h x2

I

h equalities and equation

h (0, 1, 0)

h x1 + x2 + x3 = 0,

e h

e h

x1 0, x2 0, x3 0.

e h

(1, 0, 0)e h

We see that it is an equilateral triangle.

e

…

e

x1

Figure 1.5: The regular 2-simplex

1.3.2 The previous exercise can be generalised to the case of n dimensions in

the following way. By de¬nition, the regular n-simplex is the set of solutions of

the system of simultaneous inequalities and equation

x1 + · · · + xn + xn+1 = 1

x1 0

.

.

.

xn+1 0.

14

It is the polytope in the n-dimensional a¬ne subspace A with the equation

x1 +· · ·+xn+1 = 1 bounded by the coordinate hyperplanes xi = 0, i = 1, . . . , n+1

(Figure 1.5). Prove that these hyperplanes cut A into 2n+1 ’ 1 chambers.

Hint: For a point x = (x1 , . . . , xn+1 ) in A which does not belong to any of

the hyperplanes xi = 0, look at all possible combinations of the signs + and ’

of the coordinates xi of x i = 1, . . . , n + 1.

Isometries of ARn

1.4

Now let us look at the structure of ARn as a metric space with the distance

r(a, b) = |ab|. An isometry of ARn is a map s from ARn onto ARn which

preserves the distance,

r(sa, sb) = r(a, b) for all a, b ∈ ARn .

We denote the group of all isometries of ARn by Isom ARn .

1.4.1 Fixed points of groups of isometries

The following simple result will be used later in the case of ¬nite groups of

isometries.

Theorem 1.4.1 Let W < Isom ARn be a group of isometries of ARn .

Assume that, for some point e ∈ ARn , the orbit

W · e = { we | w ∈ W }

is ¬nite. Then W ¬xes a point in ARn .

c

g

g

In the triangle abc the seg- g

g

ment cd is shorter than at g

least one of the sides ac or bc. g

g

g

a d b

Figure 1.6: For the proof of Theorem 1.4.1

Proof 3 . We shall use a very elementary property of triangles stated in

Figure 1.6; its proof is left to the reader.

3

This proof is a modi¬cation of a ¬xed point theorem for a group acting on a space

with a hyperbolic metric. J. Tits in one of his talks has attributed the proof to J. P. Serre.

A. & A. Borovik • Mirrors and Re¬‚ections • Version 01 • 25.02.00 15

Denote E = W · e. For any point x ∈ ARn set

m(x) = max r(x, f ).

f ∈E

Take the point a where m(x) reaches its minimum4 . I claim that the point

a is unique.

Proof of the claim. Indeed, if b = a is another minimal point, take

an inner point d of the segment [a, b] and after that a point c such that

r(d, c) = m(d). We see from Figure 1.6 that, for one of the points a and b,

say a,

m(d) = r(d, c) < r(a, c) m(a),

which contradicts to the minimal choice of a.

So we can return to the proof of the theorem. Since the group W

permutes the points in E and preserves the distances in ARn , it preserves

the function m(x), i.e. m(wx) = m(x) for all w ∈ W and x ∈ ARn , and

thus W should ¬x a (unique) point where the function m(x) attains its

minimum.

Structure of Isom ARn

1.4.2

Translations. For every vector ± ∈ Rn one can de¬ne the map

t± : ARn ’’ ARn ,

a ’ a + ±.

The map t± is an isometry of ARn ; it is called the translation through the

vector ±. Translations of ARn form a commutative group which we shall

denote by the same symbol Rn as the corresponding vector space.

Orthogonal transformations. When we ¬x an orthonormal coordinate

system in ARn with the origin o, a point a ∈ ARn can be identi¬ed with

its position vector ± = oa. This allows us to identify ARn and Rn . Every

orthogonal linear transformation w of the Euclidean vector space Rn , can

4

The existence of the minimum is intuitively clear; an accurate proof consists of the

following two observations. Firstly, the function m(x), being the supremum of ¬nite

number of continuous functions r(x, f ), is itself continuous. Secondly, we can search for

the minimum not all over the space ARn , but only over the set

{ x | r(x, f ) m(a) for all f ∈ E },

for some a ∈ ARn . This set is closed and bounded, hence compact. But a continuous

function on a compact set reaches its extreme values.

16

be treated as a transformation of the a¬ne space ARn . Moreover, this

transformation is an isometry because, by the de¬nition of an orthogonal

transformation w, (w±, w±) = (±, ±), hence |w±| = |±| for all ± ∈ Rn .

Therefore we have, for ± = oa and β = ob,

r(wa, wb) = |wβ ’ w±| = |w(β ’ ±)| = |β ’ ±| = r(a, b).

The group of all orthogonal linear transformations of Rn is called the or-

thogonal group and denoted On .

Theorem 1.4.2 The group of all isometries of ARn which ¬x the point o

coincides with the orthogonal group On .

Proof. Let s be an isometry of ARn which ¬xes the origin o. We have

to prove that, when we treat w as a map from Rn to Rn , the following

conditions are satis¬ed: for all ±, β ∈ Rn ,

• s(k±) = k · s± for any constant k ∈ R;

• s(± + β) = s± + sβ;

• (s±, sβ) = (±, β).

If a and b are two points in ARn then, by Exercise 1.4.3, the segment

[a, b] can be characterised as the set of all points x such that

r(a, b) = r(a, x) + r(x, b).

So the terminal point a of the vector c± for k > 1 is the only point

satisfying the conditions

r(o, a ) = k · r(0, a) and r(o, a) + r(a, a ) = r(o, a ).

If now sa = b then, since the isometry s preserves the distances and ¬xes

the origin o, the point b = sa is the only point in ARn satisfying

r(o, b ) = k · r(0, b) and r(o, b) + r(b, b ) = r(o, b ).

Hence s · k± = ob = kβ = k · s± for k > 0. The cases k 0 and 0 < k 1

require only minor adjustments in the above proof and are left to the reader

as an execise. Thus s preserves multiplication by scalars.

The additivity of s, i.e. the property s(± + β) = s± + sβ, follows, in

an analogous way, from the observation that the vector δ = ± + β can

be constructed in two steps: starting with the terminal points a and b of

A. & A. Borovik • Mirrors and Re¬‚ections • Version 01 • 25.02.00 17

the vectors ± and β, we ¬rst ¬nd the midpoint of the segment [a, b] as the

unique point c such that

1

r(a, c) = r(c, b) = r(a, b),

2

and then set δ = 2oc. A detailed justi¬cation of this construction is left to

the reader as an exercise.

Since s preserves distances, it preserves lengths of the vectors. But

from |s±| = |±| it follows that

(s±, s±) = (±, ±)

for all ± ∈ Rn . Now we apply the additivity of s and observe that

((± + β), (± + β)) = (s(± + β), s(± + β))

= ((s± + sβ), (s± + sβ))

= (s±, s±) + 2(s±, sβ) + (sβ, sβ)

= (±, ±) + 2(s±, sβ) + (β, β).

On the other hand,

((± + β), (± + β)) = (±, ±) + 2(±, β) + (β, β).

Comparing these two equations, we see that

2(s±, sβ) = 2(±, β)

and

(s±, sβ) = (±, β).

Theorem 1.4.3 Every isometry of a real a¬ne Euclidean space ARn is a

composition of a translation and an orthogonal transformation. The group

Isom ARn of all isometries of ARn is a semidirect product of the group Rn

of all translations and the orthogonal group On ,

Isom ARn = Rn On .

18

Proof is an almost immediate corollary of the previous result. Indeed,

to comply with the de¬nition of a semidirect product, we need to check

that

Isom ARn = Rn · On , Rn Isom ARn , Rn © On = 1.

and

If w ∈ Isom ARn is an arbitrary isometry, take the translation t = t±

through the position vector ± = o, wo of the point wo. Then to = wo and

o = t’1 wo. Thus the map s = t’1 w is an isometry of ARn which ¬xes

the origin o and, by Theorem 1.4.2, belongs to On . Hence w = ts and

Isom ARn = Rn On . Obviously Rn © On = 1 and we need to check only

that Rn Isom ARn . But this follows from the observation that, for any

orthogonal transformation s,

st± s’1 = ts± ,

(Exercise 1.4.5) and, consequently we have, for any isometry w = ts with

t ∈ Rn and s ∈ On ,

wt± w’1 = ts · t± · s’1 t’1 = t · ts± · t’1 = ts± ∈ Rn .

Elations. A map f : ARb ’’ ARn is called an elation if there is a

constant k such that, for all a, b ∈ ARn ,

r(f (a), f (b)) = kr(a, b).

An isometry is a partial case k = 1 of elation. The constant k is called the

coe¬cient of the elation f .

Corollary 1.4.4 An elation of ARn with the coe¬cient k is the composi-

tion of a translation, an orthogonal transformation and a map of the form

Rn ’’ Rn

± ’ k±.

Proof is an immediate consequence of Theorem 1.4.3.

Exercises

1.4.1 Prove the property of triangles in AR2 stated in Figure 1.6.

A. & A. Borovik • Mirrors and Re¬‚ections • Version 01 • 25.02.00 19

1.4.2 Barycentre. There is a more traditional approach to Theorem 1.4.1.

If F = { f1 , . . . , fk } is a ¬nite set of points in ARn , its barycentre b is a point