HYPERBOLIC PLANE AS A PATH METRIC SPACE. Contents

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1 HYPERBOLIC PLANE AS A PATH METRIC SPACE QINGCI AN Abstrct. We will study the quntities tht re invrint under the ction of Möb(H) on hyperbolic plne, nmely, the length of pths in hyperbolic plne. From these invrint elements, we construct n invrint notion of hyperbolic distnce on H nd explore some of its bsic properties. And we will prove tht Möb(H)= Isom(H, d H ). Contents 1. Introduction nd Bsic Definitions 1 2. The Generl Möbius Group nd Möb(H) 4 3. Length nd Distnce in H 7 4. Isometries in H 9 Acknowledgments 11 References Introduction nd Bsic Definitions Hyperbolic geometry is non-eucliden geometry in which the prllel postulte in Eucliden geometry does not hold. In hyperbolic geometry, for ny given line L nd point p not in L, there exist t lest two distinct lines through p tht re prllel to L. Hyperbolic spce is model for hyperbolic geometry. It could be two-dimensionl or higher. And there re severl models for hyperbolic spce, for instnce the hyperboloid model, the Klein model, the Poincré bll model nd the Poincré hlf spce model. For ny two of the bove models, there exist trnsformtions between them tht preserve geometric properties. In this pper, we use the upper hlf plne to model the hyperbolic plne nd we will explore geometries in the hyperbolic plne. The upper hlf-plne model is defined to be H = {z C Im(z) > 0}. And we will refer to the rel xis by R. We re going to define hyperbolic line in H in terms of Eucliden lines nd circles in C. Definition 1.1. A hyperbolic line in H is either the intersection of H with Eucliden line in C perpendiculr to R or the intersection of H with Eucliden circle centered on R. 1

2 2 QINGCI AN Figure 1.1: Hyperbolic lines Proposition 1.2. For ech pir of distinct points, p nd q, in H, there exists unique hyperbolic line l through p nd q. Proof. There re two cses to consider. Cse 1: Suppose Re(p) = Re(q). Then the Eucliden line L = {z C } Re(z) = Re(p)} psses through p nd q nd is perpendiculr to R. By the uniqueness of the Eucliden line, we see tht l = H L is the unique hyperbolic line through p nd q. q p Figure 1.2: Hyperbolic line determined by point p nd q Cse 2: Suppose Re(p) Re(q). Then let L pq be the Eucliden line segment joining p nd q nd let K be the perpendiculr bisector of L pq. Then K intersects R t unique point c nd c p = c q. Then let C be Eucliden circle centered t c with rdius c p. So l = H C is the desired unique hyperbolic line. p q Figure 1.3: Hyperbolic line determined by point p nd q Hyperbolic geometry behves differently from Eucliden geometry of prllel lines, lthough the definitions of prllel for Eucliden lines nd hyperbolic lines re sme. Definition 1.3. If two hyperbolic lines re disjoint, then they re prllel. Theorem 1.4. Let l be hyperbolic line in H nd let p be point in H not in l, there exist infinitely mny distinct hyperbolic lines through p tht re prllel to l.

3 HYPERBOLIC PLANE AS A PATH METRIC SPACE 3 Proof. There re two cses to consider. First, suppose tht l is contined in Eucliden line L nd p is point in H not in l. As p is not in L, p is on Eucliden line K tht is prllel to L. Since L is perpendiculr to R, K is perpendiculr to R. Then K H is hyperbolic line through p nd prllel to l. We cn construct nother such hyperbolic line by tking point x in R nd let be the hyperbolic line through p nd x. k nd re distinct becuse Re(x) Re(p). And we cn construct infinitely mny such hyperbolic lines becuse there re infinitely points on R between L nd K. Figure 1.4: Prllel hyperbolic lines The second cse is tht l is contined in Eucliden circle M. As p is in H but not in L, there is Eucliden circle N through p nd concentric to M. As concentric circles re disjoint, N H is hyperbolic line through p nd prllel to l. To construct nother such hyperbolic line, tke ny point x on R between M nd N. Let be the hyperbolic line through p nd x. Then is disjoint from l. As there re infinitely mny points between M nd N, we cn construct infinitely mny hyperbolic lines through p nd prllel to l. Figure 1.5: Prllel hyperbolic lines To determine resonble group of trnsformtions in H tht tke hyperbolic lines to hyperbolic lines, we need to unify the two types of hyperbolic lines. By stereogrphic projection, Eucliden circle cn be obtined from Eucliden line by dding single point. Definitions 1.5. By stereogrphic projection, we cn construct Riemnn sphere C. As set of points, the Riemnn sphere is the union of the complex plne C with point not in C, which we denote by, which mens C = C { }. We lso define the extended rel xis R = R { }. A circle in the Riemnn sphere C is either Eucliden circle in C or Eucliden line in C with { }.

4 4 QINGCI AN Figure 1.6: Stereogrfic projection There is clss of continuous functions from C to C tht re especilly well behved. Definitions 1.6. A function f : C C is homeomorphism if f is bijective nd both f nd f 1 re continuous. The homeomorphisms of C re the trnsformtions of C tht re of most interest of us. We denote the group of homeomorphisms by Homeo( C) = {f : C C f is homeomorphism}. By definition, the inverse of homeomorphism is lso homeomorphism nd the composition of two homeomorphisms is lso homeomorphism. And the identity homeomorphism is the function f : C C given by f(z) = z. So Homeo( C) is group. 2. The Generl Möbius Group nd Möb(H) As every hyperbolic line in H is contined in circle in C, in order to determine the trnsformtions of H tht tke hyperbolic lines to hyperbolic lines, we first determine the group of homeomorphisms of C tking circles C to circles in C. We let Homeo C ( C) denote the subset of Homeo( C) tht contins homeomorphisms of C tking circles in C to circles in C. Definition 2.1. A Möbius trnsformtion is function m : C C of the form m(z) = z + b cz + d where, b, c, d C nd d bc 0. We denote the set of ll Möbius trnsformtions by Möb +. Lemm 2.2. The element f of Homeo( C) defined by f(z) = z + b for z C nd f( ) =, where, b C nd 0, is n element of Homeo C ( C). Proof. Given circle in C A = {z C : αz z + βz + β z + γ = 0}, where α, γ R nd β C nd α = 0 if nd only if A is Eucliden line. We wnt to show tht if z A, then w = z + b stisfy similr eqution. We first consider the cse α = 0. Since w = z + b, we cn substitute z = 1 (w b) into the eqution obtining βz + β z + γ = β 1 (w b) + β 1 (w b) + γ = β w + β w 2Re(β b) + γ = 0

5 HYPERBOLIC PLANE AS A PATH METRIC SPACE 5 This shows tht w stisfies n eqution of n Eucliden line. So f tkes Eucliden lines to Eucliden lines. If α 0, we hve αz z + βz + β z + γ = α 1 (w b) 1 (w b) + β 1 (w b) + β 1 (w b) + γ = α 2 (w b)(w b) + β β (w b) + (w b) + γ = α β w + 2 α b 2 + γ β 2 α = 0 So w stisfy n eqution of Eucliden circle. Hence f tkes circles in C to circles in C. Lemm 2.3. The element g of Homeo( C) defined by g(z) = 1 z for z C {0}, g(0) =, nd g( ) = 0, is n element of Homeo C ( C). The proof of Lemm 2.3 is similr to Lemm 2.2. In fct, Möb + is generted by f nd g. Theorem 2.4. Let m be Möbius trnsformtion m(z) = z+b cz+d, where, b, c, d C nd d bc 0. If c = 0, then m(z) = d z + b d. If c 0, then m(z) = f g h(z), where h(z) = c 2 (z) + cd for z C nd h( ) =, g(z) = 1 z for z C {0}, g(0) =, nd g( ) = 0, nd f(z) = (d bc)z + c for z C nd f( ) =. The proof of Theorem 2.4 is just direct clcultion. By Theorem 2.4 nd Lemms 2.2 nd 2.3, we hve tht every Möbius trnsformtion is n element of Homeo C ( C), s it is the composition of functions in Homeo C ( C). So we hve the following theorem. Theorem 2.5. Möb + Homeo C ( C). To extend Möb + to lrger group, we consider the complex conjugtion, C : C C, C(z) = z nd C( ) =, which is not n element of Möb + but is n element of Homeo C ( C). Lemm 2.6. The function C : C C, C(z) = z nd C( ) = is n element of Homeo C ( C). The proof of Lemm 2.6 is similr to the proof of lemm 2.2. Definition 2.7. The generl Moöbius group is the group generted by Möb + nd the function C : C C, C(z) = z nd C( ) =. Let Möb denote the generl Möbius group. Theorem 2.8. Möb = Homeo C ( C) Sketch of Proof: By Lemm 2.8, we hve tht every element of Möb is n element of Homeo C ( C). Then we need to show tht Homeo C ( C) Möb. Let f be n element of Homeo C ( C)

6 6 QINGCI AN nd let n be the Möbius trnsformtion tking the triple (f(0), f(1), f( )) to the triple (0, 1, ). So n f tkes (0, 1, ) to (0, 1, ). As R is the circle in C determined by (0, 1, ) nd n f tkes circles in C to circles in C, n f( R) = R. Therefore n f(h) is either the upper-hlf plne or the lower-hlf plne. In the former cse, set m = n. In the ltter cse, set m = C n. So m f is n element of Möb such tht m f(0) = 0, m f(1) = 1, m f( ) =, nd m f(h) = H. We show tht m f is the identity by constructing dense set of points in C, ech of which is fixed by m f. Hence f = m 1 is n element of Möb. This completes the sketch of the proof of the Theorem 2.9. Nottions 2.9. Möb( R) = {m Möb m( R) = R} Möb(H) = {m Möb m(h) = H} Theorem Every element of Möb( R) is of one of the following form: 1. m(z) = z+b cz+d with, b, c, d R nd d bc = 1; 2. m(z) = z+b c z+d with, b, c, d R nd d bc = 1; 3. m(z) = z+b cz+d with, b, c, d purely imginry nd d bc = 1; 4. m(z) = z+b c z+d with, b, c, d purely imginry nd d bc = 1; Theorem Every element of Möb(H) is of one of the following form: 1. m(z) = z+b cz+d with, b, c, d R nd d bc = 1; 2. m(z) = z+b c z+d with, b, c, d purely imginry nd d bc = 1; One useful property of Möb(H) is the trnsitivity property. Theorem Möb (H) cts trnsitively on H, which mens for every pir of points x nd y in H, there exists n element m of Möb(H) such tht m(x) = y. Proof. Notice tht if for ech point x in H, there exists n element m of Möb(H) such tht m(x) = y 0 for some point in H, then Möb(H) cts trnsitively on H. Becuse for ny two points x nd z in H, if m 1 (x) = y 0 = m 2 (z), then m 1 2 m 1 (x) = z. So we just need to show tht for ny point x in H, there exists n element m of Möb(H) such tht m(x) = i. Write x = + bi, where, b R nd b > 0. We first move x to the imginry xis using γ 1 (z) = z, so m 1 (x) = bi. Then we pply γ 2 (z) = 1 b z to γ 1(x), so γ 2 (γ 1 (x)) = i. By Theorem 2.11, γ 1 nd γ 2 re elements of Möb(H), hence so is γ 2 γ 1. Theorem Möb(H) cts trnsitively on the set L of hyperbolic lines in H. Proof. It is suffice to show tht for ech hyperbolic line l, there exists n element m of Möb(H) tht tkes l to the imginry xis I. Suppose z is point in l. By Theorem 2.11, there is n element m of Möb(H) such tht m(z) = i. Let φ be the ngle between the two hyperbolic lines I nd m(l), mesured from I to m(l). For cos θz sin θ sin θz+cos θ ech θ, the Möbius trnsformtion n θ (z) = lies in Möb(H) nd fixes i. And the ngle between I nd n θ (I) t i, mesured from I to n θ (I) is 2θ. So if we tke θ = φ 2, then m(l) = n θ(i), so n 1 θ m(l) = I.

7 HYPERBOLIC PLANE AS A PATH METRIC SPACE 7 3. Length nd Distnce in H In this section, we will derive mens of mesuring lengths of pths in H tht is invrint under the ction of Möb, expressed s n invrint element of rc-length. Theorem 3.1. For every positive constnt c, the element of rc-length c Im(z) dz on H is invrint under the ction of Möb(H) Proof. Let ρ = c Im(z) b. We need to show tht ρ(f(t)) f (t) dt = b ρ(γ f(t)) γ (f(t)) f (t) dt for every piecewise C 1 pth f : [, b] H nd every element γ of Möb(H), which lso cn be written s b (ρ(f(t)) ρ(γ f(t)) γ (f(t)) ) f (t) dt = 0. So for n element γ of Möb(H), let µ γ (z) = ρ(z) ρ(γ(z)) γ (z). We just need to show tht b µ γ (z) dz = µ γ (f(t)) f (t) dt = 0. f Now consider how µ γ behves under composition with elements of Möb(H). Let γ nd φ be elements of Möb(H). Then µ γ φ = µ(z) µ((γ φ(z)) (γ φ) (z) = µ(z) µ((γ φ(z)) (γ (φ) (z)) φ (z) = µ(z) µ(φ(z)) φ (z) + µ(φ(z)) φ (z) µ((γ φ(z)) (γ (φ) (z)) φ (z) = µ φ (z) + µ γ (φ(z)) φ (z). So if µ γ = 0 for every γ in the generting set of Möb(H), then µ γ =0 for every element γ of Möb(H). A generting set for Möb(H) is Then {γ 1 = z + b, γ 2 = z, γ 3 = 1 z, γ 4 = z, b R, > 0}. µ γ1 = ρ(z) ρ(γ 1 (z)) γ 1(z) = ρ(z) ρ(z + b) = 0. Similrly for γ 2, γ 3 nd γ 4, one cn check tht µ γi = 0. This completes the proof. We cn not determine the specific vlue of c using only Möb(H). For esy clcultion, we set c =1. Definition 3.2. For piecewise C 1 pth f : [, b] H, we define the hyperbolic length of f to be length H (f)= b 1 Im(f(t)) f (t) dt. Definition 3.3. Suppose X is metric spce with metric d. We sy tht (X, d) is pth metric spce if for ech pir of x nd y in X, d(x, y) = inf{length(f) f is pth connecting x nd y}, nd there exists distnce-relizing pth f such tht d(x, y) = length(f).

8 8 QINGCI AN Definition 3.4. For ech pir of points x nd y of H, let Γ[x, y] denote the set of ll piecewise C 1 pths f : [, b] H with f() = x nd f(b) = y. We define the function d H : H H R, d H (x, y) = inf{length H (f) f Γ[x, y]}. We refer to d H (x, y) s the hyperbolic distnce between x nd y. Proposition 3.5. For every element γ of Möb(H) nd for every pir x nd y of points of H, we hve tht d H (x, y) = d H (γ(x), γ(y)). Proof. Let f : [, b] H be pth in Γ[x, y] nd γ be in Möb(H). As length H (f) is invrint under Möb(H), we hve length H (f)=length H (γ f). And observe tht for every pth f Γ[x, y], γ f Γ[x, y]. So d H (γ(x), γ(y)) = inf{ length H (g) g Γ[γ(x), γ(y)]} inf{length H (γ f) f Γ[x, y]} inf{length H (f) f Γ[x, y]} = d H (x, y). Now consider γ 1, which is lso n element of Möb(H). Then d H (x, y) = inf{length H (f) f Γ[x, y]} inf{length H (γ 1 g g Γ[γ(x), γ(y)]} inf{length H (g) g Γ[γ(x), γ(y)]} = d H (γ(x), γ(y)). Therefore, d H (x, y) = d H (γ(r), γ(y)). Theorem 3.6. (H, d H ) is pth metric spce. Proof. First, we show tht (H, d H ) is metric. As length H (f) = b 1 Im(f(t) f (t) dt is nonnegtive for ll pth f Γ(x, y), their infimum d H (x, y) is nonnegtive. Let f : [, b] H be pth in Γ[x, y], nd let h : [, b] [, b] given by h(t) = + b t. Then f h Γ[y, x] nd length H (f h) = b 1 Im(f(h(t)) f (h(t) h (t) dt = b 1 Imf(s) f (s) ds = length H(f). So every pth in Γ[x, y] cn be reprmetrized to become pth in Γ[y, x]. So {length H (f) f Γ[x, y]} = {length H (g) g Γ[y, x]}, hence they hve the sme infimum. So d H (x, y) = d H (y, x). Now we prove the tringle equlity by contrdiction. Suppose tht x, y, z re points in H such tht d H (x, z) > d H (x, y) + d H (y, z). Then let ɛ = d H (x, z) d H (x, y) d H (y, z) > 0. As d H (x, y) = inf{length H (f) f Γ[x, y]}, there exists pth f in Γ[x, y] with length H (f) d H (x, y) < 1 2 ɛ. Similrly there exists pth g in Γ[x, y] with length H(g) d H (y, z) < 1 2ɛ. Let h Γ[x, z] be the conctention of f nd g. Then d H (x, z) < length H (h) = length H (f) + length H (g) < d H (x, y) + d H (y, z) + ɛ,

9 HYPERBOLIC PLANE AS A PATH METRIC SPACE 9 which contrdicts the construction of ɛ. So the tringle equlity is stisfied by d H. Then we show tht there exists pth in H relizing the hyperbolic distnce between ny pir of points of H, which implies tht d H (x, y) > 0 if x y. So let x, y be pir of distinct point in H nd let l be the hyperbolic line pssing through x nd y. By Theorem 2.13, there exists n element γ Möb(H) so tht γ(l) is the positive imginry xis in H. We cn write γ(x) = µi nd γ(y) = λi with µ < λ. By Proposition 3.5, d H (x, y) = d H (γ(x), γ(y)). So we just need to prove tht there exist distnce-relizing pth from µi to λi.we begin by clculting the hyperbolic length of specific pth, f 0 : [µ, λ] H defined by f 0 (t) = it. So length H (f 0 ) = λ 1 µ t dt = ln λ µ.then we show tht length H(f 0 ) length H (f) for ny pth f Γ[µi, λi]. Let f : [, b] H be defined by f(t) = x(t) + y(t)i. Consider the pth g : [, b] H defined by g = y(t)i, so g Γ[µi, λi]. Since (x (t)) 2 0, length H (g) length H (f). So we reduced ourselves to showing tht if g Γ[µi, λi][ is of the form g(t) = y(t)i, then length H (f 0 ) length H (g). The imge of g is the hyperbolic line segment joining αi nd βi with α µ λ β. Define h : [α, β] H by h(t) = it. Then we hve tht length H (f 0 ) = ln λ µ ln β α = length H (h). Then we cn write g = h (h 1 g), where h 1 g : [, b] [α, β] is surjective function. Therefore length H (h) length H (g). This completes the rgument tht length H (f 0 ) length H (f) for every pth f in Γ[µi, λi]. Therefore, for every pir of distinct points x nd y in H, let l be the hyperbolic line through x nd y, nd let γ be the element of Möb(H) tking l to the imginry xis with γ(x) = µi nd γ(y) = λi where µ < λ. Define f 0 : [µ, λ] H by f 0 (t) = it. Then γ 1 f 0 is distnce-relizing pth in Γ[x, y]. 4. Isometries in H Definition 4.1. An isometry of metric spce (X, d) is homeomorphism f of X tht preserves distnce. Proposition 4.2. Let x, y nd z be distinct points in H. Then d H (x, y)+d H (y, z) = d H (x, z) if nd only if y is contined in the hyperbolic line segment joining x to z. Proof. By Theorem 2.11 nd Theorem 2.12, there exists n element m of Möb(H) such tht m(x) = i nd m(z) = βi, 1 < β. Then d H (x, z) = d H (i, βi) = ln β. Write m(y) = +bi. If y is contined in the hyperbolic line segment joining x nd z, then m(y) lies on the hyperbolic line segment joining m(x) nd m(y). So = 0, 1 < b < β nd d H (x, y) = ln b nd d H (y, z) = ln β b. Hence d H(x, y) + d H (y, z) = d H (x, z). Suppose now tht y is not contined in the hyperbolic line segment joining x to z. There re two cses. First cse: = 0, in other words, m(y) lies on the imginry xis. Then we hve either 0 < b < 1 or b > β. If 0 < b < 1, then d H (x, y) = ln b] nd d H (y, z) = ln β ln b.] As ln b < 0, d H (x, y) + d H (y, z) = d H (x, z) 2 ln b > d H (x, z). If b > β, then d H (x, y) = ln b nd d H (y, z) = ln b ln β. As ln b > d H (x, z), d H (x, y) + d H (y, z) = 2 ln b d H (x, z) < d H (x, z).

10 10 QINGCI AN The second cse is tht 0. Then d H (x, y) = d H (i, + bi) < d H (i, bi) nd d H (y, z) = d H ( + bi, βi) > d H (bi, βi). If 1 < b < β, then d H (x, z) = d H (i, β) = d H (i, bi) + d H (bi, βi) < d H (x, y) + d H (y, z). Similrly if 0 < b < 1, then d H (x, z) < d H (x, z) 2 ln b = d H (i, bi) + d H (bi, βi) < d H (x, y) + d H (y, z). If b > β, then d H (x, z) < 2 ln b d H (x, z) = d H (i, bi) + d H (bi, βi) < d H (x, y) + d H (y, z). Proposition 4.3. Let f be hyperbolic isometry nd let l be hyperbolic line, then f(l) is lso hyperbolic line. Proof. By Proposition 4.2, if y lies in the hyperbolic line segment l xz, then d H (x, y)+ d H (y, z) = d H (x, z). Since f preserves distnce, d H (f(x), f(y)) + d H (f(y), f(z)) = d H (f(x), f(z)). So f(y) lies in the hyperbolic segment l f(x)f(y) joining f(x) to f(y), hence f(l xy )= l f(x)f(y). As hyperbolic line cn be expressed s nested union of hyperbolic line segments, we hve tht hyperbolic isometries tke hyperbolic lines to hyperbolic lines. Proposition 4.4. Let (z 1, z 2 ) nd (w 1, w 2 ) be pirs of distinct points of H. If d H (z 1, z 2 ) = d H (w 1, w 2 ), then there exists n element m of Möb(H) stisfying m(z 1 ) = w 1 nd m(z 2 ) = w 2. Proof. We cn construct p nd q of Möb(H) such tht q(z 1 ) = p(w 1 ) = i nd q(z 2 ) = p(w 2 ) = e d H(z 1,z 2) i. So m = p 1 q stisfies m(z 1 ) = w 1 nd m(z 2 ) = w 2. Theorem 4.5. Let Isom(H, d H ) denote the group of isometries of (H, d H ), Then Isom(H, d H )=Möb(H). Proof. By Proposition 3.6, we hve tht Möb(H) Isom(H, d H ). Then we show the opposite inclusion using Proposition 4.3. Let f be hyperbolic isometry nd for ech pir of points p nd q of H, let l pq be the hyperbolic line segment joining p to q. Let l be the perpendiculr bisector of l pq, in other words, l = {x H d H (x, p) = d H (x, q)}. Then f(l) is the perpendiculr bisector of f(l pq ) = l f(p)f(q). Now let x nd y be points on the positive imginry xis I in H nd let H be one of the hlf-plnes in H determined by I. By Proposition 4.4, there exists n element m of Möb(H) such tht m(f(x)) = x nd m(f(y)) = y. Since m f fixes x nd y, m f tkes I to I. If necessry, replce m by B m with B(z) = z to hve m lso tke H to H. Let z be point on I. As z is uniquely determined by two hyperbolic distnce d H (x, z) nd d H (y, z) nd s m f preserves hyperbolic distnces, we hve tht m f fixes every point of I. Let w be ny point in H but not in I nd let l be the hyperbolic line through w perpendiculr to I. So l is the hyperbolic line contined in the Eucliden circle with Eucliden center 0 nd Eucliden rdius w. Let z be the intersection of l nd I. As l is the perpendiculr bisector of some hyperbolic line segment in I nd s m f fixes every point in I, m f(l)=l. As m f fixes z, s d H (z, w) = d H (m f(z), m f(w)) = d H (z, m f(w)), nd s m f preserves the two hlf-plnes determined by I, we hve tht m f fixes w.

11 HYPERBOLIC PLANE AS A PATH METRIC SPACE 11 Therefore m f is the identity. So f = m 1, hence f is n element of Möb(H). Therefore Isom(H, d H )=Möb(H). Acknowledgments. It is plesure to thnk my mentor, Kevin Csto, for his support, guidnce, dvice throughout the REU progrm. His support ws crucil to my understnding the mteril nd his ptience nd enthusism kept me motivted. I d lso like to thnk Peter My nd the Deprtment of Mthemtics t the University of Chicgo for providing me this wonderful undergrdute reserch nd study opportunity. References Jmes W. Anderson, Hyperbolic Geometry, Springer Undergrdute Mthemtics Series, 2005