3.4 Spherical motions

A spherical motion or isometry is of course just a map T : S 2 ’ S 2 preserving

spherical distance.

3.5 PROPERTIES OF S 2 LIKE E2 39

Theorem

A motion T : S 2 ’ 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 ’ Ax, where A is a 3 — 3 orthogonal

(2)

matrix.

Two points of the sphere are antipodal if and only if they are a maximum

Proof

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

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

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

(1)

given by 3 — 3 orthogonal matrixes.

40 NON-EUCLIDEAN GEOMETRIES

The motions act transitively on S 2 and on spherical lines through a given point P ∈ S 2 .

(2)

Every motion of S 2 is either a rotation Rot(P, θ), or a re¬‚ection Re¬‚(L) in a

(3)

line (= great circle) or a glide Glide(L , θ ) (the restriction of a Euclidean rotary

re¬‚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

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

(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 © 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

(2)

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

(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 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

(5)

spherical triangle. Then

∠P + ∠Q + ∠R = π + area P Q R.

3.7 PREVIEW OF HYPERBOLIC GEOMETRY 41

Σc

Q

P

Σb

Σa

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

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

2a

= area S 2 .

area (3)

a

2π

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

2π

(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 .

(1)

’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

(3)

the arc length is dx 2 + dy 2 = dθ, so that θ 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 re¬‚ection matrixes

cos θ ’ sin θ cos θ sin θ

.

and

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

(1)

˜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

(4)

’ sinh s

cosh s sinh s cosh s

.

and

’ 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.

3.9 HYPERBOLIC DISTANCE 43

H2

t

R2

(x,y)

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

Lemma

’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.

44 NON-EUCLIDEAN GEOMETRIES

This clearly follows from the stronger statement.

Proof

Given two points P = Q ∈ 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 ±, 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

2

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

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

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).

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 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)

Proof

and

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-

Proof

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

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 ∈ 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.

Theorem

1. Every hyperbolic motion preserves hyperbolic lines.

Every hyperbolic motion T : H2 ’ H2 is given in coordinates by x ’ Ax, where

2.

(a) A is a Lorentz matrix, that is

« «

’1 0 0 ’1 0 0

t

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

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 ∈ 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

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.4“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 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

1.

a set of motions O+ (1, 2).

The motions act transitively on H2 and the set of lines through a given point P ∈ H2 .

2.

Every element of O+ (1, 2) is either a rotation Rot(P, θ), a Lorentz translation

3.

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 .

and

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 ∈

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 re¬‚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 re¬‚ections; every opposite motion is a re¬‚ection or a

composite of 3 re¬‚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 © 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

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 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,

(i)

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

(ii)

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

(iii)

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

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 ˜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

3.13 THE HYPERBOLIC PLANE IS NON-EUCLIDEAN 49

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

9.4).

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

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 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

Remark

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

Proof

decomposition R3 = • ⊥ where L = © H2 . The coordinate vector p of P can

50 NON-EUCLIDEAN GEOMETRIES

L′

intersecting line M

diverging line

P

M ultraparallel line

θ

L

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.

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

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 ¬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.

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. Beltrami™s model

as the interior of an absolute conic in P2 is treated in Rees [19]; it has the great ad-

R

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

e

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,

Theorem

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

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 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

integral

P Q R = π/2 ’ a = π ’ 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 ∠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

Qθ

r

a θ

Q y=0

P

x = (cos a)t

R′

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) ’ 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

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 √= 0, y = (tan a)x, the angle

y

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

Therefore

r dr dθ

PQR = = d 1 + r 2 dθ

area

t

PQR PQR

a

cos2 a

r 2=

cos2 θ’cos2 a

= 1 + r2 dθ

r 2 =0

θ=0

«

cos2 θ

a

’1 + dθ.

=

cos2 θ ’ cos2 a

0

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

d•

= = .

(sin a)(cos •) cos2 θ ’ cos2 a

dθ

It follows that the above integral evaluates to

sin θ a

P Q R = ’a + arcsin = ’a + π/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 ∈ H2 and two ideal vertexes;

(1)

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

(2)

in¬nity. Then

P Q R = π.

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(π/2 ’ 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

= 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.

3.14 ANGULAR DEFECT 55

π’a P

a

π’c

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

which

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

16.4a.

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:

Proof

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

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(π ) = π.

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 ¬gure such as Figure 3.14f

Proof

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.

Exercises

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.

3.4

Guess and prove a formula for the area of in terms of ai . (Assume that the ¬gure

EXERCISES 57

does not overlap itself to avoid complicated explanations of how you count the

area.)

Let ±, β, γ 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 ± = 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

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 3.11“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

3.11

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

3.12

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.]

ds

(a) Let P = (1, 0, 0) ∈ H2 ; show how to parametrise the circle centre P and radius

3.13

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

3.14

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.

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

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-

3.18

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

3.19

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

3.20

(1, 1, 0) at in¬nity (see De¬nition 3.12). Verify that L c is parametrised by arc length

as

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

t

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

it

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 = .

t

60 NON-EUCLIDEAN GEOMETRIES

L2

L1

H-lines.

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

3.24

’Y + i

• : (T, X, Y ) ’ .

T’X

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 )

EXERCISES 61

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

z’

cz + d

for a real matrix

ab

cd

with ad ’ bc > 0; indirect motions are given by

a(’¯ ) + b

z

z’ .

c(’¯ ) + d

z

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

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 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,

3.26

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

n

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

thing.

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

62

4.2 BASIC PROPERTIES OF AFFINE SPACE 63

identify a vector

n

v= xi ei ∈ 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 ¬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

(I)

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.

64 AFFINE GEOMETRY

Q=P +x

’

PQ = x

P

Figure 4.2 Points, vectors and addition.