. 3
( 8)


3.4 Spherical motions
A spherical motion or isometry is of course just a map T : S 2 ’ S 2 preserving
spherical distance.


A motion T : S 2 ’ S 2 takes pairs of antipodal points to pairs of antipodal points,
and spherical lines (great circles) to spherical lines.
Any motion is given in coordinates by x ’ Ax, where A is a 3 — 3 orthogonal

Two points of the sphere are antipodal if and only if they are a maximum
distance apart (at distance πr , half a world away), so the ¬rst sentence is clear. The
rest of the proof is very similar to the Euclidean proof in Chapter 1. For (1), exactly
as in Corollary 1.7, the arcs of spherical lines [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) = 2πr or ± d(P, R) ± d(R, Q) ± d(P, Q) = 0.

Here the ¬rst 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 3.3 for some
permutation of P, Q, R. A spherical motion T preserves these equalities, so takes a
spherical line L to a spherical line L = T (L).
For (2), note ¬rst that because T : S 2 ’ S 2 takes antipodal points to antipodal
points, it extends in a unique way to a map T : R3 ’ R3 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 through the origin.
Suppose L = © S 2 and T (L) = L = © S 2 . A spherical line L = © S 2 is
parametrised by arc length: a variable point of L is cos θf1 + sin θf2 , where f1 , f2 , f3
is an orthogonal basis of R3 with f1 , f2 ∈ L, and θ equals the arc length along L.
Since T preserves distance, it preserves arc length along a spherical line, so that its
restriction TL : L ’ L is given by

T (cos θf1 + sin θf2 ) = cos θf 1 + sin θf 2 .

Here f 1 , f 2 , f 3 is a new orthogonal frame, with f 1 = T (f1 ) and f 2 = T (f2 ) ∈ L . Stated
differently, T (»f1 + µf2 ) = »f 1 + µf 2 , so T is linear. QED

Properties of S 2 like E2
The following statements are either obvious, or can be done as easy exercises. Use
them to refresh your memory of the case of E2 , or as a warm-up for the case of
the hyperbolic plane H2 . The spherical statements are if anything a little simpler:
for example, the distinction between translation and rotation disappears, and the
classi¬cation of motions comes directly from the normal form of Theorem 1.11.

The sphere S 2 is a metric geometry with a distance function d(P, Q), and motions
given by 3 — 3 orthogonal matrixes.

The motions act transitively on S 2 and on spherical lines through a given point P ∈ S 2 .
Every motion of S 2 is either a rotation Rot(P, θ), or a re¬‚ection Re¬‚(L) in a
line (= great circle) or a glide Glide(L , θ ) (the restriction of a Euclidean rotary
Given two pairs of points P, Q and P , Q , there exist exactly two motions g of S 2
such that g(P) = g(P ), g(Q) = g(Q ), of which one is a rotation and the other a
re¬‚ection or glide.
(5) Motions come in two kinds, direct and opposite. Every direct motion is the identity
or a composite of 2 re¬‚ections; every opposite motion is a re¬‚ection or a composite
of 3 re¬‚ections.
The spherical distance d(P, Q) between two points P, Q ∈ S 2 is the length of the
shortest curve C in S 2 joining P and Q.

Properties of S 2 unlike E2
(1) Incidence of lines. Any two spherical lines intersect in a pair of antipodal points.
(Proof: if L 1 = 1 © S 2 and L 2 = 2 © S 2 , consider the Euclidean line 1 © 2 in
R3 .) Therefore spherical geometry has no parallel lines.
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 de¬ned by setting the circumference of our own
parochial sphere to be 40 000 km. Another aspect of the same phenomenon is that
distances are bounded: d(P, Q) ¤ πr (=: 20 000 km).
(3) Spherical frames. If you try to de¬ne a spherical frame of reference by analogy with
the Euclidean notion, you get involved with the intrinsic distance. For example, if your
unit of measurement is very big compared to the radius of the sphere, you will end
up with your unit vector P0 Q 0 wrapping the sphere several times. Taking a small unit
of measurement, you can de¬ne a spherical frame P0 P1 P2 and prove the analogue of
Corollary 1.13 (a 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 P0 P1 to be 1/4 of
the circumference, and get a spherical frame which coincides with an orthonormal
frame of the ambient R3 , so that the result about motions and frames is contained in
Corollary 1.13.
Intrinsic curvature. To say that the sphere S 2 ‚ R3 is curved, you could calculate the
radius of curvature of lines relative to the ambient space R3 . However, the geometry
of S 2 also displays intrinsic curvature, as you can see in several ways. In E2 the
perimeter of a Euclidean circle of radius ρ is 2πρ. By contrast, a spherical circle of
radius ρ has perimeter 2π sin ρ, as discussed in Exercise 3.1.
Sum of angles in a triangle. Let S 2 be the sphere of radius r = 1, and P Q R a
spherical triangle. Then

∠P + ∠Q + ∠R = π + area P Q R.


Overlapping segments of S 2 .
Figure 3.6

Thus the sum of angles in a spherical triangle never equals 180—¦ . For very small
triangles, you can view the discrepancy as a re¬‚ection of intrinsic curvature as in the
preceding point.

I prove the last point, because it is not obvious at ¬rst 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 P Q R. Write a for
the part of S 2 contained between the two planes O P Q and O P R (that is, the union
of the two opposite segments) with a the dihedral angle between these planes, and
similarly for b and c . Then by circular symmetry, clearly

= area S 2 .
area (3)

Now I claim that a , b , c cover S 2 and overlap exactly in P Q R and its antipodal
triangle P Q R (see Figure 3.6).
Summing (3) for a , b and c gives

area S 2
area S + 4 area = area + area + area = (2a + 2b + 2c)
a b c

(points in and its antipodal triangle are covered 3 times, while the rest of S 2 is
covered once). Therefore a + b + c ’ π = (4π /area S 2 ) area = area . QED

3.7 Preview of hyperbolic geometry
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 in one dimension.

R2 with coordinates x, y and the ordinary Euclidean norm x 2 + y 2 .
’iθ ’iθ
The functions cos θ = e +e and sin θ = e ’e , which satisfy the relation
iθ iθ
(2) 2 2i
cos2 + sin2 = 1, and dθ sin θ = cos θ, dθ cos θ = ’ sin θ.
d d

The circle S 1 de¬ned by x 2 + y 2 = 1 is parametrised by x = sin θ, y = cos θ, and
the arc length is dx 2 + dy 2 = dθ, so that θ is the arc length parameter for S 1 .

(sinh s, cosh s)

(0, 1)

The hyperbola t 2 = 1 + x 2 and t > 0.
Figure 3.7

(4) Symmetries are the set O(2) of rotation and re¬‚ection matrixes

cos θ ’ sin θ cos θ sin θ
sin θ cos θ sin θ ’ cos θ

Now the ingredients of hyperbolic geometry in one dimension.
R2 with coordinates t, x and the Lorentz pseudometric ’t 2 + x 2 . Here I choose a
˜time-like™ coordinate t and a ˜space-like™ coordinate x. A vector is space-like if it
has positive squared length (for example (0, x)) and time-like if it has negative square
(for example, (t, 0) has squared length ’t 2 ).
The Lorentz space R2 is the ambient space for the hyperbola H1 de¬ned by t 2 =
1 + x 2 and t > 0 (see Figure 3.7). The tangent space to H1 at any point P0 = (t0 , x0 ) ∈
H1 is the line t = (x0 /t0 )x, which is space-like, because t0 > |x0 |. Therefore although
the Lorentz pseudometric ’t 2 + x 2 is not positive de¬nite, the geometry of H1 itself
contains only space-like directions.
’s ’s
The functions cosh s = e +e and sinh s = e ’e , which satisfy the relation
s s
(2) 2 2
cosh2 ’ sinh2 = 1, and ds sinh s = cosh s, ds cosh s = sinh s. It is useful to notice
d d

that sinh is a one-to-one map from the whole of R1 to the whole of R1 .
The hyperbola H1 de¬ned by t 2 = 1 + x 2 is parametrised by x = sinh s, t = cosh s,
(3) √
and the arc length in the Lorentz pseudometric is ’dt 2 + dx 2 = ds, so that s is the
arc length parameter for H1 .
Symmetries are the set O+ (1, 1) of Lorentz translation and re¬‚ection matrixes

’ sinh s
cosh s sinh s cosh s
’ cosh s
sinh s cosh s sinh s

3.8 Hyperbolic space
Consider R3 with the Lorentz quadratic form q L (v) = ’t 2 + x 2 + y 2 (compare B.2).
The cone {q L (v) < 0} breaks up into two subsets

{t > + x 2 + y 2 } ∪ {t < ’ x 2 + y 2 }.

I ¬x the positive choice t > 0 throughout.



Hyperbolic space H2 .
Figure 3.8

Hyperbolic space H2 ‚ R3 is the upper sheet of the hyperboloid of two sheets
given by q L (v) = ’1:

H2 = (t, x, y) ’t 2 + x 2 + y 2 = ’1 and t > 0 .

In other words, t = 1 + x 2 + y 2 (see Figure 3.8). This is the analogue of the
sphere S 2 of radius 1, which is parametrised (in the northern hemisphere) by
z = 1 ’ x 2 ’ y 2 . If you want the analogue of the sphere of radius r , just take the
hyperboloid q L (v) = ’r 2 . The coordinate t on R3 is ˜time-like™ and the coordinates
x, y are ˜space-like™ (compare 3.7).
A line L of hyperbolic geometry is the hyperbola H1 obtained as the intersection
of H2 with a 2-dimensional vector subspace ‚ R3 which is a Lorentz plane, in the
sense that it contains time-like vectors, so that L = © H2 = …; the restriction of q L
to has signature (’1, +1). It is obvious that there is a unique line P Q through any
two distinct points P, Q ∈ H2 , since the 2-dimensional vector subspace through
P, Q in R3 is unique. The analogy with the lines of S 2 is clear, and I could reasonably
call the lines of L great hyperbolas.

3.9 Hyperbolic distance
To de¬ne the hyperbolic distance function, I start with the formal analogue of formula
(1) of Remark 2 in 3.1, replacing the Euclidean inner product with the Lorentz inner
product · L (see B.2). Thus let P and Q be points of H2 given by the vectors v =
(t1 , x1 , y1 ) and w = (t2 , x2 , y2 ). I de¬ne the hyperbolic distance d(P, Q) between
two points by

’v · L w = cosh d(P, Q), d(P, Q) = arccosh(’v · L w);
so that (4)

in other words, d(P, Q) = arccosh(t1 t2 ’ x1 x2 ’ y1 y2 ).

The Lorentz inner product satis¬es

’v · L w = t1 t2 ’ x1 x2 ’ y1 y2 ≥ 1,

with equality only if P = Q. (See also Exercise 3.11.) Hence the distance d(P, Q) is
de¬ned and positive unless P = Q.

This clearly follows from the stronger statement.

Given two points P = Q ∈ H2 , there is a Lorentz basis f0 , f1 , f2 of
R3 giving rise to a new coordinate system in which P = (1, 0, 0) and Q =
(cosh ±, sinh ±, 0), with ± = d(P, Q) > 0.

This is simply Appendix B, Theorem B.3 (4), but I need one point of the proof,
so I repeat it here. Set f0 = v the position vector of P; since P ∈ H2 , this vector has
Lorentz norm ’1. The vector w = w + (w · L f0 )f0 , where w is the position vector
of Q, is orthogonal to f0 with respect to · L (just compute the product w · L f0 )), and
is nonzero because P = Q. Hence by Theorem B.3 (3), q L (w ) > 0. So I can set

f1 = w / q L (w ), and

w = cf0 + sf1 , where c = ’v · L w and s = q L (w ) > 0. (5)

I ¬nd the remaining basis element by the usual method of making an orthonormal
basis: choose u ∈ R3 not in the span of v and w, set w = u + (u · L f0 )f0 ’ (u · L f1 )f1

and ¬nally f2 = w / q L (w ).
The Lorentz basis f0 , f1 , f2 de¬nes a new coordinate system on the hyperbolic plane
H . In this coordinate system P = (1, 0, 0) and Q = (c, s, 0), the latter by the ¬rst

equality in (5). As Q ∈ H2 , c > 0 and its position vector has Lorentz norm ’1, so
’c2 + s 2 = ’1. By (5), s > 0 and hence c > 1. So c = cosh ±, s = sinh ± for some
± > 0, and in this coordinate system it is easy to compute d(P, Q) = ±. Hence the
distance function is meaningful and positive unless P = Q. QED

Compare Remark 3.1 (2) for the spherical analogy; the purist may want to reread
Remark 3.1 (1) at this point.

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 (5), s = sinh ± > 0 was part of the
construction of the vector f1 . Notice that cosh ± is a symmetric function and sinh ±
is an antisymmetric function. This is good, because I am measuring distances from
the base point P = (1, 0, 0) in terms of cosh ±, and using sinh ± to parametrise the
hyperbola by arc length ±.

3.10 Hyperbolic triangles and trig
This section is the analogue of 3.2. A hyperbolic triangle P Q R in H2 consists
of 3 vertexes P, Q, R and 3 hyperbolic lines P Q, P R, Q R joining them. Choose
coordinates as in Lemma 3.9 so that P = (1, 0, 0) and P Q is on the hyperbolic line
{y = 0}; set Q = (0, 1, 0).



Figure 3.10 Hyperbolic trig.

The hyperbolic angle a at P between the two lines P Q and P R is de¬ned to be
the dihedral angle between the two planes O P Q, O P R (see Figure 3.10). The point
is that this is a Euclidean angle, namely, the angle between two lines O Q and O R in
the space-like plane t = 0; in other words, the line P R is in the plane O P R spanned
by P and R = (0, cos a, sin a).

In a hyperbolic triangle P Q R, the
Proposition (Main formula of hyperbolic trig)
side Q R is determined by the two sides P Q and P R and the dihedral angle a: if
± = d(Q, R), β = d(P, Q), γ = d(P, R), then

cosh ± = cosh β cosh γ ’ sinh β sinh γ cos a. (6)

In the notation developed above, P = (1, 0, 0), Q = (cosh β, sinh β, 0)

R = (cosh γ , sinh γ cos a, sinh γ sin a);

here, as in (5), sinh γ > 0 is part of the de¬nition of the angle a. Thus calculating the
Lorentz dot product of the two vectors representing Q and R gives

cosh ± = cosh β cosh γ ’ sinh β sinh γ cos a. QED

d(Q, R) ¤ d(P, Q) + d(P, R), with equality if
Corollary ( Triangle inequality)
and only if P is on the interval [Q, R] (that is, the segment of line joining Q and R).

This is exactly as before: compare (6) with the standard formula of hyper-
bolic trig:

cosh(β + γ ) = cosh β cosh γ + sinh β sinh γ .

Both sinh β and sinh γ are positive, so that cosh(β + γ ) ≥ cosh ±, with equality if
and only a = π. Since cosh ± is an increasing function for ± > 0, it follows that
β + γ ≥ ±, with equality if and only if P ∈ [Q, R]. QED

An important corollary of the triangle inequality, in complete analogy
with Euclidean and spherical geometry, is the fact that the hyperbolic distance d(P, Q)

between two points P, Q ∈ H2 is the length of the shortest curve C in H2 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 1.4.

3.11 Hyperbolic motions
A hyperbolic motion T : H2 ’ H2 is a map preserving hyperbolic distance. As
before, my ¬rst aim is to get from this de¬nition to a manageable description of
T in terms of a suitable matrix. Read the homework on Lorentz matrixes in B.4“B.5,
before you continue.


1. Every hyperbolic motion preserves hyperbolic lines.
Every hyperbolic motion T : H2 ’ H2 is given in coordinates by x ’ Ax, where
(a) A is a Lorentz matrix, that is
«  « 
’1 0 0 ’1 0 0
A 0 1 0 A =  0 1 0 , and
0 01 0 01

(b) A preserves the two halves of the cone {q L (v) < 0}.

The proofs are almost the same as in the Euclidean and spherical cases
(see 1.7 and Theorem 3.4 (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 plane = R2 with arc length
parametrisation (cosh s, sinh s), it follows that T is linear when restricted to each ,
therefore linear on R3 .
More formally, I can extend T from H2 to the upper half-cone by radial extension;
write T for this extension. Give a Lorentz plane , choose a Lorentz basis f0 , f1 so
that L is parametrised as

Ps = (cosh s)f0 + (sinh s)f1 for s ∈ R;

here the time-like vector f0 is the coordinate of a point P0 ∈ L, and the space-like
vector f1 is the tangent direction to L at P0 , with s the distance function along L.
Then T takes L to the line L parametrised as Ps = (cosh s)f0 + (sinh s)f1 , so that T
is given by a linear map on . Since this holds for any line L, it follows that T is
linear within the upper half-cone (that is,

T (»u + µv) = »T (u) + µT (v)

whenever u, v and »u + µv are in the upper half-cone). Now, although T is only
de¬ned 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 R3 consisting of three vectors in the
upper half-cone). Moreover, A must be Lorentz since T preserves the Lorentz form
(compare B.4“B.5). QED

In proving Theorem 3.4, I extended T to R3 by radial extension, then
used linearity on each plane , 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 T de¬ned on the upper half-cone; my argument is
that it is linear in the upper half-cone, and so given by a matrix.

A Lorentz matrix A preserves the two halves of the cone {q L (v) < 0} if and only if
its top left entry a00 > 0; such a matrix de¬nes a Lorentz transformation of R3 . The set
O+ (1, 2) of Lorentz transformations is entirely analogous to the set Eucl(2) of motions
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 3.5).
The hyperbolic plane H2 is a metric geometry with a distance function d(P, Q) and
a set of motions O+ (1, 2).
The motions act transitively on H2 and the set of lines through a given point P ∈ H2 .
Every element of O+ (1, 2) is either a rotation Rot(P, θ), a Lorentz translation
Transl(L , ±) along an axis L, a Lorentz re¬‚ection Re¬‚(L) or a Lorentz glide. For
example, if L = {y = 0}, the translation and glide are given by
«  « 
cosh s sinh s 0 cosh s sinh s 0
 sinh s cosh s 0  sinh s cosh s 0  .
0 ’1
0 01 0

(Compare Exercise B.3.)
Given two pairs of points P, Q and P , Q , there exist exactly two motions g ∈
O+ (1, 2) such that g(P) = g(P ), g(Q) = g(Q ), of which one is a rotation or Lorentz
translation and the other a Lorentz re¬‚ection or glide.
O+ (1, 2) has two types of elements, direct and indirect. Every direct motion is the
identity or a composite of 2 re¬‚ections; every opposite motion is a re¬‚ection or a
composite of 3 re¬‚ections.

Incidence of two lines in H2
In 3.6 (1) I showed that two lines (great circles) of S 2 meet in a pair of antipodal points,
by taking L 1 = 1 © S 2 , L 2 = 2 © S 2 , then constructing the line V = 1 © 2 in
the ambient R3 , which of course meets S 2 in two points. Two familiar facts follow:
(1) the orthogonal complement V ⊥ ‚ R3 is a plane cutting out a line M = V ⊥ © 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, ±1), M

(a) Projection to the (x, y)-plane of the spherical lines y = c z. (b) Projection to the
Figure 3.12
(x, y)-plane of the hyperbolic lines y = c t.

is the equatorial plane z = 0, and the family of lines containing L 1 , L 2 is the pencil
of meridians (sin θ)x = (cos θ)y (Figure 3.12).
The same arguments apply to lines in H2 , but the conclusions are different, since
the ambient R3 is now Lorentz space: as before, let L 1 = 1 © H2 , L 2 = 2 © H2 ,
and consider the line V = v = 1 © 2 ‚ R3 . There are 3 cases.

V is space-like: q L (v) > 0. Then L 1 , L 2 are disjoint, since V © H2 = …. In this case,
the orthogonal complement V ⊥ with respect to the Lorentz inner product · 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 ⊥ © H2 is a line of H2 , 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).
V is time-like: q L (v) < 0. Then L 1 , L 2 intersect in P = V © H2 . They do not have a
common perpendicular, because the plane V ⊥ ‚ R3 is space-like, so does not meet
H2 . 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 θ)x = (cos θ)y.
V is actually on the light cone: q L (v) = 0. Then L 1 , L 2 are disjoint in H2 , but are
asymptotic, in the sense that they approach inde¬nitely at one end. For example,
V = (1, 1, 0) is the common asymptotic direction of the lines L c : (y = c(t ’ x))
with |c| < 1. The plane V ⊥ : (x = t) is tangent to the light cone along V , so does not
correspond to a line in H2 , and L 1 , L 2 do not have a common perpendicular.

I say that L 1 and L 2 diverge in case (i). A simple calculation shows
that, if L 1 and L 2 are parametrised by arc length as P1 (s), P2 (s) then d(P1 (s), P2 (s))
grows linearly in s as s 0; for details, see Exercise 3.21.
Case (iii) is the limiting case that separates (i) and (ii): although L 1 , L 2 are disjoint,
they ˜approach one another at in¬nity™. I say that L 1 , L 2 are ultraparallel. To make
this precise, it is useful to introduce the formal idea that each line L = © H2 of H2
has two ˜ends™, the two rays in which the plane intersects the null-cone q(v) = 0,
or the asymptotic lines of the hyperbola L ‚ . One views an end as an ˜ideal point™
of L or ˜point at in¬nity™, not a point of H2 , but rather an asymptotic direction.
Case (iii) above, can be described by saying that L 1 and L 2 have a common end
V = v = 1 © 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 in¬nity, as described in Exercise 3.20.

3.13 The hyperbolic plane is non-Euclidean
As discussed in the introduction to this chapter and at the end of 3.11, hyperbolic
geometry shares many features with Euclidean and spherical geometry; the differ-
ences are also striking. The incidence properties of lines in H2 just established are
qualitatively quite different from the Euclidean case. Two lines L 1 and L 2 of H2
have a common perpendicular M if and only if V = 1 © 2 is space-like, which is
clearly an 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 thus fails, as I
discuss below in more detail. The next section 3.14 treats the angular defect formula,
expressing the sum of angles in a triangle in terms of its area; this sum is always < π .
The hyperbolic non-Euclidean world also differs from the Euclidean in the exis-
tence of an intrinsic distance, by analogy with the spherical world (compare 3.6),
and the negative curvature of hyperbolic space (compare Exercise 3.13 (c) and
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 af¬ne geometry, compare 4.3); in
spherical geometry it is obviously false as there are no disjoint lines. What happens
in H2 ? A plausible attempt to ¬nd a parallel line M through a point P ∈ L is to drop
a perpendicular P Q onto L, then take M perpendicular to P Q; as we know from the
above, this is indeed a line not meeting L, but not the only one.

Let L be a hyperbolic line and P a point not lying on L. Then there
exists a unique perpendicular line P Q to L through P. Moreover,

(1) if M is orthogonal to P Q in P, then the lines L and M diverge;
there exists an angle θ < π with the property that if L is a line through P, then
(2) 2
L meets L if and only if the angle of L and P Q at P is less than θ. (See
Figure 3.13.)

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 9.1.2). In the present coordinate construction of H2 , there is nothing to
dispute: everything follows at once from the case division of 3.12. 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 9.4.

I give the coordinate proof. The line L corresponds to a Lorentz orthogonal
decomposition R3 = • ⊥ where L = © H2 . The coordinate vector p of P can

intersecting line M

diverging line


M ultraparallel line
Q R R′

The failure of the parallel postulate in H2 .
Figure 3.13

be written

p=q+v with q ∈ and v in ;

here v is nonzero and space-like, and q = 0 because p is time-like. Choosing Lorentz
coordinates in R3 with e0 the unit time-like vector proportional to q and f2 proportional
to v makes L into the line y = 0, Q = (1, 0, 0) and P = (t0 , 0, y0 ) with t0 = 1 + y0 . 2

The perpendicular line P Q is x = 0, and the line M perpendicular to it at P is y = y00 t.
The two planes of L and M intersect in the x-axis of R3 , so L and M diverge.
Any line through P = (t0 , 0, y0 ) is given by (sin •)x = (cos •)(y0 t ’ t0 y); in R3 ,
this plane intersects y = 0 in the line (tan •, y0 , 0) , which is time-like if and only
if | tan •| > y0 . This proves the claim (together with the actual value θ = arccot y0 ,
compare Exercise 3.17). QED

A second ˜proof™ in more geometric terms is much closer to the
historical context, if trickier to argue convincingly; please refer to Figure 3.13 during
the argument. The existence and uniqueness of the orthogonal P Q can be proved by
minimising the distance from P to L, as discussed in Exercise 3.15 (b); (1) follows
from the case division in 3.12, and is proved again in Exercise 3.21.
For (2), note ¬rst 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 © L = R then
there are points R along L and further away from Q, and hence further lines P R
meeting L. From this, a least upper bound argument shows that there must be a ˜¬rst™
line M (one on either side of P Q) which fails to meet L.
This proves almost all of (2); the only remaining point to clear up is the statement
that the angle θ between P Q and the ˜¬rst™ nonintersecting line M is less than π/2.
However, the line M at angle exactly π/2 diverges from L by (1), whereas M is
asymptotic to L; hence the angle θ must be less than π/2. Lines L having angle less
than θ at P with P Q are of type (i) and so intersect L; lines having angle greater than
θ are of type (ii) and are disjoint from L.

There are several alternative models of non-Euclidean geometry in
Other models
addition to the hyperbolic model in Lorentz space discussed here. Beltrami™s model
as the interior of an absolute conic in P2 is treated in Rees [19]; it has the great ad-
vantage of making the incidence of lines completely transparent. An alternative is the
Lobachevsky or Poincar´ 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; Exercises 3.23“25 lead you through the
construction of this model.

3.14 Angular defect
The remainder of this chapter discusses two proofs of the famous angular defect
formula of Gauss and Lobachevsky.

In a hyperbolic triangle P Q R with angles a, b, c,

a + b + c = π ’ area P Q R. (7)

In addition to ¬nite hyperbolic triangles P Q R with P, Q, R ∈ H2 , I generalise
the statement to allow ideal triangles, with one or more vertexes ideal points ˜at
in¬nity™. An ideal triangle has 3 sides which are lines of H2 , and any 2 sides either
intersect, or are ultraparallel in the sense of De¬nition 3.13, with every pair of sides
intersecting in distinct (ideal) points. Remember that 2 lines meeting at an ideal point
have angle 0 there.

3.14.1 There are two points in this proof.
The first
I. First, an explicit integration calculates the area of the particular triangle P Q R of
Figure 3.14a. The crucial point here is that the area of a triangle remains bounded,
even though one of its vertexes goes off to in¬nity.
II. Next, area of polygons and sum of angles of polygons have the simple additivity
property illustrated in Figure 3.14b: if you subdivide A as a union of two adjacent
polygons A = B ∪ C, then area A = area B + area C. The sum of angles also adds,
except that you subtract π if two angles coalesce to form a straight line (because the
common point is no longer viewed as a vertex).

Let a ∈ (0, π/2) be a given angle. Consider P Q R in H2 bounded
3.14.2 Proposition
by the three lines y = 0, y = (tan a)x and x = (cos a)t (see Figure 3.14a). Then
An explicit
P Q R = π/2 ’ a = π ’ angle sum( P Q R).

The triangle has two vertexes P = (1, 0, 0) and Q = sin a (1, cos a, 0) in
H2 and one ideal vertex R = (1, cos a, sin a). We know that ∠R P Q = a for the
same reason as in 3.2 and 3.10, because the angle in H2 is the dihedral angle in R3 ,
which equals the angle in the plane {t = 0}. I have drawn Figure 3.14a with symmetry


y = (tan a)x


a θ
Q y=0

x = (cos a)t

PQR with one ideal vertex.
Figure 3.14a The hyperbolic triangle

C) = area(B) + area(C)

C) =
angle sum(B
angle sum(B) + angle sum(C) ’ π

Figure 3.14b Area and angle sums are ˜additive™.

about the x-axis so that we see at once that ∠P Q R = π/2. Finally, ∠P R Q = 0 by
de¬nition. Hence

angle sum( P Q R) = π/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 P Q R. It is convenient to work in polar coordinates

x = r cos θ, y = r sin θ,
so that t = 1 + r 2 .

In these coordinates, the element of area in H2 is r dr dθ/ 1 + r 2 (see Exercise 3.22
and compare also Exercise 3.8). It is easy to integrate this element of area as an
inde¬nite integral, since

r dr
dθ = d 1 + r 2 dθ.

1+r 2

The more subtle point is to get an explicit expression for the domain of integration.
Since the two sides out of P in Figure 3.14a are given by √= 0, y = (tan a)x, the angle
θ runs through the interval [0, a]. For ¬xed θ, the point ( 1 + r 2 , r cos θ, r sin θ) runs
through the line P Q θ of Figure 3.14a. The condition to be under the hyperbola is
x ¤ (cos a)t, giving

cos2 a
r cos θ ¤ 1 + r 2 cos a =’ r 2 ¤ .
cos2 θ ’ cos2 a
r dr dθ
PQR = = d 1 + r 2 dθ
cos2 a
r 2=
cos2 θ’cos2 a
= 1 + r2 dθ
r 2 =0
« 
cos2 θ
’1 +  dθ.
cos2 θ ’ cos2 a

Now I am in luck, and the integrand is an exact differential: indeed, consider
• = arcsin(sin θ/ sin a) as a function of θ. Then differentiating the de¬ning relation
(sin a)(sin •) = sin θ gives

cos θ cos2 θ
= = .
(sin a)(cos •) cos2 θ ’ cos2 a

It follows that the above integral evaluates to

sin θ a
P Q R = ’a + arcsin = ’a + π/2. QED
sin a 0

3.14.3 The calculation of Proposition 3.14.2 implies at once the following result for ideal
Proof by triangles with two or more ideal vertexes.

Let P R R be an ideal triangle of H2 with one vertex P ∈ H2 and two ideal vertexes;
if ∠P = a then

P R R = π ’ a.
area (8)

Let P Q R be an ideal triangle of H2 with all three vertexes P, Q, R ideal points at
in¬nity. Then

P Q R = π.
area (9)


b c



Figure 3.14c The subdivision of

(1) Drop a perpendicular P Q from P onto the opposite side R R . By
Claim 3.9, I can choose coordinates such that P = (1, 0, 0) and P Q is the x-axis y =
0. This subdivides triangle P R R symmetrically about the x-axis as in Figure 3.14a
into two triangles P Q R and P Q R , each having angle a/2 at P. Thus applying
Proposition 3.14.2 to each gives

P R R = area P Q R + area P Q R = 2(π/2 ’ a/2),

as required.
(2) Choose any interior point S of the ideal triangle P Q R with 3 ideal vertexes,
and draw in the 3 hyperbolic line segments P S, Q S, R S. These subdivide P Q R into
3 triangles S P Q, S Q R, S R P of the type considered in (1), as on Figure 3.14c.
If a, b, c are the angles at S in each of these, then

P Q R = area S P Q + area S Q R + area
area SRP
= π ’ a + π ’ b + π ’ c,

which gives what I want, in view of a + b + c = 2π . QED
Starting from a ¬nite triangle P Q R, extend sides R P,
Proof of Theorem 3.14
Q R and P Q to in¬nity to get Figure 3.14d. Now the whole triangle has area equal to
π by (2) of the lemma, and it is subdivided into P Q R plus three triangles with two
ideal vertexes which have areas a, b, c by (1) of the lemma. Thus the area of P Q R
is π ’ a ’ b ’ c. QED

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

Given any two ideal triangles P Q R and P Q R having three
Lemma 1
ideal vertexes, there is a Lorentz transformation A : H2 ’ H2 taking P Q R into
P Q R.

π’a P

Qb c
π’b R

Figure 3.14d The angular defect formula.

This is an easy exercise in linear algebra: given any three distinct lines V1 , V2 , V3
of R3 contained in the cone {q L (v) = 0}, there is a Lorentz basis e0 , e1 , e2 of R3 for

V1 = e0 + e2 , V2 = e0 + e1 , V3 = e0 ’ e2 . (10)

Any ideal triangle P Q R with three ideal vertexes at in¬nity has ¬nite
Lemma 2
area π .

It follows by Lemma 1 that all ideal triangles are congruent, so the key point
is that the area is ¬nite (the π can be viewed as an arbitrary scaling factor).
There is a beautiful axiomatic geometry proof due to Gauss in Coxeter [5], Figure
Now consider an ideal triangle P Q R with P ∈ H2 , and two ideal vertexes Q, R.
Let a = ∠Q P R, and write P Q R = (a). I wish to prove that area P Q R =
π ’ a. For this purpose, de¬ne L(a) = π ’ area P Q R.

L(a) is an additive function of a, that is, if a = b + c with 0 < a, b, c <
Lemma 3
π then L(a) = L(b) + L(c).

Immediate from Figure 3.14e:

O P Q + area O Q R = area O P R + area P Q R = area O P R + π,

since all vertexes of P Q R are ideal. QED

L(a) is a monotonic function of a, that is, if a > b then L(a) > L(b).
Lemma 4
Moreover, L(0) = 0 and L(π ) = π.




Figure 3.14e Area is an additive function.

Figure 3.14f Area is a monotonic function.

There are several ways of proving that a > b in a ¬gure such as Figure 3.14f
consisting of two ideal triangles: if a ¤ b then the lines out of P and P diverge, as
discussed in Theorem 3.13. Note that as a ’ 0 the triangle (a) tends to the whole
of the ideal triangle, and as a ’ π it tends to a line. QED

It is obvious that Lemmas 3 and 4 imply that L(a) = a, so that area (a) = π ’ a
for all a ∈ (0, π). The proof then concludes as before by referring to Figure 3.14d.

In Exercises 3.1“3.10, consider the geometry of the sphere S 2 ‚ R3 of radius 1 with
the intrinsic (spherical) metric.

3.1 (a) De¬ne, by analogy with Euclidean geometry, the notions of spherical circle and
spherical disc with centre P ∈ S 2 and radius ρ.
(b) Prove that a spherical circle with radius ρ < π has circumference 2π sin ρ.
(c) Prove that a spherical disc of radius ρ < π has area 2π (1 ’ cos ρ).
[Hint: for (c), integrate (b).]
3.2 Deduce from Exercise 3.1 that there does not exist an isometric map from any region
of S 2 to a region of the Euclidean plane R2 .
3.3 (a) State and prove Pons Asinorum (1.16.1) in spherical geometry.
(b) Let P1 , P2 ∈ S 2 be distinct points. Prove that the set of points equidistant from
P1 , P2 is a spherical line (great circle). [Hint: use the ambient metric of R3 to ¬nd
the locus, and (i) to prove in terms of the intrinsic geometry of S 2 that every point
equidistant from P1 , P2 is on it.]
Let ‚ S 2 be a spherical n-gon, with internal angles a1 , . . . , an at its vertexes.
Guess and prove a formula for the area of in terms of ai . (Assume that the ¬gure

does not overlap itself to avoid complicated explanations of how you count the
Let ±, β, γ be the side lengths of a spherical triangle P Q R and a, b, c the opposite
angles. Use the main formula

cos ± = cos β cos γ ’ sin β sin γ cos a

to prove that |β ’ γ | < ± < β + γ and ± + β + γ < 2π.
Prove that every triple with ±, β, γ < π satisfying the above inequalities are the
sides of a spherical triangle.
3.6 In the same notation, prove the sine rule for spherical triangles

sin ± sin β sin γ
= = .
sin a sin b sin c

[Hint: using the notation p, q, r for the vertexes of P Q R as in 3.2, prove that the
matrix with rows p, q, r has determinant det( p, q, r ) = sin a sin β sin γ .]
3.7 Prove that if is an acute angled spherical triangle whose angles are submultiples
π/ p, π/q, π/r of π , then

( p, q, r ) = (2, 2, n) or (2, 3, 3) or (2, 3, 4) or (2, 3, 5).

is a triangle in R2 with the same properties, then the possibilities are
Prove that if

( p, q, r ) = (3, 3, 3) or (2, 4, 4) or (2, 3, 6).

[Hint: using the formula area = a + b + c ’ π , get + + > 1.]
1 1 1
p q r
3.8 Show that in polar coordinates

x = r cos θ, y = r sin θ, z= 1 ’ r2

on the sphere S 2 of unit radius, the element of area in S 2 is

r dr dθ
dA = √ .
1 ’ r2

[Hint: consider a small sector [θ, θ + δθ ] — [r, r + δr ] in R2 . Prove that the sector
of S 2 lying over it is very close to a spherical rectangle with length of sides equal to

r δθ and δr/ 1 ’ r 2 .]
3.9 Here is a general project: take any result you know in plane Euclidean geometry,
¬nd an analogue for spherical geometry, and either prove or disprove it. As concrete
exercises, prove or deny the following:
(a) the 3 medians of a triangle intersect in a point G;
(b) the 3 perpendicular bisectors of a triangle intersect in a point O;
(c) (harder) the 3 heights of a triangle intersect in a point H .
3.10 Another general project: set up de¬nitions and notation for the geometry of the n-
dimensional sphere S n . [Hint: the ambient space is Rn+1 and the distance function

comes from the Euclidean inner product.] State and prove some theorems in this more
general setting in analogy with the treatment of Chapter 1; in particular, if you feel
brave, you can classify completely motions of the 3-sphere S 3 following 1.15.
In Exercises 3.11“3.21, consider the geometry of hyperbolic plane H2 with the
hyperbolic metric.
Hyperbolic distance is de¬ned by d(P, Q) = arccosh(’v · L w). Adapt the argument
of the proof of Theorem B.3 (3) to prove directly that ’v · L w ≥ 1 for v, w ∈ H2 .
Prove that P(s) = (cosh s, sinh s) is the parametrisation of the hyperbola
H1 : (’t 2 + x 2 = ’1) ‚ R2
by arc length in the Lorentz pseudometric q = ’t 2 + x 2 ; put more simply, P(s +
ds) ’ P(s) is ds times a vector tangent to Q at P(s) of unit length for q. [Hint: if
P(s) = (cosh s, sinh s) then dP = (cosh s, sinh s), a unit space-like vector.]
(a) Let P = (1, 0, 0) ∈ H2 ; show how to parametrise the circle centre P and radius
r < π in H2 ‚ R3 . Deduce that a circle of radius r has circumference 2π sinh r ;
and that a disc with centre P of radius r < π has area 2π (1 + cosh r ). Your
formulas should be analogous to those for S 2 ‚ R3 in Exercise 3.1.
(b) Deduce from (a) that there does not exist an isometric map from any region of
H2 to a region of the Euclidean plane R2 or of the sphere S 2 .
(c) A 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?
De¬ne a re¬‚ection of H2 , and prove properties analogous to those of re¬‚ections of
R2 : there exists a re¬‚ection taking P1 to P2 , any direct motion of H2 is a composite
of 2 re¬‚ections, any opposite motion is a composite of 3 re¬‚ections, Pons Asinorum,
etc. [Hint: follow the spherical case in Exercise 3.3.]
3.15 (a) Use the main formula
cosh ± = cosh β cosh γ ’ sinh β sinh γ cos a
to prove that in a right-angled hyperbolic triangle, the hypotenuse is longer than
either of the other two sides. If L ‚ H2 is a line and P ∈ H2 a point not on L,
deduce that the length of the perpendicular dropped from P to L (if it exists) is
the shortest distance from P to L.
(b) Consider the function d(P, Q) for Q ∈ L; prove that d(P, Q) takes a minimum
value. [Hint: ¬x attention to a suitable closed ball around P and use the fact that
a function on a closed interval attains its bounds.] Deduce that a perpendicular
from P to L exists and is unique.
(c) If L , M ‚ H2 are lines not meeting in H2 and not ultraparallel, prove that L and
M have a unique common perpendicular.
3.16 Interpret the matrixes
«  « 
cosh s sinh s 0 cosh s sinh s 0
 sinh s cosh s 0 and  sinh s cosh s 0  ,
0 ’1
0 01 0

as hyperbolic translation and glide.

t0 0 y0
In Figure 3.13, let Q = (1, 0, 0) and P = (t0 , 0, y0 ), so that the matrix
3.17 010
y0 0 t0
de¬nes a hyperbolic translation taking Q to P. Show that the line L : (y = 0) goes to
M : (t0 y = y0 t), and the line y = (tan •)x through Q at angle • to L (parametrised
by (t, (cos •)r, (sin •)r ) with ’t 2 + r 2 = ’1) goes to (sin •)x = (cos •)(y0 t ’ t0 y).
Conclude that the limiting angle θ in Theorem 3.13 is given by cot θ = y0 .
(Harder) The formula cosh ± = cosh β cosh γ gives the hypotenuse ± of a right-
angled hyperbolic triangle in terms of the other two sides β, γ . Prove that this is
always longer than the corresponding Euclidean result β 2 + γ 2 .
Let ±, β, γ be the sides (lengths) of a hyperbolic triangle P Q R and a, b, c the
opposite angles. Prove the hyperbolic sine rule

sinh ± sinh β sinh γ
= = .
sin a sin b sin c
[Hint: argue as in 3.10 and Exercise 3.6.]
The hyperbolic lines L c : (y = c(t ’ x)) with |c| < 1 are ultraparallel, tending to
(1, 1, 0) at in¬nity (see De¬nition 3.12). Verify that L c is parametrised by arc length
1 1
L c : Pc (s) = t0 e’s + sinh s, sinh s, y0 e’s ,
t0 t0

where y0 = √1’c2 and t0 = √1’c2 (so that c = y00 and Pc (0) = (t0 , 0, y0 ) ∈ L c ). Cal-
c 1
culate d(Pc (s), P’c (s)) and show that the two curves L ±c approach asymptotically as
s ’ ∞.
Since L 0 : (y = 0) is sandwiched between L ±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.)
3.21 Suppose that L 1 and L 2 are divergent hyperbolic lines as in De¬nition 3.12. 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.
3.22 (a) Show that in polar coordinates

x = r cos θ, y = r sin θ, t= 1 + r 2,

the element of area in H2 is
r dr dθ
dA = √ .
1 + r2
[Hint: consider a small sector [θ, θ + δθ ] — [r, r + δr ] in the space-like Euclidean
R2 . Prove that the sector of H2 lying over √is very close to a hyperbolic rectangle
with length of sides equal to r δθ and δr/ 1 + r 2 .]
(b) By writing down the Jacobian determinant for the change of coordinates, check
that the element of area in H2 in the usual coordinates (t, x, y) is
dx dy
dA = .



Figure 3.15

The ¬nal set of exercises 3.23“3.26 aim to give an alternative model of hyperbolic
geometry, which may help you visualise some of its properties. I set up a geometry
on the complex upper half-plane H (Exercise 3.23), show that it is the same geometry
as the hyperbolic plane H2 (Exercise 3.24), and investigate the failure of the parallel
postulate in the new model (Exercise 3.25). If you want to read further on this, look
at Beardon [2], Chapter 7.
3.23 Let

H = {z = x + iy ∈ C | y > 0}

be the upper half-plane in the complex plane. De¬ne H-lines to be of two kinds (see
Figure 3.15): either vertical Euclidean half-lines L 1 = {x + iy ∈ H | x = c} for a
real constant c, or half-circles L 2 = {x + iy ∈ H | (x ’ a)2 + y 2 = c2 } with centre
(a, 0) on the real axis {y = 0}.
Show, algebraically or by drawing pictures, that
(a) two H-lines meet in at most one point;
(b) every pair of distinct points of H lies on a unique H-line.
(a) Consider the map • de¬ned by

’Y + i
• : (T, X, Y ) ’ .

Show that if (T, X, Y ) ∈ H2 then T ’ X > 0 hence • is a map from the hyperbolic
plane H2 ‚ R2,1 to the upper half-plane H.
(b) Consider the map ψ de¬ned by

1 + x 2 + y 2 ’1 + x 2 + y 2 ’x
ψ : (x + iy) ’ , , .
2y 2y y

Show that if x + iy ∈ H then its image (T, X, Y ) ∈ H2 hence ψ is a map from
H to H2 .
(c) Show that φ and ψ are inverse bijections between H and H2 .
(d) Show that the image of a hyperbolic line L ∈ H2 is an H-line and conversely.
(e) Let z 1 , z 2 ∈ H be points of the upper half-plane, and let vi = ψ(z i ) be their images
under ψ. Show, using the formulas above, that

|z 1 ’ z 2 |2
’v1 · L v2 = 1 + .
2 Im(z 1 ) Im(z 2 )

Deduce that setting

|z 1 ’ z 2 |2
dH (z 1 , z 2 ) = arccosh 1 + ,
2 Im(z 1 ) Im(z 2 )

makes (H, dH ) into a metric space isometric to (H2 , dH2 ).
Therefore H has a metric geometry, isometric to the hyperbolic plane H2 . In particu-
lar, it has its own symmetries, the H-motions. Sketch some cases like the hyperbolic
translations and re¬‚ections on a sheet of paper, starting from their geometric de¬ni-
tions. As a matter of fact, any direct H-motion is of the form
az + b
cz + d
for a real matrix


with ad ’ bc > 0; indirect motions are given by
a(’¯ ) + b
z’ .
c(’¯ ) + d
If you feel brave, try your hand at proving that these maps preserve H and its metric;
consult Beardon [2], section 7.4 for the full story.
One further point deserves special mention: although there appear to be two dif-
ferent types of H-lines, the set of H-motions acts transitively on the set of H-lines.
This holds because the analogous statement is true in H2 , and the two are the same!
(Graphical exercise) Draw a point P ∈ H and an H-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 con¬gurations are like that up to H-motions!) Draw some
lines through P meeting L. Shade the region of H covered by lines through P meeting
L. Draw the ultraparallel lines (see De¬nition 3.12) 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 E2 where this set is a line.
(Another graphical exercise) Do Exercise 3.15 (b“c) on H without any computation,
by drawing the appropriate diagrams.
4 Affine geometry

Af¬ne geometry is the geometry of an n-dimensional vector space together with
its inhomogeneous linear structure. Accordingly, this chapter covers basic material
on linear geometries and linear transformations. The inhomogeneous linear maps
that we allow as transformations of af¬ne space include translations such as (x, y) ’
(x + a, y + b), dilations such as (x, y) ’ (2x, 2y) and ˜shear™ maps such as (x, y) ’
(x, x + y). It is impossible to de¬ne an origin, distances between points, or angles
between lines in a way which makes them invariant under these transformations, or
to compare ratios of distances in different directions. However, the line P Q through
two points P and Q of An makes perfectly good sense; this is also called the af¬ne
span P, Q of P and Q. An af¬ne line is a particular case of an af¬ne linear subspace
E ‚ An ; I can view an af¬ne linear subspace as the af¬ne span P1 , . . . , Pk of a ¬nite
set of points, or as the set of solutions of a system of inhomogeneous linear equations
Mx = b. Arbitrary af¬ne linear maps take af¬ne linear subspaces into one another,
and also preserve collinearity of points, parallels and ratios of distances along parallel
lines; all of these are thus well de¬ned notions of af¬ne geometry.

4.1 Motivation for affine space
As before, I write Rn for the set of n-tuples (x1 , . . . , xn ) of real numbers and V ∼ =
R for an n-dimensional vector space over R. The rest of this chapter discusses

the same set under the name of af¬ne n-space An ; Chapter 1 called it Euclidean
n-space En . Before giving the formal de¬nitions, let me explain brie¬‚y the point of
having so many alternative names and notations for what are basically all the same
The set Rn of n-tuples (x1 , . . . , xn ) is an n-dimensional vector space over the ¬eld
R of real numbers: I can add two n-tuples and multiply an n-tuple by a real number.
These notions have a physical meaning: in mechanics, for example, you could think
of adding vectors in a parallelogram of forces or velocities. A vector space V is the
abstract structure in which the operations of linear algebra make sense: addition of
vectors and multiplication of vectors by scalars are de¬ned in V , and satisfy some
rules. Once I know that V has dimension n, I can choose a basis {e1 , . . . , en } and


identify a vector

v= xi ei ∈ V

with the n-tuple (x1 , . . . , xn ), so that V = Rn . However, there may be practical or
theoretical reasons for not wanting to ¬x a basis at the outset: a proof, or the answer
to a calculation, may turn out to be much nicer in a well chosen basis. In mechanics,
for example, you might want to distinguish forces in the direction of motion from
forces perpendicular to the motion.
Similarly, working in coordinate geometry of Rn (even R2 , of course), there may
be reasons to choose coordinates

xi = xi ’ ai for i = 1, . . . , n (1)

centred at some point P = (a1 , . . . , an ). In mechanics, for example, if two particles
at points P and Q exert forces on one other, you may want to take either P or Q as the
origin of coordinates, or you may prefer to take their centre of gravity, or some other
point. The coordinate change (1) is however not a linear map or change of basis of the
vector space V ; for example, it has the effect of changing the origin of coordinates to
make P = (0, . . . , 0). Indeed, two different choices of origin differ by a translation of
the form (1). Just as the laws of physics should not depend on the choice of origin, we
require that geometric properties of af¬ne space are invariant under af¬ne coordinate
changes, which include maps of the form (1).
The same issue commonly arises, from a slightly different point of view, in prob-
lems where we are interested in some space that is clearly linear in some sense, but has
no preferred origin. The model case is the space of solutions to a system of inhomo-
geneous linear equations Ax = b: as you know, the space of all solutions is given
by a particular solution x0 plus the general solution of the homogenised equations
Ax = 0 (the kernel of the matrix A). Solutions of the homogeneous linear equations
form a vector space; the particular solution x0 provides an identi¬cation of the set of
all solutions with a vector space U . There is no preferred particular solution x0 , and
a different particular solution x0 gives another identi¬cation of the solution set with
U , differing by a translation as in (1), with a = x0 ’ x0 .

4.2 Basic properties of affine space
This section lists basic properties that I take as the de¬nition of af¬ne space An .

Af¬ne space has a set of points P ∈ An in one-to-one correspondence with position
vectors p ∈ V in an n-dimensional vector space V over R. The one-to-one correspon-
dence P ” p between points and vectors is not ¬xed; rather, I am always allowed to
translate it by a ¬xed vector b, so that the new identi¬cation is P ” p = p + b.

Q=P +x

PQ = x

Figure 4.2 Points, vectors and addition.


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