\chapter[Non-Euclidean geometry]{Spherical and hyperbolic non-Euclidean
geometry}
\label{ch!hyp}
\index{non-Euclidean geometry|(}
\index{hyperbolic!geometry}
\index{spherical!geometry|(}
\markboth{\quad\thechapter\ Non-Euclidean geometries\hfill}{}

Together with plane Euclidean geometry, spherical and hyperbolic geometry
are 2-dimensional geo\-metries with the following properties:

\begin{enumerate}\renewcommand{\labelenumi}{(\arabic{enumi})}

\item distance, lines and angles are defined and invariant under motions;

\item the motions\index{motion} act transitively on points and directions
at a point;

\item locally, incidence\index{incidence of lines} properties are as in
plane Euclidean geometry.\index{Euclidean!geometry}
\end{enumerate}

In more detail, (2) means that if $P,P'$ are points, and $\la,\la'$
directions at these points, then there exists a motion\index{motion} $T$
taking $P$ to $P'$ and $\la$ to $\la'$; in other words, the geometry is
\begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}}
\centering
\input{hyp_homog_geom.pstex_t} \caption{Plane-like geometry}
\label{fig!plane-like}
\end{figure}
homogeneous (the same at every point) and isotropic (the same in every
direction). (3) means that in sufficiently small open sets, a line is
uniquely specified by a point and a direction, or by two points $P, Q$, and
two lines $l_i$ meet in at most one point (see Figure~\ref{fig!plane-like}).

However, the geometries also differ in several respects:
\begin{enumerate}\renewcommand{\labelenumi}{(\arabic{enumi})}

\item the global incidence\index{incidence of lines} properties of lines,
that is, the existence of parallel and nonintersecting
lines;\index{parallel!lines}

\item intrinsic curvature\index{curvature}\index{intrinsic!curvature}
properties: the perimeter of a circle, and the sum of angles in a
triangle;

\item the possibility of defining a unit of length
intrinsic\index{intrinsic!unit} to the geometry.
\end{enumerate}

Euclidean geometry\index{Euclidean!geometry} in the plane was described in
detail in Chapter~\ref{ch!eucl}. Although certainly not the same thing as
plane geometry, spherical geo\-metry\index{spherical!geometry} is still very
intuitive, because every definition and statement can be readily visualised
on the very concrete model $S^2\subset\R^3$, which you can hold in your hand
or kick around a playing field. I discuss spherical lines (great circles),
distances, angles and triangles, the classification of motions\index{motion}
in terms of rotations\index{rotation} and reflections,\index{reflection}
frames of reference\index{spherical!frame} and angular
excess.\index{angle!sum}

In contrast, plane hyperbolic geometry\index{hyperbolic!geometry} originally
arose in axiomatic geo\-metry (compare~\ref{sec!parallels}); the
coordinate model I treat in this chapter is not immediately familiar, and
was discovered many decades after axiomatic
hyper\-bolic\index{hyperbolic!geometry} geometry. Although my model of
hyperbolic geometry is not intuitive, essentially every step in my treatment
is parallel to spherical geo\-metry. Once you are sure you know what you are
doing, you can just replace $x^2+y^2=1$ by $-t^2+x^2=-1$, and the trig
functions $\sin$ and $\cos$ by the hyperbolic trig
functions\index{hyperbolic!trig} $\sinh$ and $\cosh$, and everything extends
more or less word-for-word. This is the essential content of the prophetic
suggestion by J.~H. Lambert (1728--1777) that non-Euclidean
geometry\index{non-Euclidean geometry} ``should be related to the geometry on
a sphere of radius $i=\sqrt{-1}$'' (see Coxeter \cite{Cx}, p.~299).

In Chapter~\ref{ch!eucl} on Euclidean geometry, I discussed $n$-dimensional
Euclidean space $\E^n$\index{Euclidean!space $\E^n$} along with the more
familiar planar version. There is no logical reason to discontinue this
practice, but for ease of digestion as well as notation, all definitions in
this chapter are given in two dimensions. You will benefit immensely by
generalising the definitions and, in some cases, the theorems to the higher
dimensional setup; you are explicitly encouraged to do so in
\exref{ex!Snsetup}. Higher dimensional spheres appear in later chapters
(see for example~\ref{comp} and~\ref{sec!topson}); unfortunately there is no
space in the book for a detailed treatment of higher dimensional hyperbolic
space\index{hyperbolic!space} and a discussion of its significance.

\clearpage

\section{Basic definitions of spherical geometry}
\label{defn of spherical}
\index{spherical!geometry}
\markright{\hfill \thesection\ Basic definitions of spherical
geometry\quad}
The sphere\index{sphere $S^2$} $S^2\subset\R^3$ of radius $r$ centred at the
origin $O$ is defined by the equation $x^2+y^2+z^2=r^2$. I will often refer
to points $P\in S^2$ via their position vector $\vec{OP}=\bp$. A {\em
spherical line} or {\em great circle} \index{line!spherical} in $S^2$ is the
intersection of $S^2$ with a plane $\Pi=\R^2$ through the origin; thus it is
a circle in $\Pi$ centred at $O$ and with the same radius $r$ as $S^2$. Two
points $P,Q\in S^2$ are {\em antipodal} if their position vectors $\bp, \bq$
satisfy $\bp=-\bq$. Through any two distinct points $P,Q\in S^2$ which are
not antipodal, there is a unique great circle or spherical line $L=PQ$. The
{\em (spherical)
distance}\index{distance!function}\index{spherical!geometry} $d(P,Q)$
between points $P,Q\in S^2$ is the distance measured along the shorter arc
of a great circle through $P$ and $Q$; that is, it is radius $r$ times
$\angle POQ$, the angle at $O$ between $OP$ and $OQ$, where the angle is
always interpreted as the absolute value in the range $[0,\pi]$. For ease of
notation, I usually fix the radius $r=1$ from now on.

\begin{rems}\renewcommand{\labelenumi}{(\arabic{enumi})}
\begin{enumerate}
\item If you go back to the chapter on Euclidean geometry and compare the
treatment of \ref{Rn}--\ref{En} to the one given here, you may notice that I
have been a bit sloppy here. To be consistent, I should have defined
``model'' $S^2$\index{sphere $S^2$} to be the sphere $\{x^2+y^2+z^2=r^2\}$
in $\R^3$ with its inherited spherical distance,\index{spherical!distance}
and ``abstract'' $S^2$ to be a metric space isometric to ``model'' $S^2$ but
without a fixed choice of identification. Spelling this out explicitly leads
to rather clumsy notation, but implicitly I am still following this
procedure; in particular, I reserve the right to choose different
coordinates on my ``abstract'' metric $S^2$ if so needed. This remark
applies equally well to the treatment of hyperbolic
geometry\index{hyperbolic!geometry} in \ref{sec!hdist} below.

\item The sphere $S^2$ is defined as the subset $\{x^2+y^2+z^2=1\}$ of
$\R^3$. On the northern hemisphere $\{z\ge0\}$ I can rewrite this as
$z=\sqrt{1-x^2-y^2}$. This gives a fairly good coordinate representation of
$S^2$ near the north pole, but a fairly bad one in moderate or tropical
regions. What is wrong with it? Well, if the model is the whole of $\R^2$,
it is much too big; if we take only the disk $D^2:x^2+y^2\le1$, crossing the
equator in $S^2$ corresponds to falling off the edge of the world in the
model. Furthermore, distances, angles, areas, curvature are all screwed up.

It is a basic problem in cartography to map regions of the surface of the
Earth onto a plane. However, the map based on $z=\sqrt{1-x^2-y^2}$ is one of
the most primitive and useless ways to do this. Over the course of time,
several much better ways have been invented; see the reference
under~\ref{sec!titles}, (\ref{entry!maps}) for a starting point on this.

\item The distance $d(P,Q)$ is defined as (radius times) the angle of the
$PQ$ arc, $\al=\angle POQ$. It is useful to know how to translate between
this {\em angle} and the {\em coordinates} of $P,Q$. In vector notation, the
dot product of unit vectors equals the cosine of the angle between them:
that is, if $P, Q$ have position vectors $\bp, \bq$ then $\al=\angle POQ$ is
given by
\begin{equation}
\bp\cdot \bq=\cos\al,\quad\hbox{that is,}\quad
d(P,Q)=\al=\arccos(\bp\cdot\bq).
\label{eq!arccos}
\end{equation}
(I have set $r=1$, so that $\bp$ and $\bq$ are unit vectors.) Recall that
$\arccos=\cos\1$ is the inverse function of $\cos$, so that $\al=\arccos x$
is defined by the property $x=\cos\al$; similarly for $\arcsin$. Here I
choose $\al$ in the range $[0,\pi]$.

Given $P$ and $Q$, I can choose coordinates so that $P=(0,0,1)$ and $OPQ$ is
the $(x,z)$-plane $\{y=0\}$; then $Q=(\sin\al,0,\cos\al)$. This is a
parametrisation of the great circle, with parameter $\al$. Points with $x<0$
can also be included, by allowing $\al<0$ to run through the range
$[-\pi,\pi]$, but then $d(P,Q)=|\al|$.

In fact $(\sin\al,0,\cos\al)$ is a parametrisation {\em by arc length}: if you
think of (part of) the sphere $S^2$\index{sphere $S^2$} as the graph of
$z=\sqrt{1-x^2-y^2}$ as in (2), then $d(P,Q)=\int_P^Q\dd s$ where the
infinitesimal arc length $\dd s$ is determined by $\dd s^2=\dd x^2+\dd
y^2+\dd z^2$. Thus the length of arc $PQ$ is
\[
\int_0^{\sin\al} \frac{\dd x}{\sqrt{1-x^2}}=\arcsin (\sin \al)=\al.
\]
\end{enumerate} \end{rems}

Geometers like to distinguish the {\em intrinsic} geometric properties of
$S^2$ from those related to the embedding $S^2\subset\R^3$. It is important
in this context to notice that the natural distance in spherical
geometry\index{spherical!distance} is the intrinsic distance, that is, the
length of a certain curve traced in the surface $S^2$, as opposed to the
distance in the ambient Euclidean space; you go from London to Singapore by
plane, not by tunnel.

\section{Spherical triangles and trig}
\label{s2trig}
\index{spherical!trig}
\markright{\hfill \thesection\ Spherical triangles and trig\quad}
The convention $r=1$ is still in force. A {\em spherical
triangle}\index{spherical!triangle} $\triangle PQR$ consists of 3 vertexes
$P,Q,R$ and 3 arcs of great circle $PQ$, $PR$, $QR$ joining them. These do
not have to be the shorter arcs; $P$, $Q$ are allowed to be antipodal, and
then you have to specify one of the great circles to be the arc $PQ$.

The {\em spherical angle} $a$ at $P$ between the two lines $PQ$ and $PR$ is
equal to the dihedral angle between the two planes $OPQ$, $OPR$ in $\R^3$,
in other words it is the angle between two lines cut out by the two planes
in an auxiliary plane orthogonal to $OP$. You can take this as a definition
if you like, and then you do not have to worry about how the angle between
two curves is defined. More precisely, the tangent plane to $S^2$ at $P$ is
the 2-plane $T_PS^2$ defined by $z=1$, and the tangent vectors to the two
curves $PQ$ and $PR$ are the two lines in $T_PS^2$ cut out by these two
planes. They are orthogonal\index{orthogonal!vector} to the axis $OP$, so
the angle between the two curves equals the dihedral angle $a$ between the
two planes.

\begin{prop}[main formula of spherical trig]\index{spherical!trig} The side
$QR$ of the triangle is de\-ter\-mined by the other two sides $PQ$ and $PR$
and the dihedral angle $a$. More precisely, write
\[
\al=\angle QOR=d(Q,R),\quad
\be=\angle POQ=d(P,Q),\quad
\ga=\angle POR=d(P,R).
\]
(Recall that I have fixed the radius $r=1$.) Then
\begin{equation}
\cos\al=\cos\be\cos\ga+\sin\be\sin\ga\cos a.\label{eq!sph_trig}
\end{equation}

\end{prop}

\begin{figure}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}}
\centering
\leavevmode\hbox{\epsfysize=5 true cm\epsffile{sph_trig.eps}}
\caption{Spherical trig}
\label{fig!sph_trig}
\end{figure}

\begin{pf} Although the statement looks complicated, the proof is easy 3D
coordinate geometry. In Figure~\ref{fig!sph_trig}, let $Q'$ and $R'$ be the
points on great circles at distance $\pi/2$ from $P$, so that $\vec{OQ'}$ is
orthogonal\index{orthogonal!vector} to $OP$. Choose coordinates $(x,y,z)$ so
that $P=(0,0,1)$ (the north pole), and the equator is given by $z=0$. Then
$Q'$ is a point on the equator, so I can choose $Q'=(1,0,0)$, and $R'=(\cos
a,\sin a,0)$. This determines the coordinates of all the points in the
figure; by definition of $\be,\ga$, the following relations hold between the
position vectors:
\[
\begin{array}{ccccl}
\bq&=&\cos\be \bp+\sin\be \bq'&=&(\sin\be,0,\cos\be),\\
\br&=&\cos\ga \bp+\sin\ga \br'&=&(\sin\ga\cos a,\sin\ga\sin a,\cos\ga).
\end{array}
\]
Now $\al$ is the angle between the two unit vectors $\bq$ and $\br$,
so
\[
\cos\al=\bq\cdot \br=\cos\be\cos\ga+\sin\be\sin\ga\cos a.\QED
\]
\end{pf}

\section{The spherical triangle inequality}
\label{s2trineq}
\index{spherical!triangle|(}
\index{triangle inequality}
\markright{\hfill \thesection\ The spherical triangle
inequality\quad}
\begin{cor}[Triangle inequality] In any triangle $\triangle PQR$ whose sides
are shorter arcs given by $\al,\be,\ga\le\pi$ as above,
\[
\al\le\be+\ga,
\]
with equality if and only if $PQR$ are collinear with $P$ on the shorter arc
$QR$.
\end{cor}

\begin{pf} This follows at once from the main formula (\ref{eq!sph_trig})
and calm reflection on the range of values for the angles $\al,\be,\ga$ and
$a$. Notice that $\al,\be,\ga\in[0,\pi]$ essentially by convention: in
defining distance I always take $\angle POQ$ to be the angle in the shorter
arc. If $\be$ or $\ga=0$ or $\pi$, it is easy to read off the conclusion, so
that I can assume that $\al,\be,\ga\in(0,\pi)$. On the other hand, in
Figure~\ref{fig!sph_trig}, it is clear I want to have $a\in[0,2\pi)$. Now
compare (\ref{eq!sph_trig}) with the standard trig formula
\[
\cos(\be+\ga)=\cos\be\cos\ga-\sin\be\sin\ga.
\]
We know that $\sin\be,\sin\ga\in(0,1]$; thus $\cos\al\ge\cos(\be+\ga)$, with
equality if and only if $\cos a=-1$. Now $\cos\al$ is a strictly decreasing
function in the range $[0,\pi]$, so that $\cos\al\ge\cos(\be+\ga)$ gives
$\al\le\be+\ga$. Equality holds only under the aforestated condition $\cos
a=-1$, that is, if the short arcs $PQ$ and $PR$ are opposite when viewed
from $P$. \QED \end{pf}

It is trivial that $d(P,Q)$ is symmetric, nonnegative, and positive unless
$P=Q$, so that Corollary~\ref{s2trineq} proves that $S^2$ with the spherical
distance is a metric space (see Appendix~\ref{app!metric}).
\index{spherical!triangle|)}
\index{metric!space}

\section{Spherical motions}\label{s2d}
\index{motion}
\index{spherical!motion}
\markright{\hfill \thesection\ Spherical motions\quad}

A {\em spherical motion}\index{motion} or {\em isometry} is of course just a
map $T\colon S^2\to S^2$ preserving spherical
distance.\index{spherical!distance}

\begin{thm}
\begin{enumerate}\renewcommand{\labelenumi}{(\arabic{enumi})}

\item A motion\index{motion} $T\colon S^2\to S^2$ takes pairs of antipodal
points to pairs of antipodal points, and spherical lines (great
circles)\index{spherical!line} to spherical lines.

\item Any motion\index{motion} is given in coordinates by $\bx\mapsto
A\bx$, where $A$ is a $3\times3$ orthogonal matrix.
\index{orthogonal!matrix}
\end{enumerate}
\end{thm}

\begin{pf} Two points of the sphere are antipodal if and only if they are a
maximum distance apart (at distance $\pi r$, half a world away), so the
first sentence is clear. The rest of the proof is very similar to the
Euclidean proof in Chapter~\ref{ch!eucl}. For (1), exactly as in
Corollary~\ref{sec!mcol}, the arcs of spherical lines\index{spherical!line}
$[P,Q]$ are determined purely by the metric: three points $P,Q,R$ are
collinear (that is, on a spherical line or great circle) if and only if
\[
d(P,Q)+d(Q,R)+d(R,P)=2\pi r \quad\hbox{or}\quad
\pm d(P,R)\pm d(R,Q)\pm d(P,Q)=0.
\]
Here the first equality is the statement that $P,Q,R$ are on a great circle
and not in any shorter great arc, and the second is the equality case of
Corollary~\ref{s2trineq} for some permutation of $P,Q,R$. A spherical
motion\index{spherical!motion} $T$ preserves these equalities, so takes a
spherical line\index{spherical!line} $L$ to a spherical line $L'=T(L)$.

For (ii), note first that because $T\colon S^2\to S^2$ takes antipodal
points to antipodal points, it extends in a unique way to a map
$T\colon\R^3\to\R^3$ by radial extension. I claim that $T$ is linear. For
this, it is enough to see that $T$ is linear when restricted to any plane
$\Pi$ through the origin.

Suppose $L=\Pi\cap S^2$ and $T(L)=L'=\Pi'\cap S^2$. A spherical
line\index{spherical!line} $L=\Pi\cap S^2$ is parametrised by arc length: a
variable point of $L$ is $\cos\theta \bbf_1+\sin\theta\bbf_2$, where
$\bbf_1,\bbf_2,\bbf_3$ is an orthogonal basis of $\R^3$ with
$\bbf_1,\bbf_2\in L$, and $\theta$ equals the arc length along $L$. Since
$T$ preserves distance,\index{preserves distances} it preserves arc length
along a spherical line, so that its restriction $T_L\colon L\to L'$ is given
by
\[
T(\cos\theta \bbf_1 + \sin\theta \bbf_2)
=\cos\theta \bbf'_1 +\sin\theta \bbf'_2.
\]
Here $\bbf'_1,\bbf'_2,\bbf'_3$ is a new orthogonal
frame,\index{orthogonal!frame} with $\bbf_1'=T(\bbf_1)$ and
$\bbf_2'=T(\bbf_2)\in L'$. Stated differently, $T(\la \bbf_1+\mu \bbf_2)=\la
\bbf'_1+\mu \bbf'_2$, so $T$ is linear. \QED \end{pf}

\section{Properties of $S^2$ like $\E^2$}\label{sec!s2ve2}
\markright{\hfill \thesection\ Properties of $S^2$ like $\E^2$\quad}
The following statements are either obvious, or can be done as easy
exercises. Use them to refresh your memory of the case of $\E^2$, or as a
warm-up for the case of the hyperbolic plane
$\sH^2$.\index{hyperbolic!plane $\sH^2$} The spherical statements are if
anything a little simpler: for example, the distinction between
translation\index{translation} and rotation\index{rotation} disappears,
and the classification of motions\index{motion} comes directly from the
normal form of Theorem~\ref{sec!norm}.

\begin{enumerate}

\item The sphere $S^2$\index{sphere $S^2$} is a metric geometry with
a distance function $d(P,Q)$, and motions\index{motion} given by
$3\times3$ orthogonal matrixes.\index{orthogonal!matrix}

\item The motions act transitively on $S^2$ and on spherical
lines\index{spherical!line} through a given point $P\in S^2$.

\item Every motion\index{motion} of $S^2$ is either a
rotation\index{rotation} $\Rot(P,\theta)$, or a
reflection\index{reflection} $\Refl(L)$ in a line (= great circle) or
a glide $\Glide(L,\theta)$ (the restriction of a Euclidean rotary
reflection).\index{rotary reflection}

\item Given two pairs of points $P,Q$ and $P',Q'$, there exist
exactly two motions\index{motion} $g$ of $S^2$ such that $g(P)=g(P')$,
$g(Q)=g(Q')$, of which one is a rotation\index{rotation} and the
other a reflection or glide.\index{glide}

\item Motions come in two kinds, direct and opposite. Every direct
motion is the identity or a composite of 2 reflections; every opposite
motion is a reflection or a composite of 3
reflections.\index{reflection}

\item The spherical distance $d(P,Q)$\index{spherical!distance}
between two points $P,Q\in S^2$ is the length of the shortest
curve\index{shortest distance} $C$ in $S^2$ joining $P$ and $Q$.
\end{enumerate}

\section{Properties of $S^2$ unlike $\E^2$}\label{sec!S2unlikeE2}
\markright{\hfill \thesection\ Properties of $S^2$ unlike
$\E^2$\quad}

\begin{enumerate}

\item Incidence of lines.\index{incidence of lines} Any two spherical
lines\index{spherical!line} intersect in a pair of antipodal points.
(Proof: if $L_1=\Pi_1\cap S^2$ and $L_2=\Pi_2\cap S^2$, consider the
Euclidean line $\Pi_1\cap\Pi_2$ in $\R^3$.) Therefore {\em spherical
geometry\index{spherical!geometry} has no parallel
lines}.\index{parallel!lines}

\item Intrinsic distance. If you live on $S^2$, it makes sense to take the
circumference of $S^2$ (or the length of any great circle) as a unit of
distance; recall that the kilometre, adopted during the French revolution,
was defined by setting the circumference of our own parochial
sphere\index{sphere $S^2$} to be 40,000 km. Another aspect of the same
phenomenon is that distances are bounded: $d(P,Q)\le\pi r$ $(=: 20,000\
\hbox{km})$.

\item Spherical frames:\index{spherical!frame} If you try to define a
spherical frame of reference by analogy with the Euclidean
notion,\index{Euclidean!frame} you get involved with the intrinsic
distance.\index{intrinsic!distance} For example, if your unit of
measurement is very big compared to
the radius of the sphere,\index{sphere $S^2$} you will end up with your
unit vector $P_0Q_0$ wrapping the sphere several times. Taking a small unit
of measurement, you can define a spherical frame $P_0P_1P_2$ and prove the
analogue of Corollary~\ref{sec!E2frame} (a motion\index{motion} takes any
frame into any other, and is uniquely determined by what it does to a
frame) as an easy exercise. But there is an even better solution, which
actively exploits the intrinsic distance: I can take the length $P_0P_1$ to
be 1/4 of the circumference, and get a spherical frame which coincides with
an orthonormal frame of the ambient $\R^3$, so that the result about
motions\index{motion} and frames is contained in
Proposition~\ref{sec!E2frame}.

\item Intrinsic curvature.\index{curvature}\index{intrinsic!curvature} To
say that the sphere
$S^2\subset\R^3$ is curved, you could calculate the radius of curvature of
lines relative to the ambient space $\R^3$. However, the geometry of $S^2$
also displays intrinsic curvature, as you can see in several ways. In
$\E^2$ the perimeter of a Euclidean circle of radius $\rho$ is $2\pi\rho$.
By contrast, a spherical circle of radius $\rho$ has perimeter
$2\pi\sin\rho$, as discussed in \exref{S2circle}.

\item Sum of angles in a triangle.\index{angle!sum}\index{area|(} Let $S^2$
be the sphere\index{sphere $S^2$} of radius $r=1$, and $\triangle PQR$ a
spherical triangle.\index{spherical!triangle} Then
\[
\angle P+\angle Q+\angle R=\pi+\area\triangle PQR.
\]
Thus the sum of angles in a spherical triangle\index{spherical!triangle}
{\em never} equals $180^\circ$. For very small triangles, you can view the
discrepancy as a reflection of intrinsic curvature as in the preceding
point.
\end{enumerate}

\begin{pf} I prove the last point, because it is not obvious at
first sight, and because the proof is very elegant. It is a ``Venn
diagram'' argument on the partition of $S^2$ obtained by slicing it up
along the great circles which are the sides of $\triangle PQR$. Write
$\Si_a$ for the part of $S^2$ contained between the two planes $OPQ$ and
$OPR$ (that is, the union of the two opposite segments) with $a$ the dihedral
angle between these planes, and similarly for $\Si_b$ and $\Si_c$.
Then by circular symmetry, clearly
\begin{equation}
\area\Si_a=\frac{2a}{2\pi}\cdot\area S^2.
\label{eq!segm}
\end{equation}
Now I claim that $\Si_a,\Si_b,\Si_c$ cover $S^2$ and overlap exactly in
$\triangle PQR$ and its antipodal triangle $\triangle P'Q'R'$. (See
Figure~\ref{fig!overlap}.)

\begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}}
\centering
\leavevmode\hbox{\epsfysize=5 true
cm\epsffile{sph_triangles.eps}}
\caption{Overlapping segments of $S^2$}
\label{fig!overlap}
\end{figure}

Summing (\ref{eq!segm}) for $\Si_a$, $\Si_b$ and $\Si_c$ gives
\[
\area S^2+4\area\De = \area\Si_a+\area\Si_b+\area\Si_c
=(2a+2b+2c)\cdot\frac{\area S^2}{2\pi}
\]
(points in $\De$ and its antipodal triangle are covered 3 times, while
the rest of $S^2$ is covered once). Therefore
$a+b+c-\pi=\displaystyle\frac{4\pi}{\area S^2}\area\De=\area\De$. \QED
\end{pf} \index{spherical!geometry|)}\index{area|)}

\section{Preview of hyperbolic geometry}\label{sec!prev}
\index{hyperbolic!geometry|(}
\markright{\hfill \thesection\ Preview of hyperbolic geometry\quad}
The remainder of this chapter introduces a coordinate model for hyperbolic
geometry which is entirely parallel to spherical geometry. First, I review
the ingredients of spherical geometry\index{spherical!geometry} in one
dimension:
\begin{enumerate}

\item $\R^2$ with coordinates $x,y$ and the ordinary Euclidean norm
$x^2+y^2$.

\item The functions $\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}$
and $\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$, which satisfy
the relation $\cos^2+\sin^2=1$, and $\frac{\dd\
}{\dd\theta}\sin\theta= \cos\theta$, $\frac{\dd\
}{\dd\theta}\cos\theta=-\sin\theta$.

\item The circle $S^1$ defined by $x^2+y^2=1$ is parametrised by
$x=\sin\theta$, $y=\cos\theta$, and the arc length is $\sqrt{\dd
x^2+\dd y^2}=\dd\theta$, so that $\theta$ is the arc length parameter
for $S^1$.

\item Symmetries are the set $\O(2)$ of rotation and reflection
matrixes\index{reflection!matrix}\index{rotation!matrix}
\[
\begin{pmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta
\end{pmatrix}
\quad\hbox{and}\quad
\begin{pmatrix} \cos\theta&\sin\theta\\ \sin\theta&-\cos\theta
\end{pmatrix}.
\]
\end{enumerate}

\noindent Now the ingredients of hyperbolic geometry in one dimension:
\begin{enumerate}

\item $\R^2$ with coordinates $t,x$ and the Lorentz
pseudometric\index{Lorentz!pseudometric} $-t^2+x^2$. Here I choose a
``time-like'' coordinate $t$ and a ``space-like'' coordinate $x$. A vector
is {\em space-like} if it has positive squared length (for example $(0,x)$)
and {\em time-like} if it has negative square (for example, $(t,0)$ has
squared length $-t^2$).

The Lorentz space\index{Lorentz!space} $\R^2$ is the ambient space for the
hyperbola $\sH^1$ defined by $t^2=1+x^2$ and $t>0$ (see
Figure~\ref{fig!h1}). The tangent space to $\sH^1$ at any point
$P_0=(t_0,x_0)\in\sH^1$ is the line $t=(x_0/t_0)x$, which is space-like,
because $t_0>|x_0|$. Therefore although the Lorentz
pseudometric\index{Lorentz!pseudometric} $-t^2+x^2$ is not positive
definite, the geometry of $\sH^1$ itself contains only space-like
directions.

\begin{figure}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}}
\centering
\input{hyp_space.pstex_t}
\caption{The hyperbola $t^2=1+x^2$ and $t>0$}
\label{fig!h1}
\end{figure}

\item The functions $\cosh s=\frac{e^s+e^{-s}}{2}$ and $\sinh
s=\frac{e^s-e^{-s}}{2}$, which satisfy the relation $\cosh^2-\sinh^2=1$, and
$\frac{\dd\ }{\dd s}\sinh s=\cosh s$, $\frac{\dd\ }{\dd s}\cosh s=\sinh s$.
It is useful to notice that $\sinh$ is a one-to-one map from the whole of
$\R^1$ to the whole of $\R^1$.

\item The hyperbola $\sH^1$ defined by $t^2=1+x^2$ is parametrised by
$x=\sinh s$, $t=\cosh s$, and the arc length in the Lorentz pseudometric is
$\sqrt{-\dd t^2+\dd x^2}=\dd s$, so that $s$ is the arc length parameter
for $\sH^1$.

\item Symmetries are the set $\O^+(1,1)$ of Lorentz
translation\index{Lorentz!matrix} and reflection matrixes
\[
\begin{pmatrix} \cosh s&\sinh s\\ \sinh s&\cosh s \end{pmatrix}
\quad\hbox{and}\quad
\begin{pmatrix} \cosh s&-\sinh s\\ \sinh s&-\cosh s \end{pmatrix}.
\]
\end{enumerate}

\section{Hyperbolic space}\label{sec!hspace}\index{hyperbolic!space}
\markright{\hfill \thesection\ Hyperbolic space\quad}
Consider $\R^3$ with the Lorentz\index{Lorentz!form $q_L$} quadratic
form\index{quadratic form} $q_L(\bv)=-t^2+ x^2 + y^2$
(compare~\ref{sec!euclandlor}). The cone $\{q_L(\bv)<0\}$ breaks up into two
subsets
\[
\{t>+\sqrt{x^2+y^2}\}\cup\{t<-\sqrt{x^2+y^2}\}.
\]
I fix the positive choice $t>0$ throughout.

{\em Hyperbolic space} $\sH^2\subset\R^3$ is the upper sheet of the
hyperboloid of two sheets given by $q_L(\bv)=-1$:
\[
\sH^2=\bigl\{(t,x,y)\bigm|
\hbox{$-t^2+x^2+y^2=-1$ and $t>0$}\bigr\}.
\]
In other words, $t=\sqrt{1+x^2+y^2}$ (see Figure~\ref{fig!hn}). This is the
analogue of the sphere $S^2$\index{sphere $S^2$} of radius 1, which is
parametrised (in the northern hemisphere) by $z=\sqrt{1-x^2-y^2}$. If you
want the analogue of the sphere of radius $r$, just take the hyperboloid
$q_L(\bv)=-r^2$. The coordinate $t$ on $\R^3$ is ``time-like'' and the
coordinates $x,y$ are ``space-like'' (compare~\ref{sec!prev}).

\begin{figure}[b]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}}
\centering
\leavevmode\hbox{\epsfysize=5 true cm
\epsffile{hyp_space.eps}}\put(-90, 100){$\sH^2$}\put(-35, 30){$\R^2$}
\caption{Hyperbolic space $\sH^2$}
\label{fig!hn}
\end{figure}

A {\em line} $L$ of hyperbolic geometry\index{hyperbolic!line} is the
hyperbola $\sH^1$ obtained as the inter\-section of $\sH^2$ with a
2-dimensional vector subspace $\Pi\subset\R^3$ which is a {\em Lorentz
plane}, in the sense that it contains time-like vectors, so that
$L=\Pi\cap\sH^2\ne\emptyset$; the restriction of $q_L$ to $\Pi$ has
signature $(-1,+1)$. It is obvious that there is a unique line $PQ$
through any two distinct points $P,Q\in\sH^2$, since the 2-dimensional
vector subspace $\Pi$ through $P$, $Q$ in $\R^3$ is unique. The analogy
with the lines of $S^2$ is clear, and I could reasonably call the lines
of $L$ {\em great hyperbolas}.

\section{Hyperbolic distance}\label{sec!hdist}
\index{hyperbolic!distance}
\markright{\hfill \thesection\ Hyperbolic distance\quad}
To define the hyperbolic distance\index{hyperbolic!distance} function, I
start with the formal analogue of formula (\ref{eq!arccos}) of Remark 2
in~\ref{defn of spherical}, replacing the Euclidean inner
product\index{Euclidean!inner product} with the Lorentz inner product
$\cdot_L$ (see\index{Lorentz!dot product $\cdot_L$} \ref{sec!euclandlor}).
Thus let $P$ and $Q$ be points of $\sH^2$ given by the vectors
$\bv=(t_1,x_1,y_1)$ and $\bw=(t_2,x_2,y_2)$. I define the {\em hyperbolic
distance} $d(P,Q)$ between two points by
\begin{equation}
-\bv\cdot_L \bw=\cosh d(P,Q),\quad\hbox{so that}\quad
d(P,Q)=\arccosh(-\bv\cdot_L \bw);
\label{eq!acosh}
\end{equation}
in other words, $d(P,Q)=\arccosh(t_1t_2-x_1x_2-y_1y_2)$.

\begin{lem} The Lorentz inner product\index{Lorentz!dot product $\cdot_L$}
satisfies
\[
-\bv\cdot_L \bw=t_1t_2-x_1x_2-y_1y_2\ge1,
\]
with equality only if $P=Q$. (See also \exref{ex!ge1}.) Hence the distance
$d(P,Q)$ is defined and positive unless $P=Q$. \end{lem}

\begin{pf} This clearly follows from the stronger statement.

\begin{cla} Given two points $P\ne Q\in\sH^2$, there is a Lorentz
basis\index{Lorentz!basis} $\bbf_0,\bbf_1,\bbf_2$ of $\R^3$ giving rise to a
new coordinate system in which $P=(1,0,0)$ and $Q=(\cosh\al,\sinh\al,0)$,
with $\al=d(P,Q)>0$. \end{cla}

This is simply Appendix~B, Theorem~\ref{sec!bases} (4), but I need one
point of the proof, so I repeat it here. Set $\bbf_0=\bv$ the position
vector of $P$; since $P\in \sH^2$, this vector has Lorentz
norm\index{Lorentz!norm} $-1$. The vector
$\bw'=\bw+(\bw\cdot_L\bbf_0)\bbf_0$, where $\bw$ is the position vector of
$Q$, is orthogonal\index{Lorentz!orthogonal} to $\bbf_0$ with respect to
$\cdot_L$ (just compute the product $\bw'\cdot_L\bbf_0)$), and is nonzero
because $P\ne Q$. Hence by Theorem~\ref{sec!bases} (3), $q_L(\bw')>0$. So I
can set $\bbf_1=\bw'/\sqrt{q_L(\bw')}$, and
\begin{equation}
\bw=c\bbf_0+s\bbf_1,\quad\hbox{where $c=-\bv\cdot_L\bw$ and
$s=\sqrt{q_L(\bw')}>0$.}
\label{eq!s>0}
\end{equation}
I find the remaining basis element by the usual method of making an
ortho\-normal basis: choose $\bu\in\R^3$ not in the span of $\bv$ and $\bw$,
then set $\bw''=\bu+(\bu\cdot_L\bbf_0)\bbf_0-(\bu\cdot_L\bbf_1)\bbf_1$ and
finally $\bbf_2=\bw''/\sqrt{q_L(\bw'')}$.

The Lorentz basis $\bbf_0, \bbf_1, \bbf_2$ defines a new coordinate system
on the hyperbolic plane $\sH^2$. In this coordinate system $P=(1,0,0)$ and
$Q=(c,s,0)$, the latter by the first equality in~(\ref{eq!s>0}). As
$Q\in\sH^2$, $c>0$ and its position vector has Lorentz norm $-1$, so
$-c^2+s^2=-1$. By~(\ref{eq!s>0}), $s>0$ and hence $c>1$. So $c=\cosh\al$,
$s=\sinh\al$ for some $\al>0$, and in this coordinate system it is easy to
compute $d(P,Q)=\al$. Hence the distance function is meaningful and positive
unless $P=Q$. \QED \end{pf}

Compare Remark~\ref{defn of spherical} (2) for the spherical analogy; the
purist may want to reread Remark~\ref{defn of spherical} (1) at this point.

\begin{rem} This proof illustrates the fact that in the treatment of
hyperbolic geometry given here, the methods of linear and quadratic algebra
are our main weapons of attack. The arguments are similar to their Euclidean
and spherical analogues, the only difference being the issue of the extra
sign in the Lorentz form, along with the additional care it needs.

The question of signs is important later: in (\ref{eq!s>0}), $s=\sinh\al>0$
was part of the construction of the vector $\bbf_1$. Notice that $\cosh\al$
is a symmetric function and $\sinh\al$ is an antisymmetric function. This is
good, because I am measuring distances from the base point $P=(1,0,0)$ in
terms of $\cosh\al$, and using $\sinh\al$ to parametrise the hyperbola by
arc length $\al$. \end{rem}

\section{Hyperbolic triangles and trig}\label{h2trig}
\index{hyperbolic!triangle}\index{hyperbolic!trig|(}
\markright{\hfill \thesection\ Hyperbolic triangles and trig\quad} This
section is the analogue of \ref{s2trig}. A {\em hyperbolic triangle}
$\triangle PQR$ in $\sH^2$ consists of 3 vertexes $P,Q,R$ and 3 hyperbolic
lines $PQ$, $PR$, $QR$ joining them. Choose coordinates as in
Lemma~\ref{sec!hdist} so that $P=(1,0,0)$ and $PQ$ is on the hyperbolic
line\index{line!hyperbolic} $\{y=0\}$; set $Q'=(0,1,0)$.

The {\em hyperbolic angle} $a$ at $P$ between the two lines $PQ$ and $PR$ is
defined to be the dihedral angle between the two planes $OPQ$, $OPR$ (see
Figure~\ref{fig!h_trig}). The point is that this is a Euclidean
angle,\index{Euclidean!angle} namely, the angle between two lines $OQ'$ and
$OR'$ in the space-like plane $t=0$; in other words, the line $PR$ is in the
plane $OPR'$ spanned by $P$ and $R'=(0,\cos a,\sin a)$.

\setcounter{figure}{\value{section}}
\addtocounter{figure}{-1}
\begin{figure}
\centering
\leavevmode\hbox{\epsfysize=5 true cm\epsffile{hyp_hyperbolic_points.eps}}
\caption{Hyperbolic trig}
\label{fig!h_trig}
\end{figure}

\begin{prop}[main formula of hyperbolic trig] In a hyperbolic triangle
$\triangle PQR$, the side $QR$ is determined by the two sides $PQ$ and $PR$
and the dihedral angle $a$: if $\al=d(Q,R)$, $\be=d(P,Q)$, $\ga=d(P,R)$,
then
\begin{equation}
\cosh\al=\cosh\be\cosh\ga-\sinh\be\sinh\ga\cos a.\label{eq!h_trig}
\end{equation}
\end{prop}

\begin{pf} In the notation developed above, $P=(1,0,0)$,
$Q=(\cosh\be,\sinh\be,0)$ and
\[
R=(\cosh\ga,\sinh\ga\cos a,\sinh\ga\sin a);
\]
here, as in (\ref{eq!s>0}), $\sinh\ga>0$ is part of the definition of the
angle $a$. Thus calculating the Lorentz dot product of the two vectors
representing $Q$ and $R$ gives
\[
\cosh\al=\cosh\be\cosh\ga-\sinh\be\sinh\ga\cos a.\QED
\]
\end{pf}

\begin{cor}[triangle inequality]\label{hyptriang} $d(Q,R)\le d(P,Q)+d(P,R)$,
with equality if and only if $P$ is on the interval $[Q,R]$ (that is, the
segment of line joining $Q$ and $R$). \end{cor}

\begin{pf} This is exactly as before: compare (\ref{eq!h_trig}) with the
standard formula of hyperbolic trig:
\[
\cosh(\be+\ga)=\cosh\be\cosh\ga+\sinh\be\sinh\ga.
\]
Both $\sinh\be$ and $\sinh\ga$ are positive, so that $\cosh(\be+\ga)\ge
\cosh\al$, with equality if and only $a=\pi$. Since $\cosh\al$ is an
increasing function for $\al>0$, it follows that $\be+\ga\ge\al$, with
equality if and only if $P\in [Q,R]$. \QED \end{pf}\index{hyperbolic!trig|)}

\begin{rem} An important corollary of the triangle inequality, in complete
analogy with Euclidean\index{Euclidean!geometry} and spherical
geometry,\index{spherical!geometry} is the fact that the hyperbolic distance
$d(P,Q)$ between two points $P,Q\in\sH^2$ is the length of the shortest
curve\index{shortest distance} $C$ in $\sH^2$ joining $P$ and $Q$, this
shortest curve being the hyperbolic line segment $[P,Q]$. The proof, with
the usual assumptions about the meaning of the statement, is word for word
the same as in~\ref{sec!shortest}. \end{rem}

\section{Hyperbolic motions}\label{h2d}\index{motion}
\markright{\hfill \thesection\ Hyperbolic motions\quad}

A {\em hyperbolic motion}\index{hyperbolic!motion}\index{motion}
$T\colon\sH^2\to\sH^2$ is a map preserving hyperbolic
distance.\index{hyperbolic!distance} As before, my first aim is to get from
this definition to a manageable description of $T$ in terms of a suitable
matrix. Read the homework on Lorentz matrixes\index{Lorentz!matrix|(}
in~\ref{sec!fiesymm}--\ref{sec!o&lm}, before you continue.

\begin{thm}
\begin{enumerate}

\item Every hyperbolic motion\index{hyperbolic!motion}\index{motion}
preserves hyperbolic lines.\index{hyperbolic!line}

\item Every hyperbolic motion $T\colon\sH^2\to\sH^2$ is given in
coordinates by $\bx\mapsto A\bx$, where
\begin{enumerate}
\item $A$ is a Lorentz matrix,\index{Lorentz!matrix} that is
\[
\tr A \begin{pmatrix} - 1& 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}
A = \begin{pmatrix} - 1& 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix},
\quad\hbox{and}
\]

\item $A$ preserves the two halves of the cone $\{q_L(\bv)<0\}$.
\end{enumerate} \end{enumerate} \end{thm}

\begin{pf} The proofs are almost the same as in the Euclidean and spherical
cases (see \ref{sec!mcol} and Theorem~\ref{s2d} (2)). Since lines are
determined by the distance function, a motion $T$ takes a hyperbolic line to
another hyperbolic line, proving (1). Since a hyperbolic line $L$ is a
hyperbolic arc in a Lorentz\index{Lorentz!space} plane $\Pi=\R^2$ with arc
length parametrisation $(\cosh s,\sinh s)$, it follows that $T$ is linear
when restricted to each $\Pi$, therefore linear on $\R^3$.

More formally, I can extend $T$ from $\sH$ to the upper half cone by radial
extension; write $\wT$ for this extension. Give a Lorentz plane $\Pi$,
choose a Lorentz basis $\bbf_0,\bbf_1$ so that $L$ is parametrised as
\[
P_s=(\cosh s)\bbf_0+(\sinh s)\bbf_1 \quad\hbox{for $s \in\R$;}
\]
here the time-like vector $\bbf_0$ is the coordinate of a point $P_0\in L$,
and the space-like vector $\bbf_1$ is the tangent direction to $L$ at $P_0$,
with $s$ the distance function along $L$. Then $T$ takes $L$ to the line
$L'$ parametrised as $P_s'=(\cosh s)\bbf_0'+(\sinh s)\bbf_1'$, so that $\wT$
is given by a linear map on $\Pi$. Since this holds for any line $L$, it
follows that $\wT$ is linear within the upper half cone (that is,
\[
\wT(\la\bu+\mu\bv)=\la\wT(\bu)+\mu\wT(\bv)
\]
whenever $\bu,\bv$ and $\la\bu+\mu\bv$ are in the upper half cone). Now,
although $\wT$ is only defined in the half cone, the usual linear algebra
argument shows that it is given by a matrix $A$ (just choose a basis of the
vector space $\R^3$ consisting of three vectors in the upper half cone).
Moreover, $A$ must be Lorentz\index{Lorentz!transformation|(} since $\wT$
preserves the Lorentz form\index{Lorentz!form $q_L$}
(compare~\ref{sec!fiesymm}--\ref{sec!o&lm}). \QED \end{pf}

\begin{rem} In proving Theorem~\ref{s2d}, I extended $T$ to $\R^3$ by radial
extension, then used linearity on each plane $\Pi$, which holds because the
distance function determines everything about motions in 1~dimension. In the
hyperbolic case, the awkward point is that radial extension only gives $\wT$
defined on the upper half cone; my argument is that it is linear in the
upper half cone, and so given by a matrix. \end{rem}

A Lorentz matrix\index{Lorentz!matrix|)} $A$ preserves the two halves of the
cone $\{q_L(\bv)<0\}$ if and only if its top left entry $a_{00}>0$; such a
matrix defines a {\em Lorentz transformation} of\/ $\R^3$. The set
$\O^+(1,2)$ of Lorentz transformations is entirely analogous to the set
$\Eucl(2)$ of motions\index{Euclidean!motion} of the Euclidean plane. It is
easy to state and prove the following assertions, all of which are analogues
of the corresponding statements in plane Euclidean geometry (compare
also~\ref{sec!s2ve2}):\index{Euclidean!geometry}
\begin{enumerate}

\item The hyperbolic plane $\sH^2$\index{hyperbolic!plane $\sH^2$|(} is a
metric geometry with a distance function $d(P,Q)$ and a set of motions
$\O^+(1,2)$.

\item The motions\index{motion} act transitively on $\sH^2$ and the set
of lines through a given point $P\in\sH^2$.

\item Every element of $\O^+(1,2)$ is either a rotation\index{rotation}
$\Rot(P,\theta)$, a Lorentz trans\-lation $\Transl(L,\al)$ along an axis
$L$, a Lorentz reflection\index{Lorentz!reflection} $\Refl(L)$ or a
Lorentz glide.\index{glide} For example, if $L=\{y=0\}$, the
translation\index{hyperbolic!translation} and glide are given by
\[
\begin{pmatrix} \cosh s&\sinh s&0\\ \sinh s&\cosh s&0
\\0&0&1 \end{pmatrix}
\quad\hbox{and}\quad
\begin{pmatrix} \cosh s&\sinh s&0\\ \sinh s&\cosh s&0
\\0&0&-1 \end{pmatrix}.\quad
\]
(Compare Ex.~\ref{app!q}.\ref{ex!lor_normalform}.)

\item Given two pairs of points $P,Q$ and $P',Q'$, there exist exactly
two motions\index{motion} $g\in\O^+(1,2)$ such that $g(P)=g(P')$,
$g(Q)=g(Q')$, of which one is a rotation\index{rotation} or Lorentz
translation and the other a Lorentz reflection or glide.

\item $\O^+(1,2)$ has two types of elements, direct and indirect. Every
direct motion\index{motion} is the identity or a composite of 2
reflections; every opposite motion\index{motion} is a reflection or a
composite of 3 reflections.\index{Lorentz!transformation|)}
\end{enumerate}

\section{Incidence of two lines in $\sH^2$}\label{sec!2lines}
\markright{\hfill \thesection\ Incidence of two lines in
$\sH^2$\quad}\index{hyperbolic!line|(}\index{incidence of lines}

In \ref{sec!s2ve2} (1) I showed that two lines (great circles) of $S^2$ meet
in a pair of antipodal points, by taking $L_1=\Pi_1\cap S^2$, $L_2=\Pi_2\cap
S^2$, then constructing the line $V=\Pi_1\cap\Pi_2$ in the ambient $\R^3$,
which of course meets $S^2$ in two points. Two familiar facts follow: (1)
the\index{orthogonal!complement $V^\perp$} orthogonal complement
$V^\perp\subset\R^3$ is a plane cutting out a line $M=V^\perp\cap S^2$, the
unique common perpendicular to $L_1$ and $L_2$. (2) $L_1,L_2$ generate a
pencil of lines, that pass through the same intersection points and are
perpendicular to $M$. If I choose coordinates so that $V$ is the $z$-axis,
the intersection points are the poles $(0,0,\pm1)$, $M$ is the equatorial
plane $z=0$, and the family of lines containing $L_1,L_2$ is the pencil of
meridians $(\sin\theta)x=(\cos\theta)y$.

The same arguments apply to lines in $\sH^2$, but the conclusions are
different, since the ambient $\R^3$ is now Lorentz space: as before, let
$L_1=\Pi_1\cap \sH^2$, $L_2=\Pi_2\cap \sH^2$, and consider the line
$V=\Span{\bv}=\Pi_1\cap\Pi_2\subset\R^3$. There are 3 cases:

\begin{enumerate}\renewcommand{\labelenumi}{(\roman{enumi})} \item $V$ is
space-like: $q_L(\bv)>0$. Then $L_1,L_2$ are disjoint, since
$V\cap\sH^2=\emptyset$. In this case, the orthogonal complement $V^\perp$
with\index{Lorentz!complement} respect to the Lorentz inner product
$\cdot_L$ is a Lorentz plane (the restriction of $q_L$ has signature
$(-1,+1)$, so that it contains time-like vectors), and hence
$M=V^\perp\cap\sH^2$ is a line of $\sH^2$, and is the unique common
perpendicular to $L_1,L_2$. For example, if $V$ is the $x$-axis, the lines
$L_1,L_2$ are among the meridian lines $y=ct$, having the common
perpendicular $M:(x=0)$.
\setcounter{figure}{\value{section}}
\addtocounter{figure}{-1}
\begin{figure}[ht]
\centering
\begin{picture}(100,100)(0,-8)
\put(16,50){\line(1,0){68}}
\qbezier(18,50)(50,78)(82,50)
\qbezier(18,50)(50,100)(82,50)
\qbezier(18,50)(50,0)(82,50)
\qbezier(18,50)(50,22)(82,50)
\put(50,18){\line(0,1){64}}
\put(42,-4){(a)}
\end{picture}
\qquad
\begin{picture}(100,100)(0,-8)
\put(20,50){\line(1,0){60}}
\qbezier(20,86)(50,56)(80,86)
\qbezier(20,68)(50,52)(80,68)
\qbezier(20,32)(50,48)(80,32)
\qbezier(20,14)(50,44)(80,14)
\put(50,16){\line(0,1){68}}
\put(42,-4){(b)}
\end{picture}
\caption{(a) Projection to the $(x,y)$-plane of the spherical lines $y=cz$.
(b) Projection to the $(x,y)$-plane of the hyperbolic lines $y=ct$.}
\end{figure}

\item $V$ is time-like: $q_L(\bv)<0$. Then $L_1,L_2$ intersect in
$P=V\cap\sH^2$. They do not have a common perpendicular, because the plane
$V^\perp\subset\R^3$ is space-like, so does not meet $\sH^2$. For example,
if $V$ is the $t$-axis, $L_1,L_2$ intersect at $P=(1,0,0)$ and the pencil of
lines through $P$ is $(\sin\theta)x=(\cos\theta)y$.

\index{ultraparallel lines|(}
\item $V$ is actually on the light cone: $q_L(\bv)=0$. Then $L_1,L_2$ are
disjoint in $\sH^2$, but are asymptotic, in the sense that they approach
indefinitely at one end. For example, $V=\Span{(1,1,0)}$ is the common
asymptotic direction of the lines $L_c:(y=c(t-x))$ with $|c|<1$. The plane
$V^\perp:(x=t)$ is tangent to the light cone along $V$, so does not
correspond to a line in $\sH^2$, and $L_1,L_2$ do not have a common
perpendicular.
\end{enumerate}

\begin{defn} I say that $L_1$ and $L_2$ {\em diverge} in case (i). A simple
calculation shows that, if $L_1$ and $L_2$ are parametrised by arc length as
$P_1(s)$, $P_2(s)$ then $d(P_1(s),P_2(s))$ grows linearly in $s$ as $s\gg0$;
for details, see \exref{ex!diverge}.

Case (iii) is the limiting case that separates (i) and (ii): although
$L_1,L_2$ are disjoint, they ``approach one another\index{infinity!point
at|(} at infinity''. I say that $L_1,L_2$ are {\em ultraparallel}. To make
this precise, it is useful
to introduce the formal idea that each line $L=\Pi\cap\sH^2$ of $\sH^2$ has
two ``ends'', the two rays in which the plane $\Pi$ intersects the null-cone
$q(\bv)=0$, or the asymptotic lines of the hyperbola $L\subset\Pi$. One
views an end as an ``ideal point'' of $L$ or\index{infinity!point at}
``point at infinity'' -- not a point of $\sH^2$, but rather an asymptotic
direction. Case (iii) above, can be described by saying that $L_1$ and
$L_2$ have a
common end $V=\Span{\bv}=\Pi_1\cap\Pi_2$. By convention, ultraparallel lines
$L_1$ and $L_2$ have angle $0$ at this end. All the lines $L_c:y=c(t-x)$ are
ultraparallel, with the ray $(1,1,0)$ as a common end. These lines all
approach one another arbitrarily closely as they head out to infinity, as
described in \exref{ex!asymp}.\index{infinity!point at|)}
\end{defn}

\section{The hyperbolic plane is non-Euclidean}\label{sec!nonE}
\index{hyperbolic!plane $\sH^2$|)}\index{non-Euclidean geometry}
\markright{\hfill \thesection\ The hyperbolic plane is
non-Euclidean\quad}

As discussed in the introduction to this chapter and at the end
of~\ref{h2d}, hyper\-bolic geometry shares many features with Euclidean and
spherical geometry; the differences are also striking. The incidence
properties of lines in $\sH^2$ just established are qualitatively quite
different from the Euclidean case. Two lines $L_1$ and $L_2$ of $\sH^2$ have
a common perpendicular $M$ if and only if $V=\Pi_1\cap\Pi_2$ is space-like,
which is clearly an {\em open} condition: $L_1$ and $L_2$ remain disjoint
even if we move them a little, for example, tilting one of them about a
point. The parallel postulate\index{parallel!postulate} thus fails, as I
discuss below in more detail. The next section \ref{sec!ang.def} treats the
angular defect formula, expressing the sum of angles in a triangle in terms
of its area; this sum is always $<\pi$. The hyperbolic non-Euclidean world
also differs from the Euclidean in the existence of an intrinsic
distance,\index{intrinsic!unit} by analogy with the spherical world
(compare~\ref{sec!S2unlikeE2}), and the negative\index{curvature} curvature
of hyperbolic space (compare~\exref{ex!pringle} and~\ref{sec!trichotomy}).

Euclid's parallel postulate states that given a line $L$ of the planar
geometry and a point $P$ not on it, there is one and only one line $M$
through $P$ and disjoint from $L$. This holds in plane Euclidean geometry
(and indeed in affine geometry, compare~\ref{sec!afl}); in spherical
geometry it is obviously false as there are no disjoint lines. What happens
in $\sH^2$? A plausible attempt to find a parallel
line\index{parallel!lines} $M$ through a point $P\notin L$ is to drop a
perpendicular $PQ$ onto $L$, then take $M$ perpendicular to $PQ$; as we know
from the above, this is indeed a line not meeting $L$, but not the only one.

\begin{thm} Let $L$ be a hyperbolic line and $P$ a point not lying on $L$.
Then there exists a unique perpendicular line $PQ$ to $L$ through $P$.
Moreover,
\begin{enumerate}\renewcommand{\labelenumi}{(\arabic{enumi})}
\item if $M$ is orthogonal to $PQ$ in $P$, then the lines $L$ and $M$ diverge;
\item there exists an angle $\theta<\frac{\pi}{2}$ with the property that if
$L'$ is a line through $P$, then $L'$ meets $L$ if and only if the angle of
$L'$ and $PQ$ at $P$ is {\rm less than} $\theta$. (See Figure~\ref{fig!hpll}.)
\end{enumerate}
\end{thm}
\begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}}
\centering
\input{hyp_lines.pstex_t}
\caption{The failure of the parallel postulate in $\sH^2$}
\label{fig!hpll}
\end{figure}

\paragraph{Remark} In axiomatic geometry, the logical self-consistency of
this picture was the focal point of the 2000 year old controversy concerning
Euclid's parallel postulate
(compare~\ref{sec!parallels}).\index{parallel!postulate} In the present
coordinate construction of $\sH^2$, there is nothing to dispute: everything
follows at once from the case division of \ref{sec!2lines}. Whether
Euclidean or hyperbolic geometry or some other theory is a better
approximate mathematical model for the real world in different applications
is an entirely separate question, discussed in \ref{sec!trichotomy}.

\begin{pf} I give the coordinate proof. The line $L$ corresponds to a
Lorentz orthogonal decomposition $\R^3=\Pi\oplus\Pi^\perp$ where
$L=\Pi\cap\sH^2$. The coordinate vector $\bp$ of $P$ can be written
\[
\bp=\bq+\bv \quad\hbox{with $\bq\in\Pi$ and $\bv$ in $\Pi^\perp$};
\]
here $\bv$ is nonzero and space-like, and $\bq\ne0$ because $\bp$ is
time-like. Choosing Lorentz coordinates in $\R^3$ with $\bbe_0$ the unit
time-like vector proportional to $\bq$ and $\bbf_2$ proportional to $\bv$
makes $L$ into the line $y=0$, $Q=(1,0,0)$ and $P=(t_0,0,y_0)$ with
$t_0=\sqrt{1+y_0^2}$. The perpendicular line $PQ$ is $x=0$, and the line $M$
perpendicular to it at $P$ is $y=\frac{y_0}{t_0}t$. The two planes of $L$
and $M$ intersect in the $x$-axis of $\R^3$, so $L$ and $M$ diverge.

Any line through $P=(t_0,0,y_0)$ is given by
$(\sin\fie)x=(\cos\fie)(y_0t-t_0y)$; in $\R^3$, this plane intersects $y=0$
in the line $\Span{(\tan\fie,y_0,0)}$, which is time-like if and only if
$|\tan\fie|>y_0$. This proves the claim (together with the actual value
$\theta=\arccot y_0$, compare \exref{ex!arccot}). \QED\end{pf}

\paragraph{Discussion} A second ``proof'' in more geometric terms is much
closer to the historical context, if trickier to argue convincingly; please
refer to Figure~\ref{fig!hpll} during the argument. The existence and
uniqueness of the orthogonal $PQ$ can be proved by minimising the distance
from $P$ to $L$, as discussed in~\exref{ex!hypperps} (b); (1) follows from
the case division in \ref{sec!2lines}, and is proved again in
\exref{ex!diverge}.

For (2), note first that some lines $L'$ through $P$ certainly meet $L$. On
the other hand, as (1) shows, there exists a line $M$ through $P$ that does
not meet $L$. It is also easy to see that there cannot be a ``last'' line
$L'$ through $P$ which meets $L$: if $L'\cap L=R$ then there are points $R'$
along $L$ and further away from $Q$, and hence further lines $PR'$ meeting
$L$. From this, a least upper bound argument shows that there must be a
``first'' line $\wM$ (one on either side of $PQ$) which fails to meet $L$.

This proves almost all of (2); the only remaining point to clear up is the
statement that the angle $\theta$ between $PQ$ and the ``first''
nonintersecting line $\wM$ is less than $\pi/2$. However, the line $M$ at
angle exactly $\pi/2$ diverges from $L$ by (1), whereas $\wM$ is asymptotic
to $L$; hence the angle $\theta$ must be less than $\pi/2$. Lines $L'$
having angle less than $\theta$ at $P$ with $PQ$ are of type (i) and so
intersect $L$; lines having angle greater than $\theta$ are of type (ii) and
are disjoint from $L$.\index{hyperbolic!line|)}

\paragraph{Other models} There are several alternative models of
non-Euclidean geo\-metry in addition to the hyperbolic model in Lorentz space
discussed here. Beltrami's model as the interior of an absolute conic in
$\PP^2_\R$ is treated in Rees \cite{Rees}; it has the great advantage of
making the incidence of lines completely transparent. An alternative is the
Lobachevsky or Poincar\'e model as the upper-half space in the complex
plane, which makes asymptotically converging ultraparallel lines easy to
visualise, and which is important in other mathematical contexts;
\exref{ex!defH}--\ref{ex!linesH} lead you through the construction of this
model.\index{hyperbolic!space}\index{ultraparallel lines|)}

\section{Angular defect}\label{sec!ang.def} \index{angle!sum|(}
\markright{\hfill \thesection\ Angular defect\quad} The remainder of this
chapter discusses two proofs of the famous angular defect formula of Gauss
and Lobachevsky.

\begin{thm} In a hyperbolic triangle\index{hyperbolic!triangle} $\triangle
PQR$ with angles $a,b,c$, \begin{equation} a+b+c=\pi-\area\triangle PQR.
\end{equation} \end{thm}

In addition to finite hyperbolic triangles\index{hyperbolic!triangle}
$\triangle PQR$ with $P,Q,R\in\sH^2$, I generalise the statement to allow
{\em ideal triangles},\index{ideal triangle} with one or more vertexes
ideal points ``at infinity''.\index{infinity!point at} An {\em ideal
triangle} has 3 sides which are lines of $\sH^2$, and any two sides either
intersect, or are ultraparallel in the sense of Definition~\ref{sec!nonE},
with every pair of sides intersecting in distinct (ideal) points. Remember
that two lines meeting at an ideal point have angle $0$ there.

\subsection{The first proof}

There are two points in this proof:
\begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}a}
\centering
\input{hyp_ideal_triang.pstex_t}
\caption{The hyperbolic triangle $\triangle PQR$ with one ideal vertex}
\label{fig!hyp_tri}
\end{figure}
\begin{enumerate}
\renewcommand{\theenumi}{\Roman{enumi}}

\item First, an explicit integration calculates the
area\index{area|(} of the particular triangle $\triangle PQR$ of
Figure~\ref{fig!hyp_tri}. The crucial point here is that the area of
a triangle remains bounded, even though one of its vertexes goes
off to infinity.\index{infinity!point at}

\item Next, area of polygons and sum of angles of polygons have the
simple {\em additivity} property illustrated in Figure~\ref{fig!B+C}:
if you subdivide $A$ as a union of two adjacent polygons $A=B\cup C$,
then $\area A=\area B+\area C$. The sum of angles also adds, except
that you subtract $\pi$ if two angles coalesce to form a straight line
(because the common point is no longer viewed as a vertex).
\end{enumerate}

\begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}b}
\centering
\input{hyp_sum_adds.pstex_t}
\caption{Area and angle sum are ``additive''}\index{angle!sum}
\label{fig!B+C}
\end{figure}

\subsection{An explicit integral}\label{ssec!expl}

\begin{prop} Let $a\in(0,\pi/2)$ be a given angle. Consider
$\triangle PQR$ in $\sH^2$ bounded by the three lines $y=0$, $y=(\tan
a)x$ and $x=(\cos a)t$ (see Figure~\ref{fig!hyp_tri}).
Then
\[
\area\triangle PQR=\pi/2-a = \pi-\mbox{angle sum}(\triangle PQR).
\]
\end{prop}
\begin{pf} The triangle has two vertexes $P=(1,0,0)$ and
$Q=\frac{1}{\sin a}(1,\cos a,0)$ in $\sH$ and one ideal vertex
\index{infinity!point at} $R=(1,\cos a,\sin a)$. We know that
$\angle RPQ=a$ for the same reason as in \ref{s2trig} and \ref{h2trig},
because the angle in $\sH$ is the dihedral angle in $\R^3$, which equals
the angle in the plane $\{t=0\}$. I have drawn Figure~\ref{fig!hyp_tri} with
symmetry about the $x$-axis so that we see at once that $\angle
PQR=\pi/2$. Finally, $\angle PRQ=0$ by definition. Hence
\[
\mbox{angle sum}(\triangle PQR)= \pi/2 + a
\]
which proves the second equality.

To calculate the area, I write down an element of area, and integrate it
as a double integral over the triangle $\triangle PQR$. It is convenient
to work in polar coordinates
\[
x=r\cos\theta,\quad y=r\sin\theta,\quad\hbox{so that
$t=\sqrt{1+r^2}$.}
\]
In these coordinates, the element of area in $\sH^2$
is $r\dd r\dd\theta/\sqrt{1+r^2}$ (see~\exref{ex!hyparea}
and compare also~\exref{ex!spharea}).
It is easy to integrate this element of area as an indefinite integral, since
\[
\frac{r\dd r}{\sqrt{1+r^2}}\dd\theta= \dd\sqrt{1+r^2}\dd\theta.
\]
The more subtle point is to get an explicit expression for the domain of
integration. Since the two sides out of $P$ in Figure~\ref{fig!hyp_tri}
are given by $y=0$, $y=(\tan a)x$, the angle $\theta$ runs through the
interval $[0,a]$. For fixed $\theta$, the point
$(\sqrt{1+r^2},r\cos\theta,r\sin\theta)$ runs through the line
$PQ_\theta$ of Figure~\ref{fig!hyp_tri}. The condition to be under the
hyperbola is $x\le (\cos a)t$, giving
\[
r\cos\theta\le \sqrt{1+r^2}\cos a \implies r^2\le
\frac{\cos^2a}{\cos^2\theta-\cos^2a}.
\]
Therefore
\begin{align*}
\area\triangle PQR=\iint\limits_{\triangle PQR}\frac{r\dd
r\dd\theta}{t} &\hbox{}=\iint\limits_{\triangle PQR}
\dd\sqrt{1+r^2}\dd\theta\\ &\hbox{}=\int\limits_{\theta=0}^a
\left[\sqrt{1+r^2}\right]_{r^2=0}^{r^2=\frac{\cos^2a}
{\cos^2\theta-\cos^2a}}\dd\theta\\
&\hbox{}=\int_0^a \left(-1+
\sqrt{\frac{\cos^2\theta}{\cos^2\theta-\cos^2a}}\right)
\dd\theta.
\end{align*}
Now I am in luck, and the integrand is an exact differential: indeed,
consider $\fie=\arcsin(\sin\theta/\sin a)$ as a function of $\theta$.
Then differentiating the defining relation $(\sin
a)(\sin\fie)=\sin\theta$ gives
\[
\frac{\dd\fie}{\dd\theta} =\frac{\cos\theta}{(\sin a)(\cos\fie)}=
\sqrt{\frac{\cos^2\theta}{\cos^2\theta-\cos^2a}}.
\]
It follows that the above integral evaluates to
\[
\area\triangle PQR= -a+\Bigl[\arcsin\left(\frac{\sin\theta}{\sin
a}\right)\Bigr]_0^a=-a+\pi/2.
\QED
\]
\end{pf}

\subsection{Proof by subdivision}

The calculation of Proposition~\ref{ssec!expl} implies at once the
following result for ideal triangles\index{ideal triangle|(} with two
or more ideal vertexes.

\begin{lem}
\begin{enumerate}\renewcommand{\labelenumi}{(\arabic{enumi})}

\item Let $\triangle PRR'$ be an ideal
triangle\index{ideal triangle} of $\sH^2$ with one vertex $P\in\sH^2$
and two ideal vertexes; if $\angle P=a$ then
\begin{equation}
\area \triangle PRR'=\pi-a.
\end{equation}

\item Let $\triangle PQR$ be an ideal triangle\index{ideal triangle} of
$\sH^2$ with all three vertexes $P,Q,R$ ideal points
at infinity.\index{infinity!point at} Then
\begin{equation}
\area \triangle PQR=\pi.
\end{equation}
\end{enumerate}
\end{lem}

\begin{pf} (1) Drop a perpendicular $PQ$ from $P$ onto the opposite
side $RR'$. By Claim~\ref{sec!hdist}, I can choose coordinates such
that $P=(1,0,0)$ and $PQ$ is the $x$-axis $y=0$. This subdivides
triangle $\triangle PRR'$ symmetrically about the $x$-axis as in
Figure~\ref{fig!hyp_tri} into two triangles $\triangle PQR$ and
$\triangle PQR'$, each having angle $a/2$ at $P$. Thus applying
Proposition~\ref{ssec!expl} to each gives
\[
\area \triangle PRR'=\area \triangle PQR+\area \triangle PQR'=2(\pi/2-a/2),
\]
as required.

(2) Choose any interior point $S$ of the ideal triangle\index{ideal triangle}
$\triangle PQR$ with three ideal vertexes, and draw in the 3 hyperbolic
line segments $PS$, $QS$, $RS$. These subdivide $\triangle PQR$ into 3
triangles $\triangle SPQ$, $\triangle
SQR$, $\triangle SRP$ of the type considered in (1), as on
Figure~\ref{fig!a+b+c}. If $a,b,c$ are the angles at $S$ in each of these, then
\begin{align*}
\area \triangle PQR
&\hbox{}=\area \triangle SPQ+\area \triangle SQR +\area \triangle SRP\\
&\hbox{}=\pi-a + \pi-b + \pi-c,
\end{align*}
which gives what I want, in view of $a+b+c=2\pi$. \QED
\end{pf}

\setcounter{figure}{\value{section}}
\addtocounter{figure}{-1}
\begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}c}
\centering
\input{hyp_trisect_triang.pstex_t}
\caption{The subdivision of $\triangle PQR$}
\label{fig!a+b+c}
\end{figure}

\begin{pfof}{Theorem~\ref{sec!ang.def}} Starting from a finite triangle
$\triangle PQR$, extend sides $RP$, $QR$ and $PQ$ to infinity to get
Figure~\ref{fig!h2ang}. Now the whole triangle has area equal to $\pi$ by (2)
of the lemma, and it is subdivided into $\triangle PQR$ plus three
triangles with two ideal vertexes which have areas
$a,b,c$ by (1) of the lemma. Thus the area of $\triangle PQR$ is $\pi-a-b-c$.
\QED \end{pfof}
\setcounter{figure}{\value{section}}

\begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}d}
\centering
\input{hyp_anglesum_proof.pstex_t}
\caption{The angular defect formula}\index{angle!sum}
\label{fig!h2ang}
\end{figure}


\subsection{An alternative sketch proof}
The above proof depended on an explicit integration. This dependence can be
substantially reduced, by an elegant argument making more systematic use of
the additivity of angle sums. The alternative is due to David Epstein (who
acknowledges hints from C.F. Gauss and N.I. Lobachevsky).

\begin{lemn} Given any two ideal triangles\index{ideal triangle} $\triangle
PQR$ and $\triangle P'Q'R'$ having three ideal vertexes, there is a Lorentz
transformation\index{Lorentz!transformation} $A\colon\sH^2\to\sH^2$ taking
$\triangle PQR$ into $\triangle P'Q'R'$. \end{lemn}

This is an easy exercise in linear algebra: given any three distinct lines
$V_1,V_2,V_3$ of $\R^3$ contained in the cone $\{q_L(\bv)=0\}$, there
is a Lorentz basis\index{Lorentz!basis} $\bbe_0,\bbe_1,\bbe_2$ of $\R^3$ for
which

\begin{equation}
V_1=\Span{\bbe_0+\bbe_2},\quad
V_2=\Span{\bbe_0+\bbe_1},\quad
V_3=\Span{\bbe_0-\bbe_2}.
\label{eq!idtri}
\end{equation}

\begin{lemn} Any ideal triangle\index{ideal triangle} $\triangle PQR$ with
three ideal vertexes at infinity\index{infinity!point at} has finite area
$\pi$. \end{lemn}

It follows by Lemma~1 that all ideal triangles\index{congruent triangles}
are congruent, so the key point is that the area is finite (the $\pi$ can
be viewed as an arbitrary scaling factor). There is a beautiful axiomatic
geometry proof due to Gauss in Coxeter~\cite{Cx}, Figure~16.4a.

Now consider an ideal triangle\index{ideal triangle} $\triangle PQR$ with
$P\in\sH^2$, and two ideal vertexes $Q,R$. Let $a=\angle QPR$, and write
$\triangle PQR=\De(a)$. I wish to prove that $\area\triangle PQR=\pi-a$. For
this purpose, define $L(a)=\pi-\area\triangle PQR$.

\setcounter{figure}{\value{section}}
\begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}e}
\centering
\leavevmode\hbox{\epsfysize=4 true cm\epsffile{hyp_newtriangle.eps}}
\caption{Area is an additive function}
\label{fig!add}
\end{figure}

\begin{lemn}\label{lem!add} $L(a)$ is an additive function of $a$, that is, if
$a=b+c$ with $0<a,b,c<\pi$ then $L(a)=L(b)+L(c)$.
\end{lemn}

\begin{pf} Immediate from Figure~\ref{fig!add}:
\[
\area\De OPQ+\area\De OQR=\area\De OPR +\area\De PQR = \area\De OPR +\pi,
\]
since all vertexes of $\De PQR$ are ideal. \QED
\end{pf}

\begin{lemn}\label{lem!mon} $L(a)$ is a monotonic function of $a$, that is,
if $a>b$ then $L(a)>L(b)$. Moreover, $L(0)=0$ and $L(\pi)=\pi$.
\end{lemn}

\setcounter{figure}{\value{section}}
\begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}f}
\centering
\leavevmode\hbox{\epsfysize=4 true
cm\epsffile{hyp_monotonic.eps}}
\caption{Area is a monotonic function}
\label{fig!mon}
\end{figure}

\begin{pf} There are several ways of proving that $a>b$ in a figure such
as Figure~\ref{fig!mon} consisting of two ideal triangles:\index{ideal
triangle} if $a\le b$ then the lines out of $P$ and $P'$ diverge, as
discussed in Theorem~\ref{sec!nonE}. Note that as $a\to0$ the triangle
$\De(a)$ tends to the whole of the ideal triangle, and as $a\to\pi$ it
tends to a line. \QED \end{pf}

It is obvious that Lemmas~\ref{lem!add}--\ref{lem!mon} imply that $L(a)=a$,
so that $\area\De(a)=\pi-a$ for all $a\in(0,\pi)$. The proof then concludes as
before by referring to Figure~\ref{fig!h2ang}.\index{ideal triangle|)}
\index{angle!sum|)}\index{area|)}

\section*{Exercises to Chapter~\thechapter}
\markright{\hfill Exercises to Chapter~\thechapter\quad}
\addcontentsline{toc}{section}{Exercises to Chapter~\thechapter}
\begin{enumerate}
\renewcommand{\labelenumi}{\thechapter.\theenumi.}
\addtocounter{section}{1}

\item[] In \exref{S2circle}--\ref{ex!Snsetup}, consider the geometry
of the sphere\index{sphere $S^2$} $S^2\subset\R^3$ of radius~1 with
the intrinsic (spherical) metric.\index{spherical!distance}

\item\label{S2circle}
\begin{enumerate}

\item Define, by analogy with Euclidean geometry, the notions of {\em
spherical circle} and {\em spherical disc}\index{spherical!disc} with
centre $P\in S^2$ and radius $\rho$.

\item Prove that a spherical circle with radius $\rho<\pi$ has circumference
$2\pi\cdot\sin \rho$.

\item Prove that a spherical disc of radius $\rho<\pi$ has area
$2\pi\cdot(1-\cos\rho)$.
\end{enumerate}
\begin{hint} for (c), integrate (b).\end{hint}

\item Deduce from \exref{S2circle} that there does not exist an
isometric map from any region of $S^2$ to a region of the Euclidean plane
$\R^2$.

\item\label{ex!S2refl}
\begin{enumerate}
\item State and prove {\em Pons Asinorum}
(\ref{sec!pons}) in spherical geometry.\index{spherical!geometry}

\item Let $P_1,P_2\in S^2$ be distinct points. Prove that the set of points
equi\-distant from $P_1,P_2$ is a spherical line (great circle).
\begin{hint} use the ambient metric of $\R^3$ to find the locus, and (i) to
prove in terms of the intrinsic geometry of $S^2$ that every point
equidistant from $P_1,P_2$ is on it.\end{hint}
\end{enumerate}

\item\label{ex!engon} Let $\De\subset S^2$ be a spherical $n$-gon, with
internal angles $a_1,\dots,a_n$ at its vertexes. Guess and prove a formula
for the area of $\De$ in terms of $\sum a_i$. (Assume that the figure $\De$
does not overlap itself to avoid complicated explanations of how you count
the area.)

\item\label{ex!ineqbe} Let $\al,\be,\ga$ be the side lengths of a spherical
triangle\index{spherical!triangle} $\triangle PQR$ and $a,b,c$ the
opposite angles. Use the main formula
\[
\cos\al=\cos\be\cos\ga-\sin\be\sin\ga\cos a
\]
to prove that $|\be-\ga|<\al<\be+\ga$ and $\al+\be+\ga<2\pi$.

Prove that every triple with $\al,\be,\ga<\pi$ satisfying the above
inequalities are the sides of a spherical
triangle.\index{spherical!triangle}

\item\label{ex!sphsin} In the same notation, prove the sine rule for
spherical triangles
\[
\frac{\sin\al}{\sin a}=\frac{\sin\be}{\sin b}=\frac{\sin\ga}{\sin c}.
\]
\begin{hint} using the notation $p,q,r$ for the vertexes of $\De PQR$ as in
\ref{s2trig}, prove that the matrix with rows $p,q,r$ has
determinant $\det(p,q,r)=\sin a\sin\be\sin\ga$.\end{hint}

\item Prove that if $\De$ is an acute angled spherical triangle whose angles
are submultiples $\pi/p,\pi/q,\pi/r$ of $\pi$, then
\[
(p,q,r)=(2,2,n)\quad\hbox{or}\quad(2,3,3)\quad\hbox{or}\quad(2,3,4)\quad
\hbox{or}\quad(2,3,5).
\]
Prove that if $\De$ is a triangle in $\R^2$ with the same properties,
then the possibilities are
\[
(p,q,r)=(3,3,3)\quad\hbox{or}\quad(2,4,4)\quad\hbox{or}\quad(2,3,6).
\]
\begin{hint} using the formula $\area\De=a+b+c-\pi$, get
$\frac1p+\frac1q+\frac1r>1$.\end{hint}

\item\label{ex!spharea} Show that in polar coordinates
\[
x=r\cos\theta,\quad y=r\sin\theta,\quad z=\sqrt{1-r^2}
\]
on the sphere $S^2$ of unit radius, the element of area in $S^2$ is
\[\dd A = \frac{r\dd r\dd\theta}{\sqrt{1-r^2}}.\]
\begin{hint} Consider a small sector
$[\theta,\theta+\de\theta]\times[r,r+\de r]$ in $\R^2$. Prove that the
sector of $S^2$ lying over it is very close to a spherical rectangle with
length of sides equal to $r\de\theta$ and $\de r/\sqrt{1-r^2}$.\end{hint}

\item\label{ex!8} Here is a general project: take any result you know
in plane Euclidean geometry, find an analogue for spherical
geometry,\index{spherical!geometry} and either prove or disprove it.
As concrete exercises, prove or deny the following:
\begin{enumerate}

\item the 3 medians of a triangle intersect in a point $G$;

\item the 3 perpendicular bisector\index{perpendicular bisector}s of a triangle intersect in a point $O$;

\item (harder) the 3 heights of a triangle intersect in a point $H$.
\end{enumerate}

\item\label{ex!Snsetup} Another general project: set up definitions
and notation for the geo\-metry of the $n$-dimensional sphere
$S^n$.\index{sphere $S^n$} \begin{hint} The ambient space is $\R^{n+1}$ and
the distance function comes from the Euclidean inner
product.\end{hint}\index{Euclidean!inner product} State and prove some theorems
in this more general setting in analogy with the treatment of
Chapter~\ref{ch!eucl}; in particular, if you feel brave, you can
classify completely motions\index{motion} of the 3-sphere
$S^3$\index{sphere $S^n$} following~\ref{sec!motE3}.

\item[] In \exref{ex!ge1}--\ref{ex!diverge}, consider the geometry of
hyperbolic plane\index{hyperbolic!plane $\sH^2$|(} $\sH^2$ with the
hyperbolic metric.

\item\label{ex!ge1} Hyperbolic distance\index{hyperbolic!distance}
is defined by $d(P,Q)=\arccosh(-\bv\cdot_L \bw)$. Adapt the argument
of the proof of Theorem~\ref{sec!bases} (3) to
prove directly that $-\bv\cdot_L \bw\ge1$ for $\bv,\bw\in\sH$.

\item\label{ex!hyparcs} Prove that $P(s)=(\cosh s,\sinh s)$ is the
parametrisation of the hyper\-bola $\sH^1:(-t^2+x^2=-1)\subset\R^2$ by arc
length in the Lorentz pseudometric\index{Lorentz!pseudometric} $q=-t^2+x^2$;
put more simply, $P(s+\dd s)-P(s)$ is $\dd s$ times a vector tangent to $Q$
at $P(s)$ of unit length for $q$. \begin{hint} If $P(s)=(\cosh s,\sinh s)$
then $\frac{\dd P}{\dd s}=(\cosh s,\sinh s)$, a unit space-like
vector.\end{hint}

\item\label{ex!circetal} \begin{enumerate}

\item Let $P=(1,0,0)\in \sH^2$; show how to parametrise the circle centre
$P$ and radius $r<\pi$ in $\sH^2\subset\R^3$. Deduce that a circle of
radius $r$ has circumference $2\pi\cdot\sinh r$; and that a disc with
centre $P$ of radius $r<\pi $ has area $2\pi\cdot(1+\cosh r)$. Your
formulas should be analogous to those for $S^2\subset\R^3$ in
\exref{S2circle}.

\item Deduce from (a) that there does not exist an isometric map from any
region of $\sH^2$ to a region of the Euclidean plane $\R^2$ or of the
sphere $S^2$.\index{sphere $S^2$}

\item\label{ex!pringle} A Pringle's potato chip\index{Pringle's potato
chip} is a reasonably accurate model in Euclidean 3-space of a hyperbolic
disc of radius $r=1$ (isometrically embedded). What happens if we try to
make one of radius $r=100$? \end{enumerate}

\item\label{ex!hyperrefls} Define a reflection\index{reflection} of $\sH^2$,
and prove properties analogous to those of reflections of $\R^2$: there
exists a reflection taking $P_1$ to $P_2$, any direct motion\index{motion}
of $\sH^2$ is a composite of 2 reflections,\index{composite!of reflections}
any opposite motion is a composite of 3 reflections, {\em Pons Asinorum},
etc. \begin{hint} follow the spherical case in \exref{ex!S2refl}.\end{hint}

\item\label{ex!hypperps}
\begin{enumerate}

\item Use the main formula
\[
\cosh\al=\cosh\be\cosh\ga-\sinh\be\sinh\ga\cos a
\]
to prove that in a right-angled hyperbolic
triangle,\index{hyperbolic!triangle} the hypotenuse is longer than
either of the other two sides. If $L\subset\sH^2$ is a line and
$P\in\sH^2$ a point not on $L$, deduce that the length of the
perpendicular dropped from $P$ to $L$ (if it exists) is the shortest
distance\index{shortest distance} from $P$ to $L$.

\item Consider the function $d(P, Q)$ for $Q\in L$; prove that $d(P,Q)$
takes a minimum value. \begin{hint} fix attention to a suitable
closed\index{closed} ball\index{ball} around $P$ and use the fact that a
function on a closed interval attains its bounds.\end{hint} Deduce
that a perpendicular from $P$ to $L$ exists and is unique.

\item If $L, M\subset\sH^2$ are lines not meeting in $\sH^2$ and not
ultraparallel, prove that $L$ and $M$ have a unique common perpendicular.
\end{enumerate}

\item Interpret the matrixes
\[
\begin{pmatrix} \cosh s&\sinh s&0\\ \sinh s&\cosh s&0
\\0&0&1 \end{pmatrix}
\quad\hbox{and}\quad
\begin{pmatrix} \cosh s&\sinh s&0\\ \sinh s&\cosh s&0
\\0&0&-1 \end{pmatrix},\quad
\]
as hyperbolic translation\index{hyperbolic!translation} and glide.

\item\label{ex!arccot} In Figure~\ref{fig!hpll}, let $Q=(1,0,0)$ and
$P=(t_0,0,y_0)$, so that the matrix
$\left(\begin{smallmatrix} t_0&0&y_0 \\ 0&1&0 \\ y_0&0&t_0
\end{smallmatrix}\right)$
defines a hyperbolic translation taking $Q$ to $P$. Show that the line
$L:(y=0)$ goes to $M:(t_0y=y_0t)$, and the line $y=(\tan\fie)x$ through
$Q$ at angle $\fie$ to $L$ (parametrised by $(t,(\cos\fie)r,(\sin\fie)r)$
with $-t^2+r^2=-1$) goes to $(\sin\fie)x=(\cos\fie)(y_0t-t_0y)$. Conclude
that the limiting angle $\theta$ in Theorem~\ref{sec!nonE} equals
$\cot\theta=y_0$.

\item\label{ex!hypharder} (harder) The formula $\cosh\al=\cosh\be\cosh\ga$
gives the hypotenuse $\al$ of a right-angle hyperbolic
triangle\index{hyperbolic!triangle} in terms of the other two sides
$\be,\ga$. Prove that this is always longer than the corresponding Euclidean
result $\sqrt{\be^2+\ga^2}$.

\item\label{ex!hypsin} Let $\al,\be,\ga$ be the sides (lengths) of a
hyperbolic triangle $\De PQR$ and $a,b,c$ the opposite angles. Prove the
hyperbolic sine rule\index{hyperbolic!sine rule}
\[
\frac{\sinh\al}{\sin a}=\frac{\sinh\be}{\sin b}=\frac{\sinh\ga}{\sin
c}.
\]
\begin{hint} argue as in \ref{h2trig} and \exref{ex!sphsin}.\end{hint}

\item\label{ex!asymp} The hyperbolic lines $L_c:(y=c(t-x))$ with $|c|<1$ are
ultraparallel,\index{ultraparallel lines} tending to $(1,1,0)$ at infinity
(see\index{infinity!point at} Definition~\ref{sec!2lines}). Verify that
$L_c$ is parametrised by arc length as
\[
L_c:P_c(s)=\left(t_0e^{-s}+\frac{1}{t_0}\sinh s,\frac{1}{t_0}\sinh s,y_0e^{-s}\right),
\]
where $y_0=\frac{c}{\sqrt{1-c^2}}$ and $t_0=\frac{1}{\sqrt{1-c^2}}$ (so that
$c=\frac{y_0}{t_0}$ and $P_c(0)=(t_0,0,y_0)\in L_c$). Calculate
$d(P_c(s),P_{-c}(s))$ and show that the two curves $L_{\pm c}$ approach
asymptotically as $s\to\infty$.

Since $L_0:(y=0)$ is sandwiched between $L_{\pm c}$ for any $c$ (e.g.,
$c=1/2$), it follows that $L_0$ and $L_c$ are asymptotically close. (But you
have to start the parametrisation by arc length at an appropriate point to
make the two parametrised curves converge.)

\item\label{ex!diverge} Suppose that $L_1$ and $L_2$ are divergent lines
hyperbolic lines as in Definition~\ref{sec!2lines}. Set up a parametrisation
by arc length as $L_1:P(s)$, $L_2:P'(s)$ and prove that $d(P(s),P'(s))$ must
grow at least linearly in the variable $s$.

\item\label{ex!hyparea}
\begin{enumerate}
\item Show that in polar coordinates
\[
x=r\cos\theta,\quad y=r\sin\theta,\quad t=\sqrt{1+r^2},
\]
the element of area in $\sH^2$ is
\[
\dd A= \frac{r\dd r\dd\theta}{\sqrt{1+r^2}}.
\]
\begin{hint} Consider a small sector
$[\theta,\theta+\de\theta]\times[r,r+\de r]$ in the space-like Euclidean
$\R^2$. Prove that the sector of $\sH$ lying over it is very close to a
hyperbolic rectangle with length of sides equal to $r\de\theta$ and
$\de r/\sqrt{1+r^2}$.\end{hint}

\item By writing down the Jacobian determinant for the change of coordinates,
check that the element of area in $\sH$ in the usual coordinates $(t,x,y)$ is
\[\dd A = \frac{\dd x\dd y}{t}.\]
\end{enumerate}

\item[] The final set of exercises \exref{ex!defH}--\ref{ex!hypgraphics} aim
to give an alternative {\em model} of hyperbolic
geometry,\index{hyperbolic!geometry} which may help you visualise some of
its properties. I set up a geometry on the complex upper half plane $\sH$
(\exref{ex!defH}), show that it is the same geometry as the hyperbolic plane
$\sH^2$\index{hyperbolic!plane $\sH^2$} (\exref{ex!Hidentify}), and
investigate the failure of the parallel postulate\index{parallel!postulate}
in the new model (\exref{ex!linesH}). If you want to read further on this,
look at Beardon~\cite{Bea}, Chapter~7.

\item\label{ex!defH} Let
\[
\sH=\{ z= x+iy\in\C \mid y>0\}
\]
be the {\em upper half plane} in the complex plane. Define {\em $\sH$-lines}
to be of two kinds (see Figure~\ref{H_lines}):
either vertical Euclidean half-lines
$L_1=\{x+iy\in\sH \mid x=c\}$ for a real constant $c$, or
half-circles $L_2=\{x+iy\in\sH \mid (x-a)^2+y^2=c^2\}$ with centre
$(a,0)$ on the real axis $\{y=0\}$.
\begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}}
\centering
\input{hyp_H_lines.pstex_t}
\caption{$\sH$-lines}
\label{H_lines}
\end{figure}

Show, algebraically or by drawing pictures, that
\begin{enumerate}

\item two $\sH$-lines meet in at most one point;

\item every pair of distinct points of $\sH$ lies on a unique
$\sH$-line.
\end{enumerate}

\item\label{ex!Hidentify} \begin{enumerate}

\item Consider the map $\fie$ defined by
\[
\fie\colon (T,X,Y)\mapsto \left(\frac{-Y+i}{T-X}\right).
\]
Show that if $(T,X,Y)\in\sH^2$ then $T-X>0$ hence $\fie$ is a map
from the hyperbolic plane $\sH^2\subset\R^{2,1}$ to the upper half
plane $\sH$.

\item Consider the map $\psi$ defined by
\[
\psi\colon (x+iy)\mapsto
\left(\frac{1+x^2+y^2}{2y},\frac{-1+x^2+y^2}{2y},\frac{-x}{y}\right).
\]
Show that if $x+iy\in\sH$ then its image $(T,X,Y)\in\sH^2$ hence $\psi$ is
a map from $\sH$ to $\sH^2$.

\item Show that $\phi$ and $\psi$ are inverse bijections between
$\sH$ and $\sH^2$.

\item Show that the image of a hyperbolic line\index{hyperbolic!line}
$L\in\sH^2$ is an $\sH$-line and conversely.

\item Let $z_1,z_2\in\sH$ be points of the upper half plane, and let
$\bv_i=\psi(z_i)$ be their images under $\psi$. Show, using the formulae
above, that
\[
-\bv_1\cdot_L\bv_2 = 1+ \frac{|z_1-z_2|^2}{2 \im(z_1)\im(z_2)}.
\]
Deduce that setting
\[
d_{\sH}(z_1,z_2)
=\arccosh\left(1+\frac{|z_1-z_2|^2}{2\im(z_1)\im(z_2)}\right),
\]
makes $(\sH,d_{\sH})$ into a metric space isometric to
$(\sH^2,d_{\sH^2})$.
\end{enumerate}

\item[] Therefore $\sH$ has a metric geometry, isometric to the hyperbolic
plane $\sH^2$.\index{hyperbolic!plane $\sH^2$|)} In particular, it has its
own symmetries, the $\sH$-motions\index{motion}. Sketch some cases like the
hyperbolic translations\index{hyperbolic!translation} and reflections on a
sheet of paper, starting from their geometric definitions. As a matter of
fact, any direct $\sH$-motion is of the form $z\mapsto
\displaystyle\frac{az+b}{cz+d}$ for a real matrix $\begin{pmatrix} a & b\\ c
& d\end{pmatrix}$ with $ad-bc>0$; indirect motions are given by $z\mapsto
\displaystyle\frac{a(-\bar z)+b}{c(-\bar z)+d}$. If you feel brave, try your
hand at proving that these maps preserve $\sH$ and its metric; consult
Beardon~\cite{Bea}, \S7.4 for the full story.

\item[] One further point deserves special mention: although there appear to
be two different types of $\sH$-lines, the set of
$\sH$-motions\index{motion} acts transitively on the set of $\sH$-lines.
This holds because the analogous statement is true in $\sH^2$, and the two
are the same!

\item\label{ex!linesH} (graphical exercise) Draw a point $P\in\sH$ and an
$\sH$-line $L$ not containing $P$. (To make your picture pretty, choose $L$
to be a half-circle and $P$ to be lying over its centre; of course you know
that all configurations are like that up to $\sH$-motions!)\index{motion}
Draw some lines through $P$ meeting $L$. Shade the region of $\sH$ covered
by lines through $P$ meeting $L$. Draw the ultraparallel
lines\index{ultraparallel lines} (see Definition~\ref{sec!nonE}) to $L$ from
$P$. For educational purposes, repeat the exercise with $L$ a ``vertical''
line. Now stare at your drawings and contemplate the vast regions in
hyperbolic space not contained in lines incident with $P$ and $L$, as
opposed to the case of $\E^2$ where this set is a line.

\item\label{ex!hypgraphics} (another graphical exercise)
Do~\exref{ex!hypperps}, (b--c) on $\sH$ without any computation, by drawing
the appropriate diagrams.

\end{enumerate}

\section*{Some solutions}
\markright{\hfill\ Some solutions\quad}

\paragraph{\rm \ref{ch!hyp}.\ref{ex!S2refl}} (a) If two sides of a triangle
are equal then so are the opposite angles and vice-versa (proof by finding a
reflection, \index{reflection} as in the Euclidean case). (b) The
perpendicular bisector\index{perpendicular bisector} of $P_1P_2$ in $\R^3$
intersects $S^2$ in a great circle $C$; the construction uses the ambient
notion of orthogonal and bisector\index{perpendicular bisector}. To see that
points of $C$ are equidistant from $P_1$ and $P_2$ and any point equidistant
from $P_1$ and $P_2$ is in $C$, find a reflection\index{reflection} as
in~(a).

\paragraph{\rm \ref{ch!hyp}.\ref{ex!engon}} $\sum a_i=\area\Pi+(n-2)\pi$;
proof by subdividing into triangles.

\paragraph{\rm \ref{ch!hyp}.\ref{ex!ineqbe}} The first two inequalities come from
$|\cos a|<1$; for
the third, note that if $\al,\be,\ga$ are sides of $\De PQR$ then
$\al,\pi-\be,\pi-\ga$ are sides of $\De(-P)QR$. Finally, fix $\be$ and $\ga$
and view $\al$ as an increasing function of $a$.

\paragraph{\rm \ref{ch!hyp}.\ref{ex!sphsin}} The notation in \ref{s2trig} was
\[
\bq=\bp\cos\be+\bq'\sin\be,\quad r=\bp\cos\ga+\br'\sin\ga,
\]
where $\bq',\br'$ are orthogonal to $\bp$. Now in vector
notation
\[
\det(\bp,\bq,\br)=\bp\cdot(\bq\times \br)=|\bq'\times \br'|\sin\be\sin\ga=\sin
a\sin\be\sin\ga.
\]
If you do not like vector products, you can massage the determinant into the
form
\[
\left|
\begin{matrix}
1&\cos\ga&\cos\be\\ 0&\sin\ga&\cos a\sin\be\\ 0&0&\sin a\sin\be
\end{matrix}
\right|
\]

\paragraph{\rm \ref{ch!hyp}.\ref{ex!8}} (a) True. The unit vector
proportional to $(\bu+\bv+\bw)/3$ is coplanar with the unit vector
proportional to $(\bu+\bv)/2$ and $\bw$; (b) True. The points $S$ of
perpendicular bisector\index{perpendicular bisector}s satisfy $d(S, P)=d(S,
Q)$ etc.\ (see \exref{ex!S2refl}). Hence all meet in a point $S$ which
satisfies $d(S,P)=d(S,Q)=d(S,R)$.

\paragraph{\rm \ref{ch!hyp}.\ref{ex!circetal}} (a) The points of $Q$ at
hyperbolic distance\index{hyperbolic!distance} $t$ from $P$ are
\[
(\sinh t,0,0)+(\cosh t)(0,\cos\theta,\sin\theta)
\quad\hbox{for}\quad\theta=[0,2\pi];
\]
the element of arc length is space-like, so the circumference is the same as
that of a Euclidean circle of radius $\sinh t$. For the area of the disc,
integrate. (b) Clear. (c) When $r=100$, the circumference is millions of
kilometres long.

\paragraph{\rm \ref{ch!hyp}.\ref{ex!hyperrefls}} A line $L\subset\sH^2$
corresponds to a 2-dimensional subspace $U\subset\R^3$ s.t.\ The form $q$
restricted to $U$ has type $(-1,+1)$; so $U^\perp$ is 1-dimensional, and
$s\colon\R^3\to\R^3$ defined as $+1$ on $U$, $-1$ on $U^\perp$ is the
reflection\index{reflection} in $L$. The usual Euclidean proof that every
motion\index{motion} of $\sH^2$ is a composite of $\le n+1$
reflections\index{composite!of reflections} goes through without trouble.

\paragraph{\rm \ref{ch!hyp}.\ref{ex!hypperps}} (a) $\cos a=0$, so
$\cosh\al>\cosh\be$, $\cosh\ga$. (b) If $Q\in L$ and $d(P, Q)=d$ then the
inf of $d(P, Q)$ for $Q\in L$ equals the inf for $Q\in L\cap B(d,P)$
(closed\index{closed} ball).\index{ball} This is a closed interval, so
$d(P,L)=d(P,Q_0)$; (c) same argument; the assumption is that
$d(P,Q)\to\infty$ as $P,Q\to\infty$ on $L,M$.

\paragraph{\rm \ref{ch!hyp}.\ref{ex!hypharder}} By brute force: write out
$\cosh\sqrt{y^2+z^2}$ and $\cosh y\cosh z$ term by term and compare terms;
this works because all terms are positive. There should be a good proof
based on general properties of $\cosh$. I also do not know how to prove the
corresponding spherical formula $\cos\sqrt{y^2+z^2}\le\cos y\cos z$ for
$y,z$ in an appropriate range, which says that a spherical right angled
triangle\index{spherical!trig} has smaller hypotenuse than a Euclidean one.
(The function $\cos\sqrt x$ is certainly a well defined analytic function,
but I do not know anything about it.)

\paragraph{\rm \ref{ch!hyp}.\ref{ex!hypsin}} The coordinates of $P,Q,R$ can be
taken to be
\[
\bp,\bq=(\cosh\be)\bp+(\sinh\be)\bbe,\br=(\cosh\ga)\bp+(\sinh\ga\cos a)\bbe+(\sinh\ga\sin a)\bbf,
\]
where $\bp,\bbe,\bbf$ is a Lorentz basis,\index{Lorentz!basis}; so
$\det(\bp,\bq,\br)=\sinh\be\sinh\ga\sin a$.

\index{hyperbolic!geometry|)}
\index{non-Euclidean geometry|)}

