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Step 5. Proof of the theorem
tion p(О») = det(A в€’ О»1) = 0, so that at least one real or complex eigenvalue О» exists.
Step 1 or Steps 2вЂ“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 вЉҐ is OK; the induction starts with dim W = 0
or 1. QED

Complex numbers make their п¬Ѓrst 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.
14 EUCLIDEAN GEOMETRY

P'2
P2

P' = P'
0

Q
P = P0 P1 P' Q'
P'' 1
2

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
Definition
such that d(Q 0 , Q i ) = 1 and the lines Q 0 Q i are pairwise orthogonal for 1 в‰¤ i в‰¤ n.

The point of the deп¬Ѓnition is that if Q 0 , . . . , Q n is a Euclidean frame
Remark
then it is possible to choose coordinates so that Q 0 becomes the origin 0 в€€ Rn and
в€’в€’
в€’в†’
the n vectors ei = Q 0 Q i form an orthonormal basis of Rn .

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

The correspondence is given by T в†’ T (P0 ), T (P1 ), . . . , T (Pn ). It is clear
Proof
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
1.13
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):

Corollary

Suppose that [P, Q] and [P , Q ] are two line segments in E2 of the same length
(1)
d(P, Q) = d(P , Q ) > 0. Then there exist exactly two motions T : E2 в†’ 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:
(2)

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 в†’ E2 such that T (P) = P , T (Q) = Q , T (R) = R .
1.14 CLASSIFICATION OF MOTIONS OF E2 15

О“ О“
Rot(O, Оё) v
L
O Оё О“
О“ Glide(L, v)

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

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

Every motion of E2 is a translation, rotation, reflection or glide
1.14
Let us list the motions of E2 we know, expressed in coordinates (see Figure 1.14a).
The translation Trans(b) : x в†’ x + b for b в€€ R2 .
1.
The rotation through angle Оё about a point O в€€ E2 ; if O is the origin of the coordinate
2.
system, this is written

cos Оё в€’ sin Оё
x1 x1
Rot(O, Оё ) : в†’ .
sin Оё cos Оё
x2 x2

The reп¬‚ection in a line L; if L is the x1 -axis (x2 = 0) then
3.

x1 x1
в†’ .
Reп¬‚(L) :
в€’x2
x2

4. The glide (or glide reп¬‚ection) in a line L through a vector v along L. Reп¬‚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
x1
Glide(L , v) : в†’ .
в€’x2
x2

Here v is parallel to L, and the reп¬‚ection and translation commute.
I use self-documenting notation such as Rot(O, Оё ) 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
small.

ThatвЂ™s all, folks!
Theorem
16 EUCLIDEAN GEOMETRY

Оё
Q
P'
P

Q'

Оё/2

O

Figure 1.14c Construction of rotation.

There are several ways of proving this. (Why not devise your own? See
Proof
Exercises 1.8 and 1.9 for an argument in terms of x в†’ Ax + b, and Exercise 2.11 for
an argument in terms of composing reп¬‚ections.)
The proof given here is based on the following geometric idea taken from Nikulin
and Shafarevich : 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 reп¬‚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 1вЂ“2, both of which are in
my list.

в€’в†’
в€’
в€’в†’
I п¬Ѓrst п¬Ѓnd a reп¬‚ection or glide. Write u = P Q and u = P Q . First I need
Step 1
to п¬Ѓnd the line of reп¬‚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
в€’в†’
2
arranges that the reп¬‚ection or glide reп¬‚ection in any line parallel to L takes P Q into
в€’в†’
в€’
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
в€’в†’
в€’
from P and P to L and v = A A (see Figure 1.14b). Since reп¬‚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 в†’ E2 such that P в†’ P and
Step 2
Q в†’ Q . I suppose п¬Ѓrst that P = P , and that the lines P Q and P Q intersect at a
single point in an angle Оё.
Then the (signed) angle of rotation must be Оё; the centre must be the point O of the
perpendicular bisector of the line P P determined by P O P = Оё (see Figure 1.14c).
Then by construction Rot(O, Оё ) takes P в†’ 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 в†’ Q .
1.15 CLASSIFICATION OF MOTIONS OF E3 17

L
L

Оё
v

Оё О
Rot-Refl(L, Оё, О )
Twist(L, Оё, v)

Twist (L, Оё, v) and Rot-Refl (L, Оё,
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
1.15

A motion T : E3 в†’ E3 is one of the following:
Theorem

1. Translation by a vector v.
Rotation about a directed line L as axis through an angle Оё.
2.
3. Twist: the same followed by a translation along L (Figure 1.15a).
4. Reп¬‚ection in a plane.
5. Glide: a reп¬‚ection in a plane followed by a translation by a vector in the plane.
Rotary reп¬‚ection: the rotation through Оё about a directed axis L followed by a reп¬‚ec-
6.
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 deп¬Ѓned 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
ПЂ + Оё, followed by a point reп¬‚ection in L в€© . Clearly (1)вЂ“(3) are direct motions
and (4)вЂ“(6) opposite. Notice that any motion leaves invariant a grid of parallel planes
and their orthogonal lines (Figure 1.15b).
18 EUCLIDEAN GEOMETRY

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

T2 : x в†’ Ax T1 : y в†’ y + b.
and

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

В±1 cos Оё в€’ sin Оё
A= , B= .
where
sin Оё cos Оё
B

In these coordinates, T has the form
пЈ« пЈ¶
пЈ«пЈ¶ пЈ«пЈ¶
В±x1
x1 b1
пЈ¬ пЈ· пЈ­пЈё
T : пЈ­x2 пЈё в†’ пЈ­ cos Оё в€’ sin Оё пЈё + b2 .
x2 (11)
sin Оё cos Оё x3
x3 b3

For the proof, I have to verify that this map is a motion of one of types (1)вЂ“(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 Г— t : E1 Г— E2 в†’ E1 Г— E2 , where T : E1 в†’ E1
and T : E2 в†’ E2 are given in coordinates by

cos Оё в€’ sin Оё
x2 x2 b
+ 2.
T : x1 в†’ В±x1 + b1 в†’
and T:
sin Оё cos Оё
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 Оё = 0 or rotation if Оё = 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 Оё about the point determined by

cos Оё в€’ sin Оё
x2 x2 b
+ 2,
=
sin Оё cos Оё
x3 x3 b3

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

в€’1 cos Оё в€’ 1 в€’ sin Оё
x2 b2
= .
sin Оё cos Оё в€’ 1
2 в€’ 2 cos Оё
x3 b3

The theorem follows easily on sorting out the cases. QED
1.16 SAMPLE THEOREMS OF EUCLIDEAN GEOMETRY 19

A

Оё Оё'
B O C
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 EuclidвЂ™s Elements, or the
much more extensive nineteenth century subject, but I do not want to omit to mention
it altogether. Coxeter  is very entertaining on this subject.

1.16.1 Pons asinorum, вЂ˜Bridge of assesвЂ™. Equivalent conditions on a triangle
Proposition
Pons ABC:
asinorum
d(A, B) = d(A, C);
1.
Оё = в€ ABC = Оё = в€ AC B;
2.
there exists a motion T : ABC в†’
3. AC B.

(1) в‡ђв‡’ (2) is an easy consequence of trigonometry, because in Fig-
Proof
ure 1.16a,

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

From our point of view, (3) =в‡’ (1) or (2) is obvious, and (1) or (2) =в‡’ (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 ПЂ.
1.16.2 Theorem
в€’в†’
The angle
Let ABC be a given triangle. Consider the motion T = Trans( AC) and
Proof
sum of
set A B C = T ( ABC) as in Figure 1.16b. Then because T is a motion, I get
triangles
A B C в‰Ў ABC, where в‰Ў is congruence (see Exercise 1.16). Also, since T is a
Euclidean translation, d(B, B ) = d(A, C), therefore also ABC в‰Ў B A B . Hence

О± + ОІ + Оі = в€ B CC + в€ BC B + в€ AC B = ПЂ

since the angles combine to form a straight line. QED
20 EUCLIDEAN GEOMETRY

B B'
Оі
ОІ

ОІ
Оі О±
О±
A C'
C = A'
Sum of angles in a triangle is equal to ПЂ .
Figure 1.16b

L1
P3вЂІ
P3

L2
P2вЂІ
P2

L3
P1 P1
вЂІ

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

The statement the sum of angles in a triangle equals ПЂ is equivalent to the
Remark
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
1.16.3
(compare 9.1.2). Parallel lines fall on lines in the same ratio, and conversely; they
Parallel
are also responsible for the existence of similar triangles. The following proposition
lines and
makes these statements precise.
similar
triangles
Proposition

If L 1 , L 2 , L 3 are three parallel lines in E2 , and they meet a line M in P1 , P2 , P3 ,
(1)
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
equivalent:
(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: в€ ABC = в€ AB C and в€ AC B = в€ AC B .
1.16 SAMPLE THEOREMS OF EUCLIDEAN GEOMETRY 21

A

B
C
B'

C'

Figure 1.16d Similar triangles.

A

BвЂІ CвЂІ

G

L M
C AвЂІ B
Figure 1.16e The centroid.

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

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)
1.16.4
(Figure 1.16e).
Four centres
of a triangle
(See 4.7 for a slightly different proof.) Let A , B , C be the midpoints of
Proof
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 ,
LM AC AC and LC GB

(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).
22 EUCLIDEAN GEOMETRY

A

BвЂІ CвЂІ

O

AвЂІ
C B

Figure 1.16f The circumcentre.

B

O G
H

BвЂІ
A C

Figure 1.16g The orthocentre.

This is almost obvious, since the perpendicular bisector of AB is deter-
Proof
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 .

в€’в†’
в€’ в€’в€’в†’
In vector notation, H is the point given by O H = 3 OG, where O is
Proof
the circumcentre and G the centroid. Indeed, in Figure 1.16g, B B is the median
в€’в†’
в€’
в€’в†’ в€’в†’
в€’ в€’в€’в†’
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).
1.16 SAMPLE THEOREMS OF EUCLIDEAN GEOMETRY 23

A

Q
D
BвЂІ CвЂІ
R
H

F E

AвЂІ
C P B
Figure 1.16h The Feuerbach 9-point circle.

The angle bisectors of the three angles в€ C AB, в€ ABC and
Proposition (Incentre)
в€ 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-
Proof
ercise 1.18). QED

Theorem (The Feuerbach circle)1 The following 9 points lie on a circle (see Fig-
1.16.5
ure 1.16h):
The
Feuerbach
3 feet P, Q, R of the perpendiculars dropped from a vertex to the opposite side;
9-point
3 midpoints A , B , C of the sides;
circle
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
Proof
rather easy because there are so many parallel and perpendicular lines in Figure 1.16h.
By similar triangles, the following lines are parallel:

AB DE AB and AE BD C R.

But AB вЉҐ 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, в€ A P D = 90в—¦ , which is a sufп¬Ѓcient 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 FeuerbachвЂ™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.
24 EUCLIDEAN GEOMETRY

Exercises
Redo the proof of Theorem 1.1 in detail in the cases n = 1 and n = 2.
1.1
The angle between nonzero vectors u, v в€€ Rn can be deп¬Ѓned by
1.2

cos Оё = u i v i /|u||v|.
Prove that the right-hand side is in the interval [в€’1, +1], so that its arccos is deп¬Ѓned.
The line L = xy in Rn is the set {(1 в€’ О»)x + О»y|О» в€€ R}. If z в€€ L, write y in terms of
1.3
x and z. Complete the proof of Proposition 1.8.
1.4 Show that the assumption that T is bijective in the deп¬Ѓnition of motion of Euclidean
space is superп¬‚uous; that is, a map T : En в†’ En that preserves distances is bijective,
therefore a motion. [Hint: prove that T is afп¬Ѓne 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 в€€ Rn are eigenvectors of A belonging to distinct eigenvalues О» = Вµ, prove
that e В· f = 0.
(b) If z в€€ Cn is a complex eigenvector with complex eigenvalue О» в€€ R, prove that
/
z В· z = 0. (Here x В· 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 в†’ E2 be the motion obtained by reп¬‚ecting in the x-axis then rotating
1.7
through Оё around the origin. Show that T is the reп¬‚ection in a certain line (to be
speciп¬Ѓed).
(b) Calculate the eigenvalues and eigenvectors of the reп¬‚ection matrix A =
cos Оё sin Оё
в€’ cos Оё .
sin Оё
(c) Relate (a) and (b).
(a) Let Оё be a nonzero angle and b a translation vector in the plane. Give a geometric
1.8
construction for a point P в€€ E2 such that
Rot(O, Оё )(P) = Trans(в€’b)(P).
в€’в†’
[Hint: draw a picture, to п¬Ѓnd points P, Q with b = Q P such that O is on the
perpendicular bisector of P Q and в€ P O Q = Оё.]
(b) By solving linear equations, п¬Ѓnd x, y such that
cos Оё sin Оё
x1 b x1
+1 = , A= .
A where
sin Оё в€’ cos Оё
x2 b2 x2

(c) Express the motion T : E2 в†’ E2 deп¬Ѓned in coordinates by T (x) = Ax + b in the
form T = Rot(P, Оё).
(d) Relate (a) and (b).
Let A = cos Оё в€’sin Оё Оё be the reп¬‚ection matrix of 1.11.1, and consider the motion
Оё
1.9 sin cos
T (x) = Ax + b; give a proof in coordinates that it is a glide reп¬‚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).
EXERCISES 25

Complete the proof of Step 2 in any of these cases by constructing a suitable translation
or rotation taking P в†’ P and Q в†’ Q . в€љ
Find the two motions E2 в†’ E2 taking (0, 0) в†’ (1, 2) and (0, 2 ) в†’ (2, 3). Write
1.11
each as x в†’ Ax + b. [Hint: the easy way: for the direct motion, translate then rotate;
for the opposite motion, reп¬‚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
в€’в†’ в€’ в€’ в†’
P0 = P, P0 P1 = d(P,Q) and P2 a third point; if I do the same for P , Q , there are
PQ

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 в€€ E2 be distinct noncollinear points. Show that there is a unique
1.13
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 в€€ E2 be two pairs of distinct noncollinear points
1.14
such that d(Pi , P j ) = d(Pi , P j ) for all i, j. Prove that there exists a unique mo-
tion T : E2 в†’ E2 taking Pi в†’ Pi for i = 1, 2, 3. [Hint: you know enough motions
to send P0 в†’ P0 . Then п¬Ѓxing P0 = P0 , to send P1 в†’ 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 в€€ En is uniquely
1.15
determined by its distances from all of the Pi . [Hint: take P0 as origin; the n vectors
в€’в†’
в€’ в€’в†’
в€’
ei = P0 Pi are linearly independent. The vector f = P0 Q is determined by f В· ei , so it
is enough to determine f В· ei from distances in P0 Pi Q.]
Let ABC and D E F be two triangles in E2 . Prove that the following 4 conditions
1.16
are equivalent:
(a) 3 sides are equal AB = D E, BC = E F, C A = F D;
(b) equal sideвЂ“angleвЂ“side: AB = D E, C A = F D and в€ C AB = в€ F D E;
(c) angleвЂ“sideвЂ“angle: в€ ABC = в€ D E F, BC = E F and в€ BC A = в€ E F D;
(d) there exists a motion T taking A в†’ D, B в†’ E, C в†’ F.
The triangles ABC and D E F are congruent if these conditions hold; in symbols,
ABC в‰Ў 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 в€ C AB, в€ ABC and в€ 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
ABC).
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 вЂ˜a function of a functionвЂ™, a procedure familiar from п¬Ѓrst year calculus:
if y = f (x) and z = g(y), then you can write z = g( f (x)) = (g в—¦ f )(x). The chain
rule, for example, calculates the derivative dx in terms of dy and dy .
dz dz
dx

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 в†’ Y
and g : Y в†’ Z two maps between them.

The composite of f and g is the map
Definition

gв—¦ f: X в†’ Z (g в—¦ f )(x) = g( f (x)).
deп¬Ѓned by (1)

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

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

26
2.3 COMPOSITION OF TWO REFLECTIONS OF E2 27

Composing two translations of En means adding the corresponding vectors:
1.

Trans(v) в—¦ Trans(u) = Trans(u + v).

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

Rot(Оё) в—¦ Rot(П•) = Rot(Оё + П•).

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

cos Оё в€’ sin Оё cos П• в€’ sin П• cos(Оё + П•) в€’ sin(Оё + П•)
= .
sin Оё cos Оё sin П• cos П• sin(Оё + П•) cos(Оё + П•)

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-
4.
ber z = r exp(iОё) corresponds to rotation by Оё 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 в†’ Ax + b
2.2
An afп¬Ѓne linear map T : Rn в†’ Rn is given by T (x) = Ax + b where A is an n Г— n
matrix and b is a vector (see 1.8). If T1 (x) = A1 x + b1 and T2 (x) = A2 x + b2 then

(T2 в—¦ 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 в†’ Ax + b, composition is given by the rule
T A2 ,b2 в—¦ T A1 ,b1 = T A2 A1 ,A2 b1 +b2 . Note that the п¬Ѓrst component A2 A1 is just the product,
whereas in the second component, the matrix A2 of T A2 ,b2 п¬Ѓrst acts on the translation
compare also Exercise 6.1.

Composition of two reflections of E2
2.3
Consider the reп¬‚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 Оё is the angle at P from L 1 to L 2 then
1.
Reп¬‚(L 2 ) в—¦ Reп¬‚(L 1 ) = Rot(P, 2Оё).
2. If L 1 and L 2 are parallel and v is the perpendicular vector from L 1 to L 2 then
Reп¬‚(L 2 ) в—¦ Reп¬‚(L 1 ) = Trans(2v).
28 COMPOSING MAPS

О“
Trans(2v)
Rot(P,2Оё)
О“О“ О“ P
L2
Оё
v
О“
L2
L1

О“
L1
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 reп¬‚ections Reп¬‚(L 3 ) в—¦ Reп¬‚(L 2 ) в—¦ Reп¬‚(L 1 ). As a preliminary step, a
point of set theory: suppose that X, Y, Z , T are sets, and that

f : X в†’ Y, g : Y в†’ Z, h: Z в†’ 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 в†’ f (x) в†’ g( f (x)) в†’ h(g( f (x))). (2)

The composite h в—¦ g в—¦ f is the map X в†’ T deп¬Ѓned by (2). Thus the expression
h в—¦ g в—¦ f does not admit any possible ambiguity.
In the tradition of abstract algebra, the associative law is the headache of how to
bracket h в—¦ g в—¦ 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 в—¦ (g в—¦ f ), presumably to economise on deп¬Ѓnitions. In this case, one п¬Ѓrst constructs
a map g в—¦ f : X в†’ Z , then links it with the third map to get the repeated composite
h в—¦ (g в—¦ f ) : X в†’ Z в†’ T . However, as my tautology says, whatever brackets you
put in, h в—¦ g в—¦ 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
boring.)
Another abstract algebraic notion, the вЂ˜commutative lawвЂ™, is discussed in Exer-
cise 2.4.

2.5 Decomposing motions
This section introduces the п¬Ѓrst way of decomposing a motion of En as a composite
of вЂ˜elementaryвЂ™ 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 вЂ˜elementaryвЂ™.
2.6 REFLECTIONS GENERATE ALL MOTIONS 29

An afп¬Ѓne linear subspace вЉ‚ En of Euclidean space is the image U вЉ‚ 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 в€’ 1)-dimensional afп¬Ѓne linear
subspace вЉ‚ En .

The reп¬‚ection in a hyperplane is the motion that п¬Ѓxes pointwise
Definition
and reverses orthogonal vectors to . In coordinate form, if is given by x1 = 0,
and x2 , . . . , xn are coordinates on в€ј Enв€’1 , then
=
пЈ« пЈ¶
пЈ«пЈ¶ пЈ«пЈ¶
в€’1
x1 пЈ· x1
пЈ¬ 1
пЈ¬.пЈ· пЈ¬ пЈ·пЈ¬ . пЈ·
Reп¬‚( ) : пЈ­ . пЈё в†’ пЈ¬ пЈ·пЈ­ . пЈё.
..
. пЈё.
пЈ­ .
xn xn
1

In other words, the deп¬Ѓning property of ПЃ = Reп¬‚( ) is that it п¬Ѓxes every point
of , and takes P в€€ into the point Q = ПЃ(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 Reп¬‚( ) 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 в€’ 2)-dimensional afп¬Ѓne linear subspace of En . The
Definition
rotation around the axis through (signed) angle Оё is the motion that п¬Ѓxes pointwise
and rotates by Оё 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
пЈ« пЈ¶
cos Оё в€’ sin Оё
пЈ«пЈ¶ пЈ«пЈ¶
пЈ¬ sin Оё пЈ· x1
cos Оё
x1 пЈ¬ пЈ·
пЈ¬.пЈ· пЈ¬ пЈ·пЈ¬ . пЈ·
1
Rot( , Оё) : пЈ­ . пЈё в†’ пЈ¬ пЈ·пЈ­ . пЈё.
. пЈ·.
пЈ¬ ..
пЈ­ пЈё xn
.
xn
1

Finally, there are also translations Trans(v) : x в†’ x + b for b в€€ Rn .
Every motion T of En is a composite of a translation, k reп¬‚ections and
Theorem
l rotations, where k + 2l в‰¤ n.
Convince yourself that this is really a restatement of the fact that every
Proof
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.
30 COMPOSING MAPS

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

T = ПЃ1 в—¦ ПЃ2 в—¦ В· В· В· в—¦ ПЃk , with k в‰¤ n + 1.

The rough idea is simple: if every point P в€€ En is п¬Ѓxed by T , then T =
Proof
id, so it is a composite of no reп¬‚ections at all. Otherwise, choose any P so that
T (P) = Q = P; then, by what I just said, there is a reп¬‚ection ПЃ1 taking Q back to
P, namely the reп¬‚ection in the perpendicular bisector of P Q. Then T (P) = Q and
ПЃ1 (Q) = P, so that T1 = ПЃ1 в—¦ T is a new motion п¬Ѓxing P. Now it turns out (see below)
that T1 still п¬Ѓxes any point already п¬Ѓxed by T , so that T1 п¬Ѓxes strictly more than T .
I can repeat this argument, obtaining T2 = ПЃ2 в—¦ T1 п¬Ѓxing even more points, and so on
inductively until Tk = ПЃk в—¦ Tkв€’1 п¬Ѓxes every point of En . Putting this together gives
ПЃk в—¦ В· В· В· в—¦ ПЃ1 в—¦ T = id.
Now precomposing the equation T1 = ПЃ1 в—¦ t with ПЃ1 gives

ПЃ1 в—¦ T1 = (ПЃ1 в—¦ ПЃ1 ) в—¦ T,

and since ПЃ1 в—¦ ПЃ1 = id, we get T = ПЃ1 в—¦ T1 . Arguing in the same way gives T =
ПЃ1 в—¦ T1 = ПЃ1 в—¦ ПЃ2 в—¦ T2 = В· В· В· , which concludes the proof.
To go through the argument in more detail, I assert п¬Ѓrst that the set Fix(T ) of п¬Ѓxed
points of any motion T is (either empty or) an afп¬Ѓne linear subspace of En . This
follows from Proposition 4.3 (2), and the fact that if two distinct points P, Q are п¬Ѓxed
by T , then so is any point R on the line P Q: if R в€€ [P, Q] then

d(P, R) + d(R, Q) = d(P, Q) T (P) = P, T (Q) = Q
and
=в‡’ d(P, T (R)) + d(T (R), Q) = d(P, Q),

so T (R) в€€ [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 в€’ l (for some l = 0, . . . , n) then T
Claim
is a composite of at most l reп¬‚ections.

As argued above, if T = id then I choose a point P в€€ Fix(T ), set Q = T (P) and
/
the perpendicular bisector of P Q, and let ПЃ be the reп¬‚ection in . The point of the
construction is that ПЃ(Q) = P, so that T1 = ПЃ в—¦ T п¬Ѓxes P.
is characterised as the set of points of En
Now the perpendicular bisector
equidistant from P and Q. Moreover, every point R в€€ Fix(T ) is equidistant from P
and Q, because d(P, R) = d(T (P), T (R)) = d(Q, R). Therefore Fix(T ) вЉ‚ , and
ПЃ = Reп¬‚( ) п¬Ѓxes every point of Fix(T ). It follows that Fix(T1 ) вЉѓ Fix(T ) в€Є {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 = Reп¬‚( ). Otherwise, as just proved, I can п¬Ѓnd
ПЃ so that T1 = ПЃ в—¦ T п¬Ѓxes a strictly bigger set than T , and therefore Fix(T1 ) has
2.8 PREVIEW OF TRANSFORMATION GROUPS 31

M
Q
L

v
A
B

Оё
Оё

P

Figure 2.7 Composite of a rotation and a reflection.

dimension (n в€’ l ) with l < l. By induction, I can assume the result for T1 , that is,
T1 = ПЃ1 в—¦ ПЃ2 в—¦ В· В· В· в—¦ ПЃk with k в‰¤ l so that T = ПЃ в—¦ T1 is the composite of at most
l + 1 в‰¤ l reп¬‚ections, as required. This proves the claim. If Fix(T ) = в€… then Fix(T1 )
is at least one point, so that by the claim, T1 is a composite of at most n reп¬‚ections,
and T the composite of at most n + 1 reп¬‚ections, which proves the theorem. QED

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

Every motion of E2 is the composite of at most 3 reп¬‚ections. As we saw
Proof
in 2.3, the composite of 2 reп¬‚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 reп¬‚ections ПЃ3 в—¦ ПЃ2 в—¦ ПЃ1 is a glide or reп¬‚ection. Suppose for simplicity that the axes of
ПЃ1 and ПЃ2 meet at a point P, and make an angle Оё there, so that ПЃ2 в—¦ ПЃ1 = Rot(P, 2Оё)
(see Figure 2.3). Suppose also that P в€€ L 3 (the case P в€€ L 3 is easier). The problem
/
then is to learn how to compose Rot(P, 2Оё) with ПЃ3 = Reп¬‚(L).
In Figure 2.7, L is the axis of the third reп¬‚ection ПЃ3 , and Q = ПЃ3 (P). Draw the
line M passing through the midpoint of P Q, such that the angle from M to L is Оё; 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 Reп¬‚(L) в—¦ Rot(P, 2Оё) = 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 в—¦ f is a basic, simple and
familiar idea having many useful applications. From an algebraic point of view, the
composite of Euclidean motions deп¬Ѓnes a product

Eucl(n) Г— Eucl(n) в†’ Eucl(n)
32 COMPOSING MAPS

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.

Exercises
A standard result of linear algebra identiп¬Ѓes an m Г— n matrix A = (ai j ) with a linear
2.1
map О± : Rn в†’ Rm (taking the standard basis of column vectors to the columns of
A). If B = (b jk ) is an l Г— m matrix giving a linear map ОІ : Rm в†’ Rl , verify that the
product matrix B A corresponds to the composite ОІ в—¦ О±.
The (nonzero) complex numbers can be viewed as a set of similarities of E2 :
2.2
xy
regard z = x + iy as the map Tz : R2 в†’ R2 given by the matrix в€’y x . Write
z = r exp(iОё) where r = |z| and Оё = arg z, and interpret the map Tz geometrically.
Prove that Tz is a similarity in the sense that there exists О» for which d(T (x), T (y)) =
О»d(x, y). Show how to obtain multiplication of complex numbers as composition of
similarities.
In the notation of 2.4, prove that h в—¦ g в—¦ f = (h в—¦ g) в—¦ f . Prove that for 4 consecutive
2.3
maps f, g, h, k, we have

(k в—¦ h) в—¦ (g в—¦ f ) = k в—¦ ((h в—¦ g) в—¦ 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, п¬Ѓnd the conditions for the domain and range of f, g so that the
commutative law
?
gв—¦ f = f в—¦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 п¬Ѓxed point. Show that it
does not hold for the set of all motions of Euclidean space En .
Verify by calculation that the usual deп¬Ѓnition of matrix multiplication AB = (cik =
2.5
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, afп¬Ѓne linear maps T A,b : Rn в†’ Rn compose according to the rule T A2 ,b2 в—¦
T A1 ,b1 = T A2 A1 ,A2 b1 +b2 ; verify that this formula deп¬Ѓnes an associative multiplication
rule.
Exercises in composing motions of E2 .
The half-turn about P is the rotation through 180в—¦ . Prove the following.
2.6
(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

Reп¬‚(L) and Halfturn(P) commute в‡ђв‡’ P в€€ L .
EXERCISES 33

Prove that every opposite motion of E2 is the composite of a half-turn and a reп¬‚ection.
2.7
2.8 Give a geometric treatment of the composition of a rotation with a glide, to get another
glide or reп¬‚ection. When is Glide(L , v) в—¦ Rot Оё a reп¬‚ection? [Hint: draw a diagram
similar to Figure 2.7.]
Show that any composite T1 в—¦ T2 with either T1 or T2 a reп¬‚ection or glide can be
2.9
understood by drawing a diagram like Figure 2.7. [Hint: to view g = Glide(L , v) and
its effect on a point P в€€ L, draw a rectangle with the line P T (P) as a diagonal and
/
v as a median. The best way to see g1 в—¦ 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 в—¦ t = g2 , which you can
в€’1
rewrite as T = g1 в—¦ g2 and treat similarly.]
(Harder) Use Claim 2.6 to study motions of E3 п¬Ѓxing a point O, and compare with
2.10
the conclusion of Theorem 1.11. [Hint: a composite of 2 reп¬‚ections in planes 1 , 2
through O is a rotation about a line through O. For 3 reп¬‚ections, you need to prove
that Reп¬‚( ) в—¦ Rot(L , Оё ) is a rotary reп¬‚ection, or in other words, to п¬Ѓnd 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 в‰¤ 4 reп¬‚ections of E3 , and show that they lead to the
6 cases listed in Theorem 1.15. [Hint: see Rees .]
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 вЂ˜walkвЂ™
your wardrobe anywhere in the Euclidean plane. (Ignore doors and stairs.)
Let P, Q в€€ E2 be two distinct points. Prove that every direct motion of E2 is a
composite of sufп¬Ѓciently 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 вЂ“ 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 deп¬Ѓned 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 О», О» directions at these
points, then there exists a motion T taking P to P and О» to О» ; in other words, the
geometry is homogeneous (the same at every point) and isotropic (the same in every
direction). (3) means that in sufп¬Ѓciently small open sets, a line is uniquely speciп¬Ѓed
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
triangle;
(3) the possibility of deп¬Ѓning 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 deп¬Ѓnition and statement can be readily visualised on the very concrete
model S 2 вЉ‚ R3 , which you can hold in your hand or kick around a playing п¬Ѓeld.
I discuss spherical lines (great circles), distances, angles and triangles, the classi-
п¬Ѓcation of motions in terms of rotations and reп¬‚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

34
3.1 BASIC DEFINITIONS OF SPHERICAL GEOMETRY 35

l1
P l3
T
l2
О»вЂІ
О»
P
PвЂІ
Q
(2)
(3)
Figure 3.0 Plane-like geometry.

doing, you can just replace x 2 + y 2 = 1 by в€’t 2 + x 2 = в€’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 (1728вЂ“1777) that non-Euclidean geometry вЂ˜should be related to the
в€љ
geometry on a sphere of radius i = в€’1вЂ™ (see Coxeter , 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 deп¬Ѓnitions in this
chapter are given in two dimensions. You will beneп¬Ѓt immensely by generalising the
deп¬Ѓnitions 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 signiп¬Ѓcance.

3.1 Basic definitions of spherical geometry
The sphere S 2 вЉ‚ R3 of radius r centred at the origin O is deп¬Ѓned by the equation
в€’в†’
x 2 + y 2 + z 2 = r 2 . I will often refer to points P в€€ 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 в€€ S are antipodal if their position vectors p, q satisfy
2 2

p = в€’q. Through any two distinct points P, Q в€€ 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 в€€ 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 в€ 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, ПЂ ]. For ease of notation, I usually п¬Ѓx the radius r = 1 from now on.

Remarks

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

be consistent, I should have deп¬Ѓned вЂ˜modelвЂ™ S 2 to be the sphere {x 2 + y 2 + z 2 = r 2 }
in R3 with its inherited spherical distance, and вЂ˜abstractвЂ™ S 2 to be a metric space
isometric to вЂ˜modelвЂ™ S 2 but without a п¬Ѓxed choice of identiп¬Ѓcation. 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
вЂ˜abstractвЂ™ 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 deп¬Ѓned as the subset {x 2 + y 2 + z 2 = 1} of R3 . On the northern
(2)
hemisphere {z в‰Ґ 0} I can rewrite this as z = 1 в€’ x 2 в€’ 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 в‰¤ 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 в€’ x 2 в€’ 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 deп¬Ѓned as (radius times) the angle of the P Q arc, О± =
(3)
в€ 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 О± = в€ P O Q is
given by

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

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

sin О±
dx
= arcsin(sin О±) = О±.
в€љ
1 в€’ x2
0
3.2 SPHERICAL TRIANGLES AND TRIG 37

Geometers like to distinguish the intrinsic geometric properties of S 2 from those
related to the embedding S 2 вЉ‚ 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 deп¬Ѓnition if you like, and then you do not have to worry
about how the angle between two curves is deп¬Ѓned. More precisely, the tangent plane
to S 2 at P is the 2-plane TP S 2 deп¬Ѓned 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,
write

О± = в€ Q O R = d(Q, R), ОІ = в€ P O Q = d(P, Q), Оі = в€ P O R = d(P, R).

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

cos О± = cos ОІ cos Оі + sin ОІ sin Оі cos a. (2)

Although the statement looks complicated, the proof is easy 3-dimensional
Proof
coordinate geometry. In Figure 3.2, let Q and R be the points on great circles at
в€’в†’в€’
distance ПЂ/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 п¬Ѓgure; by deп¬Ѓnition of ОІ, Оі , the
following relations hold between the position vectors:

q = cos ОІp + sin ОІq = (sin ОІ, 0, cos ОІ),
r = cos Оі p + sin Оі r = (sin Оі cos a, sin Оі sin a, cos Оі ).
38 NON-EUCLIDEAN GEOMETRIES

P = (0,0,1)

Q
R

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

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

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

cos О± = q В· r = cos ОІ cos Оі + sin ОІ sin Оі 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 О±, ОІ, Оі в‰¤ ПЂ as above,

О± в‰¤ ОІ + Оі,

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 reп¬‚ection on the
Proof
range of values for the angles О±, ОІ, Оі and a. Notice that О±, ОІ, Оі в€€ [0, ПЂ] essentially
by convention: in deп¬Ѓning distance I always take в€ P O Q to be the angle in the shorter
arc. If ОІ or Оі = 0 or ПЂ , it is easy to read off the conclusion, so that I can assume that
О±, ОІ, Оі в€€ (0, ПЂ ). On the other hand, in Figure 3.2, it is clear I want to have a в€€ [0, 2ПЂ ).
Now compare (2) with the standard trig formula

cos(ОІ + Оі ) = cos ОІ cos Оі в€’ sin ОІ sin Оі .

We know that sin ОІ, sin Оі в€€ (0, 1]; thus cos О± в‰Ґ cos(ОІ + Оі ), with equality if and
only if cos a = в€’1. Now cos О± is a strictly decreasing function in the range [0, ПЂ], so
that cos О± в‰Ґ cos(ОІ + Оі ) gives О± в‰¤ ОІ + Оі . Equality holds only under the aforestated
condition cos a = в€’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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