ISOMETRIES OF THE HYPERBOLIC PLANE

Transcription

1 ISOMETRIES OF THE HYPERBOLIC PLANE KATHY SNYDER Abstrat. In this paper I will define the hyperboli plane and desribe and lassify its isometries. I will onlude by showing how these isometries an be represented as 2 2 matries. Contents 1. Isometries of H Classifiation of the Isometries of H The Hyperboli Plane and PSL 2 (R) 1 Aknowledgments 13 Referenes Isometries of H 2 The hyperboli plane is defined as the upper half plane, without boundary, endowed with a ertain metri. We denote the set of points by H 2 = {(x, y) R 2 y > }. Definition 1.1. A path from a to b for a, b H 2 is a map γ : [; 1] H 2 suh that γ is differentiable, γ() = a, and γ(1) = b. We may also find it useful to write γ(t) = (x γ (t), y γ (t)). Definition 1.2. We define the arlength of a path as L(γ) = 1 γ (t) y γ (t) dt. This allows us to define a metri on H 2 as follows: Definition 1.3. The distane between two points, a, b H 2, is defined as d(a, b) = inf{l(γ) : γ a path from a to b}. Let us verify that this is a metri: Proof. (1) d(a, b) beause γ (t) and y γ (t) are, so therefore the integral is nonnegative as well. If d(a, b) =, then let a = (a 1, a 2 ) and b = (b 1, b 2 ), and suppose a b. Then WLOG assume a 1 b 1. Then for any path γ from a to b, x γ (t) is not onstant. Then L(γ). So it must be that a = b. Now, suppose a = b. Then the onstant path γ(t) = a is a path from a to b. However, it has length L(γ) = beause γ (t) =. So d(a, b) ; but d(a, b) is at least. So d(a, b) =. Date: August 29. 1

2 2 KATHY SNYDER (2) Every path from a to b is also a path from b to a if one takes the same path bakwards. That is, for eah 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 beomes 1 instead of, and so on. Therefore d(a, b) = d(b, a). (3) Finally we must show that d(a, ) d(a, b) + d(b, ) a, b H 2. Suppose that d(a, ) > d(a, b) + d(b, ). Then ε > suh that d(a, ) = d(a, b) + d(b, ) + ε. Then there exist paths γ 1 from a to b and γ 2 from b to suh that L(γ 1 ) < d(a, b) + ε 2 and L(γ 2) < d(b, ) + ε 2. Then we an onstrut a path γ from a to by onatenating paths γ 1 and γ 2 and reparametrizing them. Note that γ 1 and γ 2 might meet at a usp, and therefore their onatenation might not be differentiable. However, sine we an approximate their onatenation with a path that is differentiable and has arlength arbitrarily lose to the arlength of the onatenation, this is not a problem. Then we have L(γ ) = L(γ 1 )+L(γ 2 ) < d(a, b)+d(b, )+ε < d(a, ), whih is a ontradition, sine d(a, ) is the infimum of all paths from a to. Therefore, the triangle inequality must hold. We are partiuluarly interested in when the distane between two points is realized by an atual path. Definition 1.4. A geodesi is a path γ from a to b suh that L(γ) = d(a, b). Later in the paper, we will show that there exists a unique geodesi between every two points. However, at the moment we are able to show this muh: Lemma 1.5. For any two points a and b on a vertial line, the line segment between them is the unique geodesi from a to b. Proof. Without loss of generality, let our two points be of the form (1, n), (1, m), where m > n. Then let γ : t (1, n + (m n)t). Then L(γ) = 1 m n n+(m n)t dt. Let u = n + (m n)t. Then du = (m n)dt So we have L(γ) = m 1 n u du = ln m ln n = ln m n. Now I laim that this path is the unique geodesi from (1, n) to (1, m). Suppose that there exists another path µ γ up to parameterization, suh that µ has shorter arlength than γ. Suppose x µ is non-onstant. Then x µ(t) > for some nontrivial subinterval of [; 1], sine x µ is ontinuous. Then it is true that x µ(t) 2 + y µ(t) 2 > + y µ(t) 2. That is, µ (t) is larger when x µ is nononstant than it is when x µ is onstant, for any given y µ. Therefore, the integral L(µ) is also smaller when x µ is onstant. So for any path from (1, n) to (1, m), we an always find a shorter path by making the x µ oordinate onstant. Therefore we an assume that x µ (t) = 1, sine we are assuming µ is a path with shortest arlength. Now, assuming x µ is onstant, what about y µ (t)? Suppose y µ (t) a + (b a)t, up to parameterization. That is, the image y µ (t) does not simply monotonially inrease from a to b, but also dereases in some plaes. Then we define an alternate map yµ as follows: suppose after some s [; 1] suh that y µ (s) m, y µ begins to derease. Sine y µ is ontinuous, and sine y µ (1) = m, there exists some r > s, r [; 1], suh that y µ (r) = y µ (s), by the Intemediate Value Theorem. Then for all x [s; r] we define yµ(t) = y µ (s). In other words, where y µ has a valley or U shape in its image, yµ remains onstant. Let us onsider the arlength of yµ ompared to y µ. In partiular, yµ only differs from y µ on the valleys, so let

3 ISOMETRIES OF THE HYPERBOLIC PLANE 3 us onsider the arlength funtion on one of those intervals. Now, sine x µ (t) is onstant, µ (t) = y µ(t). Now d dt y µ = on the interval, but y µ(t) < for the subinterval on whih y µ is dereasing. So y µ(t) > for some subinterval of positive y µ (t) yµ (t) dt is y µ (t) y dt. Furthermore, sine y µ(t) µ(t) dereases in the measure. Therefore, in our arlength funtion, the numerator of 1 larger than the numerator of 1 valley, the denomenator of y µ (t) y µ(t) is smaller than the denominator of y µ (t) yµ (t), sine yµ(t) y µ (t) for all t in the interval we are onsidering. Then overall, the fration y µ (t) yµ (t) y µ (t) y. So µ(t) L(µ ) L(µ). However, there is one more ase to onsider. What if y µ begins to derease at some s [; 1] suh that y µ (s) > m? In this ase, y µ goes above m and then omes bak down, reating a hill over some interval [; d] [; 1]. Let us break up the arlength integral as follows (remember that sine x µ is onstant, µ (t) = y µ(t) ): L(µ) = Now I laim that y µ(t) y µ (t) dt + d y µ(t) y µ (t) dt + 1 d y µ(t) y µ (t) dt. y µ(t) 1 y µ (t) dt + y µ(t) dt d((1, n), (1, m)). d y µ (t) This is true beause we ould define a new path µ suh that y µ (t) = m on the interval [; d]. (Atually, there might be usps in this modified path, but we ould approximate this new path by a differentiable one, so this is not a problem.) Then d dt y µ(t) = on [; d], so d y µ (t) y µ(t) dt would also be. So L(µ ) L(µ). In partiular, this gives us that L(µ) L(µ ) d((1, n), (1, m)). But we an do even better than this, beause we know that the derivative y µ(t) is nonzero for some subinterval of [; d], beause we piked this interval suh that y µ was dereasing on some part of it. Therefore y µ(t) > on some subinterval of [; d], and so the integral d is positive as well. Therefore we have L(µ) d((1, n), (1, m)) + y µ(t) y µ (t) dt > d y µ(t) dt > d((1, n), (1, m)). y µ (t) So µ is not the minimal path from (1, n) to (1, m). Therefore we an onlude that γ as defined above is, up to parameterization, the unique geodesi from (1, n) to (1, m). Now we are ready to define some isometries of H 2. Definition 1.6. A translation T s : H 2 H 2 is defined by T s (x, y) = (x + s, y). Definition 1.7. For λ >, a dilation D λ : H 2 H 2 is defined by D λ (x, y) = (λx, λy). First of all, it is lear that T s and D λ are ontinuous and differentiable beause this is true in eah oordinate. Furthermore, they map paths from a to b to paths

4 4 KATHY SNYDER from the image of a to the image of b, beause the omposition of differentiable funtion is a differentiable funtion. So, if T s and D λ preserve the arlength of eah path, they will also preserve distane, sine distane is the infimum of the arlength of all paths. Theorem 1.8. T s and D λ are isometries of H 2. Proof. Let a, b H 2, and let γ be a path from a to b. (1) T s γ = (x γ (t) + s, y γ (t)). Then d dt T s γ = (x γ(t), y γ(t)) = γ (t), so L(T s γ) = 1 γ (t) y γ(t) dt = L(γ). d (2) D λ γ = (λx(t), λy(t)). Therefore, dt D λ γ = λγ (t). So L(D λ γ) = 1 λγ (t) λy dt = 1 γ (t) γ(t) y γ(t) dt, sine λ > ; but that is equal to L(γ). Then, as we observed above, T s and D λ are isometries beause they preserve arlength, and therefore distane. It is of interest to us to onsider the onjugates of these two types of isometries: T s D λ T s : (x, y) (x s, y) ((λx λs, λy) (λ(x s) + s, λy). This orresponds to a dilation by λ about the boundary point (s, ) (i.e. the vertial line (s, y) is invariant under this transformation). We also have D λ T s D 1 : (x, y) ( 1 λ λ x, 1 λ y) ( 1 λ x + s, 1 y) (x + λs, y), λ whih amounts to the translation T sλ. At this point we are equipped to show H 2 is homogeneous, meaning that no point in H 2 is speial. Lemma 1.9. For any a, b H 2, there exists an isometry that maps a to b. Proof. Let (a 1, a 2 ) (b 1, b 2 ) be two points in H 2. Then T b1 b 2 a 1 a 2 D b 2 a 2 : (a 1, a 2 ) ( b 2a 1 a 2, b 2 ) (b 1, b 2 ). Note that sine a 2, b 2 >, the dilation is well-defined. Definition 1.1. We define a map R: H 2 H 2 as follows: R(x, y) = ( x y x 2 +y, 2 x 2 +y ). 2 First let us understand what this map does. Consider any point on the unit irle. That is, any point (x, y) suh that x 2 + y 2 = 1. In this ase R takes (x, y) to ( x 1, y 1 ), so R fixes all points on the unit irle. Next, onsider any line of the form x ax (x, ax). R: (x, ax) ( (a 2 +1)x, 2 (a 2 +1)x ). Thus, these lines are invariant under R. 2 If we look at a single line (x, ax), the two sides of the line, one inside the irle, and the other outside, are exhanged by R. For this reason, in the future we will refer to R as a refletion. Theorem R is an isometry.

5 ISOMETRIES OF THE HYPERBOLIC PLANE 5 Proof. Let a, b H 2 and let γ(t) be a path from a to b. We will find it easier to onsider γ(t) as a funtion in polar oordinates, as follows: Then So we have L(R(γ)) = 1 γ(t) = (r(t) os(θ(t)), r(t) sin(θ(t))). R γ(t) = ( 1 r(t) os(θ(t)), 1 r(t) sin(θ(t)) ). ( 1 r(t) 2 r(t) os θ(t) 1 r(t) sin θ(t), 1 r(t) 2 r (t) sin θ(t) + 1 whih when we simplify, yields 1 On the other hand, we have 1 1 r(t) r (t) 2 +r(t) 2 sin θ(t) sin θ(t) dt dt. r(t) os θ(t)) (r (t os θ(t) r(t) sin θ(t), r (t) sin θ(t) + r(t) os θ(t)) L(γ) = dt r(t) sin θ(t) 1 r (t) = 2 + r(t) 2 dt. r(t) sin θ(t) So we an onlude that L(R(γ)) = L(γ). Therefore, R preserves arlength, and therefore, R preserves distane. Let us find the onjugates of R by dilations and translations. (1) T s R T s : (x, y) (x s, y) ( ) x s (x s) 2 + y 2, y (x s) 2 + y ( 2 ) (x s) (x s) 2 + y 2 + s, y (x s) 2 + y 2, whih fixes points satisfying (x s) 2 +y 2 = 1, that is, the irle of radius 1 entered about the point (s, ). (2) ) ( x D λ R D 1 : (x, y) λ λ, y ( λ ) x/λ (x 2 + y 2 )/λ 2, y/λ (x 2 + y 2 )/λ 2 = ( λ 2 x x 2 + y 2, λ 2 ) y x 2 + y 2. ( λx x 2 + y 2, ) λy x 2 + y 2. As we an see, by onjugating R by translations or dilations, we an get a refletion R about a semi-irle of any radius entered about any point on the x-axis. Note, furthermore, that one speial refletion is the one about a vertial line, for example, the y-axis, whih maps (x, y) to (x, y). This an be thought of loosely as a refletion about the irle entered at infinity.

6 6 KATHY SNYDER We define one more type of isometry, namely, a rotation. On the Eulidean plane, we an obtain any rotation by refleting aross two interseting lines. As it turns out, we will define rotations in the hyperboli plane as the omposition of two refletions as well. Sine these so-alled rotations look onsiderably more odd than those in the Eulidean plane, we start by simply looking at a small neighborhood of the point fixed by the rotation. Consider any point p H 2. Let T p H 2 denote the tangent spae of p; that is, the spae of infinitesimal vetors entered at the point p. We an think of the tangent spae as isometri to R 2 with a similarly saled metri. In R 2, a rotation of angle θ an be obtained by omposing two refletions about lines that interset at an angle of θ/2. In H 2, we take two refletions about semi-irles interseting at our point p, and in the tangent spae T p H 2, the irles look like straight lines, so their omposition gives us a rotation of the tangent spae. So let us show that for any θ, we an find two irles with enters on the x-axis that interset at our point p, and whose tangent vetors at p interset at angle θ. Consider the two semi-irles of radius 1 entered at points on the x-axis. By taking two of these irles entered at the same point, the tangent vetors at the apex of the irles are the same vetor and therefore form an angle of. If we move one of the irles ontinuously to the right, the angle formed by the tangent vetors hanges ontinuously as it approahes 18, whih it would ahieve one the irles interseted on the x-axis, exept that that point is not in H 2. Then by the Intermediate Value Theorem, every angle θ < 18 an be ahieved at the intersetion of the two irles. Then by dilations and translations, we an get these semi-irles to interset at any point, at any angle. We an onlude, therefore, that by omposing two refletions about irles, we an obtain a rotation by any degree θ in the tangent spae of the intersetion point p. Sine this map is the omposition of two isometries, the so-alled rotation ating on the whole spae is an isometry as well. However, we would like to find some way to make sense of the notion of a point far away from p being rotated around p by some angle θ. In other words, we want to understand what a rotation does to points in H 2, not just to the tangent spae. Before we an attak this problem, we need to better understand geodesis in H 2. Lemma There exists a unique geodesi between any two points a, b H 2. Proof. Let a and b be two points in H 2. If a = b, then the onstant path γ(t) = a is a geodesi. If a b but a and b are on the same vertial line, then by Lemma 1.5 there exists a unique geodesi between the points. So suppose a and b are not on the same vertial line. WLOG we may assume a = (1, ). Then we an rotate b to the y-axis in the following manner: First, perform the R isometry. Next, perform the refletion about the vertial line (, y). This time, the image of b is flipped to the opposite side of the line (, y). Call the omposition of these two refletions as ψ. Now, ψ is a rotation about the point (1, ). So there is a ontinuous family of rotations about the point (1, ) from this isometry bak to the rotation by, as we argued above. So for some rotation φ, φ(b) lies on the same vertial line as a = φ(a). Then we know there exists a unique geodesi between φ(b) and φ(a) by Lemma 1.5. Call this geodesi as γ. Now, rotations are arlength preserving isometries, as we showed in the proof of Thm 1.1 and the subsequent omments. So φ 1 preserves arlength, and so L(γ) = L(φ 1 (γ)). But φ 1 is also an isometry, and so d(φ(a), φ(b)) = d(a, b). So φ 1 (γ) realizes the distane between a and b,

7 ISOMETRIES OF THE HYPERBOLIC PLANE 7 that is, φ 1 (γ) is a geodesi. In fat, it is a unique geodesi, beause γ is unique, and isometries are injetive. So there exists a unique geodesi from a to b. Note that just beause Isometries preserve distane, does not immediately imply that they map the geodesi between two points to the geodesi between the images of those points. We need the following lemma to show that geodesis do, in fat, map to other geodesis under isometries. Lemma Isometries preserve geodesis. Proof. Let a, b be two different points in H 2 and let φ be an arbitrary isometry. Now, we already know that translations, dilations, refletions, and rotations preserve geodesis. Furthermore, as we showed in L1.12, there exists a rotation f that rotates a and b to be on the same vertial line. Let f(a) = and f(b) = d. Now onsider φ(a), φ(b). By the omposition of a rotation, translation, and dilation, we an map φ(a) to, and φ(b) to the same vertial line as (by L1.9 and L1.12). Now, there are two points along the vertial line at distane d(, d) from. One is d itself, the other point we will all e. Now, if φ(b) maps to e, let = ( 1, 2 ) and reflet aross the irle of radius 2 entered at the point ( 1, ). The vertial line ( 1, y) is invariant under this refletion, and so e maps to d. Let g be this omposition of a rotation, translation, dilation, and also the refletion if needed, suh that g(φ(a)) = and f(φ(b) = d. Then we have that ψ = g φ f 1 satisfies ψ() = and ψ(d) = d. Now, we know that g and f 1 are geodesi-preserving isometries. So if we an show that ψ preserves the geodesi between and d, then it follows that φ must preserve the geodesi between a and b. So we must show that an isometry that fixes two points also fixes the geodesi between them. Let γ be the geodesi between and d. Reall that γ is the vertial line segment between those two points. Let ξ be the image of γ under ψ. Now, every point on ξ is the image of some point on γ, and sine ψ is an isometry, the distane between any point on ξ and the points and d are preserved. If p is an arbitrary point on ξ, then p has some preimage q γ, and q is some distane r from and distane d(, d) r from d. Then p also is distane r from and distane d(, d) r from d. Assume for a moment that p / γ. Let µ 1, µ 2 be the geodesis from to p and from p to d, respetively. Now, the onatenation of µ 1 and µ 2 has arlength d(, d), but it may not be differentiable at p. However, we an find a differentiable path that approximates the onatenation of µ 1 and µ 2 and has arlength arbitrarily lose to d(, d). Furthermore, we an approximate this path with another path that is differentiable and has arlength exatly d(, d), but still has points not in γ. Sine γ is supposed to be the unique geodesi between and d, this is a ontradition. It must be, therefore, that p = q. Sine p was an arbitrary point, we an onlude that ξ = γ up to parameterization. For the next lemma, we introdue the idea of an infinite geodesi. An infinite geodesi is defined to be a urve in the plane suh that for every two points on the urve, the segment of the urve between them is the shortest path, or geodesi, between them. Lemma A point and a diretion uniquely determine an infinite geodesi. Proof. Let p H 2 and v be a tangent vetor based at p. We laim there is a unique infinite geodesi passing through p to whih v is tangent. Let us make an eduated guess. Let γ be the irle entered on the x-axis suh that p lies on the irle and

8 8 KATHY SNYDER v is tangent to the irle at p. (If v is vertial, then we define γ to be the vertial line through p, whih we may loosely think of a irle entered at infinity along the x-axis.) Then γ is indeed an infinite geodesi: Take two points, a and b on γ. Then, there exists a refletion about γ that fixes a and b. But we showed in Lemma 1.13 that if an isometry fixes two points, it also fixes the geodesi between them. But the only path from a to b that is fixed by our refletion is the segment of γ between a and b. So this must be the geodesi between them. Note that this implies that all infinite geodesis are irles with enters on the x-axis or vertial lines. Now we are ready to return to the question of how to think of a rotation outside the tangent spae. Let φ be a rotation by θ degrees about the point p. Let q be an arbitrary point that is not p. Then there exists a unique geodesi, γ from p to q. Consider the tangent vetor τ to the geodesi γ at point p. Now, φ rotates the tangent spae to p by θ degrees, and sine τ, if saled down, is in the tangent spae, τ is also rotated θ degrees. Now, there is only one geodesi going through p in the diretion of φ(τ) (L1.14). Sine φ preserves the distane between p and q, φ(q) is the point at distane d(p, q) along that geodesi, in the diretion of φ(τ). In this sense, we an say that q is rotated about p. 2. Classifiation of the Isometries of H 2 We are now ready to lassify the isometries of the hyperboli plane. As it turns out, we have already found all of them, as the following theorem shows. Theorem 2.1. If φ : H 2 H 2 is an isometry, then φ is a translation, dilation, refletion, or rotation, or some omposition of these. Proof. Case I) Suppose φ has at least two fixed points, p, q. Then the geodesi γ between p and q is fixed. That follows from the proof of L1.13. Now, rotate these two points til they are on a vertial line. Then any point on the vertial line is fixed: For any point b not between p and q, onsider the geodesi from b and the midpoint p+q 2. Our isometry fixes the part of the geodesi that is the segment between p+q 2 and one of the endpoints, so it must fix the rest of the geodesi as well, beause as we noted before (proof of L1.14), geodesis are always segments of irles or vertial lines, and not some onatenation of the two. So the vertial line is fixed. Let s all this line L. Now onsider some not on L. We want to determine what the image of under φ might be. Now, there exists some unique point m L suh that m is the losest point on the line to. This is true beause on any losed interval ε y n, the funtion d((x, y), ) has a minimum. Sine as y or y, the distane from (x, y) to approahes infinity, a global minimum must our on some suh interval. Furthermore, it an be shown, although we will not in this paper, that the minimum m = (x, y) is unique. Now, for a given m = (m 1, m 2 ) L and a distane d, what points are suh that they are distane d from m and m is the losest point on L to them? Consider the irle C defined by the equation (x m 1 ) 2 + y 2 = m 2 2. This is the irle entered at (m 1, ) with radius m 2. Note that it is an infinite geodesi. Let w, v be the points distane d away from m lying on C, one on eah side of the line L.

9 ISOMETRIES OF THE HYPERBOLIC PLANE 9 Now let f be the refletion aross the irle C. Note that f fixes C, and in partiular, the points m, v, and w and the geodesis between them. Now if the losest point to w on L was some point n m, then f(n) n sine f only fixes m out of all the points on L. However, sine f is an isometry, f(n) would also be the losest point to w. This is a ontradition sine the losest point is unique. The same holds for the point v. So m is the losest point on L to w and to v. If we let g be the refletion aross the line L, then C is invariant under g. This is true beause, onsidering the equation (x m 1 ) 2 + y 2 = m 2 2, if (x + m 1, y) C, then (m 1 x, y) C as well. And sine g is an isometry, d(g(w), m) = d(w, m), so g(w) is exatly the point at distane d along the geodesi C on the opposite side of L from, whih is just v. So w and v are refletions of eah other aross the line L. Now let us return to our onsideration of φ. Let be an arbitrary point not on the line L fixed by φ. What might φ() be? We know that there exists some losest point m L to. But φ is an isometry, and so the distanes between and every point on L are preserved. In partiular, if m is the losest point to, then it is also the losest point to φ(). But as we showed above, there are only two points distane d(, m) away from m suh that m is losest to both of them. So φ() is either itself or the refletion of aross L, whih we will denote as. Now onsider two arbitrary points, e / L. I laim that if and e are on the same side of L, then so are φ(), φ(e), and analogously, if and e are on opposite sides of L, then so are φ() and φ(e). We will break this into ases. Case 1) Suppose and e are on the same side of L. Let γ be the geodesi between them. Then I laim that γ does not interset L. Suppose that it did. Reall that geodesis are vertial lines or irles entered on the x-axis. If γ is itself a vertial line, then it will not interset L at all, as it will be parallel with L. If γ is a irle, then it intersets L exatly one, beause the upper half of a irle is a well-defined funtion over its domain and therefore every element in the domain has exatly one element as its image (that is, it passes the vertial line test). So if γ interseted L, this would mean that and e were on opposite sides of L, sine the irle is ontinuous and only rosses L one. But and e are on the same side of the line, so γ does not interset L. Therefore, φ(γ) won t interset L either. Eah point on L is fixed, so its preimage is itself, and no point of γ is in the preimage of L. However, if φ(γ) does not interset L, then φ() and φ(e) must lie on the same side of L. This is beause L separates H 2 and φ(γ) is a ontinuous path. Case 2) If and e are on different sides of L, a similar argument shows that φ() and φ(e) must also lie on different sides of L. Therefore, if for some / L, φ() =, then for any other point e, φ(e) = e, beause this is the only possible image of e that remains on the same (or opposite) side of L as, if e is on the same (or opposite) of L as. Similarly, if for some / L, φ() is the refletion of aross L, then for any other point e, the same holds. Therefore, φ is either the identity or the refletion aross the line L. Case II) Suppose that φ is an isometry with exatly one fixed point, p. Consider the tangent spae of p. Considering the tangent spae as R 2, φ ats on it fixing the origin. Then if we think of the tangent spae as isometri to R 2, the type of isometry that only fixes one point is a rotation. So φ is a rotation of the tangent

10 1 KATHY SNYDER spae, and therefore, as we showed before, a rotation of the whole plane. Case III) Suppose that φ is an isometry with no fixed points. Then let x H 2 be an arbitrary point. Then x has a unique preimage, y = φ 1 (x). By L1.9, there exists a translation and dilation mapping x to y. Let this translation omposed with a dilation be ψ. Then ψ φ is an isometry with at least one fixed point. Then this ase redues to either Case I or II. Then φ is a rotation, refletion, or the identity, omposed with a translation and dilation. 3. The Hyperboli Plane and PSL 2 (R) SL 2 (R), the group of 2 2 matries over R with determinant +1, turn out to have a orrespondene with the orientation-preserving isometries of H 2. If we onsider the omplex numbers as ordered pairs z = (x, y), where x = Re(z), y = Im(z), then we an desribe elements of the hyperboli plane as elements z C suh that Im(z) ( >. Then ) we define an ation of SL 2 (R) on C as follows: a b For a matrix and an element z C with Im(z) >, we say that d ( ) a b d z = az + b z + d. Note that sine det(a) = 1, either or d, so the denominator is nonzero. Note also, that ( ) a b d z = az + b z + d = az b z d = ( ) ( ) a b a b z = ( 1) z. d d So for the ation we defined, if A SL 2 (R), then A z = ( 1)A z. Sine we do not want to distinguish between two matries whose ations on C are the same, we define the following equivalene relation: A B iff A = ±B. We denote SL 2 (R)/ by PSL 2 (R). Let ( us onsider ) three types of matries: 1 s (1) z = 1 z+s 1 = z + s = (Re(z) + s, Im(z)). This orresponds to a translation, ( T s ) in H 2. λ (2) z = 1/λ λz 1/λ = λ2 z. This orresponds to a dilation D λ 2 in H 2. Note that sine ( λ 2 > ), this dilation is well-defined. 1 (3) z = 1 1 z. This orresponds to the refletion R omposed with the refletion about the y-axis, whih is a rotation and therefore orientation preserving, as in fat are all three of the above matries. Theorem 3.1. PSL 2 (R) is generated by the matries ( ) 1 s. 1 ( ) 1, 1 ( ) λ 1, and λ

11 ISOMETRIES OF THE HYPERBOLIC PLANE 11 ( ) a b Proof. We want to show that for an arbitrary matrix PSL d 2 (R), we an redue it to the identity matrix only by multiplying by the above three matrix types. Consider ( 1 ) ( ) ( ) ( ) ( ) 1 a 1 1 a a b d ( 1 ) ( ) ( ) ( ) = 1 a 1 b ad 1 1 d ( 1 ) ( ) ( ) = 1 a d ad 1 ( b 1 ) ( ) = ad ( ) ( b ) 1 1 = = ad b 1 beause ad b = 1, as the determinant of our matrix is 1. Note that eah matrix type was used. In terms of the hyperboli plane, this means that every matrix in PSL 2 (R) an be deomposed into a omposition of orientation-preserving isometries, and therefore that every element of PSL 2 (R) is itself an orientation-preserving isometry of H 2. And by our lassifiation of isometries, we know that every orientation-preserving isometry of H 2 is some omposition of translations, dilations, and rotations. We already know how to represent translations and dilations as matriies, so let us show that there is a matrix representation for any rotation, as well, and then we will know that there is a bijetive orrespondene between H 2 and PSL 2 (R). ( Under ) the ation defined above, the refletion R orresponds to the matrix 1. Now, this matrix is not orientation preserving, and it has determinant 1 1. However, if we reall, in setion 1, we obtained arbitrary rotations by omposing the refletion R with dilations and translations to move it around or make it larger in diameter, and then omposed it with another similarly altered refletion R. So to get an arbitrary rotation, we take some produt of translation and dilation matries, along with two of these R matries, arranged in some order. But the determinant of the produt of matries is the produt of the determinants. And sine the dilations and translations have determinant 1, and the two R matries have determinant 1, the determinant of the produt is 1, whih means that the produt is in SL 2 (R). Therefore every rotation an be represented as a matrix in PSL 2 (R). It would be onvenient if just by looking at the matrix we ould tell what kind of isometry it was. As it turns out, this an be ahieved simply by onsidering the absolute value of the trae of the matrix. ( ) a b Before, we lassified isometries by looking at fixed points. Let A = d SL 2 (R). Then z is a fixed point iff az+b z+d = z. This is equivalent to the equation = z 2 + (d a)z b.

12 12 KATHY SNYDER By the Fundamental Theorem of Algebra we know that every polynomial has a root over the omplex number. Sine roots represent fixed points, we are interested in where these roots are. Remember that the upper half plane is only those elements of C with positive imaginary part, so the fixed points may not be in the hyperboli plane at all. Let us solve this equation: z = (a d) ± (d a) 2 + 4b. 2 Whether s is real or not depends on whether (d a) 2 + 4b is nonnegative. So we have (d a) 2 + 4b = d 2 + 2ad + a 2 + 4b = d 2 + a 2 + 2ad 2ad 2ad + 4b = (d + a) 2 (4ad 4b) = (d + a) 2 4. This means that we have real roots if and only if d + a = tr(a) 2. However, note also that the harateristi polynomial of A is as follows: whih is zero when f A (t) = (t a)(t d) b = t 2 + (a + d)t + (ad b) = t 2 + (a + d)t + 1 t = (a + d) ± (a + d) This equation also has real solutions iff (a + d) 2 4 = d + a = tr(a) 2. In other words, the matrix A orresponds to an isometry of H 2 with fixed points with real solutions exatly when A has real eigenvalues. Let us break this up into ases. Case 1) tr(a) > 2; that is, we have two real eigenvalues, or two real fixed points. Call these eigenvalues λ 1, λ 2. Sine we are thinking about the hyperboli plane, these real solutions lie on the real axis, whih is not in the hyperboli plane. Now, A is a real matrix. If we have two real eigenvalues, then there exist two distint eigenvetors (in R 2 ), one for eah eigevalue. Then these vetors are linearly independent, and therefore an be a basis for R 2. Then if we think of A as a linear transformation, it is similar to the same linear transformation under a basis hange to the eigenbasis [2]. That is, for some matrix D, there exists an invertible matrix S suh that S 1 AS = D. And D will be a diagonal matrix, sine eah basis vetor gets sent to a salar multiple of itself under this transformation. Now, similar matries have the same trae and determinant [1]. So D PSL 2 (R). However, not only is D diagonal, but sine it has determinant 1, the two entries along the diagonal ( must) be inverses of eah other, sine their produt must be 1. So in fat, λ D =. Therefore, if tr(a) > 2, A is similar to a dilation matrix. 1/λ Case 2) tr(a) = 2; i.e., there is exatly one fixed real point, or one distint real eigenvalue. Call this eigenvalue as λ. In fat, if we look at the harateristi polynomial, the eigenvalue is a+d 2 = 1 when tr(a) = 2. So there exists some vetor

13 ISOMETRIES OF THE HYPERBOLIC PLANE 13 v 1 suh that Av 1 = v 1. Then ( if ) we extend v to a basis, v 1, v 2, then A is similar to 1? some matrix of the form?. But? must be 1 as well beause the matrix has ( ) 1 s determinant 1 (sine it is similar to A). So in fat, A for some s R. 1 That is, A is similar to a translation matrix. Case 3) tr(a) < 2; i.e., there are two omplex eigenvalues, or two omplex fixed points. However, if we look at the harateristi polynomial t 2 + (a + d)t + 1, it is lear that whatever the omplex roots are, their sum must be (a + d) whih is a real number. Therefore, the two omplex roots must be onjugate, that is, of the form α + iβ, α iβ, where α, β R and β >. Then α + iβ is in the hyperboli plane sine its imaginary part is positive. So we have exatly one fixed point in the plane. Then by the lassifiation of the isometries of the hyperboli plane, A ating on the hyperboli plane must be a rotation. So now we are apable of looking at a matrix in PSL 2 (R) and assoiating it with an orientation-preserving isometry of the hyperboli plane simply by alulating its trae. This shows that any matrix A in PSL 2 (R) is similar to a translation, dilation, or the rotation matrix. However, we an onlude more than that. Let A = S 1 XS, where X is one of the three types of matries in the above ases, and S is an invertible 2 2 matrix. Then sine S 1 S = I, det(s) = 1/ det(s 1 ). Now let d = ( ) p q det(s), and let S =. Then we an multiply S by a salar as follows: r s ( ) p/d q/d (1/d)S =. r/d s/d The determinant of this matrix is (1/ det(s))(ps rq) = 1. Therefore, we have A = ds 1 XS(1/d). So if we all our new matrix Q = S(1/d), we have A = Q 1 XQ and Q PSL 2 (R). This means that not only is A similar to X, but also onjugate to it by another isometry in PSL 2 (R). For example, if X is a dilation, we know that A, as an isometry, is onjugate to a dilation. In partiular, sine onjugation does not hange the essential harateristis of an isometry that is, it does not alter the angle of rotation nor the sale of a dilation or translation we an onlude that every isometry in PSL 2 (R) looks like a translation, a dilation, or a rotation. This onludes our study of the hyperboli plane and its isometries. Aknowledgments. It is my pleasure to thank my mentors Tom Churh and Katie Mann for helping me to learn this material. Referenes [1] Laszlo Babai Apprentie Course, Leture Notes, 29, p 4. [2] Laszlo Babai Apprentie Course, Leture Notes, 29, pp 9-1.