ROTATIONS OF THE RIEMANN SPHERE

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1 ROTATIONS OF THE RIEMANN SPHERE A rotation of the sphere S is a map r = r p,α described by spinning the sphere (actually, spinning the ambient space R 3 ) about the line through the origin and the point p S, counterclockwise through angle α looking at p from outside the sphere. (See figure 1.) p α Figure 1. The rotation r p,α Thus r is the linear map that fixes p and rotates planes orthogonal to p through angle α. Let q be a unit vector orthogonal to p. Then the matrix of r is (viewing p and q as column vectors) m r = p q p q 1 cosα sinα p q p q 1. sinα cosα The set Rot(S ) of such rotations forms a group, most naturally viewed as a subgroup of GL 3 (R). Showing this requires some linear algebra. Recall that if m M 3 (R), meaning that m is a 3-by-3 real matrix, then its transpose m t is obtained by flipping about the diagonal. That is, m t ij = m ji for i,j = 1,,3. The transpose is characterized by the more convenient condition where, is the usual inner product, The matrix m is orthogonal if mx,y = x,m t y for all x,y R 3, x,y = x i y i. m t m = I, or, equivalently, if m preserves inner products, mx,my = x,y for all x,y R 3. 1

2 ROTATIONS OF THE RIEMANN SPHERE The orthogonal matrices form a group O 3 (R) GL 3 (R), and the special orthogonal matrices, SO 3 (R) = {m O 3 (R) : detm = 1}, form a subgroup of index. With these facts in place it is not hard to prove that Rot(S ) forms a group, and that Theorem.1. As a subgroup of GL 3 (R), Rot(S ) = SO 3 (R). Here is a sketch of the proof. Given a rotation r = r p,α, its matrix, m r = p q p q 1 cosα sinα p q p q 1, sinα cosα is readily verified to be special orthogonal. On the other hand, take any special orthogonal matrix m. Since 3 is odd, m has a real eigenvalue λ. Any real eigenvalue λ with eigenvector p satisfies λ p,p = λp,λp = mp,mp = p,p, i.e., λ = ±1. Sincedetm = 1, andthedeterminantistheproductoftheeigenvalues, and any imaginary eigenvalues occur in conjugate pairs, m in fact has 1 for an eigenvalue with unit eigenvector p. Take any nonzero vector q perpendicular to p. Some rotation r = r p,α takes q to mq and has matrix m r SO 3 (R). Thus the matrix m 1 r m lies in SO 3 (R) and fixes both p and q. It is therefore the identity, showing that m = m r is a rotation matrix. A rotation of the Riemann sphere Ĉ is a map f : Ĉ Ĉ corresponding under stereographic projection to a true rotation r of the round sphere S. In other words, the following diagram commutes: Let S π Ĉ r S f Rot(Ĉ) denote the set of such rotations. Since Rot(S ) forms a group, Rot(Ĉ) forms an isomorphic group under r π r π 1. Since any rotation r is conformal on S, the corresponding bijection f is conformal on Ĉ and is therefore an automorphism, and so Rot(Ĉ) is a subgroup of Aut(Ĉ). With some more linear algebra we can describe Rot(Ĉ) explicitly as a subgroup of PSL (C). If m M (C) is a -by- complex matrix then its adjoint is Ĉ. m = m t, where the overbar denotes complex conjugation, i.e., m ij = m ji for i,j = 1,. The adjoint is characterized by the condition mx,y = x,m y for all x,y C, π

3 ROTATIONS OF THE RIEMANN SPHERE 3 where now, is the complex inner product x,y = x i y i. The role of the adjoint in the algebra of complex matrices is analogous to the role of the conjugate in the algebra of complex numbers. The matrix u is unitary if u u = I. (This condition generalizes the unit complex numbers.) Equivalently, ux,uy = x,y for all x,y C. The unitary matrices form a group U (C). The special unitary matrices SU (C) = {u U (C) : detu = 1} form a subgroup. A matrix is special unitary if and only if it takes the form a b u =, a + b = 1. b a The projective unitary group is PU (C) = U (C)/(U (C) C I), and the projective special unitary group is PSU (C) = SU (C)/(SU (C) C I) = SU (C)/{±I}. There is an isomorphism PU (C) = PSU (C), and the group PSU (C) can be more convenient to work with since its elements are two-element cosets {±u}. Theorem.. As a subgroup of PSL (C), Rot(Ĉ) = PSU (C). Hereisan elegant proof, which incidentally shows that Rot(Ĉ)is agroupwithout reference to SO 3 (R). We show first that any rotation lies in PSU (C), second that any element of PSU (C) is a rotation. A short calculation shows that if the antipodal pair p, p S \ {n,s} have stereographic images z,z C, then z = 1/z, where the overbar is complex conjugation. Now let f be a rotation of Ĉ induced by a rotation r of S. Let a matrix describing f be a b m f =, det(m c d f ) = 1. Since r takes antipodal pairs to antipodal pairs, f must satisfy the corresponding relation f(z ) = f(z) for all z C\{}. This condition is that for some λ C, d = λa, a = λd, c = λb, b = λc. These relations and the relation ad bc = 1 combine to show that λ = 1 and therefore m f PSU (C). For the converse, let f have matrix m f = a b b a PSU (C). If f() = then f(z) = e iα z for some α, so f is a rotation. If f() = z then some rotation f z Rot(Ĉ) PSU (C) also takes to z, and so the composition

4 4 ROTATIONS OF THE RIEMANN SPHERE g = f 1 z f PSU (C) fixes and is thus a rotation. Therefore f = f z g is also a rotation, and the proof is complete. The two theorems combine to show that PSU (C) = SO 3 (R). The next result says how to compute in PSU (C) while thinking of Rot(S ). For any rotation r p,α of S, let f π(p),α denote the corresponding rotation of Ĉ. Theorem.3. Let p = (p 1,p,p 3 ) S and let α R. Then cos α f π(p),α = +ip 3sin α p sin α +ip 1sin α p sin α +ip 1sin α cos α ip 3sin α Here is the proof. Either by geometry or by a calculation using the commutative diagram from earlier, the rotation r n,α of S induces the automorphism f,α (z) = e iα z of Ĉ, i.e., under a slight abuse of notation, e iα/ f,α = e iα/. Next consider the rotation r (,1,), of S counterclockwise about the positive x -axis through angle. We will find the corresponding rotation f i, of Ĉ. A rotation r of S takes (,1,) to n and (, 1,) to s; the corresponding rotation f of Ĉ takes i to and i to, so it takes the form f(z) = k z +i z i for some nonzero constant k. Since r (,1,), = r 1 r n, r, the corresponding result in Rot(Ĉ) is f i, = f 1 f, f, or Thus for all z Ĉ, The k cancels, leaving f f i, = f, f. k fi,(z)+i f i, (z) i = ei k z +i z i. e i/ (f i, (z)+i)(z i) = e i/ (f i, (z) i)(z +i), and some algebra gives cos f i, = sin sin cos. Now let the point p S have spherical coordinates (1,θ,), meaning that cosθ = p 1 / p 1 +p, sinθ = p / p 1 +p, cos = p 3, sin = p 1 +p. (See figure.) To carry out r p,α, move p to the north pole via rotations about the north pole and (,1,), rotate about the north pole by α, and restore p; to wit, r p,α = r n,θ r (,1,), r n,α r (,1,), r n, θ..

5 ROTATIONS OF THE RIEMANN SPHERE 5 The corresponding rotation of Ĉ is e iθ/ cos f π(p),α = e iθ/ sin e iα/ sin cos e iα/ cos sin e iθ/ sin cos e iθ/. Multiplying this out and using a little trigonometry gives the result. p θ Figure. Spherical coordinates

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