10. Classifying Möbius transformations: conjugacy, trace, and applications to parabolic transformations

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1 10. Classifying Möbius transformations: conjugacy, trace, and applications to parabolic transformations 10.1 Conjugacy of Möbius transformations Before we start discussing the geometry and classification of Möbius transformations, we introduce a notion of sameness for Möbius transformations. Definition. Let γ 1 and γ 2 be two Möbius transformations. We say that γ 1 and γ 2 are conjugate if there exists another Möbius transformation g such that γ 1 = g 1 γ 2 g. Remarks. (i) Geometrically, if γ 1 and γ 2 are conjugate then the action of γ 1 on H H is the same as the action of γ 2 on g(h H). Thus conjugacy reflects a change in coordinates of H H. (ii) If γ 2 has matrix A 2 SL(2, R) and g has matrix A SL(2, R) then γ 1 has matrix ±A 1 A 2 A. 1

2 Exercise 10.1 (i) Prove that conjugacy between Möbius transformations is an equivalence relation. (ii) Show that if γ 1 and γ 2 are conjugate then they have the same number of fixed points. Hence show that if γ 1 is hyperbolic, parabolic or elliptic then γ 2 is hyperbolic, parabolic or elliptic, respectively The trace of a Möbius transformation Recall that if A is a matrix then the trace of A is defined to be the sum of the diagonal entries of A. That is, if A = (a, b; c, d) then Trace(A) = a + d. Let γ(z) = (az + b)/(cz + d) be a Möbius transformation. By dividing the coefficients a, b, c, d by ad bc, we can always write γ in normalised form. Assume that γ is written in normalised form. Then we can associate to γ a matrix A = (a, b; c, d); as ad bc = 1 we see that A SL(2, R). However, as we saw in Lecture 9, this matrix is not unique; instead we could have associated the matrix A = ( a, b; c, d) to γ. Thus we can define a function τ(γ) = (Trace(A)) 2 = (Trace( A)) 2. Definition. Let γ Möb(H) be a Möbius transformation with γ(z) = (az+b)/(cz+d) where ad bc = 1. We call τ(γ) = (a+d) 2 the trace of γ. The following result says that conjugate Möbius transformations have the same trace. 2

3 Proposition 10.1 Let γ 1 and γ 2 be conjugate Möbius transformations. τ(γ 1 ) = τ(γ 2 ). Then Exercise 10.2 Prove the above proposition. (Hint: show that if A 1, A 2, A SL(2, R) are matrices such that A 1 = A 1 A 2 A then Trace(A 1 ) = Trace(A 1 A 2 A) = Trace(A 2 ). You might first want to show that Trace(AB) = Trace(BA) for any two matrices A, B.) We can now classify the three types of Möbius transformation hyperbolic, parabolic and elliptic in terms of the trace function. Let γ be a Möbius transformation. Suppose for simplicity that is not a fixed point (it follows that c 0). Recall from Lecture 9 that z 0 is a fixed point of γ if and only if z 0 = a d ± (a d) 2 + 4bc. 2c Thus there are two real solutions, one real solution or one complex conjugate pair of solutions depending on whether the term inside the square-root is greater than zero, equal to zero or less than zero, respectively. Using the identities it is easy to see that ad bc = 1, (a + d) 2 = τ(γ) (a d) 2 + 4bc = τ(γ) 4. When c = 0, we must have that is a fixed point. The other fixed point is given by b/(d a). Hence is the only fixed 3

4 point if a = d (in which case we must have that a = 1, d = 1 or a = 1, d = 1 as ad bc = ad = 1); hence τ(γ) = (1 + 1) 2 = 4. If a d then there are two fixed points on H and one can easily see that τ(γ) > 4. Thus we have proved: Proposition 10.2 Let γ be a Möbius transformation and suppose that γ is not the identity. Then: (i) γ is parabolic if and only if τ(γ) = 4; (ii) γ is elliptic if and only if τ(γ) [0, 4); (iii) γ is hyperbolic if and only if τ(γ) (4, ) Parabolic transformations Recall that a Möbius transformation γ is said to be parabolic if it has a unique fixed point and that fixed point lies on H. For example, the Möbius transformation γ(z) = z + 1 is parabolic. Here, the unique fixed point is. In general, a Möbius transformation of the form z z + b is called a translation. Exercise 10.3 Let γ(z) = z + b. If b > 0 then show that γ is conjugate 4

5 to γ(z) = z + 1. If b < 0 then show that γ is conjugate to γ(z) = z 1. Are z z 1, z z + 1 conjugate? Proposition 10.3 Let γ be a Möbius transformation and suppose that γ is not the identity. Then the following are equivalent (i) γ is parabolic; (ii) τ(γ) = 4; (iii) γ is conjugate to a translation; (iv) γ is conjugate either to the translation z z + 1 or to the translation z z 1. Proof. By Proposition 10.2 we know that (i) and (ii) are equivalent. Clearly (iv) implies (iii) and the exercise above implies that (iii) implies (iv). Suppose that (iv) holds. Recall that z z + 1 has a unique fixed point at. Hence if γ is conjugate to z z + 1 then γ has a unique fixed point in H, and is therefore parabolic. The same argument holds for z z 1. Finally, we show that (i) implies (iii). Suppose that γ is parabolic and has a unique fixed point at ζ H. Let g be a Möbius transformation that maps ζ to. Then gγg 1 is a Möbius transformation with a unique fixed point at. We claim that gγg 1 is a translation. Write gγg 1 (z) = az + b cz + d. 5

6 As is a fixed point of gγg 1, we must have that c = 0 (see Lecture 11). Hence gγg 1 (z) = a d z + b d, and it follows that gγg 1 has a fixed point at b/(d a). As gγg 1 has only one fixed point and the fixed point is at we must have that d = a. Thus gγg 1 (z) = z + b for some b R. Hence γ is conjugate to a translation. 6