Geometric Transformations and Wallpaper Groups

Transcription

1 and Geometric Transformations and Wallpaper of the Plane Department of Mathematics and Statistics Texas Tech University Lubbock, Texas 2010 Math Camp

2 Outline and 1 2 and 3

3 and A transformation T of the plane is an isometry if it is one-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.

4 and A matrix A is orthogonal if Ax = x for all x R 2. 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 q p q p x

5 Outline and 1 2 and 3

6 1 and Theorem A is orthogonal if and only if Ax Ay = x y for all x, y R 2. Thus, an orthogonal matrix preserves angles. Proof. (= ) ( =) Recall Ax 2 = Ax Ax = x x = x 2 2x y = x 2 + y 2 x y 2.

7 2 and 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. Theorem A matrix is orthogonal if and only if its columns are orthogonal unit vectors.

8 and Proof. (= ) Ae i = e i = 1. Ae 1 Ae 2 = e 1 e 2 = 0. 3 ( =) Let A = [u v] where u and v are orthogonal unit vectors. Note Ax = x 1 u + x 2 v. Ax 2 = Ax Ax = (x 1 u + x 2 v) (x 1 u + x 2 v) = x 2 1 u u + 2x 1 x 2 u v + x 2 2 v v = x 2 1 (1) + 2x 1 x 2 (0) + x 2 2 (1) = x x 2 2 = x 2

9 Outline and 1 2 and 3

10 1 and If A is orthogonal, Col 1 (A) = (cos(θ), sin(θ)) for some θ. So the possibilities are [ ] cos(θ) sin(θ) A = = R(θ), sin(θ) cos(θ) [ ] cos(θ) sin(θ) A = = S(θ). sin(θ) cos(θ) In the first case A = R(θ) is a rotation. What about S(θ)?

11 1 and If A is orthogonal, Col 1 (A) = (cos(θ), sin(θ)) for some θ. So the possibilities are [ ] cos(θ) sin(θ) A = = R(θ), sin(θ) cos(θ) [ ] cos(θ) sin(θ) A = = S(θ). sin(θ) cos(θ) In the first case A = R(θ) is a rotation. What about S(θ)? There are orthogonal unit vectors u and v so that S(θ)u = u and S(θ)v = v.

12 2 and In fact, let u = (cos(θ/2), sin(θ/2)) and v = ( sin(θ/2), cos(θ/2)). [ ] [ ] cos(θ) sin(θ) cos(θ/2) S(θ)u = sin(θ) cos(θ) sin(θ/2) [ ] cos(θ) cos(θ/2) + sin(θ) sin(θ/2) = sin(θ) cos(θ/2) cos(θ) sin(θ/2)

13 2 and In fact, let u = (cos(θ/2), sin(θ/2)) and v = ( sin(θ/2), cos(θ/2)). [ ] [ ] cos(θ) sin(θ) cos(θ/2) S(θ)u = sin(θ) cos(θ) 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) cos(θ/2) = = = u. sin(θ θ/2) sin(θ/2)

14 3 and [ ] [ ] cos(θ) sin(θ) sin(θ/2) S(θ)v = sin(θ) cos(θ) 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) sin(θ/2) = = = v. cos(θ/2) cos(θ/2)

15 and 4 S(θ)u = u and S(θ)v = v. If x = tu + sv, then S(θ)x = tu sv. y x sv M tu v u sv x S(θ)x

16 and 4 S(θ)u = u and S(θ)v = v. If x = tu + sv, then S(θ)x = tu sv. y x sv M tu v u sv x S(θ)x S(θ) is a reflection with its mirror line at an angle of θ/2.

17 Outline and 1 2 and 3

18 Matix 1 and S(θ) 2 = I, so S(θ) 1 = S(θ). Let A be an orthogonal matrix. Then A is a rotation (or the identity) if and only if det(a) = 1 and A is a reflection if and only if det(a) = 1. [ ] a c If A = b d [ a b A T = c d A 1 = A T. define the transpose of A by ]. Show that A is orthogonal if and only if

19 and Continued on 2 Use the addition laws for sine and cosine to verify the important identities R(θ)S(φ) = S(θ + φ), S(φ)R(θ) = S(φ θ), S(θ)S(φ) = R(θ φ)

20 Outline and 1 2 and 3

21 and For v R 2, define T v : R 2 R 2 by T v (x) = x + v. We say T v is translation by v. T u T v = T u+v and T 1 v = T v. are isometries dist(t v (x), T v (y)) = T v (x) T v (y) = (x + v) (y + v) = x + v y v = x y = dist(x, y).

22 Outline and 1 2 and 3

23 Three Distances Determine a Point and Let p 1, p 2 and p 3 be noncolinear points. A point x is uniquely determined by the three numbers r 1 = dist(x, p 1 ), r 2 = dist(x, p 2 ) and r 3 = dist(x, p 3 ). p 1 p 2 p 3

24 and If an isometry T fixes three noncolinear points p 1, p 2, p 3 then T is the identity. dist(x, p i ) = dist(t (x), T (p i )) = dist(t (x), p i ), i = 1, 2, 3, = x = T (x). Theorem Every isometry T can be written as T (x) = Ax + v where A is an orthogonal matrix, i.e., T is multiplication by an orthogonal matrix followed by a translation.

25 Proof of and Proof. 1 The points 0, e 1, e 2 are noncolinear.

26 Proof of and Proof. 1 The points 0, e 1, e 2 are noncolinear. 2 Let u = T (0). Then T u T (0) = 0.

27 Proof of and Proof. 1 The points 0, e 1, e 2 are noncolinear. 2 Let u = T (0). Then T u T (0) = 0. 3 dist(t u T (e 1 ), 0) = 1, so we can find a rotation matrix R so that RT u T (e 1 ) = e 1.

28 Proof of and Proof. 1 The points 0, e 1, e 2 are noncolinear. 2 Let u = T (0). Then T u T (0) = 0. 3 dist(t u T (e 1 ), 0) = 1, so we can find a rotation matrix R so that RT u T (e 1 ) = e 1. 4 RT u T (e 2 ) = ±e 2.

29 Proof of and Proof. 1 The points 0, e 1, e 2 are noncolinear. 2 Let u = T (0). Then T u T (0) = 0. 3 dist(t u T (e 1 ), 0) = 1, so we can find a rotation matrix R so that RT u T (e 1 ) = e 1. 4 RT u T (e 2 ) = ±e 2. 5 If RT u T (e 2 ) = e 2, let S = S(0) be reflection through the x-axis, otherwise let S = I.

30 Proof of and Proof. 1 The points 0, e 1, e 2 are noncolinear. 2 Let u = T (0). Then T u T (0) = 0. 3 dist(t u T (e 1 ), 0) = 1, so we can find a rotation matrix R so that RT u T (e 1 ) = e 1. 4 RT u T (e 2 ) = ±e 2. 5 If RT u T (e 2 ) = e 2, let S = S(0) be reflection through the x-axis, otherwise let S = I. 6 SRT u (0) = 0, SRT u T (e 1 ) = e 2, and SRT u T (e 2 ) = e 2

31 Proof of and Proof. 1 The points 0, e 1, e 2 are noncolinear. 2 Let u = T (0). Then T u T (0) = 0. 3 dist(t u T (e 1 ), 0) = 1, so we can find a rotation matrix R so that RT u T (e 1 ) = e 1. 4 RT u T (e 2 ) = ±e 2. 5 If RT u T (e 2 ) = e 2, let S = S(0) be reflection through the x-axis, otherwise let S = I. 6 SRT u (0) = 0, SRT u T (e 1 ) = e 2, and SRT u T (e 2 ) = e 2 7 SRT u T = I, so T = T u R 1 S 1

32 Proof of and Proof. 1 The points 0, e 1, e 2 are noncolinear. 2 Let u = T (0). Then T u T (0) = 0. 3 dist(t u T (e 1 ), 0) = 1, so we can find a rotation matrix R so that RT u T (e 1 ) = e 1. 4 RT u T (e 2 ) = ±e 2. 5 If RT u T (e 2 ) = e 2, let S = S(0) be reflection through the x-axis, otherwise let S = I. 6 SRT u (0) = 0, SRT u T (e 1 ) = e 2, and SRT u T (e 2 ) = e 2 7 SRT u T = I, so T = T u R 1 S 1 8 Let A = R 1 S 1 and v = u. Then T (x) = Ax + v

33 Outline and 1 2 and 3

34 Notation and Algebra and 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 1 u)

35 Rotations and Consider (R v) where R = R(θ) 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 Big Theorem.) 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).

36 and Picture of Rotation y = (R p Rp)x = p + R(x p) says that this isometry is rotation though angle θ around the point p,which is called the center of rotation or rotocenter. y y p R(x p) θ x p x x

37 with a Reflection Matrix and 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). Case: α = 0. So we have (S βv).

38 with a Reflection Matrix and 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). 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.

39 with a Reflection Matrix and 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). 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.

40 and The Line of Fixed Points The line through p parallel to u is parametrized by x = p + tu, for t R. y tu p x v u x

41 and Reflection Picture We can write a point x as x = p + tu + sv. Then y = (S 2p)x = p + tu sv. y x tu + sv sv M p tu sv tu sv y v u x

42 Reflections and Glides and (S βv) = (S 2p) is reflection through the mirror line M given by y = p + tu. Second Case: α 0. We have (S αu + βv). (S αu + βv) = (I αu)(s βv) This is a reflection followed by a translation parallel to the mirror 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.

43 and Glide Reflection Picture A glide with a horizontal glide line, and the translation vector shown in blue.

44 Theorem and Theorem Every isometry of the plane falls into on of the following five mutually 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.

45 Outline and 1 2 and 3

46 and on 1 Consider the cases for the product T 1 T 2 of two isometries T 1 and T 2. Describe what happens geometrically (e.g., rotation angle, location of the mirror line, etc.). In what cases do the isometries commute, i.e., when does T 1 T 2 = T 2 T 1 2 Project: For a rotation (R v) give a geometric description of how to find the rotocenter p = (I R) 1 v. 3 As part of the first exercise, given two rotations (with possibly different rotocenters) show that the composition is a rotation or a translation. 4 Project: Given two rotations with different rotocenters give a geometric description of how to find the rotocenter of the composition.