. 2
( 8)


Step 5. Proof of the theorem
tion p(») = det(A ’ »1) = 0, so that at least one real or complex eigenvalue » exists.
Step 1 or Steps 2“3 as appropriate gives a 1- or 2-dimensional subspace W with
AW = W on which the action of A is as indicated. By induction on the dimension, I
can assume that the action of A on W ⊥ is OK; the induction starts with dim W = 0
or 1. QED

Complex numbers make their ¬rst incursion into real geometry during the above
proof, and it is worth pondering why; quaternions also appear in a similar context
in 8.5 below.


P' = P'

P = P0 P1 P' Q'
P'' 1

The Euclidean frames P 0 , P 1 , P 2 and P 0 , P 1 , P 2 .
Figure 1.13

1.12 Euclidean frames and motions

A Euclidean frame of En is a set of n + 1 points Q 0 , Q 1 , . . . , Q n of En
such that d(Q 0 , Q i ) = 1 and the lines Q 0 Q i are pairwise orthogonal for 1 ¤ i ¤ n.

The point of the de¬nition is that if Q 0 , . . . , Q n is a Euclidean frame
then it is possible to choose coordinates so that Q 0 becomes the origin 0 ∈ Rn and
the n vectors ei = Q 0 Q i form an orthonormal basis of Rn .

If we ¬x one Euclidean frame P0 , P1 , . . . , Pn , then Euclidean motions
are in one-to-one correspondence with Euclidean frames.

The correspondence is given by T ’ T (P0 ), T (P1 ), . . . , T (Pn ). It is clear
that the image of the Euclidean frame P0 , P1 , . . . , Pn under a motion is again a
Euclidean frame. The converse, that is, the fact that two Euclidean frames are mapped
to each other by a unique motion, follows from the previous Remark and Appendix B,
Proposition B.5. QED

Frames and motions of E2
It is worth noting two useful consequences of Theorem 1.12, whose proofs are left as
easy exercises (see Figure 1.13 and Exercise 1.12):


Suppose that [P, Q] and [P , Q ] are two line segments in E2 of the same length
d(P, Q) = d(P , Q ) > 0. Then there exist exactly two motions T : E2 ’ E2 such
that T (P) = P , and T (Q) = Q .
Let P Q R and P Q R be two triangles in E2 with all sides equal:

d(P, Q) = d(P , Q ), d(P, R) = d(P , R ), d(Q, R) = d(Q , R ).

(I assume that the three vertexes of each triangle are distinct and noncollinear.) Then
there is a unique motion T : E2 ’ E2 such that T (P) = P , T (Q) = Q , T (R) = R .

“ “
Rot(O, θ) v
O θ “
“ Glide(L, v)

Rot(O, θ) and Glide(L, v).
Figure 1.14a

v L
u' Q'
Figure 1.14b Construction of glide.

Every motion of E2 is a translation, rotation, reflection or glide
Let us list the motions of E2 we know, expressed in coordinates (see Figure 1.14a).
The translation Trans(b) : x ’ x + b for b ∈ R2 .
The rotation through angle θ about a point O ∈ E2 ; if O is the origin of the coordinate
system, this is written

cos θ ’ sin θ
x1 x1
Rot(O, θ ) : ’ .
sin θ cos θ
x2 x2

The re¬‚ection in a line L; if L is the x1 -axis (x2 = 0) then

x1 x1
’ .
Re¬‚(L) :

4. The glide (or glide re¬‚ection) in a line L through a vector v along L. Re¬‚ect in L and
translate in v. If L is the x1 -axis (x2 = 0) and v = (a, 0) then this is given by:

x1 + a
Glide(L , v) : ’ .

Here v is parallel to L, and the re¬‚ection and translation commute.
I use self-documenting notation such as Rot(O, θ ) and Glide(L , v) for these mo-
tions. In each case, I have chosen coordinates in an obvious way to make the for-
mula as simple as possible. Obviously (1) and (2) are direct motions, and (3) and
(4) opposite. Note that (3) is a particular case of (4) (where the translation vec-
tor is 0). It is sometimes convenient to view (1) as a limiting case of (2), when
the centre of rotation is very far away and the angle of rotation correspondingly

That™s all, folks!





Figure 1.14c Construction of rotation.

There are several ways of proving this. (Why not devise your own? See
Exercises 1.8 and 1.9 for an argument in terms of x ’ Ax + b, and Exercise 2.11 for
an argument in terms of composing re¬‚ections.)
The proof given here is based on the following geometric idea taken from Nikulin
and Shafarevich [18]: let P, Q and P , Q be two pairs of distinct points with
d(P, Q) = d(P , Q ) = 0. By Corollary 1.13, we know that there are exactly two
motions of E2 such that T (P) = P and T (Q) = Q . In Step 1 below, I construct
a re¬‚ection or glide, and in Step 2 a rotation or translation. Now if T is any mo-
tion, pick any two distinct points P = Q, and set P = T (P), Q = T (Q). Then
T must be one of the two motions constructed in Steps 1“2, both of which are in
my list.


I ¬rst ¬nd a re¬‚ection or glide. Write u = P Q and u = P Q . First I need
Step 1
to ¬nd the line of re¬‚ection L. The direction of L and of v is the vector bisecting the
angle between u and u (that is, 1 (u + u ) if the vectors are not opposite). Doing this
arranges that the re¬‚ection or glide re¬‚ection in any line parallel to L takes P Q into

a vector parallel to P Q . Now choose L among lines with the given direction so that
d(L , P) = d(L , P ), and write A and A for the feet of the respective perpendiculars

from P and P to L and v = A A (see Figure 1.14b). Since re¬‚ection in L takes u
into a vector parallel to u by construction, and d(P, Q) = d(P , Q ), it is clear that
Glide(L , v) does what I want.

There exists a rotation or translation T : E2 ’ E2 such that P ’ P and
Step 2
Q ’ Q . I suppose ¬rst that P = P , and that the lines P Q and P Q intersect at a
single point in an angle θ.
Then the (signed) angle of rotation must be θ; the centre must be the point O of the
perpendicular bisector of the line P P determined by P O P = θ (see Figure 1.14c).
Then by construction Rot(O, θ ) takes P ’ P , and the interval [P, Q] to an interval
out of P of the same length with d(P, Q) = d(P , Q ) and the same direction as
[P Q ]; hence it takes Q ’ Q .



θ Π
Rot-Refl(L, θ, Π)
Twist(L, θ, v)

Twist (L, θ, v) and Rot-Refl (L, θ,
Figure 1.15a ).

Figure 1.15b A grid of parallel planes and their orthogonal lines.

The proof just given does not work if P = P , or if the lines P Q and P Q are
parallel, but these special cases are easy to deal with, and I leave them as exercises
(see Exercise 1.10). QED

Classification of motions of E3

A motion T : E3 ’ E3 is one of the following:

1. Translation by a vector v.
Rotation about a directed line L as axis through an angle θ.
3. Twist: the same followed by a translation along L (Figure 1.15a).
4. Re¬‚ection in a plane.
5. Glide: a re¬‚ection in a plane followed by a translation by a vector in the plane.
Rotary re¬‚ection: the rotation through θ about a directed axis L followed by a re¬‚ec-
tion in a plane perpendicular to L (Figure 1.15a).

(2) is a special case of (3), and (4) is a special case of (5). In all cases where a
motion is de¬ned as a composite of two others, these two commute. (6) is also called
a rotary inversion, because it is also the rotation around the directed axis L through
π + θ, followed by a point re¬‚ection in L © . Clearly (1)“(3) are direct motions
and (4)“(6) opposite. Notice that any motion leaves invariant a grid of parallel planes
and their orthogonal lines (Figure 1.15b).

See, for example, Exercise 1.11 or Rees [19], p. 16, Theorem 17 for a
geometric proof. I give a coordinate geometry proof based on the use of the nor-
mal form of Theorem 1.11. Let T : E3 ’ E3 be a motion expressed in coordinates
as T : x ’ Ax + b; write T = T1 —¦ T2 where Ti are given (in the same coordinate
system) by

T2 : x ’ Ax T1 : y ’ y + b.

Then by Theorem 1.11, there exists an orthogonal coordinate system such that

±1 cos θ ’ sin θ
A= , B= .
sin θ cos θ

In these coordinates, T has the form
« 
« «
x1 b1
¬ · 
T : x2  ’  cos θ ’ sin θ  + b2 .
x2 (11)
sin θ cos θ x3
x3 b3

For the proof, I have to verify that this map is a motion of one of types (1)“(6).
This can be done, for example, by a direct coordinate calculation. It is better to
argue using the following separation of variables: (11) breaks T up as a product (not
composite) of two motions T = t — t : E1 — E2 ’ E1 — E2 , where T : E1 ’ E1
and T : E2 ’ E2 are given in coordinates by

cos θ ’ sin θ
x2 x2 b
+ 2.
T : x1 ’ ±x1 + b1 ’
and T:
sin θ cos θ
x3 x3 b3

In other words, (11) separates the 3 variables in such a way that T (x) = y with
y = (y1 , y2 , y3 ), where y1 is a function of x1 only, and y2 , y3 functions of x2 , x3 only.
Now both T and T are motions in their own right. This is the real point of the
theorem. (It is easy to generalise the result to all dimensions; compare Theorem 2.5.)
T is a direct motion, and is a translation if θ = 0 or rotation if θ = 0; this follows
by Theorem 1.14, or by direct observation. In terms of coordinates (x2 , x3 ) of E2 , it
is the rotation through an angle θ about the point determined by

cos θ ’ sin θ
x2 x2 b
+ 2,
sin θ cos θ
x3 x3 b3

that is, solving for x2 , x3 by inverting a 2 — 2 matrix:

’1 cos θ ’ 1 ’ sin θ
x2 b2
= .
sin θ cos θ ’ 1
2 ’ 2 cos θ
x3 b3

The theorem follows easily on sorting out the cases. QED


θ θ'
Figure 1.16a Pons asinorum.

1.16 Sample theorems of Euclidean geometry
This chapter has mainly been concerned with the foundations of Euclidean geometry
and a description of Euclidean motions. I do not have time to give many results of
substance from Euclidean geometry, either the theory of Euclid™s Elements, or the
much more extensive nineteenth century subject, but I do not want to omit to mention
it altogether. Coxeter [5] is very entertaining on this subject.

1.16.1 Pons asinorum, ˜Bridge of asses™. Equivalent conditions on a triangle
Pons ABC:
d(A, B) = d(A, C);
θ = ∠ABC = θ = ∠AC B;
there exists a motion T : ABC ’
3. AC B.

(1) ⇐’ (2) is an easy consequence of trigonometry, because in Fig-
ure 1.16a,

d(A, O) = d(A, B) sin θ = d(A, C) sin θ .

From our point of view, (3) =’ (1) or (2) is obvious, and (1) or (2) =’ (3) follows
by Corollary 1.13. You can also directly invoke the motion of the plane consisting of
picking up the triangle and laying it down over itself so that A, B, C match up with
A, C, B in order; alternatively, you can drop a perpendicular AO from A to BC, and
argue on congruent triangles. QED

The sum of angles in a triangle is equal to π.
1.16.2 Theorem
The angle
Let ABC be a given triangle. Consider the motion T = Trans( AC) and
sum of
set A B C = T ( ABC) as in Figure 1.16b. Then because T is a motion, I get
A B C ≡ ABC, where ≡ is congruence (see Exercise 1.16). Also, since T is a
Euclidean translation, d(B, B ) = d(A, C), therefore also ABC ≡ B A B . Hence

± + β + γ = ∠B CC + ∠BC B + ∠AC B = π

since the angles combine to form a straight line. QED

B B'

γ ±
A C'
C = A'
Sum of angles in a triangle is equal to π .
Figure 1.16b



P1 P1

Figure 1.16c Parallel lines fall on lines in the same ratio.

The statement the sum of angles in a triangle equals π is equivalent to the
parallel postulate (see 3.13 and 9.1.2). The proof used translation in E2 , coming from
the coordinate model. Figure 1.16b makes sense in spherical geometry (or hyperbolic
geometry), but there d(A, A ) > d(B, B ) (respectively d(A, A ) < d(B, B )).

A distinguishing feature of Euclidean geometry is the existence of unique parallel lines
(compare 9.1.2). Parallel lines fall on lines in the same ratio, and conversely; they
are also responsible for the existence of similar triangles. The following proposition
lines and
makes these statements precise.

If L 1 , L 2 , L 3 are three parallel lines in E2 , and they meet a line M in P1 , P2 , P3 ,
then the (signed) ratio of distances d(P1 , P2 ) : d(P2 , P3 ) is independent of M (Fig-
ure 1.16c).
(2) Consider the two triangles ABC and AB C of Figure 1.16d. The following are
(a) BC is parallel to B C .
(b) Equality of ratios: d(A, B) : d(A, B ) = d(A, C) : d(A, C ).
(c) Equality of angles: ∠ABC = ∠AB C and ∠AC B = ∠AC B .




Figure 1.16d Similar triangles.


B′ C′


C A′ B
Figure 1.16e The centroid.

All this is trivial in coordinate geometry; see Exercise 1.17.

Two triangles satisfying the conditions of the second part are called similar. Cor-
responding pairs of angles of a pair of similar triangles are equal.

The three medians of a triangle ABC meet in a point G
Proposition (Centroid)
(Figure 1.16e).
Four centres
of a triangle
(See 4.7 for a slightly different proof.) Let A , B , C be the midpoints of
BC, AC, AB and let G be the point on A A with d(A, G) = 2d(G, A ). If L, M are
the midpoints of AG and C G, then by similar triangles

MA ,

(where is parallel), so that L M A C is a parallelogram, G is its centre, so M GC is
a straight line. Hence G lies on each of A A , B B , CC , so it is the centroid. QED

The three perpendicular bisectors of sides AB, BC
Proposition (Circumcentre)
and AC meet in a point O. This is the centre of the circle circumscribed around ABC
(Figure 1.16f).


B′ C′



Figure 1.16f The circumcentre.




Figure 1.16g The orthocentre.

This is almost obvious, since the perpendicular bisector of AB is deter-
mined as the locus of points equidistant from A and B, so that any two of the per-
pendicular bisectors intersect at the point O determined by d(A, O) = d(B, O) =
d(C, O). QED

The three perpendiculars dropped from a vertex onto
Proposition (Orthocentre)
the opposite side intersect in a point H .

’ ’’’
In vector notation, H is the point given by O H = 3 OG, where O is
the circumcentre and G the centroid. Indeed, in Figure 1.16g, B B is the median

’’ ’’
’ ’’’
and O B the perpendicular bisector of AC; since G B = 2 B G and G H = 2 OG,
it follows that the two triangles G B O and G B H are similar. Therefore the line
B H is perpendicular to AC, and H lies on this perpendicular. H lies on each of the
other two perpendiculars for similar reasons. QED

Note that, as a byproduct of the above proof, we also see that the centroid G lies on
the segment [O, H ] determined by the circumcentre and the orthocentre, and divides
it into the ratio (1 : 2).


B′ C′


Figure 1.16h The Feuerbach 9-point circle.

The angle bisectors of the three angles ∠C AB, ∠ABC and
Proposition (Incentre)
∠AC B meet in a point K . This is the centre of the circle inscribed into ABC.

This is exactly analogous to the case of the circumcentre above (see Ex-
ercise 1.18). QED

Theorem (The Feuerbach circle)1 The following 9 points lie on a circle (see Fig-
ure 1.16h):
3 feet P, Q, R of the perpendiculars dropped from a vertex to the opposite side;
3 midpoints A , B , C of the sides;
3 midpoints D, E, F of AH, B H, C H , where H is the orthocentre.

The intellectual achievement here is the statement, of course. The proof is
rather easy because there are so many parallel and perpendicular lines in Figure 1.16h.
By similar triangles, the following lines are parallel:


But AB ⊥ C R by construction, hence A B D E is a rectangle. Thus the circle with
diameter A D also has B E as diameter; arguing in the same way one sees that
A C D F is also a rectangle, so that the same circle with diameter A D also has C F
as diameter. Finally, ∠A P D = 90—¦ , which is a suf¬cient condition for the circle with
diameter A D to pass through P, so that this same circle passes also through the feet
of the perpendiculars. QED
1 The Feuerbach circle is alternatively called the Euler circle, because it was discovered by Poncelet and
Brianchon. The reason why the young Bavarian schoolmaster Feuerbach™s name appears in the context
is his beautiful theorem that the circle touches the inscribed circle of the triangle. Purists may prefer the
noncommital name 9-point circle.

Redo the proof of Theorem 1.1 in detail in the cases n = 1 and n = 2.
The angle between nonzero vectors u, v ∈ Rn can be de¬ned by

cos θ = u i v i /|u||v|.
Prove that the right-hand side is in the interval [’1, +1], so that its arccos is de¬ned.
The line L = xy in Rn is the set {(1 ’ »)x + »y|» ∈ R}. If z ∈ L, write y in terms of
x and z. Complete the proof of Proposition 1.8.
1.4 Show that the assumption that T is bijective in the de¬nition of motion of Euclidean
space is super¬‚uous; that is, a map T : En ’ En that preserves distances is bijective,
therefore a motion. [Hint: prove that T is af¬ne linear. Compare Exercise A.1.]
1.5 Complete the proof of Step 3 in Theorem 1.11 using the hint given in the text.
1.6 Let A be a (real) orthogonal matrix.
(a) If e, f ∈ Rn are eigenvectors of A belonging to distinct eigenvalues » = µ, prove
that e · f = 0.
(b) If z ∈ Cn is a complex eigenvector with complex eigenvalue » ∈ R, prove that
z · z = 0. (Here x · y = j x j y j is the usual inner product.) Use this to give a
better proof of Step 3 in Theorem 1.11.
(a) Let T : E2 ’ E2 be the motion obtained by re¬‚ecting in the x-axis then rotating
through θ around the origin. Show that T is the re¬‚ection in a certain line (to be
(b) Calculate the eigenvalues and eigenvectors of the re¬‚ection matrix A =
cos θ sin θ
’ cos θ .
sin θ
(c) Relate (a) and (b).
(a) Let θ be a nonzero angle and b a translation vector in the plane. Give a geometric
construction for a point P ∈ E2 such that
Rot(O, θ )(P) = Trans(’b)(P).
[Hint: draw a picture, to ¬nd points P, Q with b = Q P such that O is on the
perpendicular bisector of P Q and ∠P O Q = θ.]
(b) By solving linear equations, ¬nd x, y such that
cos θ sin θ
x1 b x1
+1 = , A= .
A where
sin θ ’ cos θ
x2 b2 x2

(c) Express the motion T : E2 ’ E2 de¬ned in coordinates by T (x) = Ax + b in the
form T = Rot(P, θ).
(d) Relate (a) and (b).
Let A = cos θ ’sin θ θ be the re¬‚ection matrix of 1.11.1, and consider the motion
1.9 sin cos
T (x) = Ax + b; give a proof in coordinates that it is a glide re¬‚ection. [Hint: you
need to turn Figure 1.14b into coordinates.]
1.10 In the proof of Theorem 1.14, Step 2, there are 3 special cases:
(a) P = P ,
(b) P Q and P Q are parallel,
(c) and P Q and P Q are opposite (that is P Q and Q P parallel).

Complete the proof of Step 2 in any of these cases by constructing a suitable translation
or rotation taking P ’ P and Q ’ Q . √
Find the two motions E2 ’ E2 taking (0, 0) ’ (1, 2) and (0, 2 ) ’ (2, 3). Write
each as x ’ Ax + b. [Hint: the easy way: for the direct motion, translate then rotate;
for the opposite motion, re¬‚ect then translate then rotate.] Express them as rotation
and glide.
1.12 Prove Corollary 1.13 (1). [Hint: as in Figure 1.13, make a Euclidean frame with
’’ ’ ’ ’
P0 = P, P0 P1 = d(P,Q) and P2 a third point; if I do the same for P , Q , there are

2 choices for P2 , one on either side of the line P Q . The statement now follows by
Theorem 1.12.]
Let P0 , P1 , P2 ∈ E2 be distinct noncollinear points. Show that there is a unique
Euclidean frame so that P0 = (0, 0), P1 = (a, 0) with a > 0 and P2 = (b, c) with
c > 0. Deduce that a motion of E2 is uniquely determined by its effect on any 3
distinct noncollinear points.
Let P0 , P1 , P2 and P0 , P1 , P2 ∈ E2 be two pairs of distinct noncollinear points
such that d(Pi , P j ) = d(Pi , P j ) for all i, j. Prove that there exists a unique mo-
tion T : E2 ’ E2 taking Pi ’ Pi for i = 1, 2, 3. [Hint: you know enough motions
to send P0 ’ P0 . Then ¬xing P0 = P0 , to send P1 ’ P1 in exactly 2 different ways.
Where does this leave P2 ?]
Let P0 , . . . , Pn be n + 1 points spanning En . Prove that a point Q ∈ En is uniquely
determined by its distances from all of the Pi . [Hint: take P0 as origin; the n vectors
’ ’’

ei = P0 Pi are linearly independent. The vector f = P0 Q is determined by f · ei , so it
is enough to determine f · ei from distances in P0 Pi Q.]
Let ABC and D E F be two triangles in E2 . Prove that the following 4 conditions
are equivalent:
(a) 3 sides are equal AB = D E, BC = E F, C A = F D;
(b) equal side“angle“side: AB = D E, C A = F D and ∠C AB = ∠F D E;
(c) angle“side“angle: ∠ABC = ∠D E F, BC = E F and ∠BC A = ∠E F D;
(d) there exists a motion T taking A ’ D, B ’ E, C ’ F.
The triangles ABC and D E F are congruent if these conditions hold; in symbols,
ABC ≡ D E F.
1.17 Prove Proposition 1.16.3 by computing in a suitably chosen coordinate system.
1.18 By analogy with the proof of Proposition 1.16.4 (Circumcentre), prove that the three
angle bisectors of angles ∠C AB, ∠ABC and ∠AC B meet in a point K . Show also
that this is the centre of the circle inscribed in ABC (a circle touching all sides of
2 Composing maps

This brief chapter takes up some examples and simple applications of composition of
maps. The aim is to clarify and review some results about motions from Chapter 1,
and to prepare some foundational points for later chapters. Composing maps is the
idea of taking ˜a function of a function™, a procedure familiar from ¬rst year calculus:
if y = f (x) and z = g(y), then you can write z = g( f (x)) = (g —¦ f )(x). The chain
rule, for example, calculates the derivative dx in terms of dy and dy .
dz dz

2.1 Composition is the basic operation
One may consider the fundamental objects in math to be numbers of various kinds;
the basic operations on them are then addition and multiplication (together with sub-
traction, division, taking roots, etc., which are in some sense the inverses of the basic
operations). There would be no point in having numbers if you could not calculate
with them. The reason that we use numbers to model the real world is precisely that it
is easier to perform operations on numbers than make the corresponding constructions
on objects out there in the wild.
However, at another level, the fundamental objects might be maps between sets.
Then the basic operation is composition of maps. Let X, Y, Z be sets, and f : X ’ Y
and g : Y ’ Z two maps between them.

The composite of f and g is the map

g—¦ f: X ’ Z (g —¦ f )(x) = g( f (x)).
de¬ned by (1)

This may look like an associative law “ but in reality it is just the de¬nition of the
left-hand side. The left-hand side is pronounced ˜g follows f , applied to x™.

The ¬rst point is that composition is a basic operation, comparable to addition and
multiplication of numbers.


Composing two translations of En means adding the corresponding vectors:

Trans(v) —¦ Trans(u) = Trans(u + v).

Indeed, either side is the operation x ’ x + u + v.
Composing two rotations of E2 (about the same centre) means adding the correspond-
ing angles (modulo 2π):

Rot(θ) —¦ Rot(•) = Rot(θ + •).

This is clear if you draw the picture; it gives the identity

cos θ ’ sin θ cos • ’ sin • cos(θ + •) ’ sin(θ + •)
= .
sin θ cos θ sin • cos • sin(θ + •) cos(θ + •)

3. In linear algebra, a matrix corresponds to a linear map; the product of two matrixes
is the composite of the corresponding linear maps (see Exercise 2.1).
One way to introduce complex numbers is as similarities of E2 : a complex num-
ber z = r exp(iθ) corresponds to rotation by θ together with a dilation by a fac-
tor r . In these terms, product of complex numbers is composite of maps (see
Exercise 2.2).

Composition of affine linear maps x ’ Ax + b
An af¬ne linear map T : Rn ’ Rn is given by T (x) = Ax + b where A is an n — n
matrix and b is a vector (see 1.8). If T1 (x) = A1 x + b1 and T2 (x) = A2 x + b2 then

(T2 —¦ T1 )(x) = A2 T1 (x) + b2 = A2 (A1 x + b1 ) + b2 = (A2 A1 )x + (A2 b1 + b2 ).

Thus if we write T A,b for the map x ’ Ax + b, composition is given by the rule
T A2 ,b2 —¦ T A1 ,b1 = T A2 A1 ,A2 b1 +b2 . Note that the ¬rst component A2 A1 is just the product,
whereas in the second component, the matrix A2 of T A2 ,b2 ¬rst acts on the translation
vector b1 before the vectors are added. I return to this composition rule in 6.5.3 below;
compare also Exercise 6.1.

Composition of two reflections of E2
Consider the re¬‚ections of E2 in two lines L 1 , L 2 . There are two cases (see Figure 2.3):

If L 1 and L 2 meet in a point P and θ is the angle at P from L 1 to L 2 then
Re¬‚(L 2 ) —¦ Re¬‚(L 1 ) = Rot(P, 2θ).
2. If L 1 and L 2 are parallel and v is the perpendicular vector from L 1 to L 2 then
Re¬‚(L 2 ) —¦ Re¬‚(L 1 ) = Trans(2v).

““ “ P


Figure 2.3 Composite of two reflections.

2.4 Composition of maps is associative
I want to consider the composite of many maps in what follows, for example the
composite of 3 re¬‚ections Re¬‚(L 3 ) —¦ Re¬‚(L 2 ) —¦ Re¬‚(L 1 ). As a preliminary step, a
point of set theory: suppose that X, Y, Z , T are sets, and that

f : X ’ Y, g : Y ’ Z, h: Z ’ T

are three maps. The associative law is the tautology that there is only one way of
getting from X to T using f, g, h in that order, namely

x ’ f (x) ’ g( f (x)) ’ h(g( f (x))). (2)

The composite h —¦ g —¦ f is the map X ’ T de¬ned by (2). Thus the expression
h —¦ g —¦ f does not admit any possible ambiguity.
In the tradition of abstract algebra, the associative law is the headache of how to
bracket h —¦ g —¦ f . It occurs if we think of the composite of only two maps as the basic
operation, and interpret a composite of three or more maps in a recursive way, such as
h —¦ (g —¦ f ), presumably to economise on de¬nitions. In this case, one ¬rst constructs
a map g —¦ f : X ’ Z , then links it with the third map to get the repeated composite
h —¦ (g —¦ f ) : X ’ Z ’ T . However, as my tautology says, whatever brackets you
put in, h —¦ g —¦ f has only one possible meaning, namely (2). You can think through
a few of these identities as exercises, see Exercise 2.3. (I warn you, it is exceedingly
Another abstract algebraic notion, the ˜commutative law™, is discussed in Exer-
cise 2.4.

2.5 Decomposing motions
This section introduces the ¬rst way of decomposing a motion of En as a composite
of ˜elementary™ motions. Although there are more powerful decompositions around
(see for example the next section), the one given here already illustrates some basic
features of any such decomposition. To start with, let us make a list of motions of En
that could reasonably be called ˜elementary™.

An af¬ne linear subspace ‚ En of Euclidean space is the image U ‚ Rn of
a vector subspace under some choice of coordinates. The dimension of is the
dimension dim U of U . (These notions will be investigated in much more detail in
4.3 below.) In particular, a hyperplane of En is an (n ’ 1)-dimensional af¬ne linear
subspace ‚ En .

The re¬‚ection in a hyperplane is the motion that ¬xes pointwise
and reverses orthogonal vectors to . In coordinate form, if is given by x1 = 0,
and x2 , . . . , xn are coordinates on ∼ En’1 , then
« 
« «
x1 · x1
¬ 1
¬.· ¬ ·¬ . ·
Re¬‚( ) :  .  ’ ¬ · . .
. .
 .
xn xn

In other words, the de¬ning property of ρ = Re¬‚( ) is that it ¬xes every point
of , and takes P ∈ into the point Q = ρ(P) such that
/ is the perpendicular
bisector of P Q. Note that if P and Q are two distinct points of En , there is a unique
hyperplane such that Re¬‚( ) takes P to Q, namely the perpendicular bisector of
P Q; this is also determined as the locus of points equidistant from P and Q.

Let be an (n ’ 2)-dimensional af¬ne linear subspace of En . The
rotation around the axis through (signed) angle θ is the motion that ¬xes pointwise
and rotates by θ in planes orthogonal to .
In coordinates, if is given by x1 = x2 = 0, then the planes orthogonal to are
described by x3 = c3 , . . . , xn = cn for c3 , . . . , cn real constants (draw a picture for
n = 3!). Hence the coordinate form is
« 
cos θ ’ sin θ
« «
¬ sin θ · x1
cos θ
x1 ¬ ·
¬.· ¬ ·¬ . ·
Rot( , θ) :  .  ’ ¬ · . .
. ·.
¬ ..
  xn

Finally, there are also translations Trans(v) : x ’ x + b for b ∈ Rn .
Every motion T of En is a composite of a translation, k re¬‚ections and
l rotations, where k + 2l ¤ n.
Convince yourself that this is really a restatement of the fact that every
orthogonal matrix has a normal form described in Theorem 1.11. QED
2.6 Reflections generate all motions
Here we aim to improve the statement of the previous section, using geometric rather
than algebraic reasoning.

Every motion T of En is a composite of at most n + 1 re¬‚ections,

T = ρ1 —¦ ρ2 —¦ · · · —¦ ρk , with k ¤ n + 1.

The rough idea is simple: if every point P ∈ En is ¬xed by T , then T =
id, so it is a composite of no re¬‚ections at all. Otherwise, choose any P so that
T (P) = Q = P; then, by what I just said, there is a re¬‚ection ρ1 taking Q back to
P, namely the re¬‚ection in the perpendicular bisector of P Q. Then T (P) = Q and
ρ1 (Q) = P, so that T1 = ρ1 —¦ T is a new motion ¬xing P. Now it turns out (see below)
that T1 still ¬xes any point already ¬xed by T , so that T1 ¬xes strictly more than T .
I can repeat this argument, obtaining T2 = ρ2 —¦ T1 ¬xing even more points, and so on
inductively until Tk = ρk —¦ Tk’1 ¬xes every point of En . Putting this together gives
ρk —¦ · · · —¦ ρ1 —¦ T = id.
Now precomposing the equation T1 = ρ1 —¦ t with ρ1 gives

ρ1 —¦ T1 = (ρ1 —¦ ρ1 ) —¦ T,

and since ρ1 —¦ ρ1 = id, we get T = ρ1 —¦ T1 . Arguing in the same way gives T =
ρ1 —¦ T1 = ρ1 —¦ ρ2 —¦ T2 = · · · , which concludes the proof.
To go through the argument in more detail, I assert ¬rst that the set Fix(T ) of ¬xed
points of any motion T is (either empty or) an af¬ne linear subspace of En . This
follows from Proposition 4.3 (2), and the fact that if two distinct points P, Q are ¬xed
by T , then so is any point R on the line P Q: if R ∈ [P, Q] then

d(P, R) + d(R, Q) = d(P, Q) T (P) = P, T (Q) = Q
=’ d(P, T (R)) + d(T (R), Q) = d(P, Q),

so T (R) ∈ [P, Q] and T (R) = R, and similarly if P, Q, R are collinear but in some
other order.
Now to get a neat induction, I add a slightly stronger clause to the theorem:

Moreover, if Fix(T ) has dimension n ’ l (for some l = 0, . . . , n) then T
is a composite of at most l re¬‚ections.

As argued above, if T = id then I choose a point P ∈ Fix(T ), set Q = T (P) and
the perpendicular bisector of P Q, and let ρ be the re¬‚ection in . The point of the
construction is that ρ(Q) = P, so that T1 = ρ —¦ T ¬xes P.
is characterised as the set of points of En
Now the perpendicular bisector
equidistant from P and Q. Moreover, every point R ∈ Fix(T ) is equidistant from P
and Q, because d(P, R) = d(T (P), T (R)) = d(Q, R). Therefore Fix(T ) ‚ , and
ρ = Re¬‚( ) ¬xes every point of Fix(T ). It follows that Fix(T1 ) ⊃ Fix(T ) ∪ {P}.
The claim now follows by induction on l. If l = 0 then T = id. If l = 1 then
Fix(T ) = is a hyperplane, and T = Re¬‚( ). Otherwise, as just proved, I can ¬nd
ρ so that T1 = ρ —¦ T ¬xes a strictly bigger set than T , and therefore Fix(T1 ) has





Figure 2.7 Composite of a rotation and a reflection.

dimension (n ’ l ) with l < l. By induction, I can assume the result for T1 , that is,
T1 = ρ1 —¦ ρ2 —¦ · · · —¦ ρk with k ¤ l so that T = ρ —¦ T1 is the composite of at most
l + 1 ¤ l re¬‚ections, as required. This proves the claim. If Fix(T ) = … then Fix(T1 )
is at least one point, so that by the claim, T1 is a composite of at most n re¬‚ections,
and T the composite of at most n + 1 re¬‚ections, which proves the theorem. QED

2.7 An alternative proof of Theorem 1.14
Every motion of E2 is a rotation, re¬‚ection, translation
Theorem (= Theorem 1.14)
or a glide.

Every motion of E2 is the composite of at most 3 re¬‚ections. As we saw
in 2.3, the composite of 2 re¬‚ections is a translation if the 2 axes are parallel, and
a rotation if they meet at a point P. It only remains to prove that the composite of
3 re¬‚ections ρ3 —¦ ρ2 —¦ ρ1 is a glide or re¬‚ection. Suppose for simplicity that the axes of
ρ1 and ρ2 meet at a point P, and make an angle θ there, so that ρ2 —¦ ρ1 = Rot(P, 2θ)
(see Figure 2.3). Suppose also that P ∈ L 3 (the case P ∈ L 3 is easier). The problem
then is to learn how to compose Rot(P, 2θ) with ρ3 = Re¬‚(L).
In Figure 2.7, L is the axis of the third re¬‚ection ρ3 , and Q = ρ3 (P). Draw the
line M passing through the midpoint of P Q, such that the angle from M to L is θ; if
we consider the rectangle P AQ B with P Q as a diagonal line, and sides P A and B Q
parallel to M, it is easy to see that Re¬‚(L) —¦ Rot(P, 2θ) = Glide(M, v) is the glide
with axis the line M and translation vector the median vector v. QED

2.8 Preview of transformation groups
As we have seen in this chapter, the composite of maps g —¦ f is a basic, simple and
familiar idea having many useful applications. From an algebraic point of view, the
composite of Euclidean motions de¬nes a product

Eucl(n) — Eucl(n) ’ Eucl(n)

on the set Eucl(n) of motions of En , which is associative (see 2.4), has an identity
element and inverses. In other words, motions form a transformation group of En .
This idea is taken up again in Chapter 6 when we are ready for serious applications.

A standard result of linear algebra identi¬es an m — n matrix A = (ai j ) with a linear
map ± : Rn ’ Rm (taking the standard basis of column vectors to the columns of
A). If B = (b jk ) is an l — m matrix giving a linear map β : Rm ’ Rl , verify that the
product matrix B A corresponds to the composite β —¦ ±.
The (nonzero) complex numbers can be viewed as a set of similarities of E2 :
regard z = x + iy as the map Tz : R2 ’ R2 given by the matrix ’y x . Write
z = r exp(iθ) where r = |z| and θ = arg z, and interpret the map Tz geometrically.
Prove that Tz is a similarity in the sense that there exists » for which d(T (x), T (y)) =
»d(x, y). Show how to obtain multiplication of complex numbers as composition of
In the notation of 2.4, prove that h —¦ g —¦ f = (h —¦ g) —¦ f . Prove that for 4 consecutive
maps f, g, h, k, we have

(k —¦ h) —¦ (g —¦ f ) = k —¦ ((h —¦ g) —¦ f ).

Generalise the statement to any number of maps and any bracketing. Please be sure
to dispose of your solution in the paper recycling bin.
2.4 In the notation of 2.4, ¬nd the conditions for the domain and range of f, g so that the
commutative law
g—¦ f = f —¦g

makes sense as a question. Show that the commutative law holds for the set of trans-
lations in En , as well as the set of rotations of E2 about a ¬xed point. Show that it
does not hold for the set of all motions of Euclidean space En .
Verify by calculation that the usual de¬nition of matrix multiplication AB = (cik =
j ai j b jk ) is associative. Use Exercise 2.1 and the associativity of maps to show that
you do not need to do the calculation.
By 2.2, af¬ne linear maps T A,b : Rn ’ Rn compose according to the rule T A2 ,b2 —¦
T A1 ,b1 = T A2 A1 ,A2 b1 +b2 ; verify that this formula de¬nes an associative multiplication
Exercises in composing motions of E2 .
The half-turn about P is the rotation through 180—¦ . Prove the following.
(a) The composite of 2 half-turns is a translation.
(b) Every translation is a composite of 2 half-turns.
(c) The composite of 3 half-turns is a half-turn.
(d) If L is a line and P a point then

Re¬‚(L) and Halfturn(P) commute ⇐’ P ∈ L .

Prove that every opposite motion of E2 is the composite of a half-turn and a re¬‚ection.
2.8 Give a geometric treatment of the composition of a rotation with a glide, to get another
glide or re¬‚ection. When is Glide(L , v) —¦ Rot θ a re¬‚ection? [Hint: draw a diagram
similar to Figure 2.7.]
Show that any composite T1 —¦ T2 with either T1 or T2 a re¬‚ection or glide can be
understood by drawing a diagram like Figure 2.7. [Hint: to view g = Glide(L , v) and
its effect on a point P ∈ L, draw a rectangle with the line P T (P) as a diagonal and
v as a median. The best way to see g1 —¦ g2 is to draw two such rectangles with a
common diagonal and the vectors v1 , v2 as respective medians. For glide composed
with rotation or translation, you guess that the answer is g1 —¦ t = g2 , which you can
rewrite as T = g1 —¦ g2 and treat similarly.]
(Harder) Use Claim 2.6 to study motions of E3 ¬xing a point O, and compare with
the conclusion of Theorem 1.11. [Hint: a composite of 2 re¬‚ections in planes 1 , 2
through O is a rotation about a line through O. For 3 re¬‚ections, you need to prove
that Re¬‚( ) —¦ Rot(L , θ ) is a rotary re¬‚ection, or in other words, to ¬nd a plane which
is rotated into itself by the composite.]
2.11 (Harder) Give a proof of Theorem 1.15 using Theorem 2.6. In other words, study the
possibilities for the composite of ¤ 4 re¬‚ections of E3 , and show that they lead to the
6 cases listed in Theorem 1.15. [Hint: see Rees [19].]
2.12 You can move a heavy piece of furniture (e.g. a bedroom wardrobe) by lifting the
front and rotating it about the two back corners. Convince yourself that you can ˜walk™
your wardrobe anywhere in the Euclidean plane. (Ignore doors and stairs.)
Let P, Q ∈ E2 be two distinct points. Prove that every direct motion of E2 is a
composite of suf¬ciently many rotations about P and Q.
[Hint: what kind of answer is required? First show that it is enough to prove that you
can carry out any translation and any rotation about P. For the translations, think how
you shift your wardrobe “ easy does it!]
3 Spherical and hyperbolic
non-Euclidean geometry

Together with plane Euclidean geometry, spherical and hyperbolic geometry are
2-dimensional geometries with the following properties:
(1) distance, lines and angles are de¬ned and invariant under motions;
(2) the motions act transitively on points and directions at a point;
(3) locally, incidence properties are as in plane Euclidean geometry.
In more detail, (2) means that if P, P are points, and », » directions at these
points, then there exists a motion T taking P to P and » to » ; in other words, the
geometry is homogeneous (the same at every point) and isotropic (the same in every
direction). (3) means that in suf¬ciently small open sets, a line is uniquely speci¬ed
by a point and a direction, or by two points P, Q, and two lines li meet in at most one
point (see Figure 3.0).
However, the geometries also differ in several respects:
(1) the global incidence properties of lines, that is, the existence of parallel and non-
intersecting lines;
(2) intrinsic curvature properties: the perimeter of a circle, and the sum of angles in a
(3) the possibility of de¬ning a unit of length intrinsic to the geometry.
Euclidean geometry in the plane was described in detail in Chapter 1. Although cer-
tainly not the same thing as plane geometry, spherical geometry is still very intuitive,
because every de¬nition and statement can be readily visualised on the very concrete
model S 2 ‚ R3 , which you can hold in your hand or kick around a playing ¬eld.
I discuss spherical lines (great circles), distances, angles and triangles, the classi-
¬cation of motions in terms of rotations and re¬‚ections, frames of reference and
angular excess.
In contrast, plane hyperbolic geometry originally arose in axiomatic geometry
(compare 9.1.2); the coordinate model I treat in this chapter is not immediately famil-
iar, and was discovered many decades after axiomatic hyperbolic geometry. Although
my model of hyperbolic geometry is not intuitive, essentially every step in my treat-
ment is parallel to spherical geometry. Once you are sure you know what you are


P l3
Figure 3.0 Plane-like geometry.

doing, you can just replace x 2 + y 2 = 1 by ’t 2 + x 2 = ’1, and the trig functions
sin and cos by the hyperbolic trig functions sinh and cosh, and everything extends
more or less word-for-word. This is the essential content of the prophetic suggestion
by J. H. Lambert (1728“1777) that non-Euclidean geometry ˜should be related to the

geometry on a sphere of radius i = ’1™ (see Coxeter [5], p. 299).
In Chapter 1 on Euclidean geometry, I discussed n-dimensional Euclidean space En
along with the more familiar planar version. There is no logical reason to discontinue
this practice, but for ease of digestion as well as notation, all de¬nitions in this
chapter are given in two dimensions. You will bene¬t immensely by generalising the
de¬nitions and, in some cases, the theorems to the higher dimensional setup; you are
explicitly encouraged to do so in Exercise 3.10. Higher dimensional spheres appear
in later chapters (see for example 7.4.2 and 8.5); unfortunately there is no space in the
book for a detailed treatment of higher dimensional hyperbolic space and a discussion
of its signi¬cance.

3.1 Basic definitions of spherical geometry
The sphere S 2 ‚ R3 of radius r centred at the origin O is de¬ned by the equation
x 2 + y 2 + z 2 = r 2 . I will often refer to points P ∈ S 2 via their position vector O P =
p. A spherical line or great circle in S 2 is the intersection of S 2 with a plane = R2
through the origin; thus it is a circle in centred at O and with the same radius
r as S . Two points P, Q ∈ S are antipodal if their position vectors p, q satisfy
2 2

p = ’q. Through any two distinct points P, Q ∈ S 2 which are not antipodal, there
is a unique great circle or spherical line L = P Q. The (spherical) distance d(P, Q)
between points P, Q ∈ S 2 is the distance measured along the shorter arc of a great
circle through P and Q; that is, it is radius r times ∠P O Q, the angle at O between
O P and O Q, where the angle is always interpreted as the absolute value in the range
[0, π ]. For ease of notation, I usually ¬x the radius r = 1 from now on.


(1) If you go back to the chapter on Euclidean geometry and compare the treatment of
1.1“1.3 to the one given here, you may notice that I have been a bit sloppy here. To

be consistent, I should have de¬ned ˜model™ S 2 to be the sphere {x 2 + y 2 + z 2 = r 2 }
in R3 with its inherited spherical distance, and ˜abstract™ S 2 to be a metric space
isometric to ˜model™ S 2 but without a ¬xed choice of identi¬cation. Spelling this
out explicitly leads to rather clumsy notation, but implicitly I am still following this
procedure; in particular, I reserve the right to choose different coordinates on my
˜abstract™ metric S 2 if so needed. This remark applies equally well to the treatment
of hyperbolic geometry in 3.9 below.
The sphere S 2 is de¬ned as the subset {x 2 + y 2 + z 2 = 1} of R3 . On the northern
hemisphere {z ≥ 0} I can rewrite this as z = 1 ’ x 2 ’ y 2 . This gives a fairly good
coordinate representation of S 2 near the north pole, but a fairly bad one in moderate
or tropical regions. What is wrong with it? Well, if the model is the whole of R2 , it is
much too big; if we take only the disc D 2 : x 2 + y 2 ¤ 1, crossing the equator in S 2
corresponds to falling off the edge of the world in the model. Furthermore, distances,
angles, areas, curvature are all screwed up.
It is a basic problem in cartography to map regions of the surface of the Earth onto
a plane. However, the map based on z = 1 ’ x 2 ’ y 2 is one of the most primitive
and useless ways to do this. Over the course of time, several much better ways have
been invented; see the references in the introduction of Chapter 9 for a starting point
on this.
The distance d(P, Q) is de¬ned as (radius times) the angle of the P Q arc, ± =
∠P O Q. It is useful to know how to translate between this angle and the coordinates
of P, Q. In vector notation, the dot product of unit vectors equals the cosine of the
angle between them: that is, if P, Q have position vectors p, q then ± = ∠P O Q is
given by

p · q = cos ±, d(P, Q) = ± = arccos(p · q).
that is, (1)

(I have set r = 1, so that p and q are unit vectors.) Recall that arccos = cos’1 is the
inverse function of cos, so that ± = arccos x is de¬ned by the property x = cos ±;
similarly for arcsin. Here I choose ± in the range [0, π].
Given P and Q, I can choose coordinates so that P = (0, 0, 1) and O P Q is the
(x, z)-plane {y = 0}; then Q = (sin ±, 0, cos ±). This is a parametrisation of the great
circle, with parameter ±. Points with x < 0 can also be included, by allowing ± < 0
to run through the range [’π, π], but then d(P, Q) = |±|.
In fact (sin ±, 0, cos ±) is a parametrisation by arc length: if you think of (part of)
the sphere S 2 as the graph of z = 1 ’ x 2 ’ y 2 as in (2), then d(P, Q) = P ds
where the in¬nitesimal arc length ds is determined by ds 2 = dx 2 + dy 2 + dz 2 . Thus
the length of arc P Q is

sin ±
= arcsin(sin ±) = ±.

1 ’ x2

Geometers like to distinguish the intrinsic geometric properties of S 2 from those
related to the embedding S 2 ‚ R3 . It is important in this context to notice that the
natural distance in spherical geometry is the intrinsic distance, that is, the length of
a certain curve traced in the surface S 2 , as opposed to the distance in the ambient
Euclidean space; you go from London to Singapore by plane, not by tunnel.

3.2 Spherical triangles and trig
The convention r = 1 is still in force. A spherical triangle P Q R consists of 3
vertexes P, Q, R and 3 arcs of great circle P Q, P R, Q R joining them. These do not
have to be the shorter arcs; P, Q are allowed to be antipodal, and then you have to
specify one of the great circles to be the arc P Q.
The spherical angle a at P between the two lines P Q and P R is equal to the
dihedral angle between the two planes O P Q, O P R in R3 , in other words it is the
angle between two lines cut out by the two planes in an auxiliary plane orthogonal to
O P. You can take this as a de¬nition if you like, and then you do not have to worry
about how the angle between two curves is de¬ned. More precisely, the tangent plane
to S 2 at P is the 2-plane TP S 2 de¬ned by z = 1, and the tangent vectors to the two
curves P Q and P R are the two lines in TP S 2 cut out by these two planes. They are
orthogonal to the axis O P, so the angle between the two curves equals the dihedral
angle a between the two planes.

The side Q R of the triangle is deter-
Proposition (Main formula of spherical trig)
mined by the other two sides P Q and P R and the dihedral angle a. More precisely,

± = ∠Q O R = d(Q, R), β = ∠P O Q = d(P, Q), γ = ∠P O R = d(P, R).

(Recall that I have ¬xed the radius r = 1.) Then

cos ± = cos β cos γ + sin β sin γ cos a. (2)

Although the statement looks complicated, the proof is easy 3-dimensional
coordinate geometry. In Figure 3.2, let Q and R be the points on great circles at
distance π/2 from P, so that O Q is orthogonal to O P. Choose coordinates (x, y, z)
so that P = (0, 0, 1) (the north pole), and the equator is given by z = 0. Then Q is a
point on the equator, so I can choose Q = (1, 0, 0), and R = (cos a, sin a, 0). This
determines the coordinates of all the points in the ¬gure; by de¬nition of β, γ , the
following relations hold between the position vectors:

q = cos βp + sin βq = (sin β, 0, cos β),
r = cos γ p + sin γ r = (sin γ cos a, sin γ sin a, cos γ ).

P = (0,0,1)


R' = (cos a, sin a, 0)

Q' = (1,0,0)
Figure 3.2 Spherical trig.

Now ± is the angle between the two unit vectors q and r, so

cos ± = q · r = cos β cos γ + sin β sin γ cos a. QED

3.3 The spherical triangle inequality
In any triangle P Q R whose sides are shorter
Corollary (Triangle inequality)
arcs given by ±, β, γ ¤ π as above,

± ¤ β + γ,

with equality if and only if P Q R are collinear with P on the shorter arc Q R.

This follows at once from the main formula (2) and calm re¬‚ection on the
range of values for the angles ±, β, γ and a. Notice that ±, β, γ ∈ [0, π] essentially
by convention: in de¬ning distance I always take ∠P O Q to be the angle in the shorter
arc. If β or γ = 0 or π , it is easy to read off the conclusion, so that I can assume that
±, β, γ ∈ (0, π ). On the other hand, in Figure 3.2, it is clear I want to have a ∈ [0, 2π ).
Now compare (2) with the standard trig formula

cos(β + γ ) = cos β cos γ ’ sin β sin γ .

We know that sin β, sin γ ∈ (0, 1]; thus cos ± ≥ cos(β + γ ), with equality if and
only if cos a = ’1. Now cos ± is a strictly decreasing function in the range [0, π], so
that cos ± ≥ cos(β + γ ) gives ± ¤ β + γ . Equality holds only under the aforestated
condition cos a = ’1, that is, if the short arcs P Q and P R are opposite when viewed
from P. QED
It is trivial that d(P, Q) is symmetric, nonnegative, and positive unless P = Q,
so that Corollary 3.3 proves that S 2 with the spherical distance is a metric space (see
Appendix A).


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