{2}
    \psline[linestyle=dotted](-2.8,0)(2.8,0)
    \psarc{o-o}(0,0){\myrad}0{180}
    \psline{o->}(0,0)(0,4)
    \uput[u](0,4){$\infty$}
    \uput[d](-\myrad,0){$-1$}
    \uput[d](\myrad,0){$1$}
    \uput[d](0,0){$0$}
    \uput[ur](0,\myrad){$O$}
  \end{pspicture}
  }
  \caption{Line of ends}
  \label{fig:end-circ}
\end{figure}

\subsection{Ordering}

On the set of all ends,
we define a quaternary relation $Q$
so that $Q(\alpha,\beta,\gamma,\delta)$
if and only if $\myline{\alpha,\gamma}$
crosses $\myline{\beta,\delta}$.
If $\alpha$, $\beta$, $\gamma$, and $\delta$
are distinct ends, then
\begin{equation*}
  Q(\alpha,\beta,\gamma,\delta)
  \iff
   Q(\alpha,\beta,\delta,\gamma),
\end{equation*}
and also
\begin{equation*}
  Q(\alpha,\beta,\gamma,\delta)
  \iff
   Q(\alpha^{\sigma},\beta^{\sigma},\gamma^{\sigma},\delta^{\sigma}),
\end{equation*}
where $\sigma$ is an even permutation of the set
$\{\alpha,\beta,\gamma,\delta\}$.
We use $Q$ to order $K$ by defining
\begin{equation*}
  \alpha<0\iff Q(\alpha,0,1,\infty).
\end{equation*}
If neither of $\alpha$ and $\beta$
is $0$,
we define $\alpha<\beta$ if and only if
one of the following holds:
\begin{gather*}
  \alpha<0\And Q(\infty,\alpha,\beta,0),\\
  \alpha<0\And 0<\beta,\\
  0<\alpha\And Q(0,\alpha,\beta,\infty).
\end{gather*}
The \textbf{positive} ends
are those finite ends $\alpha$ for which $0<\alpha$.


\subsection{Addition}

Our definition of addition on $K$
will be motivated by an analogy.
In $\C$, when $\alpha\in\R$, we can define
$\refl{\alpha}$
as reflection about
$z+\conj z=2\alpha$,
so that
\begin{equation*}
  \refl{\alpha}(z)
  =\alpha-(z-\alpha)
  =2\alpha-z.
\end{equation*}
Then
\begin{align*}
  \refl{\alpha}\refl0\refl{\beta}(z)
  &=\refl{\alpha}\refl0(2\beta-z)\\
  &=\refl{\alpha}(z-2\beta)\\
  &=2\alpha+2\beta-z\\
  &=\refl{\alpha+\beta}(z).
\end{align*}
In our hyperbolic plane,
when $\alpha$ in $K$,
we define $\refl{\alpha}$
to be reflection about $\myline{\alpha,\infty}$,
so that, if
\begin{equation*}
  \refl{\alpha}(P)=P'
\end{equation*}
this means either
\begin{compactitem}
\item
  $P$ is on $\myline{0,\infty}$, and $P'=P$, or
\item
  $\myline{0,\infty}$ is the perpendicular bisector of $PP'$.
\end{compactitem}


\begin{theorem}
  Reflection about a line
  is an automorphism of the hyperbolic plane.
\end{theorem}

\begin{proof}
  We have to show that reflection preserves lengths and angles.
  This follows from Side-Angle-Side.
\end{proof}

The following corresponds to Lobachevski's Theorem 30
and is illustrated by \figref{fig:4}.
\begin{figure}
  \centering
  \begin{pspicture}(10,7)
%    \psgrid
    \psset{PointSymbol=none}
    \pstGeonode[PointName=none](0.6,0.6){dl}(9.4,0.6){dr}(0.6,6.4){ul}
    \pstHomO[HomCoef=0.4,PosAngle=180]{dl}{ul}[O]
    \pstTranslation{dl}{dr}O[B]
    \pstHomO[HomCoef=0.71,PosAngle=-135]OB[A]
    \psset{PointName=none}
    \pstMiddleAB AOa
    \pstMiddleAB OBb
    \pstMiddleAB ABd
    \pstProjection {dl}{dr}{a,b,d,O}
    \pstArcOAB {a'}AO
    \pstArcOAB {b'}BO
    \pstArcOAB {d'}BA
    \pstTranslation{dl}{ul}{a',b',d',O'}
    \psline[linestyle=dotted](dl)(dr)
    \psset{arrows=o->}
    \ncline{dl}{ul}
    \ncline{a'}{a''}\ncline{b'}{b''}\ncline{d'}{d''}\ncline{O'}{O''}
    \uput[u](ul){$\infty$}
    \uput[d](a'){$\alpha$}\uput[u](a''){$\infty$}
    \uput[d](b'){$\beta$}\uput[u](b''){$\infty$}
    \uput[d](O'){$\gamma$}\uput[u](O''){$\infty$}
    \uput[d](d'){$\delta$}
    \uput[u](d''){$\infty$}
  \end{pspicture}
  \caption{Parallel perpendicular bisectors}
  \label{fig:4}
\end{figure}

\begin{lemma}\label{lem:4}
  In any triangle $AOB$,
  if the perpendicular bisectors of $AO$ and $OB$
  are parallel,
  then they are parallel to that of $AB$.
\end{lemma}

\begin{proof}
  Hartshorne \cite{MR1761093}
  improves on the proofs of Lobachevski and Hilbert
  by using Lemma \ref{lem:2}, as follows.
  If the perpendicular bisectors of $AO$ and $AB$
  \begin{compactdesc}
  \item[meet,]
    that of $OB$ meets them at the same point,
    by the usual proof;
  \item[neither meet nor are parallel,]\sloppy
    the perpendiculars dropped
    to their common perpendicular
    from $O$, $A$, and $B$
    are equal,
    so the perpendicular bisector of $OB$
    must also be perpendicular to that common perpendicular;
  \item[are parallel,]
    $OB$ can only be parallel to them.\qedhere
  \end{compactdesc}
\end{proof}

\begin{lemma}\label{lem:5}
  If $\alpha$, $\beta$, and $\gamma$ belong to $K$,
    then for some $\delta$ in $K$,
  \begin{equation*}
    \refl{\beta}\refl{\gamma}\refl{\alpha}=\refl{\delta}.
  \end{equation*}
\end{lemma}

\begin{proof}
  Hilbert has three cases,
  but we need only two,
  by the method of Hartshorne.
  (Hartshorne omits the easier case.)
  \begin{asparaenum}
    \item
  If the three finite ends are distinct as in \figref{fig:4},
  let $O$ be a point on $\myline{\gamma,\infty}$,
  and let
  \begin{align*}
    \refl{\alpha}O&=A,
    &\refl{\beta}O&=B.
  \end{align*}
  The perpendicular bisector of $AB$
  is $\myline{\delta,\infty}$ for some $\delta$, by Lemma \ref{lem:4}.
  The operation
  \begin{equation*}
    \refl{\delta}\refl{\beta}\refl{\gamma}\refl{\alpha}
  \end{equation*}
  fixes $A$ and the line through it with end $\infty$;
  being also the product of an even number of reflections,
  the operation is the identity.
\item
  If the three ends are not distinct,
  the only possibility that is not completely trivial
  is that $\alpha$ and $\beta$ coincide.
  In this case,
  we let
  \begin{equation*}
    \delta=\refl{\alpha}\gamma.\qedhere
  \end{equation*}
  \end{asparaenum}
\end{proof}

We can now define addition in $K$ by the rule
\begin{equation}\label{eqn:a+b}
  \refl{\alpha}\refl0\refl{\beta}=\refl{\alpha+\beta}.
\end{equation}

\begin{theorem}
  The addition just defined
  makes $K$ an ordered abelian group.
\end{theorem}

\begin{proof}
  Addition is
  \begin{compactitem}
    \item
      associative, since composition is associative;
    \item
      commutative,
      since composition of automorphisms having order $2$
      is commutative.
  \end{compactitem}
  The additive inverse $-\alpha$ is defined
  so that the image of $\myline{\alpha,\infty}$ under $\refl0$
  is $\myline{-\alpha,\infty}$.
  That the sum of positive ends is positive
  can be read from \figref{fig:4}
  when $\gamma=0$, so that $\delta=\alpha+\beta$.
\end{proof}

For a finite end $\alpha$
we define
\begin{equation*}
  \infty+\alpha=\infty=\alpha+\infty.
\end{equation*}
It is not now clear whether an automorphism of the hyperbolic plane
must preserve parallelism
and therefore induce a permutation of the ends.
However, some automorphisms do this.

\begin{theorem}\label{thm:a-aut}
  Addition of a finite end $\alpha$
  is a permutation of $K\cup\{\infty\}$
  induced by an automorphism of the plane.
\end{theorem}

\begin{proof}
  That automorphism is $\refl{\alpha/2}\refl0$.
\end{proof}

\subsection{Multiplication and Inversion}

The choice of $O$ on $\myline{0,\infty}$
makes the points of this line into an abelian group under addition.
We make the positive ends into a multiplicative abelian group
isomorphic to this one under the map that takes $\alpha$
to the point where $\myline{\alpha,-\alpha}$
intersects $\myline{0,\infty}$;
the intersection is at right angles.
If $A$ is a point on $\myline{0,\infty}$
corresponding to a positive end $\alpha$,
then the segment $OA$ has a midpoint $B$,
corresponding to a positive end $\beta$,
and in this case
\begin{equation*}
  \beta=\surd\alpha.
\end{equation*}
We extend the multiplication to
all of $K\setminus\{0\}$
by the usual rules of signs.
We extend multiplication partially to $K\cup\{\infty\}$
as in \eqref{eqn:infty} in \S\ref{sect:hyp-aut}.

\begin{theorem}\label{thm:m-aut}
  \begin{compactenum}
    \item
    Multiplication by a positive end
    is a permutation of $K\cup\{\infty\}$
    induced by an automorphism of the plane.
\item
    Multiplication by $-1$ is induced by $\refl0$.
\item
    Inversion of ends is induced by reflection about $\myline{-1,1}$.
  \end{compactenum}
\end{theorem}

\begin{proof}
  Multiplication by a positive end $\gamma$
  is induced by the automorphism such that,
  as in \figref{fig:perm},
  \begin{figure}
    \centering
    \begin{pspicture}(10,7)
%          \psgrid
      \psset{PointSymbol=none,PointName=none}
      \pstGeonode(5,0.6){zero}(8,0.6){one}
      \pstHomO[HomCoef=2]{zero}{one}[lr]
      \pstHomO[HomCoef=1.6]{zero}{one}[gamma]
      \pstRotation[RotAngle=90]{zero}{one,gamma,lr}[O,C,mt]
      \pstRotation[RotAngle=180]{zero}{one,gamma}[mone,mgamma]
      \pstHomO[HomCoef=0.8]{zero}O[Q]
      \pstHomO[HomCoef=0.8]{zero}C[Q']
      \pstRotation[RotAngle=-55,PosAngle=180]{zero}{Q,Q'}[P,P']
      \psline[linestyle=dotted](mgamma)(gamma)
      \pstArcnOAB[linewidth=2.4pt]{zero}{Q'}{P'}
      \pstArcnOAB[linewidth=2.4pt]{zero}QP
      \psset{arrows=o-o}
      \pstArcOAB{zero}{one}{mone}
      \pstArcOAB{zero}{gamma}{mgamma}
      \psset{arrows=o->}
      \psline[linewidth=2.4pt](zero)(mt)
      \uput[d](zero){$0$}
      \uput[d](one){$1$}
      \uput[d](gamma){$\gamma$}
      \uput[d](mone){$-1$}
      \uput[d](mgamma){$-\gamma$}
      \uput[ul](O){$O$}
      \uput[ul](C){$C$}
      \uput[l](Q){$Q$}
      \uput[l](Q'){$Q'$}
      \uput[dr](P){$P$}
      \uput[dr](P'){$P'$}
      \uput[u](mt){$\infty$}
    \end{pspicture}
    \caption{Automorphism}
    \label{fig:perm}
      \end{figure}
  $O\cdot\gamma$ is the intersection of $\myline{0,\infty}$
  and $\myline{\gamma,-\gamma}$,
  and if $Q$ is the foot of the perpendicular from a point $P$
  dropped to $\myline{0,\infty}$,
  then $Q\cdot\gamma$ is the foot of the perpendicular from $P\cdot\gamma$
  dropped to $\myline{0,\infty}$, and, as segments,
  \begin{align*}
    (P\cdot\gamma)(Q\cdot\gamma)&=PQ,
    &Q(Q\cdot\gamma)&=O(O\cdot\gamma).\qedhere
  \end{align*}
\end{proof}

  As a special case of \eqref{eqn:a+b},
  \begin{equation*}%\label{eqn:ac+bc}
    \refl{\alpha\cdot\gamma+\beta\cdot\gamma}
    =\refl{\alpha\cdot\gamma}\refl 0\refl{\beta\cdot\gamma}.
  \end{equation*}
  Applying the automorphism of Theorem \ref{thm:m-aut} to \eqref{eqn:a+b},
  we obtain
  \begin{equation*}
    \refl{\alpha\cdot\gamma}\refl 0\refl{\beta\cdot\gamma}
    =\refl{(\alpha+\beta)\cdot\gamma}.
  \end{equation*}
  Thus
  multiplication distributes over addition on $K$.

Given a hyperbolic plane,
we have obtained an ordered field $K$.
It is straightforward that,
as Hartshorne shows
 \cite[43.1, p.\ 416--8]{MR1761093},
\begin{compactenum}[1)]
\item
  $K$ is independent (up to isomorphism)
  of the choice of $(\infty,0,1)$,
  since an automorphism of the plane will take one choice to another;
\item
  two hyperbolic planes with isomorphic fields
  are themselves isomorphic,
  since an isomorphism of lines determines an isomorphism of points.
\end{compactenum}
  
  \section{Coordinates}

  Having obtained
    an ordered field $K$
from a hyperbolic plane and
  a choice of ends denoted respectively
  $\infty$, $0$, and $1$,
  we shall describe the plane in terms of $K$.
  First we note which lines pass through $O$.

  \begin{theorem}\label{thm:-1}
    In the hyperbolic plane,
    \begin{compactenum}[1)]
      \item
      the line $\myline{\xi,\infty}$ contains $O$
      if and only if
      \begin{equation*}
        \xi=0,
      \end{equation*}
      \item
      the line $\myline{\xi,\eta}$, neither end infinite,
      contains $O$ if and only if
      \begin{equation*}%\label{eqn:xy=-1}
        \xi\cdot\eta=-1.
      \end{equation*}
    \end{compactenum}
  \end{theorem}

  \begin{proof}
    \begin{asparaenum}
      \item
      Clear.
    \item
      Reflection of the plane about
      $\myline{-1,1}$, and then $\myline{0,\infty}$,
      fixes
      \begin{compactitem}
      \item
        $O$, and only this, of all points of the plane;
      \item
        $\myline{\alpha,-\alpha\inv}$,
        and only this, of all lines $\myline{\alpha,\beta}$,
        whenever $\alpha\in K\setminus\{0\}$.
      \end{compactitem}
      Since the line does cross $\myline{0,\infty}$,
      it must cross at the point.\qedhere
    \end{asparaenum}
\end{proof}

\begin{theorem}\label{thm:Oa}
  The point $O\cdot\alpha$ lies on
   $\myline{0,\infty}$ and on
  every line
  $\myline{-\alpha\cdot\omega,\alpha\cdot\omega\inv}$,
  but no other line.
\end{theorem}

\begin{proof}
    The point is,
  by Theorem \ref{thm:m-aut},
  the intersection of $\myline{0,\infty}$
  and
  $\myline{-\alpha,\alpha}$.
  The rest follows also from Theorem \ref{thm:-1}.
\end{proof}


We want to know which lines pass through an arbitrary point $P$.
For this,
we seek an automorphism of the plane that takes $O$ to $P$.

\begin{theorem}
  In the hyperbolic plane,
  a line $\myline{\xi,\eta}$ that is not $\myline{0,\infty}$
  contains the intersection of
  $\myline{-\nu,\nu}$ and $\myline{\mu,\infty}$
  if and only if the point
  \begin{equation*}
    (\xi\eta,\xi+\eta)
  \end{equation*}
  in $K^2$ lies on the line defined by
\begin{equation}\label{eqn:xy}
  x-\mu y+\nu^2=0.
\end{equation}
\end{theorem}

\begin{proof}
  We show first that the intersection is $O\cdot\alpha+\mu$
  for \emph{some} $\alpha$,
  as in \figref{fig:P-aut}.
  \begin{figure}
    \centering
    \begin{pspicture}(10,6)
%      \psgrid
      \psset{PointSymbol=none,PointName=none}
      \pstGeonode(5,0.6){zero}(10,0.6){dr}(8,3)P
      \pstRotation[RotAngle=90]{zero}{dr}[mt]
      \pstProjection{zero}{dr}P[mu]
      \pstProjection{zero}{mt}P[A]
      \pstInterLC{zero}{mt}{zero}PnN
      \pstInterLC{zero}{dr}{zero}P{mnu}{nu}
      \pstTranslation{zero}{mt}{mu}[mui]
      \ncline[linestyle=dashed] AP
      \psset{arrows=<->}
      \pstArcOAB{zero}{nu}{mnu}
      \ncline{mu}{mui}
      \ncline{zero}{mt}
      \uput[d](zero){$0$}
      \uput[d](mu){$\mu$}
      \uput[d](nu){$\nu$}
      \uput[l](A){$O\cdot\alpha$}
      \uput[ur](P){$O\cdot\alpha+\mu$}
      \uput[ul](N){$O\cdot\nu$}
      \uput[u](mui){$\infty$}
      \uput[u](mt){$\infty$}
    \end{pspicture}
    \caption{Lines through an arbitrary point}
    \label{fig:P-aut}
  \end{figure}
  By applying the automorphisms of Theorems \ref{thm:a-aut} and \ref{thm:m-aut}
  to Theorem \ref{thm:Oa},
  we know that any point $O\cdot\alpha+\mu$
  is the intersection
  with $\myline{\mu,\infty}$
  of no other line, but every line
  \begin{equation*}
    \myline{-\alpha\cdot\omega+\mu,\alpha\cdot\omega\inv+\mu}.
  \end{equation*}
  To find which of these is orthogonal to $\myline{0,\infty}$,
  we solve
  \begin{gather*}
    -(-\alpha\cdot\omega+\mu)=\alpha\cdot\omega\inv+\mu,\\
    \alpha\cdot\omega^2-2\mu\cdot\omega-\alpha=0,\\
    \omega=\frac{\mu\pm\sqrt{\mu^2+\alpha^2}}{\alpha},\qquad
    \omega\inv=\omega-\frac{2\mu}{\alpha}.
  \end{gather*}
  Thus
    $O\cdot\alpha+\mu$
  is the intersection of
  $\myline{\mu,\infty}$ and
  \begin{equation*}
      \myline{-\sqrt{\mu^2+\alpha^2},\sqrt{\mu^2+\alpha^2}}.
  \end{equation*}
  Again by Theorem \ref{thm:Oa},
  $O\cdot\alpha+\mu$
  lies on $\myline{\xi,\eta}$
  if and only if
  \begin{gather*}
    \frac{\xi-\mu}{\alpha}\cdot
    \frac{\eta-\mu}{\alpha}=-1,\\
    \xi\eta-\mu(\xi+\eta)+\mu^2+\alpha^2=0.\qedhere
  \end{gather*}
\end{proof}

Three lines $\myline{\alpha,\beta}$,
different from $\myline{0,\infty}$,
have a common point if and only if the three points
$(\alpha\beta,\alpha+\beta)$
are collinear in $K^2$.
Thus we recover the hyperbolic plane from $K$.

Every line defined by an equation \eqref{eqn:xy}
corresponds to a point of the hyperbolic plane,
provided
\begin{equation*}
  -\nu<\mu<\nu.
\end{equation*}

Hilbert refers to $\bigl(\xi\eta,(\xi+\eta)/2\bigr)$
as the \textbf{coordinates} of the line $\myline{\xi,\eta}$,
and he concludes,
\begin{quote}
  Having seen that the equation of a point in line coordinates is linear
it is easy to deduce the special case of Pascal's Theorem for a pair of
lines and Desargues' Theorem for perspectively situated triangles as well
as the other theorems of projective geometry.
\end{quote}
``Pascal's Theorem for a pair of lines'' is Pappus's Theorem,
which, along with Desargues's Theorem,
holds in the projective plane over $K$.
Therefore the duals also hold,
and so the theorems themselves hold in the hyperbolic plane.

Hilbert continues:
\begin{quote}
The familiar formulas of
Bolyai-Lobachevskian geometry can then also be derived with no
difficulty and the development of this geometry has been thus
completed with the aid of Axioms I-IV alone.
\end{quote}
Hartshorne remarks \cite[p.\ 394]{MR1761093},
\begin{quote}
  we have found that calculations seem to work out better
  if we continue to think of a line as given by coordinates
  \lips
  and a point as given by an equation.
  This is the opposite of the analytic geometry
  we are used to\lips
\end{quote}
He derives the construction of parallels in the next section,
and some formulas,
though without discussing Pascal and Desargues,
as far as I can tell,
except for a brief mention of the former
\cite[p.\ 426]{MR1761093},
\begin{compactenum}[1)]
\item
  as proved by Hjelmslev in the projective plane,
  in which he embeds an arbitrary Hilbert plane;
\item
  as shown by Hilbert to be equivalent to commutativity
  of the field of segment arithmetic.
\end{compactenum}
This is in the context of an historical sketch of doing geometry
without continuity,
which ``reached its modern form in the book of Artin \cite{Artin}.''

\section{Bolyai's Parallel Construction}\label{sect:Bolyai}

We establish a ruler-and-compass construction of parallels,
assuming they exist.

\begin{theorem}
  In the hyperbolic plane,
  if
  \begin{compactitem}
  \item
    perpendiculars $PS$ and $QR$ to $PQ$ are erected,
    \item
      the perpendicular dropped from $R$ to $PS$ meets this at $S$,
      \item
    the parallel to $QR$ from $P$ meets $SR$ at $T$,
  \end{compactitem}
  then
  \begin{equation*}
    PT=QR.
  \end{equation*}
\end{theorem}

\begin{proof}
  Since $PQ\perp PS$,
  we may assume, as in \figref{fig:bpc},
  \begin{figure}
    \centering
    \begin{pspicture}(-4,-0.6)(6,6)
      \psset{PointSymbol=none}
      \newcommand{\myu}{2.5}
      \newcommand{\mya}{1.6}
      \newcommand{\myb}{2.4}
      \pstGeonode[PointName={none,none,default,default,none,none,none},
        PosAngle={0,0,-90,-90,0,0,0}]
      (0,0){zero}
      (! \myu\space 0){one}
      (! \myu\space \mya\space mul 0){a}
      (! \myu\space \myb\space mul 0){b}
      (! \myu\space \mya\space div neg 0){mai}
      (! \myu\space \myb\space div 0){bi}
      (0,5.4){infty}
      \uput[u](infty){$\infty$}
      \uput[d](zero){$0$}
      \uput[d](one){$1$}
      \uput[d](mai){$-a\inv$}
      \uput[d](bi){$b\inv$}
      \psset{PointName=none}
      \pstMiddleAB{mai}{a}{amid}
      %    \pstArcOAB{amid}{a}{mai}
      \pstTranslation{one}{zero}{zero}[mone]
      \uput[d](mone){$-1$}
      \pstTranslation{a}{zero}{zero}[ma]
      \uput[d](ma){$-a$}
      \pstArcOAB{zero}{one}{mone}
      \pstArcOAB{zero}{a}{ma}
      \pstMiddleAB{b}{bi}{bmid}
      \psset{PointName=default}
      \pstInterCC[PointNameA={},PosAngleB=75]{bmid}{bi}{zero}{a}{R'}{R}
      \pstArcOAB{bmid}{b}{bi}
      \pstRotation[RotAngle=90,PosAngle=135]{zero}{a}[Q]
      \ncline[arrows=->]{zero}{infty}
      \pstInterLC[PointNameA={},PosAngleB=135]{zero}Q{zero}{one}{P'}{P}
      \pstInterCC[PointNameA={},PosAngleB=180]{bmid}{bi}{zero}{one}{S'}{S}
      \pstInterCC[PointNameA={},PosAngleB=-90]{bmid}{bi}{amid}{a}{T'}{T}
      \psset{linewidth=2.4pt}
      \pstArcOAB{zero}SP
      \pstArcOAB{zero}RQ
      \ncline PQ
      \pstArcOAB{bmid}RS
      \pstArcOAB[linestyle=dotted]{amid}{a}{mai}
    \end{pspicture}
    \caption{Parallel construction}
    \label{fig:bpc}  
  \end{figure}
  \begin{align*}
    PQ&=\myline{0,\infty},
    &PS&=\myline{-1,1},
    &P&=O.
  \end{align*}
  Since also $PQ\perp QR$,
  for some finite end $a$,
  \begin{equation*}
    QR=\myline{-a,a}.
  \end{equation*}
  Since $PS\perp RS$, for some finite end $b$,
  \begin{equation*}
    RS=\myline{b,b\inv}.
  \end{equation*}
  Then finally, since $PT$ passes through $O$
  and is parallel to  $QR$, by Theorem \ref{thm:Oa},
  \begin{equation*}
    PT=\myline{-a\inv,a}.
  \end{equation*}
  Rotation of the plane about $P$
  so that $a$ goes to $0$
  is effected by the automorphism
  \begin{equation*}
    x\mapsto\frac{x-a}{ax+1},
  \end{equation*}
  since
  \begin{equation*}
    \frac{-x\inv-a}{-ax\inv+1}
    =-\frac{1+ax}{-a+x}.
  \end{equation*}
  First translating $Q$ to $P$,
  then rotating about $P$
  so that $1$ goes to $0$,
  is effected by the automorphism
  \begin{equation*}
    x\mapsto\frac{x-a}{x+a}.
  \end{equation*}
  These send $\myline{b,b\inv}$
  to lines that cross $PQ$ at the same point,
  since the products of their ends are the same:
  \begin{align*}
    \frac{x-a}{ax+1}\cdot\frac{x\inv-a}{ax\inv+1}
    &=\frac{x-a}{ax+1}\cdot\frac{1-ax}{a+x}\\
    &=\frac{x-a}{a+x}\cdot\frac{1-ax}{1+ax}
    =\frac{x-a}{x+a}\cdot\frac{x\inv-a}{x\inv+a}.\qedhere
  \end{align*}
\end{proof}

According to Hartshorne, by the classification of Hilbert planes
(satisfying Axioms I--III) by Pejas,
the construction must work in any such plane
satisfying Archimedes's Axiom and also
the axiom of intersection of two circles
(they intersect
if each encircles a point of the other).

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\end{document}
