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3.4 Spherical motions
A spherical motion or isometry is of course just a map T : S 2 в†’ S 2 preserving
spherical distance.
3.5 PROPERTIES OF S 2 LIKE E2 39

Theorem

A motion T : S 2 в†’ S 2 takes pairs of antipodal points to pairs of antipodal points,
(1)
and spherical lines (great circles) to spherical lines.
Any motion is given in coordinates by x в†’ Ax, where A is a 3 Г— 3 orthogonal
(2)
matrix.

Two points of the sphere are antipodal if and only if they are a maximum
Proof
distance apart (at distance ПЂr , half a world away), so the п¬Ѓrst 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) = 2ПЂr or В± d(P, R) В± d(R, Q) В± d(P, Q) = 0.

Here the п¬Ѓrst 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 п¬Ѓrst that because T : S 2 в†’ S 2 takes antipodal points to antipodal
points, it extends in a unique way to a map T : R3 в†’ 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 = в€© S 2 and T (L) = L = в€© S 2 . A spherical line L = в€© S 2 is
parametrised by arc length: a variable point of L is cos Оёf1 + sin Оёf2 , where f1 , f2 , f3
is an orthogonal basis of R3 with f1 , f2 в€€ L, and Оё equals the arc length along L.
Since T preserves distance, it preserves arc length along a spherical line, so that its
restriction TL : L в†’ L is given by

T (cos Оёf1 + sin Оёf2 ) = cos Оёf 1 + sin Оёf 2 .

Here f 1 , f 2 , f 3 is a new orthogonal frame, with f 1 = T (f1 ) and f 2 = T (f2 ) в€€ L . Stated
differently, T (О»f1 + Вµf2 ) = О»f 1 + Вµf 2 , so T is linear. QED

Properties of S 2 like E2
3.5
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
classiп¬Ѓcation 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
(1)
given by 3 Г— 3 orthogonal matrixes.
40 NON-EUCLIDEAN GEOMETRIES

The motions act transitively on S 2 and on spherical lines through a given point P в€€ S 2 .
(2)
Every motion of S 2 is either a rotation Rot(P, Оё), or a reп¬‚ection Reп¬‚(L) in a
(3)
line (= great circle) or a glide Glide(L , Оё ) (the restriction of a Euclidean rotary
reп¬‚ection).
Given two pairs of points P, Q and P , Q , there exist exactly two motions g of S 2
(4)
such that g(P) = g(P ), g(Q) = g(Q ), of which one is a rotation and the other a
reп¬‚ection or glide.
(5) Motions come in two kinds, direct and opposite. Every direct motion is the identity
or a composite of 2 reп¬‚ections; every opposite motion is a reп¬‚ection or a composite
of 3 reп¬‚ections.
The spherical distance d(P, Q) between two points P, Q в€€ S 2 is the length of the
(6)
shortest curve C in S 2 joining P and Q.

Properties of S 2 unlike E2
3.6
(1) Incidence of lines. Any two spherical lines intersect in a pair of antipodal points.
(Proof: if L 1 = 1 в€© S 2 and L 2 = 2 в€© S 2 , consider the Euclidean line 1 в€© 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
(2)
the length of any great circle) as a unit of distance; recall that the kilometre, adopted
during the French revolution, was deп¬Ѓned 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) в‰¤ ПЂr (=: 20 000 km).
(3) Spherical frames. If you try to deп¬Ѓne 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 deп¬Ѓne 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 вЉ‚ R3 is curved, you could calculate the
(4)
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 ПЃ is 2ПЂПЃ. By contrast, a spherical circle of
radius ПЃ has perimeter 2ПЂ sin ПЃ, 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
(5)
spherical triangle. Then

в€ P + в€ Q + в€ R = ПЂ + area P Q R.
3.7 PREVIEW OF HYPERBOLIC GEOMETRY 41

ОЈc
Q
P
ОЈb
ОЈa
R

Overlapping segments of S 2 .
Figure 3.6

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

I prove the last point, because it is not obvious at п¬Ѓrst sight, and because the
Proof
proof is very elegant. It is a вЂ˜Venn diagramвЂ™ 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

2a
= area S 2 .
area (3)
a
2ПЂ

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)
2
a b c
2ПЂ

(points in and its antipodal triangle are covered 3 times, while the rest of S 2 is
covered once). Therefore a + b + c в€’ ПЂ = (4ПЂ /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 .
(1)
в€’iОё в€’iОё
The functions cos Оё = e +e and sin Оё = e в€’e , which satisfy the relation
iОё iОё
(2) 2 2i
cos2 + sin2 = 1, and dОё sin Оё = cos Оё, dОё cos Оё = в€’ sin Оё.
d d

The circle S 1 deп¬Ѓned by x 2 + y 2 = 1 is parametrised by x = sin Оё, y = cos Оё, and
(3)
the arc length is dx 2 + dy 2 = dОё, so that Оё is the arc length parameter for S 1 .
42 NON-EUCLIDEAN GEOMETRIES

t
(sinh s, cosh s)

(0, 1)

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

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

cos Оё в€’ sin Оё cos Оё sin Оё
.
and
sin Оё cos Оё sin Оё в€’ cos Оё

Now the ingredients of hyperbolic geometry in one dimension.
R2 with coordinates t, x and the Lorentz pseudometric в€’t 2 + x 2 . Here I choose a
(1)
вЂ˜time-likeвЂ™ coordinate t and a вЂ˜space-likeвЂ™ 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 в€’t 2 ).
The Lorentz space R2 is the ambient space for the hyperbola H1 deп¬Ѓned by t 2 =
1 + x 2 and t > 0 (see Figure 3.7). The tangent space to H1 at any point P0 = (t0 , x0 ) в€€
H1 is the line t = (x0 /t0 )x, which is space-like, because t0 > |x0 |. Therefore although
the Lorentz pseudometric в€’t 2 + x 2 is not positive deп¬Ѓnite, the geometry of H1 itself
contains only space-like directions.
в€’s в€’s
The functions cosh s = e +e and sinh s = e в€’e , which satisfy the relation
s s
(2) 2 2
cosh2 в€’ 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 deп¬Ѓned by t 2 = 1 + x 2 is parametrised by x = sinh s, t = cosh s,
(3) в€љ
and the arc length in the Lorentz pseudometric is в€’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 reп¬‚ection matrixes
(4)

в€’ sinh s
cosh s sinh s cosh s
.
and
в€’ cosh s
sinh s cosh s sinh s

3.8 Hyperbolic space
Consider R3 with the Lorentz quadratic form q L (v) = в€’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 } в€Є {t < в€’ x 2 + y 2 }.

I п¬Ѓx the positive choice t > 0 throughout.
3.9 HYPERBOLIC DISTANCE 43

H2
t

R2
(x,y)

Hyperbolic space H2 .
Figure 3.8

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

H2 = (t, x, y) в€’t 2 + x 2 + y 2 = в€’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 в€’ x 2 в€’ y 2 . If you want the analogue of the sphere of radius r , just take the
hyperboloid q L (v) = в€’r 2 . The coordinate t on R3 is вЂ˜time-likeвЂ™ and the coordinates
x, y are вЂ˜space-likeвЂ™ (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 вЉ‚ R3 which is a Lorentz plane, in the
sense that it contains time-like vectors, so that L = в€© H2 = в€…; the restriction of q L
to has signature (в€’1, +1). It is obvious that there is a unique line P Q through any
two distinct points P, Q в€€ 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 deп¬Ѓne 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 В· 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 deп¬Ѓne the hyperbolic distance d(P, Q) between
two points by

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

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

The Lorentz inner product satisп¬Ѓes
Lemma

в€’v В· L w = t1 t2 в€’ x1 x2 в€’ y1 y2 в‰Ґ 1,

with equality only if P = Q. (See also Exercise 3.11.) Hence the distance d(P, Q) is
deп¬Ѓned and positive unless P = Q.
44 NON-EUCLIDEAN GEOMETRIES

This clearly follows from the stronger statement.
Proof

Given two points P = Q в€€ H2 , there is a Lorentz basis f0 , f1 , f2 of
Claim
R3 giving rise to a new coordinate system in which P = (1, 0, 0) and Q =
(cosh О±, sinh О±, 0), with О± = 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 в€€ H2 , this vector has
Lorentz norm в€’1. The vector w = w + (w В· L f0 )f0 , where w is the position vector
of Q, is orthogonal to f0 with respect to В· L (just compute the product w В· L f0 )), and
is nonzero because P = Q. Hence by Theorem B.3 (3), q L (w ) > 0. So I can set
в€љ
f1 = w / q L (w ), and
в€љ
w = cf0 + sf1 , where c = в€’v В· L w and s = q L (w ) > 0. (5)

I п¬Ѓnd the remaining basis element by the usual method of making an orthonormal
basis: choose u в€€ R3 not in the span of v and w, set w = u + (u В· L f0 )f0 в€’ (u В· L f1 )f1
в€љ
and п¬Ѓnally f2 = w / q L (w ).
The Lorentz basis f0 , f1 , f2 deп¬Ѓnes 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 п¬Ѓrst
2

equality in (5). As Q в€€ H2 , c > 0 and its position vector has Lorentz norm в€’1, so
в€’c2 + s 2 = в€’1. By (5), s > 0 and hence c > 1. So c = cosh О±, s = sinh О± for some
О± > 0, and in this coordinate system it is easy to compute d(P, Q) = О±. 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
Remark
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
The question of signs is important later: in (5), s = sinh О± > 0 was part of the
construction of the vector f1 . Notice that cosh О± is a symmetric function and sinh О±
is an antisymmetric function. This is good, because I am measuring distances from
the base point P = (1, 0, 0) in terms of cosh О±, and using sinh О± to parametrise the
hyperbola by arc length О±.

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).
3.10 HYPERBOLIC TRIANGLES AND TRIG 45

R Q

a
P
Q'
R'

Figure 3.10 Hyperbolic trig.

The hyperbolic angle a at P between the two lines P Q and P R is deп¬Ѓned 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
О± = d(Q, R), ОІ = d(P, Q), Оі = d(P, R), then

cosh О± = cosh ОІ cosh Оі в€’ sinh ОІ sinh Оі cos a. (6)

In the notation developed above, P = (1, 0, 0), Q = (cosh ОІ, sinh ОІ, 0)
Proof
and

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

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

cosh О± = cosh ОІ cosh Оі в€’ sinh ОІ sinh Оі cos a. QED

d(Q, R) в‰¤ 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-
Proof
bolic trig:

cosh(ОІ + Оі ) = cosh ОІ cosh Оі + sinh ОІ sinh Оі .

Both sinh ОІ and sinh Оі are positive, so that cosh(ОІ + Оі ) в‰Ґ cosh О±, with equality if
and only a = ПЂ. Since cosh О± is an increasing function for О± > 0, it follows that
ОІ + Оі в‰Ґ О±, with equality if and only if P в€€ [Q, R]. QED

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

between two points P, Q в€€ 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 в†’ H2 is a map preserving hyperbolic distance. As
before, my п¬Ѓrst aim is to get from this deп¬Ѓnition to a manageable description of
T in terms of a suitable matrix. Read the homework on Lorentz matrixes in B.4вЂ“B.5,
before you continue.

Theorem

1. Every hyperbolic motion preserves hyperbolic lines.
Every hyperbolic motion T : H2 в†’ H2 is given in coordinates by x в†’ Ax, where
2.
(a) A is a Lorentz matrix, that is
пЈ« пЈ¶ пЈ« пЈ¶
в€’1 0 0 в€’1 0 0
tпЈ­
A 0 1 0пЈё A = пЈ­ 0 1 0пЈё , 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
Proof
(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 в€€ R;

here the time-like vector f0 is the coordinate of a point P0 в€€ 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 (О»u + Вµv) = О»T (u) + ВµT (v)

whenever u, v and О»u + Вµv are in the upper half-cone). Now, although T is only
deп¬Ѓned in the half-cone, the usual linear algebra argument shows that it is given by a
3.12 INCIDENCE OF TWO LINES IN H2 47

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.4вЂ“B.5). QED

In proving Theorem 3.4, I extended T to R3 by radial extension, then
Remark
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 deп¬Ѓned 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 deп¬Ѓnes 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
1.
a set of motions O+ (1, 2).
The motions act transitively on H2 and the set of lines through a given point P в€€ H2 .
2.
Every element of O+ (1, 2) is either a rotation Rot(P, Оё), a Lorentz translation
3.
Transl(L , О±) along an axis L, a Lorentz reп¬‚ection Reп¬‚(L) or a Lorentz glide. For
example, if L = {y = 0}, the translation and glide are given by
пЈ« пЈ¶ пЈ« пЈ¶
cosh s sinh s 0 cosh s sinh s 0
пЈ­ sinh s cosh s 0пЈё пЈ­ sinh s cosh s 0 пЈё .
and
0 в€’1
0 01 0

(Compare Exercise B.3.)
Given two pairs of points P, Q and P , Q , there exist exactly two motions g в€€
4.
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 reп¬‚ection or glide.
O+ (1, 2) has two types of elements, direct and indirect. Every direct motion is the
5.
identity or a composite of 2 reп¬‚ections; every opposite motion is a reп¬‚ection or a
composite of 3 reп¬‚ections.

Incidence of two lines in H2
3.12
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 в€© S 2 , L 2 = 2 в€© S 2 , then constructing the line V = 1 в€© 2 in
the ambient R3 , which of course meets S 2 in two points. Two familiar facts follow:
(1) the orthogonal complement V вЉҐ вЉ‚ R3 is a plane cutting out a line M = V вЉҐ в€© 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, В±1), M
48 NON-EUCLIDEAN GEOMETRIES

(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 Оё)x = (cos Оё)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 в€© H2 , L 2 = 2 в€© H2 ,
and consider the line V = v = 1 в€© 2 вЉ‚ R3 . There are 3 cases.

V is space-like: q L (v) > 0. Then L 1 , L 2 are disjoint, since V в€© H2 = в€…. In this case,
(i)
the orthogonal complement V вЉҐ with respect to the Lorentz inner product В· L is a
Lorentz plane (the restriction of q L has signature (в€’1, +1), so that it contains time-
like vectors), and hence M = V вЉҐ в€© 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 в€© H2 . They do not have a
(ii)
common perpendicular, because the plane V вЉҐ вЉ‚ 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 Оё)x = (cos Оё)y.
V is actually on the light cone: q L (v) = 0. Then L 1 , L 2 are disjoint in H2 , but are
(iii)
asymptotic, in the sense that they approach indeп¬Ѓnitely at one end. For example,
V = (1, 1, 0) is the common asymptotic direction of the lines L c : (y = c(t в€’ x))
with |c| < 1. The plane V вЉҐ : (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
Definition
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 вЂ˜approach one another at inп¬ЃnityвЂ™. 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 = в€© H2 of H2
has two вЂ˜endsвЂ™, the two rays in which the plane intersects the null-cone q(v) = 0,
or the asymptotic lines of the hyperbola L вЉ‚ . One views an end as an вЂ˜ideal pointвЂ™
of L or вЂ˜point at inп¬ЃnityвЂ™, 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 в€© 2 . By convention, ultraparallel lines L 1 and L 2 have angle 0 at
this end. All the lines L c : y = c(t в€’ x) are ultraparallel, with the ray (1, 1, 0) as a
3.13 THE HYPERBOLIC PLANE IS NON-EUCLIDEAN 49

common end. These lines all approach one another arbitrarily closely as they head
out to inп¬Ѓnity, 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 в€© 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 < ПЂ .
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
9.4).
EuclidвЂ™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 afп¬Ѓne 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 п¬Ѓnd a parallel line M through a point P в€€ 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
Theorem
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 Оё < ПЂ 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 Оё. (See
Figure 3.13.)

In axiomatic geometry, the logical self-consistency of this picture was
Remark
the focal point of the 2000 year old controversy concerning EuclidвЂ™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
Proof
decomposition R3 = вЉ• вЉҐ where L = в€© H2 . The coordinate vector p of P can
50 NON-EUCLIDEAN GEOMETRIES

LвЂІ
intersecting line M

diverging line

P

M ultraparallel line
Оё
L
Q R RвЂІ

The failure of the parallel postulate in H2 .
Figure 3.13

be written

вЉҐ
p=q+v with q в€€ 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.
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 П•)x = (cos П•)(y0 t в€’ t0 y); in R3 ,
this plane intersects y = 0 in the line (tan П•, y0 , 0) , which is time-like if and only
if | tan П•| > y0 . This proves the claim (together with the actual value Оё = arccot y0 ,
compare Exercise 3.17). QED

A second вЂ˜proofвЂ™ in more geometric terms is much closer to the
Discussion
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 п¬Ѓrst 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 вЂ˜lastвЂ™ line L through P which meets L: if L в€© 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 вЂ˜п¬ЃrstвЂ™
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 Оё between P Q and the вЂ˜п¬ЃrstвЂ™ nonintersecting line M is less than ПЂ/2.
However, the line M at angle exactly ПЂ/2 diverges from L by (1), whereas M is
asymptotic to L; hence the angle Оё must be less than ПЂ/2. Lines L having angle less
than Оё at P with P Q are of type (i) and so intersect L; lines having angle greater than
Оё are of type (ii) and are disjoint from L.
3.14 ANGULAR DEFECT 51

There are several alternative models of non-Euclidean geometry in
Other models
addition to the hyperbolic model in Lorentz space discussed here. BeltramiвЂ™s model
as the interior of an absolute conic in P2 is treated in Rees ; it has the great ad-
R
vantage of making the incidence of lines completely transparent. An alternative is the
Lobachevsky or PoincarВґ model as the upper half-space in the complex plane, which
e
makes asymptotically converging ultraparallel lines easy to visualise, and which is
important in other mathematical contexts; Exercises 3.23вЂ“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,
Theorem

a + b + c = ПЂ в€’ area P Q R. (7)

In addition to п¬Ѓnite hyperbolic triangles P Q R with P, Q, R в€€ H2 , I generalise
the statement to allow ideal triangles, with one or more vertexes ideal points вЂ˜at
inп¬ЃnityвЂ™. An ideal triangle has 3 sides which are lines of H2 , and any 2 sides either
intersect, or are ultraparallel in the sense of Deп¬Ѓnition 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
proof
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 inп¬Ѓnity.
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 в€Є C, then area A = area B + area C. The sum of angles also adds,
except that you subtract ПЂ if two angles coalesce to form a straight line (because the
common point is no longer viewed as a vertex).

Let a в€€ (0, ПЂ/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
integral
P Q R = ПЂ/2 в€’ a = ПЂ в€’ angle sum( P Q R).
area

The triangle has two vertexes P = (1, 0, 0) and Q = sin a (1, cos a, 0) in
1
Proof
H2 and one ideal vertex R = (1, cos a, sin a). We know that в€ 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
52 NON-EUCLIDEAN GEOMETRIES

R

y = (tan a)x

QОё
r

a Оё
Q y=0
P

x = (cos a)t

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

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

C) =
angle sum(B
B C
angle sum(B) + angle sum(C) в€’ ПЂ

Figure 3.14b Area and angle sums are вЂ˜additiveвЂ™.

about the x-axis so that we see at once that в€ P Q R = ПЂ/2. Finally, в€ P R Q = 0 by
deп¬Ѓnition. Hence

angle sum( P Q R) = ПЂ/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
в€љ
x = r cos Оё, y = r sin Оё,
so that t = 1 + r 2 .
в€љ
In these coordinates, the element of area in H2 is r dr dОё/ 1 + r 2 (see Exercise 3.22
and compare also Exercise 3.8). It is easy to integrate this element of area as an
indeп¬Ѓnite integral, since

r dr
dОё = d 1 + r 2 dОё.
в€љ
1+r 2
3.14 ANGULAR DEFECT 53

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 в€љ= 0, y = (tan a)x, the angle
y
Оё runs through the interval [0, a]. For п¬Ѓxed Оё, the point ( 1 + r 2 , r cos Оё, r sin Оё) runs
through the line P Q Оё of Figure 3.14a. The condition to be under the hyperbola is
x в‰¤ (cos a)t, giving

cos2 a
r cos Оё в‰¤ 1 + r 2 cos a =в‡’ r 2 в‰¤ .
cos2 Оё в€’ cos2 a
Therefore
r dr dОё
PQR = = d 1 + r 2 dОё
area
t
PQR PQR
a
cos2 a
r 2=
cos2 Оёв€’cos2 a
= 1 + r2 dОё
r 2 =0
Оё=0
пЈ« пЈ¶
cos2 Оё
a
пЈ­в€’1 + пЈё dОё.
=
cos2 Оё в€’ cos2 a
0

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

cos Оё cos2 Оё
dП•
= = .
(sin a)(cos П•) cos2 Оё в€’ cos2 a
dОё

It follows that the above integral evaluates to

sin Оё a
P Q R = в€’a + arcsin = в€’a + ПЂ/2. QED
area
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.
subdivision
Lemma

Let P R R be an ideal triangle of H2 with one vertex P в€€ H2 and two ideal vertexes;
(1)
if в€ P = a then

P R R = ПЂ в€’ a.
area (8)

Let P Q R be an ideal triangle of H2 with all three vertexes P, Q, R ideal points at
(2)
inп¬Ѓnity. Then

P Q R = ПЂ.
area (9)
54 NON-EUCLIDEAN GEOMETRIES

R Q

a
b c

S

P

PQR.
Figure 3.14c The subdivision of

(1) Drop a perpendicular P Q from P onto the opposite side R R . By
Proof
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(ПЂ/2 в€’ a/2),
area

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
= ПЂ в€’ a + ПЂ в€’ b + ПЂ в€’ c,

which gives what I want, in view of a + b + c = 2ПЂ . QED
Starting from a п¬Ѓnite triangle P Q R, extend sides R P,
Proof of Theorem 3.14
Q R and P Q to inп¬Ѓnity to get Figure 3.14d. Now the whole triangle has area equal to
ПЂ 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 ПЂ в€’ a в€’ b в€’ 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 в†’ H2 taking P Q R into
P Q R.
3.14 ANGULAR DEFECT 55

ПЂв€’a P
a

ПЂв€’c
Qb c
ПЂв€’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
which

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

Any ideal triangle P Q R with three ideal vertexes at inп¬Ѓnity has п¬Ѓnite
Lemma 2
area ПЂ .

It follows by Lemma 1 that all ideal triangles are congruent, so the key point
is that the area is п¬Ѓnite (the ПЂ can be viewed as an arbitrary scaling factor).
There is a beautiful axiomatic geometry proof due to Gauss in Coxeter , Figure
16.4a.
Now consider an ideal triangle P Q R with P в€€ H2 , and two ideal vertexes Q, R.
Let a = в€ Q P R, and write P Q R = (a). I wish to prove that area P Q R =
ПЂ в€’ a. For this purpose, deп¬Ѓne L(a) = ПЂ в€’ 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
ПЂ then L(a) = L(b) + L(c).

Immediate from Figure 3.14e:
Proof

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

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(ПЂ ) = ПЂ.
56 NON-EUCLIDEAN GEOMETRIES

b+c
0

bc

P
R
Q

Figure 3.14e Area is an additive function.
P'
P
b
a

Figure 3.14f Area is a monotonic function.

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

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

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

3.1 (a) Deп¬Ѓne, by analogy with Euclidean geometry, the notions of spherical circle and
spherical disc with centre P в€€ S 2 and radius ПЃ.
(b) Prove that a spherical circle with radius ПЃ < ПЂ has circumference 2ПЂ sin ПЃ.
(c) Prove that a spherical disc of radius ПЃ < ПЂ has area 2ПЂ (1 в€’ cos ПЃ).
[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 в€€ 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 п¬Ѓnd
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 вЉ‚ S 2 be a spherical n-gon, with internal angles a1 , . . . , an at its vertexes.
3.4
Guess and prove a formula for the area of in terms of ai . (Assume that the п¬Ѓgure
EXERCISES 57

does not overlap itself to avoid complicated explanations of how you count the
area.)
Let О±, ОІ, Оі be the side lengths of a spherical triangle P Q R and a, b, c the opposite
3.5
angles. Use the main formula

cos О± = cos ОІ cos Оі в€’ sin ОІ sin Оі cos a

to prove that |ОІ в€’ Оі | < О± < ОІ + Оі and О± + ОІ + Оі < 2ПЂ.
Prove that every triple with О±, ОІ, Оі < ПЂ 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 О± sin ОІ sin Оі
= = .
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 ОІ sin Оі .]
3.7 Prove that if is an acute angled spherical triangle whose angles are submultiples
ПЂ/ p, ПЂ/q, ПЂ/r of ПЂ , 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 в€’ ПЂ , get + + > 1.]
1 1 1
p q r
3.8 Show that in polar coordinates

x = r cos Оё, y = r sin Оё, z= 1 в€’ r2

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

r dr dОё
dA = в€љ .
1 в€’ r2

[Hint: consider a small sector [Оё, Оё + ОґОё ] Г— [r, r + Оґr ] 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
в€љ
r ОґОё and Оґr/ 1 в€’ r 2 .]
3.9 Here is a general project: take any result you know in plane Euclidean geometry,
п¬Ѓnd 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 deп¬Ѓnitions and notation for the geometry of the n-
dimensional sphere S n . [Hint: the ambient space is Rn+1 and the distance function
58 NON-EUCLIDEAN GEOMETRIES

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 3.11вЂ“3.21, consider the geometry of hyperbolic plane H2 with the
hyperbolic metric.
Hyperbolic distance is deп¬Ѓned by d(P, Q) = arccosh(в€’v В· L w). Adapt the argument
3.11
of the proof of Theorem B.3 (3) to prove directly that в€’v В· L w в‰Ґ 1 for v, w в€€ H2 .
Prove that P(s) = (cosh s, sinh s) is the parametrisation of the hyperbola
3.12
H1 : (в€’t 2 + x 2 = в€’1) вЉ‚ R2
by arc length in the Lorentz pseudometric q = в€’t 2 + x 2 ; put more simply, P(s +
ds) в€’ 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.]
ds
(a) Let P = (1, 0, 0) в€€ H2 ; show how to parametrise the circle centre P and radius
3.13
r < ПЂ in H2 вЉ‚ R3 . Deduce that a circle of radius r has circumference 2ПЂ sinh r ;
and that a disc with centre P of radius r < ПЂ has area 2ПЂ (1 + cosh r ). Your
formulas should be analogous to those for S 2 вЉ‚ 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 PringleвЂ™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?
Deп¬Ѓne a reп¬‚ection of H2 , and prove properties analogous to those of reп¬‚ections of
3.14
R2 : there exists a reп¬‚ection taking P1 to P2 , any direct motion of H2 is a composite
of 2 reп¬‚ections, any opposite motion is a composite of 3 reп¬‚ections, Pons Asinorum,
etc. [Hint: follow the spherical case in Exercise 3.3.]
3.15 (a) Use the main formula
cosh О± = cosh ОІ cosh Оі в€’ sinh ОІ sinh Оі cos a
to prove that in a right-angled hyperbolic triangle, the hypotenuse is longer than
either of the other two sides. If L вЉ‚ H2 is a line and P в€€ 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 в€€ L; prove that d(P, Q) takes a minimum
value. [Hint: п¬Ѓx 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 вЉ‚ 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
пЈ« пЈ¶ пЈ« пЈ¶
cosh s sinh s 0 cosh s sinh s 0
пЈ­ sinh s cosh s 0пЈё and пЈ­ sinh s cosh s 0 пЈё ,
0 в€’1
0 01 0

as hyperbolic translation and glide.
EXERCISES 59

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
deп¬Ѓnes 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 П•)x through Q at angle П• to L (parametrised
by (t, (cos П•)r, (sin П•)r ) with в€’t 2 + r 2 = в€’1) goes to (sin П•)x = (cos П•)(y0 t в€’ t0 y).
Conclude that the limiting angle Оё in Theorem 3.13 is given by cot Оё = y0 .
(Harder) The formula cosh О± = cosh ОІ cosh Оі gives the hypotenuse О± of a right-
3.18
angled hyperbolic triangle in terms of the other two sides ОІ, Оі . Prove that this is
always longer than the corresponding Euclidean result ОІ 2 + Оі 2 .
Let О±, ОІ, Оі be the sides (lengths) of a hyperbolic triangle P Q R and a, b, c the
3.19
opposite angles. Prove the hyperbolic sine rule

sinh О± sinh ОІ sinh Оі
= = .
sin a sin b sin c
[Hint: argue as in 3.10 and Exercise 3.6.]
The hyperbolic lines L c : (y = c(t в€’ x)) with |c| < 1 are ultraparallel, tending to
3.20
(1, 1, 0) at inп¬Ѓnity (see Deп¬Ѓnition 3.12). Verify that L c is parametrised by arc length
as
1 1
L c : Pc (s) = t0 eв€’s + sinh s, sinh s, y0 eв€’s ,
t0 t0

where y0 = в€љ1в€’c2 and t0 = в€љ1в€’c2 (so that c = y00 and Pc (0) = (t0 , 0, y0 ) в€€ L c ). Cal-
c 1
t
culate d(Pc (s), Pв€’c (s)) and show that the two curves L В±c approach asymptotically as
s в†’ в€ћ.
Since L 0 : (y = 0) is sandwiched between L В±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 Deп¬Ѓnition 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 Оё, y = r sin Оё, t= 1 + r 2,

the element of area in H2 is
r dr dОё
dA = в€љ .
1 + r2
[Hint: consider a small sector [Оё, Оё + ОґОё ] Г— [r, r + Оґr ] in the space-like Euclidean
R2 . Prove that the sector of H2 lying over в€љis very close to a hyperbolic rectangle
it
with length of sides equal to r ОґОё and Оґr/ 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 = .
t
60 NON-EUCLIDEAN GEOMETRIES

L2

L1

H-lines.
Figure 3.15

The п¬Ѓnal set of exercises 3.23вЂ“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 , Chapter 7.
3.23 Let

H = {z = x + iy в€€ C | y > 0}

be the upper half-plane in the complex plane. Deп¬Ѓne H-lines to be of two kinds (see
Figure 3.15): either vertical Euclidean half-lines L 1 = {x + iy в€€ H | x = c} for a
real constant c, or half-circles L 2 = {x + iy в€€ H | (x в€’ 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 П• deп¬Ѓned by
3.24

в€’Y + i
П• : (T, X, Y ) в†’ .
Tв€’X

Show that if (T, X, Y ) в€€ H2 then T в€’ X > 0 hence П• is a map from the hyperbolic
plane H2 вЉ‚ R2,1 to the upper half-plane H.
(b) Consider the map П€ deп¬Ѓned by

1 + x 2 + y 2 в€’1 + x 2 + y 2 в€’x
П€ : (x + iy) в†’ , , .
2y 2y y

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

|z 1 в€’ z 2 |2
в€’v1 В· L v2 = 1 + .
2 Im(z 1 ) Im(z 2 )
EXERCISES 61

Deduce that setting

|z 1 в€’ 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 reп¬‚ections on a sheet of paper, starting from their geometric deп¬Ѓni-
tions. As a matter of fact, any direct H-motion is of the form
az + b
zв†’
cz + d
for a real matrix

ab
cd

with ad в€’ bc > 0; indirect motions are given by
a(в€’ВЇ ) + b
z
zв†’ .
c(в€’ВЇ ) + d
z
If you feel brave, try your hand at proving that these maps preserve H and its metric;
consult Beardon , 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 в€€ H and an H-line L not containing P. (To make
3.25
your picture pretty, choose L to be a half-circle and P to be lying over its centre; of
course you know that all conп¬Ѓgurations 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 Deп¬Ѓnition 3.12) to L from P. For educational
purposes, repeat the exercise with L a вЂ˜verticalвЂ™ 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 (bвЂ“c) on H without any computation,
3.26
by drawing the appropriate diagrams.
4 Affine geometry

Afп¬Ѓne 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 afп¬Ѓne space include translations such as (x, y) в†’
(x + a, y + b), dilations such as (x, y) в†’ (2x, 2y) and вЂ˜shearвЂ™ maps such as (x, y) в†’
(x, x + y). It is impossible to deп¬Ѓne 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 afп¬Ѓne
span P, Q of P and Q. An afп¬Ѓne line is a particular case of an afп¬Ѓne linear subspace
E вЉ‚ An ; I can view an afп¬Ѓne linear subspace as the afп¬Ѓne span P1 , . . . , Pk of a п¬Ѓnite
set of points, or as the set of solutions of a system of inhomogeneous linear equations
Mx = b. Arbitrary afп¬Ѓne linear maps take afп¬Ѓne 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 deп¬Ѓned notions of afп¬Ѓne 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 в€ј =
R for an n-dimensional vector space over R. The rest of this chapter discusses
n

the same set under the name of afп¬Ѓne n-space An ; Chapter 1 called it Euclidean
n-space En . Before giving the formal deп¬Ѓnitions, let me explain brieп¬‚y the point of
having so many alternative names and notations for what are basically all the same
thing.
The set Rn of n-tuples (x1 , . . . , xn ) is an n-dimensional vector space over the п¬Ѓeld
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 deп¬Ѓned in V , and satisfy some
rules. Once I know that V has dimension n, I can choose a basis {e1 , . . . , en } and

62
4.2 BASIC PROPERTIES OF AFFINE SPACE 63

identify a vector

n
v= xi ei в€€ V
i=1

with the n-tuple (x1 , . . . , xn ), so that V = Rn . However, there may be practical or
theoretical reasons for not wanting to п¬Ѓx 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 в€’ 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 afп¬Ѓne space are invariant under afп¬Ѓne 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 identiп¬Ѓcation 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 identiп¬Ѓcation of the solution set with
U , differing by a translation as in (1), with a = x0 в€’ x0 .

4.2 Basic properties of affine space
This section lists basic properties that I take as the deп¬Ѓnition of afп¬Ѓne space An .

Afп¬Ѓne space has a set of points P в€€ An in one-to-one correspondence with position
(I)
vectors p в€€ V in an n-dimensional vector space V over R. The one-to-one correspon-
dence P в†” p between points and vectors is not п¬Ѓxed; rather, I am always allowed to
translate it by a п¬Ѓxed vector b, so that the new identiп¬Ѓcation is P в†” p = p + b.
64 AFFINE GEOMETRY

Q=P +x

в†’
PQ = x

P
Figure 4.2 Points, vectors and addition.

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