Models of hyperbolic geometry

MINISTRY OF EDUCATION AND TRAINING HANOI PEDAGOGICAL UNIVERSITY N2 DEPARTMENT OF MATHEMATICS HOANG MAI PHUONG MODELS OF HYPERBOLIC GEOMETRY BACHELOR THESIS Major: Geometry HANOI - 2019 MINISTRY OF EDUCATION AND TRAINING HANOI PEDAGOGICAL UNIVERSITY N2 DEPARTMENT OF MATHEMATICS HOANG MAI PHUONG MODELS OF HYPERBOLIC GEOMETRY BACHELOR THESIS Major: Geometry Supervisor: Assoc Prof Dr NGUYEN THAC DUNG HANOI - 2019 Bachelor thesis HOANG MAI PHUONG Thesis Acknowledgement I would like to express my gratitude to the teachers in the geometry group as well as all the teachers of the Department of Mathematics, Hanoi Pedagogical University No.2 They have facilitated, supported me during the time I have done the thesis In particular, I would like to express my deep respect and gratitude to Assoc Prof Dr Nguyen Thac Dung, who has directed guidance, conscientiously counseled, helped me complete this thesis Due to limitation of time, capacity and conditions, so the thesis can not avoid any mistakes Thus, I am looking forward to receiving valuable comments from teachers and friends Thank you so much Hanoi, May 5, 2019 Student Hoang Mai Phuong i Bachelor thesis HOANG MAI PHUONG Thesis Assurance The thesis is completed after self-study and synthetic process with the guidance of Assoc Prof Dr Nguyen Thac Dung It is written by following the textbook entitled “Elementary Differential Geometry” by Andrew Pressley Most of the contents of this thesis are taken from the chapter 11: “Hyperbolic geometry” in the mentioned textbook While the thesis is completed, I consulted some documents which have been mentioned in the bibliography section I assure that this thesis is not copied from any other theses I certify that these statements are true and I will be responsible for their correctness Hanoi, May 5, 2019 Student Hoang Mai Phuong ii Contents Preface 1 Preliminaries 1.1 Smooth surfaces and the first fundamental form 1.2 The Măobius transformation and the inversion Some models of hyperbolic geometry 11 2.1 Upper half-plane model 11 2.2 Isometries of H 20 2.3 Poincar´e disc model 27 2.4 Hyperbolic parallels 35 2.5 Beltrami - Klein model 40 Conclusion 50 Bibliography 51 iii Bachelor thesis HOANG MAI PHUONG Preface Rationale Hyperbolic Geometry as known as non - Euclidean Geometry, is an important mathematical object It is a geometry that discards one of Euclid’s axioms and indicates that the geometry of space in which Euclid’s parallel axiom fails Hyperbolic geometry is closely connected to many other parts mathematics such as: differntial geometry, complex analysis, topology, etc So, It has attracted many international mathematical researchers and gained many significant achievements They have focused on not only researching some axioms, postulates but also exploiting the special models of Hyperbolic Geometry and using its important properties to solve the problems Desiring to study more about the properties, some related problems about models in Hyperbolic Geometry and its importance, I chosen the topic:“Models of Hyperbolic geometry” as the subject of undergraduate thesis Aims of the study and Research questions Starting acquainted with the scientific research and learn more about the models of hyperbolic geometry Research methods - Researching on textbook and materials related to problem’s reseach - Analysis and synthesis of knowledge Main contents Chapter 1: “Preliminaries” Chapter 2: “Some models of hyperbolic geometry” Chapter Preliminaries 1.1 Smooth surfaces and the first fundamental form Definition 1.1.1 A subset S of R3 is a surface if, for every point p ∈ S, there is an open set U in R2 and an open set W in R3 containing p such that S ∩ W is homeomorphic to U A subset of a surface S of the form S ∩ W , where W is an open subset of R3 , is called an open subset of S A homeomorphism σ : U → S ∩ W as in this definition is called a surface patch or parametrization of the open subset S ∩ W of S Definition 1.1.2 A surface patch σ : U → R3 is called regular if it smooth and the vectors σu and σv are linearly independent at all points (u, v) ∈ U If S is a surface, an allowable patch for S is a regular surface patch σ : U → R3 such that σ is a homeomorphism from U to an open subset of S A smooth surface is a surface S such that, for any point p ∈ S, there is an allowable surface patch σ as above such that p ∈ σ(U ) Example 1.1.2.1 The plane is given by formula: σ (u, v) = a + up + vq is a smooth surface It’s clearly smooth and σu = p and σv = q are linearly independent because p and q were chosen to be perpendicular unit vectors Bachelor thesis HOANG MAI PHUONG Definition 1.1.3 A tangent vector to a smooth surface S at a point p ∈ S is the tangent vector at p of a smooth curve in S passing through p The tangent space Tp S of S at p is the set of all tangent vectors to S at p Definition 1.1.4 Let p be a point of a smooth surface S The first fundamental form of S at p associates to tangent vectors v, w ∈ Tp S the scalar v, w p,S = v · w Example 1.1.4.1 Suppose that σ(u, v) is a surface patch of S Then, any tangent vector to S at a point p in the image of σ can be expressed uniquely as a linear combination of σu and σv Define maps du : Tp S → R and dv : Tp S → R by du(v) = λ dv(v) = µ if v = λσu + µσv , for some λ, µ ∈ R We can see that du and dv are linear maps Then, using the fact that , is a symmetric bilinear form, we have v, v = λ2 σu , σv + 2λµ σu , σv + µ2 σu , σv Writing E = σu , F = σu · σ v , G = σv , this becomes v, v = Eλ2 + 2F λµ + Gµ2 = Edu(v)2 + 2F du(v)dv(v) + Gdv(v)2 Traditionally, the expression Edu2 + 2F dudv + Gdv is called the first fundamental form of the surface patch σ(u, v) Bachelor thesis HOANG MAI PHUONG Example 1.1.4.2 For the plane σ (u, v) = a + up + vq with p and q being perpendicular unit vectors, we have σu = p, σv = q, so E = σu = p = 1, F = σu · σv = p · q = 0, G = σv = q = Hence, the first fundamental form is du2 + dv Definition 1.1.5 If S1 and S2 are surfaces, a conformal map f : S1 → S2 is a local diffeomorphism such that, if γ1 and γ˜1 are any two curves on S1 that intersect, say at a point p ∈ S1 , and if γ2 and γ˜2 are their images under f , the angle of intersection of γ1 and γ˜1 at p is equal to the angle of intersection of γ2 and γ˜2 at f (p) Proposition 1.1.6 A local diffeomorphism f : S1 → S2 is conformal if and only if, for any surface patch σ of S1 , the first fundamental forms of the patches σ of S1 and f ◦ σ of S2 are proportional In particular, a surface patch σ(u, v) is conformal if and only if its first fundamental form is λ(du2 + dv ) for some smooth function λ(u, v) Example 1.1.6.1 The catenoid can be parametrized by σ(u, v) = ((cosh u) cos v, (cosh u) sin v, u) is conformal Indeed, now we will find the first fundamental form of σ(u, v) We have: σu = (sinh u cos v, sinh u sin v, 1) σv = (− cosh u sin v, cosh u cos v, 0) These imply, E = σu = cosh2 u, F = 0, G = σv = cosh2 u Bachelor thesis HOANG MAI PHUONG So, the first fundamental form is cosh2 u(du2 + dv ) By Proposition 1.1.6, σ(u, v) is conformal Example 1.1.6.2 We consider the unit sphere S If q is any point of S other than the north pole n = (0, 0, 1), the straight line joining n and q intersects the xy-plane at some point p The map that takes q to p is called stereographic projection from S to the plane, and we denote it by Π We are going to show that Π is conformal Let p = (u, v, 0), q = (x, y, z) Since, p, q, n lie on a straight line, there is a scalar ρ such that q − n = ρ(p − n), and hence (x, y, z) = (0, 0, 1) + ρ((u, v, 0) − (0, 0, 1)) = (ρu, ρv, − ρ) Hence, ρ = − z, u = (1.1) x y ,v = and we have 1−z 1−z Π(x, y, z) = x y , ,0 1−z 1−z On the other hand, from Eq 1.1 and x2 + y + z = we get ρ = and hence u2 + v + q= 2u 2v u2 + v − , , u2 + v + u2 + v + u2 + v + If we denote the right-hand side by σ1 (u, v), then σ1 is a parametrization of S with the north pole removed Parametrizing the xy-plane by σ2 (u, v) = (u, v, 0), we then have Π(σ1 (u, v)) = σ2 (u, v) Bachelor thesis HOANG MAI PHUONG We can therefore assume that l is the real axis and that a = ir where r = d The circle m through a that touches the real axis at has centre c = + iR and radius R for some R > 0, and so has equation |z − − iR| = R Since m passes through ir, we have |−1 + i(r − R)| = R, which gives R= + r2 2r In the right-angled (Euclidean) triangle with vertices a, iR and c, the hypotenuse in perpendicular to m, so the angle of the triangle at a is π − Π Hence, by Euclidean trigonometry, R sin Π = R − r and we get − tanh2 21 d r 2r2 − r2 sin Π = − = − = = = R + r2 + r2 + 21 d cosh d 37 Bachelor thesis HOANG MAI PHUONG The proof is complete One characterization of parallel lines in Euclidean plane geometry is that such lines have a common perpendicular In hyperbolic geometry, this property characterizes ultra-parallels: Proposition 2.4.3 Two hyperbolic lines in DP are ultra-parallel if and only if they have a common perpendicular line (i.e., a hyperbolic line that intersects them both at right angles) In that case, the common perpendicular is unique In Euclidean plane geomertry, two parallel lines have infinitely many common perpendiculars Proof Suppose that l, m are two hyperbolic lines in D that are ultraparallel Assume l is the real axis and m is a circle with centre a, radius r We have: |a|2 = r2 + (2.7) −1 < Re(a) < (2.8) Then, 38 Bachelor thesis HOANG MAI PHUONG Indeed, m intersect l at point v if and only if |v − a| = r From (2.7), we get: v − 2vRe(a) + = (2.9) If |Re(a)| > 1, then (2.9) has two distinct real roots whose product is equal to 1, then one root v satisfies −1 < v < This implies that l, m intersect in DP (contradict with assume) If |Re(a)| = 1, (2.9) has two roots ±1, so l touches m at or −1 on the boundary of DP (contradict with assume) Thus, (2.8) must holds Conversely, suppose that l, m are two hyperbolic lines in DP such that they have a common perpendicular n that intersect them at two points a, b We assume that l, n are real and imaginary axis, respectively and that a is the origin Then m is a part of the circle with centre iR on the imaginary axis (|R| > 1) Since, m ∩ C at right angle, the radius r of C satisfies R2 = r + In particularly, |R| > r, then m doesn’t intersect real axis Therefore, l and m are ultra-parallel Now, we consider a circle with centre b on the real axis and radius s That circle intersect both m and C at right angle if and only if  b2 = s2 + |b − a|2 = r2 + s2 , s = Re(a)−2 − Re(a) and the corresponding circle n is the unique common perpendicular to If Re = 0, these equations has roots b = l and m If Re = 0, it is clear that the imaginary axis is the unique common perpendicular 39 Bachelor thesis 2.5 HOANG MAI PHUONG Beltrami - Klein model This model is constructed by using two projections of the unit sphere S The first projection is the stereographic projection map Π from S to the xy-plane It defines a diffeomorphism from the ‘southern hemisphere’ S−2 = (x, y, z) ∈ S |z < to the unit disc D = (x, y, 0) ∈ R3 |x2 + y < The second projection is the ‘vertical’ projection of R3 onto the xy-plane: pr(x, y, z) = (x, y, 0) This also defines a diffeomorphism from S−2 to D Hence, the composite map K = pr ◦ Π−1 : D → D is a diffeomorphism From Example 1.1.6.2, we have −1 Π (v, w) = 2v 2w v + w2 − , , , v + w2 + v + w2 + v + w2 + so, K(v, w) = 2(v, w) v + w2 + Then it is easy to see that, if we identify the xy-plane with C by (x, y, 0) → x + iy as usual, we have K(z) = 2z , |z|2 + 40 z ∈ D (2.10) Bachelor thesis HOANG MAI PHUONG Definition 2.5.1 The Beltrami-Klein model DK of non - Euclidean geometry is the disc D equipped with the first fundamental form for which the diffeomorphism K : DP → DK is an isometry Proposition 2.5.2 The first fundamental form of DK is (1 − w)2 dv + 2vwdvdw + − v dw2 (1 − v − w2 )2 The proof of this proposition was computed in Exercise 8.3.1(iii) in textbook [1] The Beltrami - Klein model has the following remarkable property: Proposition 2.5.3 The hyperbolic lines in Beltrami-Klein model are the (Euclidean) straight line segments contained in the side DK Proof Let be a line segment passing through two points a, b on C The curve on S which corresponds to under projection pr is the intersec- tion of S with the plane perpendicular to xy-plane that contains This is semicircle m, it intersects C at two right angles at two points a, b 41 Bachelor thesis HOANG MAI PHUONG Since Π is conformal map that takes circle in S to lines and circles in the xy-plane Π(m) is an arc of a circle in D which intersects with boundary of D at right angles or Π(m) is a hyperbolic line in DP Hence, every line segment in DK is a hyperbolic line Since a line segment passing through any point of DK in any given direction, these must be all of the hyperbolic lines in DK Corollary 2.5.4 DK is not a conformal model of hyperbolic geometry Proof Consider a hyperbolic triangle in DK Following Proposition 2.5.4, we refer that it is a triangle Euclidean Then, the sum of angles in this triangle (Euclidean) is equal to π But by Theorem 2.1.6, the sum of its internal hyperbolic angles is less than π That’s contradictory The proof is complete 42 Bachelor thesis HOANG MAI PHUONG If a ∈ C and |a| > 1, define the perspectivity Πa : DK → DK with centre a as follows Let z ∈ DK and let l be any hyperbolic line in DK passing through z Thus, l is a (Euclidean) line segment that intersects C at two points, say p and q Let the lines through a and p and through a and q intersect C at r and s, respectively (if the line through a and p happens to be tangent to C at p, then r = p; and similarly for the line through a and q) Then, Πa (z) is defined to be the point of intersection of the line through a and z with the line through r and s (see the diagram above) Proposition 2.5.5 With above notion, Πa = K ◦ Ia,r ◦ K−1 where r = |a|2 − In particular, Πa is an isometry of DK To prove this we need: 43 Bachelor thesis HOANG MAI PHUONG Lemma 2.5.6 Let l and m be hyperbolic lines in DK and suppose that these line intersect C at the point b, c and d, e, respectively Suppose that the tangents to C at b and c intersect at a, and that the extension of m passes through a Then, l and m intersect at right angles in the hyperbolic sense Proof Hyperbolic lines K−1 (l), K−1 (m) in DP corresponding to l and m are circular arcs that intersect C at right angles at the points b, c and d, e, respectively Take I is the inversion of circle of which K−1 (l) is an arc So that I is an isometry of DP Now, I takes K−1 (m) to circle arc such that it intersect C at right angles and it interchanges the point d, e It follows that I preseves K−1 (m) This means that K−1 (l), K−1 (m) are perpendicular in Euclidean Since DP is a conformal model, then K−1 (l), K−1 (m) are perpendicular in the hyperbolic sense Moreover, K : DP → DK is an isometry, this follows that l, m are perpendicular in the hyperbolic sense Returning proof Proposition 2.5.5: 44 Bachelor thesis HOANG MAI PHUONG Proof Let the tangents from a to C touch C at b, c and m be a line segment through b, c Take a line o passing through a and z that intersects C at t, u Let l be an any line through z and l intersects C at p, q Take r, s are two points of C such that the line passing through p, r and q, s passing through a Let n be a line segment through r, s Since z is the intersection of l and o then K−1 (z) is the intersection of K−1 (l) and K−1 (o) Similarly, K−1 (Πa (z)) is the intersection of K−1 (n) and K−1 (o) By Lemma 2.5.6, m and o are perpendicular in DK , so K−1 (m) and K−1 (o) are perpendicular in DP This means that Ia,r fixes K−1 (o) Since, Ia,r takes p → r and q → s, so it takes K−1 (l) → K−1 (n) Hence, Ia,r takes K−1 (z) to K−1 (Πa (z)) : Ia,r (K−1 (z)) = K−1 (Πa (z)) This is what we wanted to prove Definition 2.5.7 If a, b, c, d are distinct complex numbers, their crossratio is (a, b; c, d) = (a − c)(b − d) (a − d)(b − c) Proposition 2.5.8 Suppose that the points a, b, c, d lie on a line and that a and b are between c and d Then, (a, b; c, d) > Moreover, if p is a point distinct from a, b, c, d and the lines through p and each of the 45 Bachelor thesis HOANG MAI PHUONG points a, b, c, d intersect another line at a , b , c , d , then (a, b; c, d) = (a , b; c , d ) This result is expressed by saying that the cross-ratio is a “projective invariant”: The cross-ratio of four points on a line is unchanged when they are “projected” from some point p onto another line Proof Let l be the line that containing a, b, c, d Since a, b lie on the same side with c, arg(a − c) = arg(b − c), so a − c |a − c| = b−c |b − c| Similarly, b−d |b − d| = a − d |a − d| Hence, (a, b; c, d) = |a − c| |b − d| |a − d| |b − c| In particular, this cross-ratio is a positive number Let apb is an angle between the line passes p, a and the line passes p, b 46 Bachelor thesis HOANG MAI PHUONG Applying Sine’s Theorem in the triangle (Euclidean) We get: |a − c| |p − c| |a − d| |p − d| = ; = , sin apc sin pac sin apd sin pad |b − c| |p − c| |b − d| |p − d| = ; = sin bpc sin pbc sin bpd sin pbd Thus, (a, b; c, d) = sin apc sin bpd sin apd sin bpc Otherwise, we have a pc = apc Therefore, (a, b; c, d) = sin a pc sin b pd = (a , b ; c , d ) sin a pd sin b pc The proof is complete In particular, the cross ratio (a, b; c, d) with a, b, c, d ∈ DK is unchanged if a, b, c, d are subjected to any perspectivity or any rotation about the origin Theorem 2.5.9 Let a, b ∈ DK and c, d be the points of intersection of the line through a, b with C Then, the Beltrami-Klein distance between a, b is |ln(a, b; c, d)| Proof We use a suitable isometry of DK to reduce to the case in which dD K (a, b) = a and b are real Let l be the line through c and 1, and m be the line through d and −1 We consider two cases, according to whether l and m are parallel (in the Euclidean sense) or not If l and m are parallel, the line joing c and d passes through the origin, and a suitable rotation about the origin will take c to 1, d to −1 and a, b to points a , b on the real axis If l and m intersect at a point p If |p| > 1, the perspectivity Πp takes c to 1, d to −1 and a, b to point a , b on the line going −1 and 1,i.e., the 47 Bachelor thesis HOANG MAI PHUONG real axis If |p| < 1, the line l joining c and and m joining d and intersect at a point p with |p | = and the perspectivity Πp takes c to −1, d to and a, b to point a , b on the real axis We compute the distance dDK (a, b) = dDK (a , b ) by transfering to DP using the isometry K : DP → DK So, dDK (a , b ) = dDP (K−1 (a ), K−1 (b )) Using Proposition 2.3.2, we get: K−1 (b ) dDK (a , b ) = (1 + K−1 (b ))(1 − K−1 (a )) 2dv = ln − v2 (1 + K−1 (a ))(1 − K−1 (a )) (2.11) K−1 (a ) Using the formular (2.10) for K, we find that: K−1 (λ) = 1− λ − λ2 , λ ∈ D, which implies that + K−1 (λ) = − K−1 (λ) 1+λ 1−λ Using this, (2.11) becomes: dDK (a , b ) = (1 + b )(1 − a ) ln (1 − b )(1 + a ) (2.12) On the other hand, we have seen that there is a perspectivity or a rotation about the origin that takes (a, b, c, d) to (a , b , 1, −1) or (a , b , −1, 1) with a , b ∈ R, and that these transformation of DK leave the cross-ratio unchanged In the first case, (a, b; c, d) = (a , b ; 1, −1) = (1 − a )(1 + b ) , (1 + a )(1 − b ) and in the second case, (a, b; c, d) = (a , b ; −1, 1) = 48 (1 + a )(1 − b ) , (1 − a )(1 + b ) Bachelor thesis HOANG MAI PHUONG so in both cases, we have a result as was to be shown dDK (a, b) = dDK (a , b ) = |ln(a, b; c, d)| The proof is complete 49 Bachelor thesis HOANG MAI PHUONG Conclusion In this thesis, we have presented systematically the following results: Recall the definition and some properties about smooth surface and the first fundamental form Introduce about The Măobius transformation and its connection with some models of Hyperbolic geometry Introduce about three models of Hyperbolic geometry, there are Upper half-plane model, Poincar´e disc model, Beltrami - Klein model and presented some properties related to them 50 Bibliography [1] Andrew Pressley, Elementary Differential Geometry, second ed., Springer Undergraduate Mathematics Series, Springer, 2010 [2] Matthew Harvey, Geometry Illuminated: An Illustrated Introduction to Euclidean and Hyperbolic Plane Geometry, The Mathematical Association of America, 2015 51 ...MINISTRY OF EDUCATION AND TRAINING HANOI PEDAGOGICAL UNIVERSITY N2 DEPARTMENT OF MATHEMATICS HOANG MAI PHUONG MODELS OF HYPERBOLIC GEOMETRY BACHELOR THESIS Major: Geometry Supervisor: Assoc Prof Dr... Rationale Hyperbolic Geometry as known as non - Euclidean Geometry, is an important mathematical object It is a geometry that discards one of Euclid’s axioms and indicates that the geometry of space... special models of Hyperbolic Geometry and using its important properties to solve the problems Desiring to study more about the properties, some related problems about models in Hyperbolic Geometry