MATH 132 (19F) (TA) A. Zhou Complex Analysis for Applications A THE RIEMANN SPHERE A.1 STEREOGRAPHIC PROJECTION The points of the complex plane can be (mostly) associated with points on the sphere. To make 2 2 2 2 3 this more precise, consider the unit sphere S = (x1; x2; x3) x1 + x2 + x3 = 1 sitting in R , and identify the plane x3 = 0 with the complex plane C via (x1; x2; 0) $ x1 + x2i. Let N = (0; 0; 1) denote the north pole of the unit sphere. Given a point P 2 S2 with P 6= N, we obtain a complex 3 number φN (P ) by intersecting line NP with C ⊂ R . Definition A.1.1: Stereographic projection 2 The map P 7! φN (P ) from S nfNg to C is stereographic projection through the north pole. This map is invertible, with inverse given by taking a complex number z, drawing the line through N and z, and taking the intersection point of this line with S2 that is not N. The inverse map is also called stereographic projection; it will be clear from context which direction is meant. 2 Given P = (x1; x2; x3) 2 S with P 6= N, the line NP has parametric form (tx1; tx2; tx3 + (1 − t)). This intersects C when tx3 + (1 − t) = 0, so t = 1=(1 − x3). (This is defined, because P 6= N means that x3 6= 1.) It thus follows that x1 x2 φN (P ) = + · i: (A.1) 1 − x3 1 − x3 −1 For the inverse map, let z = x + yi 2 C. If φN (z) = (x1; x2; x3), then x1=(1 − x3) = x and −1 x2=(1 − x3) = y, so x1 = x(1 − x3) and x2 = y(1 − x3). Since φN (z) lies on the unit sphere, we 2 2 2 have x1 + x2 + x3 = 1, so 2 2 2 2 2 x (1 − x3) + y (1 − x3) + x3 = 1; (A.2) which rearranges to 2 2 2 2 2 2 2 (x + y + 1)x3 − 2(x + y )x3 + (x + y − 1) = 0: (A.3) One solution is x3 = 1, corresponding to the fact that the line through N and z does intersect the unit sphere at N, and the other point of intersection therefore has x2 + y2 − 1 jzj2 − 1 x = = : (A.4) 3 x2 + y2 + 1 jzj2 + 1 Hence 2x z + z x = x(1 − x ) = = (A.5) 1 3 jzj2 + 1 jzj2 + 1 and 2y z − z x = y(1 − x ) = = ; (A.6) 2 3 jzj2 + 1 i(jzj2 + 1) so we have the formula z + z z − z jzj2 − 1 φ−1(z) = ; ; : (A.7) N jzj2 + 1 i(jzj2 + 1) jzj2 + 1 17 A.1 Stereographic projection MATH 132 (19F) 2 Both maps are continuous, so φN gives a homeomorphism of S nfNg and C. If P ! N, then 2 2 2 2 x1 + x2 1 − x3 1 + x3 jφN (P )j = 2 = 2 = ! 1 (A.8) (1 − x3) (1 − x3) 1 − x3 − 2 because x3 ! 1 , so φN (P ) ! 1. Therefore, we can extend φN continuously to all of S by introducing a point at infinity, denoted 1, and declaring φN (N) = 1. Definition A.1.2: Extended complex plane and Riemann sphere The extended complex plane is the set C1 = C [ f1g, where 1 denotes a point at infinity. 2 1 The stereographic projection φN : S ! C is defined as above. In the context of being identified with the extended complex plane, we call S2 the Riemann sphere. 1 2 The topology on C is borrowed from S so that φN remains a homeomorphism. More concretely, the open neighborhoods of 1 correspond to open neighborhoods of N 2 S2. The complements of open neighborhoods of N are closed subsets of S2, hence compact, and they do not contain N, so their images are compact subsets of C. Therefore, the open neighborhoods of 1 2 C1 are complements of compact subsets of C ⊂ C1. Problems 1. Let S denote the south pole of the Riemann sphere. What is φN (S)? 2. Show that the circles on the Riemann sphere correspond bijectively to the circles and lines in C. Which circles on the Riemann sphere correspond to lines? 3. Let denote stereographic projection through the south pole S, defined in the same way as stereographic projection through the north pole but with lines through S instead of N. Show −1 that ( ◦ φN )(z) = 1=z. Since conjugation poses problems for complex analysis, we usually compose stereographic projection through the south pole with conjugation to give the map −1 φS = , and then (φS ◦ φN )(z) = 1=z is holomorphic for z 6= 0. 18.