Mathematics 2Q - Groups, Symmetry and Fractals

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1 ?,?th August, 2002? to? //euler/texmf/dvips/unigla.eps EXAMINATION FOR THE DEGREES OF M.A. AND B.Sc. Mathematics 2Q - Groups, Symmetry and Fractals Candidates must not attempt more than THREE questions. 1. (i) Determine the geometric effect of the isometry R 2 R 2 represented by the Seitz symbol (A t), where A [ ] 3/2 1/2 1/2, t 3/ For the matrix B, determine the Seitz symbol of the composition 1 0 (A t)(b 0) and state what kind of isometry it represents (i.e., a translation, a rotation, or a glide reflection). 3 (ii) For a real number α, let L α {(x, y) R 2 : x sin α y cos α 0} R 2 and let (S α 0) be the Seitz symbol of the reflection Refl α in the line L α. Write down the value of the matrix S α. 3 If α, β R, evaluate the matrix S α S β and hence prove that Refl α Refl β is a rotation about the origin (i) Working in the symmetric group S 7, (a) evaluate the following product of permutations: ( ) ( ) ; (b) determine the disjoint cycle decomposition and the sign of the permutation ( ) [OVER

2 2 (ii) Let Γ be the group of symmetries of the regular octagon centred at the origin O and with vertices A, B, C, D, E, F, G, H. D E C F O B A H G By identifying Γ with a group of permutations of the vertices, describe the elements of Γ which are rotations both geometrically and by using permutation notation. Write down the permutation corresponding to reflection in the line AE. 8 By composing permutations, determine the effect of anti-clockwise rotation about O through π/4 followed by reflection in the line AE (i) Define the Euclidean group (Euc(2), ). 2 Show that the subset O(2) Euc(2) consisting of all the isometries which fix the origin is a subgroup of the Euclidean group (Euc(2), ). 3 (ii) Suppose that F : R 2 R 2 is an isometry with Seitz symbol (A t). If x, y R 2 and s R, show that F (sx + (1 s)y) sf (x) + (1 s)f (y). 5 (iii) Define a dilation of the plane from a point and explain what it means for two isometries of the plane to be similar. 3 Show that the rotations through the angle θ in the anti-clockwise direction about two points P and Q are similar (i) For the frieze pattern F, part of which is shown in the diagram below, indicate a generator of the group of translational symmetries and a glide reflection symmetry. Describe the elements of the symmetry subgroup Euc(2) F Euc(2) in terms of these generators. 6 Are there any points in the plane fixed by all the elements of Euc(2) F? 2 If the symbols and were replaced with and respectively, what extra symmetries would be introduced? 3 [OVER

3 3 (ii) Explain why the matrix 3/2 0 1/2 R /2 0, 3/2 corresponds to a rotation of R 3 about a line through the origin and determine its axis and angle of rotation. 9 END]

4 ?,?th August, 2002? to? 2Q Resit exam 2002 Solutions 1. (i) We have cos( π/6) sin( π/6) A, det A cos 2 ( π/6) + sin 2 ( π/6) 1, sin( π/6) cos( π/6) so (A t) represents a clockwise rotation through the angle π/6. The centre of 3 rotation is the point with position vector [ ] 1 [ ] 1 1 3/2 1/2 c (I A) 1 t 1/ / [ ] (2 3) [ ] [ ] (2 3) 1 [ ] 1/2 (1 +. 3/2) So the centre of rotation is at c (1/2, (1 + 3/2)). 5 The Seitz symbol of the composition is where AB (A t)(b 0) (AB t), [ ] 3/2 1/2 1/ /2 1 0 [ ] 1/2 3/2. 3/2 1/2 Since det(ab) 1, this represents a glide reflection in some line. 3 (ii) We have cos 2α sin 2α S α. 3 sin 2α cos 2α The matrix product is cos 2α sin 2α cos 2β sin 2β S α S β sin 2α cos 2α sin 2β cos 2β cos 2α cos 2β + sin 2α sin 2β cos 2α sin 2β sin 2α cos 2β sin 2α cos 2β cos 2α sin 2β sin 2α sin 2β + cos 2α cos 2β cos 2(α β) sin 2(α β), sin 2(α β) cos 2(α β) and the Seitz symbol (S α S β 0) represents rotation about the origin through the angle 2(α β). Since this is the Seitz symbol of Refl α Refl β, this a rotation about the origin through 2(α β). 6

5 2 2. (i) (a) We have ( ) ( ) (b) We have Also, σ ( ) ( ) (1 3 2)(4)(5 7)(6) (1 3 2)(5 7) sgn σ sgn(1 3 2) sgn(5 7) ( 1) 2 ( 1) ( 1) (ii) There are eight rotations taken anti-clockwise: through 0 the identity ι, through 2π/8 π/4 (A B C D E F G H), through 4π/8 π/2 (A C E G)(B D F H), through 6π/8 3π/4 (A D G B E H C F ), through 8π/8 π (A E)(B F )(C G)(D H), through 10π/8 5π/4 (A F C H E B G D), through 12π/8 3π/2 (A G E C)(B H F D), through 14π/8 7π/4 (A H G F E D C B). Reflection in AE corresponds to the permutation (B H)(C G)(D F ). 8 The symmetries being composed are (A B C D E F G H) and (B H)(C G)(D F ) so we have (B H)(C G)(D F )(A B C D E F G H) (A H)(B G)(C F )(D E), which is corresponds to reflection in the line joining the midpoints of the edges AH and DE (i) (Euc(2), ) consists of Euc(2), the set of all isometries R 2 R 2, under composition of functions. 2 For (A 0), (B 0) O(2) we have and (A 0)(B 0) (AB 0) (AB) T (AB) (B T A T )(AB) B T (A T A)B B T I 2 B B T B I 2. So (A 0)(B 0) O(2). Also, (I 2 0) O(2) and since A 1 A T and (A 0) 1 (A 1 0) O(2) (A T ) T (A T ) AA T AA 1 I 2, hence A 1 is orthogonal. 3

6 3 (ii) We have (A t)(sx + (1 s)y) A(sx + (1 s)y) + t A(sx) + A((1 s)y) + t sax + A(1 s)y + st + (1 s)t s(a t)x + (1 s)(a t)y, giving the desired result. 5 (iii) A dilation of the plane is a function H : R 2 R 2 which has the form H(x) δ(x c) + c, where δ > 0 is the dilation factor and c is the position vector of the centre of the dilation. 1 Two isometries of the plane F, G are similar if there is a similarity transformation H (δa t) with δ > 0, A orthogonal and t R 2, for which G H F H F H 1. 2 Let t P Q q p. Then Trans t Rot P,θ Trans 1 t Trans t Rot P,θ Trans t is another rotation through θ since its Seitz symbol is (I t)(r u)(i t) (I t)(r u Rt) (R u Rt + t), where Rot P,θ (R u). Since P is the centre of rotation, Rot P,θ (p) p. Now applying this isometry to the position vector of Q we obtain Trans t Rot P,θ Trans t (q) t + Rot P,θ (q t) (q p) + Rot P,θ (q (q p)) (q p) + Rot P,θ (p) (q p) + p q, hence Trans t Rot P,θ Trans t fixes Q which must therefore be its centre of rotation. Hence we have Trans t Rot P,θ Rot Q,θ (i) There is a translation by t parallel to the dashed line and shifting the whole pattern one step the right. There is also a glide reflection whose effect of is indicated below, with square equal to Trans t. There are no rotations or reflections in lines perpendicular to the dashed line. The symmetry group of F is infinite and cyclic with generator, Euc(2) F { n : n Z}. 6 t

7 4 There are no points simultaneously fixed by all the elements of this symmetry group since a non-trivial translation has no fixed points. 2 There would be reflections in the vertical lines through the midpoints of the and and half turn rotations about the points on the dashed line midway between these vertical lines. 3 (ii) By expanding along the middle row we obtain det R 3/2 1/2 1/2 (3 + 1) 3/2 4 1, so R represents a rotation about an axis through the origin. Since Re 2 e 2, the axis of rotation is the y-axis. 4 Now let w e 2, u e 3, v w u e 2 e 3 e 1. So u, v, w is a right handed orthonormal system. We have Ru ( 3/2)u + ( 1/2)v cos(5π/6)u sin(5π/6)v, Rv (1/2)u + ( 3/2)v sin(5π/6)u + cos(5π/6)v, so the angle of rotation is 5π/6 and the sense is that of a right handed screw driver pointing along the positive y-axis. 5 END]