# 程序代写案例-MATH32051

ook is probably the best source for the second half of the course (sections 19–42). Beardon’s book contains everything in the course, and much more. 4 MATH32051 2. Where we are going 2. Where we are going: introduction and motivation §2.1 Introduction One purpose of this course is to provide an introduction to some aspects of hyperbolic geom- etry. Hyperbolic geometry is one of the richest areas of mathematics, with connections not only to geometry but also to dynamical systems, chaos theory, number theory, relativity, and many other areas of mathematics and physics. Unfortunately, it would be impossible to discuss all of these aspects of hyperbolic geometry within the confines of a single lecture course. Instead, we will develop hyperbolic geometry in a way that emphasises the similar- ities and (more interestingly!) the many differences with Euclidean geometry (that is, the ‘real-world’ geometry that we are all familiar with). §2.2 Euclidean geometry Euclidean geometry is the study of geometry in the Euclidean plane R2, or more generally in n-dimensional Euclidean space Rn. This is the geometry that we are familiar with from the real world. For example, in a right-angled triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides; this is Pythagoras’ Theorem. But what makes Euclidean geometry ‘Euclidean’? And what is ‘geometry’ anyway? One convenient meta-definition is due to Felix Klein (1849-1929) in his Erlangen programme (1872), which we paraphrase here: given a set with some structure and a group of trans- formations that preserve that structure, geometry is the study of objects that are invariant under these transformations. For 2-dimensional Euclidean geometry, the set is the plane R2 equipped with the Euclidean distance function (the normal way of defining the distance between two points) together with a group of transformations (such as rotations, transla- tions) that preserve the distance between points. A rotation or translation a triangle is still a triangle, so triangles are objects that are invariant under these transformations; in terms of the Erlangen programme, this means that studying triangles forms part of the study of Euclidean geometry. Similarly, as a rotation or translation of a circle is still a circle, this means that the study of circles falls under the remit of Euclidean geometry. We will define hyperbolic geometry in a similar way: we take a set, define a notion of distance on it, and study the transformations which preserve this distance. §2.3 Distance in the Euclidean plane Consider the Euclidean plane R2. Take two points x, y ∈ R2. What do we mean by the distance between x and y? If x = (x1, x2) and y = (y1, y2) then one way of calculating the distance between x and y is by using Pythagoras’ Theorem: distance(x, y) = ‖x− y‖ = √ (y1 − x1)2 + (y2 − x2)2; (3.1) this is the length of the straight line drawn in Figure 2.1. Writing d(x, y) for distance(x, y) we can see that there are some natural properties satisfied by this formula for distance: 1 MATH32051 2. Where we are going x1 x2 y2 y1 x y y2 − x2 y1 − x1 Figure 2.1: The (Euclidean) distance from x to y is the length of the ‘straight’ line joining them. (i) d(x, y) ≥ 0 for all x, y with equality if and only if x = y, (ii) d(x, y) = d(y, x) for all x, y, (iii) d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z. Thus, condition (i) says that the distance between any pair of distinct points is positive, condition (ii) says that the distance from x to y is the same as the distance from y to x, and condition (iii) says that that distance between two points is increased if we go via a third point. This is often called the triangle inequality and is illustrated in Figure 2.2. z y x Figure 2.2: The triangle inequality: d(x, z) ≤ d(x, y) + d(y, z). In mathematics, it is often fruitful to pick out useful properties of known objects and abstract them. If we have a set X and a function d : X ×X → R that behaves in the way that we expect distance to behave (that is, d satisfies conditions (i), (ii) and (iii) above), then we call X a metric space and we call d a distance function or a metric. Because of our familiarity with Euclidean geometry, there are often issues surrounding our definitions that we do not realise need to be proved. For example, we define the 2 MATH32051 2. Where we are going distance between x, y ∈ R2 by (3.1) and recognise that the straight line drawn from x to y in Figure 2.1 represents the shortest ‘path’ from x to y: any other path drawn from x to y would have a longer length. However, this needs proof. Note also that we have said that this straight line is ‘the’ shortest path; there are two statements here, firstly that there is a path of shortest length between x and y, and secondly that there is only one such path. These statements again need to be proved. Consider the surface of the Earth, thought of as the surface of a sphere. See Figure 2.3. The paths of shortest length are arcs of great circles. Between most pairs of points, there is a unique path of shortest length; in Figure 2.3 there is a unique path of shortest length from A to B. However, between pairs of antipodal points (such as the ‘north pole’ N and ‘south pole’ S) there are infinitely many paths of shortest length. Moreover, none of these paths of shortest length are ‘straight’ lines in R3. This indicates that we need a more careful approach to defining distance and paths of shortest length. A B N S Figure 2.3: There is just one path of shortest length from A to B, but infinitely many from N to S. The way that we shall regard distance as being defined is as follows. Because a priori we do not know what form the paths of shortest length will take, we need to work with all paths and be able to calculate their length. We do this by means of path integrals. Having done this, we now wish to define the distance d(x, y) between two points x, y. We do this defining d(x, y) to be the minimum of the lengths of all paths from x to y. In hyperbolic geometry, we begin by defining the hyperbolic length of a path. The hyperbolic distance between two points is then defined to be the minimum of the hyperbolic lengths of all paths between those two points. We then prove that this is indeed a metric, and go on to prove that given any pair of points there is a unique path of shortest length between them. We shall see that in hyperbolic geometry, these paths of shortest length are very different to the straight lines that form the paths of shortest length in Euclidean geometry. In order to avoid saying ‘straight line’ we instead call a path of shortest length a geodesic. §2.4 Groups and isometries of the Euclidean plane §2.4.1 Groups Recall that a group G is a set of elements together with a group structure: that is, there is a group operation such that any two elements of G can be ‘combined’ to give another 3 MATH32051 2. Where we are going element of G (subject to the ‘group axioms’). If g, h ∈ G then we denote their ‘combination’ (or ‘product’, if you prefer) by gh. The group axioms are: (i) associativity: if g, h, k ∈ G then (gh)k = g(hk); (ii) existence of an identity: there exists an identity element e ∈ G such that ge = eg = g for all g ∈ G; (iii) existence of inverses: for each g ∈ G there exists g−1 ∈ G such that gg−1 = g−1g = e. A subgroup H ⊂ G is a subset of G that is in itself a group. §2.4.2 Isometries An isometry is a map that preserves distances. There are some obvious maps that preserve distances in R2 using the Euclidean distance function. For example: (i) the identity map e(x, y) = (x, y) (trivially, this preserves distances); (ii) a translation τ(a1,a2)(x, y) = (x+ a1, y + a2) is an isometry; (iii) a rotation of the plane is an isometry; (iv) a reflection (for example, reflection in the y-axis, (x, y) 7→ (−x, y)) is an isometry. One can show that the set of all isometries of R2 form a group, and we denote this group by Isom(R2). We shall only be interested in orientation-preserving isometries. (We will not define orientation-p
ound of [1, 2] as 0 ≤ a for all a ∈ [1, 2]. Similarly −10 is a lower bound, as is 1. Indeed, any number less than are equal to 1 is a lower bound. Hence the greatest lower bound is of [1, 2] is 1, so that inf[1, 2] = 1. (ii) Similarly, if one considers the open interval (3, 4) ⊂ R then any number less than or equal to 3 is a lower bound. Hence inf[3, 4] = 3. (iii) Examples (i) and (ii) show that the infimum of a subset A, if it exists, may or may not be an element of A. This is why we use the term infimum of A rather than the (perhaps more familiar) term minimum of A. (iv) The infimum of a given subset A ⊂ R need not exist. This happens when A when A is unbounded from below. For example, the set (−∞, 0) ⊂ R does not have an infimum. We now define the hyperbolic distance between two points in H. Definition. Let z, z′ ∈ H. We define the hyperbolic distance dH(z, z′) between z and z′ to be dH(z, z ′) = inf{lengthH(σ) | σ is a piecewise continuously differentiable path with end-points z and z′}. 3 MATH32051 4. Length and distance in the upper half-plane Remark. Thus we consider all piecewise continuously differentiable paths between z and z′, calculate the hyperbolic length of each such path, and then take the shortest. Later we will see that this infimum is achieved by a path (a geodesic), and that this path is unique. 4 MATH32051 5. Circles and lines in the complex plane 5. Circles and lines in the complex plane §5.1 Introduction We are interested in the following problem: given two points w, z in H, what is the path of shortest hyperbolic length between them? (A path achieving the shortest length is called a geodesic.) The purpose of this section is to give a useful method for simultaneously treating circles and lines in the complex plane. This will provide a useful device for calculating and working with the geodesics in H. Recall that we can identify R2 with C by identifying the point (x, y) ∈ R2 with the complex number x+ iy ∈ C. We are familiar with the equations for a straight line and for a circle in R2; how can we express these equations in C? §5.2 Lines First consider a straight (Euclidean) line L in R2. Then the equation of L has the form: ax+ by + c = 0 (2.1) for some choice of a, b, c ∈ R. Write z = x+ iy. Recalling that the complex conjugate of z is given by z¯ = x− iy it is easy to see that x = 1 2 (z + z¯), y = 1 2i (z − z¯). Substituting these expressions into (2.1) we have a ( 1 2 (z + z¯) ) + b ( 1 2i (z − z¯) ) + c = 0, and simplifying gives 1 2 (a− ib)z + 1 2 (a+ ib)z¯ + c = 0. Let β = (a− ib)/2. Then the equation of L is βz + β¯z¯ + c = 0. (2.2) §5.3 Circles Now let C be a circle in R2 with centre (x0, y0) and radius r. Then C has the equation (x − x0)2 + (y − y0)2 = r2. Let z = x + iy and z0 = x0 + iy0. Then C has the equation |z − z0|2 = r2. Recalling that |w|2 = ww¯ for a complex number w ∈ C, we can write this equation as (z − z0)(z − z0) = r2. 1 MATH32051 5. Circles and lines in the complex plane Expanding this out (and recalling that z − z0 = z¯ − z¯0) we have that zz¯ − z¯0z − z0z¯ + z0z¯0 − r2 = 0. Let β = −z¯0 and γ = z0z¯0 − r2 = |z0|2 − r2. Then C has the equation zz¯ + βz + β¯z¯ + γ = 0. (3.1) Remark. Observe that if we multiply an equation of the form (2.2) or (3.1) by a non-zero constant then the resulting equation determines the same line or circle. We can combine (2.2) and (3.1) as follows: Proposition 5.3.1 Let A be either a circle or a straight line in C. Then A has the equation αzz¯ + βz + β¯z¯ + γ = 0, (3.2) where α, γ ∈ R and β ∈ C. Remark. Thus equations of the form (3.2) with α = 0 correspond to straight lines, and equations of the form (3.2) with α 6= 0 correspond to circles. In the latter case, we can always divide equation (3.2) by α to obtain an equation of the form (3.1). §5.4 Geodesics in H A particularly important class of circles and lines in C are those for which all the coefficients in (3.2) are real. By examining the above analysis, we have the following result. Proposition 5.4.1 Let A be a circle or a straight line in C satisfying the equation αzz¯ + βz + β¯z¯ + γ = 0. Suppose β ∈ R. Then A is either (i) a circle with centre on the real axis, or (ii) a vertical straight line. R Figure 5.1: Circles and lines with real coefficients in (3.2). Later in the course we will see that the geodesics (the paths of shortest hyperbolic length) in the upper half-plane model of hyperbolic space are precisely the intersections 2 MATH32051 5. Circles and lines in the complex plane of the circles and lines appearing in Proposition 5.4.1 with the upper half-plane. Note that a circle in C with a real centre meets the real axis orthogonally (meaning: at right- angles); hence the intersection of such a circle with the upper half-plane H is a semi-circle. Instead of saying ‘circles in C with real centres’ we shall often say ‘circles in C that meet R orthogonally’. Definition. Let H denote the set of semi-circles orthogonal to R and vertical lines in the upper half-plane H. 3 MATH32051 6. Mobius transformations of H 6. Mo¨bius transformations of H §6.1 Introduction An isometry is a distance preserving transformation. Examples of isometries in Euclidean geometry include: rotations, translations and reflections in straight lines. In spherical ge- ometry, the isometries include rotations and reflections in great circles. In general, having a large set of isometries means that there is a lot of symmetry in the underlying geometry, and therefore the geometry is likely to be interesting. Hyperbolic geometry has a particu- larly nice set of isometries, known as Mo¨bius transformations; the goal of this section is to introduce these. §6.2 Mo¨bius transformations of H Definition. Let a, b, c, d ∈ R be such that ad− bc > 0 and define the map γ : H→ H by γ(z) = az + b cz + d . Transformations of H of this form are called Mo¨bius transformations of H. Proposition 6.2.1 Let γ ∈ Mo¨b(H) be a Mo¨bius transformation of H. Then γ is a well-defined map γ : H→ H (that is, if z ∈ H then γ(z) ∈ H). Moreover, γ maps H to itself bijectively and γ−1 is also a Mo¨bius transformation. Proof. This is an exercise (see Exercise 6.1). ✷ Recall that a group is a setG together with a map G×G→ G (denoted by juxtaposition) such that the following axioms hold: (i) associativity: g1(g2g3) = (g1g2)g3, (ii) existence of an identity element: there exists e ∈ G such that eg = ge = g for all g ∈ G, (iii) existence of inverses: for all g ∈ G there exists g−1 ∈ G such that gg−1 = g−1g = e. One of the main aims of this course is to study the set of Mo¨bius transformations on H. We have the following important result. Proposition 6.2.2 Let Mo¨b(H) denote the set of all Mo¨bius transformations of H. Then Mo¨b(H) is a group under composition. Proof. This is an exercise (see Exercise 6.2). ✷ 1 MATH32051 6. Mobius transformations of H Remark. The group operation is composition: given two Mo¨bius transformations γ1, γ2 ∈ Γ, we denote by γ1γ2 the composition γ1 ◦ γ2. (Important note! This is not multiplication of the two complex numbers γ1(z)γ2(z); it is the composition γ1(γ2(z)).) Examples of Mo¨bius transformations of H include: dilations z 7→ kz (k > 0), transla- tions z 7→ z + b, and the inversion z 7→ −1/z. Let H ∈ H be one of our candidates for a geodesic in H, namely H is either a semi-circle or a straight line orthogonal to the real axis. We show that a Mo¨bius transformation of H maps H to another such candidate. Proposition 6.2.3 Let H be either (i) a semi-circle orthogonal to the real axis, or (ii) a vertical straight line. Let γ be a Mo¨bius transformation of H. Then γ(H) is either a semi-circle orthogonal to the real axis or a vertical straight line. Proof. By Proposition 6.2.1 we know that Mo¨bius transformations of H map the upper half-plane to itself bijectively. Hence it is sufficient to show that γ maps vertical straight lines in C and circles in C with real centres to vertical straight lines and circles with
ing this is to show that one of our candidate geodesics can be mapped onto the imaginary axis by a Mo¨bius transformation of H. Remark. Our candidates for the geodesics can be described uniquely by their end points in ∂H. Semi-circles orthogonal to R have two end points in R, and vertical lines have one end point in R and the other at ∞. Lemma 9.2.1 Let H ∈ H and let z0 be a poing on H. Then there exists a Mo¨bius transformation of H that maps H to the imaginary axis and z0 to i. Proof. We first show how to move H to the imaginary axis. Suppose that H is the vertical line Re(z) = a. Then the translation γ1(z) = z − a is a Mo¨bius transformation of H that maps H to the imaginary axis Re(z) = 0. Now suppose that H be a semi-circle with end points a, b ∈ R, with a < b. First note that, by the remark above, the imaginary axis is characterised as the unique element of H with end-points at 0 and ∞. Consider the map γ1(z) = z − b z − a. 1 MATH32051 9. The geodesics of H As −a + b > 0 this is a Mo¨bius transformation of H. By Proposition 6.2.3 we know that γ1(H) ∈ H. Clearly γ1(b) = 0 and γ1(a) =∞, so γ1(H) must be the imaginary axis. In either case, we have found γ1, a Mo¨bius transformation of H, that maps H to the imaginary axis. Now consider γ1(z0). This is a point on the imaginary axis. Let γ1(z0) = i/k for some k > 0. Let γ2(z) = kz be a dilation by a factor of k. Then γ2 ∈ Mo¨b(H) and maps the imaginary axis to itself. Moreover γ1(z0) = i. Let γ = γ2 ◦ γ1. Then γ maps H to the imaginary axis and maps the point z0 to i. ✷ We can now prove that the geodesics in the upper half-plane are the vertical straight lines in H and the semi-circles in H with a real centre. Theorem 9.2.2 The geodesics in H are the semi-circles orthogonal to the real axis and the vertical straight lines. Moreover, given any two points in H ∪ ∂H there exists a unique geodesic passing through them. Proof. Let z, w ∈ H∪∂H. Then we can always find a unique element of H ∈ H containing z, w: if z and w have the same real part then H will be a vertical straight line, otherwise H will be a semi-circle with a real centre. Let z, w ∈ H. Let H ∈ H be the unique vertical straight line or semi-circle with real centre passing through z and w. Let γ ∈ Mo¨b(H) be a Mo¨bius transformation of H that maps H to the imaginary axis; this exists by Lemma 9.2.1. Let [z, w] denote the arc of H from z to w. Let σ denote any other path from z to w. By Proposition 6.2.3 we know that γ([z, w]) is the arc of imaginary axis from γ(z) to γ(w). Moreover, γ ◦ σ is a path from γ(z) to γ(w). By Proposition 8.1.1 we know that lengthH(γ([z, w])) ≤ lengthH(γ ◦ σ) with equality if and only if γ ◦ σ is equal to the arc of imaginary axis from γ(z) to γ(w). Applying Proposition 7.2.1 we have that lengthH([z, w]) ≤ lengthH(γ ◦ σ) with equality if and only if σ is equal to [z, w]. Hence [z, w] is the unique path of shortest length from z to w. ✷ 2 MATH32051 10. Area, angles and the parallel postulate 10. Some loose ends: angles, area and the parallel postulate §10.1 Introduction In this section we cover a few topics that will be important later on, namely angles and area in hyperbolic geometry. We also discuss briefly Euclid’s parallel postulate. The parallel postulate will not play a role later in the course; however, it has huge importance in the history of the subject and in geometry more widely. §10.2 Angles Suppose that we have two paths σ1 and σ2 that intersect at the point z ∈ H. By choosing appropriate parametrisations of the paths, we can assume that z = σ1(0) = σ2(0). The angle between σ1 and σ2 at z is defined to be the angle between their tangent vectors at the point of intersection and is denoted by ∠σ′1(0), σ ′ 2(0), θ θ σ1 σ2(i) (ii) Figure 10.1: (i) The angle between two vectors, (ii) The angle between two paths at a point of intersection. It will be important for us to know that Mo¨bius transformations preserve angles. That is, if σ1 and σ2 are two paths that intersect at z with angle θ, then the paths γσ1 and γσ2 intersect at γ(z) also with angle θ. If a transformation preserves angles, then it is called conformal. Proposition 10.2.1 Let γ ∈ Mo¨b(H) be a Mo¨bius transformation of H. Then γ is conformal. We can use the (Euclidean) cosine rule to calculate the angle between two geodesics. Proposition 10.2.2 Let C1 and C2 be two circles in R 2 with centres c1, c2 and radii r1, r2, respectively. Suppose C1 and C2 intersect. Let θ denote the internal angle at the point of intersection (see Figure 10.2.2). Then cos θ = |c1 − c2|2 − (r21 + r22) 2r1r2 . Proof. This is Exercise 10.1. ✷ 1 MATH32051 10. Area, angles and the parallel postulate θ Figure 10.2: The internal angle between two circles. In the case where one geodesic is vertical and the other is a semicircle, the following result tells us how to calculate the angle between them. Proposition 10.2.3 Suppose that two geodesics intersect as illustrated in Figure 10.3. Then sin θ = 2ab a2 + b2 , cos θ = b2 − a2 a2 + b2 . θ ib a Figure 10.3: The angle between two geodesics in the case where one is a vertical straight line. Proof. This is Exercise 10.2. ✷ §10.3 Area Let A ⊂ H be a subset of the upper half-plane. The hyperbolic area of A is defined to be the double integral AreaH(A) = ∫ ∫ A 1 y2 dx dy = ∫ ∫ A 1 Im(z)2 dz. (3.1) Again, it will be important for us to know that Mo¨bius transformations of H preserve area. This is contained in the following result. 2 MATH32051 10. Area, angles and the parallel postulate Proposition 10.3.1 Let A ⊂ H and let γ ∈ Mo¨b(H) be a Mo¨bius transformation of H. Then AreaH(γ(A)) = AreaH(A). Remark. There are some measure-theoretic technicalities at work here that we have chosen, for simplicity, to ignore. It turns out that it is not possible to define the area of every subset A of H. One needs to assume, for example, that A is a Borel set. All open subset and all closed subsets of H (see §22.2) are Borel sets, and so it makes sense to talk about the area of these. If you do a course on Fractal Geometry or on Lebesgue Integration then you will learn more about the technicalities you need to be aware of. §10.4 Euclid’s parallel postulate fails Recall from §2.7 that Euclid’s parallel postulate states that: given any infinite straight line and a point not on that line, there exists a unique infinite straight line through that point and parallel to the given line. This is true in Euclidean geometry but false in hyperbolic geometry. We can now see why Euclid’s parallel postulate fails in H. Indeed, given any geodesic and any point not on that geodesic there exist infinitely many geodesics through that point that do not intersect the given geodesic (see Figure 10.4). P H Figure 10.4: There are infinitely many geodesics through P that do not intersect the geodesic H. §10.5 Appendix: Towards Riemannian geometry §10.5.1 Introduction The aim of this appendix is to explain why hyperbolic angles and Euclidean angles are the same, why Mo¨bius transformations of H are conformal, why we define hyperbolic area as we do, and why Mo¨bius transformations of H are area-preserving. This is somewhat outside the scope of the course as it is best explained using ideas that lead on to a more general construction called Riemannian geometry of which hyperbolic geometry is one particular case. §10.5.2 Angles We defined angles in the upper half-plane model of hyperbolic geometry to be the same as angles in Euclidean geometry. To see why this is the case (and, indeed, to see how the 3 MATH32051 10. Area, angles and the parallel postulate concept of ‘angle’ is actually defined) we need to make a slight diversion and recall some facts from linear algebra. Let us first describe how angles are defined in the Euclidean plane R2. Let (x, y) ∈ R2 and suppose that v = (v1, v2) and w = (w1, w2) are two vectors at the point (x, y). We (x, y) w v θ Figure 10.5: The angle between two vectors v,w and the point (x, y). define an inner product 〈·, ·〉 between tw
easy to check that h maps the upper half-plane H bijectively to the Poincare´ disc D. One can also check that h maps ∂H to ∂D bijectively. §12.2 Distances in the Poincare´ disc We give a formula for the distance between two points in the Poincare´ model of the hyper- bolic plane. We do this (as we did in the upper half-plane) by first defining the length of a (piecewise continuously differentiable) path, and then defining the distance between two points to be the infimum of the lengths of all such paths joining them. Let g(z) = h−1(z). Then g maps D to H and has the formula g(z) = −z + i −iz + 1 . Let σ : [a, b] → D be a path in D (strictly, this is a parametrisation of a path). Then g ◦ σ : [a, b]→ H is a path in H. The length of g ◦ σ is given by: lengthH(g ◦ σ) = ∫ b a |(g ◦ σ)′(t)| Im(g ◦ σ(t)) dt = ∫ b a |g′(σ(t))||σ′(t)| Im(g ◦ σ(t)) dt, 1 MATH32051 12. The Poincare´ disc model using the chain rule. It is easy to calculate that g′(z) = −2 (−iz + 1)2 and Im(g(z)) = 1− |z|2 | − iz + 1|2 . Hence lengthH(g ◦ σ) = ∫ b a 2 1− |σ(t)|2 |σ ′(t)| dt. (2.1) We define the length of the path σ in D by (2.1): lengthD(σ) = ∫ b a 2 1− |σ(t)|2 |σ ′(t)| dt = ∫ σ 2 1− |z|2 . In the upper half-plane, we integrate 1/ Im(z) along a path to obtain its length; in the Poincare´ disc, we integrate 2/(1 − |z|2) instead. The distance between two points z, z′ ∈ D is then defined by taking the length of the shortest path between them: dD(z, z ′) = inf{lengthD(σ) | σ is a piecewise continuously differentiable path from z to z′}. As we have used h to transfer the distance function on H to a distance function on D we have that dD(h(z), h(w)) = dH(z, w), (2.2) where dH denotes the distance in the upper half-plane model H. Proposition 12.2.1 Let x ∈ [0, 1). Then dD(0, x) = log ( 1 + x 1− x ) . Moreover, the real axis is the unique geodesic joining 0 to x. §12.3 Mo¨bius transformations of the Poincare´ disc Let γ ∈ Mo¨b(H). Then we obtain an isometry of the Poincare´ disc D by using the map h to transform γ into a map of D. To see this, consider the map hγh−1 : D → D. Then for any u, v ∈ D dD(hγh −1(u), hγh−1(v)) = dH(γh−1(u), γh−1(v)) = dH(h −1(u), h−1(v)) = dD(u, v), where we have used (2.2) and the fact that γ is an isometry of H. Hence hγh−1 is an isometry of D. One can show (this is Exercise 12.1) that z 7→ hγh−1(z) is a map of the form z 7→ αz + β β¯z + α¯ , α, β ∈ C, |α|2 − |β|2 > 0. This suggests the following definition. 2 MATH32051 12. The Poincare´ disc model Definition. We call a map of the form αz + β β¯z + α¯ , α, β ∈ C, |α|2 − |β|2 > 0. a Mo¨bius transformation of D. The set of all Mo¨bius transformations of D forms a group, which we denote by Mo¨b(D). Examples of Mo¨bius transformations of D include the rotations. Take α = eiθ/2, β = 0. Then |α|2 − |β|2 = 1 > 0 so that γ(z) = eiθ/2z/e−iθ/2 = eiθz is a Mo¨bius transformation of D. Observe that this map is a rotation of the unit circle in C. §12.4 Geodesics in the Poincare´ disc The geodesics in the Poincare´ disc are the images under h of the geodesics in the upper half-plane H. Figure 12.1: Some geodesics in the Poincare´ disc D. Proposition 12.4.1 The geodesics in the Poincare´ disc are the diameters of D and the arcs of circles in D that meet ∂D at right-angles. Proof (sketch). One can show (using the same arguments as in §10.5) that h is confor- mal, i.e. h preserves angles. Using the characterisation of lines and circles in C as solutions to αzz¯+βz+ β¯z¯+γ = 0 one can show that h maps circles and lines in C to circles and lines in C. Recall that h maps ∂H to ∂D. Recall that the geodesics in H are the arcs of circles and lines that meet ∂H orthogonally. As h is conformal, the image in D of a geodesic in H is a circle or line that meets ∂D orthogonally. ✷ In the upper half-plane model H we often map a geodesic H to the imaginary axis and a point z0 on that geodesic to the point i. The following is the analogue of this result in the Poincare´ disc model. Proposition 12.4.2 Let H be a geodesic in D and let z0 ∈ H. Then there exists a Mo¨bius transformation of D that maps H to the real axis and z0 to 0. 3 MATH32051 12. The Poincare´ disc model Proof (sketch). Consider h−1(H); this is a geodesic in H. Moreover, the point h−1(z0) is a point on this geodesic. By Lemma 9.2.1 we can find a Mo¨bius transformation γ ∈ Mo¨b(H) such that γ(h−1(H)) is the imaginary axis and γ(h−1(z0)) = i. Note that hγh−1 ∈ Mo¨b(D) maps H to the imaginary axis and z0 to 0. Finally we apply the rotation z 7→ e−iπ/2z so that the imaginary axis in D is mapped to the real axis and maps 0 to itself. ✷ Recall that an arbitrary straight line in C or an arbitrary circle in C has the equation αzz¯ + βz + β¯z¯ + γ where α, γ ∈ R and β ∈ C. We saw that the special case of vertical straight lines and circles with real centres have equations of the form αzz¯ + βz + βz¯ + γ where α, β, γ ∈ R and that these describe the geodesics in H. Can we identify the equations that correspond to the geodesics in D? The following result gives the answer to this. Proposition 12.4.3 Straight lines that pass through the origin (and so are diameters of D) and circles that meet the unit circle at right angles have equations of the form αzz¯ + βz + β¯z¯ + α = 0 where α ∈ R and β ∈ C. Proof. See Exercise 12.4. ✷ §12.5 Area in D Recall that the area of a subset A ⊂ H is defined to be AreaH(A) = ∫ ∫ A 1 (Im z)2 dz. We can again use h to transfer this definition to D. Indeed, one can check that if A ⊂ D then AreaD(A) = ∫ ∫ A 4 (1− |z|2)2 dz. §12.6 A dictionary Upper half-plane Poincare´ disc H = {z ∈ C | Im(z) > 0} D = {z ∈ C | |z| < 1} Boundary ∂H = R ∪ {∞} ∂D = {z ∈ C | |z| = 1} Length of a path σ ∫ b a 1 Imσ(t) |σ′(t)| dt ∫ b a 2 1− |σ(t)|2 |σ ′(t)| dt Area of a subset A ∫ ∫ A 1 (Im z)2 dz ∫ ∫ A 4 (1− |z|2)2 dz Orientation-preserving γ(z) = az + b cz + d , γ(z) = αz + β β¯z + α¯ , isometries a, b, c, d ∈ R, α, β ∈ C, ad− bc > 0 |α|2 − |β|2 > 0 Geodesics vertical half-lines diameters of D and semi-circles and arcs of circles orthogonal to ∂H that meet ∂D orthogonally Angles Same as Euclidean Same as Euclidean angles angles 4 MATH32051 13. The Gauss-Bonnet Theorem 13. The Gauss-Bonnet Theorem §13.1 Hyperbolic polygons In Euclidean geometry, an n-sided polygon is a subset of the Euclidean plane bounded by n straight lines. Thus the edges of a Euclidean polygon are formed by segments of Euclidean geodesics. A hyperbolic polygon is defined in an analogous manner. Definition. Let z, w ∈ H ∪ ∂H. Then there exists a unique geodesic that passes through both z and w. We denote by [z, w] the part of this geodesic that connects z and w. We call [z, w] the segment or arc of geodesic between z and w. Definition. Let z1, . . . , zn ∈ H ∪ ∂H. Then the hyperbolic n-gon P with vertices at z1, . . . , zn is the region of H bounded by the geodesic segments [z1, z2], . . . , [zn−1, zn], [zn, z1]. (i) (ii) Figure 13.1: A hyperbolic triangle (i) in the upper half-plane model, (ii) in the Poincare´ disc. Remark. Notice that we allow some of the vertices to lie on the boundary of the hyper- bolic plane. Such a vertex is called an ideal vertex. If all the vertices lie on ∂H then we call P an ideal polygon. Notice that the angle at an ideal vertex is zero; this is because all geodesics meet ∂H at right-angles and so the angle between any two such geodesics is zero. 1 MATH32051 13. The Gauss-Bonnet Theorem (i) (ii) Figure 13.2: An ideal triangle (i) in the upper half-plane model, (ii) in the Poincare´ disc. §13.2 The Gauss-Bonnet Theorem for a triangle The Gauss-Bonnet Theorem can be stated in a wide range of contexts and at many lev- els of generality, far beyond the setting we describe here. In hyperbolic geometry, the Gauss-Bonnet Theorem gives a formula for the area of a hyperbolic polygon in term
s of its angles—a result that has no analogue in Euclidean geometry. We will use the Gauss-Bonnet Theorem to study tessellations of the hyperbolic plane by regular polygons, and we will see that there are infinitely many distinct tessellations using regular polygons (whereas in Euclidean geometry there are only finitely many: equilateral triangles, squares, and regular hexagons). Theorem 13.2.1 (Gauss-Bonnet Theorem for a hyperbolic triangle) Let ∆ be a hyperbolic triangle with internal angles α, β and γ. Then AreaH(∆) = π − (α+ β + γ). (2.1) Remarks. 1. In Euclidean geometry it is well-known that the sum of the internal angles of a Euclidean triangle is equal to π (indeed, this is equivalent to the parallel postulate). In hyperbolic geometry, (2.1) implies that the sum of the internal angles of a hyperbolic triangle is strictly less than π. 2. The equation (2.1) implies that the area of a hyperbolic triangle is at most π. The only way that the area of a hyperbolic triangle can be equal to π is if all the internal angles are equal to zero. This means that all of the vertices of the triangle lie on the circle at infinity, i.e. the triangle is an ideal triangle. 3. In Euclidean geometry, the angles of a triangle do not determine the triangle’s area (this is clear: scaling a triangle changes its area but not its angles). This is not the case in hyperbolic geometry. Proof. Let ∆ be a hyperbolic triangle with internal angles α, β and γ. We first study the case when at least one of the vertices of ∆ belongs to ∂H, and hence the angle at this vertex is zero. Recall that Mo¨bius transformations of H are conformal (that is, they preserve angles) and area-preserving. By applying a Mo¨bius transformation of H, we can map the vertex on the boundary to ∞ without altering the area or the angles. 2 MATH32051 13. The Gauss-Bonnet Theorem By applying the Mo¨bius transformation z 7→ z + b for a suitable b we can assume that the circle joining the other two vertices is centred at the origin in C. By applying the Mo¨bius transformation z 7→ kz we can assume it has radius 1. Hence (see Figure 13.3) AreaH(∆) = ∫ ∫ ∆ 1 y2 dx dy = ∫ b a (∫ ∞ √ 1−x2 1 y2 dy ) dx = ∫ b a (−1 y ∣∣∣∣ ∞ √ 1−x2 ) dx = ∫ b a 1√ 1− x2 dx = ∫ β π−α −1 dθ substituting x = cos θ = π − (α+ β). This proves (2.1) when one of the vertices of ∆ lies on ∂H. ∆ α β α β 0a b π − α Figure 13.3: The Gauss-Bonnet Theorem with one vertex of ∆ at ∞. Now suppose that ∆ has no vertices in ∂H. Let the vertices of ∆ be A, B and C, with internal angles α, β and γ, respectively. Apply a Mo¨bius transformation of H so that the side of ∆ between vertices A and C lies on a vertical geodesic. Let δ be the angle at B between the side CB and the vertical. This allows us to construct two triangles, each with one vertex at ∞: triangle AB∞ and triangle CB∞. See Figure 13.4. AreaH(∆) = AreaH(ABC) = AreaH(AB∞)−AreaH(BC∞). Now AreaH(AB∞) = π − (α+ (β + δ)) AreaH(BC∞) = π − ((π − γ) + δ). Hence AreaH(ABC) = π − (α+ (β + δ))− (π − ((π − γ) + δ)) = π − (α+ β + γ). 3 MATH32051 13. The Gauss-Bonnet Theorem ✷ A α γ π − γ C β δ B Figure 13.4: The Gauss-Bonnet Theorem for the triangle ABC with no vertices on ∂H. We can generalise Theorem 13.2.1 to give a formula for the area of an n-sided polygon. Theorem 13.2.2 (Gauss-Bonnet Theorem for a hyperbolic polygon) Let P be an n-sided hyperbolic polygon with vertices v1, . . . , vn and internal angles α1, . . . , αn. Then AreaH(P ) = (n− 2)π − (α1 + · · ·+ αn). (2.2) Proof (sketch). Cut up P into triangles. Apply Theorem 13.2.1 to each triangle and then sum the areas. ✷ 4 MATH32051 14. Tiling the hyperbolic plane 14. Tiling the hyperbolic plane §14.1 Introduction Recall that a regular n-gon is an n-gon where all n sides have the same length and all internal angles are equal. We are interested in the following problem: when can we tile the plane using regular n-gons with k polygons meeting at each vertex? It is known that the only possible tessellations of R2 are given by: equilateral triangles (with 6 triangles meeting at each vertex), squares (with 4 squares meeting at each vertex), and by regular hexagons (with 3 hexagons meeting at each vertex). In hyperbolic geometry, the situation is far more interesting: there are infinitely many different tessellations by regular polygons! §14.2 Tiling the hyperbolic plane using regular hyperbolic polygons Theorem 14.2.1 There exists a tessellation of the hyperbolic plane by regular hyperbolic n-gons with k polygons meeting at each vertex if and only if 1 n + 1 k < 1 2 . (2.1) Proof. We only prove that if there is a tessellation then n, k satisfy (2.1), the converse is harder. Let α denote the internal angle of a regular n-gon P . Then as k such polygons meet at each vertex, we must have that α = 2π/k. As the area of the polygon P must be positive, substituting α = 2π/k into (2.2) and re-arranging we have: 1 n + 1 k < 1 2 , as required. ✷ Figures 14.1, 14.2 and 14.3 illustrate some tilings of the hyperbolic plane. In Figure 14.1, the Poincare´ disc is tiled by regular hyperbolic octagons, with 4 octagons meeting at each vertex. In Figure 14.2, the Poincare´ disc is tiled by regular hyperbolic pentagons, with 4 pentagons meeting at each vertex. In Figure 14.3, the Poincare´ disc is tiled by regular hyperbolic quadrilaterals (hyperbolic squares), with 8 quadrilaterals meeting at each ver- tex. All of the hyperbolic octagons (respectively pentagons, quadrilaterals) in Figure 14.1 (respectively Figure 14.2, Figure 14.3) have the same hyperbolic area and the sides all have the same hyperbolic length. They look as if they are getting smaller as they approach the boundary of the hyperbolic plane because we are trying to represent all of the hyperbolic plane in the Euclidean plane, and necessarily some distortion must occur. You are already familiar with this: when one tries to represent the surface of the Earth on a sheet of (Eu- clidean!) paper, some distortion occurs as one tries to flatten out the sphere; in Figure 14.4, Greenland appears unnaturally large compared to Africa when the surface of the Earth is projected onto the plane. 1 MATH32051 14. Tiling the hyperbolic plane Figure 14.1: A tessellation of the Poincare´ disc with n = 8, k = 4. Remark. Here is a tiling of the hyperbolic plane that you can scroll around in: https://www.math.univ-toulouse.fr/ cheritat/AppletsDivers/Escher/index. You can make your own tilings of the hyperbolic plane here: http://www.malinc.se/m/ImageTiling.php. Remark. Here are two games, playable online for free, that take place in hyperbolic space: • ‘Hyperbolic Maze’: http://www.madore.org/~david/math/hyperbolic-maze.html • ‘Hyperrogue’: http://www.roguetemple.com/z/hyper. One technical point that we have glossed over is the existence of regular n-gons in hyperbolic geometry. To see that such polygons exist we quote the following result. Proposition 14.2.2 Let α1, . . . , αn be such that (n− 2)π − n∑ k=1 αk > 0. Then there exists a polygon with internal angles αk. Proof. See Theorem 7.16.2 in Beardon’s book. ✷ Remark. One can show that if the internal angles of a hyperbolic polygon are all equal then the lengths of the sides are all equal. (This is not true in Euclidean geometry: a rectangle has right-angles for all of its internal angles, but the sides are not all of the same length.) 2 MATH32051 14. Tiling the hyperbolic plane Figure 14.2: A tessellation of the Poincare´ disc with n = 5, k = 4. Figure 14.3: A tessellation of the Poincare´ disc with n = 4, k = 8. 3 MATH32051 14. Tiling the hyperbolic plane Figure 14.4: When projected onto a (Euclidean) plane using the Mercator projection, the surface of the Earth is distorted. 4 MATH32051 15. Hyperbolic trigonometry 15. Hyperbolic trigonometry §15.1 Right-angled triangles In Euclidean geometry there are many well-known relationships between the sides and the angles of a right-angled triangle. For example, Pythagora
s’ Theorem gives a relationship between the three sides. Here we study the corresponding results in hyperbolic geometry. Throughout this section, ∆ will be a right-angled triangle. The internal angles will be α, β, π/2, with the opposite sides having lengths a, b, c. §15.2 Two sides, one angle For a right-angled triangle in Euclidean geometry there are well-known relationships be- tween an angle and any of two of the sides, namely ‘sine = opposite / hypotenuse’, ‘cosine = adjacent / hypotenuse’ and ‘tangent = opposite / adjacent’. Here we determine similar relationships in the case of a hyperbolic right-angled triangle. Proposition 15.2.1 Let ∆ be a right-angled triangle in H with internal angles α, β, π/2 and opposing sides with lengths a, b, c. Then (i) sinα = sinh a/ sinh c, (ii) cosα = tanh b/ tanh c, (iii) tanα = tanh a/ sinh b. Proof. As in the proof of Theorem 11.4.1, we can apply a Mo¨bius transformation of H to ∆ and assume without loss in generality that the vertices of ∆ are at i, ki and s+ it, where s+ it lies in the unit circle centred at the origin and the right-angle occurs at i. The vertices at ik and s + it lie on a unique geodesic. This geodesic is a semi-circle with centre x ∈ R. The (Euclidean) straight line from x to ik is inclined at angle α from the real axis. See Figure 15.1. The line from x to ik is a radius of this semi-circle, as is the line from x to s+ it. Calculating the lengths of these radii, we see that k2 + x2 = (s+ x)2 + t2 so that k2 = 1 + 2sx, (2.1) using the fact that s2 + t2 = 1. By considering the Euclidean triangle with vertices at x, ik, 0, we see that tanα = k x = 2ks k2 − 1 , (2.2) where we have substituted for x from (2.1). 1 MATH32051 15. Hyperbolic trigonometry 0 a i b ik c α β s+ it α x Figure 15.1: The point x is the centre of the semi-circle corresponding to the geodesic through ik and s+ it. Using the facts that cosh2− sinh2 = 1 and tanh = sinh / cosh it follows from (4.2) and (4.3) that sinh b = k2 − 1 2k , tanh a = s. Combining this with (2.2) we see that tanα = tanh a sinh b , proving statement (iii) of the proposition. The other two statements follow by using trigonometric identities, relationships between sinh and cosh, and the hyperbolic version of Pythagoras’ Theorem. ✷ §15.3 The angle of parallelism Consider the special case of a right-angled triangle with one ideal vertex. (Recall that a vertex is said to be ideal if it lies on the boundary.) In this case, the internal angles of the triangle are α, 0 and π/2 and the only side with finite length is that between the vertices with internal angles α and π/2. The angle of parallelism is a classical term for this angle expressed in terms of the side of finite length. Proposition 15.3.1 Let ∆ be a hyperbolic triangle with angles α, 0 and π/2. Let a denote the length of the only finite side. Then (i) sinα = 1cosh a ; (ii) cosα = 1coth a ; (iii) tanα = 1sinh a . Proof. The three formulæ for α are easily seen to be equivalent. Therefore we need only prove that (i) holds. 2 MATH32051 15. Hyperbolic trigonometry After applying a Mo¨bius transformation of H, we can assume that the ideal vertex of ∆ is at ∞ and that the vertex with internal angle π/2 is at i. The third vertex is then easily seen to be at cosα+ i sinα. See Figure 15.2. i a α α cosα+ i sinα Figure 15.2: The angle of parallelism. Recall that cosh dH(z, w) = 1 + |z − w|2 2 Im(z) Im(w) . Applying this formula with z = i and w = cosα+ i sinα we see that cosh a = cosh dH(z, w) = 1 + 2(1 − sinα) 2 sinα = 1 sinα . ✷ §15.4 Non-right-angled triangles: the sine rule Recall that in Euclidean geometry the sine rule takes the following form. In a triangle (not necessarily right-angled) with internal angles α, β and γ and side lengths a, b and c we have sinα a = sin β b = sin γ c . The hyperbolic version of this is the following. Proposition 15.4.1 Let ∆ be a hyperbolic triangle with internal angles α, β and γ and side lengths a, b, c. Then sinα sinh a = sin β sinh b = sin γ sinh c . §15.5 Non-right-angled triangles: cosine rules §15.5.1 The cosine rule I Recall that in Euclidean geometry we have the following cosine rule. Consider a triangle (not necessarily right-angled) with internal angles α, β and γ and sides of lengths a, b and c, with side a opposite angle α, etc. Then c2 = a2 + b2 − 2ab cos γ. 3 MATH32051 15. Hyperbolic trigonometry The corresponding hyperbolic result is. Proposition 15.5.1 Let ∆ be a hyperbolic triangle with internal angles α, β and γ and side lengths a, b, c. Then cosh c = cosh a cosh b− sinh a sinh b cos γ. Proof. See Anderson’s book. ✷ §15.5.2 The cosine rule II The second cosine rule is the following. Proposition 15.5.2 Let ∆ be a hyperbolic triangle with internal angles α, β and γ and side lengths a, b, c. Then cosh c = cosα cos β + cos γ sinα sin β . Proof. See Anderson’s book. ✷ Remark. The second cosine rule has no analogue in Euclidean geometry. Observe that the second cosine rule implies the following: if we know the internal angles α, β, γ of a hyperbolic triangle, then we can calculate the lengths of its sides. In Euclidean geometry, the angles of a triangle do not determine the lengths of the sides. 4 MATH32051 16. Fixed points of Mo¨bius transformations 16. Fixed points of Mo¨bius transformations §16.1 Introduction Recall that a transformation γ : H→ H of the form γ(z) = az + b cz + d where a, b, c, d ∈ R, ad − bc > 0, is called a Mo¨bius transformation of H. The aim of the next few sections is to classify the different types of behaviour that Mo¨bius transforma- tions exhibit. We will see that there are three different classes of Mo¨bius transformation: parabolic, elliptic and hyperbolic. To do this classification, we first look at the number and possible locations of fixed points for a given Mo¨bius transformation. §16.1.1 Fixed points of Mo¨bius transformations Let γ be a Mo¨bius transformation of H. We say that a point z0 ∈ H ∪ ∂H is a fixed point of γ if γ(z0) = az0 + b cz0 + d = z0. (1.1) Our initial classification of Mo¨bius transformations is based on how many fixed points a given Mo¨bius transformation has and whether they lie in H or on the circle at infinity ∂H. Clearly the identity map is a Mo¨bius transformation which fixes every point. Through- out this section, we will assume that γ is not the identity. Let us first consider the case when ∞ ∈ ∂H is a fixed point. Recall that we calculate γ(∞) by writing γ(z) = a+ b/z c+ d/z and noting that as z →∞ we have 1/z → 0. Hence γ(∞) = a/c. Thus ∞ is a fixed point of γ if and only if γ(∞) =∞, and this happens if and only if c = 0. Suppose that∞ is a fixed point of γ so that c = 0. What other fixed points can γ have? Observe that now γ(z0) = a d z0 + b d . Hence γ also has a fixed point at z0 = b/(d − a). Note that if a = d then this point will be ∞; in this case, ∞ is the only fixed point. However, if a 6= d then b/(d − a) is a real number and so we obtain a second fixed point on the boundary ∂H. Thus if∞ ∈ ∂H is a fixed point for γ then γ has at most one other fixed point, and this fixed point also lies on ∂H. Now let us consider the case when ∞ is not a fixed point of γ. In this case, c 6= 0. Multiplying (1.1) by cz0+d (which is non-zero as z0 6= −d/c) we see that z0 is a fixed point if and only if cz20 + (d− a)z0 − b = 0. (1.2) 1 MATH32051 16. Fixed points of Mo¨bius transformations This is a quadratic in z0 with real coefficients. Hence there are either (i) one or two real solutions, or (ii) two complex conjugate to (1.2). In the latter case, only one solution lies in H ∪ ∂H. Thus we have proved: Proposition 16.1.1 Let γ be a Mo¨bius transformation of H and suppose that γ is not the identity. Then γ has either: (i) two fixed points in ∂H and none in H; (ii) one fixed point in ∂H and none in H; (iii) no fixed points in ∂H and one in H. Corollary 16.1.2 Suppose γ is a Mo¨bius transformation of H with three or more fixed points. Then γ is the
Mo¨bius transformation of H of the form z 7→ z + b is called a translation. We need the following lemma. Lemma 18.3.1 Let γ(z) = z+ b. If b > 0 then show that γ is conjugate to γ(z) = z+1. If b < 0 then show that γ is conjugate to γ(z) = z − 1. Are z 7→ z − 1, z 7→ z + 1 conjugate? Proof. See Exercise 18.3. ✷ We can now classify all parabolic Mo¨bius transformations. Proposition 18.3.2 Let γ be a Mo¨bius transformation of H 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 7→ z + 1 or to the translation z 7→ z − 1. Proof. By Proposition 17.3.1 we know that (i) and (ii) are equivalent. Clearly (iv) implies (iii) and Lemma 18.3.1 implies that (iii) implies (iv). Suppose that (iv) holds. Recall that z 7→ z+1 has a unique fixed point at∞. Hence if γ is conjugate to z 7→ z+1 then γ has a unique fixed point in ∂H, and is therefore parabolic. The same argument holds for z 7→ 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 Mo¨bius transformation of H that maps ζ to ∞. Consider the Mo¨bius transformation gγg−1. This is conjugate to γ as γ = g−1(gγg−1)g. Moreover, gγg−1 has a unique fixed point at∞. To see this, suppose that z0 is a fixed point of gγg−1. Then gγg−1(z0) = z0 if and only if γ(g−1(z0)) = g−1(z0). Thus g−1(z0) is a fixed point of γ. As γ has a unique fixed point at ζ, it follows that g−1(z0) = ζ, i.e. z0 = g(ζ) = ∞. Hence gγg−1 has a unique fixed point at ∞. We claim that gγg−1 is a translation. Write gγg−1(z) = az + b cz + d . As ∞ is a fixed point of gγg−1, we must have that c = 0 (see Lecture 16). 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. Let b′ = b/d so that gγg−1(z) = z+b′. Hence γ is conjugate to a translation. ✷ 2 MATH32051 18. Parabolic, hyperbolic and elliptic Mo¨bius transformations §18.4 Hyperbolic transformations Recall that a Mo¨bius transformation of H is said to be hyperbolic if it has exactly two fixed points on ∂H. For example, let k > 0 and suppose that k 6= 1. Then the Mo¨bius transformation γ(z) = kz of H is hyperbolic. The two fixed points are 0 and ∞. In general, a Mo¨bius transformation of the form z 7→ kz where k 6= 1 is called a dilation. It is not the case that any two dilations are conjugate. Indeed, we have the following result. Lemma 18.4.1 Two dilations z 7→ k1z, z 7→ k2z are conjugate (as Mo¨bius transformations of H) if and only if k1 = k2 or k1 = 1/k2. Proof. See Exercise 18.4. ✷ We can now classify hyperbolic Mo¨bius transformations. Proposition 18.4.2 Let γ ∈ Mo¨b(H) be a Mo¨bius transformation of H. Then the following are equivalent: (i) γ is hyperbolic; (ii) τ(γ) > 4; (iii) γ is conjugate to a dilation, i.e. γ is conjugate to a Mo¨bius transformation of H of the form z 7→ kz, for some k > 0. Proof. We have already seen in Proposition 17.3.1 that (i) is equivalent to (ii). Suppose that (iii) holds. Then γ is conjugate to a dilation. We have already seen that a dilation has two fixed points in ∂H, namely 0 and ∞. Hence γ also has exactly two fixed points in ∂H. Hence (i) holds. Finally, we prove that (i) implies (iii). We first make the remark that if γ fixes both 0 and ∞ then γ is a dilation. To see this, write γ(z) = az + b cz + d where ad− bc > 0. As ∞ is a fixed point of γ, we must have that c = 0 (see Lecture 16). Hence γ(z) = (az + b)/d. As 0 is fixed, we must have that b = 0. Hence γ(z) = (a/d)z so that γ is a dilation. Suppose that γ is a hyperbolic Mo¨bius transformation of H. Then γ has two fixed points in ∂H; denote them by ζ1, ζ2. First suppose that ζ1 =∞ and ζ2 ∈ R. Let g(z) = z−ζ2. Then the Mo¨bius transforma- tion gγg−1 is conjugate to γ; this is because γ = g−1(gγg−1)g. Moreover, gγg−1 has fixed points at 0 and∞. To see this, note that gγg−1(z0) = z0 if and only if γ(g−1(z0)) = g−1(z0), that is g−1(z0) is a fixed point of γ. Hence z0 = g(ζ1) or z0 = g(ζ2), i.e. z0 = 0 or ∞. By the above remark, gγg−1 is a dilation. Now suppose that ζ1 ∈ R and ζ2 ∈ R. We may assume that ζ1 < ζ2. Let g be the transformation g(z) = z − ζ2 z − ζ1 . 3 MATH32051 18. Parabolic, hyperbolic and elliptic Mo¨bius transformations As −ζ1 + ζ2 > 0, this is a Mo¨bius transformation of H. Moreover, as g(ζ1) = ∞ and g(ζ2) = 0, we see that gγg −1 has fixed points at 0 and ∞ and is therefore a dilation. Hence γ is conjugate to a dilation. ✷ §18.5 Elliptic transformations To understand elliptic isometries it is easier to work in the Poincare´ disc D. Recall that Mo¨bius transformations of D have the form γ(z) = αz + β β¯z + α¯ (5.1) where α, β ∈ C and |α|2 − |β|2 > 0. Again, we can normalise γ (by dividing the numerator and denominator in (5.1) by √|α|2 − |β|2) so that |α|2 − |β|2 = 1. We have the same classification of Mo¨bius transformations, but this time in the context of D, as before: (i) γ is hyperbolic if it has 2 fixed points on ∂D and 0 fixed points in D, (ii) γ is parabolic if it has 1 fixed point on ∂D and 0 fixed points in D, (iii) γ is elliptic if it has 0 fixed points on ∂D and 1 fixed point in D. We can again classify Mo¨bius transformations of D by using the trace. If γ is a Mo¨bius transformation of D and is written in normalised form (5.1) then we define τ(γ) = (α+ α¯)2. It is then easy to prove that: (i) γ is hyperbolic if and only if τ(γ) > 4; (ii) γ is parabolic if and only if τ(γ) = 4; (iii) γ is elliptic if and only if τ(γ) ∈ [0, 4). There are two ways in which we can prove this. Firstly, we could solve the quadratic equa- tion γ(z0) = z0 and examine the sign of the discriminant (as in Section 16). Alternatively, we can use the map h : H → D, h(z) = (z − i)/(iz − 1) we introduced in Section 12 as follows. Recall that Mo¨bius transformations of D have the form hγh−1 where γ is a Mo¨bius transformation of H. We can think of h as a ‘change of co-ordinates’ (from H to D). As in Section we can see that γ is hyperbolic, parabolic, elliptic if and only if hγh−1 is hyper- bolic, parabolic, elliptic, respectively. By considering traces of matrices, we can also see that τ(hγh−1) = τ(γ). Let γ be an elliptic Mo¨bius transformation of D, so that there is a unique fixed point in D. As an example of an elliptic transformation of the Poincare´ disc D, let θ ∈ (0, 2π) and consider the map γ(z) = eiθz. This is a Mo¨bius transformation of D (take α = eiθ/2 and β = 0 in (5.1)). It acts on D by rotating the Poincare´ disc around the origin by an angle of θ. Proposition 18.5.1 Let γ ∈ Mo¨b(D) be a Mo¨bius transformation of D. The following are equivalent: (i) γ is elliptic; 4 MATH32051 18. Parabolic, hyperbolic and elliptic Mo¨bius transformations (ii) τ(γ) ∈ [0, 4); (iii) γ is conjugate to a rotation z 7→ eiθz. Proof. We have already seen in Proposition 17.3.1 and the discussion above that (i) is equivalent to (ii). Suppose that (iii) holds. A rotation has a unique fixed point (at the origin). If γ is conjugate to a rotation then it must also have a unique fixed point, and so is elliptic. Finally, we prove that (i) implies (iii). Suppose that γ is elliptic and has a unique fixed point at ζ ∈ D. Let g be a Mo¨bius transformation of D that maps ζ to the origin 0. Then gγg−1 is a Mo¨bius transformation of D that is conjugate to γ and has a unique fixed point at 0. Suppose that gγg−1(z) = αz + β β¯z + α¯ where |α|2 − |β|2 > 0. As 0 is a fixed point, we must have that β = 0. Write α in polar form as α = reiθ. Then gγg−1(z) = α α¯ z = reiθ re−iθ z = e2iθz so that γ is conjugate to a rotation. ✷ Remark. What do rotations look like in H? Recall the map h
roposition follows from the following (fairly easily proved) fact: the sequence nθ mod 1 is a discrete subset of [0, 1] if and only if θ is rational, say θ = k/m. As Γ is a Fuchsian group, the subgroup {γn | n ∈ Z} is also a Fuchsian group, and therefore discrete. Hence γ is conjugate to a rotation by 2kπ/m. Hence γm is conjugate to a rotation through 2kπ, i.e. γm is the identity. ✷ Let Γ be a Fuchsian group with Dirichlet polygon D. Let v be a vertex of D with elliptic cycle transformation γv,s ∈ Γ. Then by Proposition 30.2.1, there exists an integer m ≥ 1 such that γmv,s = id. The order of γv,s is the least such integer m. We would like to be able to define the order of an elliptic cycle E . Suppose we start with a (vertex, side)-pair used in the construction of E ; then we obtain an elliptic cycle transformation and so can, in principle, calculate its order. However, if we had started with a different (vertex, side)-pair on the same elliptic cycle E then we would have obtained 1 MATH32051 30. Elliptic cycles: angle sums and orders a different elliptic cycle transformation and so, perhaps, the order of this elliptic cycle transformation will be different. However, this is not the case. Recall from Section 29 that two elliptic cycle transformations arising from the same elliptic cycle are either inverses of each other, or conjugate. If two transformations are inverses of each other or are conjugate then they have the same order; this follows from the following proposition. Proposition 30.2.2 (i) Suppose that γ1 has order m and γ2 is conjugate to γ1. Show that γ2 also has order m. (ii) Show that if γ has order m then so does γ−1. Proof. This is Exercise 30.1. ✷ It follows from Proposition 30.2.2 that the order does not depend on which vertex we choose in an elliptic cycle, nor does it depend on whether we start at (v, s) or (v, ∗s). Hence for an elliptic cycle E we write mE for the order of γv,s where v is some vertex on the elliptic cycle E and s is a side with an endpoint at v. We call mE the order of E . §30.3 Angle sum Let ∠v denote the internal angle of D at the vertex v. Consider the elliptic cycle E = v0 → v1 → · · · → vn−1 of the vertex v = v0. We define the angle sum to be sum(E) = ∠v0 + · · ·+ ∠vn−1. Clearly, the angle sum of an elliptic cycle does not depend on which vertex we start at. Hence we can write sum(E) for the angle sum along an elliptic cycle. The following result relates the order of an elliptic cycle and its angle sum. Proposition 30.3.1 Let Γ be a Fuchsian group with Dirichlet polygon D with all vertices in H and let E be an elliptic cycle. Then there exists an integer mE ≥ 1 such that mE sum(E) = 2π. Moreover, mE is the order of E . Proof. See Katok. ✷ Remark. Recall that we say that an elliptic cycle E is accidental if the associated elliptic cycle transformation is the identity. Clearly the identity has order 1. Hence if E is an accidental cycle then it has order mE = 1 and sum(E) = 2π. 2 MATH32051 31. Generators of a group 31. Generators of a group §31.1 Introduction In general, a Fuchsian group may contain infinitely many (albeit countably many) elements. However, one can often find a finite set of group elements that ‘generate’ the group. For example, the group Γ = {γn | γn(z) = 2nz, n ∈ Z} of dilations by powers of 2 is clearly infinite. However, note that γn(z) = γ n 1 (z). Hence every element of Γ can be expressed in terms of powerds of γ1; in this sense, γ1 generates Γ. Given a Fuchsian group Γ, we would like to be able to find a finite set of group elements in Γ that generate Γ. Before we do this, we will discuss generators of a group more generally. §31.2 Generators of a group Definition. Let Γ be a group. We say that a subset S = {γ1, . . . , γn} ⊂ Γ is a set of generators if every element of Γ can be written as a composition of elements from S and their inverses. We write Γ = 〈S〉. Examples. 1. Consider the additive group Z. Then Z is generated by the element 1: then every positive element n > 0 of Z can be written as 1+ · · ·+1 (n times), and every negative element −n, n > 0 of Z, can be written (−1) + · · ·+ (−1) (n times). 2. The additive group Z2 = {(n,m) | n,m ∈ Z} is generated by {(1, 0), (0, 1)}. 3. The multiplicative group of pth roots of unity {1, ω, . . . , ωp−1}, ω = e2πi/p, is gener- ated by ω. Remark. A given group Γ will, in general, have many different generating sets. For example, the set {2, 3} generates the additive group of integers. (To see this, note that 1 = 3− 2 hence n = 3 + · · ·+ 3 + (−2) + · · ·+ (−2) where there are n 3s and n −2s.) §31.3 The side-pairing transformations generate a Fuchsian group Let Γ be a Fuchsian group and let D(p) be a Dirichlet polygon for Γ. In Section 28 we saw how to associate to D(p) a set of side-pairing transformations. The importance of side-pairing transformations comes from the following result. Theorem 31.3.1 Let Γ be a Fuchsian group. Suppose that D(p) is a Dirichlet polygon with AreaH(D(p)) < ∞. Then the set of side-pairing transformations of D(p) generate Γ. Proof. See Katok’s book. ✷ 1 MATH32051 31. Generators of a group Example. Consider the modular group Γ = PSL(2,Z). We have seen that a fundamental domain for Γ is given by the set D(p) = {z ∈ H | −1/2 < Re(z) < 1/2, |z| > 1}, where p = ik for any k > 1. We saw in Section 28 that the side-pairing transformations are z 7→ z + 1 (and its inverse z 7→ z − 1) and z 7→ −1/z. It follows from Theorem 31.3.1 that the modular group PSL(2,Z) is generated by the transformations z 7→ z+1 and z 7→ −1/z. We write PSL(2,Z) = 〈z 7→ z + 1, z 7→ −1/z〉. 2 MATH32051 32. Presentations of groups 32. Presentations of groups §32.1 Introduction In Section 31 we saw how to take a Fuchsian group Γ and find a set of generators for Γ: a finite set S = {γ1, . . . , γn} ⊂ Γ such that every element of Γ can be written as a composition of elements in S and their inverses. However, knowing a set of generators does not tell us everything about the group. For example, we saw in Section 31 that the modular group PSL(2,Z) is generated by γ1(z) = z + 1 and γ2(z) = −1/z. Thus every element in PSL(2,Z) can be expressed as a composition of γ1s and γ2s. However, this expression is not necessarily unique. For example, let γ(z) = z/(z + 1) ∈ PSL(2,Z). Then one can check that γ = γ1γ2γ1 and γ = γ2γ −1 1 γ2. To make progress here it will be useful to think how to define a large family of groups in a different, more combinatorial, way. §32.2 Groups abstractly defined in terms of generators and relations Generators and relations provide a useful and widespread combinatorial method for de- scribing a group. Although generators and relations can be set up formally, we prefer for simplicity to take a more heuristic approach here. §32.3 Free groups Let S be a finite set of k symbols. If a ∈ S is a symbol then we introduce another symbol a−1 and denote the set of such symbols by S−1. We look at all finite concatenations of symbols chosen from S ∪ S−1, subject to the condition that concatenations of the form aa−1 and a−1a are removed. Such a finite con- catenation of n symbols is called a word of length n. Let Wn = {all words of length n} = {wn = a1 · · · an | aj ∈ S ∪ S−1, aj±1 6= a−1j }. We let e denote the empty word (the word consisting of no symbols) and, for consistency, let W0 = {e}. Note that the order that the symbols appear in a given word matters. For example, suppose we have three symbols a, b, c. The words abc and acb are different words. If wn and wm are words then we can form a new word wnwm of length at most n+m by concatenation: if wn = a1 · · · an and wm = b1 · · · bm then wnwm = a1 · · · anb1 · · · bm. If b1 = a −1 n then we delete the term anb1 from this product (and then we have to see if b2 = a −1 n−1; if so, then we delete the term an−1b2, etc). 1 MATH32051 32. Presentations of groups Definition. Let S be a finite set of k elements. We define Fk = ⋃ n≥0 Wn, the c
ciated to the side ∗s1. Then γ2(∗s1) is a side s2 of D and γ2(v1) = v2, a vertex of D. (v) Repeat the above inductively. Thus we obtain a sequence of pairs of vertices and sides:( v0 s0 ) γ1→ ( v1 s1 ) ∗→ ( v1 ∗s1 ) γ2→ ( v2 s2 ) ∗→ · · · γi→ ( vi si ) ∗→ ( vi ∗si ) γi+1→ ( vi+1 si+1 ) ∗→ · · · . Again, as there are only finitely many pairs (v, s), this process of applying a side-pairing transformation followed by applying ∗ must eventually return to the initial pair (v0, s0). Let n be the least integer n > 0 for which (vn, ∗sn) = (v0, s0). Definition. The sequence of vertices E = v0 → v1 → · · · → vn−1 is called an elliptic cycle. The transformation γnγn−1 · · · γ2γ1 is called an elliptic cycle transformation. Again, as there are only finitely many pairs of vertices and sides, we see that there are only finitely many elliptic cycles and elliptic cycle transformations. Definition. Let v be a vertex of the hyperbolic polygon D and let s be a side of D with an end-point at v. We denote the elliptic cycle transformation associated to the pair (v, s) by γv,s. Definition. Let ∠v denote the internal angle of D at the vertex v. Consider the elliptic cycle E = v0 → v1 → · · · → vn−1 of the vertex v = v0. We define the angle sum to be sum(E) = ∠v0 + · · ·+ ∠vn−1. Definition. We say that an elliptic cycle E satisfies the elliptic cycle condition if there exists an integer m ≥ 1, depending on E such that m sum(E) = 2π. Remark. Observe that if one vertex on a vertex cycle satisfies the elliptic cycle condition, then so does any other vertex on that vertex cycle. Thus it makes sense to say that an elliptic cycle satisfies the elliptic cycle condition. 2 MATH32051 33. Poincare´’s Theorem: no boundary vertices We can now state Poincare´’s Theorem. Put simply, it says that if each elliptic cycle satisfies the elliptic cycle condition then the side-pairing transformations generate a Fuch- sian group. Moreover, it also tells us how to write the group in terms of generators and relations. Theorem 33.2.1 (Poincare´’s Theorem) Let D be a convex hyperbolic polygon with finitely many sides. Suppose that all vertices lie inside H and that D is equipped with a collection G of side-pairing Mo¨bius transformations. Suppose that each side-pairing transformation satisfies the half-plane hypothesis. Suppose that no side of D is paired with itself. Suppose that the elliptic cycles are E1, . . . , Er. Suppose that each elliptic cycle Ej of D satisfies the elliptic cycle condition: for each Ej there exists an integer mj ≥ 1 such that mj sum(Ej) = 2π. Then: (i) The subgroup Γ = 〈G〉 generated by G is a Fuchsian group; (ii) The Fuchsian group Γ has D as a fundamental domain. (iii) The Fuchsian group Γ can be written in terms of generators and relations as follows. Think of G as an abstract set of symbols. For each elliptic cycle Ej , choose a cor- responding elliptic cycle transformation γj = γv,s (for some vertex v on the elliptic cycle E); this is a word in symbols chosen from G ∪G−1. Then Γ is isomorphic to the group with generators γs ∈ G (i.e. we take G to be a set of symbols), and relations γ mj j : Γ = 〈γs ∈ G | γm11 = γm22 = · · · = γmrr = e〉. Proof. See Katok or Beardon. ✷ Remark. The relations in (iii) appear to depend on which pair (v, s) on the elliptic cycle Ej is used to define γj . In fact, the relation γmjj is independent of the choice of (v, s). This follows from Section 30, particularly Proposition 30.2.2 and the discussion around it. If (v′, s′) is another (vertex,side)-pair on the same elliptic cycle as (v, s) then the elliptic cycle transformations we obtain are either conjugate, or one is conjugate to the inverse of the other. Remark. The hypothesis that D does not have a side that is paired with itself is not a real restriction: if D has a side that is paired with itself then we can introduce another vertex on the mid-point of that side, thus dividing the side into two smaller sides which are then paired with each other. We saw how to do this in Section 28 but for completeness we repeat the idea here. Suppose that s is a side with side-pairing transformation γs that pairs s with itself. Suppose that s has end-points at the vertices v0 and v1. Introduce a new vertex v2 at the mid-point of [v0, v1]. Notice that γs(v2) = v2. We must have that γs(v0) = v1 and γs(v1) = v0 (for otherwise γs would fix three points in H and hence would be the identity, by Corollary 16.1.2). Let s1 be the side [v0, v2] and let s2 be the side [v2, v1]. Then γs(s1) = s2 and γs(s2) = s1. Hence γs pairs the sides s1 and s2. Notice that the internal angle at the vertex v2 is equal to π. See Figure 33.2. 3 MATH32051 33. Poincare´’s Theorem: no boundary vertices v0 v1 v2 s1 s2 γs Figure 33.2: The side s is paired with itself; by splitting it in half, we have two distinct sides that are paired. 4 MATH32051 34. A hyperbolic octagon 34. An example of Poincare´’s Theorem: a hyperbolic octagon §34.1 Introduction In this section we use Poincare´’s Theorem (Theorem 33.2.1 ) to find a new example of a Fuchsian group. This is a typical example of how to use Poincare´’s Theorem in practice. §34.2 A hyperbolic octagon From Exercise 14.1 we know that there exists a regular hyperbolic octagon with each internal angle equal to π/4. Label the vertices of such an octagon anti-clockwise v1, . . . , v8 and label the sides anti- clockwise s1, . . . , s8 so that side sj occurs immediately after vertex vj . See Figure 34.1. As P is a regular octagon, each of the sides sj has the same length. s1s8 v7 v8 v1 v2 v3 v4 v5 v6 s4 s3 s2s7 s6 s5 π/4 γ1 γ4 γ3 γ2 Figure 34.1: A regular hyperbolic octagon with internal angles π/4 and side-pairings indicated. ( v1 s1 ) γ1→ ( v4 s3 ) ∗→ ( v4 s4 ) γ2→ ( v3 s2 ) ∗→ ( v3 s3 ) γ−1 1→ ( v2 s1 ) ∗→ ( v2 s2 ) γ−1 2→ ( v5 s4 ) ∗→ ( v5 s5 ) 1 MATH32051 34. A hyperbolic octagon γ3→ ( v8 s7 ) ∗→ ( v8 s8 ) γ4→ ( v7 s6 ) ∗→ ( v7 s7 ) γ−1 3→ ( v6 s5 ) ∗→ ( v6 s6 ) γ−1 4→ ( v1 s8 ) ∗→ ( v1 s1 ) . Thus there is just one elliptic cycle: E = v1 → v4 → v3 → v2 → v5 → v8 → v7 → v6. with associated elliptic cycle transformation: γ−14 γ −1 3 γ4γ3γ −1 2 γ −1 1 γ2γ1 As the internal angle at each vertex is π/4, the angle sum is 8 π 4 = 2π. Hence the elliptic cycle condition holds (with mE = 1). Thus by Poincare´’s Theorem, the group generated by the side-pairing transformations γ1, . . . , γ4 generate a Fuchsian group. Moreover, we can write this group in terms of generators and relations as follows: 〈γ1, γ2, γ3, γ4 | γ−14 γ−13 γ4γ3γ−12 γ−11 γ2γ1 = e〉. 2 MATH32051 35. Poincare´’s Theorem: boundary vertices 35. Poincare´’s Theorem: the case of boundary vertices §35.1 Introduction In Section 33 we studied groups generated by side-pairing transformations defined on a hyperbolic polygon D with no vertices on the boundary. Here we consider what happens if the hyperbolic polygon has vertices on the boundary ∂H. §35.2 Poincare´’s Theorem in the case of boundary vertices Recall that a convex hyperbolic polygon can be described as the intersection of a finite number of half-planes and that this definition allows the possibility that the polygon has an edge lying on the boundary. (Such edges are called free edges.) We will assume that this does not happen, i.e. all the edges of D are arcs of geodesics. See Figure 35.1. D D (i) (ii) Figure 35.1: (i) A polygon in D with no free edges, (ii) a polygon in D with a free edge. Let D be a convex hyperbolic polygon with no free edges and suppose that each side s of D is equipped with a side-pairing transformation γs. We will continue to require that side-pairing transformations must satisfy the half-plane hypothesis. In particular, γs cannot be the identity. Notice that as Mo¨bius transformations of H act on ∂H and indeed map ∂H to itself, each side-pairing transformation maps a boundary vertex to another boundary vertex. Let v = v0 be a bo
undary vertex of D and let s = s0 be a side with an end-point at v0. Then we can repeat the procedure described in Section 33 (using the same notation) starting at the pair (v0, s0) to obtain a finite sequence of boundary vertices P = v0 → · · · → vn−1 and an associated Mo¨bius transformation γv,s = γn · · · γ1. Definition. Let v = v0 be a boundary vertex of D and let s = s0 be a side with an end-point at v. We call P = v0 → · · · → vn−1 a parabolic cycle with associated parabolic cycle transformation γv,s = γn · · · γ1. As there are only finitely many vertices and sides, there are at most finitely many parabolic cycles and parabolic cycle transformations. 1 MATH32051 35. Poincare´’s Theorem: boundary vertices Example. Consider the polygon described in Figure 35.2 with the side-pairings indicated. Then following the procedure as described in Section 33 starting at the pair (A, s1) we have: A B CD E F s1 s2 s3 s4 s5 s6 γ1 γ2 γ3 Figure 35.2: A polygon with 2 boundary vertices and with side-pairings indicated. ( A s1 ) γ1→ ( D s3 ) ∗→ ( D s4 ) γ−1 2→ ( A s6 ) ∗→ ( A s1 ) . Hence we have a parabolic cycle A → D with associated parabolic cycle transformation γ−12 γ1. There is also an elliptic cycle:( B s2 ) γ3→ ( F s5 ) ∗→ ( F s6 ) γ2→ ( E s4 ) ∗→ ( E s5 ) γ−1 3→ ( C s2 ) ∗→ ( C s3 ) γ−1 1→ ( B s1 ) ∗→ ( B s2 ) . Hence we have the elliptic cycle B → F → E → C with associated elliptic cycle transfor- mation γ−11 γ −1 3 γ2γ3. Remarks. 1. Suppose instead that we had started at (v, ∗s) instead of (v, s). Then we would have obtained the parabolic cycle transformation γ−1v,s . 2 MATH32051 35. Poincare´’s Theorem: boundary vertices 2. Suppose instead that we had started at (vi, ∗si) instead of (v0, s0). Then we would have obtained the parabolic cycle transformation γvi,∗si = γiγi−1 · · · γ1γn · · · γi+2γi+1, i.e. a cyclic permutation of the maps involved in defining the parabolic cycle trans- formation associated to (v0, s0). Moreover, it is easy to see that γvi,∗si = (γi · · · γ1)γv0,s0(γi · · · γ1)−1 so that γvi,∗si and γv0,s0 are conjugate Mo¨bius transformations. Let v be a boundary vertex of D and let s be a side with an end-point at v. The associated parabolic cycle transformation is denoted by γv,s. Observe that γv,s is a Mo¨bius transformation with a fixed point at the vertex v ∈ ∂H. In Section 16 we saw that if a Mo¨bius transformation has at least one fixed point in ∂H then it must be either parabolic, hyperbolic or the identity. Thus each parabolic cycle transformation is either a parabolic or hyperbolic Mo¨bius transformation or the identity. Definition. We say that a parabolic cycle P satisfies the parabolic cycle condition if for some (hence all) vertex v ∈ P, the parabolic cycle transformation γv,s is either a parabolic Mo¨bius transformation or the identity Remark. Let γ ∈ Mo¨b(H)\{id}. Recall that γ is parabolic if and only if the trace, τ(γ), is 4. Also note that if γ = id then τ(γ) = 4. Hence a parabolic cycle P satisifes the parabolic cycle condition if for some (hence all) vertex v ∈ P, the parabolic cycle transformation γv,s has τ(γv,s) = 4. Remark. Observe that γv0,s0 is parabolic (or the identity) if and only if γvi,si is parabolic (or the identity) for any other vertex vi on the parabolic cycle containing v0. Also observe that γv,s is parabolic (or the identity) if and only if γv,∗s is parabolic (or the identity). Thus it makes sense to say that a parabolic cycle P satisfies the parabolic cycle condition. We can now state Poincare´’s Theorem in the case when D has boundary vertices (but no free edges). Theorem 35.2.1 (Poincare´’s Theorem) Let D be a convex hyperbolic polygon with finitely many sides, possibly with boundary vertices (but with no free edges). Suppose that D is equipped with a collection G of side- pairing Mo¨bius transformations such that the half-plane hypothesis holds and such that no side is paired with itself. Let the elliptic cycles be E1, . . . , Er and let the parabolic cycles be P1, . . . ,Ps. Suppose that: (i) each elliptic cycle Ej satisfies the elliptic cycle condition, and (ii) each parabolic cycle Pj satisfies the parabolic cycle condition. Then: (i) The subgroup Γ = 〈G〉 generated by G is a Fuchsian group, 3 MATH32051 35. Poincare´’s Theorem: boundary vertices (ii) The Fuchsian group Γ has D as a fundamental domain. (iii) The Fuchsian group Γ can be written in terms of generators and relations as follows. Think of G as an abstract set of symbols. For each elliptic cycle Ej , choose a cor- responding elliptic cycle transformation γj = γv,s (for some vertex v on the elliptic cycle Ej); this is a word in symbols chosen from G ∪G−1. Then Γ is isomorphic to the group with generators γs ∈ G (i.e. we take G to be a set of symbols), and relations γ mj j , where mj sum Ej = 2π: Γ = 〈γs ∈ G | γm11 = · · · = γmrr = e〉. Remark. The hypothesis that D does not have a side that is paired with itself is not a real restriction: if D has a side that is paired with itself then we can introduce another vertex on the mid-point of that side, thus dividing the side into two smaller sides which are then paired with each other. See Section 33 . 4 MATH32051 36. The modular group 36. An example of Poincare´’s Theorem: the modular group §36.1 Introduction In this section we use the version of Poincare´’s Theorem stated in Theorem 35.2.1 to check that the modular group PSL(2,Z) is indeed a Fuchsian group. More importantly, we use Theorem 35.2.1 to give a presentation of the modular group in terms of generators and relations. §36.2 The modular group Consider the polygon in Figure 36.1; here A = (−1 + i√3)/2 and B = (1 + i√3)/2. The side pairing transformations are given by γ1(z) = z + 1 and γ2(z) = −1/z. Notice that γ2(A) = B and γ2(B) = A. γ2 s3 s1 s2 γ1 A B Figure 36.1: Side pairing transformations for the modular group. The side [A,B] is paired with itself by γ2. We need to introduce an extra vertex C = i at the mid-point of [A,B]; see the discussion in Section 33. This is illustrated in Figure 36.2. Note that the internal angle at C is equal to π. We calculate the elliptic cycles. We first calculate the elliptic cycle containing the vertex A: ( A s1 ) γ1→ ( B s2 ) ∗→ ( B s4 ) γ−1 2→ ( A s3 ) ∗→ ( A s1 ) . 1 MATH32051 36. The modular group γ2 s1 s2 γ1 A B s3 s4 C Figure 36.2: Introduce an extra vertex at C so that the side s3 is paired with the side s4. Hence A → B is an elliptic cycle E1 which has elliptic cycle transformation γ−12 γ1(z) = (−z − 1)/z. The angle sum of this elliptic cycle satisfies 3(∠A+ ∠B) = 3(π/3 + π/3) = 2π. Hence the elliptic cycle condition holds with m1 = 3. Now calculate the elliptic cycle containing the vertex C:( C s3 ) γ2→ ( C s4 ) ∗→ ( C s3 ) . Hence we have an elliptic cycle E2 = C with elliptic cycle transformation γ2. The angle sum of this elliptic cycle satisfies 2∠C = 2π. Hence the elliptic cycle condition holds with m2 = 2. We now calculate the parabolic cycles. There is just one parabolic cycle, the cycle that contains the vertex ∞. We have( ∞ s1 ) γ1→ ( ∞ s2 ) ∗→ ( ∞ s1 ) , so that we have a parabolic cycle ∞ with parabolic cycle transformation γ1(z) = z+1. As γ1 has a single fixed point at ∞ it is parabolic. Hence the parabolic cycle condition holds. By Poincare´’s Theorem, we see that the group generated by γ1 and γ2 is a Fuchsian group. Let a = γ1, b = γ2. Then we can use Poincare´’s Theorem to write the group generated by γ1, γ2 in terms of generators and relations, as follows: PSL(2,Z) = 〈a, b | (b−1a)3 = b2 = e〉. Remark. The above example illustrates why we need to assume that D does not have any sides that are paired with themselves. If we had not introduced the vertex C, then we would not have got the relation b2 = e. 2 MATH32051 37. Hyperbolic surfaces 37. Hyperbolic surfaces §37.1 Introduction Let Γ be a Fuchsian group and letD(p) be a Diri
chlet polygon and suppose that AreaH(D) < ∞. We equip D with a set of side-pairing transformations, subject to the condition that a side is not paired with itself. We can construct a space H/Γ by gluing together the sides that are paired by side-pairing transformations. This space is variously called a quotient space, an identification space or an orbifold. Before giving some hyperbolic examples, let us give a Euclidean example. Consider the square in Figure 37.1(i) with the sides paired as indicated. We first glue together the horizontal sides to give a cylinder; then we glue the vertical sides to give a torus. See Figure 37.1(ii). (i) (ii) γ2 γ1 Figure 37.1: (i) A square with horizontal and vertical sides paired as marked, and (ii) the results of gluing first the horizontal and then the vertical sides together. In the above Euclidean example, the angles at the vertices of D glued together nicely (in the sense that they glued together to form total angle 2π) and we obtained a surface. For a general Fuchsian group the situation is slightly more complicated due to the possible presence of cusps and marked points. Consider a Fuchsian group Γ with Dirichlet polygon D. Let us describe how one con- structs the space H/Γ. Let E be an elliptic cycle in D. All the vertices on this elliptic are glued together. The angles at these vertices are glued together to give total angle sum(E). This may or may not be equal to 2π. The angle sum is equal to 2π if and only if the elliptic cycle E is an accidental cycle. (Recall that an elliptic cycle is said to be accidental if the elliptic 1 MATH32051 37. Hyperbolic surfaces cycle transformation is equal to the identity; equivalently in the elliptic cycle condition msum(E) = 2π we have m = 1.) Definition. Let E be an elliptic cycle and suppose that sum(E) 6= 2π. Then the vertices on this elliptic cycle are glued together to give a point on H/Γ with total angle less than 2π. This point is called a marked point. A marked point on H/Γ is a point where the total angle is less than 2π. Thus they look like ‘kinks’ in the surface H/Γ. Definition. It follows from Proposition 30.3.1 that there exists an integer mE such that mEsum(E) = 2π. We call mE the order of the corresponding marked point. Now let P be a parabolic cycle. Vertices along a parabolic cycle are glued together. Each parabolic cycle gives rise to a cusp on H/Γ. These look like ‘funnels’ that go off to infinity. Topologically, the space H/Γ is determined by its genus (the number of ‘holes’) and the numbers of cusps. Figure 37.2: A hyperbolic surface of genus 2 with 3 cusps. If there are no marked points, then we call H/Γ a hyperbolic surface. For example, consider the Fuchsian group Γ generated by the hyperbolic octagon de- scribed in Section 34. All the internal angles are equal to π/4. The octagon D in Section 34 is a fundamental domain for Γ. If we glue the edges of D together according to the indicated side-pairings then we obtain a hyperbolic surface H/Γ. Notice that there is just one elliptic cycle E and that sum(E) = 2π; hence there are no marked points. This surface is a torus of genus 2, i.e. a torus with two holes. See Figure 37.3. Figure 37.3: Gluing together the sides of D gives a torus of genus 2. 2 MATH32051 38. Euler characteristic and genus 38. Euler characteristic and genus §38.1 Introduction Given a 2-dimensional space X, one of the most important topological invariants that we can associate to X is its Euler characteristic χ(X). This is a topological invariant of the space X. We also state how to relate the Euler characteristic to the genus of the space. Roughly speaking, the genus of a space is the number of ‘holes’ in the space. In the next section we shall show how to calculate the genus of the hyperbolic surface H/Γ and relate it to algebraic properties of Γ. §38.2 The Euler characteristic Here we define the Euler characteristic. Definition. Let X be a 2-dimensional space. Then X can be triangulated into finitely many polygons. Suppose that in this triangulation we have V vertices, E edges and F faces (i.e. the number of polygons). Then the Euler characteristic is given by χ(X) = V − E + F. Examples. (i) Consider the triangulation of the space illustrated in Figure 38.1; this is formed by gluing eight triangles together. This space is homeomorphic (meaning: topologically the same as) to the surface of a sphere. There are V = 6 vertices, E = 12 edges and F = 8 faces. Hence the Euler characteristic is χ = 6− 12 + 8 = 2. (ii) Consider the triangulation of a torus illustrated in Figure 38.2. There is just one polygon (so F = 1) and just one vertex (so V = 1). There are two edges, so E = 2. Hence χ = 0. Definition. Let X be a 2-dimensional surface. The genus g of X is given by χ(X) = 2− 2g. Thus, a sphere has genus 0 and a torus has genus 1. Topologically, the genus of a surface is the number of ‘handles’ that need to be attached to a sphere to give the surface. One can also think of it as the number of ‘holes’ through the surface. 1 MATH32051 38. Euler characteristic and genus Figure 38.1: A triangulation of the surface of a sphere; here V = 6, E = 12, F = 8, so that χ = 2. Figure 38.2: A triangulation of the surface of a torus; here V = 1, E = 2, F = 1, so that χ = 0. 2 MATH32051 39. The signature of a Fuchsian group 39. The signature of a Fuchsian group §39.1 Introduction In this section we define the signature of a cocompact Fuchsian group Γ. This is a finite set of integers, arising from the orders of the non-accidental elliptic cycles and the genus of H/Γ; we will see in the next section how the signature determines many properties of the Fuchsian group Γ. §39.2 The signature of a Fuchsian group We begin with the following definition. Definition. Let Γ be a Fuchsian group. Suppose that Γ has a finite-sided Dirichlet polygon D(p) with all vertices in H and none in ∂H. Then we say that Γ is cocompact. Let Γ be a cocompact Fuchsian group. The signature of Γ is a set of geometric data that is sufficient to reconstruct Γ as an abstract group. The signature will also allow us to generate infinitely many different cocompact Fuchsian groups. Let D(p) be a Dirichlet polygon for Γ. Then D(p) has finitely many elliptic cycles, and the order of each elliptic cycle transformation is finite. Definition. Let Γ be a cocompact Fuchsian group. Let g be the genus of H/Γ. Sup- pose that there are k elliptic cycles E1, . . . , Ek. Suppose that Ej has order mEj = mj so that mjsum(Ej) = 2π. Suppose that E1, . . . , Er are non-accidental and Er+1, . . . , Ek are accidental. The signature of Γ is defined to be sig(Γ) = (g;m1, . . . ,mr). (That is, we list the genus of H/Γ together with the orders of the non-accidental elliptic cycles.) If all the elliptic cycles are accidental cycles, then we write sig(Γ) = (g;−). 1 MATH32051 40. The area of a fundamental domain 40. The area of a fundamental domain for a cocompact Fuchsian group §40.1 Introduction In the last section we defined the signature of a cocompact Fuchsian group Γ. Recall that the signature sig(Γ) = (g;m1, . . . ,mr) where g is the genus of H/Γ and the mj are the orders of the non-accidental elliptic cycles. In this section we relate the signature of Γ to the hyperbolic area of any fundamental domain for Γ. §40.2 Relating the signature to the hyperbolic area of a fundamental domain Let Γ be a cocompact Fuchsian group. We can use the data given by the signature of Γ to give a formula for the hyperbolic area of any fundamental domain of Γ. (Recall from Proposition 23.2.1 that, for a given Fuchsian group, any two fundamental domains have the same hyperbolic area.) Proposition 40.2.1 Let Γ be a cocompact Fuchsian group with signature sig(Γ) = (g;m1, . . . ,mr). Let D be a fundamental domain for Γ. Then AreaH(D) = 2π  (2g − 2) + r∑ j=1 ( 1− 1 mj ) . (2.1) Proof. By Proposition 23.2.1 it is sufficient to prove that the formula (2.1) holds for a Dirichlet polygon D. As in Section 33, we can add extra vertices if necessary to assume that no side is paired with itself. Suppose that D has n vertices (hence n sides). We use the Gauss-Bonnet Theorem (Theorem 13.2.1). Let E1, . . . , Er be the non- accidental elliptic cycles. By Proposition 30.3.1, the angle sum along the elliptic cycle Ej is sum(Ej) = 2π mj Suppose that there are s accidental cycles. (Recall that an elliptic cycle is said to be accidental if the corresponding elliptic cycle transformation is the identity, and in particular has order 1.) By Proposition 30.3.1, the internal angle sum along an accidental cycle is 2π. Hence the internal angle sum along all accidental cycles is 2πs. As each vertex must belong to some elliptic cycle (either an elliptic cycle with order at least 2, or to an accidental cycle) the sum of all the internal angles of D is given by 2π   r∑ j=1 1 mj + s   . 1 MATH32051 40. The area of a fundamental domain By the Gauss-Bonnet Theorem (Theorem 13.2.1), we have AreaH(D) = (n− 2)π − 2π   r∑ j=1 1 mj + s   . (2.2) Consider now the space H/Γ. This is formed by taking D and gluing together paired sides. The vertices along each elliptic cycle are glued together; hence each elliptic cycle in D gives one vertex in the triangulation of H/Γ. Hence D gives a triangulation of H/Γ with V = r + s vertices. As paired sides are glued together, there are E = n/2 edges (notice that we are assuming here that no side is paired with itself). Finally, as we only need the single polygon D, there is only F = 1 face. Hence 2− 2g = χ(H/Γ) = V − E + F = r + s− n 2 + 1 which rearranges to give n− 2 = 2((r + s)− (2− 2g)). (2.3) Substituting (2.3) into (2.2) we see that AreaH(D) = 2π  r + s− (2− 2g) − r∑ j=1 1 mj − s   = 2π  (2g − 2) + r∑ j=1 ( 1− 1 mj ) . ✷ We can use Proposition 40.2.1 to find a lower bound for the area of a Dirichlet polygon for a Fuchsian group. Proposition 40.2.2 Let Γ be a cocompact Fuchsian group (so that the Dirichlet polygon D(p) has no vertices on the boundary). Then AreaH(D) ≥ π 21 . Proof. By Proposition 40.2.1 this is equivalent to proving that if 2g − 2 + r∑ j=1 ( 1− 1 mj ) ≥ 0 then 2g − 2 + r∑ j=1 ( 1− 1 mj ) ≥ 1 42 . (2.4) Notice that 1− 1/mj is always positive. If g > 1 then 2g − 2 > 1. Hence the left-hand side of (2.4) is greater than 1, and the result certainly holds. Suppose that g = 1 (so that 2g − 2 = 0). Now m1 ≥ 2 so that 1 − 1/m1 ≥ 1/2, which is greater than 1/42. So the result holds. 2 MATH32051 40. The area of a fundamental domain Suppose that g = 0 (so that 2g − 2 = −2). As in the previous paragraph, we see that for each j = 1, . . . , r we have 1 − 1/mj ≥ 1/2. If r ≥ 5 then the left-hand side of (2.4) is at least 1/2, so the result holds. When r = 4 the positive minimum of the left-hand side of (2.4) occurs for signature (0; 2, 2, 2, 3); in this case 2g − 2 + r∑ j=1 ( 1− 1 mj ) ≥ −2 + 1 2 + 1 2 + 1 2 + ( 1− 1 3 ) = 1 6 . When g = 0 and r = 2 we have that 2g − 2 + r∑ j=1 ( 1− 1 mj ) = −2 + ( 1− 1 m1 ) + ( 1− 1 m2 ) = − ( 1 m1 + 1 m2 ) < 0. Similarly, when g = 0 and r = 1 we have that 2g − 2 + r∑ j=1 ( 1− 1 mj ) = −2 + ( 1− 1 m1 ) = −1− 1 m1 < 0. Hence there are no Fuchsian groups with signature sig(Γ) = (0;m1,m2) or (0;m1). It remains to treat the case g = 0, r = 3. In this case, we must prove that 2g − 2 + r∑ j=1 ( 1− 1 mj ) = 1− ( 1 m1 + 1 m2 + 1 m3 ) ≥ 1 42 provided that the left-hand side is positive. Let s(k, l,m) = 1− ( 1 k + 1 l + 1 m ) for k, l,m ≥ 2. We prove that if s(k, l,m) > 0 then s(k, l,m) ≥ 1/42. We assume that k ≤ l ≤ m. Suppose k = 3 then s(3, 3, 3) = 0 and s(3, 3, 4) = 1/12 > 1/42. Hence s(3, l,m) ≥ 1/12 so the result holds. Hence we need only concern ourselves with k = 2. Note that s(2, 2,m) < 0, s(2, 4, 4) = 0, s(2, 4, 5) = 1/20 > 1/42, s(2, 4,m) ≥ 1/20. Hence we need only concern ourselves with l = 3. Now s(2, 3,m) = 1 6 − 1 m which achieves the minimum 1/42 when m = 7. ✷ Remark. In Section 41 we shall show that if (g;m1, . . . ,mr) is an (r+1)-tuple of integers such that the right-hand side of (2.1) is positive, then there exists a Fuchsian group Γ with sig(Γ) = (g;m1, . . . ,mr). 3 MATH32051 41. Constructing a Fuchsian group of a given signature 41. Constructing a Fuchsian group of a given signature §41.1 Introduction In Section 39 we defined the signature sig(Γ) = (g;m1, . . . ,mr) of a cocompact Fuchsian group. We saw that if D is a fundamental domain for Γ then AreaH(D) = 2π  (2g − 2) + r∑ j=1 ( 1− 1 mj ) . As this quantity must be positive, the condition that (2g − 2) + r∑ j=1 ( 1− 1 mj ) > 0 (1.1) is a necessary condition for (g;m1, . . . ,mr) to be the signature of a Fuchsian group. The pur- pose of this section is to sketch a proof of the converse of this statement: if (g;m1, . . . ,mr) satisfies (1.1) then there exists a cocompact Fuchsian group with signature (g;m1, . . . ,mr). This gives a method for constructing infinitely many examples of cocompact Fuchsian groups. (Recall that a Fuchsian group Γ is said to be cocompact if it has a Dirichlet polygon with all its vertices inside H.) §41.2 Existence of a Fuchsian group with a given signature Theorem 41.2.1 Let g ≥ 0 and mj ≥ 2, 1 ≤ j ≤ r be integers. (We allow the possibility that r = 0, in which case we assume that there are no mjs.) Suppose that (2g − 2) + r∑ j=1 ( 1− 1 mj ) > 0. (2.1) Then there exists a cocompact Fuchsian group Γ with signature sig(Γ) = (g;m1, . . . ,mr). Remark. In particular, for each g ≥ 2 there exists a Fuchsian group Γg with signature sig(Γg) = (g;−). Thus for each g ≥ 2 we can find a Fuchsian group Γg such that H/Γg is a torus of genus g. Remark. The proof of Theorem 41.2.1 consists of constructing a polygon and a set of side-pairing transformations satisfying Poincare´’s Theorem. There are two phenomena that we want to capture in this polygon. 1 MATH32051 41. Constructing a Fuchsian group of a given signature Figure 41.1: Glueing together the sides paired gives a handle. (i) We need to generate handles. By considering the example of a regular hyperbolic octagon in Section 34, we see that the part of a polygon illustrated in Figure 41.1 with the side-pairing illustrated will generate a handle. (ii) We need to generate marked points. By considering the discussion in Section 36 of how the modular group satisfies Poincare´’s Theorem, we see that the part of a polygon illustrated in Figure 41.2 with the side pairing illustrated will generate a marked point arising from an elliptic cycle of order m. 2π/m Figure 41.2: Glueing together the sides paired gives a marked point of order m. Proof. The proof is essentially a long computation using Poincare´’s Theorem. We con- struct a convex polygon, equip it with a set of side-pairing transformations, and apply Poincare´’s Theorem to show that these side-pairing transformations generate a Fuchsian group. Finally, we show that this Fuchsian group has the required signature. We work in the Poincare´ disc D. Consider the origin 0 ∈ D. Let θ denote the angle θ = 2π 4g + r . Draw 4g + r radii, each separated by angle θ. Fix t ∈ (0, 1). On each radius, choose a point at (Euclidean) distance t from the origin. Join successive points with a hyperbolic geodesic. This gives a regular hyperbolic polygon M(t) with 4g + r vertices. Starting at an arbitrary point, label the vertices clockwise v1, v2, . . . , vr, v1,1, v1,2, v1,3, v1,4, v2,1, . . . , v2,4, v3,1, . . . , vg,1, . . . , vg,4. 2 MATH32051 41. Constructing a Fuchsian group of a given signature Figure 41.3: The polygon M(t) is a regular hyperbolic (4g + r)-gon. On each of the first r sides of M(t) we construct an isosceles triangle, external to M(t). We label the vertex at the ‘tip’ of the jth isosceles triangle by wj and construct the triangle in such a way so that the internal angle at wj is 2π/mj . If mj = 2 then 2π/mj = π and we have a degenerate triangle, i.e. just an arc of geodesic constructed in the previous paragraph and wj is the midpoint of that geodesic. Call the resulting polygon N(t). See Figure 41.4. v2,1 v2,2 v2,4 v2