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ρ— = ’ .
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
 
 

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