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16 the Complex Projective Line 16 The complex projective line Now we will to study the simplest case of a complex projective space: the complex projective line. We will see that even this case has already very rich geometric interpretations. The close relation of complex arithmetic operation allows us to express geometric properties by nice algebraic structures. In par- ticular, this case will be the first example of a projective space in which we will be properly able to deal with circles. 1 16.1 CP Let us recall how we introduced the real projective line. We took the one- dimensional space R considered it as a euclidean line and added one point at infinity. Topologically we obtained a circle. The best way to express the elements of the real projective line algebraically was to introduce homogeneous coordinates. For this we associated to each number x R the vector (x, 1)T and identified non-zero scalar multiples. Finally we ident∈ified the vector (1, 0)T (and all its non-zero multiples) with the point at infinity. We ended up with the space 2 T 1 R (0, 0) RP = − { }. R 0 − { } We will do exactly the same for the complex numbers! To obtain the complex projective line we start with all the numbers in C. We associate every number z C with the vector (z, 1)T and identify non-zero scalar multiples. By this we∈ associate all vectors of the form (a, b)T ; b = 0 to a number in # C. What is left is the vector (1, 0)T and all its non-zero multiples. They will 1 represent a unique point at infinity. All in all we obtain the space CP defined by 2 T 1 C (0, 0) CP = − { }. C 0 − { } 248 16 The complex projective line This space is isomorphic to all all complex numbers together with one point at infinity. What is the dimension of this space? In a sense this depends on the point of few. From the perspective of real numbers already the complex plane C is a two-dimensional object, since it requires two real parameters to specify the 1 objects of C. Hence one would say that also CP is a real-two-dimensional object since it differs from C just by one point. On the other hand from a complex perspective C is just a one-dimensional object. It contains just one (complex) parameter. As R is the real number line one could consider C as 1 the complex number line. Thus C is complex-one-dimensional, and so is CP . 1 This is the reason why we call the space CP the complex projective line! There is another issue important to mention in this context. Identifying vectors that only differ by a non-zero multiple this time also includes mul- tiplication by complex numbers. Thus (1, 2)T , (3i, 6i)T , (2 + i, 4 + 2i)T all represent the same point. For every point represented by (a, b)T with b = 0 # we can reconstruct the corresponding number of C by multiplication with 1/b. The dehomogenized number is then a/b C. We will also frequently identify 1 ∈ CP with the space C := C . As in the case of real numbers we use the standard rules for arithmeti∪c {o∞pe}rations with : ∞ ! 1/ = 0; 1/0 = ; 1 + = ; + = . ∞ ∞ ∞ ∞ ∞ ∞ ∞ 16.2 Testing geometric properties We will know consider identify the finite part of the complex projective (i.e. C) with the Euclidean plane R2 and investigate how certain geometric properties can be expressed in terms of algebraic expressions in C. Most of these tests will however not correspond to projectively invariant properties. We will first express the property that two vectors associated to two com- plex numbers z1 and z2 point into the same (or in the opposite) direction. For this we simply have to calculate the quotient z1/z2. Using polar coordinates we get (if z = 0) 2 # iψ1 iψ2 i(ψ1 iψ2) z1/z2 = (r1e /r2e ) = (r1/r2)e − . If the two vectors point in to the same direction we have ψ1 = ψ2 and the above expression is a positive real number. If they point in opposite directions iπ we have ψ = π + ψ and (since e− = 1) the above number is real and 1 2 − negative. In other word we could say that in the complex plane 0 z1 and z2 are collinear if the quotient z1/z2 is real (provided z2 = 0). Using complex conjugation we can even turn this into an equality, since# z is real if and only if z = z. We get z and z point into the same or opposite direction z /z = z /z . 1 2 ⇐⇒ 1 2 1 2 16.2 Testing geometric properties 249 C PSfrag replacements B PSfrag replacements z2 A B z1 C z1 z2 z = r eiψ A · Fig. 16.1. Testing simple geometric properties. We can use a similar test to decide whether three arbitrary distinct num- bers correspond to collinear points. Let A, B and C be three points in the complex plane. We represent these points by their corresponding complex numbers. Then these three points are collinear if the quotient (B A)/(C A) is real (provided the denominator does not vanish). This can is a−n immed−iate consequence of our previous considerations since this quotient simply com- pares the directions of the vectors B A and C A. Analogously to the last statement we get: − − A, B, C are collinear (B A)/(C A) = (B A)/(C A). ⇐⇒ − − − − Unfortunately, the last two expressions do not fit into our concepts of projective invariants. The next one, however, will. We will describe whether four points lie commonly on a circle. For this we first need a well known theorem of elementary geometry: the peripheral angle theorem. This theorem states that if we have a circle and a secant from A to B on this circle. Then all points on the circle on one side of the secant “see” the secant under the same angle. The two (invariant) angles on the left and on the right side of the secant sum up to an angle of π. Figure 16.2 illustrates this theorem. We omit a proof here (it can be found in many textbooks of elementary geometry) but we formulate the theorem on the level of complex numbers. For this we denote by ∠A(B, C) the counterclockwise angle between the vectors B A and C A. We can summarize both cases of the peripheral angle theorem−by measuri−ng angle differences modulo multiples of π. In such a version the peripheral angle theorem reads: Theorem 16.1. Let A and B, C, D be four points on a circle embedded in the complex plane. Then the angle difference ∠C(A, B) ∠D(A, B) is a multiple of π − We postpone a proof of this version of the peripheral angle theorem until we have the possibility to prove it by a simple projective argument. If C and D 250 16 The complex projective line α α C α D α PSfrag replacements PSfrag replacements A α B π α − A C π α − π α B D − Fig. 16.2. The peripheral angle theorem. are on the same side of the secant through A and B both angles are equal and the difference is 0 π. If they are on opposite sides of the secant the difference is +π or π depen· ding on the orientation. We wi−ll interpret this theorem in the light of complex numbers. The angle ∠C (A, B) can be calculated as the angle ψ1 of the following complex number C A − = r eiψ1 . C B 1 − Similarly the angle ∠D(A, B) can be calculated as angle of D A − = r eiψ2 . D B 2 − We can get the difference of the angles simply by dividing these two numbers. We get C A D A i(ψ ψ ) − − = (r /r )e 1− 2 . C B D B 1 2 − − Since the angle difference is"a multiple of π by the peripheral angle theorem this number must be real. Now, the amazing fact is: the expression on the left is nothing else but a cross-ratio in the complex projective plane. Thus we can say that if the four points are on a circle, this cross-ratio (A, B; C, D) is a real number. The converse is also “almost” true we only have to include the special case that the circle may have infinite radius and degenerates to a line. It is easy to check that if A, B, C, D are collinear the cross-ratio is real, as well. In our considerations we have to consider as real number as well. The cross-ratio assumes this value if either C = B o∞r D = A. All in all we obtain the following beautiful theorem that highlights the close relation of complex projective geometry and the geometry of circles. 1 Theorem 16.2. Four points in CP are cocircular or collinear if and only if the cross-ratio (A, B; C, D) is in R . ∪ {∞} 16.3 Projective transformations 251 16.3 Projective transformations 1 A projective transformation in CP can (as in the real case) be expressed T 2 by a matrix multiplication. If the vector (z1, z2) C represents a point in 1 ∈ CP by homogeneous coordinates. Then a projective transformation can be expressed as 1 1 τ: CP CP → z1 a b z1 z2 )→ c d z2 # $ # $# $ As usual the matrix must be non-degenerate.
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