j

ri,1 ... ri,j’1 ri,j+1 . . . ri,n

det R i

. . . . .

. . . . .

. . . . .

rn,1 . . . rn,j’1 rn,j+1 . . . rn,n

n

Notice that in the case n = 3 we come to the formula

1 1 1

ρ— = ’ ρ2 —ρ3 , ρ— = ’ ρ3 —ρ1 , ρ— = ’ ρ1 —ρ2 ,

1 2 3

(ρ1 , ρ2 , ρ3 ) (ρ1 , ρ2 , ρ3 ) (ρ1 , ρ2 , ρ3 )

where ( , , ) denotes the scalar triple product and — the cross (or vector)

product of vectors.

1.5.4 Let “ be a cone in Rn spanned by a set Π of m linearly independent

vectors ρ1 , . . . , ρm , with m < n. Let U be the vector subspace spanned by Π.

Then “ is a simplicial cone in U ; we denote its dual in U as “ , and set “—

to be the dual cone for “ in V . Let also Π = { ρ1 , . . . , ρm } be the basis in U

dual to the basis Π. We shall use in the sequel the following properties of the

cone “— .

1. For any set A ∈ Rn , de¬ne

A⊥ = { χ ∈ Rn | (χ, ±) = 0 for all γ ∈ A }.

Check that A⊥ is a linear subspace of Rn . Prove that dim “⊥ = n ’ m.

Hint: “⊥ = U ⊥ .

2. “— is the intersection of the closed half spaces de¬ned by the inequalities

(χ, ρi ) 0, i = 1, . . . , m.

3. “— = “ + “⊥ ; this set is, by de¬nition,

“ + “⊥ = { κ + χ | κ ∈ “ , χ ∈ “⊥ }.

4. (“— )— = “.

5. Let Hi and Hi— be the hyperplanes in V given by the equations (χ, ρi ) = 0

and (χ, ρi ) = 0, correspondingly. Denote I = { 1, . . . , m } and set, for

J ⊆ I,

Hj , “— = “ — © — —

“J = “ © Hj and “J = “ © Hj .

J

j∈J j∈J j∈J

Prove that “— = “J + “⊥ .

J

A. & A. Borovik • Mirrors and Re¬‚ections • Version 01 • 25.02.00 29

6. The cones “J and “— are called faces of the cones “ and “— , corre-

J

spondingly. There is a one-to-one correspondence between the set of k-

dimensional faces of “, k = 1, . . . , m’1, and n’k dimensional faces of “— ,

de¬ned by the rule “— = “ © “⊥ . If we treat “ as its own m-dimensional

J IrJ

face “… , then it corresponds to “— = “⊥ .

I

30

Chapter 2

Mirrors, Re¬‚ections, Roots

2.1 Mirrors and re¬‚ections

Mirrors and re¬‚ections. Recall that a re¬‚ection in an a¬ne real Eu-

clidean space ARn is a nonidentity isometry s which ¬xes all points of

some a¬ne hyperplane (i.e. a¬ne subspace of codimension 1) H of ARn .

The hyperplane H is called the mirror of the re¬‚ection s and denoted Hs .

Conversely, the re¬‚ection s will be sometimes denoted as s = sH .

Lemma 2.1.1 If s is a re¬‚ection with the mirror H then, for any point

± ∈ ARn ,

• the segment [s±, ±] is normal to H and H intersects the segment in

its midpoint;

• H is the set of points ¬xed by s;

• s is an involutary transformation1 , that is, s2 = 1.

In particular, the re¬‚ection s is uniquily determined by its mirror H, and

vice versa.

Proof. Choose some point of H for the origin o of a orthonormal coor-

dinate system, and identify the a¬ne space ARn with the underlying real

Euclidean vector space Rn . Then, by Theorem 1.4.2, s can be identi¬ed

with an orthogonal transformation of Rn . Since s ¬xes all points in H, it

has at least n ’ 1 eigenvalues 1, and, since s is non identity, the only possi-

bility for the remaining eigenvalue is ’1. In particular, s is diagonalisable

and has order 2, that is s2 = 1 and s = 1. It also follows from here that H

is the set of all point ¬xed by s.

1

A non-identity element g of a group G is called an involution if it has order 2. Hence

s is an involution.

31

32

t

tH ' H

s t± = ts± s±

t

t' t

˜ ˜

˜ ˜

˜ ˜

˜ ˜

˜t ˜t

˜ ˜

˜ ˜

˜ ˜

˜ ˜

˜ ˜

˜ t' ˜t

t

˜ ˜

t± ±

Figure 2.1: For the proof of Lemma 2.1.3: If s is the re¬‚ection in the mirror H

and t is an isometry then the re¬‚ection s in the mirror tH can be found from

the condition s t = ts hence s = tst’1 .

If now we consider the vector s± ’ ± directed along the segment [s±, ±]

then

s(s± ’ ±) = s2 ± ’ s± = ± ’ s±,

which means that the vector s± ’± is an eigenvector of s for the eigenvalue

1

’1. Hence the segment [s±, ±] is normal to H. Its midpoint 2 (s± + ±) is

s-invariant, since

1 1 1

s (s± + ±) = (s2 ± + s±) = (s± + ±),

2 2 2

hence belongs to H.

In the course of the proof of the previous lemma we have also shown

Lemma 2.1.2 Re¬‚ections in ARn which ¬x the origin o are exactly the or-

thogonal transformations of Rn with n’1 eigenvalues 1 and one eigenvalue

’1; their mirrors are eigenspaces for the eigenvalue 1.

We say that the points s± and ± are symmetric in H. If X ‚ A then

the set sX is called the re¬‚ection or the mirror image of the set X in the

mirror H.

Lemma 2.1.3 If t is an isometry of ARn , s the re¬‚ection in the mirror

H and s is the re¬‚ection in tH then s = tst’1 .

A. & A. Borovik • Mirrors and Re¬‚ections • Version 01 • 25.02.00 33

Proof. See Figure 2.1. Alternatively, we may argue as follows.

We need only show that tst’1 is a non-identity isometry which ¬xes

tH. Since tst’1 is a composition of isometries, it is clearly an isometry. If

± ∈ tH, then t’1 ± ∈ H, hence s ¬xes t’1 ±, hence tst’1 ± = ±. If ± ∈ tH,

then t’1 ± ∈ H, hence s does not ¬x t’1 ±, hence tst’1 ± = ±.

Exercises

Re¬‚ections and rotations in R2 .

2.1.1 Prove that every 2 orthogonal matrix A over R can be written in one of

the forms

cos θ ’ sin θ

sin θ cos θ

or

cos θ sin θ

,

sin θ ’ cos θ

depending on whether A has determinant 1 or ’1.

2.1.2 Prove that if, in the notation of the previous Exercise, det A = 1 then A is

the matrix of the rotation through the angle θ about the origin, counterclockwise.

2.1.3 Prove that if det A = ’1 then A is the matrix of a re¬‚ection.

2.1.4 Check that

cos φ/2 ’ sin φ/2

u= and v =

sin φ/2 cos φ/2

are eigenvectors with the eigenvalues 1 and ’1 for the matrix

cos φ sin φ

.

sin φ ’ cos φ

2.1.5 Use trigonometric identities to prove that

cos φ ’ sin φ cos ψ ’ sin ψ cos(φ + ψ) ’ sin(φ + ψ)

· = .

sin φ cos φ sin ψ cos ψ sin(φ + ψ) cos(φ + ψ)

Give a geometric interpretation of this fact.

34

Finite groups of orthogonal transformations in 2 dimensions.

2.1.6 Prove that any ¬nite group of rotations of the Euclidean plane R2 about

the origin is cyclic.

Hint: It is generated by a rotation through the smallest angle.

2.1.7 Prove that if r is a rotation of R2 and s a re¬‚ection then sr is a re¬‚ection,

in particular, |sr| = 2. Deduce from this the fact that s inverts r, i.e. srs’1 =

r’1 .

2.1.8 If G is a ¬nite group of orthogonal transformations of the 2-dimensional

Euclidean space R2 then the map

det : G ’’ { 1, ’1 }

A ’ det A

is a homomorphism with the kernel R consisting of all rotations contained in G.

If R = G then |G : R| = 2 and all elements in G R are re¬‚ections.

2.1.9 Prove that the product of two re¬‚ections in R2 (with a common ¬xed

point at the origin) is a rotation through twice the angle between their mirrors.

Involutary orthogonal transformations in three dimensions.

2.1.10 In R3 there are three, up to conjugacy of matrices, involutive orthog-

onal transformations, with the eigenvalues 1, 1, ’1 (re¬‚ections), 1, ’1, ’1 and

’1, ’1, ’1. Give a geometric interpretation of the two latter transformations.

2.2 Systems of mirrors

Assume now that we are given a solid ∆ ‚ ARn . Consider the set Σ of all

mirrors of symmetry of ∆, i.e. the mirrors of re¬‚ections which send ∆ to

∆. The reader can easily check (Exercise 2.2.1) that Σ is a closed system

of mirrors in the sense of the following de¬nition: a system of hyperplanes

(mirrors) in A is called closed if, for any two mirrors H1 and H2 in Σ, the

mirror image of H2 in H1 also belongs to Σ (see Figure 2.)

Slightly abusing language, we shall call a ¬nite closed system Σ of

mirrors simply a system of mirrors.

Systems of mirrors are the most natural objects. The reader most likely

has seen them when looking into a kaleidoscope2 ; and, of course, everybody

2

My special thanks are due to Dr. Robert Sandling who lent me, for demonstration

to my students, a fascinating old fashioned kaleidoscope. It contained three mirrors

arranged as the side faces of a triangular prism with an equilateral base and produced

˜

the mirror system of type A2 .

A. & A. Borovik • Mirrors and Re¬‚ections • Version 01 • 25.02.00 35

Σ

The system Σ of mirrors of

symmetry of a geometric body

∆ is closed: the re¬‚ection of a