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Classification of Isometries

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Classification of Isometries Classification of isometries 1.1 Isometries of the Euclidean plane Definition 1.1 Isometry is a distance-preserving bijective mapping of the space to itself. Claim 1.2 The isometries of the space form a non-commutative group. Theorem 1.3 Every isometry is a collineation. Proof. Let A, B and C be three collinear points, and their image be A0, B0 and C0. Assume, that A0B0C0 form a triangle. But AC ∼= A0C0, BC ∼= B0C0 and AB ∼= A0B0 ⇒ |AC| + |BC| = |AB| = |A0B0| < |A0C0| + |B0C0|. ⇒⇐ Theorem 1.4 If, an isometry has two fix points, then the determined line is also fixed point-by-point (axis). Proof. Let A = A0 and B = B0 be true and P be a point on this line. Then P 0 must be also on the line and AP ∼= AP 0 ∧ BP ∼= BP 0 ⇒ P = P 0. Theorem 1.5 If, an isometry has three fix points, then the determined plane is also fixed point-by-point (plane). B Definition 1.6 The reflection of the plane in a line t assign the P 0 point to P if PP 0⊥t and PT ∼= P 0T , A where T = PP 0 ∩ t. Theorem 1.7 Reflection is an isometry. C D Proof. Let C and D be the footpoints of the per- , pendiculars to t from A and B respectively. Then A BCD4 ∼ B0CD4 ⇒ BC ∼ B0C∧BCD B0CD ⇒ = = \ \ , ∼ 0 0 ∼ 0 0 ∼ ACB\ = A CB \ ⇒ ABC4 = A B C4 ⇒ AB = B 0 0 A B Figure 1.1: Hypercycle model Claim 1.8 The composition of a reflection to itself is the identical mapping. 1 Theorem 1.9 If, a non-identical isometry has exactly two fix points, then it is a reflection in the determined line. Proof. Let A and B be the fix points and P be an arbitrary point. Then AP ∼= AP 0 ∧ BP ∼= BP 0 ⇒ P = P 0, or AB is the perpendicular bisector of PP 0 ⇒ d(P, AB) = d(P 0, AB) and PP 0⊥AB. Theorem 1.10 If, a non-identical R isometry has exactly one fix point, then it is the composition of two reflections, through this fix point. Proof. Let P be an arbitrary point and A be the fix point. Then AP ∼= AP 0, therefore 0 A lies on the perpendicular bisector (t1) of PP . The composition of the original isometry and the reflection in this line has two fix points, therefore it is a reflection in a line t2. −1 Then R ◦ t1 = t2 ⇒ R = t2 ◦ t1 = t2 ◦ t1. Theorem 1.11 If, a non-identical isometry T has no fix point, then it is the composition of either two or three reflections. 0 Proof. Let P be an arbitrary point and P be its image. Then T ◦ t1 has either one, or 0 two fix points, where t1 is the reflection to the perpendicular bisector of PP . −1 1. T ◦ t1 = R = t2 ◦ t3 ⇒ T = t2 ◦ t3 ◦ t1 = t2 ◦ t3 ◦ t1 −1 2. T ◦ t1 = t2 ⇒ T = t2 ◦ t1 = t2 ◦ t1 Theorem 1.12 Any isometry of the plane is the composition of at most three reflections in a line. Proof. It follows from the above theorems trivially, since any isometry has either 0, 1, or 2 fix points or all the points on the plane is fixed point-by-point. Lemma 1.13 The composition of two reflections can be considered as either a rotation (intersecting axis) or a translation (parallel axis). Proof. ∼ – Intersecting axis: Let A be the intersection of the two lines t1 and t2. Then P At1\ = 0 0 ∼ 00 00 P At1\ and P At2\ = P At2\, therefore P AP \ = 2t1t2\ = 2α ⇒ Rotation around A by 2α. 0 0 00 – Parallel axis: Now, we can say that d(P, t1) = d(P , t1) and d(P , t2) = d(P , t2) 0 00 00 and P, P ,P are collinear points, therefore d(P, P ) = d(t1, t2) = 2d ⇒ Translation, perpendicular to ti with 2d. Remark 1.14 In the case of the rotation, only the angle of the lines matters, in the case of the translation, only the distance and the direction matters. 2 Theorem 1.15 Every isometry of the Euclidean plane belongs to one of the following 5 groups: 1. Identity (0) 3. Rotation (2a) 5. Glide reflection (3) 2. Reflection (1) 4. Translation (2b) Proof. We have a full discussion for 2 reflections. t 1. case t1 ∩ t2 6= ∅ 2 ' t t t t' t3 1 3 1 2 (a) t1 ∩ t2 = t2 ∩ t3 Then we can consider the t2 ◦ t3 as a rotation around this point, and we can rotate the lines 0 around this center to get t1 = t2. Then t1 ◦ t2 ◦ 0 0 −1 0 0 t3 = t1 ◦ t2 ◦ t3 = t1 ◦ t1 ◦ t3 = t3. Figure 1.2: Case 1. (a) t 3 t, t 2 (b) t1 ∩ t2 6= t2 ∩ t3 2 t t3 1 First, we rotate t1 and t2 around their intersec- K t, 1 0 0 O tion O until t2⊥t3. Then we rotate t2 and t3 O 00 0 around their intersection K until t2 k t1. Then t,, 0 00 0 00 0 3 t1 ◦ t2 is a translation and t1 ◦ t2 ◦ t3 is a glide reflection. ,, K t2 t , 2. case t1 ∩ t2 = ∅ ⇒ t1 k t2 1 (a) t1 k t2 k t3 Figure 1.3: Case 1. (b) Then we can translate t2 and t3 along their per- 0 pendicular line until t1 = t2 then their compo- , t3 t sition will be the identity, therefore only t re- 3 , 3 t t 2 1 mains, so this is a translation. t t 1 2 (b) t1 k t2 , t3 t,, First, we rotate t2 and t3 around their intersec- 2 0 0 t, tion until t2⊥t1. Then we rotate t1 and t2 around 3 their intersection until t0 k t0 . Finally, we rotate 1 3 t,, 00 0 0 000 1 t2 and t3 around their intersection, until t1 k t2 . 0 000 00 Then t1 ◦ t2 ◦ t3 is a glide reflection. Figure 1.4: Case 2. (b) Remark 1.16 Two reflections are commutative to each other if their axis are orthogonal to each other. 3 1.2 Isometries of the Euclidean space Theorem 1.17 Any isometry of the space is the composition of at most four reflections in a plane. Theorem 1.18 Every isometry of the Euclidean space belongs to one of the following 7 groups: 1. Identity (0) 4. Translation (2b) 7. Screw displacement (4) 2. Reflection (1) 5. Glide reflection (3a) 3. Rotation (2a) 6. Improper rotation (3b) where 5. Glide reflection: composition of a translation and a reflection 6. Improper rotation: composition of a rotation and a reflection 7. Screw displacement: composition of a rotation and a translation Definition 1.19 The reflection in a line in the space means a rotation around this line by 180◦. This can be represented as a reflection in two orthogonal planes through the line. Lemma 1.20 The composition of two reflections in a line is either rotation or translation or screw displacement, if the two lines are intersecting, parallel or skew lines. ,,,,, Proof. P , P ,,,, Assume that e = t1 ◦ t2, f = t3 ◦ t4. P ,, 1. case: The lines e and f determine a plan P e f We choose the positions of ti such that, t2 and t3 coin- e f cide the determined plane and t1 and t4 are orthogonal ,,, P to it. Then t2 ◦ t3 is identity and if e k f, then t1 k t4, otherwise t1 , t4. Figure 1.5: 1. case 2. case: The lines e and f are skew lines Let n be the perpendicular transverse of e and f. n t1 := (e, n) ⇒ t2⊥n, t3 := (f, n) ⇒ t4⊥n. Since t2 k t4 ∧t3⊥t4 ⇒ t3⊥t2 ⇒ t1 ◦t2 ◦t3 ◦t4 = (t1 ◦t3)◦(t2 ◦t4). e But n lies both on t1 and t3 therefore they determine a rotation, and t2 k t4 then they determine a trans- f lation. The composition of these isometries is a skew displacement. Figure 1.6: 2. case 4 Lemma 1.21 The composition of two translations is also translation. Proof. Let t1 k t2 and t3 k t4 be two translation. If, t2 k t3, then it is trivial. Otherwise ◦ 0 0 t2 ∩ t3 6= ∅ ⇒ t2 ∩ t3 = e. If, we rotate t2 and t3 around e by 90 , then t2⊥t1 and t3⊥t4. 0 0 Then = t1, t2 and t3, t4 determines two reflections in a line with parallel axis, therefore it is a translation. Lemma 1.22 The composition of two rotations with orthogonal axis is a screw displace- ment. Proof. Let t0 and t00 be the axis of the rotations and n n be their perpendicular transverse line. Then t0⊥(t00, n) e , 00 0 and t ⊥(t , n). Let t1 and t2 be reflections such that t 0 0 ,, t1 ∩ t2 = t and t2 = (t , n), furthermore t3 and t4 be t 00 00 reflections such that t3 ∩t4 = t and t3 = (t , n). Since t2⊥t3 ⇒ t1 ◦ t2 ◦ t3 ◦ t4 = (t1 ◦ t3) ◦ (t2 ◦ t4). Let e and f be the intersections of t1, t3 and t2, t4 respectively. f But t1⊥t3 and t2⊥t4, therefore this isometry is can be represented as the composition of two reflections in a line, which axis are in skew position ⇒ screw displacement. Figure 1.7: Proof of Lemma 1.22 Proof of Theorem 1.18 We have a full discussion for 2 reflections.
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