Math 3153 { The Euclidean Group E(3) Prof. Philip Pennance1 -Version: Feb 19, 2016 1. F : IR3 ! IR3 is called: (b) AAT = I (a) An isometry (written F 2 E(3)) if Proof: Suppose that F 2 O(3). Recall that jF (p) − F (q)j = jp − qj: the i'th column of A is given by 3 for all p; r 2 IR . i T A = F (ei) : (b) An orthogonal transformation (written F 2 O(3)) if Hence • F linear. δ = he ; e i •h F (p);F (q)i = hp; qi ij i j = hF (ei);F (ej)i for all p; r 2 IR3. = h(Ai)T ; (Aj)T i (c)A rotation about the origin written = (Ai)T Aj + F 2 O (3) if T j = (A )iA • F is orthogonal. T = (A A)ij • det F = 1. T 2. Clearly O+(3) ⊆ O(3) ⊆ E(3). It will Hence A A = I. Conversely note be shown that E(3) is a group (the that the preceeding argument can be re- Euclidean group under composition and versed. that O+(3) ⊆ O(3) are subgroups. 7. Corollaries: Let F 2 O(3). Let A be 3. Exercise: The composition of two the matrix of F relative to an orthonor- isometries is an isometry. mal basis. 4. Exercise: Let u be a vector in IR3. De- (a) F is invertible. 3 3 fine Tu : IR ! IR by (b) A−1 = AT and so AAT = I Tu(x) = x + u (c) det A = ±1 (d) The columns and rows of A are or- Show that Tu is an isometry. thonormal vectors. 5. Exercise: Show that 8. Exercise: Show that O(3) is a group. 3 T (3) = fTu : u 2 IR g 9. Lemma: Show that T (3) is a group (the transla- Let F 2 E(3). If F (0) = 0 then tion subgroup). F 2 O(3) Proof: 6. Claim: Let F 2 IR3 ! IR3 be linear. We must show: Let A be the matrix of F with respect to an orthonormal basis (e1; e2; e2). The (a) F linear. following are equivalentl (b) hF (p);F (q)i = hp; qi, (a) F 2 O(3) for all p; q 2 IR3. 1 http://pennance.us Proof of (b). 10. Main Claim: Let F 2 E(3), there exist hF (p) − F (q);F (p) − F (q)i T 2 T (3) and M 2 O(3), both unique, such that = jF (p) − F (q)j2 = jp − qj2 F = T ◦ M (2) = hp − q; p − qi Proof Hence Let F 2 O(3). Consider the translation −1 jF (p)j2 + jF (q)j2 − 2hF (p);F (q)i T = TF (0). Then T ◦ F (0) = 0 By lemma (9), M := T −1 ◦ F 2 O(3) and = jpj2 + jqj2 − 2hp; qi so F = T ◦ M (1) Uniqueness: But F (0) = 0 and so for all p 2 IR3 Suppose jF (p)j = jF (p) − F (0)j = jp − 0j F = T ◦ M = jpj = T 0 ◦ M 0 and so (1) yields 0 0 hF (p);F (q)i = hp; qi where T;T 2 T (3) and M; M 2 O(3) then Proof of (a): T −1 ◦ F (0) = 0 Let (e1; e2; e3) be an orthonormal basis. By part (a) F preserves the inner prod- T 0−1 ◦ F (0) = 0 uct and so (F (e1);F (e2);F (e3)) is also orthonormal. Hence if x; y 2 IR3 and and so T (0) = T 0(0): Since T and T 0 c 2 IR are translations it must be that T = T 0 from which M = M 0. X F (x) = hF (x);F (ei) iF (ei) 11. Exercise: Show that E(3) is a group. X = hx; ei iF (ei) Hint: Let F 2 E(3). If F = T ◦ M, then F −1 = M −1 ◦ T −1. Show that and similarly F −1 2 E(3). X 12. Exercise: T(3) is a normal subgroup of F (y) = hx; ei iF (ei) X E(3), i.e., for any translation T and any F x + y) = hx + y; ei iF (ei) isometry F , we have F −1TF 2 T (3). X F (cx) = hcx; ei iF (ei) 13. Remark: Uniqueness in (2) shows that E(3) is the semidirect product of the Linearity of F now follows from the bi- translation subgroup T (3) and the or- linearity of the scalar product. thogonal group O(3). 2.