\input macros.tex
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\vskip 2in
\centerline{\bf Introduction}
\medskip

In this thesis I will look at some of the methods that are employed to
gain a better understanding of complete three-dimensional manifolds. I will
explain the use of geometry in classifying manifolds. The focus will then be
exclusively on those manifolds which admit a hyperbolic
geometry. I will look into the volume of hyperbolic 3-manifolds as a useful
invariant and survey some useful properties of this invariant. I conclude with
a study of the two manifolds with the lowest known volumes.
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\centerline{\bf Thurston's Geometrisation Conjecture}
\medskip
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\beginsection Two-Dimensional Manifolds

To get an idea of the concepts involved we start by looking at two-dimensional
complete orientable manifolds. The machinery we will use is somewhat more
powerful than we need, but this is necessary to prepare us for the
three-dimensional case.

We attempt to gain a better understanding of a manifold by giving it a Riemannian
metric. We require this metric to be homogeneous which means that for any two
points of the space there must be at least one isometry taking one to the other.
In addition, we require the metric to be isotropic which means that for any two
frames at a point there is an isometry fixing the point and taking one frame to
another. Roughly speaking, the metric looks the same at every point and in every
direction. The dual conditions of homogeneity and isotropy are actually very
restrictive. In any dimension there are only three essentially different
geometries satisfying these conditions: those with zero sectional curvature,
those with constant positive sectional curvature, and those with constant
negative sectional curvature. After scaling, the curvature can be taken to be 0,
1 or -1. The corresponding geometries are called Euclidean, spherical and
hyperbolic geometry, respectively.

Euclidean geometry is the easiest to grasp intuitively since it closely
approximates the geometry of the real world. The two-dimensional Euclidean
geometry can be modelled by endowing the plane $\R2$ with the Euclidean metric
$d\rho=\sqrt{{\rm d}x^2+{\rm d}y^2}$.

The two-dimensional spherical geometry is
also fairly easy to visualise since it can be thought of as geometry on the
surface of a sphere. Geodesics are great circles while angles and distances are
inherited from the three-dimensional Euclidean space.

Hyperbolic geometry is the least familiar to most people, but also the most
important for two- and three-dimensional topology. In order to help us
understand hyperbolic geometry we construct models by which we
identify hyperbolic space with some object in Euclidean space. To do this
we must distort the space so that objects represented in the model may appear
stretched in some places, shrunk in others, bent, twisted or otherwise distorted
from its true nature. Different models distort in different
ways and are useful for different purposes. The one we will be using is the upper
half-space model which is the easiest to do computational work in and also
fairly easy to grasp intuitively.

Here is a brief description of the upper half-plane model for two-dimensional
hyperbolic space. We shall discuss the three-dimensional equivalent in more
detail later.

We identify two-dimensional hyperbolic space with the upper
half-plane in {$\R2$} $$\{(x,y):y>0\}$$ and endow it with the metric
$ds={d\rho\over y}$ where $d\rho$ is the Euclidean metric. Observe that with
this metric, the line $\{(x,0)\}$ is infinitely far away from any point in the
space. An object approaching this line at constant speed in hyperbolic space
will appear in this model to decrease in size and speed in such a way that it
never reaches the boundary.

Geodesics are given by those Euclidean semicircles and lines orthogonal to the
boundary $y=0$. If we consider each point $z=(x,y),\ y>0$ as the complex number
$x+iy$ then the hyperbolic isometries are the M\"obius transformations:
$$g(z)={az+b\over cz+d}; \ a,b,c,d\in {\bf R}, ad-bc>0.$$ It is fairly easy to
check that the condition $ad-bc>0$ ensures that g(z) is in the upper half plane.
%??map is not constant and is defined everywhere except at $-d/c$ if $c\ne0$. We
extend this action to $\Chat={\bf C}\cup\{\infty\}$. If $c=0$ we define
$g(\infty)=\infty$ while if $c\ne0$ we define $g(\infty)=a/c$ and
$g(-d/c)=\infty$. This gives us a 1-1 map of $\Chat$ onto itself.

\beginsection Tiling the universal cover

We can cut any 2-manifold along a sequence of curves to obtain a polyhedron.
Thus any two-manifold can be thought of as a polyhedron whose edges are
identified in some way. We will use copies of these polyhedra to obtain a
tiling of the universal cover. We will see that the nature of this
tiling determines the geometry of the manifold.

Perhaps the simplest example of this process is given by the torus T, which can
be cut open to form a square with opposite sides identified. Place a square in
the Euclidean plane. For convenience assume its sides have length one and are
parallel to the $x$ and $y$ axes. A path leaving through one side of the square
should enter through the opposite side so we place four new copies of the square
adjacent to each edge of the original square. A point on any of these squares is
to be identified with the corresponding point in the original square. This
identification is made by a horizontal or vertical Euclidean translation of
length one.

We continue adding squares in this way until we obtain a tiling of the Euclidean
plane. Points in the plane are identified if one can be obtained from the other
by a sequence of horizontal or vertical Euclidean translations of length one,
that is if one can be obtained from the other by some element of the group
$\Gamma$ of Euclidean isometries generated by the horizontal translation of
length one and the vertical translation of length one. We see that $\Gamma$ is a
subgroup of the group of Euclidean isometries and is isomorphic to
$\pi_1(T)={\bf Z}\oplus{\bf Z}$. Conversely given a group of isometries we can
reconstruct the manifold by identifying any two points in the plane that are
mapped one to the other by some element of the group.

Every point in the torus is identified with an infiinite set of points in the
Euclidean plane. We have an induced metric on the torus by defining the
distance between two points to be the shortest distance between two of their
corresponding points in the Euclidean plane. This metric is locally Euclidean.
We say that the torus admits a Euclidean geometry.

This technique is a very useful tool for gaining information about a manifold.
As well as studying topological properties of the torus we can now also study
properties of its Euclidean metric and its group of Euclidean isometries.

Let us try the same technique on the genus two surface. This can be obtained by
identifying appropriate edges on the octagon. In this identification, all eight
vertices are identified so that if we wanted to obtain a tiling of the plane as
we did with the torus, we would require eight octagons to fit around every
vertex. Clearly this is impossible to do in the Euclidean plane; the interior
angle of the octagon is just too big. However if we consider an octagon in a
negatively curved space we see that the negative curvature means that the
interior angles are less than the usual $135^\circ$. The larger the octagon is,
the greater the effect of this negative curvature and the smaller the angles.
For a regular octagon of just the right size, the internal angles are all
$45^\circ$ and it is possible to fit eight around each vertex. Thus we can
obtain a tiling of hyperbolic space. We can consider the genus two surface
as the hyperbolic plane with points identified by elements of some group of
hyperbolic isometries. From this tiling the genus two surface inherits a
hyperbolic metric.

We saw in the above example that the octagon had to be exactly the right size
for the tiling to work. In fact the area of the octagon is an invariant of the
manifold and does not depend on the way it is cut open. Also no two distinct
hyperbolic 2-manifolds can have the same area. Thus we can use this area
function to completely classify all 2-dimensional hyperbolic manifolds.

Of the two-dimensional orientable manifolds, the sphere can be given a
spherical geometry, the torus can be given a Euclidean geometry and all of the
rest are hyperbolic. Thus having classified all hyperbolic 2-manifolds it is a
short step to classifying all 2-manifolds.

\beginsection Three-dimensional geometries

In order to better understand 3-manifolds, we would like to follow a similar
program as the one we have oulined for two dimensions. In
addition to the three homogenous isotropic geometries $\E3$, $\S3$ and $\H3$ we
also require five additional geometries which are homogeneous but not isotropic.
These look the same at every point but not in every direction. They are
called $\S2\otimes\E1$, $\H2\otimes\E1$, PSL(2,{\bf R}), Nil and Solve.

One problem that arises in attempting to assign a geometry to each orientable
3-manifolds is that it is possible to glue together manifolds with different
geometries, resulting in a manifold which will not permit a single
geometry. We resolve this by first subdividing our manifold into simpler pieces
by the following methods
\item{i} Sphere splitting: Cut the
manifold along a two sphere and fill the wound with a ball.
\item{ii} Torus splitting: Cut the manifold along an incompressible torus.

A manifold in which no further sphere or torus splitting is possible is called
a simple manifold.

We can cut up a simple manifold to get a polyhedron with faces identified.
We want to use copies of this polyhedron to tile one of the eight geometries
listed above. The following conjecture states that we can always do this.

\proclaim Thurston's Geometrisation Conjecture. Every simple 3-manifold admits
one of the eight geometries listed above.

>From Thurston's work it appears that as in the two-dimensional case the generic
3-manifold is hyperbolic. Hyperbolic manifolds display the richest variety and
the greatest complexity. Thus the task of understanding hyperbolic 3-manifolds
is one of great importance. For the rest of this thesis, manifolds will be
assumed to be 3-dimensional, orientable and hyperbolic unless otherwise stated.

We will pause now to give a brief account of hyperbolic geometry.

%\bye
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\centerline{\bf Hyperbolic Geometry}
\medskip
%\input macros
The following account of hyperbolic geometry is intended only as a brief
summary. Many facts are given without proof and I will only cover the concepts
which are required in the rest of this thesis.
% For a fuller account of hyperbolic geometry see ??HELP

\beginsection The upper half-space model

There are many different models of hyperbolic space, each with its own
advantages and disadvantages. For our purposes the upper half-space model is
the easiest to work with. We identify the hyperbolic space $\H3$ with the upper
half space in $\R3$, namely those points $(x,y,z)$ with $z>0$. We endow this with
the metric $ds={d\rho\over z}$ where $d\rho$ is the Euclidean metric
$\sqrt{{\rm d}x^2+{\rm d}y^2+{\rm d}z^2}$.

The $x$-$y$ plane together with a special point $\infty$ is called the sphere
at infinity. This is not actually a part of the hyperbolic space, being
infinitely far away from any point. When we talk about behavior on the sphere at
infinity, we are really talking about limiting behavior.

Between any two distinct points there is a unique curve between them
with a shortest length. In other words there is a unique curve $\gamma$
from one point to the other which minimises the integral $\smallint_\gamma ds$.
Such curves are in fact arcs of Euclidean circles which are orthogonal to the
$x$-$y$ plane. Thus the geodesics are Euclidean semi-circles
which are orthogonal to the bounding plane. We also consider vertical lines
on which $x$ and $y$ are constant to be ``degenerate'' semicircles.
\beginsection Hyperbolic isometries

We start with a qualitative description of the isometries
of hyperbolic space which should help to give an intuitive understanding. We will
then give an algebraic description which is most useful in performing
calculations. Note that the isometries described here are only those which
preserve orientation. Isometries which do not preserve orientation can be
obtained by simply reflecting an orientation preserving isometry along some
plane in hyperbolic space. Since I will be dealing primarily with orientable
manifolds, isometries will usually be assumed to be orientation preserving.

First observe some important properties hyperbolic isometries must have.
Distances must by definition be preserved. From this it follows that geodesics
must be mapped to geodesics, and spheres and planes must be mapped to spheres
and planes.

Lef $f:\H3\to\H3 $ be an orientation-preserving isometry other than the
identity. We define an {\sl axis} of $g$ to be a line that is 
left invariant under $g$ and on which $g$ acts as a translation (possibly
trivial). This axis can be shown to be unique, for example see [Th2].

If a non-trivial orientation-preserving isometry $g$ of $\H3$ has an
axis that is fixed pointwise by $g$ then it is called an {\sl elliptic}
isometry and acts as a rotation about its axis.

If $g$ has an axis which is translated non-trivially then we will call it
{\sl loxodromic}, although some authors call these hyperbolic. Such an isometry
can be a pure translation or it can be a translation combined with a rotation
about the axis, a sort of screw motion.

If $g$ has no axis then it is called {\sl parabolic}. For example, any isometry
of $\H3$ which acts as a non-trivial Euclidean translation parallel to the
bounding plane in the upper half-space model is parabolic.

If $g$ fixes two points on the sphere at infinity then the unique geodesic
joining these points must be mapped to itself and must therefore be an axis of
$g$. Since $g$ cannot have more than one axis, $g$ can have no more than two
fixed points on the sphere at infinity. Also $g$ must have at least one fixed
point on the sphere at infinity since any map of a sphere onto itself must have
a fixed point. If $g$ has a fixed point  inside hyperbolic space then it must
fix a geodesic joining that point to some fixed point on the sphere at infinity
so $g$ must be elliptic. From this discussion we have the following
equivalent definitions to those given above:
\item{i} $g$ is parabolic if it has a unique fixed point
\item{ii} $g$ is loxodromic if it has exactly two fixed points
\item{iii} $g$ is elliptic if it has infinitely many fixed points.

\noindent In the first two cases the fixed points will be on the sphere at
infinity.

The following result is well known.
\proclaim Theorem. Let $z_1,z_2,z_3$ and $w_1,w_2,w_3$ be two triples of
distinct points on the sphere at infinity. Then there is a unique isometry
mapping $z_1,z_2,z_3$ to $w_1,w_2,w_3$ respecively.

Note that uniqueness can be proved by observing that if $f$ and $g$ are two
such isometries then $fg^{-1}$ fixes three points on the sphere at infinity and
thus is the identity map.
\smallskip
We now turn to an
algebraic description of the hyperbolic isometries. One of the advantages of the
upper half-space model is that this will take a particlarly simple form. We
start by associating each point $(x,y,z),\ z>0$ with the quaternion $x+yi+zj$.
We first look at isometries on the sphere at infinity which we associate with
the complex numbers together with the point $\infty$. We then extend these
actions into the upper half-space.

Consider a map acting on {\bf C} given by $$g(z)={az+b\over cz+d}$$
where $a,b,c,d\in{\bf C}$ and $ad-bc\ne0$. This latter condition ensures that
$g$ is not a constant map. It also ensures we do not have $c=d=0$, so $g$ is
defined for all ${z\in\bf C}$ except if $c\ne0$ and $z=-d/c$. We extend the
function $g$ to $\Chat={\bf C}\cup\{\infty\}$ as follows: if $c=0$ we define
$$g(\infty)=\infty,$$ and if $c\ne0$ we define $$g(-d/c)=\infty,{\rm\ and\ }
g(\infty)=a/c.$$ With these definitions it is not hard to show that the
function given by $$g^\prime(z)={dz-b\over -cz+a}$$ is the inverse of $g$, so $g$
is a one-to-one map of $\Chat$ onto itself.

This gives a complete description of the action of all M\"obius
transformations on the $x$-$y$ plane in the upper half-space model. Given
a function $$g(z)={az+b\over cz+d},\ ad-bc=1$$ acting on $\Chat$ there is a
unique way of extending this to an isometry in the upper half-space.
This is called the Poincar\'e extension and is given by
$g(w)=(aw+b)(cw+d)^{-1}$ where $w=z+tj$ for some positive $t$.
\beginsection Matrix representation

Define a map $\Phi:\GL\to{\bf M}$ by $\Phi(A)=g_A$ where
$$A=\left(\matrix{a&b\cr c&d}\right),\ g_A(z)={az+b\over cz+d}.$$We say that
the matrix $A$ represents the M\"obius transformation $g$. It can be shown that
the composition of two M\"obius transformations is represented by the product
of their matrices. That is,$$g_A(g_B(z))=g_{AB}(z)$$for all $z\in\Chat$
and $A,B\in\GL$. Thus $\Phi$ is a homomorphism. Now the matrix $A$ is in the
kernel of $\Phi$ if and only if the map$$g_A(z)={az+b\over cz+d}$$is the
identity map in $\Chat$. Clearly this is only the case when $b=c=0$ and $a=d$,
that is if $A=aI$ where $I$ is the identity matrix. In other words, the
function $g_A$ determines the matrix $A$ to within a non-zero multiple. Thus
the group of M\"obius transformations is isomorphic to the group P\SL. We will
usually treat M\"obius transformations as elements of \SL, with the
understanding that a matrix and its negative are to be considered identical.

Two important properties of a matrix $A\in\SL$ are its norm and its trace, given
by $$\tr(A)=a+d$$ and $$\|A\|=\sqrt{|a|^2+|b|^2+|c|^2+|d|^2}$$ respectively. For
a given M\"obius transformation $g$ the corresponding matrix $A\in\SL$ is
determined up to a factor of $\pm1$. However the values of
$\tr^2(A)=\left(\tr(A)\right)^2$ and $\|A\|$ are determined uniquely by $g$ so
we can write without ambiguity $\tr^2(g)=\tr^2(A)$ and $\|g\|=\|A\|$. The
importance of these properties is shown by the following two theorems. \proclaim
Theorem. For each M\"obius transformation g we have
$$\|g\|^2=2\cosh\rho(j,gj).$$ where $\rho$ is the hyperbolic metric.

Thus knowing $\|g\|^2$ immediately tells us how far the point (0,0,1) is
moved by $g$. This is Theorem 4.2.1 in [Be].
\proclaim Theorem. Let $f$ and $g$ be
M\"obius transformations, neither equal to the identity. Then $f$ and $g$ are
conjugate if and only if $tr^2(f)=tr^2(g)$.

Before proving this we introduce certain M\"obius transformations with a
particularly simple form. For each non-zero $k$ we define $m_k$
by$$m_k(z)=kz\ (k\ne1)$$and$$m_1(z)=z+1.$$ Note that for every $k$, including 1,
$\tr^2(m_k)=k+(1/k)+2$. Also, for all $k$ not equal to 1 or 0 we can conjugate
$m_k$ with $\left(\matrix{0&1\cr-1&0}\right)$ to obtain $m_{1/k}$. So for all
$k\ne0$, $m_k$ is conjugate to $m_{1/k}$.
\proclaim Lemma. Every M\"obius transformation $f$ not equal to the identity is
conjugate to some $m_k$.

{\sl Proof of lemma:} We have seen that $f$ has either one
or two fixed points in $\Chat$.

If $f$ has a unique fixed point $x\in\Chat$ then choose arbitrary
$y\in\Chat$ other than $x$ and let $t$ be a M\"obius transformation such that
\settabs 4\columns
\+&$t(x)=\infty$&$t(y)=0$&$t\left(f(y)\right)=1.$\cr
\noindent We see $tft^{-1}$ has a unique fixed point at $\infty$ and sends 0 to
1. From this we conclude $tft^{-1}=m_1$.

If $f$ has two fixed points $x,y\in\Chat$ then let $t$ be a M\"obius
transformation such that
\settabs 3\columns
\+&$t(y)=0$&$t(x)=\infty$\cr
\noindent Then $tft^{-1}$ has two fixed points at 0 and $\infty$. From this we
conclude $tft^{-1}=m_k$ for some $k\ne1$.

{\sl Proof of theorem:} Firstly, it is easy to show that for any
matrices $A$ and $B$, $\tr(AB)=\tr(BA)$ and thus
$\tr(A^{-1}BA)=\tr(BAA^{-1})=\tr(B)$.

For the ``if'' direction we choose $p,q\in\Chat$ such that $f$ is conjugate
to $m_p$ and $g$ is conjugate to $m_q$. Since
$\tr^2(f)=\tr^2(g)$ and trace is preserved under conjugation we conclude
$\tr^2(m_p)=\tr^2(m_q)$ so $p+(1/p)+2=q+(1/q)+2$ whence $p=q$ or $p=1/q$. In
either case, $m_p$ is conjugate to $m_q$. Thus $f$ is conjugate to $g$.
\smallskip
%\bye
\vfill\eject
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\centerline{\bf Well-ordering of volumes of hyperbolic
3-manifolds} \medskip
%\input macros
\beginsection The volume of a hyperbolic manifold

We saw in the earlier example of the genus two surface that the octagon had to
be exactly the right size in order to form a tesselation. The same sort of result
applies in three dimensions. Recall that we can represent a manifold as a
polyhedron with faces identified. In the case of a hyperbolic manifold we can
embed this polyhedron in $\H3$ in such a way that the dihedral angles and solid
angles at vertices all add up correctly. This then gives us a tiling
of the hyperbolic space. If we think of $\H3$ as the universal cover of the
manifold then the fundamental group is given by transformations of one lift to
another, that is hyperbolic isometries from one copy of the polyhedron to
another. The following definition describes a way of recovering a polyhedron
from this group of isometries.
\proclaim Definition. For a given group of hyperbolic isometries, the Dirichlet
domain with center $x$ is the set of points in hyperbolic space which are closer
to $x$ than to any image of $x$ under the action of the group.

The following theorem can be used to show that the volume of such a polyhedron is
an invariant of the manifold and does not depend on such things as the way
we originally cut up the manifold or the choice of the basepoint $x$ in the
above definition.

\proclaim Mostow's rigidity theorem. Let $M$ and $N$ be complete hyperbolic
3-manifolds with finite volumes. If there is an isomorphism $f:\pi_1(M)\to
\pi_1(N)$ then there is an isometry $\phi:M\to N$ so that the induced map
$\phi_*:\pi_1(M)\to \pi_1(N)$ is equal to $f$.

In particular the hyperbolic structure on a 3-manifold is unique. We can deduce
that any geometric invariant of the hyperbolic metric is actually a topological
invariant. In particular, the volume of a Dirichlet domain for a manifold is an
invariant and is called the volume of the manifold. This turns out to be a
tremendously important invariant. Jeff Weeks has designed a computer program
which will calculate volumes of manifolds. All the available computer evidence
shows this invariant to be very useful for distinguishing manifolds, although
there are examples of distinct manifolds which have the same volume. Many nice
properties of the volume function are being discovered, one of which we outline
here.
\beginsection Volumes are well-ordered

Let $\F$ be the set of all finite volumes of complete orientable
hyperbolic 3-manifolds.
\proclaim Thurston's theorem. The set $\F$ forms a closed
non-discrete set on the real line. It is well ordered with ordinal type
$\omega^\omega$. Also, there are only finitely many manifolds with a given
volume.

Our aim for the rest of this chapter is to review the basic ingredients which go
into proving this. In the process we will meet some very useful concepts
which have applications beyond the scope of this theorem.

For $\F$ to be well ordered means that any subset of $\F$ has a smallest element. In
particular $\F$ itself has a smallest element and for any given element there
is a next smallest element. The ordinal type $\omega^\omega$ corresponds to the
natural ordering of all polynomials in $\omega$ with integral coefficients
$\ge0$. Here is a schematic picture of the values of $vol(M)$.
\bigskip
\newcount\n
\def\twothirdsn{\multiply\n2\divide\n3}
\def\tick#1{\vrule height #1true pt depth #1true pt width.1true pt}
\def\om#1#2{\n=#1{}\ifnum\n>20000%
  \hbox to\n true sp{\tick{#2}\hfil\twothirdsn\om{\n}{#2}}\fi}
\def\Om#1#2#3{\n=#1%
  \hbox to\n true sp{\tick{#3}\hfil\twothirdsn\om{\n}{#2}}}
\def\omsq#1#2#3{\n=#1{}\ifnum\n>20000%
  \hbox{\Om{\n}{#2}{#3}\twothirdsn\omsq{\n}{#2}{#3}}\fi}

\centerline{{\raise 1true pt\vbox to50true
pt{\omsq{3000000}{0}{1}\hbox{\kern-.3em1}\vfil}%
\raise 4true pt\vbox to50true
pt{\omsq{2000000}{1}{4}\hbox{\kern-.3em$\omega$}\vfil}%
\raise 8true pt\vbox to50true
pt{\omsq{2000000}{4}{8}\hbox{\kern-.3em$\omega^2$}\vfil}%
\raise12true pt\vbox to50true
pt{\omsq{2000000}{8}{12}\hbox{\kern-.3em$\omega^3$}\vfil}%
\raise16true pt\vbox to50true
pt{\hbox{\tick{16}}\hbox{\kern -.3em$\omega^4$}\vfil}%
\raise 6true pt\vbox to50true
pt{\hbox{etc$\ldots$}\vfil}}}
In this diagram each vertical line represents a volume of some hyperbolic
manifold. The smallest vertical lines represent isoleted points which we
shall see correspond to compact manifolds. The second smallest vertical lines
represent limit points which we shall see correspond to volumes of manifolds
with one cusp. These limit points themselves have limit points which are
represented by larger lines and correspond to the volumes of two-cusped
manifolds. These larger lines have limits represented by still larger lines, and
so on. Note that some of the limiting sequences have been omitted for obvious
reasons of space.
\beginsection The thick-thin decomposition

Let $M$ be a complete hyperbolic 3-manifold and take any $x\in M$. Because the
group of covering transformations is discrete, there is a minimum distance $d$
between any two lifts of $x$ in the universal cover $\H3$. This is also the
shortest possible length of a noncontractible loop based at $x$. An open ball of
radius $r=d/2$ about $x$ is embedded since all its lifts are disjoint. This
radius is called the {\it injectivity radius\/} of $M$ at $x$ written $r=i_r(x)$

For any $\epsilon>0$ we can decompose $M$ into two components,
$M=\Mthick\cup\Mthin$, where $$\Mthick=\{x\in M:i_r(x)\ge\epsilon\}$$ and
$$\Mthin=\{x\in M:i_r(x)\le\epsilon\}$$We see that the points in
$\H3$ that lie above $\Mthin$ are those points which are moved a small distance
by some non-trivial element of $\pi_1(M)$.
\proclaim Proposition. If $M$ is a complete hyperbolic manifold of finite
volume then $\Mthick$ is compact.

{\sl Proof:} A ball of radius $\epsilon$ in $M$ about a point in $\Mthick$
must be embedded and hence has the same volume as a ball of radius $\epsilon$ in
hyperbolic space. Since $M$ has finite volume, there is an upper bound to the
number of $\epsilon$ balls which can be disjointly embedded in $M$. If we take a
maximal set of disjointly embedded $\epsilon$ balls and then double their radii
we obtain a covering of $\Mthick$ with a finite number of balls of radius
$2\epsilon$. Thus $\Mthick$ is compact.
\beginsection The Margulis lemma

The Margulis lemma basically tells us that for small enough $\epsilon$, the
$\epsilon$-thin component of a complete hyperbolic manifold is not too
complicated. We will use this fact to ultimately gain a complete description of
the $\Mthin$. The Margulis lemma as originally formulated is very general and
applies to all dimensions. We are only concerned with the special case of
hyperbolic 3-manifolds.

\proclaim Theorem (Special case of the Margulis lemma). There exists a constant
$\mu$ such that for each orientable hyperbolic 3-manifold $\tilde M$ and each
$x\in M$ the subgroup $\Ge(x)$ of $\pi_1(M)$ generated by elements which move
$x$ distance $\le 2\mu$ has an abelian subgroup of finite index.

{\sl Sketch of proof:} We define a subgroup $\Gep(x)$ consisting of
elements of $\Ge(x)$ with derivative at $x$ $\epsilon$-close to the identity.
%??derivative??

We show that there is a constant $\delta$ such that the commutator of two small
elements is bounded above by $\delta$ times the product of their sizes. Since
$\Gep(x)$ is generated by small elements it
follows that the in the sequence
$$\Gep\supset[\Gep,\Gep]\supset[\Gep,[\Gep,\Gep]]\supset[\Gep,[\Gep,[\Gep,\Gep]]]\supset\ldots$$
the size of elements is continually decreasing. Since $\Gep(x)$ is discrete,
there can be no infinite sequence of elements of decreasing volume, so it follows
that the sequence must be finite. In other words, $\Gep(x)$ is nilpotent. By
considering the geometric classification of hyperbolic isometries it can be shown
that nilpotent groups of isometries are in fact abelian.

We then choose an $\epsilon_1$ so that ${\Gamma^\prime}_{\epsilon_1}$ is abelian
and show that for small enough $\epsilon$, products of generators of
$\Ge(x)$ will lie in $\Gamma_{\epsilon_1}$.

\proclaim Corollary. There is a $\mu>0$ such that for any complete oriented
hyperbolic 3-manifold $M$ and for every $\epsilon\le\mu$, each component of
$\Mthin$ is either \item{i} A sphere centered at infinity, called a horoball, or
\item{ii} A solid tube around some a geodesic.

{\sl Proof:} Take $x\in\Mthin$ and let $\tilde x\in\H3$ be some lift of $x$.
There must be a $\gamma\in\pi_1(M)$ which moves $\tilde x$ a distance less than
or equal to $\epsilon$.

Suppose $\gamma$ is loxodromic with axis $a$. Then the distance a point
is moved by $\gamma$ depends only upon the distance of that point from the
axis $a$. Moreover, the closer a point is
to $a$ the less it will be affected by the rotational component of $\gamma$
and the shorter the distance moved. It follows that the set of points moved a
distance less than $\epsilon$ by $\gamma$ must be a tube $B_r(a)$
containing the point $\tilde x$.

Suppose on the other hand $\gamma$ is parabolic with a fixed point $p$ on the
sphere at infinity. First consider the case $p=\infty$ in the upper half-space
model. Then $\gamma$ acts as a Euclidean translation parallel to the $x-y$
plane. The distance a point is moved by such a translation is inversely
proportional to the height of that point above the $x$-$y$ plane. Thus the set of
points moved a distance less than $\epsilon$ by $\gamma$ must be equal to the
set of points above some plane parallel to the $x$-$y$ plane. This is called a
horoball with center $\infty$. In general, when $p\ne\infty$ we obtain a
horoball centered at $p$. This will be the image of a
horoball centered at $\infty$ under some isometry taking $\infty$ to $p$. Because
hyperbolic isometries take spheres and planes to spheres and planes, this
horoball can be thought of as a sphere of infinite radius centered at $p$. In the
upper half-space model it looks like a Euclidean sphere tangent to the
$x$-$y$ plane at the point $p$. Note that because of the homogeneity of
hyperbolic space there is nothing special about the horoballs centered at
$\infty$, it was just easier to describe them first.

We have shown that the lift of $\Mthin$ into the universal cover
$\H3$ is a union of horoballs and solid cylinders. Whenever two of
these are not disjoint, their corresponding group elements $\gamma_1$ and
$\gamma_2$ both move some point a distance less than or equal to $\epsilon$. We
can use the Margulis lemma to show that $\gamma_1$ and $\gamma_2$ must commute.
It is not hard to see that this can only happen if $\gamma_1$ and $\gamma_2$
have the same set of fixed points on the sphere at infinity. Thus the
corresponding horoballs or solid tubes must be concentric. It follows that 
the lift of $\Mthin$ into $\H3$ is a union of disjoint horoballs and solid tubes.
\smallskip
We have now shown that any finite volume hyperbolic manifold can be decomposed
into a compact part bounded by tori and a finite number of cusps and solid tubes
glued to these bounding tori. A cusp can be thought of as $T\times [0,\infty)$
where the torus gets exponentially smaller as we approach infinity. A tube can
be thought of as a solid torus with whose boundary is glued to the bounding
torus. Such an identification is called Dehn surgery. There are many ways of
performing this surgery, which basically correspond to many ways the boundary of
the solid torus can be twisted before identification. Another way of thinking of
this is that a short geodesic in the boundary of the solid torus can be
identified with many different loops in the boundary of $\Mthick$. We see then
that a choice of Dehn surgery corresponds to a choice of group element of ${\bf
Z}\oplus{\bf Z}$, the fundamental group of the torus bounding $\Mthick$.

The following theorem, due to J{\o}rgensen gives a description of all complete
hyperbolic manifolds with volume smaller than any given constant $C$. The full
version of the theorem which we shall see later shows that the concept of Dehn
surgeries can be used to give a very elegant description of all finite volume
hyperbolic 3-manifolds.
\proclaim J{\o}rgensen's Theorem (first version). Let $C>0$ be any constant. Among
all complete hyperbolic 3-manifolds with volume $\le C$ there are only finitely
many homeomorphism types of $\Mthick$.

{\sl Proof:} Choose a maximal set of points $V$ in $\Mthick$ such that no two points
in $V$ are closer than $\epsilon=\mu/2$. The balls of radius $\epsilon/4$ about
points in $V$ are disjoint while the balls of radius $\epsilon/2$ about points
in $V$ cover $\Mthick$. The combinatorial pattern of intersections
of the set of $\epsilon/2$ balls about points in $V$ determines $\Mthick$ up to
diffeomorphism. The upper bound $C$ on the volume of $M$ gives an upper
bound to the number of points in $V$. This in turn gives an upper bound to the
number of possible patterns of intersections of these $\epsilon/2$-balls.
\smallskip
If two hyperbolic manifolds $M$ and $M^\prime$ have $\Mthick$=$\Mthick^\prime$
then one can be obtained from the other by performing Dehn surgery on the thin
parts. Thus J{\o}rgensen's theorem yields that all hyperbolic manifolds with
volume $\le C$ can be obtained from a finite set of manifolds by Dehn surgery.
Each member of this set can be obtained by Dehn surgery on some link complement
in S$^3$, so all hyperbolic manifolds of volume $\le C$ can be obtained by Dehn
surgery on a finite set of links in S$^3$

\beginsection Convergence of Manifolds

The full version of J{\o}rgensen's theorem uses the concept of two manifolds
being geometrically near. Basically, $M$ and $M^\prime$ are geometrically near
if for some small $\epsilon$ there is a diffeomorphism from $M$ to
$M^\prime$ which closely approximates an isometry from $\Mthick$ to
$\Mthick^\prime$. This definition gives us a topology on the set of isometry
classes of complete hyperbolic manifolds of finite volume.

For two metric spaces $X,Y$ and a map $f:X\to Y$ we set $$L(f) = \sup_{x_1\neq
x_2\in X} \left|\log{dist(x_1,x_2)\over dist(f(x_1),f(x_2))}\right|$$ Put
simply, $L(f)$ is the logarithm of the largest factor by which a distance is
stretched or shrunk by $f$.

Consider a sequence of metric spaces $X^i,i=1,2,3,\ldots$ with basepoints
$x_i\in X^i$. We say the sequence $(X^i,x_i)$ {\it converges} to $(Y,y)$ if for
arbitrarily small $\epsilon>0$ and arbitrarily large $r$ there is a number $j$
such that for each $i\ge j$ there exists a map $f:B_r(x_i)\to Y$ satisfying:
\smallskip
\item{(a)}$f(x_i)=y$
\item{(b)}the image of $B_r(x_i)$ contains $B_{r-\epsilon}(y)$
\item{(c)}$L(f)\le\epsilon$.

This notion of convergence saisfies the important property that if a sequence
of manifolds converge to $M$ then their sequence of volumes converge to
$vol(M)$.

The following result is the second part of J{\o}rgensen's theorem
\proclaim J{\o}rgensen's Theorem (Extension). Every sequence of hyperbolic
3-manifolds with bounded volume has a convergent subsequence.

{\sl Proof:} For each $M$ in the sequence we let $V$ be a maximal set
of points in $\Mthick$ such that no two elements of $V$ are closer that
$\epsilon/2$. Then the set of balls of radius $\epsilon/2$ about points in
$V$ covers $\Mthick$. We choose a set of isometries of these balls in
hyperbolic space. The set of possible ways of glueing these balls together forms
a compact subset of Isom($\H3$), so any sequence of glueing maps has a convergent
subsequence. It is clear that in the limit, the glueing maps still give a
hyperbolic structure.

We now have a sequence of manifolds whose thick parts converge. We know that
the thin parts of this sequence are either cusps or tubes. We know that a tube
can be obtained by performing a Dehn surgery on a cusp and that the nature
of this surgery is determined by a geodesic on the bounding torus. If the
lengths of the geodesics of the Dehn surgeries are bounded then they clearly
have a convergent subsequence. If these geodesics are not bounded then we
have a subsequence whose solid tubes have geodesics of length approaching
infinity. Now if a solid tube is very large then its core geodesic must be
short. In the limit as this core geodesic approaches a point, the solid tube
approaches a cusp.

Thus we have a
subsequence whose thick parts converge to some $\Mthick^\prime$. To show that
this extends to a complete hyperbolic manifold we note that if $\Mthin$ has a
very short geodesic then its corresponding solid tube is very large. We need to
show that this extends to $\Mthin$. 

As an immediate corollary to this and the fact that the volume function is
continuous we have
\proclaim Corollary. The set $\F$ forms a closed set of real numbers.

\beginsection Closing cusps

I have already described how a sequence of Dehn surgeries can approach a cusp.
Thurston proved that in this cusp opening procedure, the volume is always
increased. Roughly, as we get more and more like a cusp we move further and
further out from the torus bounding $\Mthick$ and the volume gets larger and
larger approaching some limit.

The proof of this is very difficult. Thurston tackled the problem by
reversing the cusp opening process by gluing tori in place of cusps. The
problem is that such a gluing will not yield a hyperbolic manifold since the
natural metric will have constant negative curvature outside the bounding torus,
but on the torus the metric will be singular. To overcome this problem Thurston
showed that it was possible to deform the metric in the tube and in $\Mthick$
and that the resulting manifold was hyperbolic and had lower volume.

The upshot of this result is that the volume of an $n$-cusped manifold is a
limit from below of volumes of $n-1$-cusped manifolds. From this we see that
the order type of $\F$ is indeed as described earlier, with limit points
limiting to limit points and so on.
%\bye
\vfill\eject
%===============================================================
\centerline{\bf Volume estimates}
\medskip
%\input macros
We have seen that the volumes of hyperbolic 3-manifolds are well-ordered and in
particular that there is a manifold with the lowest volume. The manifold with
the lowest known volume is the Weeks manifold, which is obtained by
(5,1),(5,2)-Dehn surgery on the Whitehead link and has volume approximately
0.9427... (see [We]).
We would like to obtain an explicit lower bound on the volume of
hyperbolic 3-manifolds. All such estimates to date either fall woefully short
of this lowest known volume or require special conditions on the manifold. In
this chapter I will review some results in this direction.

\beginsection Short Geodesics

In our discussion on the well ordering of volumes of hyperbolic 3-manifolds we
saw that short geodesics have an embedded tubular neighborhood and the shorter
the geodesic the closer the tube approximates a cusp and the larger its volume.
This fact can be used to obtain a lower bound for the volume of complete compact
hyperbolic 3-manifolds. The idea is that if a manifold has a short geodesic
then it has an embedded tubular neighborhood with large volume, while if it gas
a long geodesic then it must have a large
embedded ball. In other words, for any $r$ the manifold must have either an
embedded ball of radius $r$ or a geodesic of length less than $2r$. For a given
value of $r$ we can calculate volume estimates for a manifold with a geodesic
length less than $2r$ and for a manifold with an embedded ball of radius $r$.
In any manifold, one of these must hold so the minimum of the two results gives
a lower bound for the volume of any hyperbolic 3-manifold. Our strategy then is
to choose an $r$ for which the two possibilities give roughly the same
volume estimate. This ``trade off'' value affords the best volume estimate.
For example, choosing a smaller $r$ would give a larger volume in the solid tube
case but a lower volume for the embedded ball case thus giving a lower overall
estimate.

Meyerhoff  in [Me1] showed that if we take $r=0\cdot053475$ then an embedded ball
of radius $r$ has volume at least $0\cdot00064$ while a geodesic of length at
most $2r$ has an embedded tubular neighborhood of volume at least $0\cdot00068$.
Thus the volume of a closed hyperbolic 3-manifold must be greater than
$0\cdot00064$.

Meyerhoff was able to improve this result in [Me2]. An embedded sphere in a
manifold will invariably leave gaps which are ignored in the calculation of the
volume of the embedded sphere. Meyerhoff picked up some of the volume in these
gaps by using sphere packing results of B\"or\"oczky [B\"o].

An embedded sphere in the manifold lifts to a sphere packing in hyperbolic
space. The fraction of the volume of the manifold covered by the sphere is equal
to the local density of the sphere packing in hyperbolic space. For an exact
definition of ``local density'' see [Me2], but it is sufficient to rely on your
intuitive guess at its meaning. The best way to pack four spheres of equal
radius in hyperbolic space is in a tetrahedral pattern so that each sphere
touches the other three. B\"or\"oczky showed that the densest packing of
hyperbolic 3-space is to extend this tetrahedral packing. Ironically, the densest
packing of Euclidean 3-space, known to all greengrocers, is not known to
mathematicians.

\proclaim Theorem (B\"or\"oczky). Consider 4 pairwise tangent balls of radius $r$
in $\H3$. Their centers determine a regular tetrahedron $T$ of edge length $2r$
and dihedral angles $2\alpha$ where $\sec(2\alpha)=2+\sech(2r)$. Let $S$ be the
union of the four balls. Then any radius $r$ sphere-packing has local density
less than or equal to the density of the $S\cap T$ in the tetrahedron $T$, which
we denote $d(r)$. It can be shown that $$d(r)={vol(S\cap T)\over
vol(T)}={(6\alpha-\pi)(\sinh(2r)-2r)\over vol(t)}.$$ In the special case of
horosphere packings the centers of the horoballs determine an ideal regular
tetrahedron T. The volume of an ideal tetrahedron is $v=1\cdot01494\ldots$ (this
is calculated in [Mi]) and $vol(S\cap T)$ can be shown to equal $\sqrt3/2$
giving a local density of $\sqrt3/(2v)\approx0\cdot853$.

With this theorem in hand we see that the volume of a manifold with embedded
sphere of radius $r$ is greater than or equal to the volume of a ball of radius
$r$ in hyperbolic space divided by our function $d(r)$. If we keep
$r=0\cdot053475$ as before then the volume estimate if an embedded ball of
radius $r$ sits in $M$ is improved to $0\cdot00082$, but the solid-tube
contribution remains at $0\cdot00068$ and our overall volume estimate is not
greatly improved. However we can take a smaller value of $r$ and improve the
solid-tube volume while only slightly decreasing the embedded-ball volume.
This time, a good ``trade-off'' value is $r=0\cdot053463$ which yields a
solid-tube volume and a embedded ball volume both greater than $0\cdot00082$.
Thus all hyperbolic manifolds have volume greater than $0\cdot00082$.

\beginsection J\o rgensen's Inequality

We have seen that every hyperbolic manifold can be uniquely associated with a
subgroup of the isometry group of $\H3$. Estimates on the minimum volume of a
complete hyperbolic manifold can be obtained by finding restructions that such a
group must satisfy if it is to define a manifold.

\proclaim Definition. A group of isometries is discrete if the identity map is
isolated.

This means that if a sequence $A_n$ of group elements approaches the identity map
then almost all of the $A_n$ are equal to the identity map. Apparently, this is
equivalent to the statement that for a given basepoint there is a minimum
distance by which it is moved by any element. Clearly a group must satisfy this
if it is to have a nontrivial Dirichlet domain.

\proclaim Definition. A group of isometries is elementary if it has an abelian
subgroup of finite index.

If a group of isometries is abelian then it must either consist of parabolic
isometries with a common fixed point or elliptic and loxodromic isometries with
a  common axis. Both of these cases would result in a Dirichlet domain with
infinite volume. In fact it can be shown that if a group of isometries is
elementary then the Dirichlet domain must have infinite volume.

We now have two restrictions on the group of hyperbolic isometries corresponding
to a finite volume hyperbolic manifold. Namely, it must be a discrete,
nonelementary group. Roughly speaking, J{\o}rgensen's inequality says that if two
matrices in \SL generate a discrete non-elementary group then they cannot be too
close to the identity matrix. The idea is to use this inequality to find a $d$
such that for every discrete non-elementary group there is some point which is
moved by a distance at least $d$ by every group element. The Dirichlet region of
such a point will contain an inscribed ball of radius $r=d/2$. Thus the volume
of this ball in hyperbolic space gives a lower bound for the volume of any
hyperbolic 3-manifold. Sphere packing arguments can be used to improve this
result.

Let $$A=\left(\matrix{a&b\cr c&d}\right){\rm\ \ and\ \ }
B=\left(\matrix{\alpha&\beta\cr\gamma&\delta}\right)$$ be matrices in \SL
representing the  M\"obius transformations $f$ and $g$ respectively. These
matrices are determined up to a factor of $\pm1$ by $f$ and $g$. We write
$\tr(f)=\tr(A)$ and $\tr(g)=\tr(B)$, also determined up to a factor of $\pm1$.
We see however that the commutator matrix $[A,B]=ABA^{-1}B^{-1}$ is uniquely
determined by $f$ and $g$. Thus the three complex numbers
$$\beta(f)=\tr^2(f)-4=\tr^2(A)-4$$
$$\beta(g)=\tr^2(g)-4=\tr^2(B)-4$$and$$\gamma(f,g)=tr([f,g])-2=tr([A,B])-2$$
are uniquely determined by $f$ and $g$.

\proclaim Theorem (J\o rgensen's inequality). Suppose that the M\"obius
transformations $f$ and $g$ generate a discrete non-elementary group. Then
$$|\beta(f)|+|\gamma(f,g)|\ge1$$and$$|\beta(g)|+|\gamma(f,g)|\ge1.$$This lower
bound is the best possible.

For a proof, see [Be].

In particular if one of the generators, say $f$, is close to the identity map
then its matrix $A$ will be close to the identity matrix so $\beta(f)$ will be
small. Also $f$ and $g$ will ``almost commute'' so $\gamma(f,g)$ will also be
small and the theorem tells us that $f$ and $g$ do not generate a discrete
non-elementary group. In order to make this precise we need to introduce a
norm function which explicitly measures how far a matrix is from the identity
matrix.

Perhaps the most obvious such norm is given by
$$\|A-I\|=\sqrt{|a-1|^2+|b|^2+|c|^2+|d-1|^2}.$$ The following is my own
estimate of $\max\{\|A-I\|),\|B-I\|\}$ based on the methods used in
[Be]: $$\max\{\|A\|,\|B\|\}>0\cdot146.$$

Suppose $A,B\in \SL$ generate a discrete non-elementary group. First set $A=X+I$,
$A^{-1}=X^*+I$, $B=Y+I$ and $B^{-1}=Y^*+I$. We aim to derive J\o rgensen's
inequality a lower bound on the value of $\max\{\|X\|,\|Y\|\}$. First observe the
following immediate consequences of these definitions.\smallskip
\item{(i)}$X+X^*+XX^*=0$ and $Y+Y^*+YY^*=0$
\item{(ii)}$XX^*=X^*X=\det(X)I$ and $YY^*=Y^*Y=\det(Y)I$
\item{(iii)}$\tr(X)=-\det(X)$
\item{(iv)}$\|X\|=\|X^*\|$ and $\|Y\|=\|Y^*\|$.
\item{(v)}$\tr^2(A)=tr(A^2)+2$\smallskip
Also, for any $$C=\left(\matrix{p&q\cr
r&s}\right)$$and$$D=\left(\matrix{t&u\cr v&w}\right)$$in GL(2,{\bf C})we
have:
\item{(vi)}$\|CD\|\le\|C\|\,\|D\|$
\item{(vii)}$\tr (CD)\le\|C\|\,\|D\|$;
\item{(viii)}$\tr(C)\le\sqrt{2}\|C\|$ and
\item{(ix)}$|\det(C)|\le{1\over2}\|C\|^2$.

Of these, (vi), (vii) and (ix) are proved using the Cauchy-Schwartz
inequality and (viii) is obtained by setting $D=I$ in (vii).

We can use (i) to obtain

$\eqalign{[A,B]-I&=(I+X)(I+Y)(I+X^*)(I+Y^*)-I\cr
&=XYX^*+(XX^*+X^*)Y^*+XX^*YY^*+YX^*+YX^*Y^*\cr
&=XYX^*-XY^*+XX^*YY^*+YX^*+YX^*Y^*}$

We now use the fact that trace is preserved under conjugation to cyclically
permute terms:

$$\eqalign{|\tr([A,B]-I)|&=|\tr(-XY^*+YX^*+YXX^*+YX^*Y^*+XYX^*Y^*)|\cr
&=|\tr(-XY^*-XY+YX^*Y^*+XYX^*Y^*)|\cr
&=|\tr(XYY^*+X^*Y^*Y+XYX^*Y^*)|\cr
&=|\tr(XYY^*+X^*YY^*+XYX^*Y^*)|\cr
&=|\tr(-XX^*YY^*+XYX^*Y^*)|\cr
&\le|\tr(XX^*YY^*)|+|\tr(XYX^*Y^*|\cr
&\le\|XX^*\|\,\|YY^*\|+\|X\|\,\|Y\|\,\|X^*\|,\|Y^*\|\cr
&=\|\det(X)I\|\,\|\det(Y)I\|+(\|X\|\,\|Y\|)^2\cr
&=2\left(|\det(X)|\,|\det(Y)|\right)+(\|X\|\,\|Y\|)^2\cr
&\le2({1\over2}\|X\|^2)({1\over2}\|Y\|^2)+(\|X\|\,\|Y\|)^2\cr
&={3\over2}\|X\|^2\|Y\|^2}$$

Thus $\gamma(f,g)\le {3\over2}\|A-I\|^2\|B-I\|^2$.

Now
$$\eqalign{|tr^2A-4|&=|\tr(A^2-2)|\cr
&=|\tr(A^2-I)|\cr
&=|\tr(X^2+2X)|\cr
&\le|\tr(X^2)|+2|\tr(X)|\cr
&\le\|X\|^2+\|X\|^2\cr
&=2\|X\|^2}$$
Thus $\beta(f)\le2\|A-I\|^2$

Combining these results with J{\o}rgensen's
inequality yields:
$$1\le\beta(f)+\gamma(f,g)\le2\|A-I\|^2+{3\over2}\|A-I\|^2\|B-I\|^2$$
If $\max\{\|A-I\|,\|B-I\|\}<\epsilon$ then this
yields $$2\epsilon^2+{3/2}\epsilon^4\ge1$$ so $\epsilon>0\cdot622$. Thus
$\max\{\|A-I\|,\|B-I\|\}>0\cdot622$. Waterman [Wa] obtained the stronger result
$\|A-I\|\,\|B-I\|>\sqrt2-1$ by a somewhat laborious calculation after
a preliminary conjugation designed to put $A$ and $B$ in a special form without
increasing their norms.

We have an estimate on the maximum distance of a group generator from the
identity map. What we really want however is an estimate on the minimum distance
of any group element from the identity. We can do this by conjugating the group
by some appropriate M\"obius transformation. Waterman showed that it is
possible to conjugate the group in such a way that $\|A-I\|^2>\sqrt2-1$ for
every group element $A$

Waterman now uses this result to obtain an inscribed ball in the
manifold. We can use this lower bound on $\|A\|$ to deduce
constraints on the complex translational length of $A$. The real
part of the translational length tells us the translation along
the axis and the imaginary part tells us the rotation. Geometrically speaking, we
find that the rotation and translation components of $A$ are not both small.
What we really want is a lower bound on the translation component of all
group elements $A$.

To reduce the impact of this rotational component we look for a power
$A^k$ of $A$ which has a small rotational component. Since $\|A^k-I\|$ is large,
yet $A^k$ has a small rotational component, we can deduce a lower bound on its
translational component. To obtain an estimate on the translational component of
$A$ we simply divide this by $k$. To this end, Waterman divides the range of
possible angles into intervals which are close to some fraction of a full
circle. Each of these fractions has a denominator less than or equal to 7. Thus
we can raise $A$ to some power less than or equal to 7 to obtain a map whose
rotational component is close to a whole number of complete rotations, and thus
close to zero. This yields an estimate on the minimum translation component of a
non-trivial element of the fundamental group. Since every point is moved a
distance greater than this translational distance by every element of the
fundamental group, we can inscribe a ball with diameter equal to this
translation distance. Waterman obtained an inscribed ball of radius $1\over300$,
and if the group has no primitive torsion element of order less than or equal to
six, he showed that this radius could be improved to $1\over 25$. The volume of
this ball gives a lower bound for the volume of a complete orientable hyperbolic
3-manifold.

Gehring and Martin in [Ge] use similar techniques to obtain new
estimates for the volume of an
tube in a hyperbolic 3-manifold with a shortest geodesic of length $2l$.
The ``trade off'' argument can be used to find a new lower bound for the
volume of all hyperbolic 3-manifolds. I will now outline briefly the techniques
used by Gehring and Martin in [Ge]. A set of notes by Meyerhoff [Me3]
considerably helped me in understanding the Gehring-Martin approach.

Gehring and Martin analyse three different norms and obtain
restrictions on the values these norms can take in a discrete non-elementary
subgroup of Isom($\H3$). Instead of the norm $\|A-I\|$, Gehring and Martin used
$m(f)=\|A-A^{-1}\|$ where $A$ is a matrix representation of $f$. This norm is
still a measure of how close a map is to the identity, and results obtained
using $m(f)$ parallel results that can be obtained with the $\|A-I\|$ norm. The
$m(f)$ norm is however in many ways more appealing. For example it does not
depend on the matrix chosen for $f$ ad the geometry of $A-A^{-1}$ seems to
relate more closely to the geometry of $A$. The group can then be conjugated in
such a way that $\|A-A^{-1}\|^2\ge4(\sqrt2-1)$ for every group element $A$.

If $g$ has complex translational length $l+i\theta$ then it can be shown
$$\beta(g)=\sinh^2(t)+\sin^2(\theta).$$
Also $\gamma(f,g)$ relates closely to the distance between the axes of $f$ and
$g$. Thus J{\o}rgensen's inequality tells us that if $\beta(g)$ is small then
the axis of $g$ is a long way from the axis of any conjugate to $f$. This
enables us to embed a reasonably large solid tube around the geodesic
corresponding to $g$. As in the Waterman paper we try to ``reduce the impact'' of
$\theta$ by looking at powers of $g$. The function
$$\inf_k\{\sinh^2(kt)+\sin^2(k\theta)\}$$ is basically the result of choosing
the best possible $k$. In order to allow for the worst possible choice of
$\theta$ Gehring and Martin introduce the function
$$s(t)=\sup_\theta\{\inf_k\{\sinh^2(kt)+sin^2(k\theta)\}\}.$$
By studying this function they obtain better estimates for the size of an
embedded tube. Again we choose a length such that this estimate is the same as
the usual estimate by finding an inscribed ball and using sphere packing
arguments. The resulting volume estimate is given as $0\cdot00115$ but due to a
small arithmetic error discovered by Meyerhoff in [Me3] the true result should
be $0\cdot00101$.

\beginsection The lowest volume noncompact manifold

In this section, manifolds are not assumed to be orientable. Many results
relating to orientable manifolds can be proven in the nonorientable case by
using the fact that every nonorientable hyperbolic 3-manifold is double-covered by an orientable
hyperbolic 3-manifold. In particular, the volume of all
hyperbolic 3-manifolds is a topological invariant and the finite volumes of
all hyperbolic 3-manifolds are well ordered. In particular there is a lowest
volume hyperbolic 3-manifold and a lowest volume noncompact hyperbolic
3-manifold.

It is worth noting some results which do not carry over well to the
nonorientable case. Recall that in the orientable case we were able to cut the
manifold along two-sided tori leaving a compact component and a finite number of
cusps each homeomorphic to $T^2\times[0,1)$. In a nonorientable manifold a
similar decomposition can be performed except that we are required to cut along
two-sided Klein bottles as well as two-sided tori, leaving a compact component
and a finite set of cusps, each homeomorphic to either $T^2\times[0,1)$ or
$K^2\times[0,1)$. There are many Dehn surgeries which can be performed on
orientable cusps but they all result in a hyperbolic manifold with lower volume.
There is however only one way to glue a solid Klein bottle to a nonorientable
cusp $K\times[0,1)$ and the resulting manifold need not be hyperbolic. Thus
unlike in the orientable case we do not know that for a given $n$-cusped
hyperbolic nonorientable 3-manifold there is a $n-1$-cusped nonorientable
hyperbolic 3-manifold of smaller volume.

We will describe a nonorientable noncompact hyperbolic 3-manifold called the
Gieseking manifold and by examining the volumes of cusps we will outline
Adam's proof [Ad] that it is the unique noncompact hyperbolic 3-manifold of
minimal volume.

Let $T$ be an ideal regular tetrahedron in $\H3$, that is a tetrahedron with all
four vertices on the sphere at infinity and all dihedral angles equal to
$\pi/3$. Then identify pairs of faces with orientations as indicated in the
diagram. These identifications correspond to orientation-reversing hyperbolic
isometries which generate a discrete torsion free subgroups of the groups of
all isometries of $\H3$. After identification, all six edges are identified and
the total angle around this edge is equal to $6\times\pi/3=2\pi$. The resulting
manifold is called the Gieseking manifold and has volume $1\cdot01494\ldots$,
equal to the volume of an ideal regular tetrahedron.
\beginsection Maximal cusp volumes

Let $M$ be a noncompact finite volume hyperbolic 3-manifold. Consider first the
case where $M$ has exactly one cusp. The lift of that cusp into hyperbolic
space is an infinite set of disjoint 
horoballs. The boundaries of these horoballs are called horospheres and can be
thought of as spheres of infinite radius in hyperbolic space with their centers
on the sphere at infinity. In the upper half-space model they look like Euclidean
spherical balls touching the $x$-$y$ plane. The point at which a sphere touches
the $x$-$y$ plane is actually the center of the horosphere. In the case of a ball
centered at $\infty$ the horosphere is a Euclidean plane parallel to the $x$-$y$
plane.

Note that any horoball is the image of any other horoball under the action
of some element of the fundamental group $\pi_1(M)$ .

We can expand the cusp in $M$ until two of the horoballs first become tangent.
The result is called a maximal cusp of $M$ and we denote its volume $v_C$.

\proclaim Theorem. For any cusp C in a finite volume hyperbolic manifold $v_C\ge
\sqrt3/4$.

{\sl Proof:} We can conjugate to ensure that two tangent horoballs have centers
at 0 and $\infty$, say $H_1$ is centered at $\infty$ and $H_2$ is tangent to
$H_1$, centered about 0 and has Euclidean diameter $h$. Now let $P$ be the
subgroup of $\pi_1(M)$ which fixes $\infty$ and thus fixes $H_1$. Note that any
horoball tangent to $H_1$ must be mapped by $P$ to other horoballs tangent to
$H_1$.

Now $H_1$ actually consists of an
infinite number of lifts of the maximal cusp, mapped to each other by elements
of $P$. Thus the volume of the maximal cusp is the fundamental region in
$H_1$ for the action on $H_1$ by $P$.

Now $P$ is either the fundamental group of a torus or a Klein bottle there
exists a fundamental region for its action on the $x-y$ plane which is a
parallelogram with one vertex at 0 and all four vertices identified by $P$.
Now every element of $P$ sends the center of $H_2$ to the center of another
horoball tangent to $H_1$. Thus each of the vertices is the center of a horoball
with Euclidean diameter $h$. Since the horoballs have disjoint interiors, it is
possible to place a disk of radus $h/2$ about each vertex of $D$ such that the
interiors of the disks are disjoint. Since the images of $D$ under the action of
$P$ tile the plane, the images of these disks under the action of $P$ form a
disk-packing of the plane. The densest such disk-packing is the hexagonal packing
so the Euclidean area of $D$ is at most $2\sqrt{3}/\pi$ times the Euclidean area
of a disk, that is $\sqrt{3}h^2/2$.

Now the fundamental region for the action of $P$ on $H_1$ is shaped like a
``semi-infinite prism'' made up of those points in $H_1$ which lie directly
above the aformentioned parallelogram. Now if $A$ is the Euclidean area of the
parallelogram then the volume of a thin layer of this prism at height $z$ and
width d$z$ is given by $(A/z^3){\rm d}z$. Thus the total volume is given by
$$\eqalign{v_C&=\int_h^\infty{A\over z^3}{\rm d}z\cr
&={A\over 2h^2}\cr
&\ge{\sqrt{3}\over4}}$$
which is the result we required.
\smallskip
The basic method employed here was to show that the fundamental region of $P$ on
the $x$-$y$ plane contained the equivalent of one disk with radius $h$. Adams
goes on to show that the fundamental region of $P$ on the $x$-$y$ plane must in
fact contain the equivalent of two disks, thus improving the result by a factor
of two. The proof involves showing that the maximal cusp touches itself in two
places.
\proclaim Theorem. For any cusp C in a finite volume hyperbolic manifold
$v_C\ge \sqrt3/2$.

{\sl Proof:} Again, set $H_1$ and $H_2$ to be tangent horoballs centered
at $\infty$ and 0 respectively, $P$ the subgroup of $\pi_1(M)$ which
fixes $\infty$ and $D$ the fundamental domain for the action of $P$ on the
$x-y$ plane which is a parallelogram with one vertex at 0.

Since $H_1$ and $H_2$ are both covers of $C$ there must exist a $g\in\pi_1(M)$
sending $H_2$ to $H_1$ and hence sending 0 to $\infty$. Suppose
$g(\infty)=p(0)$ for some $p\in P$. Then $p^{-1}g(\infty)=0$ and
$p^{-1}g(0)=\infty$. Thus $p^{-1}g$ sends the geodesic from 0 to $\infty$ to
itself with reversed orientation. Thus $p^{-1}g$ must fix some point,
contradicting the fact that it is a nontrivial covering transformation. It
follows that $g(\infty)$ cannot be contained in $P(0)$.

Now $H_1$ is tangent to $H_2$ so $g(H_1)$ is tangent to $g(H_2)=H_1$. Thus
$g(H_1)$ is a horoball tangent to $H_1$ but not contained in $P(H_2)$. Since 
$D$ tiles the $x-y$ plane, $D$ contains a copy of $g(H_1)$ under the action
of $P$ with center at $y$, say. We have horospheres centered at the vertices of
$D$ and at $y$, all of diameter $h$ and all with disjoint interiors.
Thus we can place disks of radius $h/2$ about each vertex and about $y$ and let
$P$ act on these disks to obtain a disk-packing of the plane. Each $D$ now
contains the equivalent of two disks. Hence as in the previous proof $D$ has
area at least $\sqrt3h^2$ whence $v_C\ge\sqrt3/2$.

We now apply the theorem due to B\"or\"oczky which states that any packing
of horoballs in $\H3$ must have local density less than or equal to
$\sqrt{3}/2v$ where $v$ is the volume of an ideal regular tetrahedron.
Meyerhoff [Me2] used this to prove that for a 1-cusp hyperbolic 3-manifold
$M$, the total volume of $M$ must be at least $2v/\sqrt3$ times the volume
of its maximal cusp. Thus $vol(M)\ge v_C(2v/\sqrt3)\ge v$. The Gieseking
manifold realizes this lower bound. Note that for a manifold $M$ to realise this
lower bound it must satisfy $v_C(M)=\sqrt3/2$ for every cusp $C$ in $M$.
\smallskip
We now prove that the Giesekig manifold is the unique noncompact
hyperbolic manifold with this volume.
\proclaim Theorem. If $M$ is a noncompact hyperbolic 3-manifold of minimal
volume then $M$ is the Gieseking manifold.

{\sl Proof:} Suppose $M$ has volume $v$. Let $C$ be a maximal cusp in $M$.
Then $C$ must have volume $\sqrt3/2$. Let $H_1$ be the horoball centered at
$\infty$ and let $P$, $D$, and $y$ be as in the previous proof. In order to
attain the lowest volume, the images of the disks of radius $h/2$ centered about 
the $y$ and the vertices of $D$ must exhibit hexagonal packing. From this
hexagonal array we obtain a tiling of the plane by equilateral
triangles of edge length $h$ and vertices at the centers of these disks.

Let $\{t_1,t_2,...,t_n\}$ be the finite set of these triangles whose interiors
meet $D$. Let $D^\prime=\cup_{i=1}^nt_i$. For each $i$ let $T_i$ be the ideal
regular tetrahedron with its four vertices at each of the three vertices of
$t_i$ and at $\infty$. Let $R=\cup_{i=1}^nT_i$. We claim that the images of $R$
under $\pi_1(M)$ cover all of $\H3$.

It is clear that the images of $R$ cover
$H_1$ under the action of $P$. Hence the images of $R$ under $pi_1(M)$ cover all
the maximall horoballs, since all of the maximal horoballs are identified by
$pi_1(M)$. We now show that the bottom face $f_i$ of each $T_i$ is identified to
some face on $T_j$ for some $j$. Since the vertices of $f_i$ are centers
of horoballs, there must be a $\mu\in\pi_1(M)$ sending one of these vertices of
to $\infty$. Since the vertices of $f_i$ are the centers of three tangential
horoballs, their image under $\mu$ must also be the centers of three tangential
horoballs. Hence the remaining two vertices must be sent to the centers of two
adjacent horoballs of Euclidean diameter $h$ corresponding to $C$. Hence $\mu$
sends $f_i$ to a side face of the tetrahedron $p(T_j)$ for some $p\in P$ and
some $j$. Thus $p^{-1}\mu$ identifies the bottom face $f_i$ to a face of some
$T_j$. Thus every face of a $T_i$ is identified with a face of some $T_j$.
From this discussion it is clear that the images of $R$ under $\pi_1(M)$ cover
all of $\H3$.

We also have that no interior point of a tetrahedron $T_i$ is mapped to any
other point of $T_i$. if there was such a map it would permute the vertices of
$T_i$ in which case it can be shown to have a fixed point in $T_i$ which is not
allowed in a non-trivial covering transformation.

The volume of $M$ is by assumption equal to the volume of any one of these
ideal regular tetrahedra. Thus its fundamental region must be a single ideal
regular tetrahedron with faces identified. It is not hard to check that the
only identifications which yield a manifold yield the Gieseking manifold.
 %\bye
\vfill\eject
%===============================================================
\centerline{\bf The two lowest known volume
hyperbolic 3-manifolds } \medskip
%\input macros
In this chapter I examine closely the topology of the two lowest known volume
manifolds. These are the Weeks manifold which has volume $0.9427\ldots$ and the
Meyerhoff manifold with volume $0.9813\ldots$. I have used data obtained from the
computer program ``Snappea'' to find group presentations for their fundamental
groups. From these presentations I was able to find information about possible
matrix generators in \SL.
\beginsection The Weeks manifold

The Dirichlet domain given by Snappea is a polyhedron with 18 faces, 42 edges
and 26 vertices. Of the 18 faces, 12 are pentagons and 6 are quadrilaterals.

I examined these face identifications. Using the edge lengths supplied by
Snappea and the fact that the manifold is orientable I was able to deduce which
which edges and which vertices were identified. This revealed that after
identification there were a total of 9 faces, 14 edges and 6 vertices

Now consider the fundamental group of this manifold. Pick a basepoint $x$ in the
polyhedron and now think of hyperbolic space being tiled with copies of this
polyhedron, each containing a lift of the basepoint. An element of the funamental
group can then be thought of as a path from one lift of $x$ to another. Such a
path represents a trivial group element if and only if it is a loop in $\H3$.

Now given a path from one lift of $x$ to another we deform it slightly if
necessary to ensure that it does not pass through any edges or vertices, only
faces. Now each time the path passes through a face into a new polyhedron we can
insert a short detour to the basepoint in that polyhedron. The path is now the
product of short paths from one lift of $x$ through one face of the polyhedron
to a neighboring lift of $x$. Thus we have generators
$a,b,c,\ldots,i,A,B,\ldots,I$ corresponding to each of the faces.

We now look for relations in these generators. First note that words 
$aA,bB,\ldots,iI$ correspond to paths through one face and then
straight back again so that identified faces correspond to inverse pairs. Also
we have paths that loop around an edge of the polyhedron and come back to the
same basepoint. Thus each edge of the polyhedron gives us a relation. I claim
that these are the only relations we need, that is any trivial word can be
shown to be trivial using the relations I have described.

Suppose we have a loop in $\H3$ corresponding to a trivial word in the
generators. We contract this slowly and see how this affects the corresponding
word. The word is altered when the path changes the sequence of faces it cuts
through. This can only happen if the path is pushed into or lifted out of a
face, or if the path is pulled through an edge. In the former case the the word
is altered by adding or removing an inverse pair while in the latter the word is
altered by applying one of the edge relations. After a finite number of such
alterations the loop is contracted to a point. Thus the word can be reduced to
the identity after a finite number of applications of the relations indicated.

The 14 relations obtained are:
\settabs 7 \columns
\+&$f=hd$&&$g=fi$&&$h=ge$\cr
\+&$f=ce$&&$g=ad$&&$h=bi$\cr
\+&$e=cd$&&$d=ai$&&$i=be$\cr
\+&$h=cc$&&$f=aa$&&$g=bb$\cr
\+&&$abc=1$&&$edi=1$\cr
\noindent where 1 denotes the identity element. These relations display an order
three automorphism which cycles $(a,b,c)$, $(e,d,i)$ and $(f,g,h)$. This
automorphism reflects the fact that the diagram above can be rotated through
$120^\circ$ and left essentially unchanged. Thus the automorphism corresponds to
conjugation of the group by a $120^\circ$ rotation.

We can use the relations to eliminate $c,d,e,f,g,h,i,$ leaving us with the two
generators $a$ and $b$ and two relations $$a^2=b^2abab^2$$and$$b^2=a^2baba^2.$$
Thus we have the presentation: $$\pi_1(M_0)=\langle
a,b|a^2=b^2abab^2,\,b^2=a^2baba^2\rangle.$$ This brings to light another
automorphism which swaps the generators $a$ and $b$. This does not correspond to
a symmetry in the polyhedron. However there is another basepoint whose
Dirichlet domain does display this symmetry.

Mostow's  rigidity theorem implies that the automorphism
interchanging $a$ and $b$ corresponds to a hyperbolic isometry. This means that
their matrices are conjugate in \SL, although they are not conjugate in 
$\pi_1(M_0)$. Thus their matrices have the same trace.

\proclaim Lemma. If $a$ and $b$ are conjugate matrices in \SL then there exists
a $t\in\SL$ such that $$t^{-1}at=\left(\matrix{u&v\cr0&1/u}\right){\rm\ 
and\ }t^{-1}bt=\left(\matrix{u&0\cr v&1/u}\right).$$

{\sl Proof:} First note that we can conjugate $a$ and $b$ by the same
element to make $a$ diagonal. So we will assume
$$a=\left(\matrix{u&0\cr 0&1/u}\right){\rm\ and\ }
b=\left(\matrix{\alpha&\beta\cr \gamma&\delta}\right).$$ Our
next step is to find a matrix $r$ such that $r^{-1}ar$ is upper triangular and
$r^{-1}br$ is lower triangular. If $\beta=0$ then we have finished. If
$\gamma=0$ then $$r=\left(\matrix{0&-1\cr1&0}\right)$$ does the job. Otherwise
put $$r=\left(\matrix{1&\lambda\cr0&1}\right).$$ This gives:
$$r^{-1}ar=\left(\matrix{u&\lambda/u\cr0&1/u}\right){\rm\ and\ }
r^{-1}br=\left(\matrix{\alpha+\gamma\lambda&-\gamma\lambda^2+(\delta-\alpha)\lambda+\beta\cr
\gamma&-\gamma\lambda+\delta}\right).$$
Because $\gamma\ne0$ we can choose
$\lambda$ to be a root of $-\gamma\lambda^2+(\delta-\alpha)\lambda+\beta$
giving the desired result.

Now assume we have $$a^\prime=\left(\matrix{u&x\cr0&1/u}\right){\rm\ and\ }
b^\prime=\left(\matrix{u&0\cr y&1/u}\right).$$Let $s$ be the diagonal matrix
$\left(\matrix{\mu&0\cr0&1/\mu}\right)$. Then $$s^{-1}a^\prime s
=\left(\matrix{u&x\mu^2\cr0&1/u}\right){\rm\ and\ }
b^\prime=\left(\matrix{u&0\cr y/\mu^2&1/u}\right).$$ Clearly we can choose $\mu$
so that $x\mu^2=y/\mu^2$ and the proof is complete.
\smallskip
Now we assume the matrices $a$ and $b$ are of this form and use the group
relations to calculate their trace, $u+1/u$. I used Mathematica to evaluate the
matrices $(a^2-b^2abab^2)$ and $(b^2-a^2baba^2)$. Both of these are required to
be zero, giving us a total of eight polynomials in $u$ and $v$. Most of these
are redundant, leaving only two equations in the variables $u$ and $v$. Using
Mathematica I eliminated $v$ and obtained the following expression for u:
$$(u-1)^2u^{36}(1+u)^2(1+u^2)^6(1+2u^2-u^3+2u^4+u^6)^2=0.$$ Clearly $u\ne0$.
Also it is clear that $a$ and $b$ are hyperbolic so $|u|\ne1$. Thus we are left
with:$$1+2u^2-u^3+2u^4+u^6=0.$$ Because of the symmetry of this polynomial we
can reduce it to a third order polynomial in $(u+1/u)$. In fact,
$$\eqalign{1+2u^2-u^3+2u^4+u^6&=u^3\left((u^3+u^{-3})+2(u^2+u^{-2})+1\right)\cr
&=(t^3-t-1)u^3}$$ where $t=u+(1/u)$ is the trace of $a$ and $b$. So we see that
the trace satisfies the very simple monic equation $t^3=t+1$.

There are three solutions to
this cubic equation, one real and one complex conjugate pair. Using Mathematica
I found these to be approximately $$1\cdot32472$$
$$-0\cdot662359+0\cdot56228i$$  and $$-0\cdot662359-0\cdot56228i.$$
Of these, the real solution matches exactly the value of the trace given by
Snappea. The other two probably do not yield a complete hyperbolic 3-manifold.

\beginsection The Meyerhoff manifold

The manifold with the second lowest known volume is obtained by (5,1) Dehn
surgery on the figure eight knot. We apply the same calculations to this manifold
as we did to the previous example, this time
dwelling a little less on details.

I used Snappea to obtain data on a Dirichlet domain. The result was a
polyhedron with 16 faces, 30 edges and 16 vertices. Four of the faces
were triangular, the other 12 being quadrilateral. 

I found that the edges were identified in groups of three so as in the previous
example, three copies of the polyedron meet around each edge. After
identification there were a total of 8 faces, 10 edges and 3 vertices.
As before, the 8 faces give us 8 generators $a,b,\ldots,h$ with inverses
$A,B,\ldots,H$ and the 10 edges give us ten relations, namely
$$a=cg=fh$$
$$b=hg$$
$$c=ge=ba$$
$$d=ec=ha$$
$$f=de=bd$$
$$h=eb$$
\noindent I reduced these to two relations in the generators e and b and thus
obtained the group presentation: $$\pi_1(M_1)=\langle
b,e|BEbeBE=eBEbe^3,\,eBEbe=beBEb\rangle.$$ Note that the last relation gives
us $(eBEbe)e=b(eBEbe)$ demonstrating that our generators are in fact conjugate.
Thus their corresponding matrices must have the same trace and can be taken to
be of the form $$b=\left(\matrix{u&v\cr0&1/u}\right){\rm\ and\ }
e=\left(\matrix{u&0\cr v&1/u}\right).$$ I again used Mathematica to solve the
defining relations for $u$ and obtained:
$$1-3u+5u^2-6u^3+7u^4-6u^5+5u^6-3u^7+u^8=0$$ whence: the
trace $t$ of $b$ and $e$ satisfies $$-1+3t+t^2-3t^3+t^4=0$$
This can be solved to give
$$0\cdot328956$$
$$-0\cdot905166$$
$$1\cdot7881+0\cdot401358i$$ and
$$1\cdot7881-0\cdot401358i$$
Again, the correct choice of solution is the one which matches the value given
by Snappea, namely $1\cdot7881-0\cdot401358i$.

\beginsection The trace field

In the Weeks manifold example we saw that the traces of both generators were
roots of the cubic polynomial $t^3-t-1$. The {\sl trace field} of the group is
the set of all the traces of all the elements of the group. It turns out that
in addition to the trace of the generators we also need to know the trace of
the product of the generators. From these we can deduce the trace field of the
whole group.

The trace of $ab$ is equal to $u^2+v^2+u^{-2}$. Thus if we set $t=\tr(ab)$ we
have $u^4+v^2u^2-u^2t=0$. I used Mathematica to solve this simultaneously with
the two polynomials relating $u$ and $v$. This showed that $t^3-t-1$ gave a
possible solution. Other solutions did not match the numerical result obtained
by direct substitution of our known value for $u+1/u$. Thus the trace of $ab$
is also a root of $t^3-t-1$.

Now
observe the easily verified fact that for any two matrices $A$ and $B$ in $\SL$
we have $\tr(AB)+\tr(A^{-1}B)=\tr(A)\tr(B)$. Replacing $B$ by $A^nB$ yields
$\tr(A^{n+1}B+\tr(A^{n-1}B)=\tr(A)\tr(A^nB)$. Using this relation inductively
yields an expression for $\tr(A^nB)$ in terms of $\tr(A)$, $\tr(B)$ and
$\tr(AB)$. Such techniques can be used to reduce the trace of any word in $A$
and $B$ to a polynomial in $\tr(A)$, $\tr(B)$ and $\tr(AB)$. Thus the trace
field is equal to ${\bf Q}(\zeta)$ where $\zeta$ is a root of $t^3-t-1$.

The results for the Meyerhoff manifold were much the same, although the
polynomials involved were rather more complicated. As we look at higher volumes
we expect more complicated manifolds and the sort of calculations I have
performed will rapidly become unmanageable.

It is possible to use the fact that the trace field is algebraic to verify that
the group is discrete. This is a simple check that our results do indeed yield
a manifold.



 %\bye
\vfill\eject
%===============================================================
\centerline{\bf Bibliography}
\medskip
\noindent[Ad] Adams, C.C., {\sl The Noncompact Hyperbolic 3-Manifold of Minimal Volume},
Proceedings of the American Mathematical Society, Vol. 100 Number 4, (August
1987).

\noindent[Be] Beardon, A.F., {\sl The Geometry of Discrete Groups},
Graduate Texts in Mathematics 91, Springer-Verlag, New York, (1983).

\noindent[B\"o] B\"or\"oczky, K., {\sl Packing of Spheres in Spaces of Constant
Curvature}, Acta Mathematica Academiae Scientiarum Hungaricae, (1978) 243-261.

\noindent[BP] Benedetti, R. and Petronio, C., {\sl Lectures on Hyperbolic
Geometry}, Universitext, Springer-Verlag, Berlin Heidelberg, 1992.

\noindent[Ge] Gehring, F.W. and Martin, G.J., {\sl Inequalities for
M\"obius transformations and discrete groups}, J. reine angew. Math. 418
(1991) 31-37

\noindent[Gr] Gromov, M., {\sl Hyperbolic Manifolds According to Thurston
and J{\o}rgensen}, S\'eminaire Bourbaki, (Nov. 1979).

\noindent[Me1] Meyerhoff, R.,
{\sl A Lower Bound for the Volume of Hyperbolic 3-Manifolds}, Can. J. Math.
Vol. XXXIX, No. 5, (1987), 1038-1056.

\noindent[Me2] Meyerhoff, R., {\sl Sphere packing and volume
in hyperbolic 3-space}, Comment. Math. Helvetici, (1986),
271-278.

\noindent[Me3] Meyerhoff, R., {\sl A New Lower Bound for the Volume of Hyperbolic 3-Manifolds},
printed notes, as yet unpublished.

\noindent[Mi] Milnor, J., {\sl Hyperbolic geometry: the first 150 years},
Bull. Amer. Math. Soc. 6 (1982), 9-24.

\noindent[Th] Thurston, W., {\sl The geometry and topology of 3-manifolds},
Lecture notes from Princeton University, 1977/78.

\noindent[Th2] Thurston, W., {\sl Three-dimensional Topology and Geometry} (pending publication).%???

\noindent[Wa] Waterman, P.L., {\sl An inscribed ball for Kleinian groups},
Bull. London Math. Soc. 16 (1984), 525-530.

\noindent[We] Weeks, J., PhD Thesis Princeton University
(1985).
\bye