13 Circles and Cross Ratio

Following up Klein’s Erlangen Program, after deﬁning the group of linear transformations, we study the invariant properites of subsets in the Riemann Sphere under this group. Appolonius’ Circle Recall that a line or a circle on the complex plane, their corresondence in the Riemann Sphere are all circles. We would like to prove that any linear transformation f maps a circle or a line into another circle or line. To prove this, we may separate 4 cases: f maps a line to a line; f maps a line to a circle; f maps a circle to a line; and f maps a circle to a circle. Also, we observe that f is a composition of several simple linear transformations of three type: 1 f1(z)= az, f2(z)= a + b and f3(z)= z because az + b g(z) f(z)= = = g(z)φ(h(z)) (56) cz + d h(z)

1 where g(z) = az + b, h(z) = cz + d and φ(z) = z . Then it suﬃces to separate 8 cases: fj maps a line to a line; fj maps a line to a circle; fj maps a circle to a line; and fj maps a circle to a circle for j =1, 2, 3. To simplify this proof, we want to write a line or a circle by a single equation, although usually equations for a line and for a circle are are quite diﬀerent. Let us consider the following single equation:

z − z1 = k, (57) z − z2

for some z1, z2 ∈ C and 0

z − z1 z ∈ C = k z − z2

the Appolonius’s circle. Remark Apollonius (262 B.C.-190 B.C.) was a Greek geometer and astronomer. His major mathematical work on the theory of conic sections had a very great inﬂuence on the development of mathematics and his famous book Conics introduced the terms parabola, ellipse and hyperbola. Apollonius’ theory of conics was so admired that it was he, rather than Euclid, who in antiquity earned the title the Great Geometer. Let us consider two cases: k = 1 and k =6 1.

88 2 2 • If k = 1, the equation (57) gives |z − z1| = |z − z2| .

=⇒ (z − z1)(z − z1)=(z − z2)(z − z2),

2 2 2 2 i.e., |z| − z1z − zz1 + |z1| = |z| − z2z + z2z + |z2| ,

2 2 i.e., −z1z − zz1 + |z1| = −z2z + z2z + |z2| ,

2 2 i.e., (z2 − z1)z +(z2 − z1)z + |z1| −|z2| =0. Substituting z = x + iy and z = x − iy into the equation, we obtain

2 2 (z2 − z1 + z2 − z1)x +(iz2 − iz1 − iz2 + iz1)y + |z1| −|z2| =0 (58)

z−z Since z + z = Re(z) and 2i = Im(z), the above coﬃcients of x and y are all real numbers. Recall Ax + By + C =0 with A,B,C ∈ R gives a line, the equation (58) gives a line in the real plane and hence in the complex space C = R2. 2 2 2 • If k =6 1, the equation (57) gives |z − z1| = k |z − z2| . 2 =⇒ (z − z1)(z − z1)= k (z − z2)(z − z2),

2 2 2 i.e., (1 − k)|z| +(−z1 + kz2)z +(−z1 + kz2)z + |z1| −|z2| =0, i.e.,

2 2 2 2 (1−k)(x +y )+(−z1 +kz2 −z1 +kz2)x+(−iz1 +ikz2 +iz1 −ikz2)y +|z1| −|z2| =0, (59)

Recall that an equation a(x2 + y2)+ bx + cy + d =0, with real numbers a =6 0,b,c and d, gives either a circle or the empty set. Geometricall by considering distance, the equation |z − z1| = k|z − z2| must have solutions. As a result, the equation (59) gives a circle in R2 or in C. Invariance of Circles Under Linear Transformations From now, by a circle in CP1, we mean either a line or a circle in C. The following theorem means that any circle in CP1, under a linear transformation, is mapped into again a circle in CP1.

89 Theorem 13.1 Any linear transformation f maps a circle of Cˆ into a circle of Cˆ.

Proof: By Remarks below (55), it suﬃcient to prove this theorem for special linear trans- 1 formations: f(z)= z + b, f(z)= az and f(z)= z . (i) Consider f(z) = z + b, i.e., z 7→ w = f(z) = z + b. It is equivalent to z = w − b. z−z1 k Substituting this into the Applollonia equation (57) z−z2 = , we obtain

w − b − z1 = k, w − b − z2

i.e., w − (b + z1) = k w − (b + z2)

which is again Applonia equation. Therefore, as we have proved, this equation gives a circle in Cˆ. w (ii) Consider f(z)= az, i.e., z 7→ w = f(z)= az. It is equivalent to z = a . Substituting z−z1 k this into the Applollonia equation (57) z−z2 = , we obtain

w a − z1 w = k, − z2 a i.e., w − az1 = k w − az2

which is again Applonia equation. Therefore, as we have proved, this equation gives a circle in Cˆ. 1 1 w (iii) Consider f(z)= z , i.e., z 7→ w = f(z)= z . It is equivalent to z = a . Substituting this into the Applollonia equation (57), we obtain 1 w − z1 1 = k, − z2 w i.e., 1 − z1w = k. 1 − z2w

If z1 =6 0 and z2 =6 0, we get 1 w z1 − |z2| 1 = k, − w |z1| Z2

90 which is again Applonia equation. If z1 = 0(⇒ z2 =6 0), we have

1 w 1 = k, − z2 w i.e., 1 1 1 = k =⇒ w − = . |1 − z2w| z2 k|z2|

This is a circle in C by considering distance. If z2 = 0(⇒ z1 =6 0), we have

1 − z1 w = k, 1 w i.e., 1 k |1 − z1w| = k =⇒ w − = . z1 |z1|

This is a circle in C by considering distance.

Cross ratio For any four distinct points z1, z2, z3, z4 ∈ Cˆ, the cross ratio of z1, z2, z3 and z4 is deﬁned by (z1 − z3)(z2 − z4) (z1, z2, z3, z4) := . (z1 − z4)(z2 − z3)

z2−z4 z1−z3 z1 z1, z2, z3, z4 z2 z1, z2, z3, z4 Notice: If = ∞, then ( ) := z2−z3 . If = ∞, then ( ) := z1−z4 . If z2−z3 z1−z3 z3 z1, z2, z3, z4 z4 z1, z2, z3, z4 = ∞, then ( ) := z1−z4 . If = ∞, then ( ) := z2−z3 .

Proposition 13.2 For any linear transformation f, and for any four distinct points z1, z2, z3, z4 ∈ Cˆ, we have (z1, z2, z3, z4)=(f(z1), f(z2), f(z3), f(z4)). Namely, cross ratio is invariant under linear transformation.

Proof: Deﬁne a linear transformation

(z − z3)(z2 − z4) g(z) := (z, z2, z3, z4)= . (z − z4)(z2 − z3) and (f(z) − f(z3))(f(z2) − f(z4)) h(z) := (f(z), f(z2), f(z3), f(z4)) = . (f(z) − f(z4))(f(z2) − f(z3))

91 Since g(z2)= h(z2)=1,g(z3)= h(z3) = 0 and g(z4)= h(z4)= ∞, by Theorem 12.2 and its uniqueness, we conclude g = h so that g(z1)= h(z1), i.e.,

(z1, z2, z3, z4)=(f(z1), f(z2), f(z3), f(z4)).

Remark: From the above proof, given any three distinct numbers z2, z3, z4 ∈ Cˆ, the function g(z) := (z, z2, z3, z4) deﬁnes a linear transformation satisfying g(z2)=1,g(z3) = 0 and g(z4)= ∞. This is a way to construct linear transformations.

Proposition 13.3 For any four distinct points z1, z2, z3, z4 ∈ Cˆ, they lie in a circle (i.e., a stright line or a circle in C) if and only if the cross tratio (z1, z2, z3, z4) is a real number.

Proof: Since the image of circle by a linear transformtaiton is still a circle, choose a linear transformation f such that f(z2)=1, f(z3)=0, f(z4)= ∞. Then

z1, z2, z3, z4 lie in a circle ⇐⇒ f(z1), f(z2), f(z3), f(z4) lie in the x-axis. ⇐⇒ f(z1) is real ⇐⇒ (f(z1), f(z2), f(z3), f(z4)) is real ⇐⇒ (z1, z2, z3, z4) is real

The last equivalence comes from Proposition 13.2. . Non-Euclidean geometry In the Elements, Euclid began with a limited number of assumptions (23 definitions, five common notions, and five postulates) and sought to prove all the other results (propositions) in the work. The most famous part of The Elements is the following five postulates:

1. A straight line segment can be drawn joining any two points.

2. Any straight line segment can be extended indeﬁnitely in a straight line.

3. Given any straight line segment, a circle can be drawn having the segment as radius and one endpoint as center.

4. That all right angles are equal to each other.

92 5. If two lines are drawn which intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect each other on that side if extended far enough (see above picture). This postulate is equivalent to what is known as the parallel postulate, that is, given a traight line L and a point A not in the line, there exists a unique straight line L′ that is parallel to L and contains the point A.

It is apparent that the fifth postulate, which can also be called “Euclid’s Fifth Postulate”, is radically different from the first four. Euclid was not satisfied with it, so he tried to avoid using it in The Elements. In fact, the first 28 theorems in The Elements are proved without using the fifth postulate. Many mathematicians attempted to find a simpler form of this property, or to prove the fifth postulate from the other four. Over a period of at least a thousand years, regardless of the forms of the postulate that have been found, it consistently appears to be more complicated than Euclid’s other postulates. János Bolyai(1802-1860), Nicolai Lobachevsky(1792-1856) and Carl Friedrich Gauss(1777- 1855) discovered that, without Euclid’s Fifth Postulate, there is still a geometry. It is now called the Non-Euclidean geometry. In modern language, Euclidean geometry is a space with curvature zero, and non-Euclidean geometry is a space with curvature −1. A model of non-Euclidean geometry in the interior of a circle was developed by Henri Poincaréin 1882. In this model, straight lines are represented by arcs of circles that are orthogonal to the boundary circle. Parallel lines are then represented by circular arcs that intersect at the boundary. This model has the advantage that angles between circles are measured in the Euclidean way. It is this model that convinced mathematicians by the end of the century that non-Euclidean geometry was as valid as Euclid’s. Now we consider the space to be the unit disk ∆(1). The group of automorphisms of the unit disk is a subgroup of the group of linear transformations. Fixing three distinct points z1, z2, z3 in the circle ∂∆(1), for any other triple of distinct three points (w1,w2,w3) in ∂∆, by Theorem 12.2, there is a unique linear transformation which sends the circle to itself. These linear transformations form the automorphism group of ∆(1). By Poincaré’s model, a “straight line” is mapped into another “straight line”, under such linear transformation. Here we used a property, which will be proved late, that the angle of two circles will be preserved under linear transformation. Thus since a “straight line” is a circle perpendicular to the circle ∂∆(1), its image is also a circle perpendicular to the circle ∂∆(1). Therefore the image of a “stright line” is a “straight lines” again under automorphisms of ∆(1).