ISOMETRIES OF THE HYPERBOLIC PLANE KATHY SNYDER Abstract. In this paper I will define the hyperbolic plane and describe and classify its isometries. I will conclude by showing how these isometries can be represented as 2 × 2 matrices. Contents 1. Isometries of H2 1 2. Classification of the Isometries of H2 8 3. The Hyperbolic Plane and PSL2(R) 10 Acknowledgments 13 References 13 1. Isometries of H2 The hyperbolic plane is defined as the upper half plane, without boundary, en- dowed with a certain metric. We denote the set of points by 2 2 H = f(x; y) 2 R j y > 0g: Definition 1.1. A path from a to b for a; b 2 H2 is a map γ : [0; 1] ! H2 such that γ is differentiable, γ(0) = a, and γ(1) = b. We may also find it useful to write γ(t) = (xγ (t); yγ (t)). Definition 1.2. We define the arclength of a path as Z 1 jγ0(t)j L(γ) = dt: 0 yγ (t) This allows us to define a metric on H2 as follows: Definition 1.3. The distance between two points, a; b 2 H2, is defined as d(a; b) = inffL(γ): γ a path from a to bg. Let us verify that this is a metric: 0 Proof. (1) d(a; b) ≥ 0 because jγ (t)j and yγ (t) are ≥ 0, so therefore the integral is nonnegative as well. If d(a; b) = 0, then let a = (a1; a2) and b = (b1; b2), and suppose a 6= b. Then WLOG assume a1 6= b1. Then for any path γ from a to b, xγ (t) is not constant. Then L(γ) ≥ 0. So it must be that a = b. Now, suppose a = b. Then the constant path γ(t) = a is a path from a to b. However, it has length L(γ) = 0 because jγ0(t)j = 0. So d(a; b) ≤ 0; but d(a; b) is at least 0. So d(a; b) = 0. Date: August 2009. 1 2 KATHY SNYDER (2) Every path from a to b is also a path from b to a if one takes the same path backwards. That is, for each h in the image of a path, with preimage t, we define a new path by taking the preimage of h to be 1 − t. So the preimage of a becomes 1 instead of 0, and so on. Therefore d(a; b) = d(b; a). (3) Finally we must show that d(a; c) ≤ d(a; b) + d(b; c) 8 a; b 2 H2. Suppose that d(a; c) > d(a; b) + d(b; c). Then 9 " > 0 such that d(a; c) = d(a; b) + d(b; c) + ". Then there exist paths γ1 from a to b and γ2 from b to c " " such that L(γ1) < d(a; b) + 2 and L(γ2) < d(b; c) + 2 . Then we can 0 construct a path γ from a to c by concatenating paths γ1 and γ2 and reparametrizing them. Note that γ1 and γ2 might meet at a cusp, and therefore their concatenation might not be differentiable. However, since we can approximate their concatenation with a path that is differentiable and has arclength arbitrarily close to the arclength of the concatenation, this is 0 not a problem. Then we have L(γ ) = L(γ1)+L(γ2) < d(a; b)+d(b; c)+" < d(a; c), which is a contradiction, since d(a; c) is the infimum of all paths from a to c. Therefore, the triangle inequality must hold. We are particuluarly interested in when the distance between two points is real- ized by an actual path. Definition 1.4. A geodesic is a path γ from a to b such that L(γ) = d(a; b). Later in the paper, we will show that there exists a unique geodesic between every two points. However, at the moment we are able to show this much: Lemma 1.5. For any two points a and b on a vertical line, the line segment between them is the unique geodesic from a to b. Proof. Without loss of generality, let our two points be of the form (1; n), (1; m), R 1 jm−nj where m > n. Then let γ : t 7! (1; n + (m − n)t). Then L(γ) = 0 n+(m−n)t dt. R m 1 Let u = n + (m − n)t. Then du = (m − n)dt So we have L(γ) = n u du = m ln m − ln n = ln n . Now I claim that this path is the unique geodesic from (1; n) to (1; m). Suppose that there exists another path µ 6= γ up to parameterization, such that µ has shorter 0 arclength than γ. Suppose xµ is non-constant. Then xµ(t) > 0 for some nontrivial q 0 2 0 2 subinterval of [0; 1], since xµ is continuous. Then it is true that xµ(t) + yµ(t) > q 0 2 0 0 + yµ(t) . That is, jµ (t)j is larger when xµ is nonconstant than it is when xµ is constant, for any given yµ. Therefore, the integral L(µ) is also smaller when xµ is constant. So for any path from (1; n) to (1; m), we can always find a shorter path by making the xµ coordinate constant. Therefore we can assume that xµ(t) = 1, since we are assuming µ is a path with shortest arclength. Now, assuming xµ is constant, what about yµ(t)? Suppose yµ(t) 6= a + (b − a)t, up to parameterization. That is, the image yµ(t) does not simply monotonically increase from a to b, but also decreases in some places. Then we define an alternate ∗ map yµ as follows: suppose after some s 2 [0; 1] such that yµ(s) ≤ m, yµ begins to decrease. Since yµ is continuous, and since yµ(1) = m, there exists some r > s, r 2 [0; 1], such that yµ(r) = yµ(s), by the Intemediate Value Theorem. Then ∗ for all x 2 [s; r] we define yµ(t) = yµ(s). In other words, where yµ has a valley ∗ or U shape in its image, yµ remains constant. Let us consider the arclength of ∗ ∗ yµ compared to yµ. In particular, yµ only differs from yµ on the valleys, so let ISOMETRIES OF THE HYPERBOLIC PLANE 3 us consider the arclength function on one of those intervals. Now, since xµ(t) is 0 0 d ∗ 0 constant, jµ (t)j = jyµ(t)j. Now dt yµ = 0 on the interval, but yµ(t) < 0 for the 0 subinterval on which yµ is decreasing. So jyµ(t)j > 0 for some subinterval of positive ∗0 R 1 jyµ (t)j measure. Therefore, in our arclength function, the numerator of ∗ dt is 0 yµ(t) 0 R 1 jyµ(t)j larger than the numerator of dt. Furthermore, since yµ(t) decreases in the 0 yµ(t) 0 ∗0 jyµ(t)j jyµ (t)j valley, the denomenator of is smaller than the denominator of ∗ , since yµ(t) yµ(t) ∗ yµ(t) ≥ yµ(t) for all t in the interval we are considering. Then overall, the fraction ∗0 0 jyµ (t)j jyµ(t)j ∗ ∗ ≤ . So L(µ ) ≤ L(µ). yµ(t) yµ(t) However, there is one more case to consider. What if yµ begins to decrease at some s 2 [0; 1] such that yµ(s) > m? In this case, yµ goes above m and then comes back down, creating a hill over some interval [c; d] ⊂ [0; 1]. Let us break up the 0 0 arclength integral as follows (remember that since xµ is constant, jµ (t)j = jyµ(t)j): Z c jy0 (t)j Z d jy0 (t)j Z 1 jy0 (t)j L(µ) = µ dt + µ dt + µ dt: 0 yµ(t) c yµ(t) d yµ(t) Now I claim that Z c jy0 (t)j Z 1 jy0 (t)j µ dt + µ dt ≥ d((1; n); (1; m)): 0 yµ(t) d yµ(t) ∗ This is true because we could define a new path µ such that yµ(t) = m on the interval [c; d]. (Actually, there might be cusps in this modified path, but we could approximate this new path by a differentiable one, so this is not a problem.) Then jy0 (t)j j d y∗(t)j = 0 on [c; d], so R d µ dt would also be 0. So L(µ∗) ≤ L(µ). In dt µ c yµ(t) particular, this gives us that L(µ) ≥ L(µ∗) ≥ d((1; n); (1; m)). But we can do even 0 better than this, because we know that the derivative yµ(t) is nonzero for some subinterval of [c; d], because we picked this interval such that yµ was decreasing 0 on some part of it. Therefore jyµ(t)j > 0 on some subinterval of [c; d], and so the integral Z d jy0 (t)j µ dt > 0 c yµ(t) is positive as well. Therefore we have Z d jy0 (t)j L(µ) ≥ d((1; n); (1; m)) + µ dt > d((1; n); (1; m)): c yµ(t) So µ is not the minimal path from (1; n) to (1; m). Therefore we can conclude that γ as defined above is, up to parameterization, the unique geodesic from (1; n) to (1; m). Now we are ready to define some isometries of H2. 2 2 Definition 1.6.