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ПЃв€— = в€’ .
j
ri,1 ... ri,jв€’1 ri,j+1 . . . ri,n
det R i
. . . . .
. . . . .
. . . . .
rn,1 . . . rn,jв€’1 rn,j+1 . . . rn,n
n

Notice that in the case n = 3 we come to the formula
1 1 1
ПЃв€— = в€’ ПЃ2 Г—ПЃ3 , ПЃв€— = в€’ ПЃ3 Г—ПЃ1 , ПЃв€— = в€’ ПЃ1 Г—ПЃ2 ,
1 2 3
(ПЃ1 , ПЃ2 , ПЃ3 ) (ПЃ1 , ПЃ2 , ПЃ3 ) (ПЃ1 , ПЃ2 , ПЃ3 )
where ( , , ) denotes the scalar triple product and Г— the cross (or vector)
product of vectors.
1.5.4 Let О“ be a cone in Rn spanned by a set О  of m linearly independent
vectors ПЃ1 , . . . , ПЃm , with m < n. Let U be the vector subspace spanned by О .
Then О“ is a simplicial cone in U ; we denote its dual in U as О“ , and set О“в€—
to be the dual cone for О“ in V . Let also О  = { ПЃ1 , . . . , ПЃm } be the basis in U
dual to the basis О . We shall use in the sequel the following properties of the
cone О“в€— .
1. For any set A в€€ Rn , deп¬Ѓne
AвЉҐ = { П‡ в€€ Rn | (П‡, О±) = 0 for all Оі в€€ A }.
Check that AвЉҐ is a linear subspace of Rn . Prove that dim О“вЉҐ = n в€’ m.
Hint: О“вЉҐ = U вЉҐ .
2. О“в€— is the intersection of the closed half spaces deп¬Ѓned by the inequalities
(П‡, ПЃi ) 0, i = 1, . . . , m.
3. О“в€— = О“ + О“вЉҐ ; this set is, by deп¬Ѓnition,
О“ + О“вЉҐ = { Оє + П‡ | Оє в€€ О“ , П‡ в€€ О“вЉҐ }.

4. (О“в€— )в€— = О“.
5. Let Hi and Hiв€— be the hyperplanes in V given by the equations (П‡, ПЃi ) = 0
and (П‡, ПЃi ) = 0, correspondingly. Denote I = { 1, . . . , m } and set, for
J вЉ† I,
Hj , О“в€— = О“ в€— в€© в€— в€—
О“J = О“ в€© Hj and О“J = О“ в€© Hj .
J
jв€€J jв€€J jв€€J

Prove that О“в€— = О“J + О“вЉҐ .
J
A. & A. Borovik вЂў Mirrors and Reп¬‚ections вЂў Version 01 вЂў 25.02.00 29

6. The cones О“J and О“в€— are called faces of the cones О“ and О“в€— , corre-
J
spondingly. There is a one-to-one correspondence between the set of k-
dimensional faces of О“, k = 1, . . . , mв€’1, and nв€’k dimensional faces of О“в€— ,
deп¬Ѓned by the rule О“в€— = О“ в€© О“вЉҐ . If we treat О“ as its own m-dimensional
J IrJ
face О“в€… , then it corresponds to О“в€— = О“вЉҐ .
I
30
Chapter 2

Mirrors, Reп¬‚ections, Roots

2.1 Mirrors and reп¬‚ections
Mirrors and reп¬‚ections. Recall that a reп¬‚ection in an aп¬ѓne real Eu-
clidean space ARn is a nonidentity isometry s which п¬Ѓxes all points of
some aп¬ѓne hyperplane (i.e. aп¬ѓne subspace of codimension 1) H of ARn .
The hyperplane H is called the mirror of the reп¬‚ection s and denoted Hs .
Conversely, the reп¬‚ection s will be sometimes denoted as s = sH .

Lemma 2.1.1 If s is a reп¬‚ection with the mirror H then, for any point
О± в€€ ARn ,
вЂў the segment [sО±, О±] is normal to H and H intersects the segment in
its midpoint;
вЂў H is the set of points п¬Ѓxed by s;
вЂў s is an involutary transformation1 , that is, s2 = 1.
In particular, the reп¬‚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 aп¬ѓne space ARn with the underlying real
Euclidean vector space Rn . Then, by Theorem 1.4.2, s can be identiп¬Ѓed
with an orthogonal transformation of Rn . Since s п¬Ѓxes all points in H, it
has at least n в€’ 1 eigenvalues 1, and, since s is non identity, the only possi-
bility for the remaining eigenvalue is в€’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 п¬Ѓxed 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 tО± = tsО± sО±
t
t' t
В˜ В˜
В˜ В˜
В˜ В˜
В˜ В˜
В˜t В˜t
В˜ В˜
В˜ В˜
В˜ В˜
В˜ В˜
В˜ В˜
В˜ t' В˜t
t
В˜ В˜
tО± О±
 

 
 
 

Figure 2.1: For the proof of Lemma 2.1.3: If s is the reп¬‚ection in the mirror H
and t is an isometry then the reп¬‚ection s in the mirror tH can be found from
the condition s t = ts hence s = tstв€’1 .

If now we consider the vector sО± в€’ О± directed along the segment [sО±, О±]
then
s(sО± в€’ О±) = s2 О± в€’ sО± = О± в€’ sО±,
which means that the vector sО± в€’О± is an eigenvector of s for the eigenvalue
1
в€’1. Hence the segment [sО±, О±] is normal to H. Its midpoint 2 (sО± + О±) is
s-invariant, since

1 1 1
s (sО± + О±) = (s2 О± + sО±) = (sО± + О±),
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 Reп¬‚ections in ARn which п¬Ѓx the origin o are exactly the or-
thogonal transformations of Rn with nв€’1 eigenvalues 1 and one eigenvalue
в€’1; their mirrors are eigenspaces for the eigenvalue 1.

We say that the points sО± and О± are symmetric in H. If X вЉ‚ A then
the set sX is called the reп¬‚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 reп¬‚ection in the mirror
H and s is the reп¬‚ection in tH then s = tstв€’1 .
A. & A. Borovik вЂў Mirrors and Reп¬‚ections вЂў Version 01 вЂў 25.02.00 33

Proof. See Figure 2.1. Alternatively, we may argue as follows.
We need only show that tstв€’1 is a non-identity isometry which п¬Ѓxes
tH. Since tstв€’1 is a composition of isometries, it is clearly an isometry. If
О± в€€ tH, then tв€’1 О± в€€ H, hence s п¬Ѓxes tв€’1 О±, hence tstв€’1 О± = О±. If О± в€€ tH,
then tв€’1 О± в€€ H, hence s does not п¬Ѓx tв€’1 О±, hence tstв€’1 О± = О±.

Exercises
Reп¬‚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 Оё в€’ sin Оё
sin Оё cos Оё
or
cos Оё sin Оё
,
sin Оё в€’ cos Оё

depending on whether A has determinant 1 or в€’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 Оё about the origin, counterclockwise.

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

2.1.4 Check that

cos П†/2 в€’ sin П†/2
u= and v =
sin П†/2 cos П†/2

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

cos П† sin П†
.
sin П† в€’ cos П†

2.1.5 Use trigonometric identities to prove that

cos П† в€’ sin П† cos П€ в€’ sin П€ cos(П† + П€) в€’ sin(П† + П€)
В· = .
sin П† cos П† sin П€ cos П€ sin(П† + П€) cos(П† + П€)

Give a geometric interpretation of this fact.
34

Finite groups of orthogonal transformations in 2 dimensions.

2.1.6 Prove that any п¬Ѓnite 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 reп¬‚ection then sr is a reп¬‚ection,
in particular, |sr| = 2. Deduce from this the fact that s inverts r, i.e. srsв€’1 =
rв€’1 .

2.1.8 If G is a п¬Ѓnite group of orthogonal transformations of the 2-dimensional
Euclidean space R2 then the map

det : G в€’в†’ { 1, в€’1 }
A в†’ 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 reп¬‚ections.

2.1.9 Prove that the product of two reп¬‚ections in R2 (with a common п¬Ѓxed
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, в€’1 (reп¬‚ections), 1, в€’1, в€’1 and
в€’1, в€’1, в€’1. Give a geometric interpretation of the two latter transformations.

2.2 Systems of mirrors
Assume now that we are given a solid в€† вЉ‚ ARn . Consider the set ОЈ of all
mirrors of symmetry of в€†, i.e. the mirrors of reп¬‚ections which send в€† to
в€†. The reader can easily check (Exercise 2.2.1) that ОЈ is a closed system
of mirrors in the sense of the following deп¬Ѓnition: a system of hyperplanes
(mirrors) in A is called closed if, for any two mirrors H1 and H2 in ОЈ, the
mirror image of H2 in H1 also belongs to ОЈ (see Figure 2.)
Slightly abusing language, we shall call a п¬Ѓnite closed system ОЈ 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
Лњ
the mirror system of type A2 .
A. & A. Borovik вЂў Mirrors and Reп¬‚ections вЂў Version 01 вЂў 25.02.00 35

В
ОЈВ
В
В
The system ОЈ of mirrors of
В
В  В  В
В  symmetry of a geometric body
В  В  В
В
В
В
В
В
В  В
В
В  В  в€† is closed: the reп¬‚ection of a
В
В
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