By Vladimir Kanovei

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Then the point in P with position vector c = c2 v2 + c3 v3 satisfying (R | t)c = c is found by solving the equation Rc + t = c, or equivalently (I3 − R)c = t. In matrix form this becomes      0 0 0 0 0 v1 v2 v3 0 1 − cos θ sin θ  c2  = v1 v2 v3 t2  , 0 − sin θ 1 − cos θ c3 t3 2. ISOMETRIES OF 3-DIMENSIONAL SPACE 59 or equivalently, 1 − cos θ sin θ − sin θ 1 − cos θ c2 t = 2 . c3 t3 Since det 1 − cos θ sin θ = (1 − cos θ)2 + sin2 θ = 2(1 − cos θ), − sin θ 1 − cos θ this determinant is non-zero if R = I3 , and the equation then has the unique solution c2 1 − cos θ sin θ = c3 − sin θ 1 − cos θ −1 t2 , t3 using essentially the same algebra in the 2-dimensional situation of rotation about a point.

We need to make this notion more precise. A dilation or scaling of the plane is a function H : R2 −→ R2 which has the form H(x) = δ(x − c) + c, where δ > 0 is the dilation factor and c is the centre of the dilation. It is easy to see that H(c) = c and |H(x) − c| = δ|x − c|, so the effect of H is the expand or contract the plane radially from the point c. We can rewrite the above formula to give H(x) = δx + (1 − δ)c, so we can express H as the Seitz symbol (δI | (1 − δ)c). Of course, if δ = 1 then this is just the identity function, otherwise it is not an isometry.

7. FRIEZE PATTERNS AND THEIR SYMMETRY GROUPS 33 We also have ρ4 ◦ Transt4 ◦ρ4 = Trans−t4 , so the symmetry group here is again a dihedral group with generators α = Transt4 , ··· • • ρ4 × β = ρ4 . t4 • × ρ4 G ··· Pattern 5. Let t5 be a smallest translation vector. There is a glide reflection γ5 whose square is γ5 ◦ γ5 = Transt5 . There is also a reflection symmetry σ5 in the vertical line L. Let Lv denote the line obtained by translation of L by a vector v. Then the half rotation ρ5 about one of the points marked • at (−1/4)t5 away from L agrees with the composition ρ5 = γ5 ◦ σ5 .

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