Starting with numerical algorithms resulting in new kinds of amazing fractal patterns on the sphere, this book describes the theory underlying these phenomena and indicates possible future applications. The book also explores the following questions:

q(x) q(x) = (x1 )2 + (x2 )2 + (x3 )2 − (x4 )2 (2.47) = −(x ) + (x ) + (x ) + (x ) . 0 2 1 2 2 2 3 2 The form q(x) can be also written in a matrix form as q(x) = xT η x, where, if x denotes a column vector, then xT = (x1 , x2 , x3 , x4 ), ⎛ ⎞ 1 0 0 0 ⎜0 1 0 0 ⎟ ⎟ η=⎜ ⎝0 0 1 0 ⎠ . 0 0 0 −1 (2.48) (2.49) We say that the metric tensor η has signature (1, 1, 1, −1) or (3, 1). More popular among physicists is the opposite convention: one plus, three minuses. This other choice is particularly

ξ) = (A∗ ξ, XA∗ ξ) = (ξ , Xξ ), where ξ = A∗ ξ, is also positive. Therefore X is positive deﬁnite as well. Since det(X ) = 0 det(X) = 0, it follows that the trace of X is positive, therefore x > 0. Topology of SO+ (3, 1). The image of SL(2, C) under the group homomorphism SL(2, C) → SO+ (3, 1) not only preserves the time direction, but also does not contain space inversions, that is that all the matrices L(A) have determinant +1. This property follows from the fact that SL(2, C), as a topological

where 0 < k < 1, n = (n1 , n2 , n3 ) is a unit vector, n2 = 1, and s = (s1 , s2 , s3 ) are Pauli’s spin matrices — cf. Eq. (2.76). Proof. Let us write X = xμ sμ . Then, owing to the fact that X is positive, Tr (X) = 2x0 > 2, we have that x0 > 2. From 1 = det(X) = (x0 )2 − x2 it follows that x (2.263) n= 0 (x )2 − 1 is a unit vector. Let us deﬁne k by the formula: k= (x0 )2 − 1 . x0 (2.264) 1 . 1 − k2 (2.265) Then 0 < k < 1 and x0 = √ The rest is a simple algebra veriﬁcation. Remark 2.13.

or w(i)1 a + w(i)2 b + w(i)3 c + 2d ≥ 0. (2.344) Consider ﬁrst the case √ Eq. (2.344), we get d ≥ a+b. √ of i = 1. For i = 1, form Since d > 0 and d = a2 + b2 + c2 , we have a2 + b2 + c2 ≥ a+b, therefore a2 + b2 + c2 ≥ (a + b)2 , or c2 ≥ 2ab. On the other hand application of the inverse matrix, h(6) = h(1)−1 , leads to d ≥ a − b, which, by the same reasoning, implies c2 ≥ −2ab. Therefore c2 ≥ 2|ab|. Doing the same for the pairs (h(2), h(5)) and (h(3), h(4)) we obtain the following system of

straightforward calculation to ﬁnd out that substituting x in 1 + k 2 + 2k · x for w−1 (x) from Eq. (3.35) gives (1 − k2 )2 /(1 + k 2 − 2k · x). Therefore we get: pi wi−1 (x) = 1 1 − k2 · . n(1 + k 2 ) 1 + k 2 − 2ki · x (3.44) pg. 128/3 May 29, 2014 10:6 World Scientific Book - 9in x 6in QuantumFractals3 Examples 129 On the other hand, substituting −k for k in Eq. (3.43), we get dμ wi −1 (x) = dμ(x) 1 − k2 1 + k 2 − 2k · x 2 . (3.45) Therefore we obtain the following result: