Introduction Orthogonal Geometric Transformations and...

WallpaperGroups

Lance Drager

Introduction

OrthogonalMatrices

OrthogonalMatrices andDot Product

ClassifyingOrthogonalMatrices

Exercises

ClassifyingIsometries

Translations

FirstClassification

FinalClassification

Exercises

Geometric Transformations and WallpaperGroups

Isometries of the Plane

Lance Drager

Department of Mathematics and StatisticsTexas Tech University

Lubbock, Texas

2010 Math Camp

WallpaperGroups

Lance Drager

Introduction

OrthogonalMatrices



Exercises


Translations

FirstClassification

FinalClassification

Exercises

Outline

1 Introduction

2 Orthogonal MatricesOrthogonal Matrices and Dot ProductClassifying Orthogonal MatricesExercises

3 Classifying IsometriesTranslationsFirst ClassificationFinal ClassificationExercises

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Introduction

OrthogonalMatrices



Exercises


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FirstClassification

FinalClassification

Exercises

Isometries

• A transformation T of the plane is an isometry if it isone-to-one and onto and

dist(T (p),T (q)) = dist(p, q), for all points p and q.

• If S and T are isometries, so is ST , where(ST )(p) = S(T (p)).

• If T is an isometry, so is T−1.

• Our goal is to find all the isometries.

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FirstClassification

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Exercises

Orthogonal Matrices

• A matrix A is orthogonal if ‖Ax‖ = ‖x‖ for all x ∈ R2.

• An orthogonal matrix gives an isometry of the plane since

dist(Ap,Aq) = ‖Ap − Aq‖= ‖A(p − q)‖ = ‖p − q‖ = dist(p, q)

• If A and B are orthogonal, so is AB.

• If A is orthogonal, it is invertible and A−1 is orthogonal.

• Rotation matrices are orthogonal.

y

x

‖p− q‖

p

q

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Outline

1 Introduction



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Dot Product 1

TheoremA is orthogonal if and only if Ax · Ay = x · y for all x , y ∈ R2.Thus, an orthogonal matrix preserves angles.

Proof.(=⇒)

‖Ax‖2 = Ax · Ax = x · x = ‖x‖2

(⇐=) Recall

2x · y = ‖x‖2 + ‖y‖2 − ‖x − y‖2.

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Dot Product 2

Proof Continued.So, we have

2Ax · Ay = ‖Ax‖2 + ‖Ay‖2 − ‖Ax − Ay‖2

= ‖Ax‖2 + ‖Ay‖2 − ‖A(x − y)‖2 (distributive law)

= ‖x‖2 + ‖y‖2 − ‖x − y‖2 (A is orthogonal)

= 2x · y ,

and dividing by 2 gives the result.

TheoremA matrix is orthogonal if and only if its columns are orthogonalunit vectors.

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Dot Product 3

Proof.(=⇒)

‖Aei‖ = ‖ei‖ = 1.

Ae1 · Ae2 = e1 · e2 = 0.

(⇐=) Let A = [u | v ] where u and v are orthogonal unitvectors. Note Ax = x1u + x2v .

‖Ax‖2 = Ax · Ax

= (x1u + x2v) · (x1u + x2v)

= x21 u · u + 2x1x2u · v + x2

2 v · v= x2

1 (1) + 2x1x2(0) + x22 (1)

= x21 + x2

2 = ‖x‖2

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Outline

1 Introduction



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Classifying Orthogonal Matrices 1

• If A is orthogonal, Col1(A) = (cos(θ), sin(θ)) for some θ.So the possibilities are

A =

[cos(θ) − sin(θ)sin(θ) cos(θ)

]= R(θ),

A =

[cos(θ) sin(θ)sin(θ) − cos(θ)

]= S(θ).

• In the first case A = R(θ) is a rotation.

• What about S(θ)?

• There are orthogonal unit vectors u and v so thatS(θ)u = u and S(θ)v = −v .

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• In fact, let u = (cos(θ/2), sin(θ/2)) andv = (− sin(θ/2), cos(θ/2)).

•

S(θ)u =


] [cos(θ/2)sin(θ/2)

]=

[cos(θ) cos(θ/2) + sin(θ) sin(θ/2)sin(θ) cos(θ/2)− cos(θ) sin(θ/2)

]

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• In fact, let u = (cos(θ/2), sin(θ/2)) andv = (− sin(θ/2), cos(θ/2)).

•

S(θ)u =


] [cos(θ/2)sin(θ/2)

]=

[cos(θ) cos(θ/2) + sin(θ) sin(θ/2)sin(θ) cos(θ/2)− cos(θ) sin(θ/2)

]=

[cos(θ − θ/2)sin(θ − θ/2)

]=

[cos(θ/2)sin(θ/2)

]= u.

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•

S(θ)v =


] [− sin(θ/2)

cos(θ/2)

]=

[− cos(θ) sin(θ/2) + sin(θ) cos(θ/2)− sin(θ) sin(θ/2)− cos(θ) cos(θ/2)

]=

[sin(θ − θ/2)

− cos(θ − θ/2)

]=

[sin(θ/2)

− cos(θ/2)

]= −

[− sin(θ/2)

cos(θ/2)

]= −v .

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• S(θ)u = u and S(θ)v = −v . If x = tu + sv , thenS(θ)x = tu − sv .

x

y

u

tu

v

sv

−sv

M

x

S(θ)x

• S(θ) is a reflection with its mirror line at an angle of θ/2.

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Outline

1 Introduction



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Orthogonal Matix Exercises 1

Exercises

• S(θ)2 = I , so S(θ)−1 = S(θ).

• Let A be an orthogonal matrix. Then A is a rotation (orthe identity) if and only if det(A) = 1 and A is a reflectionif and only if det(A) = −1.

• If A =

[a cb d

]define the transpose of A by

AT =

[a bc d

]. Show that A is orthogonal if and only if

A−1 = AT .

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Exercises on Orthogonal Matrices2

Exercises Continued

• Use the addition laws for sine and cosine to verify theimportant identities

R(θ)S(φ) = S(θ + φ),

S(φ)R(θ) = S(φ− θ),

S(θ)S(φ) = R(θ − φ)

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Outline

1 Introduction



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Translations

• For v ∈ R2, define Tv : R2 → R2 by Tv (x) = x + v . Wesay Tv is translation by v.

• TuTv = Tu+v and T−1v = T−v .

• Translations are isometries

dist(Tv (x),Tv (y)) = ‖Tv (x)− Tv (y)‖= ‖(x + v)− (y + v)‖= ‖x + v − y − v‖= ‖x − y‖= dist(x , y).

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Outline

1 Introduction



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Three Distances Determine a Point

• Let p1, p2 and p3 be noncolinear points. A point x isuniquely determined by the three numbers r1 = dist(x , p1),r2 = dist(x , p2) and r3 = dist(x , p3).

p1 p2

p3

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First Classification

• If an isometry T fixes three noncolinear points p1, p2, p3

then T is the identity.

dist(x , pi ) = dist(T (x),T (pi )) = dist(T (x), pi ), i = 1, 2, 3,

=⇒ x = T (x).

TheoremEvery isometry T can be written as T (x) = Ax + v where A isan orthogonal matrix, i.e., T is multiplication by an orthogonalmatrix followed by a translation.

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Proof of First Classification

Proof.

1 The points 0, e1, e2 are noncolinear.

2 Let u = −T (0). Then TuT (0) = 0.

3 dist(TuT (e1), 0) = 1, so we can find a rotation matrix Rso that RTuT (e1) = e1.

4 RTuT (e2) = ±e2.

5 If RTuT (e2) = −e2, let S = S(0) be reflection throughthe x-axis, otherwise let S = I .

6 SRTu(0) = 0, SRTuT (e1) = e2, and SRTuT (e2) = e2

7 SRTuT = I , so T = T−uR−1S−1

8 Let A = R−1S−1 and v = −u. Then T (x) = Ax + v

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Notation and Algebra

• If T (x) = Ax + v write T = (A | v).

• Calculate the product (composition) in this notation

(A | u)(B | v)x = (A | u)(Bx + v)

= A(Bx + v) + u

= ABx + Av + u

= (AB | u + Av)x .

• (A | u)(B | v) = (AB | u + Av)

• The identity transformation is (I | 0). Translation by v is(I | v).

• (A | u)−1 = (A−1 | − A−1u)

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Rotations

• Consider (R | v) where R = R(θ) 6= I is a rotation.

• Look for a fixed point p, i.e., (R | v)p = p

Rp + v = p

⇐⇒ p − Rp = v

⇐⇒ (I − R)p = v .

• If (I − R) is invertible, there is a unique solution p.

• (I − R) is invertible because(I − R)x = 0 ⇐⇒ x = Rx ⇐⇒ x = 0. (Use the BigTheorem.)

• There is a unique fixed point p, and we can write(R | v) = (R | p − Rp).

• (R | p − Rp)x = Rx + p − Rp = p + R(x − p).

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Picture of Rotation

• y = (R | p − Rp)x = p + R(x − p) says that this isometryis rotation though angle θ around the point p,which iscalled the center of rotation or rotocenter.

x

y

p

xx− pθ

R(x− p)

y

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Isometries with a Reflection Matrix

• Consider (S |w) where S = S(θ) is a reflection matrix.

• Let u and v be the vectors with Su = u and Sv = −v .We can write w = αu + βv , so (S |w) = (S |αu + βv).

• First Case: α = 0. So we have (S |βv).

• The point p = βv/2 is fixed.

(S |βv)p = S(βv/2) + βv = −βv/2 + βv = βv/2 = p.

• The fixed points are exactly x = p + tu.

(S | 2p)(p + tu) = Sp + tSu + 2p = −p + tu + 2p = p + tu.

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The Line of Fixed Points

• The line through p parallel to u is parametrized byx = p + tu, for t ∈ R.

x

y

x

p

tu

v u

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Reflection Picture

• We can write a point x as x = p + tu + sv . Theny = (S | 2p)x = p + tu − sv .

x

y

M

p

x

y

tu

sv

−sv

tu + sv

tu− sv

v u

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Reflections and Glides

• (S |βv) = (S | 2p) is reflection through the mirror line Mgiven by y = p + tu.

• Second Case: α 6= 0. We have (S |αu + βv).

(S |αu + βv) = (I |αu)(S |βv)

This is a reflection followed by a translation parallel to themirror line. This is called a glide refection or just a glide.The mirror line of the reflection is called the glide line.

• A glide has no fixed points.

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Glide Reflection Picture

• A glide with a horizontal glide line, and the translationvector shown in blue.

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Classification Theorem

TheoremEvery isometry of the plane falls into on of the following fivemutually exclusive classes.

1 The identity.

2 A translation (not the identity).

3 A rotation about some point (not the identity);

4 A reflection through some mirror line.

5 A glide along some glide line.

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Exercises on Isometries

Exercises

1 Consider the cases for the product T1T2 of two isometriesT1 and T2. Describe what happens geometrically (e.g.,rotation angle, location of the mirror line, etc.). In whatcases do the isometries commute, i.e., when doesT1T2 = T2T1

2 Project: For a rotation (R | v) give a geometric descriptionof how to find the rotocenter p = (I − R)−1v .

3 As part of the first exercise, given two rotations (withpossibly different rotocenters) show that the compositionis a rotation or a translation.

4 Project: Given two rotations with different rotocentersgive a geometric description of how to find the rotocenterof the composition.

Introduction Orthogonal Geometric Transformations and...

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