- January 16, 2021

MATH32051 1. Course structure 1. Course structure §1.1 Welcome! Welcome to MATH32051 Hyperbolic Geometry. This document will explain how this course unit is organised, what you need to do to succeed in the course unit, what I will expect from you, and what you can expect of me. §1.2 Contact details The lecturer is Dr Charles Walkden, email: [email protected] Pronouns he/him/his. If you want to talk to me then please email me and we can arrange a mutually convenient time to talk, either in person or via Zoom, Collaborate, or Skype. If you email me then please include a meaningful subject line (eg: ‘MATH32051: ques- tion about Mobius transformations’). Please remember to sign-off in the way that you prefer to be addressed. You are welcome to call me Charles (or Dr Walkden, if you prefer to be more formal). It may also be more appropriate for you to ask your question via the discussion board, see below. §1.3 Course structure MATH32051 Hyperbolic Geoemtry is a 10 credit course unit. In common with all other course units, it will be delivered via blended learning. This puts greater emphasis on you directing your own study. However, there will be increased time for interactive sessions (rather than passive lectures) compared to previous years. This document contains all the key information about how the course unit is organised, what I will expect from you, and what you can expect from me. §1.3.1 Blended learning Activities in this course unit can be broadly catagorised as asynchronous (directed study that can be performed individually, at your own pace and at a time of your choosing) and synchronous (activities that we do together at a particular time), as we now explain. Each week you will get a learning plan for that week. The learning plan will detail what you need to do in that week. Here are the activities that you will do in a typical week week. • Asynchronous material: – Short videos. You will be expected to watch (in your own time) 3–5 short videos, totalling around 45–60 minutes. These videos are the primary way of delivering new mathematical information to you. 1 MATH32051 1. Course structure – Blackboard quizzes. Most videos have a short quiz or set of short answer ques- tions on Blackboard. You should do these quizzes to check that you have got the basics of what the video was about. These quizzes are purely formative; they do not count towards your mark for the course. – More detailed notes. Each video has an accompanying set of notes. In some places, the notes provide a slightly different account of the same material; in other places, the notes go into slightly more detail or contain additional expla- nation or motivation. – Exercise sheets. Each week there is a set of exercises. The exercises are numbered n.m. You can do exercise n.m once you have watched video n. The exercises are a key part of the course. • Synchronous classes: – Online review class. Each week there is an online review class. This is where I will go through any questions about the course material that you have posted on Blackboard or asked me. I will also use these sessions to summarise the main ideas in that week’s material, go through common misunderstandings of the material, talk through the Blackboard quizzes and to discuss and work through additional examples. – Tutorials. Each week there is a tutorial class. The tutorial classes are a key part of the course. I will try to make them as interactive as possible by getting you to revise material that will be useful in the course or getting you to work through some of the exercises, perhaps with additional hints. Please point out any errors or typos in the videos, lecture notes, exercises and solutions. (I can’t promise to correct the videos due to the time it takes to produce them (but I will highlight any errors on the Blackboard discussion forum). I will fix any errors in the written material.) The review session will be recorded and I will upload it onto the Blackboard page. The tutorial will probably not be recorded. This is for two reasons. Firstly, most benefit will be gained by attending the tutorial live. Secondly, I want you to be able to interact with each other and with me in the tutorials; recording this creates GDPR difficulties. Any material I produce for the tutorials will be put on Blackboard. §1.3.2 Accessibility and DASS As soon as I receive them, I read all of the DASS support statements for students on this course. I am very aware that this is the first time we have taught courses via blended learning; if there is anyway in which accessibility of the learning materials for this course unit can be improved then please let me know. Where possible, the videos are stored both on the University’s Video Portal and on Youtube. The subtitling is generally better on Youtube, but I appreciate that Youtube is not easily available in some countries. If the subtitling is a problem then please let me know. §1.3.3 Discussion forum There is a discussion board on the course Blackboard space. If you have a question about any aspect of the course then please consider asking it via the appropriate thread on the 2 MATH32051 1. Course structure discussion forum: if you are unsure about something then others may be wondering about exactly the same thing. I will aim to respond to all (reasonable!) questions posted on the discussion forum within 3 working days, either on the forum or in the weekly review session. The discussion forum also allows you to communicate with each other about the course. As ground rules for the discussion boards: (i) check if your question has been asked/answered elsewhere before posting, (ii) be respectful of others in your posts, (iii) do not post off-topic. §1.3.4 Learning outcomes Learning outcomes are the skills that you will have gained after completing the course unit. Here are the learning outcomes for this course unit. On successfully completing the course you will be able to: ILO1 calculate the hyperbolic distance between and the geodesic through points in the hyperbolic plane, ILO2 compare different models (the upper half-plane model and the Poincare´ disc model) of hyperbolic geometry, ILO3 prove results (Gauss-Bonnet Theorem, angle formulæ for triangles, etc as listed in the syllabus) in hyperbolic trigonometry and use them to calcu- late angles, side lengths, hyperbolic areas, etc, of hyperbolic triangles and polygons, ILO4 classify Mo¨bius transformations in terms of their actions on the hyperbolic plane, ILO5 calculate a fundamental domain and a set of side-pairing transformations for a given Fuchsian group, ILO6 define a finitely presented group in terms of generators and relations, ILO7 use Poincare´’s Theorem to construct examples of Fuchsian groups and calculate presentations in terms of generators and relations, ILO8 relate the signature of a Fuchsian group to the algebraic and geometric properties of the Fuchsian group and to the geometry of the corresponding hyperbolic surface. §1.4 Coursework and end-of-semester assessment There will be a piece of coursework worth 20% of the course unit with a deadline likely to be Week 6 (TBC). The end-of-semester assessment takes place in January and is worth 80% of the course unit. §1.5 Feedback You will get weekly feedback on your understanding by completing the short quizzes on Blackboard. You can also gain feedback during the Kahoot quizzes in the tutorial, during the tutorial, from the coursework, and from me during my office hour. 3 MATH32051 1. Course structure §1.6 Recommended texts The following are the recommended texts for this course unit. You probably do not need to buy any book and can rely solely on the videos and lecture notes. J. Anderson, Hyperbolic Geometry, 1st ed., Springer Undergraduate Mathematics Series, Springer-Verlag, Berlin, New York, 1999. S. Katok, Fuchsian Groups, Chicago Lecture Notes in Mathematics, Chicago University Press, 1992. A. Beardon, The Geometry of Discrete Groups, Springer-Verlag, Berlin, New York, 1983. The book by Anderson is the most suitable for the first half of the course (sections 2–18). Katok’s book 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-presering rigorously. The intuitive idea is as follows. Take a triangle (or any other geometerical object) and label the vertices in an anti-clockwise direction. Now apply the isometry to the triangle. If the labelling remains in an anti-clockwise direction then the isometry is orientation-preserving; if the labelling changes so as to be in a clockwise direction then the isometry is not orientation-presering. It is easy to convince yourself that the first three examples of isometries of R2 above are orientation-preserving, but that a reflection is not orientation-preserving.) We denote the set of orientation preserving isometries of R2 by Isom+(R2). Note that Isom+(R2) is a subgroup of Isom(R2). §2.5 Tiling the Euclidean plane A regular n-gon is a polygon with n sides, each side being a geodesic and all sides having the same length, and with all internal angles equal. Thus, a regular 3-gon is an equilateral triangle, a regular 4-gon is a square, and so on. For what values of n can we tile the Euclidean plane by regular n-gons? (By a tiling, or tessellation, we mean that the plane can be completely covered by regular n-gons of the same size, with no overlapping and no gaps, and with vertices only meeting at vertices.) It is easy to convince oneself that this is only possible for n = 3, 4, 6. Thus in Euclidean geometry, there are only three tilings of the plane by regular n-gons. Hyperbolic geometry is, as we shall see, far more interesting—there are infinitely many such tilings! This is one reason why hyperbolic geometry is studied: the hyperbolic world is richer in structure than the Euclidean world! Notice that we can associate a group of isometries to a tiling: namely the group of isometries that preserves the tiling. Thus, given a geometric object (a tiling) we can asso- ciate to it an algebraic object (a subgroup of isometries). Conversely, as we shall see later, we can go in the opposite direction: given an algebraic object (a subgroup of isometries sat- isfying some technical hypotheses) we can construct a geometric object (a tiling). Thus we establish a link between two of the main areas of pure mathematics: algebra and geometry. 4 MATH32051 2. Where we are going Figure 2.4: Tiling the Euclidean plane by regular 3-, 4- and 6-gons. §2.6 Where we are going There are several different, but equivalent, ways of constructing hyperbolic geometry. These different constructions are called ‘models’ of hyperbolic geometry. The model that we shall primarily study is the upper half-plane model H. We shall explain how one calculates lengths and distances in H and we shall describe all isometries of H. Later we will study another model of hyperbolic geometry, namely the Poincare´ disc model. This has some advantages over the upper half-plane model, for example pictures are a lot easier to draw! We then study trigonometry in hyperbolic geometry. We shall study analogues of famil- iar results from Euclidean geometry. For example, we shall derive the hyperbolic version of Pythagoras’ Theorem which gives a relationship between the lengths of the sides of a right-angled hyperbolic triangle. We shall also discuss the Gauss-Bonnet Theorem. This is a very beautiful result that can be used to study tessellations of the hyperbolic plane; in particular, we shall prove that there are infinitely many tilings of the hyperbolic plane by regular hyperbolic n-gons. We will then return to studying and classifying isometries of the hyperbolic plane. We shall see that isometries can be classified into three distinct types (elliptic, parabolic and hyperbolic) and we shall explain the differences between them. As we shall see, the collection of all (orientation preserving) isometries of the hyperbolic plane form a group. We will describe the orientation preserving isometries in terms of Mo¨bius transformation, and denote the group of such by Mo¨b(H). Certain subgroups of Mo¨b(H) called Fuchsian groups have very interesting properties. We shall explain how one can start with a Fuchsian group and from it construct a tessellation of the hyerbolic plane. Conversely, (with mild and natural conditions) one can start with a tessellation and construct a Fuchsian group. This gives an attractive connection between algebraic structures (Fuchsian groups) and geometric structures (tessellations). To establish this connection we have to use some analysis, so this course demonstrates how one may tie together the three main subjects in pure mathematics into a coherent whole. §2.7 Appendix: a historical interlude There are many ways of constructing Euclidean geometry. Klein’s Erlangen programme can be used to define it in terms of the Euclidean plane, equipped with the Euclidean distance function and the set of isometries that preserve the Euclidean distance. An alternative way of defining Euclidean geometry is to use the definition due to the Greek mathematician Euclid (c.325BC–c.265BC). In the first of his thirteen volume set ‘The Elements’, Euclid systematically developed Euclidean geometry by introducing definitions of geometric terms 5 MATH32051 2. Where we are going (such as ‘line’ and ‘point’), five ‘common notions’ concerning magnitudes, and the following five postulates: (i) a straight line may be drawn from any point to any other point; (ii) a finite straight line may be extended continuously in a straight line; (iii) a circle may be drawn with any centre and any radius; (iv) all right-angles are equal; (v) if a straight line falling on two straight lines makes the interior angles on the same side less than two right-angles, then the two straight lines, if extended indefinitely, meet on the side on which the angles are less than two right-angles. α β Figure 2.5: Euclid’s fifth postulate: here α+ β < 180◦. The first four postulates are easy to understand; the fifth is more complicated. It is equiv- alent to the following, which is now known as the parallel postulate: 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. Euclid’s Elements has been a standard text on geometry for over two thousand years and throughout its history the parallel postulate has been contentious. The main criticism was that, unlike the other four postulates, it is not sufficiently self-evident to be accepted without proof. Can the parallel postulate be deduced from the previous four postulates? Another surprising feature is that most of plane geometry can be developed without using the parallel postulate (it is not used until Proposition 29 in Book I); this suggested that the parallel postulate is not necessary. For over two thousand years, many people attempted to prove that the parallel postulate could be deduced from the previous four. However, in the first half of the 19th century, Gauss (1777–1855) proved that this was impossible: the parallel postulate was independent of the other four postulates. He did this by making the remarkable discovery that there exist consistent geometries for which the first four postulates hold, but the parallel postulate fails. In 1824, Gauss wrote, in a letter to Taurinus, ‘The assumption that the sum of the three sides [of a triangle] is smaller than 180 degrees leads to a geometry which is quite different from our (Euclidean) geometry, but which is in itself completely consistent.’ (One can show that the parallel postulate holds if and only if the angle sum of a triangle is always equal to 180 degrees.) This was the first example of a non-Euclidean geometry. 6 MATH32051 2. Where we are going Gauss never published his results on non-Euclidean geometry. (You can read many of the letters that Gauss sent to other mathematicians on non-Euclidean geometry here: www.math.uwaterloo.ca/~snburris/htdocs/noneucl.pdf.) However, it was soon redis- covered independently by Lobachevsky in 1829 and by Bolyai in 1832. Today, the non- Euclidean geometry of Gauss, Lobachevsky and Bolyai is called hyperbolic geometry and any geometry which is not Euclidean is called non-Euclidean geometry. 7 MATH32051 3. The upper half-plane and the boundary 3. The upper half-plane and the boundary §3.1 The upper half-plane There are several different ways of constructing hyperbolic geometry. These different con- structions are called ‘models’. In this section we will discuss one particularly simple and convenient model of hyperbolic geometry, namely the upper half-plane model. Remark. Throughout this course we will often identify R2 with C, by noting that the point (x, y) ∈ R2 can equally well be thought of as the point z = x+ iy ∈ C. Definition. The upper half-plane H is the set of complex numbers z with positive imag- inary part: H = {z ∈ C | Im(z) > 0}. Definition. The circle at infinity or boundary of H is defined to be the set ∂H = {z ∈ C | Im(z) = 0} ∪ {∞}. That is, ∂H is the real axis together with the point ∞. Remark. What does ∞ mean? It’s just a point that we have ‘invented’ so that it makes sense to write things like 1/x→∞ as x→ 0 and have the limit as a bona fide point in the space. (If this bothers you, remember that you are already used to ‘inventing’ numbers; for example irrational numbers such as √ 2 have to be ‘invented’ because rational numbers need not have rational square roots.) Remark. We will use the conventions that, if a ∈ R and a 6= 0 then a/∞ = 0 and a/0 =∞, and if b ∈ R then b+∞ =∞. We leave 0/∞,∞/0,∞/∞, 0/0,∞±∞ undefined. Remark. We call ∂H the circle at infinity because (at least topologically) it is a circle! We can see this using a process known as stereographic projection. Let K = {z ∈ C | |z| = 1} denote the unit circle in the complex plane C. Define a map π : K → R ∪ {∞} as follows. For z ∈ K \ {i} let Lz be the (Euclidean) straight line passing through i and z; this line meets the real axis at a unique point, which we denote by π(z). We define π(i) = ∞. The map π is a homeomorphism from K to R ∪ {∞}; this is a topological way of saying the K and R ∪ {∞} are ‘the same’. See Figure 3.1. Remark. We call ∂H the circle at infinity because (as we shall see below) points on ∂H are at an infinite ‘distance’ from any point in H. 1 MATH32051 3. The upper half-plane and the boundary R π(z) z Lz i Figure 3.1: Stereographic projection. Notice how as z approaches i, the image π(z) gets large; this motivates defining π(i) =∞. 2 MATH32051 4. Length and distance in the upper half-plane 4. Length and distance in the upper half-plane §4.1 Introduction The aim of this section is to define the hyperbolic distance between two points z, z′ ∈ H. We will do this as follows. We first look at a path σ in H that starts at the point z and ends at the point z′. We define the hyperbolic length of σ and we do this by way of a path integral. To define the distance between z and z′ we consider all paths between z and z′, calculate their hyperbolic lengths and then take the infimum. We first need to recall how to calculate path integrals in C (equivalently, in R2). §4.2 Path integrals By a path σ in the complex plane C, we mean the image of a continuous function σ(·) : [a, b]→ C, where [a, b] ⊂ R is an interval. We will assume that σ is differentiable and that the derivative σ′ is continuous. Thus a path is, heuristically, the result of taking a pen and drawing a curve in the plane. We call the points σ(a), σ(b) the end-points of the path σ. We say that a function σ : [a, b] → C whose image is a given path is a parametrisation of that path. Notice that a path will have lots of different parametrisations. Example. Define σ1 : [0, 1] → C by σ1(t) = t + it and define σ2 : [0, 1] → C by σ2(t) = t2 + it2. Then σ1 and σ2 are different parametrisations of the same path in C, namely the straight (Euclidean) line from the origin to 1 + i. Let f : C → R be a continuous function. Then the integral of f along a path σ is defined to be: ∫ σ f = ∫ b a f(σ(t))|σ′(t)| dt; (2.1) here | · | denotes the usual modulus of a complex number, in this case, |σ′(t)| = √ (Re σ′(t))2 + (Imσ′(t))2. Remark. To calculate the integral of f along the path σ we have to choose a parametri- sation of that path. So it appears that our definition of ∫ σ f depends on the choice of parametrisation. One can show, however, that this is not the case: any two parametrisa- tions of a given path will always give the same answer. For this reason, we shall sometimes identify a path with its parametrisation. So far we have assumed that σ is differentiable and has continuous derivative. It will be useful in what follows to allow a slightly larger class of paths. Definition. A path σ with parametrisation σ(·) : [a, b] → C is piecewise continuously differentiable if there exists a partition a = t0 < t1 < · · · < tn−1 < tn = b of [a, b] such that σ : [a, b] → C is a continuous function and, for each j, 0 ≤ j ≤ n − 1, σ : (tj , tj+1) → C is differentiable and has continuous derivative. 1 MATH32051 4. Length and distance in the upper half-plane (Roughly speaking this means that we allow the possibility that the path σ has finitely many ‘corners’.) For example, the path σ(t) = (t, |t|), −1 ≤ t ≤ 1 is piecewise continuously differentiable: it is differentiable everywhere except at the origin, where it has a ‘corner’. To define ∫ σ f for a piecewise continuously differentiable path σ we merely write σ as a finite union of differentiable sub-paths, calculating the integrals along each of these subpaths, and then summing the resulting integrals. §4.3 Distance in hyperbolic geometry We are now is a position to define the hyperbolic metric in the upper half-plane model of hyperbolic space. To do this, we first define the length of an arbitrary piecewise continuously differentiable path in H. Definition. Let σ : [a, b]→ H be a path in the upper half-plane H = {z ∈ C | Im(z) > 0}. Then the hyperbolic length of σ is obtained by integrating the function f(z) = 1/ Im(z) along σ, i.e. lengthH(σ) = ∫ σ 1 Im(z) = ∫ b a |σ′(t)| Im(σ(t)) dt. Examples. 1. Consider the path σ(t) = a1 + t(a2 − a1) + ib, 0 ≤ t ≤ 1 between a1 + ib and a2 + ib. Then σ′(t) = a2 − a1 and Im(σ(t)) = b. Hence lengthH(σ) = ∫ 1 0 |a2 − a1| b dt = |a2 − a1| b . 2. Consider the points −2 + i and 2 + i. By the example above, the length of the horizontal path between them is 4. 3. Now consider a different path from −2+ i to 2+ i. Consider the piecewise linear path that goes diagonally up from −2+ i to 2i and then diagonally down from 2i to 2 + i. A parametrisation of this path is given by σ(t) = { (2t− 2) + i(1 + t), 0 ≤ t ≤ 1, (2t− 2) + i(3− t), 1 ≤ t ≤ 2. Then σ′(t) = { 2 + i, 0 ≤ t ≤ 1, 2− i, 1 ≤ t ≤ 2 so that |σ′(t)| = { |2 + i| = √5, 0 ≤ t ≤ 1, |2− i| = √5, 1 ≤ t ≤ 2, and Im(σ(t)) = { 1 + t, 0 ≤ t ≤ 1, 3− t, 1 ≤ t ≤ 2. Hence lengthH(σ) = ∫ 1 0 √ 5 1 + t dt+ ∫ 2 1 √ 5 3− t dt = √ 5 log(1 + t) ∣∣∣1 0 − √ 5 log(3− t) ∣∣∣2 1 = 2 √ 5 log 2, 2 MATH32051 4. Length and distance in the upper half-plane which is approximately 3.1. Note that the path from −2+ i to 2+ i in the third example has a shorter hyperbolic length than the path from −2 + i to 2 + i in the second example. This suggests that the geodesic (the paths of shortest length) in hyperbolic geometry are very different to the geodesics we are used to in Euclidean geometry. -2+i 2+i -2+i 2+i Figure 4.1: The first path has hyperbolic length 4, the second path has hyperbolic length approximately 3.1. §4.4 Hyperbolic distance We are now in a position to define the hyperbolic distance between two points in H. We first recall the following definitions. Definition. Let A ⊂ R. A lower bound of A is any number b ∈ R such that b ≤ a for all a ∈ A. A lower bound ℓ is called the infimum of A or greatest lower bound of A if it is greater than, or equal to, any other lower bound; that is, b ≤ ℓ for all lower bounds b of A. We write inf A for the infimum of A, if it exists. Remarks. (i) Consider, for example, the closed interval [1, 2] ⊂ R. Then 0 is a lower bound 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 real centres. A vertical line or a circle with a real centre in C is given by an equation of the form αzz¯ + βz + βz¯ + γ = 0 (2.1) for some α, β, γ ∈ R. Let w = γ(z) = az + b cz + d . Then z = dw − b −cw + a. Substituting this into (2.1) we have: α ( dw − b −cw + a )( dw¯ − b −cw¯ + a ) + β ( dw − b −cw + a ) + β ( dw¯ − b −cw¯ + a ) + γ = 0. Hence α(dw − b)(dw¯ − b) + β(dw − b)(−cw¯ + a) +β(dw¯ − b)(−cw + a) + γ(−cw + a)(−cw¯ + a) = 0. Expanding this out and gathering together terms gives (αd2 − 2βcd+ γc2)ww¯ + (−αbd+ βad+ βbc− γac)w +(−αbd+ βad+ βbc− γac)w¯ + (αb2 − 2βab+ γa2) = 0. (2.2) Let α′ = αd2 − 2βcd+ γc2 β′ = −αbd+ βad+ βbc− γac γ′ = αb2 − 2βab+ γa2. Hence (2.2) has the form α′ww¯ + β′w + β′w¯ + γ′ with α′, β′, γ′ ∈ R, which is the equation of either a vertical line or a circle with real centre. ✷ 2 MATH32051 7. Mo¨bius transformations of H are isometries of H 7. Mo¨bius transformations of H are isometries of H §7.1 Introduction Let γ ∈ Mo¨b(H) be a Mo¨bius transformation of H. Then γ has the form γ(z) = (az + b)/(cz + d) where a, b, c, d ∈ R and ad − bc > 0. We have seen that γ maps H to itself bijectively. In this section we study the action of γ on ∂H. We also prove that Mo¨bius transformations are isometries. §7.2 Mo¨bius transformations and the boundary of H Recall that we have defined the upper half-plane to be the set H = {z ∈ C | Im(z) > 0} and the boundary of H is defined to be ∂H = {z ∈ C | Im(z) = 0} ∪ {∞}. Let a, b, c, d ∈ R be such that ad− bc > 0. Recall that a map of the form γ(z) = az + b cz + d is called a Mo¨bius transformation of H. Recall thatMo¨bius transformations of H form a group (under composition) which we denote by Mo¨b(H). We can extend the action of a Mo¨bius transformation of H to the circle at infinity ∂H of H as follows. Clearly γ maps R to itself, except at the point z = −d/c where the denominator is undefined. We define γ(−d/c) =∞. To determine γ(∞) we write γ(z) = a+ b/z c+ d/z and notice that 1/z → 0 as z →∞. Thus we define γ(∞) = a/c. (Note that if c = 0 then, as ad − bc > 0, we cannot have either a = 0 or d = 0. Thus we can make sense of the expressions a/c and −d/c when c = 0 by setting a/0 =∞ and −d/0 =∞.) The following two important results say that Mo¨bius transformations of H preserve distance. A bijective map that preserve distance is called an isometry. Thus Mo¨bius transformations of H are isometries of H. Proposition 7.2.1 Let γ be a Mo¨bius transformation of H. Let z, z′ ∈ H and let σ be a path from z to z′. Then lengthH(γ ◦ σ) = lengthH(σ). Proof. Let γ(z) = (az + b)/(cz + d) where a, b, c, d ∈ R and ad − bc > 0. It is an easy calculation to check that for any z ∈ H |γ′(z)| = ad− bc|cz + d|2 and Im(γ(z)) = (ad− bc) |cz + d|2 Im(z). 1 MATH32051 7. Mo¨bius transformations of H are isometries of H Let σ : [0, 1]→ H be a parametrisation of σ. Then, by using the chain rule, lengthH(γ ◦ σ) = ∫ 1 0 |(γ ◦ σ)′(t)| Im(γ ◦ σ)(t) dt = ∫ 1 0 |γ′(σ(t))||σ′(t)| Im(γ ◦ σ)(t) dt = ∫ 1 0 ad− bc |cσ(t) + d|2 |σ ′(t)| |cσ(t) + d| 2 ad− bc 1 Im(σ(t)) dt = ∫ 1 0 |σ′(t)| Im(σ(t)) dt = lengthH(σ). ✷ Proposition 7.2.2 Let γ be a Mo¨bius transformation of H. Then γ is an isometry of H. That is, for any z, z′ ∈ H we have dH(γ(z), γ(z ′)) = dH(z, z′). Proof. We first note that if σ is a path from z to z′ then γ ◦ σ is a path from γ(z) to γ(z′). Moreover, any path from γ(z) to γ(z′) arises in this way. By Proposition 7.2.1 we have that lengthH(γ ◦ σ) = lengthH(σ). Taking the infimum over all paths from z to z′ proves the proposition. ✷ Remark. We have shown that Mo¨bius transformations are isometries of H. Are there any other isometries that are not of this form? Exercise 7.3 asks you to consider the map γ(x + iy) = −x + iy. This is the reflection in the imaginary axis and is not a Mo¨bius transformation. However Exercise 7.3 shows that γ is an isometry of H. Mo¨bius transformations are orientation-preserving transformations. Roughly speak- ing, orientation-preserving means the following: Let ∆ be a triangle in H with vertices A,B,C, labelled anticlockwise. Then γ is orientation-preserving if γ(∆) has vertices at γ(A), γ(B), γ(C) and these are still labelled anti-clockwise). A transformation γ is orientation-reversing if, instead, γ(A), γ(B), γ(C) are now labelled clockwise. Note that γ(z) = −x+iy reflects the point z in the imaginary axis and so is orientation-reversing. We will prove in Proposition 11.5.1 that all orientation-preserving isometries of H are Mo¨bius transformations of H, and all orientation-reversing isometries of H are the composition of a Mo¨bius transformation of H and the reflection in the imaginary axis. 2 MATH32051 8. The imaginary axis is a geodesic 8. The imaginary axis is a geodesic §8.1 Introduction We are now in a position to calculate the geodesics—the paths of shortest distance—in H. Our first step is to prove that the imaginary axis is a geodesic. Proposition 8.1.1 Let a ≤ b. Then the hyperbolic distance between ia and ib is log b/a. Moreover, the vertical line joining ia to ib is the unique path between ia and ib with length log b/a; any other path from ia to ib has length strictly greater than log b/a. Proof. Let σ(t) = it, a ≤ t ≤ b. Then σ is a path from ia to ib. Clearly |σ′(t)| = 1 and Imσ(t) = t so that lengthH(σ) = ∫ b a 1 t dt = log b/a. Now let σ(t) = x(t) + iy(t) : [0, 1] → H be any path from ia to ib. Then lengthH(σ) = ∫ 1 0 √ x′(t)2 + y′(t)2 y(t) dt ≥ ∫ 1 0 |y′(t)| y(t) dt (1.1) ≥ ∫ 1 0 y′(t) y(t) dt (1.2) = log y(t)|10 = log b/a. Hence any path joining ia to ib has hyperbolic length at least log b/a, with equality precisely when both (1.1) and (1.2) are equalities. We see that (1.1) is an equality precisely when x′(t) = 0; this happens precisely when x(t) is constant, i.e. σ is a vertical line joining ia to ib. For (1.2) to be an equality, we require |y′(t)| = y′(t), i.e. y′(t) is positive for all t. This latter condition means that the path σ travels ‘straight up’ the imaginary axis from ia to ib without doubling back on itself. Thus we have shown that lengthH(σ) ≥ log b/a with equality precisely when σ is the vertical path joining ia to ib. ✷ 1 MATH32051 9. The geodesics of H 9. The geodesics of H §9.1 Introduction We can now complete the proof 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. Before we do that, we explore the hyperbolic version of the following well-known proce- dure in Euclidean geometry. Suppose you have been given a triangle in the Euclidean plane and asked to calculate something about that triangle. Often, you would first simplify this by arranging it so that one side of the triangle is horizontal, and you would also normally assume that one vertex is at the origin; doing this will often help simplify the calculations. What you are, in effect, doing is applying an isometry of the Euclidean plane so that, with- out loss of generality, one side lies along your ‘favourite’ geodesic in the Euclidean plane (namely, the x-axis) and that one point is at your ‘favourite’ point in the Euclidean plane (namely, the origin). We will often apply the same trick in hyperbolic geometry. Our ‘favourite’ geodesic will be the imaginary axis and our favourite point on the imaginary axis will be the point i. Often, when asked to calculate something in hyperbolic geometry, we will apply a Mo¨bius transformation so that one side lies along the imaginary axis and one point is at i. That we can always do this is proved below. §9.2 Mapping to the imaginary axis So far we have seen that the imaginary axis is a geodesic. We claim that any vertical straight line and any circle meeting the real axis orthogonally is also a geodesic. The first step in proving 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 two vectors v,w that meet at the point (x, y) by 〈v,w〉(x,y) = v1w1 + v2w2. We also define the norm of a vector v at the point (x, y) by ‖v‖(x,y) = √ 〈v, v〉(x,y) = √ v21 + v 2 2 . The Cauchy Schwartz inequality says that |〈v,w〉(x,y)| ≤ ‖v‖(x,y)‖w‖(x,y). We define the (Euclidean) angle θ = ∠v,w between the vectors v,w meeting at the point (x, y) by cos θ = 〈v,w〉(x,y) ‖v‖(x,y)‖w‖(x,y) . (Note that we are not interested in the sign of the angle: for our purposes angles can be measured either clockwise or anti-clockwise so that ∠v,w = ∠w, v.) In the upper half-plane, we have a similar definition of angle, but we use a different inner product. Suppose z ∈ H is a point in the upper half-plane. Let v,w be two vectors that meet at z. We define the inner product of v,w at z by 〈v,w〉z = 1 Im(z)2 (v1w1 + v2w2) (that is, the usual Euclidean inner product but scaled by a factor of 1/ Im(z)2.) We also define the norm of the vector v at z by ‖v‖z = √ 〈v, v〉z = 1 Im(z) √ v21 + v 2 2 . 4 MATH32051 10. Area, angles and the parallel postulate The Cauchy-Schwartz inequality still holds and we can define the angle θ = ∠v,w between two vectors v,w meeting at z by cos θ = 〈v,w〉z ‖v‖z‖w‖z . (5.1) Notice that, as the terms involving Im(z) cancel, this definition of angle coincides with the Euclidean definition. Suppose that we have two paths σ1, σ2 that intersect at the point z = σ1(0) = σ2(0). Then we define the angle between σ1, σ2 to be ∠σ′1(0), σ ′ 2(0), that is, the angle between two paths is the angle between their tangent vectors at the point of intersection. θ θ σ1 σ2(i) (ii) Figure 10.6: (i) The angle between two vectors, (ii) The angle between two paths at a point of intersection. §10.5.3 Conformal transformations Definition. A map γ : H → H is said to be conformal if it preserves angles between paths. That is, if σ1, σ2 intersect at z with angle θ, then the angle between the intersection of the paths γσ1, γσ2 at γ(z) is also θ. We will see that Mo¨bius transformations of H are conformal. To see this, we need to recall the following standard result from complex analysis. Proposition 10.5.1 (Cauchy-Riemann equations) Let f : C → C be a (complex) differentiable function. Write f as f(x + iy) = u(x, y) + iv(x, y). Then ∂u ∂x = ∂v ∂y , ∂u ∂y = −∂v ∂x . Proposition 10.5.2 Let γ ∈ Mo¨b(H) be a Mo¨bius transformation of H. Then γ is conformal. Proof (sketch). Let γ be a Mo¨bius transformation of H and write γ in terms of its real and imaginary parts as γ(x + iy) = u(x, y) + iv(x, y). Regarding H as a subset of C, we can view γ as a map R2 → R2. The matrix of partial derivatives of γ is given by Dγ(z) = ( ux uy vx vy ) where we write ux = ∂u/∂x. 5 MATH32051 10. Area, angles and the parallel postulate Let σ1, σ2 be paths that intersect at z = σ1(0) = σ2(0) with tangent vectors σ ′ 1(0), σ ′ 2(0). Then γσ1 and γσ2 are paths that intersect at γ(z) with tangent vectorsDγ(z)σ ′ 1(0),Dγ(z)σ ′ 2(0) where Dγ(z) denotes the matrix of partial derivatives of γ at z. Let v = (v1, v2), w = (w1, w2) be two vectors at the point z. By (5.1) it is sufficient to prove that 〈Dγ(v),Dγ(w)〉γ(z) ‖Dγ(v)‖γ(z)‖Dγ(w)‖γ(z) = 〈v,w〉z ‖v‖z‖w‖z . Notice that 〈Dγ(v),Dγ(w)〉γ(z) = 1 Im(γ(z)) 〈v, (Dγ)TDγ(w)〉 where (Dγ)T denotes the transpose of Dγ. Using the Cauchy-Riemann equations, we see that (Dγ)TDγ = ( ux vx uy vy )( ux uy vx vy ) = ( u2x + u 2 y 0 0 u2x + u 2 y ) , a scalar multiple of the identity matrix. It is straight-forward to see that this implies the claim. ✷ Remark. In fact, we have proved that any complex differentiable function with non- vanishing derivative is conformal. §10.5.4 Hyperbolic area Before we define hyperbolic area, let us motivate the definition by recalling how the hyper- bolic length of a path is defined. Let σ : [a, b]→ H be a path. Then the hyperbolic length of σ is given by lengthH(σ) = ∫ σ 1 Im(z) = ∫ b a |σ′(t)| Im(σ(t)) dt. In light of the above discussion, we can write this as lengthH(σ) = ∫ b a ‖σ′(t)‖σ(t) dt; Intuitively, we are approximating the path σ by vectors of length ‖σ(t)′‖σ(t) and then integrating. Let A ⊂ H be a subset of the upper half-plane. How can we intuitively define the area of A? If we take a point z ∈ A then we can approximate the area near z by taking a small rectangle with sides dx, dy. The area of this rectangle is given by the product of the lengths of the sides, namely 1 Im(z)2 dx dy. This suggests that we define the hyperbolic area of a subset A ⊂ H to be AreaH(A) = ∫ ∫ A 1 Im(z)2 dz = ∫ ∫ A 1 y2 dx dy. By definition, isometries of the hyperbolic plane H preserve lengths. However, it is not clear that they also preserve area. That they do is contained in the following result: 6 MATH32051 10. Area, angles and the parallel postulate σ(t) σ′(t) Figure 10.7: The path σ can be approximated at the point σ(t) by the tangent vector σ′(t). Adxz dy Figure 10.8: The area of A can be approximated at the point z by a small rectangle with sides dx, dy. Proposition 10.5.3 Let A ⊂ H and let γ ∈ Mo¨b(H) be a Mo¨bius transformation of H. Then AreaH(γ(A)) = AreaH(A). Proof (sketch). Let γ(z) = (az + b)/(cz + d), ad− bc > 0 be a Mo¨bius transformation. Let h : R2 → R and recall the change-of-variables formula:∫ ∫ γ(A) h(x, y) dx dy = ∫ ∫ A h ◦ γ(x, y)|det(Dγ)| dx dy (5.2) where Dγ is the matrix of partial derivatives of γ. Using the Cauchy-Riemann equations (and brute force!), one can check that det(Dγ) = (ad− bc)2 ((cx+ d)2 + c2y2)2 . The hyperbolic area of A is determined by setting h(x, y) = 1/y2 in (5.2). In this case, we have that h ◦ γ(x, y) = ( (cx+ d)2 + c2y2 (ad− bc)y )2 7 MATH32051 10. Area, angles and the parallel postulate and it follows that AreaH(γ(A)) = ∫ ∫ γ(A) h(x, y) dx dy = ∫ ∫ A h ◦ γ(x, y)|det(Dγ)| dx dy = ∫ ∫ A ( (cx+ d)2 + c2y2 (ad− bc)y )2( ad− bc (cx+ d)2 + c2y2 )2 dx dy = ∫ ∫ A 1 y2 dx dy = AreaH(A). ✷ 8 MATH32051 11. The hyperbolic distance between two points 11. The hyperbolic distance between two points: a formula and some applications §11.1 Introduction In Euclidean geometry we have a formula for the Euclidean distance between two arbitrary points; namely, if x = (x1, x2), y = (y1, y2) ∈ R2 then d(x, y) = √ (y1 − x1)2 + (y2 − x2)2. So far we do not have an analogous formula in hyperbolic geometry. From Proposition 8.1.1 we have a formula for the hyperbolic distance between two points on the imaginary axis. Indeed, if 0 < a < b then dH(ia, ib) = log b/a. But we do not have a formula for the distance between two arbitrary points in H. The goal of this section is to find a formula for the hyperbolic distance between two arbitrary points in H. We will also see some applications of this formula. §11.2 The distance between arbitrary points So far, we only have a formula for the hyperbolic distance between points of the form ia and ib. We can now give a formula for the distance between any two points in H. We will need the following (easily proved) lemma. Lemma 11.2.1 Let γ be a Mo¨bius transformation of H. Then for all z, w ∈ H we have |γ(z)− γ(w)| = |z − w| |γ′(z)|1/2|γ′(w)|1/2. We can now prove the following result. Proposition 11.2.2 Let z, w ∈ H. Then cosh dH(z, w) = 1 + |z − w|2 2 Im(z) Im(w) . (2.1) Proof. The idea is the following. We take two arbitrary points z, w ∈ H and look at the geodesic that passes through them. We move this geodesic to the imaginary axis and z, w to two points on the imaginary axis and check that (2.1) holds there. Then we map the imaginary axis back to the original geodesic. Exercise 11.1 leads you through this argument. ✷ In the remainder of this section, we will see some applications of the formula (2.1). Several of the proofs below also use the trick of mapping an arbitrary geodesic to the imaginary axis, proving the result there, and then mapping back again. 1 MATH32051 11. The hyperbolic distance between two points §11.3 Hyperbolic circles A Euclidean circle in the complex plane with centre z0 and radius r > 0 is the set of all points z that are Euclidean distance r from z0. We can make an analogous definition of a hyperbolic circle. Definition. Let z0 ∈ H and let r > 0. The hyperbolic circle with centre z0 and radius r is defined to be the set C := {z ∈ H | dH(z, z0) = r}. Hyperbolic circles are the same (as sets) as Euclidean circles. Proposition 11.3.1 Let C be a hyperbolic circle with centre z0 = x0 + iy0 ∈ H and radius r. Then C is a Euclidean circle with centre (x0, y0 cosh r) and Euclidean radius y0 sinh r. Proof. This is Exercise 11.2. ✷ §11.4 Pythagoras’ Theorem In Euclidean geometry, Pythagoras’ Theorem gives a relationship between the three side lengths of a right-angled triangle. Here we prove an analogous result in hyperbolic geometry using Proposition 11.2.2. Theorem 11.4.1 (Pythagoras’ Theorem for hyperbolic triangles) Let ∆ be a right-angled triangle in H with internal angles α, β, π/2 and opposing sides with lengths a, b, c. Then cosh c = cosh a cosh b. (4.1) Remark. If a, b, c are all very large then approximately we have c ≈ a + b − log 2. Thus in hyperbolic geometry (and in contrast with Euclidean geometry), the length of the hypotenuse is not substantially shorter than the sum of the lengths of the other two sides. Proof. Let ∆ be a triangle satisfying the hypotheses of the theorem. By applying a Mo¨bius transformation of H, we may assume that the vertex with internal angle π/2 is at i and that the side of length b lies along the imaginary axis. It follows that the side of length a lies along the geodesic given by the semi-circle centred at the origin with radius 1. Therefore, the other vertices of ∆ can be taken to be at ik for some k > 0 and at s+ it, where s+ it lies on the circle centred at the origin and of radius 1. See Figure 11.1. Recall from Proposition 11.2.2 that for any z, w ∈ H cosh dH(z, w) = 1 + |z − w|2 2 Im(z) Im(w) . Applying this formula to the the three sides of ∆ we have: cosh a = cosh dH(s+ it, i) = 1 + |s+ i(t− 1)|2 2t = 1 + s2 + (t− 1)2 2t = 1 t , (4.2) cosh b = cosh dH(ik, i) = 1 + (k − 1)2 2k = 1 + k2 2k , (4.3) 2 MATH32051 11. The hyperbolic distance between two points 0 a i b ik c α β s+ it Figure 11.1: Without loss of generality, we can assume that ∆ has vertices at i, ik and s+ it. cosh c = cosh dH(s+ it, ik) = 1 + |s+ i(t− k)|2 2tk = 1 + s2 + (t− k)2 2tk = 1 + k2 2tk , (4.4) where to obtain (4.2) and (4.4) we have used the fact that s2 + t2 = 1, as s+ it lies on the unit circle. Combining (4.2), (4.3) and (4.4) we see that cosh c = cosh a cosh b, proving the theorem. ✷ §11.5 Isometries of the upper half-plane We have already seen that Mo¨bius transformations of H are isometries of H. Are there any others? First let us recall the Euclidean case. In Section 2 we stated that the isometries of the Euclidean plane R2 are: (i) translations of the form τ(a1,a2)(x, y) = (x+ a1, y + a2), (iii) rotations of the plane, (iv) reflections in a straight line (for example, reflection in the y-axis, (x, y) 7→ (−x, y)), together with the identity. Translations and rotations are orientation-preserving whereas reflections are orientation-reversing. In Proposition 7.2.2 we saw that Mo¨bius transformations of H are isometries. Exer- cise 7.3 shows that there are other isometries. However, note that Mo¨bius transformations of H are also orientation-preserving (roughly this means the following: Let ∆ be a hyperbolic triangle in H with vertices A,B,C, labelled anticlockwise. Then γ is orientation-preserving if γ(∆) has vertices at γ(A), γ(B), γ(C) and these are still labelled anti-clockwise). Note that γ(z) = −x+ iy reflects the point z in the imaginary axis, and is orientation-reversing. We prove the following result. 3 MATH32051 11. The hyperbolic distance between two points Proposition 11.5.1 Let γ : H → H be an isometry of the upper half-plane. Then γ is either (i) a Mo¨bius transformation of H, or (ii) the composition of a Mo¨bius transformation of H and reflection in the imaginary axis. Case (i) in Proposition 11.5.1 corresponds to orientation preserving isometries. Case (ii) corresponds to orientation reversing isometries. We first need the following lemma; we omit the proof. Lemma 11.5.2 Let γ be any isometry of the upper half-plane. Then γ maps geodesics to geodesics. We can now prove Proposition 11.5.1. Proof of Proposition 11.5.1. Let γ be an isometry of the upper half-plane. Then γ maps the imaginary axis in H to another geodesic in H; call this geodesic H. By Lemma 9.2.1 we can find g ∈ Mo¨b(H) such that g maps H back to the imaginary axis and the point γ(i) ∈ H back to i. Hence g ◦ γ maps the imaginary axis to itself and g(γ(i)) = i. Recall that the imaginary axis has end points at 0,∞. Note that g ◦ γ maps the end points of the imaginary axis to the end points of the imaginary axis. If g ◦ γ swaps the end points over (i.e. g(γ(0)) = ∞ and g(γ(∞)) = 0) then we further compose g with the transformation z 7→ −1/z. Hence we can assume that g(γ(0)) = 0, g(γ(∞)) = ∞, g(γ(i)) = i. We claim that, further, g ◦ γ fixes every point on the imaginary axis. To see this, note that g◦γ maps the arc of imaginary axis from i to∞ to itself; g◦γ also maps the arc of imaginary axis from 0 to i to itself. Consider the point ki with k > 1. Then g(γ(ki)) = ℓi for some ℓ > 1. Then log k = dH(i, ki) = dH(g(γ(i)), g(γ(ki))) = dH(i, ℓi) = log ℓ. Hence k = ℓ. The same argument works if k ∈ (0, 1). Let z = x+ iy ∈ H. Let u+ iv = g(γ(x+ iy)) ∈ H. We will show that v = y and either u = x (in which case g ◦ γ is the identity, so γ = g−1, a Mo¨bius transformation of H) or u = −x (in which case γ is the composition of g−1 and reflection in the imaginary axis). Let t > 0 be arbitrary. Then dH(x+ iy, it) = dH(gγ(z), gγ(it)) = dH(u+ iv, it) as g ◦ γ fixes every point on the imaginary axis. Applying cosh to both sides and using Proposition 11.2.2 we have 1 + |(x+ iy)− it|2 2yt = 1 + |(u+ iv)− it|2 2vt . This simplifies to (x2 + (y − t)2)v = (u2 + (v − t)2)y. (5.1) Dividing this by t2 gives( x2 t2 + (y t − 1 )2) v = ( u2 t2 + (v t − 1 )2) y. (5.2) As (5.2) is valid for all t ∈ R, t > 0, we can let t→∞. This shows that v = y. Substituting v = y into (5.1) gives x2 = u2, hence u = ±x. Hence gγ(x + iy) = x + iy or gγ(x + iy) = −x+ iy; note also that g−1 ∈ Mo¨b(H). In the first case, we have that γ(z) = g−1(z), so that γ ∈ Mo¨b(H). In the second case, we have that γ is the composition of reflection in the imaginary axis with g−1. ✷ 4 MATH32051 12. The Poincare´ disc model 12. The Poincare´ disc model §12.1 Introduction So far we have studied the upper half-plane model of hyperbolic geometry. There are several other ways of constructing hyperbolic geometry; here we describe the Poincare´ disc model. Definition. The disc D = {z ∈ C | |z| < 1} is called the Poincare´ disc. The circle ∂D = {z ∈ C | |z| = 1} is called the circle at ∞ or boundary of D. One advantage of the Poincare´ disc model over the upper half-plane model is that the unit disc D is a bounded subset of the Euclidean plane. Thus we can view all of the hyperbolic plane easily on a sheet of paper (we shall see some pictures of this in the next lecture). One advantage of the upper half-plane model over the Poincare´ disc model is the ease with which Cartesian co-ordinates may be used in calculations. The geodesics in the Poincare´ disc model of hyperbolic geometry are the arcs of circles and diameters in D that meet ∂D orthogonally. We could define a distance function and develop an analysis analogous to that of the upper half-plane H in lectures 2–5, but it is quicker and more convenient to transfer the results from the upper half-plane H directly to this new setting. To do this, consider the map h : H→ D defined by h(z) = z − i iz − 1 . (1.1) (Note that h is not a Mo¨bius transformation of H; it does not satisfy the condition that ad − bc > 0.) It is 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 terms 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, Pythagoras’ 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 identity (and so fixes every point). Definition. Let γ be a Mo¨bius transformation of H. We will say that (i) γ is hyperbolic if it has two fixed points on ∂H and none in H, (ii) γ is parabolic if it has one fixed point on ∂H and none in H, (iii) γ is elliptic if it has no fixed points on ∂H and one in H. 2 MATH32051 17. Classifying Mo¨bius transformations via trace 17. Classifying Mo¨bius transformations via trace §17.1 Introduction We have seen how to classify Mo¨bius transformations in terms of the number and location of fixed points. Recall that if γ ∈ Mo¨b(H) is not the identity then • if γ has two fixed points on ∂H and none in H then we say γ is hyperbolic, • if γ has one fixed point on ∂H and none in H then we say γ is parabolic, • if γ has no fixed points on ∂H and one in H then we say γ is elliptic. The same classification works for Mo¨bius transformations of D. We would like a way to be able to read off whether a given Mo¨bius transformation is hyperbolic, parabolic or elliptic. To do this, we need to explore the connection between Mo¨bius transformations and matrices. §17.2 A matrix representation Let γ1 and γ2 be the Mo¨bius transformations of H given by γ1(z) = a1z + b1 c1z + d1 , γ2(z) = a2z + b2 c2z + d2 , respectively. Then the composition γ2 ◦ γ1 is the Mo¨bius transformation of H given by γ2γ1(z) = a2 ( a1z+b1 c1z+d1 ) + b2 c2 ( a1z+b1 c1z+d1 ) + d2 = (a2a1 + b2c1)z + (a2b1 + b2d1) (c2a1 + d2c1)z + (c2b1 + d2d1) . (2.1) Observe the connection between the coefficients in (2.1) and the matrix product( a2 b2 c2 d2 )( a1 b1 c1 d1 ) = ( a2a1 + b2c1 a2b1 + b2d1 c2a1 + d2c1 c2b1 + d2d1 ) . Thus we can calculate the coefficients of the composition of two Mo¨bius transformations γ1, γ2 by multiplying the 2× 2 matrices of the coefficients of γ1, γ2. We now further explore the connections between Mo¨bius transformations of H and matrices. Notice that if γ(z) = az + b cz + d 1 MATH32051 17. Classifying Mo¨bius transformations via trace is a Mo¨bius transformation of H, then z 7→ λaz + λb λcz + λd gives the same Mo¨bius transformation of H (provided λ 6= 0). Thus, by taking λ = 1/ √ (ad− bc) we can always assume, without loss of generality, that ad− bc = 1. Definition. The Mo¨bius transformation γ(z) = (az + b)/(cz + d) of H is said to be in normalised form (or normalised) if ad− bc = 1. We introduce the following group of matrices. Definition. The set of matrices SL(2,R) = { A = ( a b c d ) | a, b, c, d ∈ R, det A = ad− bc = 1 } is called the special linear group of R2. Hence if A ∈ SL(2,R) is a matrix with entries (a, b; c, d) then we can associate a nor- malised Mo¨bius transformation γA ∈ Mo¨b(H) by defining γA(z) = (az + b)/(cz + d). However, distinct matrices in SL(2,R) can give the same Mo¨bius transformation of H. To see this, notice that the two matrices( a b c d ) , ( −a −b −c −d ) where ad− bc = 1 are both elements of SL(2,R) but give the same Mo¨bius transformation of H. This, however, is the only way that distinct matrices in SL(2,R) can give the same Mo¨bius transformation of H. Remark. Thus we can think of Mo¨b(H) as the group of matrices SL(2,R) with two matrices A,B identified iff A = −B. Sometimes Mo¨b(H) is denoted by PSl(2,R), the projective special linear group. §17.3 The trace of a Mo¨bius transformation Recall that if A is a matrix then the trace of A is defined to be the sum of the diagonal entries of A. That is, if A = (a, b; c, d) then Trace(A) = a+ d. Let γ(z) = (az+ b)/(cz+ d) be a Mo¨bius transformation of H, ad− bc > 0. By dividing the coefficients a, b, c, d by √ ad− bc, we can always write γ in normalised form. Assume that γ is written in normalised form. Then we can associate to γ a matrix A = (a, b; c, d); as ad − bc = 1 we see that A ∈ SL(2,R). However, as we saw in §17.2, this matrix is not unique; instead we could have associated the matrix −A = (−a,−b;−c,−d) to γ. Thus we can define a function τ(γ) = (Trace(A))2 = (Trace(−A))2. Definition. Let γ ∈Mo¨b(H) be a Mo¨bius transformation of H with γ(z) = (az+b)/(cz+ d) where ad− bc = 1. We abuse notation slightly and we will call τ(γ) = (a+ d)2 the trace of γ. 2 MATH32051 17. Classifying Mo¨bius transformations via trace We can now classify the three types of Mo¨bius transformation—hyperbolic, parabolic and elliptic—in terms of the trace function. Let γ ∈ Mo¨b(H) be a Mo¨bius transformation of H. Suppose first∞ is not a fixed point (it follows that c 6= 0). Recall from Section 16 that z0 is a fixed point of γ if and only if z0 = a− d±√(a− d)2 + 4bc 2c . Thus there are two real solutions, one real solution or one complex conjugate pair of solu- tions depending on whether the term inside the square-root is greater than zero, equal to zero or less than zero, respectively. Using the identities ad− bc = 1, (a+ d)2 = τ(γ) we see that (a− d)2 + 4bc = a2 − 2ad+ d2 + 4(ad− 4) = a2 + 2ad+ d2 − 4 = τ(γ)− 4. Now suppose that ∞ is a fixed point; equivalently suppose that c = 0. We have already seen that γ has another fixed point at b/(d − a). There are two cases: (i) a = d, and (ii) a 6= d. In case (i), ∞ is the only fixed point (and so γ is parabolic). As 1 = ad− bc = ad = a2, we must have that either a = d = 1 or a = d = −1. Hence τ(γ) = (1+1)2 = (−1− 1)2 = 4. In case (ii), γ has two fixed points on the boundary, and so is hyperbolic. As∞ is a fixed point we have c = 0. Hence 1 = ad − bc = ad, so that d = 1/a. Hence τ(γ) = (a+ 1/a)2. By completing the square we see that (x+ 1/x)2 ≥ 4 if and only if (x2 − 1)2 + 2 ≥ 0 with equality when x = ±1. As the latter is always true, we have that (a + 1/a)2 > 4 when a 6= d. Thus we have proved: Proposition 17.3.1 Let γ ∈ Mo¨b(H) be a Mo¨bius transformation of H and suppose that γ is not the identity. Then: (i) γ is parabolic if and only if τ(γ) = 4; (ii) γ is elliptic if and only if τ(γ) ∈ [0, 4); (iii) γ is hyperbolic if and only if τ(γ) ∈ (4,∞). 3 MATH32051 18. Parabolic, hyperbolic and elliptic Mo¨bius transformations 18. Parabolic, hyperbolic and elliptic Mo¨bius transformations §18.1 Introduction We have seen how to classify Mo¨bius transformations into parabolic, hyperbolic and elliptic transformations. We now study each of these different types in turn. §18.2 Conjugacy of Mo¨bius transformations Before we start discussing the geometry and classification of Mo¨bius transformations, we introduce a notion of ‘sameness’ for Mo¨bius transformations. Definition. Let γ1, γ2 ∈ Mo¨b(H) be two Mo¨bius transformations of H. We say that γ1 and γ2 are conjugate if there exists another Mo¨bius transformation g ∈ Mo¨b(H) such that γ1 = g −1 ◦ γ2 ◦ g. Remarks. (i) Geometrically, if γ1 and γ2 are conjugate then the action of γ1 on H∪ ∂H is the same as the action of γ2 on g(H ∪ ∂H). Thus conjugacy reflects a change in coordinates of H ∪ ∂H. (ii) If γ2 has matrix A2 ∈ SL(2,R) and g has matrix A ∈ SL(2,R) then γ1 has matrix ±A−1A2A. (iii) We can define conjugacy for Mo¨bius transformations of D in exactly the same way: two Mo¨bius transformations γ1, γ2 ∈ Mo¨b(D) of D are conjugate if there exists g ∈ Mo¨b(D) such that γ1 = g −1 ◦ γ2 ◦ g. We will use the fact that conjugacy is an equivalence relation. Proposition 18.2.1 Conjugacy is an equivalence relation between Mo¨bius transformations. Proof. See Exercise 18.1. ✷ The following result says that conjugate Mo¨bius transformations of H have the same trace. Proposition 18.2.2 Let γ1 and γ2 be conjugate Mo¨bius transformations of H. Then τ(γ1) = τ(γ2). Proof. See Exercise 18.2. ✷ 1 MATH32051 18. Parabolic, hyperbolic and elliptic Mo¨bius transformations §18.3 Parabolic transformations Recall that a Mo¨bius transformation γ ∈ Mo¨b(H) is said to be parabolic if it has a unique fixed point and that fixed point lies on ∂H. For example, the Mo¨bius transformation of H given by γ(z) = z + 1 is parabolic. Here, the unique fixed point is ∞. In general, a 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 : H → D used to transfer results between H and D. If γ(z) = eiθz ∈ Mo¨b(D) is a rotation of the Poincare´ disc D then h−1γh ∈ Mo¨b(H) is a Mo¨bius transformation of H of the form h−1γh(z) = cos(θ/2) z + sin(θ/2) − sin(θ/2) z + cos(θ/2) (5.2) This map has a unique fixed point at i ∈ H. Maps of the form (5.2) are often called rotations of H. 5 MATH32051 19. Discreteness in metric spaces 19. Discreteness in metric spaces §19.1 Introduction For the remainder of the course we will be studying groups of Mo¨bius transformations. Recall that the set Mo¨b(H) of Mo¨bius transformations of H, and the set Mo¨b(D) of Mo¨bius transformations of D, form a group under composition. There are many subgroups of these groups. We will study a particularly interesting class of subgroups of Mo¨bius transformations known as Fuchsian groups. These allow us to study the symmetries of tilings/tessellations of the hyperbolic plane. Because the definition of a Fuchsian group is so important, we give it here. Definition. A Fuchsian group is a discrete subgroup of either Mo¨b(H), the Mo¨bius trans- formations of H, or Mo¨b(D), the Mo¨bius transformations of D. In this section, we will study what ‘discrete’ means; it makes sense to do this in a more abstract setting first. In the next section, we will see how to study discreteness of subgroups of Mo¨bius transformations. §19.2 Discreteness The concept of discreteness plays an important role in many areas of geometry, topology and metric spaces. A metric space is, roughly speaking, a mathematical space on which it is possible to define the distance between two points in the space. The concept of distance must satisfy some fairly natural assumptions: (i) the distance from a point to itself is zero, (ii) the distance from x to y is equal to the distance from y to x, and (iii) the triangle inequality: d(x, y) ≤ d(x, z) + d(z, y). Examples of metric spaces include: (i) Rn with the Euclidean metric d((x1, . . . , xn), (y1, . . . , yn)) = ‖(x1, . . . , xn)− (y1, . . . , yn)‖ = √ |x1 − y1|2 + · · · + |xn − yn|2, (2.1) (ii) the upper half-plane H with the metric dH that we defined in Section 4.4. Let (X, d) be a metric space. Heuristically, a subset Y ⊂ X is discrete if every point y ∈ Y is isolated, i.e., the other points of Y do not come arbitrarily close to y. More formally: Definition. We say that a point y ∈ Y is isolated if: there exists δ > 0 such that if y′ ∈ Y and y′ 6= y then d(y, y′) > δ. That is, a point y in a subset Y is isolated if, for some δ > 0, there are no other points of Y within distance δ of y. 1 MATH32051 19. Discreteness in metric spaces Definition. A subset Y is said to be discrete if every point y ∈ Y is isolated. Examples. 1. In any metric space, a single point {x} is discrete. 2. The set of integers Z forms a discrete subset of the real line R. To see this, let n ∈ Z be an integer and choose δ = 1/2. Then if |m− n| < δ, we see that m is an integer a distance at most 1/2 from n; this is only possible if m = n, as integers lie distance 1 apart. 3. The set of rationals Q is not a discrete subgroup of R: there are infinitely many distinct rationals arbitrarily close to any given rational. 4. The subset Y = {1, 1/2, 1/3, . . . , 1/n, . . .} of R is discrete. To see this, take y = 1/n. Choose δ = 1/n(n + 1). Then if y′ ∈ Y satisfies |y′ − y| < 1/n(n + 1) then y′ = 1/n (draw a picture!). 5. The subset Y = {1, 1/2, 1/3, . . . , 1/n, . . . , } ∪ {0} is not discrete. This is because 0 is not isolated: points of the form 1/n can come arbitrarily close to 0. No matter how small we choose δ, there are non-zero points of Y lying within δ of 0. 2 MATH32051 20. Fuchsian groups 20. Fuchsian groups §20.1 Introduction Recall the following definition: Definition. A Fuchsian group is a discrete subgroup of either Mo¨b(H) or Mo¨b(D). In the previous section, we saw what it meant for a subset of a metric space to be discrete. In order to study Fuchsian groups, we need to have a metric on the set Mo¨b(H) and on the set Mo¨b(D). We will then be able to say what it means for a subset of Mo¨b(H) or of Mo¨b(D) to be discrete. §20.2 A metric on Mo¨b(H) Here we discuss how to construct a metric on Mo¨b(H). Often, when studying metric spaces, the actual formula for the distance is not the most important consideration. Often it is more important to have an understanding of what it means for two points in the metric space to be ‘close’ to each other (here ‘close’ means a small, but unspecified, distance apart). For example, intuitively in R2 with the usual Euclidean metric, one might regard the two points (1, 1), (1.1, 0.8) ∈ R2 as being close, but the two points (1, 1), (100,−100) ∈ R2 as not being close. Intuitively, it is clear what we mean for two Mo¨bius transformations of H to be close: two Mo¨bius transformations of H are close if the coefficients (a, b; c, d) defining them are close. However, things are not quite so simple because, as we have seen in Section 17.2, different coefficients (a, b; c, d) can give the same Mo¨bius transformation. To get around this problem, we can insist on writing all Mo¨bius transformations of H in a normalised form. Recall that the Mo¨bius transformation γ(z) = (az + b)/(cz + d) is normalised if ad− bc = 1. However, there is still some ambiguity because if γ(z) = az + b cz + d is normalised, then so is γ(z) = −az − b −cz − d. This, however, is the only ambiguity (see Section 17.2). Thus we will say that the (normalised) Mo¨bius transformations of H given by γ1(z) = (a1z + b1)/(c1z + d1) and γ2(z) = (a2z + b2)/(c2z + d2) are close if either (a1, b1, c1, d1), (a2, b2, c2, d2) are close or (a1, b1, c1, d1), (−a2,−b2,−c2,−d2) are close. If we wanted to make this precise and in particular have a formula, then we could define a metric on the space Mo¨b(H) of Mo¨bius transformations of H by setting dMo¨b(γ1, γ2) = min{‖(a1, b1, c1, d1)− (a2, b2, c2, d2)‖, ‖(a1, b1, c1, d1)− (−a2,−b2,−c2,−d2)‖}. (2.1) 1 MATH32051 20. Fuchsian groups (Here ‖ · ‖ is the Euclidean metric in R4 defined by (2.1) in the case n = 4.) However, we will never use an explicit metric on Mo¨b(H) and prefer instead to think of Mo¨bius transformations of H being close if they ‘look close’ (secure in the knowledge that we could fill in the details using the metric given above if we had to). We can also define a metric on Mo¨b(D) in exactly the same way. Again, we will never need to use the formula for this metric explicitly; instead, two Mo¨bius transformations of D are ‘close’ if (up to normalisation and sign) the coefficients defining them are close. §20.3 Examples of Fuchsian groups We list below some examples of Fuchsian groups. Examples. 1. Any finite subgroup of Mo¨b(H) or Mo¨b(D) is a Fuchsian group. This is because any finite subset of any metric space is discrete. 2. As a specific example in the upper half-plane, let γθ(z) = cos(θ/2) z + sin(θ/2) − sin(θ/2) z + cos(θ/2) be a rotation around i. Let q ∈ N. Then {γ2πj/q | 0 ≤ j ≤ q− 1} is a finite subgroup. In D, this is the group {z 7→ e2πj/qz | j = 0, 1, . . . , q−1} of rotations about 0 through angles that are multiples of 2π/q. 3. The subgroup of integer translations {γn(z) = z + n | n ∈ Z} is a Fuchsian group. The subgroup of all translations {γb(z) = z + b | b ∈ R} is not a Fuchsian group as it is not discrete. 4. The subgroup Γ = {γn(z) = 2nz | n ∈ Z} is a Fuchsian group. 5. The subgroup Γ = {id} containing only the identity Mo¨bius transformation is a Fuchsian group. We call it the trivial Fuchsian group. 6. If Γ is a Fuchsian group and Γ1 < Γ is a subgroup then Γ1 is a Fuchsian group. 7. One of the most important Fuchsian groups is the modular group PSL(2,Z). This is the group given by Mo¨bius transformations of H of the form γ(z) = az + b cz + d , a, b, c, d ∈ Z, ad− bc = 1. 8. Let q ∈ N. Define Γq = { γ(z) = az + b cz + d | a, b, c, d ∈ Z, ad− bc = 1, b, c are divisible by q } . This is called the level q modular group or the congruence subgroup of order q. 2 MATH32051 20. Fuchsian groups §20.4 A criterion for a subgroup to be a Fuchsian group Recall that a subset is discrete if every point is isolated. The following results tells us that, if we want to check that a subgroup is discrete, then it is sufficient to check that the identity is isolated. Proposition 20.4.1 Let Γ be a subgroup of Mo¨b(H). The following are equivalent: (i) Γ is a discrete subgroup of Mo¨b(H) (i.e. Γ is a Fuchsian group); (ii) the identity element of Γ is isolated. Remark. The same statement holds in the case of the Poincare´ disc model D. Proof. Clearly (i) implies (ii). The proof of (ii) implies (i) is straight-forward, but requires knowledge of concepts from metric spaces. The idea is to show that (i) the image of an isolated point under a continuous map is isolated, and (ii) the map defined on Mo¨b(H) by compisition by a fixed element of Mo¨b(H) is continuous. Then if the identity id is isolated, by considering the image of id under composition by γ ∈ Γ, we see that γ is isolated. As γ is arbitrary, we are done. ✷ 3 MATH32051 21. Orbits 21. Orbits §21.1 Introduction In the previous section we defined what is meant by a Fuchsian group. Recall that a Fuchsian group Γ is a discrete subgroup of Mo¨b(H) or Mo¨b(D). If we are given a subgroup of Mo¨b(H) or Mo¨b(D) and we want to check whether it is a Fuchsian group or not then we have two option: (i) use our intuition as to what it means for two Mo¨bius transformations to be close to each other, and then convince ourselves that the subset is or is not a discrete subset of Mo¨b(H) or Mo¨b(D), or (ii) use the formula for distance between two Mo¨bius transformations given in Section 20 and the definition of discreteness given in Section 19 to check this rigorously. Normally, our intuition will be good enough and will give the correct answer. However, this intuition is based on discreteness of subsets of R4, and thinking in 4 dimensions might not be easy for some people! Instead, we would like a criterion for a subset of Mo¨bius transformations to be a Fuchsian group that is easier to check. This leads to the notion of the ‘orbit’ of a point. Orbits turn out to be very important later on in the course. §21.2 Orbits and a criterion for discreteness Let Γ be a subgroup of Mo¨b(H). Definition. Let z ∈ H. The orbit Γ(z) of z under Γ is the set of all points of H that we can reach by applying elements of Γ to z: Γ(z) = {γ(z) | γ ∈ Γ}. The following result says that for subgroups of isometries of the hyperbolic plane, dis- creteness of the group is the same as discreteness of every orbit. Proposition 21.2.1 Let Γ be a subgroup of Mo¨b(H). Then the following are equivalent: (i) Γ is a Fuchsian group; (ii) For each z ∈ H, the orbit Γ(z) is a discrete subset of H. Remark. The same statement holds in the case of the Poincare´ disc model D. Proof. Omitted. See §2.2 in Katok’s book. ✷ Example. Let Γ = {γn | γn(z) = 2nz, n ∈ Z}. Fix z ∈ H. Then the orbit of z is Γ(z) = {2nz | n ∈ Z}. We will show directly that Γ(z) is a discrete subgroup of H. To see this, first observe that the points 2nz lie on the (Euclidean) straight line through the origin inclined at angle arg(z). Fix 2nz and let δ = 2n−1|z|. It is easy to see that |2mz− 2nz| ≥ δ whenever m 6= n. Hence Γ(z) is discrete. 1 MATH32051 21. Orbits Remark. A subgroup Γ of Mo¨b(H) also acts on ∂H. However, the orbit Γ(z) of a point z ∈ ∂H under Γ need not be discrete, even if the group Γ itself is discrete. For example, consider the modular group: PSL(2,Z) = { γ(z) = az + b cz + d | a, b, c, d ∈ Z, ad− bc = 1 } . This is a Fuchsian group (and therefore discrete). However, the orbit of the point 0 ∈ ∂H under PSl(2,Z) is the set {b/d | ad − bc = 1}. It is easy to see that this set is equal to Q∪{∞}, which is not a discrete subset of ∂H (because an irrational point on R can always be arbitrarily well approximated by rationals). 2 MATH32051 22. Fundamental domains: the definition 22. Fundamental domains: the definition §22.1 Introduction In the last section we defined what is meant by a Fuchsian group. Fuchsian groups are subgroups of the group of Mo¨bius transformations of H or D—they are algebraic objects. We want to study how Fuchsian groups act geometrically on the hyperbolic plane. We will do this via fundamental domains. Here is an example of this construction in Euclidean geometry. Consider the set of (Euclidean) isometries of R2 given by Γ := {γn,m | γn,m(x, y) = (x+ n, y +m), m, n ∈ Z}. That is, Γ is the set of all translations, horizontally and vertically, of R2 by integers; this is a discrete subgroup of the group of (Euclidean) isometries of R2. Now consider the tiling of R2 by unit squares, illustrated in Figure 22.1 below. Figure 22.1: The square lattice in R2. This tiling covers the whole of the Euclidean plane (in the sense that every point in R2 is in at least one tile; moreover the tiles do not overlap, except along their edges. This tiling is also invariant under Γ in the following sense: if we look at the image of this tiling under any element of Γ then the tiling remains unchanged. Alternatively, we could start with the unit square F = {(x, y) ∈ R2 | 0 ≤ x, y ≤ 1}. Then the image of F under γn,m ∈ Γ is γn,m(F ) = {(x, y) ∈ R2 | n ≤ x ≤ n + 1,m ≤ y ≤ m+ 1}. The union over all of these images is the Euclidean plane R2:⋃ γ∈Γ γ(F ) = R2. However, the images are not disjoint: two adjacent images overlap along their shared edge. We could eliminate this problem by consider a ’half-open’ square, or tile, of the form F = {(x, y) ∈ R2 | 0 ≤ x, y < 1}; then we have bigcupγ∈Γγ(F ) = R2 but (for example) the right-hand edge of F and the left-hand side 1 MATH32051 22. Fundamental domains: the definition of γ1,0(F ) no longer overlap. For more complicated groups Γ or more complicated tiles F (which, for the moment, we assume to be polygons), it may not be clear which sides of the tile to make open and which to make closed. We will avoid this problem by only using tiles that are open sets; this allows us to say that no two tiles overlap in the sense that γ1(F ) and γ2(F ) are disjoint for γ1, γ2 ∈ Γ. However, the union ⋃ γ∈Γ γ(F ) will not, in general, be equal to the plane, as the edges of the tiles will not be included; we work around this by looking at a notion called the closure of F . In the following sections we address the following problem: given a Fuchsian group Γ, how do we find a subset F ⊂ H or D such that the images of F under elements of Γ tile the hyperbolic plane? §22.2 Open and closed subsets We will need to say what it means for a subset of H to be open or closed. (For convenience we work in H but everything below has an analogue in D.) Definition. A subset Y ⊂ H is said to be open if for each y ∈ Y there exists ε > 0 such that the open ball Bε(y) = {z ∈ H | dH(z, y) < ε} of radius ε and centre y is contained in Y . A subset Y ⊂ H is said to be closed if its complement H \ Y is open. Remark. Recall from Section 11 that hyperbolic circles are Euclidean circles (albeit with different radii and centres). Thus to prove a subset Y ⊂ H is open it is sufficient to find a Euclidean open ball around each point that is contained in Y . In particular, the open subsets of H are the same as the open subsets of the (Euclidean) upper half-plane. Examples. 1. The subset {z ∈ H | 0 < Re(z) < 1} is open. 2. The subset {z ∈ H | 0 ≤ Re(z) ≤ 1} is closed. 3. The subset {z ∈ H | 0 < Re(z) ≤ 1} is neither open nor closed. Definition. Let Y ⊂ H be a subset. Then the closure of Y is the smallest closed subset containing Y . We denote the closure of Y by cl(Y ). For example, the closure of {z ∈ H | 0 < Re(z) < 1} is {z ∈ H | 0 ≤ Re(z) ≤ 1}. §22.3 Fundamental domains Definition. Let Γ be a Fuchsian group. A fundamental domain F ⊂ H for Γ is an open subset of H such that (i) ⋃ γ∈Γ γ(cl(F )) = H, (ii) the images γ(F ) are pairwise disjoint; that is, γ1(F )∩γ2(F ) = ∅ if γ1, γ2 ∈ Γ, γ1 6= γ2. Remark. Notice that in (i) we have written γ(cl(F )) (i.e. we first take the closure, then apply γ). We could instead have written cl(γ(F )) (i.e. first apply γ, now take the closure). These two sets are equal. This follows from the fact that both γ and γ−1 are continuous maps. (See any text on metric spaces for more details.) 2 MATH32051 22. Fundamental domains: the definition Thus F is a fundamental domain if every point lies in the closure of some image γ(F ) and if two distinct images do not overlap. We say that the images of F under Γ tessellate H. Remark. Some texts require fundamental domains to be closed. If this is the case then condition (i) is replaced by the assumption that ⋃ γ∈Γ γ(F ) = H, and condition (ii) requires the set γ(int(F )) to be pairwise disjoint (here int(F ) denotes the interior of F , the largest open set contained inside F ). 3 MATH32051 23. Fundamental domains: examples and first results 23. Fundamental domains: examples and first results §23.1 Introduction Let Γ be a Fuchsian group acting on H or D. A fundamental domain F ⊂ H or D is an open set whose images under elements of Γ tile the hyperbolic plane so that they cover the hyperbolic plane and no two images overlap. More specifically, we say that the open set F is a fundamental domain if (i) ⋃ γ∈Γ γ(cl(F )) = H or D, (ii) γ1(F ) ∩ γ2(F ) = ∅ if γ1, γ2 ∈ Γ, γ1 6= γ2. In this section, we give some examples of fundamental domains and study some simple properties of them. §23.2 Examples of fundamental domains Example. Consider the subgroup Γ of Mo¨b(H) given by integer translations: Γ = {γn | γn(z) = z + n, n ∈ Z}. This is a Fuchsian group. Consider the set F = {z ∈ H | 0 < Re(z) < 1}. This is an open set. Clearly if Re(z) = a then Re(γn(z)) = n+ a. Hence γn(F ) = {z ∈ H | n < Re(z) < n+ 1} and γn(cl(F )) = {z ∈ H | n ≤ Re(z) ≤ n+ 1}. Hence H = ⋃ n∈Z γn(cl(F )). It is also clear that if γn(F ) and γm(F ) intersect, then n = m. Hence F is a fundamental domain for Γ. See Figure 23.1. Example. Consider the subgroup Γ = {γn | γn(z) = 2nz, n ∈ Z} of Mo¨b(H). This is a Fuchsian group. Let F = {z ∈ H | 1 < |z| < 2}. This is an open set. Clearly, if 1 < |z| < 2 then 2n < |γn(z)| < 2n+1. Hence γn(F ) = {z ∈ H | 2n < |z| < 2n+1} and γn(cl(F )) = {z ∈ H | 2n ≤ |z| ≤ 2n+1}. Hence H = ⋃ n∈Z γn(cl(F )). It is also clear that if γn(F ) and γm(F ) intersect, then n = m. Hence F is a fundamental domain for Γ. See Figure 23.2. 1 MATH32051 23. Fundamental domains: examples and first results F Re(z)=−1 Re(z)=0 Re(z)=1 Re(z)=2 γ1(F ) γ2(F )γ−1(F ) Figure 23.1: A fundamental domain and tessellation for Γ = {γn | γn(z) = z + n}. -4 -2 -1 0 1 2 4 γ1(F ) γ−1(F ) F Figure 23.2: A fundamental domain and tessellation for Γ = {γn | γn(z) = 2nz}. Suppose that Γ = {id}, the trivial group containing just one element. In this case, H is a fundamental domain for Γ; indeed H is the only fundamental domain for Γ. Now suppose that Γ 6= {id}. For a non-trivial Fuchsian group, fundamental domains are not unique. That is, for a given non-trivial Fuchsian group there will be many different fundamental domains. For example, Figure 23.3 gives an example of a different fundamental domain for the Fuchsian group Γ = {γn | γn(z) = z + n, n ∈ Z}. However, we have the following result which (essentially) says that any two fundamental regions have the same area. To state it precisely, we need to take care to avoid some pathological fundamental domains. Recall that the boundary ∂F of a set F is defined to be the set cl(F ) \ int(F ); here cl(F ) is the closure of F and int(F ) is the interior of F . Proposition 23.2.1 Let F1 and F2 be two fundamental domains for a Fuchsian group Γ, with AreaH(F1) Assume that AreaH(∂F1) = 0 and AreaH(∂F2) = 0. Then AreaH(F1) = AreaH(F2). Proof. First notice that, for i = 1, 2, we have that AreaH(cl(Fi)) = AreaH(Fi) for i = 1, 2 2 MATH32051 23. Fundamental domains: examples and first results Re(z)=−1 Re(z)=0 Re(z)=1 Re(z)=2 Re(z)=3 γ−1(F ) F γ1(F ) γ2(F ) Figure 23.3: Another fundamental domain and tessellation for Γ = {γn | γn(z) = z + n}. as AreaH(∂Fi) = 0. Now cl(F1) ⊃ cl(F1) ∩ ⋃ γ∈Γ γ(F2) = ⋃ γ∈Γ (cl(F1) ∩ γ(F2)) . As F2 is a fundamental domain, the sets cl(F1) ∩ γ(F2) are pairwise disjoint. Hence, using the facts that (i) the area of the union of disjoint sets is the sum of the areas of the sets, and (ii) Mo¨bius transformations of H preserve area we have that AreaH(cl(F1)) ≥ ∑ γ∈Γ AreaH(cl(F1) ∩ γ(F2)) = ∑ γ∈Γ AreaH(γ −1(cl(F1)) ∩ F2) = ∑ γ∈Γ AreaH(γ(cl(F1)) ∩ F2). Since F1 is a fundamental domain we have that⋃ γ∈Γ γ(cl(F1)) = H. Hence ∑ γ∈Γ AreaH(γ(cl(F1)) ∩ F2) ≥ AreaH ⋃ γ∈Γ γ(cl(F1)) ∩ F2 = AreaH(F2). Hence AreaH(F1) = AreaH(cl(F1)) ≥ AreaH(F2). Interchanging F1 and F2 in the above gives the reverse inequality. Hence AreaH(F1) = AreaH(F2). ✷ Remarks. (i) In the above proof, we assumed that we could calculate the area of any set. In fact, there are sets for which the notion of ‘area’ does not make sense. These 3 MATH32051 23. Fundamental domains: examples and first results are called non-measurable sets. It is beyond the scope of this course to discuss the intricacies of what is called measure theory; you can find out more in the Fractal Geometry or Fourier Series and Lebesgue Integration course units. (ii) So far, all of the examples of fundamental domains that we have seen have infinite hyperbolic area. We will see later that the modular group PSL(2,Z) has a fundamental domain with finite hyperbolic area. We will also construct many examples of other Fuchsian groups with fundamental domains with finite hyperbolic area. So far, we do not yet know that there exists a fundamental domain for a given Fuchsian group. There are several methods of constructing fundamental domains and we discuss one such method in the next few lectures. 4 MATH32051 24. Dirichlet polgons: the preliminaries 24. Dirichlet polygons: the preliminaries §24.1 Introduction Let Γ be a Fuchsian group. Recall that a Fuchsian group is a discrete subgroup of Mo¨b(H) or of Mo¨b(D). In Section 22, we defined the notion of a fundamental domain F . Recall that a subset F ⊂ H is a fundamental domain if, essentially, the images γ(F ) of F under the Mo¨bius transformations γ ∈ Γ tessellate (or tile) the upper half-plane H. In Section 23 we saw some specific examples of fundamental domains. For example, we saw that the set {z ∈ H | 0 < Re(z) < 1} is a fundamental domain for the group of integer translations {γn(z) = z + n | n ∈ Z}. However, we do not yet know that each Fuchsian group possesses a fundamental domain. The purpose of the next four sections is to give a method for constructing a fundamental domain for a given Fuchsian group. The fundamental domain that we construct is called a Dirichlet polygon. It is worth remarking that the construction that we give below works in far more general circumstances than those described here. We also remark that there are other methods for constructing fundamental domains that, in general, give different fundamental domains than a Dirichlet polygon; such an example is the Ford fundamental domain which is described in Katok’s book. The construction given below is written in terms of the upper half-plane H. The same construction works in the Poincare´ disc D. §24.2 Convex polygons as intersections of half-planes In Section 13 we defined a polygon as the region bounded by a finite set of geodesic segments. It will be useful to slightly modify this definition. Definition. Let C be a geodesic in H. Then C divides H into two components. These components are called half-planes. For example, the imaginary axis determines two half-planes: {z ∈ H | Re(z) < 0} and {z ∈ H | Re(z) > 0}. The geodesic given by the semi-circle of unit radius centred at the origin also determines two half-planes (although they no longer look like Euclidean half-planes): {z ∈ H | |z| < 1} and {z ∈ H | |z| > 1}. We define a convex hyperbolic polygon as follows. Definition. A convex hyperbolic polygon is the intersection of a finite number of half- planes. One difference between this definition of a hyperbolic polygon and the more naive definition given in Section 13 is that we now allow for the possibility of an edge of a hyperbolic polygon to be an arc of the circle at infinity. See Figure 24.1. 1 MATH32051 24. Dirichlet polgons: the preliminaries (i) (ii) P P Figure 24.1: A polygon with one edge on the boundary (i) in the upper half-plane, (ii) in the Poincare´ disc. 2 MATH32051 25. Dirichlet polygons: perpendicular bisectors 25. Dirichlet polygons: perpendicular bisectors §25.1 Introduction Let Γ be a Fuchsian group. We want to be able to construct a fundamental domain for Γ. To do this, we need a brief discussion on perpendicular bisectors. §25.2 Perpendicular bisectors Let z1, z2 ∈ H. Recall that [z1, z2] is the segment of the unique geodesic from z1 to z2. The perpendicular bisector of [z1, z2] is defined to be the unique geodesic perpendicular to [z1, z2] that passes through the midpoint of [z1, z2]. z1 z2 [z1, z2] Figure 25.1: The perpendicular bisector of [z1, z2]. Proposition 25.2.1 Let z1, z2 ∈ H. The set of points {z ∈ H | dH(z, z1) = dH(z, z2)} that are equidistant from z1 and z2 is the perpendicular bisector of the line segment [z1, z2]. Proof. By applying a Mo¨bius transformation of H, we can assume that both z1 and z2 lie on the imaginary axis and z1 = i. Write z2 = ir 2 for some r > 0 and there is no loss in generality (by applying the Mo¨bius transformation z 7→ −1/z if required) that r > 1. By using Proposition 8.1.1 it follows that the mid-point of [i, ir2] is at the point ir. It is clear that the unique geodesic through ir that meets the imaginary axis at right-angles is given by the semi-circle of radius r centred at 0. Recall from Proposition 11.2.2 that cosh dH(z, w) = 1 + |z − w|2 2 Im z Imw . In our setting this implies that |z − i|2 = |z − ir 2|2 r2 . This easily simplifies to |z| = r, i.e. z lies on the semicircle of radius r, centred at 0. ✷ 1 MATH32051 26. Dirichlet polygons: the construction 26. Dirichlet polygons: the construction §26.1 Introduction From now on, we will normally assume that Γ is a non-trivial Fuchsian group. Recall that the trivial Fuchsian group is the group Γ = {id} consisting of just the identity. In this case, a fundamental domain for Γ is given by the hyperbolic plane H (or D, if we are working in the Poincare´ disc model) and this is the only fundamental domain for Γ. Let Γ be a non-trivial Fuchsian group. We are now in a position to describe a method to construct a fundamental domain for Γ. Before we do that, we need to discuss a technical result about Fuchsian groups. §26.2 A technical result We need to state the following technical result: Lemma 26.2.1 Let Γ be a non-trivial Fuchsian group. Then there exists a point p ∈ H that is not a fixed point for any non-trivial element of Γ. (That is, γ(p) 6= p for all γ ∈ Γ \ {id}.) Proof. This relies on the fact that Mo¨bius transformations have few fixed points, but the details involve very techincal arguments. See Lemma 2.2.5 in Katok. ✷ §26.3 Dirichlet polygons: the construction Let Γ be a Fuchsian group and let p ∈ H be a point such that γ(p) 6= p for all γ ∈ Γ \ {id}. Let γ be an element of Γ and suppose that γ is not the identity. The set {z ∈ H | dH(z, p) < dH(z, γ(p))} consists of all points z ∈ H that are closer to p than to γ(p). We define the Dirichlet region to be: D(p) = {z ∈ H | dH(z, p) < dH(z, γ(p)) for all γ ∈ Γ \ {id}}, Thus the Dirichlet region is the set of all points z that are closer to p than to any other point in the orbit Γ(p) = {γ(p) | γ ∈ Γ} of p under Γ. To better describe the Dirichlet region consider the following procedure: (i) Choose p ∈ H such that γ(p) 6= p for all γ ∈ Γ \ {id}. (ii) For a given γ ∈ Γ \ {id} construct the geodesic segment [p, γ(p)]. (iii) Take Lp(γ) to be the perpendicular bisector of [p, γ(p)]. 1 MATH32051 26. Dirichlet polygons: the construction (iv) Let Hp(γ) be the half-plane determined by Lp(γ) that contains p. (Thus by Proposi- tion 25.2.1 Hp(γ) consists of all points z ∈ H that are closer to p than to γ(p).) (v) Then D(p) = ⋂ γ∈Γ\{id} Hp(γ). Theorem 26.3.1 Let Γ be a Fuchsian group and let p be a point not fixed by any non-trivial element of Γ. Then the Dirichlet region D(p) is a fundamental domain for Γ. Moreover, if AreaH(D(p)) < ∞ then D(p) is a convex hyperbolic polygon (in the sense of §24.2); in particular it has finitely many edges. Remarks. 1. There are many other hypotheses that ensure thatD(p) is a convex hyperbolic polygon with finitely many edges; requiring D(p) to have finite hyperbolic area is probably the simplest. Fuchsian groups that have a convex hyperbolic polygon with finitely many edges as a Dirichlet region are called geometrically finite. 2. If D(p) has finitely many edges then we refer to D(p) as a Dirichlet polygon. Notice that some of these edges may be arcs of ∂H. If there are finitely many edges then there are also finitely many vertices (some of which may be on ∂H). 3. The Dirichlet polygon D(p) depends on p. If we choose a different point p, then we may obtain a different polygon with different properties, such as the number of edges. Given a Fuchsian group Γ, Beardon (Theorem 9.4.5) describes the properties that a Dirichlet polygon D(p) will have for a typical point p. Proof. There are two things to show here: namely, that D(p) is a convex hyperbolic polygon, and that D(p) is a fundamental domain. Both of these facts rely on technical properties of Fuchsian groups that we have chosen to avoid, and we do not go into them here. See Theorem 6.17 in Anderson or §§2,3 in Katok. ✷ §26.4 The group of all integer translations We work through the procedure above to construct a Dirichlet polygon for the group of integer translations of H. Proposition 26.4.1 Let Γ be the Fuchsian group {γn | γn(z) = z + n, n ∈ Z}. Then D(i) = {z ∈ H | −1/2 < Re(z) < 1/2}. Proof. Let p = i. Then clearly γn(p) = i+n 6= p so that p is not fixed by any non-trivial element of Γ. As γn(p) = i + n, it is clear that the perpendicular bisector of [p, γn(p)] is the vertical straight line with real part n/2. Hence Hp(γn) = { {z ∈ H | Re(z) < n/2} if n > 0, {z ∈ H | Re(z) > n/2} if n < 0. 2 MATH32051 26. Dirichlet polygons: the construction n/2 Hp(γn) p p+ n [p, γn(p)] Lp(γn) Figure 26.1: The half-plane determined by the perpendicular bisector of the geodesic segment [p, p + n]. Hence D(p) = ⋂ γ∈Γ\{id} Hp(γ) = Hp(γ1) ∩Hp(γ−1) = {z ∈ H | −1/2 < Re(z) < 1/2}. ✷ §26.5 Groups of rotations We work through the procedure above to construct a Dirichlet polygon for a finite group of rotations of D. Proposition 26.5.1 Fix n > 0 and let Γ be the discrete group of Mo¨bius transformations of D given by Γ = {γk | γk(z) = e2πik/nz, k = 0, 1, . . . , n − 1}. Let p = 1/2. Then D(p) = {z ∈ D | −π/n < arg z < π/n}. Proof. Clearly the only fixed point of γn is the origin, so that we may take any p ∈ D\{0}. Let us take p = 1/2. Then γk(p) = (e 2πik/n)/2. The geodesic segment [p, γk(p)] is an arc of semi-circle and it is easy to see that the perpendicular bisector Lp(γk) of this arc is the diameter of D inclined at angle (2πk/n)/2 = πk/n. See Figure 26.2. Hence Hp(γk) is a sector of the unit disc bounded by the diameter Lp(γk). Taking the intersection, we see that D(p) = {z ∈ D | −π/n < arg z < π/n}. The tessellation of the Poincare´ disc in the case n = 7 is illustrated in Figure 26.3. ✷ 3 MATH32051 26. Dirichlet polygons: the construction p Hp(γk) γk(p) Lp(γk) 2πk/n Figure 26.2: The half-plane determined by the perpendicular bisector of the geodesic segment [p, e2πik/np]. 2π/7 Figure 26.3: The tessellation of D in the case when n = 7. 4 MATH32051 27. Dirichlet polygons: the modular group 27. Dirichlet polygons: the modular group §27.1 The modular group In this section we use the procedure described in Section 26 to find a fundamental domain for the modular group PSL(2,Z). Recall that the modular group is defined to be PSL(2,Z) = { az + b cz + d | a, b, c, d ∈ Z, ad− bc = 1 } . Proposition 27.1.1 Let k > 1 and let p = ki. Then a Dirichlet polygon for the modular group PSL(2,Z) is given by: D(p) = {z ∈ H | |z| > 1, −1/2 < Re(z) < 1/2}. p D(p) Figure 27.1: A Dirichlet polygon for PSL(2,Z). Proof. It is easy to check that if p = ik for k > 1 then p is not fixed under any non-trivial element of PSL(2,Z). Consider the Mo¨bius transformations of H given by γ1(z) = z + 1 and γ2(z) = −1/z. Clearly these lie in PSL(2,Z). The perpendicular bisector of [p, γ1(p)] = [p, p + 1] is easily seen to be the vertical line Re(z) = 1/2. Hence Hp(γ1) = {z ∈ H | Re(z) < 1/2}. Similarly, Hp(γ−11 ) = {z ∈ H | Re(z) > −1/2}. The geodesic segment [p, γ2(p)] is the arc of imaginary axis between ik and i/k. By using Proposition 8.1.1 it follows that the perpendicular bisector of [p, γ2(p)] is the semi-circle of radius 1 centred at the origin. Hence Hp(γ2) = {z ∈ H | |z| > 1}. 1 MATH32051 27. Dirichlet polygons: the modular group Let F = Hp(γ1) ∩Hp(γ−11 ) ∩Hp(γ2). Then D(p) = ⋂ γ∈PSL(2,Z)\{id} Hp(γ) ⊂ ⋂ γ=γ1,γ −1 1 ,γ2 Hp(γ) = F. It remains to show that this inclusion is an equality, i.e. D(p) = F . Suppose for a contradiction that this is not the case, i.e. D(p) ⊂ F but D(p) 6= F . Then as D(p) is a fundamental domain, there exists a point z0 ∈ D(p) ⊂ F and γ ∈ PSL(2,Z) \ {id} such that γ(z0) ∈ F . We show that this can not happen. To see this, write γ(z) = az + b cz + d where a, b, c, d ∈ Z and ad− bc = 1. Then |cz0 + d|2 = c2|z0|2 + 2Re(z0)cd + d2 > c2 + d2 − |cd| = (|c| − |d|)2 + |cd|, since |z0| > 1 and Re(z0) > −1/2. This lower bound is a non-negative integer. Moreover, it is non-zero, for it were zero then both c = 0 and d = 0 which contradicts the fact that ad− bc = 1. Thus the lower bound is at least 1, so that |cz0 + d|2 > 1. Hence Im(γ(z0)) = Im z0 |cz0 + d|2 < Im z0. Repeating the above argument with z0 replaced by γ(z0) and γ replaced by γ −1 we see that Im z0 < Im(γ(z0)), a contradiction. ✷ 2 MATH32051 27. Dirichlet polygons: the modular group 0 1-1 2 F γ1(F ) γ 2 1(F ) γ1γ2(F )γ2(F ) γ2γ1(F ) Figure 27.2: The tessellation of H determined by the Dirichlet polygon given in Proposi- tion 27.1.1 for the modular group. 3 MATH32051 28. Side-pairing transformations 28. Side-pairing transformations §28.1 Introduction Let Γ be a Fuchsian group and suppose that Γ has a Dirichlet polygon D(p). In this section we will see how, for each side s of D(p), we can associate an element γ of Γ to s in such a way that γ maps the side s to another side of D(p). These so-called side-pairing transformations will prove to be extremely useful later in the course. §28.2 Side-pairing transformations Let D be a hyperbolic polygon. A side s ⊂ H of D is an edge of D in H equipped with an orientation. That is, a side of D is an edge which starts at one vertex and ends at another. Let Γ be a Fuchsian group and let D(p) be a Dirichlet polygon for Γ. We assume that D(p) has finitely many sides. Let s be a side of D. Suppose that for some γ ∈ Γ \ {id}, we have that γ(s) is also a side of D(p). Note that γ−1 ∈ Γ \ {id} maps the side γ(s) back to the side s. Definition. We say that the sides s and γ(s) are paired and call γ a side-pairing trans- formation. (As we shall see, it can happen that s and γ(s) are the same side, albeit with opposing orientations; in this case, we say that s is paired with itself.) Given a side s of a Dirichlet polygon D(p), we can explicitly find a side-pairing transfor- mation associated to it. By the way in which D(p) is constructed, we see that s is contained in the perpendicular bisector Lp(g) of the geodesic segment [p, g(p)], for some g ∈ Γ \ {id}. We will show that the Mo¨bius transformation γ = g−1 maps s to another side of D(p). See Figure 28.1. We often denote by γs the side-pairing transformation associated to the side s. p D(p) s γ(s) g(p) Figure 28.1: The transformation γ = g−1 is associated to the side s of D(p). Proposition 28.2.1 Let Γ ⊂ Mo¨b(H) be a Fuchsian group and let D(p) be a Dirichlet polygon for Γ. Let s be a side of D(p), so that s ⊂ Lp(g) for some g ∈ Γ \ {id}. Then g−1(s) is also a side of D(p). 1 MATH32051 28. Side-pairing transformations Proof. Let z be a point on s and suppose that s ⊂ Lp(γ).. Then z is characterised by: (i) dH(z, p) = d(z, γ(p)) and (ii) dH(z, p) ≤ dH(z, g(p)) for all g ∈ Γ. Condition (i) ensures that z is on Lp(γ) and condition (ii) ensures that z ∈ cl(D(p)). Let w = γ−1(z) and note that z = γ(w). We need to show that (i) d(w, p) = d(w, γ−1(p)) (this ensures that w ∈ Lp(γ−1)), and (ii) d(w, p) ≤ d(w, g(p)) for all g ∈ Γ (this ensures that w ∈ cl(D(p))). We prove (i). Note that, using the fact that γ is an isometry, dH(w, p) = dH(γ −1(z), p) = dH(z, γ(p)) = dH(z, p) (2.1) where, for the last equality, we used the fact that z ∈ Lp(γ). Hence dH(w, p) = dH(z, p) = dH(γ(w), p) = dH(w, γ −1(p)). This proves (i). We prove (ii). Let g ∈ Γ. Then, using (2.1) we have dH(w, γ(p)) = d(γ −1(z), g(p)) = d(z, γg(p)) ≥ d(z, p) = d(w, p). ✷ §28.3 Examples of side-pairing transformations Let us calculate some examples of side-pairing transformations. Example. Let Γ = {γn | γn(z) = z + n, n ∈ Z} be the Fuchsian group of integer transla- tions. We have seen that if p = i then D(p) = {z ∈ H | −1/2 < Re(z) < 1/2} is a Dirichlet polygon for Γ. Let s be the side s = {z ∈ H | Re(z) = −1/2}. Let g(z) = z − 1. Then s is the perpendicular bisector of [p, p − 1] = [p, g(p)]. Hence γs(z) = g−1(z) = z + 1 so that γs(s) = s ′ where s′ is the side s′ = {z ∈ H | Re(z) = 1/2}. s γs(s) = s ′γs D(p) Figure 28.2: A side-pairing map for Γ = {γn | γn(z) = z + n}. 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 2 MATH32051 28. Side-pairing transformations p = ik for any k > 1. This polygon has three sides: s1 = {z ∈ H | Re(z) = −1/2, |z| > 1} s2 = {z ∈ H | Re(z) = 1/2, |z| > 1} s3 = {z ∈ H | |z| = 1, −1/2 < Re(z) < 1/2}. As in the above example, it is clear that γs1(z) = z + 1 so that γs1 pairs s1 with s2. The side pairing transformation associated to the side s2 is γs2(z) = z − 1. Consider s3. This is the perpendicular bisector of [p,−1/p] = [p, γ−1s3 (p)] where γs3(z) = −1/z. Hence γs3(z) = −1/z is a side-pairing transformation and pairs s3 with itself; note however that γs3 reverses the orientation of s3. γs3 s3 s1 s2 γs1 γs2 Figure 28.3: Side-pairing transformations for the modular group: s1 is paired with s2 and s3 is paired with itself. §28.4 Representing the side-pairing transformations in a diagram Often it is convenient to indicate which sides of D(p) are paired and how the side-pairing transformations act by recording the information in a diagram such as in Figure 28.4. Here, the sides with an equal number of arrowheads are paired and the pairing preserves s γs(s) γs Figure 28.4: The side s′ is mapped to the side s by γs. The sides with an equal number of arrowheads are paired. the direction of the arrows denoting the orientation of the sides. 3 MATH32051 28. Side-pairing transformations §28.4.1 Sides that are paired with themselves It can happen that a side-pairing transformation pairs a side with itself, albeit swapping the end-points over. We saw an example of this above with the modular group PSL(2,Z). In some occasions, it will be useful to assume that no side of D(p) is paired with itself. We can always assume that no side of D(p) is paired with itself by introducing additional vertices, if necessary. Suppose that s is a side with side-pairing transformation γs that pairs s with itself. Suppose that s has end-points v0, v1. Then γs(v0) = v1 and γs(v1) = v0. Introduce a new vertex v2 at the midpoint of s. Then γs(v2) = v2. 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 28.5. v0 v1 v2 s1 s2 γs Figure 28.5: The side s is paired with itself; by splitting it in half, we have two distinct sides that are paired. 4 MATH32051 29. Elliptic cycles: the construction 29. Elliptic cycles: the construction §29.1 Introduction Let Γ be a Fuchsian group and suppose that Γ has a Dirichlet polygon D(p) with no vertices on the boundary. We can equip D(p) with a set of side-pairing transformations. In this section and the next we use the side-pairing transformations to construct elliptic cycles and elliptic cycle transformations, their orders and angle sums. These will prove useful in linking the algebra of the group Γ with the geometry of the polygon D(p). §29.2 Elliptic cycles From now on we shall switch automatically between the upper half-planeH and the Poincare´ disc, depending on which model is most appropriate to use in a given context. Often we shall refer to H, but draw pictures in D; we can do this as the map constructed in Section 12 allows us to switch between the two at will. Let Γ be a Fuchsian group and let D = D(p) be a Dirichlet polygon for Γ. We assume that all the vertices of D(p) lie inside H (later we shall see many examples where this happens). In Section 28 we saw how to associate to each side s of D(p) a side-pairing transformation γs ∈ Γ \ {id} that pairs s with another side γs(s) of D(p). Recall that we indicate which sides of D are paired and how the side-pairing transfor- mations act by recording the information in a diagram as follows: the sides with an equal number of arrowheads are paired and the pairing preserves the direction of the arrows denoting the orientation of the sides. Notice that each vertex v of D is mapped to another vertex of D under a side-pairing transformation associated to a side with end point at v. Each vertex v of D has two sides s and ∗s of D with end points at v. Let the pair (v, s) denote a vertex v of D and a side s of D with an endpoint at v. We denote by ∗(v, s) the pair comprising of the vertex v and the other side ∗s that ends at v. Consider the following procedure: (i) Let v = v0 be a vertex of D and let s0 be a side with an endpoint at v0. Let γ1 be the side-pairing transformation associated to the side s0. Thus γ1 maps s0 to another side s1 of D. (ii) Let s1 = γ1(s0) and let v1 = γ1(v0). This gives a new pair (v1, s1). (iii) Now consider the pair ∗(v1, s1). This is the pair consisting of the vertex v1 and the side ∗s1 (i.e. the side of D other than s1 with an endpoint at v1). (iv) Let γ2 be the side-pairing transformation associated 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. See Figure 29.1. 1 MATH32051 29. Elliptic cycles: the construction D v0 v1v2 s0 s1 ∗s1 s2 ∗s2 γ1 γ2 γ3 Figure 29.1: The pair (v0, s0) is mapped to (v1, s1), which is mapped to (v1, ∗s1), which is mapped to (v2, s2), etc. 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 ) ∗→ · · · . As there are only finitely many pairs (v, s), this process of applying a side-pairing transfor- mation 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. 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. Example. Consider the polygon in Figure 29.2. Notice that we can label the side-pairing transformations in any way we choose. Thus in Figure 29.2 the map γ2 is an isometry that maps the side s6 = AF to the side s4 = DE. Notice the orientation: γ2 maps the vertex A to the vertex D, and the vertex F to the vertex E. We follow the procedure described above: ( A s1 ) γ1→ ( F s5 ) ∗→ ( F s6 ) γ2→ ( E s4 ) ∗→ ( E s5 ) 2 MATH32051 29. Elliptic cycles: the construction DEF A s1 B s2 C s3 s4 s5 s6 γ1 γ2 γ3 Figure 29.2: An example of a polygon with sides, vertices and side-pairing transformations labelled. γ−1 1→ ( B s1 ) ∗→ ( B s2 ) γ−1 3→ ( D s3 ) ∗→ ( D s4 ) γ−1 2→ ( A s6 ) ∗→ ( A s1 ) . Thus we have the elliptic cycle A → F → E → B → D with associated elliptic cycle transformation γ−12 γ −1 3 γ −1 1 γ2γ1. However, the vertex C is not part of this elliptic cycle and so must be part of another elliptic cycle:( C s3 ) γ3→ ( C s2 ) ∗→ ( C s3 ) . Thus we have another elliptic cycle C with associated elliptic cycle transformation γ3. Definition. Let v be a vertex of the hyperbolic polygon D. We denote the elliptic cycle transformation associated to the vertex v and the side s by γv,s. Remarks. 1. Suppose instead that we had started at (v, ∗s) instead of (v, s). The procedure above gives an elliptic cycle transformation γv,∗s. One can easily check that γv,∗s = γ−1v,s . 2. Suppose instead that we had started at (vi, ∗si) instead of (v0, s0). Then we would have obtained the elliptic cycle transformation γvi,∗si = γiγi−1 · · · γ1γn · · · γi+2γi+1, i.e. a cyclic permutation of the maps involved in defining the elliptic cycle transfor- mation 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 vertex of D with associated elliptic cycle transformation γv,s. Then γv,s is a Mo¨bius transformation fixing the vertex v. In Section 16 we saw that if a Mo¨bius transformation has a fixed point in H then it must be either elliptic or the identity. Thus each elliptic cycle transformation is either an elliptic Mo¨bius transformation or the identity. 3 MATH32051 29. Elliptic cycles: the construction Definition. If an elliptic cycle transformation is the identity then we call the elliptic cycle an accidental cycle. 4 MATH32051 30. Elliptic cycles: angle sums and orders 30. Elliptic cycles: angle sums and orders §30.1 Introduction Let Γ be a Fuchsian group and suppose that Γ has a Dirichlet polygon D(p) with no vertices on the boundary. We can equip D(p) with a set of side-pairing transformations. In the previous section we constructed elliptic cycles and elliptic cycle transformations. In this section we define the order of an elliptic cycle and its angle sum. §30.2 The order of an elliptic cycle Definition. Let γ ∈ Mo¨b(H) or Mo¨b(D) be a Mo¨bius transformation. We say that γ has finite order if there exists an integer m > 0 such that γm = id. We call the least such integer m > 0 the order of γ. Example. Working in D, the rotation γ(z) = e2πiθz has finite order if and only if θ is a rational. Indeed, if θ = p/q where p, q ∈ Z, q 6= 0 and p, q are coprime, then γ has order q. More generally, if γ is conjugate to a rotation through a rational multiple of 2π then γ has finite order. Indeed, this is the only way in which elements of finite order can arise, Thus, if γ has finite order then γ must be elliptic. For elements of a Fuchsian group, the converse is also true: elliptic elements must also be of finite order (and therefore conjugate to a rotation through a rational multiple of 2π). Proposition 30.2.1 Let Γ be a Fuchsian group and let γ ∈ Γ be an elliptic element. Then there exists an integer m ≥ 1 such that γm = id. Proof (sketch). Recall that an elliptic Mo¨bius transformation γ is conjugate to a ro- tation, say through angle 2πθ where θ ∈ [0, 1]. Consider the elements γn ∈ Γ; these are conjugate to rotations through angle 2πnθ mod 1 (that is, we take nθ and ignore the integer part). The proposition 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 collection of all finite words (subject to the condition that symbol a never follows or is followed by a−1), to be the free group on k generators. Let us check that this is a group where the group operation is concatenation of words. (i) The group operation is well-defined: as we saw above, the concatenation of two words is another word. (ii) Concatenation is associative (intuitively this is clear, but it is suprisingly difficult to prove rigorously). (iv) Existence of an identity: the empty word e (the word consisting of no symbols) is the identity element; if w = a1 · · · an ∈ Fn then we = ew = w. (iii) Existence of inverses: if w = a1 · · · an is a word then the word w−1 = a−1n · · · a−11 is such that ww−1 = w−1w = e. §32.4 Generators and relations We call Fk a free group because the group product is free: there is no cancellation between any of the symbols (other than the necessary condition that aa−1 = a−1a = e for each symbol a ∈ S). We can obtain a wide range of groups by introducing relations. A relation is a word that we declare to be equal to the identity. When writing a group element as a concatenation of symbols then we are allowed to cancel any occurrences of the relations. Definition. Let S = {a1, . . . , ak} be a finite set of symbols and let w1, . . . , wm be a finite set of words. We define the group Γ = 〈a1, . . . , ak | w1 = . . . = wm = e〉 (4.1) to be the set of all words of symbols from S ∪ S−1, subject to the conditions that (i) any subwords of the form aa−1 or a−1a are deleted, and (ii) any occurrences of the subwords w1, . . . , wm are deleted. Thus any occurrences of the words w1, . . . , wm can be replaced by the empty word e, i.e. the group identity. We call the above group Γ the group with generators a1, . . . , ak and relations w1, . . . , wm. It is an important and interesting question to ask when a given group can be written in the above form. Recall the following definition. Definition. Let Γ1,Γ2 be two groups. We say that a map φ : Γ1 → Γ2 is an isomorphism if φ is a bijection and φ(γ1γ2) = φ(γ1)φ(γ2) for all γ1, γ2 ∈ Γ1. We say that Γ1,Γ2 are isomorphic. We say that a group Γ is finitely presented if it can be written in the form (4.1). Definition. We say that a group Γ is finitely presented if it is isomorphic to a group in the form (4.1), with finitely many generators and finitely many relations. We call an expression of the form (4.1) a presentation of Γ. 2 MATH32051 32. Presentations of groups Examples. (i) Trivially, the free group on k generators is finitely presented (there are no relations). (ii) Let ω = e2πi/p. The group Γ = {1, ω, ω2, . . . , ωp−1} of pth roots of unity is finitely presented. Using the group isomorphism ω 7→ a, we can write it in the form 〈a | ap = e〉. (iii) The (additive) group Z of integers is finitely presented. Indeed, it is the free group on 1 generator: 〈a〉 = {an | n ∈ Z}. Notice that an+m = anam, so that 〈a〉 is isomorphic to Z under the isomorphism an 7→ n. (iv) The (additive, abelian) group Z2 = {(n,m) | n,m ∈ Z} is finitely presented. This is because it is isomorphic to Γ = 〈a, b | a−1b−1ab = e〉. If we take a word in the free group 〈a, b〉 on 2 generators, then it will be of the form an1bm1an2 · · · anℓbmℓ . In particular, the free group 〈a, b〉 is not abelian because ab 6= ba. However, adding the relation a−1b−1ab allows us to make the group abelian. To see this, note that ba = bae = ba(a−1b−1ab) = b(aa−1)b−1ab = beb−1ab = bb−1ab = ab. Thus, in the group 〈a, b | a−1b−1ab = e〉 we can use the relation a−1b−1ab to write the word an1bm1an2 · · · anℓbmℓ . as an1+···+nkbm1+···+mk . Hence 〈a, b | a−1b−1ab = e〉 = {anbm | n,m ∈ Z} which, using the group isomorphism (n,m) 7→ anbm, is seen to be isomorphic to Z2. (v) Consider the group 〈a, b | a4 = b2 = (ab)2 = e〉. One can easily check that there are exactly 8 elements in this group, namely: e, a, a2, a3, b, ab, a2b, a3b. This is the symmetry group of a square (it is also called the dihedral group). The element a corresponds to an anti-clockwise rotation through a right-angle; the element b corresponds to reflection in a diagonal. 3 MATH32051 32. Presentations of groups (vi) We shall see in Section 36 that the group 〈a, b | a2 = (ab)3 = e〉 is isomorphic to the modular group PSL(2,Z). The symbol a corresponds to the Mo¨bius transformation z 7→ −1/z; the symbol b corresponds to the Mo¨bius transfor- mation z 7→ z + 1. In the above, the group in example (v) is finite and one can easily write down all of the elements in the group. However, the group 〈a, b | a5 = b3 = (ab)2 = e〉 has 60 elements and it is not at all easy to write down all 60 elements. More generally, there are many important and open questions about writing a group in terms of a finite set of generators and relations. For example: (i) Let Γ be a countable group (that is, a group with countably many elements). Is it possible to decide if Γ is finitely presented? (ii) Suppose that Γ is finitely presented. Is it possible to decide if Γ is a finite group? (iii) Given two sets of generators and relations, is it possible to decide if the two groups are isomorphic? (iv) Suppose that Γ is finitely presented and H is a subgroup of Γ. When is it true that H is finitely presented? Questions like these are typically extremely hard to answer and require techniques from logic and computer science to be able to answer; often the answer is that the problem is ‘undecidable’ ! 4 MATH32051 33. Poincare´’s Theorem: no boundary vertices 33. Poincare´’s Theorem: the case of no boundary vertices §33.1 Introduction In Sections 22–28 we started with a Fuchsian group and then constructed a Dirichlet polygon and a set of side-pairing transformations. In this section, we study the reverse. Namely, we start with a convex hyperbolic polygon and a set of side-pairing transformations and ask: when do these side-pairing transforma- tions generate a Fuchsian group? In general, the group generated will not be discrete and so will not be a Fuchsian group. However, under natural conditions, the group will be discrete. This is Poincare´’s theorem and it will allow us to construct a large number of new examples of Fuchsian groups. §33.2 Poincare´’s Theorem Let D be a convex hyperbolic polygon. In this section we shall assume that all the vertices of D are in H, that is, there are no vertices on the boundary ∂H. We assume that D is equipped with a set of side-pairing transformations. That is, to each side s, we have a Mo¨bius transformation γs associated to s such that γs(s) = s ′, another side of D. We will need the following technical hypothesis on the side-pairing transformations. Definition. We say that the side-pairing transformation γs satisfies the half-plane hypoth- esis if the following holds: locally, the half-plane bounded by s containing D is mapped by γs to the half-plane bounded by γs(s) but opposite D. In practice, this means that paired sides have the opposite orientation. For example, all the side-pairing transformations in Figure 33.1 satisfy the half-plane hypothesis. In particular, s γs(s) γs Figure 33.1: Sides with an equal number of arrow-heads have opposite orientations. Here all the side-pairing transformations satisfy the half-plane hypothesis. if the side-pairing trasnformation γs satisfies the half-plane hypothesis then it cannot be the identity. We follow the procedure in Section 29 to construct elliptic cycles, namely: 1 MATH32051 33. Poincare´’s Theorem: no boundary vertices (i) Let v = v0 be a vertex of D and let s0 be a side with an endpoint at v0. Let γ1 be the side-pairing transformation associated to the side s0. Thus γ1 maps s0 to another side s1 of D. (ii) Let s1 = γ1(s0) and let v1 = γ1(v0). This gives a new pair (v1, s1). (iii) Now consider the pair ∗(v1, s1). This is the pair consisting of the vertex v1 and the side ∗s1 (i.e. the side of D other than s1 with an endpoint of v1. (iv) Let γ2 be the side-pairing transformation associated 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 boundary 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 Dirichlet 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 v2,3 v1 w1 v2 w2 v3 w3 v4 w4 v1,1 v1,2 v1,3 v1,4 Figure 41.4: Illustrating N(t) in the case g = 2, r = 4 with m1,m2,m3 > 2 and m4 = 2. The solid dots indicate vertices of N(t) (note the degenerate triangle with vertex at w4). Consider the vertices vj, wj , vj+1 (1 ≤ j ≤ r). Pair the sides as illustrated in Figure 41.5 and call the side-pairing transformation γj. Note that γj is a rotation about wj through angle 2π/mj . For each ℓ = 1, 2, . . . , g, consider the vertices vℓ,1, vℓ,2, vℓ,3, vℓ,4. Pair the sides as illustrated in Figure 41.6 and call the side-pairing transformations γℓ,1, γℓ,2. We label the sides of N(t) by s(vj), s(vℓ,j), s(wj) where the side s(v) is immediately clockwise of vertex v. 3 MATH32051 41. Constructing a Fuchsian group of a given signature vj wj vj+1 γj 2π/nj s(vj) s(wj) Figure 41.5: Pairing the sides between vertices vj , wj , vj+1 (1 ≤ j ≤ r). vℓ,2 vℓ+1,1 vℓ,4 vℓ,3 vℓ,1 γℓ,1 γℓ,2 s(vℓ,4) s(vℓ,3) s(vℓ,2) s(vℓ,1) Figure 41.6: Pairing the sides between vertices vℓ,1, vℓ,2, vℓ,3, vℓ,4. We will now apply Poincare´’s Theorem to the polygon N(t). Our aim is to show how to choose t ∈ (0, 1) so that the side-pairing transformations above generate a Fuchsian group with the required signature. First we calculate the elliptic cycles. For each j = 1, . . . , r, consider the pair (wj , s(vj)). Then( wj s(vj) ) γj→ ( wj s(wj) ) ∗→ ( wj s(vj) ) . Hence we have an elliptic cycle wj with corresponding elliptic cycle transformation γj . The angle sum is given by the internal angle at wj , namely sum(wj) = 2π/mj . Hence mjsum(wj) = 2π so that the elliptic cycle condition holds. Consider the pair (vℓ,1, s(vℓ,1)). Using Figure 41.6, we see that we get the following segment of an elliptic cycle: · · · → vℓ,1 → vℓ,4 → vℓ,3 → vℓ,2 → vℓ+1,1 → · · · with corresponding segment of elliptic cycle transformation · · · γ−1ℓ,2 γ−1ℓ,1 γℓ,2γℓ,1 · · · which we denote by [γℓ,2, γℓ,1]. (Here we use the notational convention that vg+1,1 = v1.) Now consider the pair (vj , s(vj)). The elliptic cycle through this pair contains the following: · · · → ( vj s(vj) ) γj→ ( vj+1 s(wj) ) ∗→ ( vj+1 s(vj+1) ) → · · · 4 MATH32051 41. Constructing a Fuchsian group of a given signature Thus, starting at the pair (v1,1, s(v1,1)), we have the elliptic cycle E v1,1 → v1,4 → v1,3 → v1,2 → v2,1 → · · · → vg−1,2 → vg,1 → vg,4 → vg,3 → vg,2 → v1 → v2 → · · · → vr with corresponding elliptic cycle transformation γrγr−1 · · · γ1[γg,2, γg,1] · · · [γ1,2, γ1,1]. Let 2α(t) denote the internal angle of each vertex in the polygonM(t). Let βj(t) denote the internal angle at each vertex at the base of the jth isosceles triangle that is added to the polygon M(t) to form the polygon N(t), that is βj(t) is the angle ∠wjvjvj+1, and is also the angle ∠wjvj+1vj . See Figure 41.7. Then the angle sum along the elliptic cycle E is given by sum(E) = 8gα(t) + 2 r∑ j=1 (α(t) + βj(t)). α α α α β β Figure 41.7: Labelling the angles α(t), βj(t) in the polygon N(t). We show that t (and hence the polygon N(t)) can be chosen so that sum(E) = 2π. Then the elliptic cycle condition holds, E is an accidental cycle, and we can apply Poincare´’s Theorem. One can prove the following: lim t→1 α(t) = 0, lim t→1 βj(t) = 0, lim t→0 α(t) = π 2 − 1 2 2π 4g + r , lim t→0 βj(t) = π 2 − π mj To see this, notice that as t → 1 the vertices approach the boundary. Hence the internal angles converge to 0. 5 MATH32051 41. Constructing a Fuchsian group of a given signature Consider the hyperbolic triangle with a vertex at 0 and two other vertices at two con- secutive vertices on M(t). The internal angles are 2π/(4g + r), α(t) and α(t). By the Gauss-Bonnet Theorem the hyperbolic area of this triangle is π − ( 2π 4g + r + 2α(t) ) . As t→ 0, this triangle tends to a single point. Hence lim t→0 π − ( 2π 4g + r + 2α(t) ) = 0. Re-arranging this shows that lim t→0 α(t) = π 2 − 1 2 2π 4g + r . The argument to calculate limt→0 βj(t) is similar, using the triangle with vertices v1, v2, w1. Now lim t→1 8gα(t) + 2 r∑ j=1 (α(t) + βj(t)) = 0 and (after some rearrangement) lim t→0 8gα(t) + 2 r∑ j=1 (α(t) + βj(t)) = 2π (2g − 2) + r∑ j=1 ( 1− 1 mj )+ 2π. (2.2) The term inside the large brackets on the right-hand side of (2.2) is positive by the assump- tions of the theorem. Hence lim t→0 8gα(t) + 2 r∑ j=1 (α(t) + βj(t)) > 2π. As the quantities α(t) and βj(t) vary continuously in t, by the Intermediate Value Theorem there exists t0 ∈ (0, 1) such that sum(E) = 2π. Hence, for the polygon N(t0), the elliptic cycle condition holds. By Poincare´’s Theorem, the side-pairing transformations generate a Fuchsian group Γ. It remains to check that the group Γ has the required signature. The group Γ has r elliptic cycles corresponding to each of the wj. The elliptic cycle transformation associated to the elliptic cycle wj has order mj. Consider the space H/Γ. This is formed by taking N(t0) and gluing together the paired sides. Thus H/Γ has a triangulation using a single polygon (so F = 1) with V = r + 1 vertices (as there are r+1 elliptic cycles) and E = 2g+ r edges. Let h denote the genus of H/Γ. Then by the Euler formula, 2− 2h = V − E + F = (r + 1)− (2g + r) + 1 = 2− 2g. Hence h = g. Hence Γ has signature sig(Γ) = (g;m1, . . . ,mr). ✷ 6 MATH32051 42. Where we could go next 42. Where we could go next §42.1 Introduction In this section we briefly indicate some of the further directions one could take in studying hyperbolic geometry. §42.2 Compact surfaces In Section 41 we saw how, given g ≥ 2, we could construct a Fuchsian group Γ such that H/Γ is a torus of genus g (i.e. we constructed a Fuchsian group with signature (g,−)). Thus we can generate a large number of surfaces using hyperbolic geometry. The following two theorems say that, in some sense, most surfaces arise from hyperbolic geometry. Below, you may think of ‘compact’ as meaning ‘closed and bounded’. Moreover, the boundary of the surface can be thought of as its ‘edge’; thus the cylinder [0, 1]×S1 has a boundary (the two circles at the ends), whereas a torus does not have a boundary. Theorem 42.2.1 (Mo¨bius Classification Theorem (1863)) Let S be a compact orientable surface and suppose that S does not have a boundary. Then S is either: (i) a sphere, (ii) a torus of genus 1, or (iii) a torus of genus g, g ≥ 2. In particular, all but two compact orientable surfaces without boundary arise from hyper- bolic geometry. You may have met curvature in other courses, such as Differential Geometry. Curvature measures the extent to which space is curved, and in which direction it is curved. The following, known either as Diguet’s formula or the Bertrand-Diguet-Puiseux Theorem, gives a formula for the curvature of a surface at a point. We define κ(x) = lim r→0 12 π ( πr2 −Area(B(x, r)) r4 ) where B(x, r) = {y ∈ S | d(x, y) < r} denotes a ball of radius r in S. Then one can see that (with the usual notion of distance) a sphere has curvature +1, a torus of genus 1 has curvature 0, etc. The following theorem gives a more precise description of all surfaces with constant curvature. Theorem 42.2.2 (Poincare´-Koebe Uniformisation Theorem (1882, 1907)) Let S be a compact orientable surface of constant curvature and without boundary. Then there exists a covering space M with a suitable distance function and a discrete group of isometries Γ of M such that S is homeomorphic to M/Γ. Moreover, 1 MATH32051 42. Where we could go next (i) if S has positive curvature then M is a sphere, (ii) if S has zero curvature then M is the plane R2, (iii) if S has negative curvature then M is the hyperbolic plane H. The generalisation of this result to studying 3-dimensional ‘surfaces’ is called the Thurston Uniformisation Conjecture. It is an important open problem in mathematics that is a topic of major current research interest. §42.3 Higher-dimensional hyperbolic space Throughout this course we have studied the hyperbolic plane. Thus we have studied two- dimensional hyperbolic geometry. We could go on to study higher-dimensional hyperbolic geometry. We can define n-dimensional hyperbolic space as follows. Let Hn = {(x1, x2, . . . , xn) ∈ Rn | xn > 0} and let ∂Hn = {(x1, x2, . . . , xn−1, 0) ∈ Rn | x1, . . . , xn−1 ∈ R} ∪ {∞}. Thus Hn is an n-dimensional hyperbolic analogue of the upper half-plane H and ∂Hn is an n-dimensional analogue of the boundary of H. We can again define distance in Hn by first defining the length of a (piecewise con- tinuously differentiable) path, and then defining the distance between two points as the infimum of the length of all (piecewise continuously differentiable) paths between them. If σ = (σ1, . . . , σn) : [a, b]→ Hn is a piecewise continuously differentiable path then we define lengthHn(σ) = ∫ b a ‖σ′(t)‖ σn(t) dt where ‖σ′(t)‖ =√σ′1(t)2 + · · · + σ′n(t)2. Then for z, w ∈ Hn define dHn(z, w) = inf{lengthHn(σ) | σ is a piecewise continuously differentiable path from z to w}. We can then go on to study and classify the higher-dimensional Mo¨bius transformations. We can study discrete subgroups of these groups, and formulate a version of Poincare´’s Theorem. We could also go on to study higher dimensional hyperbolic ‘surfaces’ by taking Hn and quotienting it by a discrete group. This gives us an extremely powerful method of constructing a very large class of geometric spaces with many interesting properties, many of which are still topics of current research. 2 MATH32051 43. What do you need to know for this course? 43. What do you need to know for this course? §43.1 Introduction These notes contain some information about the assessment in January and what is and is not examinable. §43.2 How to pass Hyperbolic Geometry What is the final assessment for this course unit? The final assessment for this course is divided into two components: • A timed online test with a time limit of one hour. Once the test is started you must complete it in one sitting. (This will be similar in style to the coursework test.) • A written assessment, to be scanned and submitted online. The advised time to spend on the written component is 1 hour. (This will be similar in style to parts of a traditional exam paper.) Thus the total amount of time to be spent on the assessment is 2 hours. How much is the assessment worth? In total, the final assessment counts for 80% of your final grade in this course (the coursework test counts for 20%). Each component above counts for 40%. What will be tested? The course will be assessed against the Intended Learning Out- comes (ILOs) of course unit. These state that after successful completion of this course you will be able to: ILO1 calculate the hyperbolic distance between and the geodesic through points in the hyperbolic plane, ILO2 compare different models (the upper half-plane model and the Poincare´ disc model) of hyperbolic geometry, ILO3 prove results (Gauss-Bonnet Theorem, angle formulæ for triangles, etc as listed in the syllabus) in hyperbolic trigonometry and use them to calcu- late angles, side lengths, hyperbolic areas, etc, of hyperbolic triangles and polygons, ILO4 classify Mo¨bius transformations in terms of their actions on the hyperbolic plane, ILO5 calculate a fundamental domain and a set of side-pairing transformations for a given Fuchsian group, ILO6 define a finitely presented group in terms of generators and relations, 1 MATH32051 43. What do you need to know for this course? ILO7 use Poincare´’s Theorem to construct examples of Fuchsian groups and calculate presentations in terms of generators and relations, ILO8 relate the signature of a Fuchsian group to the algebraic and geometric properties of the Fuchsian group and to the geometry of the corresponding hyperbolic surface. The weekly learning plans state explicitly how each week’s work is related to the ILOs. When will the exam take place? The exam will take place during the January exam period (18th January – 29th January). The exact date of the examination will be commu- nicated to you separately. What can I use in the assessment? The examination is open book. You may look at the notes for the course, any notes that you’ve made, exercise sheets, solutions etc, during the exam. You can also use a calculator if you wish. What can’t I use in the assessment? Work submitted for assessment MUST be your own. You MUST NOT discuss the problems with others during the assessment window. Can I ask questions during the exam? During the exam it is not permitted to ask questions about how to interpret or solve any of the problems. However, if you suspect that there may be an error on an examination paper, you can query this. The mechanism for doing so will be explained to you in advance of the exam. It is important to remember that I will not be able to respond to your emails during the exam. What should I do if I have technical issues in the exam? We understand that it is frustrating when things do not work as they should. If something goes wrong, please alert us. You should email: [email protected] Please remember that this mail box is monitored by human beings; you should not expect to receive an instant reply. When will I get my results? You will get your results during Semester 2, once we have marked and moderated the exams. In particular, you will not get the mark for the online component immediately or shortly after you have taken the online test. Is there a mock assessment? Yes. I will provide a copy of the last three years’ closed book exams. I will also indicate how this may have looked had it been given as an open book assessment. How can I prepare for the exam? In addition to previous year’s exam papers, you have lots of exercises and quizzes to practice with. What do I need to know? Anything we covered in the videos or in the review sessions should be considered to be examinable unless I explicitly stated otherwise. The exceptions are: • Anything in the detailed pdf notes that isn’t covered in the videos. This is anything that appears in the detailed notes as an appendix. 2 MATH32051 43. What do you need to know for this course? • The proof covered in Part 2 of the video in Section 41. • Anything in Section 42. Do I need to memorise the proofs from the course material? No. You should aim to read and understand any proofs that we have done in the course, but I will not ask you to reproduce any of these proofs in the exam. The reason for covering proofs in the course is that they show you tricks and techniques that are commonly used for proving results in hyperbolic geometry, and some of these may help you to answer some of the questions on the exam. Can you provide more resources? I doubt that you need any further resources. You have a large bank of quiz questions, exercise sheets and tutorial problems. If, for example, this is a request for solutions to past exam papers, note that you have the feedback docu- ments for the last three years of exams. These include a description of common mistakes that students made and, often, solutions to the questions. Will you be available to answer my questions over the Christmas break? The University is closed between 19th December and 3rd January. I will be able to respond to email (or the Discussion Forum on Blackboard or Discord), but please be mindful that I also need a break and I may not be able to respond as quickly as you might like. I would also very strongly recommend that you take some time off from your studies and relax and refresh yourself after what has been an extremely challenging semester for all of us. §43.3 How to fail Hyperbolic Geometry Here are some of the most common misunderstandings and mistakes that occur almost every year. The section headings contain the most common mistakes students make. If you make these mistakes in the exam then you are throwing marks away—you have been warned! The upper half-plane is the right half-plane. No it isn’t! In Section 2 we defined the upper half-plane H to be the set H = {z ∈ C| Im(z) > 0}. It is the region of the complex plane that lies above the real axis. It is not the region of the complex plane that lies to the right of the imaginary axis. Draw a picture! Every circle in the complex plane has a centre on the real line and every straight line is vertical. This isn’t true! In Section 5 we saw that an arbitrary straight line or circle in C has an equation of the form αzz¯ + βz + β¯z¯ + γ = 0 (3.1) where α, γ ∈ R and β ∈ C. (Straight lines arise from the case α = 0, circles from α 6= 0.) Some (but clearly not all) straight lines are vertical, and some (but clearly not all) circles have real centres. Vertical straight lines and circles with real centres correspond to equations of the form (3.1) in the special case when β ∈ R. Note that when β ∈ R then β¯ = β. Hence they have equations of the form αzz¯ + βz + β¯z¯ + γ = 0 where α, β, γ ∈ R 3 MATH32051 43. What do you need to know for this course? Mo¨bius transformations are usually multiplied together. No they aren’t! Mo¨bius transformations are composed. Let γ1, γ2 be two Mo¨bius transformations. When we write γ1γ2, we always mean the composition γ1 ◦ γ2 (i.e. γ1γ2(z) = γ1(γ2(z))) and not the mul- tiplication of the two complex number γ1(z) × γ2(z). We saw in Section 6 that Mo¨bius transformations form a group under composition. For example, if γ1(z) = (2z + 1)/(z + 1) and γ2(z) = z + 3 then γ1γ2 denotes the transformation γ1γ2(z) = γ1(γ2(z)) = γ1(z + 3) = 2(z + 3) + 1 (z + 3) + 1 = 2z + 7 z + 4 . It does not denote the multiplication γ1(z)γ2(z) = 2z + 1 z + 1 × (z + 3) = 2z 2 + 7z + 3 z + 1 , which isn’t even a Mo¨bius transformation. In short: if you get a z2 somewhere then you’ve multiplied and not composed. Note also that composition is not necessarily commutative. In the above example we saw that γ1γ2(z) = (2z + 7)/(z + 4). However γ2γ1(z) = γ2 ( 2z + 1 z + 1 ) = 2z + 1 z + 1 + 3 = 5z + 4 z + 1 6= γ1γ2(z). There’s no need to check that a Mo¨bius transformation is a Mo¨bius transforma- tion. Yes there is! A Mo¨bius transformation of H is a map of the form z 7→ (az+b)/(cz+d) where ad− bc > 0. You do need to check that ad− bc > 0. There are two places where this is particularly important: (i) When checking that Mo¨bius transformations of H form a group, you need to check that the inverse of a Mo¨bius transformation of H is a Mo¨bius transformation of H and the composition of two Mo¨bius of H transformations is a Mo¨bius transformation of H. (ii) When moving an arbitrary geodesic to the imaginary axis, you need to check that the transformation used is a Mo¨bius transformation of H. (See Lemma 9.2.1 in the notes.) With regard to (ii), suppose that we want to move the geodesic with end- points α < β ∈ ∂H\{∞} to the imaginary axis. The end-points of the imaginary axis are 0,∞, so we want to move α to either 0 or ∞ and β to either ∞ or 0. That is, we need a transformation of the following form: z 7→ z − α z − β , z 7→ z − β z − α. Note that if we calculate ’ad− bc’ for these two transformations we obtain −β + α for the first transformation and −α+ β for the second. As α < β we have −α+ β > 0 and so only the second transformation is a Mo¨bius transformation of H. 4 MATH32051 43. What do you need to know for this course? There is no point in normalising Mo¨bius transformations. Yes there is! Recall from Section 17 that a Mo¨bius transformation γ(z) = (az+ b)/(cz+d) of H is normalised if ad− bc = 1. We can always normalise a Mo¨bius transformation by dividing the coefficients by √ ad− bc. The trace τ(γ) of a Mo¨bius transformation γ of H is defined to be (a + d)2 where γ(z) = (az + b)/(cz + d) is in normalised form. We saw that γ is elliptic, parabolic or hyperbolic according to whether τ(γ) ∈ [0, 4), τ(γ) = 4, or τ(γ) > 4, respectively. If we do not normalise, then this classification does not work. For example, consider γ(z) = (z − 1)/(4z + 5). Here ad− bc = 1× 5− (−1)× 4 = 9, so this isn’t normalised. Dividing by √ 9 = 3 we can write γ(z) in normalised form as γ(z) = 1 3z − 13 4 3z + 5 3 . (Note that this is now normalised: ad − bc = 13 × 53 − −13 × 43 = 99 − 1.). This has trace τ(γ) = (1/3+5/3)2 = 4 and so is parabolic. (This can be checked directly by showing that γ has a unique fixed point at z = −1/2 ∈ ∂H.) However, if we hadn’t normalised γ then working out (a+ d)2 would give us (1 + 5)2 = 36 and we would have incorrectly concluded that γ was hyperbolic. Hyperbolic, parabolic and elliptic Mo¨bius transformations are easily confused. No they aren’t! Learn the following table. No. of No. of Trace Conjugate to fixed points fixed points in H/D in H/D hyperbolic 0 2 τ(γ) > 4 a dilation z 7→ kz, k 6= 1 parabolic 0 1 τ(γ) = 4 a translation z 7→ z + b, b ∈ R elliptic 1 0 τ(γ) ∈ [0, 4) a rotation Two MSc students once suggested the following mnemonic: • The number of fixed points in H/D is the number of ‘t’s in the words hyperbolic, parabolic, elliptic, respectively. • The number of fixed points in ∂H/∂D is the number of ‘y’s plus the number of ‘bolic’s in the words hyperbolic, parabolic, elliptic, respectively. • The trace corresponds the number of letters—either more than, or equal, to 4— before ‘bolic’ in hyperbolic, parabolic (with elliptic transformations corresponding to the remaining possible values of the trace). (It’s probably easier to just learn the table above…) The terms fundamental domain, Dirichlet region and Dirichlet polygon are interchangeable. No they aren’t! Let Γ be a Fuchsian group. A fundamental domain F for Γ is (loosely speaking) an open subset of H whose images under Γ tile H. (In fact, in Section 22 we said that an open subset F is a fundamental domain for Γ if 5 MATH32051 43. What do you need to know for this course? (i) ⋃ γ∈Γ γ(cl(F )) = H, (ii) γ1(F ) ∩ γ2(F ) = ∅ for all γ1, γ2 ∈ Γ, γ1 6= γ2. A given Fuchsian group Γ may have lots of fundamental domains (see Section 22 for some examples). Let Γ be a Fuchsian group. It is not, in general, clear how to write down a fundamental domain for Γ. An algorithm that will generate a fundamental domain is given in Sections 24–26, and a fundamental domain generated in this way is called a Dirichlet region, usually denoted in the course by D(p). Now D(p) is defined to be ⋂ γ∈Γ\{id} Hp(γ), i.e. as an intersection of possibly infinitely many half-planes. In the vast majority (but not all—although it is beyond the scope of the course to give an example) of cases, this intersection is actually the intersection of just finitely many half-planes. (See the worked examples in Section 26.) In this case, D(p) is a hyperbolic polygon and we call it a Dirichlet polygon. ’If and only if ’ statements only need proving in one direction. No, they need proving in both directions. For example, recall that a Mo¨bius transformation is hyperbolic if it has two fixed points on the boundary. If you’re asked to discuss the proof that a Mo¨bius transformation γ is hyperbolic if, and only if, it is conjugate to a dilation then this means you have to discuss two things: (i) if γ is hyperbolic then it is conjugate to a dilation, and (ii) if γ is conjugate to a dilation then it is hyperbolic. It is not enough just to do one implication! Poincare´’s Theorem is impossible to remember. No it isn’t, although I admit that the statement of Poincare´’s Theorem is rather long and complicated. The statement breaks down into 3 section: (i) defining notation, (ii) the hypotheses, (iii) the conclusions. Also, remember to check whether you’re working in the case of no boundary vertices (Section 33), or boundary vertices but no free edges (Section 35). The part that most people forget is the hypothesis that no side of the hyperbolic polygon is paired with itself. This is important because if this assumption is omitted then it’s possible to miss out one of the relations in the group. For example, in Section 36 when we illustrate Poincare´’s Theorem in the case of the modular group, if we hadn’t introduced the extra vertex at C = i then we would not have obtained the relation b2 = e. The remaining hypotheses are that the Elliptic Cycle Condition holds and, in the case of boundary vertices (but no free edges), the Parabolic Cycle Condition holds. There are essentially three points to remember in the conclusions: (i) the side-pairing transformations generate a Fuchsian group, (ii) the polygon is a fundamental domain, and (iii) the elliptic cycles can be used to give a presentation of the group in terms of generators and relations. 6 MATH32051 43. What do you need to know for this course? It’s possible to do well in the hyperbolic geometry exam without engaging with course material (the videos, review sessions, tutorials, etc) and without doing any of the exercise. Sadly not 😉 7 MATH32051 Exercises for Week 1 Exercises for Week 1 Exercises are numbered n.m. You can do Exercise n.m after watching the video/studying the notes for Section n. Exercises marked with a flat sign (♭) are there for completeness. If you are pushed for time then you can regard them as optional. Exercise 2.1 Let Rθ denote the 2 × 2 matrix that rotates R2 clockwise about the origin through angle θ ∈ [0, 2π). Then Rθ has matrix ( cos θ sin θ − sin θ cos θ ) . Let a = (a1, a2) ∈ R2. Define the transformation Tθ,a : R 2 → R2 by Tθ,a ( x y ) = ( cos θ sin θ − sin θ cos θ )( x y ) + ( a1 a2 ) . Thus Tθ,a first rotates the point (x, y) about the origin through an angle θ and then trans- lates by the vector a. Let G = {Tθ,a | θ ∈ [0, 2π), a ∈ R2}. (i) Let θ, φ ∈ [0, 2π) and let a, b ∈ R2. Find an expression (in the form Tψ,c) for the composition Tθ,a ◦ Tφ,b. Hence show that G is a group under composition of maps (i.e. show that this product is (a) well-defined (i.e. the composition of two elements of G gives another element of G), (b) associative (hint: don’t do unnecessary work: you already know that composition of functions is associative), (c) that there is an identity element, and (d) that inverses exist). (ii) Show that the set of all rotations about the origin is a subgroup of G. (iii) Show that the set of all translations is a subgroup of G. (One can show that G is actually the group Isom+(R2) of orientation preserving isometries of R2 with the Euclidean matrices.) Exercise 3.1 Let γ(z) = (3z + 1)/(z + 2). Show that γ maps H to H. Exercise 4.1 Consider the points i and ai where 0 < a < 1. (i) Consider the path σ between i and ai that consists of the arc of imaginary axis between them. Find a parametrisation of this path. 1 MATH32051 Exercises for Week 1 (ii) Show that lengthH(σ) = log 1/a. (Notice that as a → 0, we have that log 1/a → ∞. Thus, as ai approaches the real axis, the distance from i to ai tends to infinity. This motivates why we call R ∪ {∞} the circle at infinity.) Exercise 4.2 Show that dH satisfies the triangle inequality : dH(z1, z3) ≤ dH(z1, z2) + dH(z2, z3), ∀ z1, z2, z3 ∈ H. That is, the distance between two points is increased if one goes via a third point. Exercise 5.1 Let L be a straight line in C with equation αzz¯ + βz + β¯z¯ + γ = 0 where α, γ ∈ R and β ∈ C. Find a formula for its gradient and intersections with the real and imaginary axes in terms of α, β, γ. Exercise 5.2 Let C be a circle in C with equation αzz¯ + βz + β¯z¯ + γ = 0 where α, γ ∈ R and β ∈ C. Find a formula for the centre and radius of C in terms of α, β, γ. 2 MATH32051 Exercises for Week 2 Exercises for Week 2 Exercise 6.1 Let γ be a Mo¨bius transformation of H. Show that γ is a well-defined map γ : H→ H (that is, if z ∈ H then γ(z) ∈ H). Show that γ maps H to itself bijectively and give an explicit expression for the inverse map. (This proves Proposition 6.2.1 in the notes.) Exercise 6.2 Prove the Mo¨b(H) forms a group under composition. (To do this, you must: (i) show that the composition γ1γ2 of two Mo¨bius transformations of H is a Mo¨bius transformation of H, (ii) check associativity (hint: don’t do unnecessary work: you already know that composition of maps is associative), (iii) show that the identity map z 7→ z is a Mo¨bius transformation of H, and (iv) show that if γ ∈ Mo¨b(H) is a Mo¨bius transformation, then γ−1 exists and is a Mo¨bius transformation of H.) (This proves Proposition 6.2.2 in the notes.) Exercise 6.3 Let A be either an arc of circle (not necessarily with a real centre) in H or part of a straight line (not necessarily vertical) in H. Let γ ∈ Mo¨b(H). Show that γ(A) is also either an arc of circle in H or part of a straight line in H. Exercise 7.1 Show that if ad− bc 6= 0 then γ maps ∂H to itself bijectively. (Note that we only need the weaker condition that ad− bc 6= 0 rather than ad− bc > 0 here.) Exercise 7.2♭ Let γ(z) = (az + b)/(cz + d) ∈ Mo¨b(H). Show that: |γ′(z)| = ad− bc|cz + d|2 , Im(γ(z)) = (ad− bc) |cz + d|2 Im(z). Exercise 7.3 Let z = x+ iy ∈ H and define γ(z) = −x+ iy. (Note that γ is not a Mo¨bius transformation of H.) (i) Show that γ maps H to H bijectively. (ii) Let σ : [a, b]→ H be a differentiable path. Show that lengthH(γ ◦ σ) = lengthH(σ). Hence conclude that γ is an isometry of H. 1 MATH32051 Exercises for Week 3 Exercises for Week 3 Exercise 9.1 Let H1,H2 ∈ H and let z1 ∈ H1, z2 ∈ H2. Show that there exists a Mo¨bius transformation γ ∈ Mo¨b(H) such that γ(H1) = H2 and γ(z1) = z2. In particular, conclude that given z1, z2 ∈ H, one can find a Mo¨bius transformation γ ∈ Mo¨b(H) such that γ(z1) = z2. (Don’t do unnecessary work! If your answer is more than 2 lines long then you’re probably over-thinking this!) Exercise 10.1 (i) 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 inter- section (see Figure 46). Show that cos θ = |c1 − c2|2 − (r21 + r22) 2r1r2 . θ Figure 46.1: The internal angle between two circles. (ii) Consider the geodesic between −6 and 6 and the geodesic between 4√2 and 6√2, as illustrated in Figure 46). Both of these geodesics are semi-circles. Find the centre and radius of each semi-circle. Hence use the result in (i) to calculate the angle φ. Exercise 10.2 Suppose that two geodesics intersect as illustrated in Figure 10.3. Show that sin θ = 2ab a2 + b2 , cos θ = b2 − a2 a2 + b2 . 1 MATH32051 Exercises for Week 3 φ −6 4√2 6 6√2 Figure 46.2: Two geodesics intersecting with angle φ. θ ib a Figure 46.3: The angle between two geodesics in the case where one is a vertical straight line. Exercise 10.3 Prove Proposition 11.2.2 using the following steps. For z, w ∈ H let LHS(z, w) = cosh dH(z, w) RHS(z, w) = 1 + |z − w|2 2 Im(z) Im(w) denote the left- and right-hand sides of (2.1) [the formula for cosh dH(z, w)] respectively. We want to show that LHS(z, w) = RHS(z, w) for all z, w ∈ H. (i) Let γ ∈ Mo¨b(H) be a Mo¨bius transformation of H. Using the fact that γ is an isometry, prove that LHS(γ(z), γ(w)) = LHS(z, w). Using Exercise 7.2 and Lemma 11.2.1, prove that RHS(γ(z), γ(w)) = RHS(z, w). (ii) Let H denote the geodesic passing through z, w. By Lemma 9.2.1 there exists a Mo¨bius transformation γ of H that maps H to the imaginary axis. Let γ(z) = ia and γ(w) = ib. Prove, using the fact that dH(ia, ib) = log b/a if a < b, that for this choice of γ we have LHS(γ(z), γ(w)) = RHS(γ(z), γ(w)). (iii) Conclude that LHS(z, w) = RHS(z, w) for all z, w ∈ H. 2 MATH32051 Exercises for Week 3 Exercise 11.2 A hyperbolic circle C with centre z0 ∈ H and radius r > 0 is defined to be the set of all points of hyperbolic distance r from z0. Show, using (2.1), show that a hyperbolic circle is a Euclidean circle (i.e. an ordinary circle) but with a different centre and radius. (Hint: substitute z0 = x0 + iy0, z = x + iy into (2.1), expand this out, and then complete the square.) Exercise 11.3 Recall that we defined the hyperbolic distance by first defining the hyperbolic length of a piecewise continuously differentiable path σ: lengthH(σ) = ∫ |σ′(t)| Im(σ(t)) dt = ∫ σ 1 Im(z) . (0.1) We then saw that the Mo¨bius transformations of H are isometries. Why did we choose the function 1/ Im z in (0.1)? In fact, one can choose any positive function and use it to define the length of a path, and hence the distance between two points. However, the geometry that one gets may be very complicated (for example, there may be many geodesics between two points); alternatively, the geometry may not be very interesting (for example, there may be very few symmetries, i.e. the group of isometries is very small). The group of Mo¨bius transformations of H is, as we shall see, a very rich group with lots of interesting structure. The point of this exercise is to show that if we want the Mo¨bius transformations of H to be isometries then we must define hyperbolic length by (0.1). Let ρ : H → R be a continuous positive function. Define the ρ-length of a path σ : [a, b]→ H to be lengthρ(σ) = ∫ σ ρ = ∫ b a ρ(σ(t))|σ′(t)| dt. (i) Suppose that lengthρ is invariant under Mo¨bius transformations of H, i.e. if γ ∈ Mo¨b(H) then lengthρ(γ ◦ σ) = lengthρ(σ). Prove that ρ(γ(z))|γ′(z)| = ρ(z). (0.2) (Hint: you may use the fact that if f is a continuous function such that ∫ σ f = 0 for every path σ then f = 0.) (ii) By taking γ(z) = z+ b in (0.2), deduce that ρ(z) depends only on the imaginary part of z. Hence we may write ρ as ρ(y) where z = x+ iy. (iii) By taking γ(z) = kz in (0.2), deduce that ρ(y) = c/y for some constant c > 0. Hence, up to a normalising constant c, we see that if we require the Mo¨bius transformations of H to be isometries, then the distance in H must be given by the formula we introduced in Section 4. 3 MATH32051 Exercises for Week 4 Exercises for Week 4 Exercise 12.1♭ Check some of the assertions that we made about the map h : H → D defined by h(z) = (z − i)/(iz − 1). For example: (i) Show that h maps H to D bijectively. Show that h maps ∂H to ∂D bijectively. (ii) Let g(z) = h−1(z). Find a formula1 for g(z) and show that g′(z) = −2 (−iz + 1)2 , Im(g(z)) = 1− |z|2 | − iz + 1|2 . Exercise 12.2 Mimic the proof of Proposition 8.1.1 to show that the real axis is the unique geodesic joining 0 to x ∈ (0, 1) and that dD(0, x) = log ( 1 + x 1− x ) . Exercise 12.3 Let h : H → D be given by h(z) = (z − i)/(iz − 1). Show 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. Exercise 12.4 Let C either be a straight line in C that passes through the origin or a circle that meets the unit circle in C at right angles. Show that C can be described by an equation of the form αzz¯ + βz + β¯z¯ + α = 0 where α ∈ R and β ∈ C. (Hint: use the map h.) Exercise 12.5 Let C = {w ∈ D | dD(z0, w) = r} be a hyperbolic circle in D with centre z0 and radius r > 0. Calculate the (hyperbolic) circumference and (hyperbolic) area of C. [Hints: First move C to the origin by using a Mo¨bius transformation of D. Use the formula dD(0, x) = log(1 + x)/(1 − x) to show that this is a Euclidean circle, but with a different radius. To calculate area, use polar co-ordinates.] Exercise 13.1 Consider the hyperbolic triangle in H with vertices at 0, (−1 + i√3)/2, (1 + i√3)/2 as illustrated in Figure 47.1. 1If you carefully compare the formulæ for g and for h then you might notice a similarity! However, remember that they are different functions: g maps D to H whereas h maps H to D. 1 MATH32051 Exercises for Week 4 θ1 θ2 −1+i √ 3 2 1+i √ 3 2 0 Figure 47.1: A hyperbolic triangle with vertices at 0, (−1 + i√3)/2, (1 + i√3)/2. (i) Determine the geodesics that comprise the sides of this triangle. (ii) Use Exercise 10.1 to calculate the internal angles of this triangle. Hence use the Gauss-Bonnet Theorem to calculate the hyperbolic area of this triangle. Exercise 13.2 Assuming Theorem 13.2.1 but not Theorem 13.2.2, prove that the area of a hyperbolic quadrilateral with internal angles α1, α2, α3, α4 is given by 2π − (α1 + α2 + α3 + α4). Exercise 14.1 Let n ≥ 3. By explicit construction, show that there exists a regular n-gon with internal angle equal to α if and only if α ∈ [0, (n − 2)π/n). (Hint: Work in the Poincare´ disc D. Let ω = e2πi/n be an nth root of unity. Fix r ∈ (0, 1) and consider the polygon D(r) with vertices at r, rω, rω2, . . . , rωn−1. This is a regular n-gon (why?). Let α(r) denote the internal angle of D(r). Use the Gauss-Bonnet Theorem to express the area of D(r) in terms of α(r). Examine what happens as r → 0 and as r→ 1. (To examine limr→0AreaHD(r), note that D(r) is contained in a hyperbolic circle C(r), and use Exercise 11.2 to calculate limr→0AreaHC(r).) You may use without proof the fact that α(r) depends continuously on r.) In particular, conclude that there there exists a regular n-gon with each internal angle equal to a right-angle whenever n ≥ 5. This is in contrast with the Euclidean case where, of course, the only regular polygon with each internal angle equal to a right-angle is the square. 2 MATH32051 Exercises for Week 5 Exercises for Week 5 Exercise 15.1 Let ∆ be a hyperbolic right-angled triangle with internal angles α, β, γ and side of hyper- bolic length a, b, c. Assuming that tanα = tanh a/ sinh b, prove that sinα = sinh a/ sinh c and cosα = tanh b/ tanh c. Exercise 15.2 We have relationships involving: (i) three angles (the Gauss-Bonnet Theorem), (ii) three sides (Pythagoras’ Theorem) and (iii) two sides, one angle (SOH CAH TOA). Prove the following relationships between one side and two angles: cosh a = cosα cosec β, cosh c = cotα cot β. What are the Euclidean analogues of these identities? Exercise 15.3 Let ∆ be a right-angled hyperbolic triangle with one ideal vertex, one side of finite hyper- bolic length a and internal angles 0, π/2, α. Assuming that sinα = 1/ cosh a, check using standard trigonometric and hyperbolic trigonometric identities that cosα = 1/ coth a and tanα = 1/ sinh a. Exercise 15.4 Let ∆ be a right-angled hyperbolic triangle with one ideal vertex and internal angles α, 0, π/2. Suppose that the side of ∆ of finite hyperbolic length has hyperbolic length log(2 + √ 3). Use the angle of parallelism formula and the Gauss-Bonnet Theorem to calculate AreaH(∆). Exercise 15.5 Prove the Hyperbolic Sine Rule in the case when ∆ is acute (the obtuse case is a simple modification of the argument, and is left for anybody interested…). (Hint: label the vertices A,B,C with angle α at vertex A, etc. Drop a perpendicular from vertex B meeting the side [A,C] at, say, D to obtain two right-angled triangles ABD, BCD. Use Pythagoras’ Theorem and Proposition 15.2.1 in both of these triangles to obtain an expression for sinα.) Exercise 16.1 Find the fixed points in H ∪ ∂H of the following Mo¨bius transformations of H: γ1(z) = 2z + 5 −3z − 1 , γ2(z) = 7z + 6, γ3(z) = − 1 z , γ4(z) = z z + 1 . In each case, state whether the map is parabolic, elliptic or hyperbolic. Exercise 17.1 Normalise the Mo¨bius transformations of H: γ1(z) = 2z + 5 −3z − 1 , γ2(z) = 7z + 6, γ3(z) = − 1 z , γ4(z) = z z + 1 . 1 MATH32051 Exercises for Week 5 In each case, calculate the trace and hence decide whether the map is parabolic, elliptic or hyperbolic. Exercise 18.1 (i) Prove that conjugacy between Mo¨bius transformations of H is an equivalence relation. (ii) Show that if γ1 and γ2 are conjugate then they have the same number of fixed points. Hence show that if γ1 is hyperbolic, parabolic or elliptic then γ2 is hyperbolic, parabolic or elliptic, respectively. Exercise 18.2 Let γ1 and γ2 be conjugate Mo¨bius transformations of H. Show that τ(γ1) = τ(γ2). (Hint: show that if A1, A2, A ∈ SL(2,R) are matrices such that A1 = A−1A2A then Trace(A1) = Trace(A −1A2A) = Trace(A2). You might first want to show that Trace(AB) = Trace(BA) for any two matrices A,B.) Exercise 18.3 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? Exercise 18.4 Show that 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. Exercise 18.5 Let γ ∈ Mo¨b(H) be a hyperbolic Mo¨bius transformation of H. By Proposition 18.4.2, we know that γ is conjugate to a dilation z 7→ kz. Find a relationship between τ(γ) and k. Exercise 18.6 Let γ ∈Mo¨b(D) be a elliptic Mo¨bius transformation of D. By Proposition 18.5.1, we know that γ is conjugate to a rotation z 7→ eiθz. Find a relationship between τ(γ) and θ. 2 MATH32051 Exercises for Week 6 Exercises for Week 6 Exercise 20.1 Let b ∈ R and let γb(z) = z + b. Then γb ∈ Mo¨b(H). What do you think happens to γb as b→ 0? Use the formula given in the course to check that your intuition is correct. Exercise 20.2 Let q ∈ N. Define Γq = { γ(z) = az + b cz + d | a, b, c, d ∈ Z, ad− bc = 1, b, c are divisible by q } . Show that Γq is indeed a subgroup of Mo¨b(H). Exercise 20.3 Fix k > 0, k 6= 1. Consider the subgroup of Mo¨b(H) generated by the Mo¨bius transforma- tions of H given by γ1(z) = z + 1, γ2(z) = kz. Is this a Fuchsian group? (Hint: consider γ−n2 γ m 1 γ n 2 (z).) Exercise 21.1 Let Γ = {id, γ1, γ2, γ3} ⊂ Mo¨b(D) where γ1, γ2, γ3 denote rotations through angles 90, 180 and 270 degrees, respectively. Let z ∈ D. Determine the orbit Γ(z). Exercise 23.1 Figures 49.1 and 49.2 illustrate two tessellations of H arising from fundamental domains of the group of integer translations and the group of dilations by powers of 2, respectively. What do these tessellations look like in the Poincare´ disc D? F Re(z)=−1 Re(z)=0 Re(z)=1 Re(z)=2 γ1(F ) γ2(F )γ−1(F ) Figure 49.1: A fundamental domain and tessellation for Γ = {γn | γn(z) = z + n}. 1 MATH32051 Exercises for Week 6 -4 -2 -1 0 1 2 4 γ1(F ) γ−1(F ) F Figure 49.2: A fundamental domain and tessellation for Γ = {γn | γn(z) = 2nz}. 2 MATH32051 Exercises for Week 7 Exercises for Week 7 Exercise 24.1♭ (Included for completeness only.) Show that a convex hyperbolic polygon actually is convex. To do this, first show that a half-plane is convex. Then show that the intersection of a finite number of convex sets is itself convex. Exercise 25.1 (i) Write z1 = x1 + iy1, z2 = x2 + iy2, z1, z2 ∈ H. Show that the perpendicular bisector of [z1, z2] can also be written as {z ∈ H | y2|z − z1|2 = y1|z − z2|2}. (ii) Use the result in (i) to show that the perpendicular bisector of the arc of geodesic between i and i+ 1 is the vertical straight line with real part equal to 1/2. (iii) Use the result in (i) to find the perpendicular bisector of the arc of geodesic between 1 + 2i and (6 + 8i)/5. Exercise 25.2 Does the result of Proposition 25.2.1 have an analogue in Euclidean geometry? If so, what is it and what does it say? Exercise 26.1 Let Γ = {γn | γn(z) = 2nz, n ∈ Z}. This is a Fuchsian group. Choose a suitable p ∈ H and construct a Dirichlet polygon D(p). 1 MATH32051 Exercises for Week 8 Exercises for Week 8 Exercise 27.1 Let p = ki where k > 1. Show that γ(p) 6= p for any γ ∈ PSL(2,Z) \ {id}. Exercise 28.1 Take Γ = {γn | γn(z) = 2nz, n ∈ Z}. Calculate the side-pairing transformations for the Dirichlet polygon calculated in Exercise 26.1. Exercise 30.1♭ (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. 1 MATH32051 Exercises for Week 9 Exercises for Week 9 Exercise 32.1 Let Γ = 〈a, b | a4 = b2 = (ab)2 = e〉. Show that Γ contains exactly 8 elements. Exercise 33.1 Take a hyperbolic quadrilateral such that each pair of opposing sides have the same length. Define two side-pairing transformation γ1, γ2 that pair each pair of opposite sides. See Figure 52.1. Show that there is only one elliptic cycle and determine the associated elliptic cycle transformation. When do γ1 and γ2 generate a Fuchsian group? γ2 γ1 Figure 52.1: A hyperbolic quadrilateral with opposite sides paired. 1 MATH32051 Exercises for Week 10 Exercises for Week 10 Exercise 35.1 Consider the polygon in Figure 53.1. The side-pairing transformations are: γ1(z) = z + 2, γ2(z) = z 2z + 1 . What are the elliptic cycles? What are the parabolic cycles? Use Poincare´’s Theorem to show that the Fuchsian group generated by γ1, γ2 is discrete and has the polygon in Figure 53.1 as a fundamental domain. Use Poincare´’s Theorem to show that the group generated by γ1, γ2 is the free group on 2 generators. γ2 γ1 −1 0 1 Figure 53.1: A fundamental domain for the free group on 2 generators. Exercise 35.2 Consider the hyperbolic quadrilateral with vertices A = − ( 1 + √ 2 2 ) , B = i √ 2 2 , C = ( 1 + √ 2 2 ) , and ∞ and a right-angle at B, as illustrated in Figure 53.2. (i) Verify that the following Mo¨bius transformations are side-pairing transformations: γ1(z) = z + 2 + √ 2, γ2(z) = √ 2 2 z − 12 z + √ 2 2 . (ii) By using Poincare´’s Theorem, show that these side-pairing transformations generate a Fuchsian group. Give a presentation of Γ in terms of generators and relations. 1 MATH32051 Exercises for Week 10 (1 + √ 2 2 )−(1 + √ 2 2 ) 0 γ1 γ2 i √ 2 2 Figure 53.2: A hyperbolic quadrilateral. Exercise 36.1 (i) Consider the regular hyperbolic decagon in Figure 53.3 below with each internal angle equal to π/9 and with the sides paired as illustrated (you may assume that such a hyperbolic decagon exists). Show that there are two elliptic cycles and determine their orders. (ii) By using Poincare´’s Theorem, show that the side pairing transformations generate a co-compact Fuchsian group Γ. Give a presentation of Γ in terms of generators and relations.) Figure 53.3: Each internal angle is π/9 and the sides are paired as indicated. 2 MATH32051 Exercises for Week 10 Exercise 38.1 Consider the regular dodecagon in Figure 53.3. In Exercise 36.1 we saw that the side- pairing transformations generate a Fuchsian group Γ. Determine the Euler characteristic and genus of the quotient surface H/Γ. How many marked points does H/Γ have and what are their orders? 3 MATH32051 Exercises for Week 11 Exercises for Week 11 Exercise 39.1 Consider the hyperbolic polygon illustrated in Figure 54.1 with the side-pairing transforma- tions as indicated (note that one side is paired with itself). Assume that θ1+ θ2 + θ3 = 2π (one can show that such a polygon exists). 2π/7 θ2 2π/3 θ1θ3 Figure 54.1: A hyperbolic polygon with sides paired as indicated. (i) Show that there are 3 non-accidental cycles and 1 accidental cycle. (ii) Show that the side-pairing transformations generate a Fuchsian group Γ and give a presentation of Γ in terms of generators and relations. (iii) Calculate the signature of Γ. Exercise 39.2 Consider the regular hyperbolic octagon with each internal angle equal to θ and the sides paired as indicated in Figure 54.2. Use Exercise 14.1 to show that such an octagon exists provided θ ∈ [0, 3π/4). For which values of θ do γ1, γ2, γ3, γ4 generate a Fuchsian group Γθ? In each case when Γθ is a Fuchsian group write down a presentation of Γθ, determine the signature sig(Γθ) and briefly describe geometrically the quotient space H/Γθ. Exercise 40.1 This exercise works through how one would define signature and its applications to cal- culating the hyperbolic area of a Fuchsian group in the case when we allow the Dirichlet polygon to have vertices on the boundary (but no free edges). 1 MATH32051 Exercises for Week 11 θ Figure 54.2: See Exercise 39.2. Let Γ be a Fuchsian group and let D be a Dirichlet polygon for D. We allow D to have vertices on ∂H, but we assume that D has no free edges (so that no arcs of ∂H are edges). We also assume that no side of D is paired with itself. The space H/Γ then has a genus (heuristically, the number of handles), possibly some marked points, and cusps. The cusps arise from gluing together the vertices on parabolic cycles and identifying the sides on each parabolic cycle. (i) Convince yourself that the H/PSL(2,Z) has genus 0, one marked point of order 3, one marked point of order 2, and one cusp. (Hint: remember that a side is not allowed to be paired to itself.) Suppose that H/Γ has genus g, r marked points of order m1, . . . ,mr, and c cusps. We define the signature of Γ to be sig(Γ) = (g;m1, . . . ,mr; c). (ii) Using the Gauss-Bonnet Theorem, show that AreaH(D) = 2π (2g − 2) + r∑ j=1 ( 1− 1 mj ) + c . (iii) Show that if c ≥ 1 then AreaH(D) ≥ π 3 and that this lower bound is achieved for just one Fuchsian group (which one?). Exercise 42.1♭ Recall that the Bertrand-Diguet-Puiseux Theorem gives a formula for the curvature of a surface at a point, namely κ(x) = lim r→0 12 π ( πr2 −Area(B(x, r)) r4 ) 2 MATH32051 Exercises for Week 11 where B(x, r) = {y ∈ S | d(x, y) < r}. Using the Euclidean metric, check that κ(x) = 0 for all x ∈ R2. Using the hyperbolic metric, check that κ(x) = −1 for all x ∈ H. 3 MATH32051 Solutions for Week 1 Solutions for Week 1 I will trust you to have made a serious attempt at the exercises before looking at the solutions. Solution 2.1 We write Tθ,a(x, y) in the form Rθ(x, y) + (a1, a2) where Rθ denotes the 2× 2 matrix that rotates the plane about the origin by angle θ. (i) (a) Let Tθ,a, Tθ′,a′ ∈ G. We have to show that the composition Tθ,aTθ′,a′ ∈ G. Now Tθ,aTθ′,a′(x, y) = Tθ,a(Tθ′,a′(x, y)) = Tθ,a(Rθ′(x, y) + (a ′ 1, a ′ 2)) = Rθ(Rθ′(x, y) + (a ′ 1, a ′ 2)) + (a1, a2) = RθRθ′(x, y) + (Rθ(a ′ 1, a ′ 2) + (a1, a2)) = Tθ+θ′,Rθ(a′1,a′2)+(a1,a2)(x, y) where we have used the fact that Rθ is linear (because it is a matrix, and ma- trices act linearly on vectors) and the observation that RθRθ′ = Rθ+θ′ (if you rotate around the origin by θ and then by θ′ then you have rotated by θ + θ′). As Tθ+θ′,Rθ(a′1,a′2)+(a1,a2) ∈ G, the composition of two elements of G is another element of G, hence the group operation is well-defined. (b) This is trivial: composition of functions is already known to be associative. (c) The identity map on R2 is the map that leaves every point alone. We choose θ = 0 and a = (0, 0). T0,(0,0)(x, y) = R0(x, y) + (0, 0). As R0 is the rotation through angle 0, it is clearly the identity matrix, so that R0(x, y) = (x, y). Hence T0,(0,0)(x, y) = (x, y). Hence G has an identity element. (d) Let Tθ,a ∈ G. We want to find an inverse for Tθ,a and show that it lies in G. Write Tθ,a(x, y) = (u, v). Then (u, v) = Rθ(x, y) + (a1, a2) and some re-arrangement, together with the fact that R−1θ = R−θ, shows that (x, y) = R−θ(u, v) −R−θ(a1, a2). Hence T−1θ,a = T−θ,−R−θ(a1,a2), which is an element of G. (ii) The rotations about the origin have the form Tθ,0. It is easy to check that Tθ,0Tθ′,0 = Tθ+θ′,0 so that the composition of two rotations is another rotation. The identity map is a rotation (through angle 0). The inverse of rotation by θ is rotation by −θ. Hence the set of rotations is a subgroup of G. 1 MATH32051 Solutions for Week 1 (iii) The translations have the form T0,a where a ∈ R2. It is easy to see that T0,aT0,a′ = T0,a+a′ so that the composition of two translations is another translation. The identity map is a translation (by (0, 0)). The inverse of translation by (a1, a2) is translation by (−a1,−a2). Hence the set of translations is a subgroup of G. Solution 3.1 Let γ(z) = (3z + 1)/(z + 2). Show that γ maps H to H. We must show that if z ∈ H then γ(z) ∈ H. Let z ∈ H and write z = x+ iy. Note that y > 0 as z ∈ H. Then γ(z) = 3(x+ iy) + 1 (x+ iy) + 2 = (3x+ 1) + 3iy (x+ 2) + iy = (3x+ 1) + 3iy (x+ 2) + iy × (x+ 2)− iy (x+ 2)− iy = (3x+ 1)(x + 2) + 3y2 (x+ 2)2 + y2 + i 5y (x+ 2)2 + y2 (where we have used the standard trick of multiplying numerator and denominator by the complex conjugate of the denominator in order to ‘realise’ the denominator). As y > 0 and the denominator is positive, we have that γ(z) ∈ H. Solution 4.1 (i) Choose σ : [a, 1] → H given by σ(t) = it. Then clearly σ(a) = ia and σ(1) = i (so that σ(·) has the required end-points) and σ(t) belongs to the imaginary axis. (Note there are many choices of parametrisations, your answer is correct as long as your parametrisation has the correct end-points and belongs to the imaginary axis.) (ii) For the parametrisation given above, |σ′(t)| = 1 and Im(σ(t)) = t. Hence lengthH(σ) = ∫ 1 a 1 t dt = log t|1a = − log a = log 1/a. Solution 4.2 The idea is simple: The distance between two points is the infimum of the (hyperbolic) lengths of (piecewise continuously differentiable) paths between them. Only a subset of these paths pass through a third point; hence the infimum of this subset is greater than the infimum over all paths. Let z1, z2, z3 ∈ H. Let σz1,z2 : [a, b]→ H be a path from z1 to z2 and let σz2,z3 : [b, c]→ H be a path from z2 to z3. Then the path σz1,z3 : [a, c]→ H formed by defining σz1,z3(t) = { σz1,z2(t) for t ∈ [a, b] σz2,z3(t) for t ∈ [b, c] is a path from z1 to z3 and has length equal to the sum of the lengths of σz1,z2 , σz2,z3 . Hence dH(z1, z3) ≤ lengthH(σz1,z3) = lengthH(σz1,z2) + lengthH(σz2,z3). 2 MATH32051 Solutions for Week 1 Taking the infima over path from z1 to z2 and from z2 to z3 we see that dH(z1, z) ≤ dH(z1, z2) + dH(z2, z3). Solution 5.1 For a straight line we have α = 0, i.e. βz + β¯z¯ + γ = 0. Recall that the line ax+ by+ c = 0 has gradient −a/b, x-intercept −c/a and y-intercept −c/b. Let z = x + iy so that x = (z + z¯)/2 and y = (z − z¯)/2i. Substituting these into ax+ by + c we see that β = (a− ib)/2 and γ = c. Hence the gradient is Re(β)/ Im(β), the x-intercept is at −γ/2Re(β) and the y-intercept is at γ/2 Im(β). Solution 5.2 A circle with centre z0 and radius r has equation |z − z0|2 − r2 = 0. Multiplying this out (see the proof of Proposition 5.3.1) we have: zz¯ − z¯0z − z0z¯ + |z0|2 − r2 = 0 and multiplying by α ∈ R we have αzz¯ − αz¯0z − αz0z¯ + α|z0|2 − αr2 = 0. Comparing the coefficients of this with αzz¯ + βz + β¯z¯ + γ = 0 we see that β = −αz¯0 and γ = α|z0|2 − αr2. Hence the centre of the circle is z0 = −β¯/α and the radius is given by r = √ |z0|2 − γ α = √ |β|2 α2 − γ α . 3 MATH32051 Solutions for Week 2 Solutions for Week 2 Solution 6.1 We first show that γ maps H to itself, i.e. if z ∈ H then γ(z) ∈ H. To see this, let z = u + iv ∈ H. Then Im(z) = v > 0. Let γ(z) = (az + b)/(cz + d) be a Mo¨bius transformation of H. Then γ(z) = a(u+ iv) + b c(u+ iv) + d = (au+ b+ iav) (cu+ d+ icv) (cu+ d− icv) (cu+ d− icv) , which has imaginary part 1 |cz + d|2 (−cv(au + b) + (cu+ d)av) = 1 |cz + d|2 (ad− bc)v which is positive. Hence γ maps H to itself. If γ(z) = (az+ b)/(cz+ d) then letting w = (az+ b)/(cz+ d) and solving for z in terms of w shows that γ−1(z) = (dz − b)/(−cz + a). Hence γ−1 exists and so γ is a bijection. Solution 6.2 (i) If γ1 = (a1z+ b1)/(c1z+ d1) and γ2 = (a2z+ b2)/(c2z+ d2) then their composition is γ2γ1(z) = a2 ( a1z+b1 c1z+d1 ) + b2 c2 ( a1z+b1 c1z+d1 ) + d2 = (a2a1 + b2c1)z + (a2b1 + b2d1) (c2a1 + d2c1)z + (c2b1 + d2d1) , which is a Mo¨bius transformation of H as (a2a1 + b2c1)(c2b1 + d2d1)− (a2b1 + b2d1)(c2a1 + d2c1) = (a1d1 − b1c1)(a2d2 − b2c2) > 0. (ii) Composition of functions is associative. (iii) The identity map z 7→ z is a Mo¨bius transformation of H (take a = d = 1, b = c = 0). (iv) It follows from the solution to Exercise 6.1 that if γ is a Mo¨bius transformation of H then so is γ−1. Solution 6.3 Let γ ∈ Mo¨b(H). We know that A is contained in either a circle or straight line in C, and so can be described as A = {z ∈ H | αzz¯ + βz + β¯z¯ + γ = 0} 1 MATH32051 Solutions for Week 2 for some α, γ ∈ R and β ∈ C. We need to show that γ(A) = {z ∈ H | α′zz¯+β′z+ β¯′z¯+γ′ = 0} for (possibly different) α′, γ′ ∈ R and β′ ∈ C. We know that γ maps H to H. Hence it is sufficient to prove that if z solves αzz¯+βz+ β¯z¯ + γ = 0 then γ(z) solves α′zz¯ + β′z + β¯′z¯ + γ′ = 0. Write γ(z) = (az+ b)/(cz+ d) where a, b, c, d ∈ R and ad− bc > 0. Let w = γ(z). Then z = γ−1(w) = (dw − b)/(−cw + a). Suppose that z solves αzz¯ + βz + β¯z¯ + γ = 0. Then w solves α ( dw − b −cw + a )( dw¯ − b −cw¯ + a ) + β ( dw − b −cw + a ) + β¯ ( dw¯ − b −cw¯ + a ) + γ = 0. Hence α(dw − b)(dw¯ − b) + β(dw − b)(−cw¯ + a) +β¯(dw¯ − b)(−cw + a) + γ(−cw + a)(−cw¯ + a) = 0. Expanding this out and gathering together terms gives (αd2 − (β + β¯)cd+ γc2)ww¯ + (−αbd+ βad+ β¯bc− γac)w +(−αbd+ β¯ad+ βbc− γac)w¯ + (αb2 − (β + β¯)ab+ γa2) = 0. (0.1) Let α′ = αd2 − (β + β¯)cd + γc2 β′ = −αbd+ βad+ β¯bc− γac γ′ = αb2 − (β + β¯)ab+ γa2. Recall that β + β¯ = 2Re(β), a real number. Hence α′, γ′ are real. Hence w satisfies an equation of the form α′ww¯ + β′w + β′w¯ + γ′ with α′, β′, γ′ ∈ R, which is the equation of either a vertical line or a circle with real centre. Solution 7.1 To see that γ maps ∂H to itself bijectively, it is sufficient to find an inverse. Notice that γ−1(z) = (dz − b)/(−cz + a) (defined appropriately for z = ∞, namely we set γ−1(∞) = −d/c) is an inverse for γ. Solution 7.2 Let γ(z) = (az + b)/(cz + d). Then γ′(z) = (cz + d)a− (az + b)c (cz + d)2 = ad− bc (cz + d)2 so that |γ′(z)| = ad− bc|cz + d|2 . To calculate the imaginary part of γ(z), write z = x+ iy. Then γ(z) = a(x+ iy) + b c(x+ iy) + d = (ax+ b+ iay) (cx+ d+ icy) (cx+ d− icy) (cx+ d− icy) , 2 MATH32051 Solutions for Week 2 which has imaginary part Im γ(z) = 1 |cz + d|2 (−cy(ax+ b) + (cx+ d)ay) = 1 |cz + d|2 (ad− bc)y = 1 |cz + d|2 (ad− bc) Im(z). Solution 7.3 Let z = x+ iy and define γ(z) = −x+ iy. (i) Suppose that γ(z1) = γ(z2). Write z1 = x1 + iy1, z2 = x2 + iy2. Then −x1 + iy1 = −x2 + iy2. Hence x1 = x2 and y1 = y2, so that z1 = z2. Hence γ is injective. Let z = x+ iy ∈ H. take w = −x+ iy. Then γ(w) = −(−x) + iy = x+ iy = z. Hence γ is surjective. Hence γ is a bijection. (ii) Let σ(t) = σ1(t) + iσ2(t) : [a, b] → H be a piecewise continuously differentiable path in H. Note that γ ◦ σ(t) = −σ1(t) + iσ2(t). Hence lengthH(γ ◦ σ) = ∫ b a 1 Im γ ◦ σ(t) √ (−σ′1(t))2 + (σ′2(t))2 dt = ∫ b a 1 σ2(t) |σ′(t)| dt = lengthH(σ). Let z, w ∈ H. Note that σ is a piecewise continuously differentiable path from z to w if and only if γ ◦ σ is a piecewise continuously differentiable path from γ(z) to γ(w). Hence dH(γ(z), γ(w)) = inf{lengthH(γ ◦ σ) | σ is a piecewise continuously differentiable path from z to w} = inf{lengthH(σ) | σ is a piecewise continuously differentiable path from z to w} = dH(z, w). Hence γ is an isometry of H. 3 MATH32051 Solutions for Week 3 Solutions for Week 3 Solution 9.1 Let γ1 ∈ Mo¨b(H) map H1 to the imaginary axis and the point z1 to i. Let γ2 ∈ Mo¨b(H) map H2 to the imaginary axis and the point z2 to i. Let γ = γ −1 2 γ1. Then γ ∈ Mo¨b(H), γ maps H1 to H2, and γ(z1) = z2. Solution 10.1 (i) Draw in the tangent lines to the circles at the point of intersection; then θ is the angle between these two tangent lines. Draw the (Euclidean!) triangle with vertices at the point of intersection and the two centres. See Figure 57.1. The internal angle of this triangle at the point of intersection is split into three; the middle part is equal to θ. Recall that a radius of a circle meets the tangent to a circle at right-angles. Hence both the remaining two parts of the angle in the triangle at the point of intersection is given by π/2−θ. Hence the triangle has angle π/2− θ + θ + π/2 − θ = π − θ at the vertex corresponding to the point of intersection. c1 c2 r1 r2θ Figure 57.1: The Euclidean triangle with vertices at c1, c2 and the point of intersection. The cosine rule gives the required formula (recall that cos π − θ = − cos θ). (ii) The points −6 and 6 clearly lie on the semi-circle with centre 0 and radius 6. Similarly, the points 4 √ 2 and 6 √ 2 clearly lies on the semi-circle with centre 5 √ 2 and radius√ 2. (If you can’t determine the geodesic by considering the geometry then you can find it as follows. We know geodesics have equations of the form αzz¯ + βz + βz¯ + γ = 0 where α, β, γ ∈ R. The geodesic between 4√2 and 6√2 is clearly a semi-circle, and 1 MATH32051 Solutions for Week 3 so α 6= 0; we divide through by α to assume that α = 1. Hence we are looking for an equation of the form zz¯ + βz + βz¯ + γ = 0. Substituting first z = 4 √ 2 and then z = 6 √ 2 we obtain the simultaneous equations 32+8 √ 2β+γ = 0, 72+12 √ 2β+γ = 0. Solving these gives β = −5√2, γ = 48. Putting z = x+ iy we thus have the equation x2 + y2 − 10√2x+ 48 = 0. Completing the square gives (x− 5√2)2 + y2 = (√2)2, so that we have a semi-circle in C with centre 5 √ 2 and radius √ 2.) Part (i) allows us to calculate the angle ψ in Figure 57.2. Substituting c1 = 0, r1 = 6, −6 4√2 6 6√2 ψ Figure 57.2: The angle ψ. c2 = 5 √ 2, r2 = √ 2 into the result from (i) shows that cosψ = 1/ √ 2 so that ψ = π/4. The angle in Figure 46 that we want to calculate is φ = π − ψ = 3π/4. Solution 10.2 Suppose that the semi-circular geodesic has centre at x ∈ R and radius r. Construct the (Euclidean) right-angled triangle with vertices x, 0, ib, as illustrated in Figure 57.3. As θ ib a θ x r r − a br θ Figure 57.3: The (Euclidean) triangle with vertices at x, 0, ib. the radius of the semicircle is r, we have that |x − ib| = r and |x − a| = r; hence the base of the right-angled triangle has length r − a. By Pythagoras’ Theorem, we have that (r − a)2 + b2 = r2. Expanding this out and simplyfying it we have r = (a2 + b2)/2a. From Figure 57.3 we also have that sin θ = b r , cos θ = r − a r and the result follows after substituting in r = (a2 + b2)/2a. Solution 11.1 2 MATH32051 Solutions for Week 3 (i) Let γ be a Mo¨bius transformation of H. As γ is an isometry, by Proposition 7.2.2 we know that cosh dH(γ(z), γ(w)) = cosh dH(z, w). Hence LHS(γ(z), γ(w)) = LHS(z, w). By Exercise 7.2 we know that if γ is a Mo¨bius transformation then Im(γ(z)) = |γ′(z)| Im(z). By Lemma 11.2.1 it follows that 1 + |γ(z)− γ(w)|2 2 Im(γ(z)) Im(γ(w)) = 1 + |z − w|2|γ′(z)||γ′(w)| 2|γ′(z)| Im(z)|γ′(w)| Im(w) = 1 + |z − w|2 2 Im(z) Im(w) . Hence RHS(γ(z), γ(w)) = RHS(z, w). (ii) Let H be the geodesic passing through z and w. Then by Lemma 9.2.1 there exists a Mo¨bius transformation γ of H mapping H to the imaginary axis. Let γ(z) = ia and γ(w) = ib. By interchanging z and w if necessary, we can assume that a < b. Then LHS(γ(z), γ(w)) = cosh dH(γ(z), γ(w)) = cosh dH(ia, ib) = cosh log b/a = elog b/a + elog a/b 2 = b/a+ a/b 2 = b2 + a2 2ab . Moreover, RHS(γ(z), γ(w)) = RHS(ia, ib) = 1 + |ia− ib|2 2ab = 1 + (b− a)2 2ab = b2 + a2 2ab . Hence LHS(γ(z), γ(w)) = RHS(γ(z), γ(w)). (iii) For any two points z, w let H denote the geodesic containing both z, w. Choose a Mo¨bius transformation γ of H that maps H to the imaginary axis. Then LHS(z, w) = LHS(γ(z), γ(w)) = RHS(γ(z), γ(w)) = RHS(z, w). Solution 11.2 Let C = {z ∈ H | dH(z, z0) = r} be a hyperbolic circle with centre z ∈ H and radius r > 0. Recall cosh dH(z, z0) = 1 + |z − z0|2 2 Im(z) Im(z0) . 3 MATH32051 Solutions for Week 3 Let z0 = x0 + iy0 and z = x+ iy. Then cosh r = 1 + (x− x0)2 + (y − y0)2 2y0y which can be simplified to (x− x0)2 + (y − y0 cosh r)2 + y20 − y20 cosh2 r = 0 which is the equation of a Euclidean circle with centre (x0, y0 cosh r) and radius y0 √ cosh2 r − 1 = y0 sinh r. Solution 11.3 (i) Let σ : [a, b] → H be any piecewise continuously differentiable path. As we are assuming lengthρ(σ) = lengthρ(γ ◦ σ) we have∫ b a ρ(σ(t))|σ′(t)| dt = lengthρ(σ) = lengthρ(γ ◦ σ) = ∫ b a ρ(γ(σ(t)))|(γ(σ(t)))′ | dt = ∫ b a ρ(γ(σ(t)))|γ′(σ(t))||σ′(t)| dt where we have used the chain rule to obtain the last equality. Hence ∫ b a ( ρ(γ(σ(t)))|γ′(σ(t))| − ρ(σ(t))) |σ′(t)| dt = 0. Using the hint, we see that ρ(γ(z))|γ′(z)| = ρ(z) (0.1) for all z ∈ H. (ii) Take γ(z) = z + b in (0.1). Then |γ′(z)| = 1. Hence ρ(z + b) = ρ(z) for all b ∈ R. Hence ρ(z) depends only the imaginary part of z. Write ρ(z) = ρ(y) where z = x+ iy. (iii) Take γ(z) = kz in (0.1). Then |γ′(z)| = k. Hence kρ(ky) = ρ(y). Setting y = 1 and letting c = ρ(1) we have that ρ(k) = ρ(1)/k = c/k. Hence ρ(z) = c/ Im(z). 4 MATH32051 Solutions for Week 4 Solutions for Week 4 Solution 12.1 (i) First note that h is a bijection from H to its image because it has an inverse g(z) = (−z + i)/(−iz + 1). Similarly, h is a bijection from ∂H to its image. We now show that h(H) = D. Note that |h(z)| = |z − i||iz − 1| = |z − i| |i||z + i| = |z − i| |z + i| . We claim that if z ∈ H then |z− i| < |z+ i|. To see this, let z = x+ iy and note that y > 0. Then |z−i|2 = x2+(y−1)2 and |z+i|2 = x2+(y+1)2. Hence |z−i|2 < |z+i|2 if and only if x2+(y−1)2 < x2+(y+1)2; expanding this out we see that this happens if and only if −y < y, which is true as y > 0. Hence |z − i|/|i||z + i| < 1. Hence h maps H inside D. Now suppose that z ∈ ∂H. Note that the above calculation also shows that if y = 0 then |h(z)| = 1. Clearly |h(∞)| = 1. Hence h maps ∂Hto∂D. As h is continuous it follows that h maps H to D bijectively and ∂H to ∂D bijectively. (ii) We have already seen that g(z) = h−1(z) = (−z + i)/(−iz + 1). Calculating g′(z) is easy. To calculate Im(g(z)) write z = u+ iv and realise the denominator. Solution 12.2 Let σ(t) = t, 0 ≤ t ≤ x. Then σ is a path from 0 to x and it has length∫ σ 2 1− |z|2 = ∫ x 0 2 1− t2 dt = ∫ x 0 1 1− t + 1 1 + t dt = log 1 + x 1− x. To show that this is the optimal length of a path from 0 to x (and thus that the real-axis is a geodesic) we have to show that any other path from 0 to x has a larger length. Let σ(t) = x(t) + iy(t), a ≤ t ≤ b be a path from 0 to x. Then it has length ∫ b a 2 1− |σ(t)|2 |σ ′(t)| dt = ∫ b a 2 1− (x(t)2 + y(t)2) √ x′(t)2 + y′(t)2 dt 1 MATH32051 Solutions for Week 4 ≥ ∫ b a 2 1− x(t)2x ′(t) dt = ∫ b a x′(t) 1− x(t) + x′(t) 1 + x(t) dt = log 1 + x(t) 1− x(t) ∣∣∣∣ b a = log 1 + x 1− x, with equality precisely when y′(t) = 0 and y(t) = 0, i.e. with equality precisely when the path lies along the real axis. Solution 12.3 Recall h(z) = (z − i)/(iz − 1) and h−1(z) = (−z + i)/(−iz + 1). Let γ(z) = (az + b)/(cz + d), ad − bc > 0, be a Mo¨bius transformation of H. We claim that hγh−1 is a Mo¨bius transformation of D. To see this, first note that (after a lot of algebra!) hγh−1(z) = [a+ d+ i(b− c)]z + [−(b+ c)− i(a− d)] [−(b+ c) + i(a− d)]z + [a+ d− i(b− c)] = αz + β β¯z + α¯ . Finally, we must check that |α|2 − |β|2 > 0 which is a simple calculation, using the fact that ad− bc > 0. Solution 12.4 The map h : H→ D defined in (1.1) maps geodesics in H to geodesics in D. Suppose that z ∈ H lies on a geodesic. Then z lies on either a horizontal straight line or semi-circle with real centre with an equation of the form αzz¯ + βz + βz¯ + γ = 0. Let w = h(z). Then z = −w + i −iw + 1 so that z¯ = −w¯ − i iw¯ + 1 . Hence w satisfies an equation of the form α −w + i −iw + 1 −w¯ − i iw¯ + 1 + β −w + i −iw + 1 + β −w¯ − i iw¯ + 1 + γ = 0. Equivalently, w satisfies an equation of the form α(−w + i)(−w¯ − i) + β(−w + i)(iw¯ + 1) + β(−w¯ − i)(−iw + 1) + γ(−iw + 1)(iw¯ + 1) = 0. Multiplying this out and collecting terms we see that w satisfies an equation of the form (α+ γ)ww¯ + (−2β + i(α− γ))w + (−2β − i(α − γ))w + (α+ γ) = 0. 2 MATH32051 Solutions for Week 4 Let α′ = α + γ, β′ = −2β + i(α − γ). Then α′ ∈ R and β′ ∈ C. Moreover w satisfies an equation of the form α′ww¯ + β′w + β¯′w¯ + α′ = 0. Solution 12.5 By applying a Mo¨bius transformation of D, we can move the circle so that its centre is at the origin 0 ∈ D. (This uses the additional facts that (i) a hyperbolic circle is a Euclidean circle (but possibly with a different centre and radius), and (ii) Mo¨bius transformations of D map circles to circles.) As Mo¨bius transformations of D preserve lengths and area, this does not change the circumference nor the area. Let Cr = {w ∈ D | dD(0, w) = r}. By Proposition 12.2.1 and the fact that a rotation is a Mo¨bius transformation of D, we have that Cr is a Euclidean circle with centre 0 and radius R where 1 +R 1−R = e r. Hence R = (er − 1)/(er + 1) = tanh(r/2). Now circumference(Cr) = ∫ σ 2 1− |z|2 where σ(t) = Reit, 0 ≤ t ≤ 2π is a path that describes the Euclidean circle of radius R, centred at 0. Now circumference(Cr) = ∫ σ 2 1− |z|2 = ∫ 2π 0 2 1− |σ(t)|2 |σ ′(t)| dt = ∫ 2π 0 2R 1−R2 dt = 4πR 1−R2 and substituting for R in terms of r gives that the circumference of Cr is 2π sinh r. Similarly, the area of Cr is given by AreaD(Cr) = ∫ ∫ Dr 4 (1− |z|2)2 dz where Dr = {w ∈ D | dD(0, w) ≤ r} is the disc of hyperbolic radius r with centre 0. Now Dr is the Euclidean disc of radius R = tanh(r/2) centred at 0. Recall that when integrating using polar co-ordinates, the area element is ρ dρ dθ. Then AreaD(Cr) = ∫ 2π θ=0 ∫ R ρ=0 4 (1− ρ2)2ρ dρ dθ = 4π 1 1− ρ2 ∣∣∣∣ R ρ=0 = 4π R2 1−R2 = 4π sinh2 r/2. 3 MATH32051 Solutions for Week 4 Solution 13.1 (i) Clearly both (−1 + i√3)/2 and (1 + i√3)/2 lie on the unit circle in C with centre 0 and radius 1. One can determine the other two geodesics by recognition; alternatively one can argue as follows. Consider 0, (−1 + i√3)/2. These two points lie on a geodesic given by a semi-circle with equation zz¯ + βz + βz¯ + γ = 0. Substituting these two values of z in gives the simultaneous equations γ = 0, 1 − β + γ = 0. Hence β = 1, γ = 0. Hence the equation of the geodesic through 0, (−1 + i√3)/2 is given by zz¯ + z + z¯ = 0. Writing z = x + iy this becomes x2 + y2 + 2x = 0. Completing the square gives (x + 1)2 + y2 = 1. Hence 0, (−1 + i√3)/2 lie on the circle in C with centre −1 and radius 1. A similar calculation shows that 0, (1 + i √ 3)/2 lie on the circle in C with centre 1 and radius 1. (ii) As the vertex 0 is on the boundary of H, the internal angle is 0. We can calculate the angles ψ1, ψ2 in Figure 58.1 using Exercise 10.1. We obtain ψ1 ψ2 −2 −1 0 1 2 Figure 58.1: The angles ψ1, ψ2. cosψ1 = (0− 1)2 − (12 + 12) 2 = −1 2 so that ψ1 = 2π/3. As θ1 = π − ψ1 we have θ1 = π/3. Similarly, θ2 = π/3. By the Gauss-Bonnet Theorem, the area of the triangle is π− (0+π/3+π/3) = π/3. Solution 13.2 Let Q be a hyperbolic quadrilateral with vertices A,B,C,D (labelled, say, anti-clockwise) and corresponding internal angles α, β, γ, δ. Construct the geodesic from A to C, creating triangles ABC (with internal angles α1, β, γ1) and CDA (with internal angles γ2, δ, α2), where α1 + α2 = α and γ1 + γ2 = γ. By the Gauss-Bonnet Theorem AreaH(Q) = AreaH(ABC) + AreaH(CDA) = π − (α1 + β + γ1) + π − (α2 + β + γ2) = 2π − (α+ β + γ + δ). Solution 14.1 Let D(r) be the hyperbolic polygon with vertices at r, rω, . . . , rωn−1. Let αj(r) denote the 4 MATH32051 Solutions for Week 4 internal angle at vertex rωj. For each 0 ≤ k ≤ n − 1, consider the Mo¨bius transformation of D given by γk(z) = w kz; this rotates the polygon so that vertex vi is mapped to vertex vi+k. Thus γk(D(r)) = D(r). As Mo¨bius transformations of D preserve angles, this shows that the internal angle at vertex v1 is equal to the internal angle at vertex v1+k. By varying k, we see that all internal angles are equal. By the Gauss-Bonnet Theorem, we see that AreaHD(r) = (n− 2)π − nα(r). Notice that D(r) is contained in C(r), the hyperbolic disc with hyperbolic centre 0 and Euclidean radius r. By (the solution to) Exercise 11.2, we see that lim r→0 AreaHD(r) ≤ lim r→0 AreaHC(r) = lim r→0 4πr2 1− r2 = 0. Hence lim r→0 α(r) = (n− 2)π n . As r → 1, each vertex rωk → ωk ∈ ∂D. The internal angle at a vertex on the boundary is equal to 0. Hence limr→1 α(r) = 0. Hence given any α ∈ [0, (n − 2)π/n), we can find a value of r for which α = α(r), and hence construct a regular n-gon with internal angle α. Conversely, suppose that D is a regular hyperbolic polygon with each internal angle α ≥ (n− 2)π/n. Then we have that nα ≥ (n− 2)π. By the Gauss-Bonnet Theorem, AreaHD = (n− 2)π − nα ≤ (n− 2)π − (n− 2)π = 0. As area must be positive, this is a contradiction. 5 MATH32051 Solutions for Week 5 Solutions for Week 5 Solution 15.1 First note that cos2 α = 1 1 + tan2 α = 1 1 + tanh 2 a sinh2 b = sinh2 b sinh2 b+ tanh2 a . Now using the facts that cosh c = cosh a cosh b and tanh2 a = 1− 1/ cosh2 a we see that tanh2 a = 1− cosh 2 b cosh2 c . Substituting this into the above equality gives cos2 α = sinh2 b sinh2 b+ 1− cosh2 b cosh2 c = tanh2 b tanh2 c (after some manipulation, using the fact that cosh2− sinh2 = 1). To see that sin β = sinh b/ sinh c we multiply the above equation and the equation given in Proposition 15.2.1 together to obtain sinα = tanh b tanh c tanh a sinh = sinh b cosh b cosh c sinh c sinh a cosh a 1 sinh b = sinh a sinh c , using the fact that cosh c = cosh a cosh b. Solution 15.2 We prove the first identity. By Proposition 15.2.1 we know that cosα = tanh b tanh c , sin β = sinh b sinh c . Hence cosα sin β = tanh b tanh c sinh c sinh b = cosh c cosh b = cosh a, 1 MATH32051 Solutions for Week 5 using the hyperbolic version of Pythagoras’ Theorem. We prove the second identity. By Proposition 15.2.1 we have that tanα = tanh a sinh b , tan β = tanh b sinh a . Hence cotα cot β = sinh a tanh b sinh b tanh a = cosh a cosh b = cosh c, by the hyperbolic version of Pythagoras’ Theorem. Take a Euclidean right-angled triangle with sides of length a, b and c, with c being the hypotenuse. Let α be the angle opposite a and β opposite b. Then cosα = b/c and sin β = b/c so that cosα cosec β = 1. As in a Euclidean triangle the angles sum to π, we must have that β = π/2−α. Hence the above identity says that sin(π/2− α) = cosα. Similarly, we have that tanα = a/b and tan β = b/a. Hence cotα cot β = 1. Again, this can be re-written as tan(π/2− α) = 1/ tan α. Solution 15.3 Note that cosα = √ 1− sin2 α = √ 1− 1 cosh2 a = sinh a cosh a = 1 coth a . Hence tanα = sinα cosα = 1 cosh a cosh a sinh a = 1 sinh a . Solution 15.4 By the angle of parallelism formula, we can find the angle α. We have cosh(log(2 + √ 3)) = elog(2+ √ 3) + e−(log(2+ √ 3)) 2 = 2 + √ 3 + 1 2+ √ 3 2 = (2 + √ 3)(2 + √ 3) + 1 2(2 + √ 3) = 4 + 4 √ 3 + 3 + 1 2(2 + √ 3) = 4(2 + √ 3) 2(2 + √ 3) = 2. Hence sinα = 1/2 so that α = π/6. By the Gauss-Bonnet Theorem, the hyperbolic area of this triangle is π− (π/2+π/6) = π/3. Solution 15.5 Label the vertices A,B and C so that the angle at A is α, etc. By applying a Mo¨bius 2 MATH32051 Solutions for Week 5 transformation of H we may assume that none of the sides of ∆ are segments of vertical lines. Construct a geodesic from vertex B to the geodesic segment [A,C] in such a way that these geodesics meet at right-angles. This splits ∆ into two right-angled triangles, BDA and BDC. Let the length of the geodesic segment [B,D] be d, and suppose that BDA has internal angles β1, π/2, α and side lengths d, b1, c, as in the figure. Label BDC similarly. See Figure 59.1. b2 b a c b1 α γd β β2β1 A C B D Figure 59.1: The sine rule. From Proposition 15.2.1 we know that sinβ1 = sinh b1 sinh c , cos β1 = tanh d tanh c , sinβ2 = sinh b2 sinh a , cos β2 = tanh d tanh a . By the hyperbolic version of Pythagoras’ Theorem we know that cosh c = cosh b1 cosh d, cosh a = cosh b2 cosh d. Hence sin β = sin(β1 + β2) = sin β1 cos β2 + sinβ2 cos β1 = sinh b1 sinh c sinh d cosh d cosh a sinh a + sinh b2 sinh a sinh d cosh d cosh c sinh c = sinh b1 sinh d sinh c sinh a cosh b2 + sinh b2 sinh d sinh a sinh c cosh b1 = sinh d sinh a sinh c (sinh b1 cosh b2 + sinh b2 cosh b1) = sinh d sinh a sinh c sinh(b1 + b2) = sinh b sinh d sinh a sinh c . Using Proposition 15.2.1 again, we see that sinα = sinh d/ sinh c and sin γ = sinh d/ sin a. Substituting these into the above equality proves the result. 3 MATH32051 Solutions for Week 5 Solution 16.1 γ1 has one fixed point in H at (−3 + i √ 51)/6 and so is elliptic. γ2 has fixed points at ∞ and −1 and so is hyperbolic. γ3 has one fixed point at i and so is elliptic. γ4 has one fixed point at 0 and so is parabolic. Solution 17.1 We have γ1(z) = 2√ 13 z + 5√ 13 −3√ 13 z + −1√ 13 , γ2(z) = 7√ 7 z + 6√ 7 1√ 7 , and γ3 and γ4 are already normalised. We have τ(γ1) = ( 2√ 13 + −1√ 13 )2 = 1 13 ∈ [0, 4) τ(γ2) = ( 7√ 7 + 1√ 7 )2 = 64 7 > 4 τ(γ3) = 0 ∈ [0, 4) τ(γ4) = (1 + 1) 2 = 4. Hence γ1, γ3 are elliptic, γ2 is hyperbolic and γ4 is parabolic. Solution 18.1 (i) Recall that the Mo¨bius transformation γ1 of H is conjugate to the Mo¨bius trans- formation γ2 of H if there exists a Mo¨bius transformation g ∈ Mo¨b(H) such that gγ1g −1 = γ2. Clearly γ is conjugate to itself (take g = id). If γ2 = gγ1g −1 then γ1 = g−1γ2g so that γ2 is conjugate to γ1 if γ1 is conjugate to γ2. If γ2 = gγ1g −1 and γ3 = hγ2h−1 then γ3 = (hg)γ1(hg)−1 so that γ3 is conjugate to γ1. (ii) Let γ1 and γ2 be conjugate. Write γ2 = gγ1g −1 where g ∈ Mo¨b(H). Then γ1(x) = x ⇔ g−1γ2g(x) = x ⇔ γ2(g(x)) = g(x) so that x is a fixed point of γ1 if and only if g(x) is a fixed point of γ2. Hence g maps the set of fixed points of γ1 to the set of fixed points of γ2. As g is a Mo¨bius transformation of H and therefore a bijection, we see that γ1 and γ2 have the same number of fixed points. Solution 18.2 Let A = (aij), B = (bij) be two matrices. We first show that trace(AB) = trace(BA). Recall that the trace of a matrix is the sum of the diagonal elements. Hence trace(AB) = ∑ i (AB)ii = ∑ i ∑ j aijbji = ∑ j ∑ i bjiaij = ∑ j (BA)j = trace(BA) 4 MATH32051 Solutions for Week 5 where (AB)ij denotes the (i, j)th entry of AB. Let γ1(z) = a1z + b1 c1z + d1 , γ2(z) = a2z + b2 c2z + d2 be two conjugate Mo¨bius transformations of H. Let A1 = ( a1 b1 c1 d1 ) , A2 = ( a2 b2 c2 d2 ) , be their corresponding (normalised) matrices. Let g be a Mo¨bius transformation of H such that γ1 = g −1γ2g. Suppose that g has matrix A. By replacing A by −A if necessary, we have that A1 = A −1A2A. Hence trace(A1) = trace(A −1A2A) = trace(A2AA−1) = trace(A2). Hence τ(γ1) = trace(A1) 2 = trace(A2) 2 = τ(γ2). Solution 18.3 Let γ1(z) = z + b where b > 0 and let γ2(z) = z + 1. As both γ1 and γ2 have fixed points at ∞ and a conjugacy acts a ‘change of co-ordinates’, we look for a conjugacy from γ1 to γ2 that fixes ∞. We will try g(z) = kz for some (to be determined) k > 0. Now g−1γ1g(z) = g−1γ1(kz) = g−1(kz + b) = z + b/k. So we choose k = b. Now let γ1(z) = z − b where b > 0 and let γ2(z) = z − 1. Again, let g(z) = kz for some k > 0. Then g−1γ1g(z) = g−1γ(kz) = g−1(kz − b) = z − b/k. So again we choose k = b. Suppose that γ1(z) = z + 1 and γ2(z) = z − 1 are conjugate. Then there exists g(z) = (az+ b)/(cz + d) ∈ Mo¨b(H) such that γ1g(z) = gγ2(z). In terms of matrices, this says that( 1 1 0 1 )( a b c d ) = ± ( a b c d )( 1 −1 0 1 ) . That is, ( a+ c b+ d c d ) = ( a −a+ b c −c+ d ) . Comparing coefficient in the ‘+’ case, we see that c = 0 and d = −a. Hence ad − bc = −a2 < 0, a contradiction. In the ‘−’ case, we see that c = 0, d = 0, so that ad − bc = 0, again a contradiction. Hence γ1, γ2 are not conjugate in Mo¨b(H). Solution 18.4 Let γ1(z) = k1z and γ2(z) = k2z where k1, k2 6= 1. Suppose that γ1 is conjugate to γ2. Then there exists a Mo¨bius transformation of H, γ(z) = (az + b)/(cz + d), such that γγ1(z) = γ2γ(z). Explicitly: ak1z + b ck1z + d = k2 ( az + b cz + d ) . Multiplying out and equating coefficients gives ack1 = ack1k2, adk1 + bc = k2ad+ k1k2bc, bd = k2bd. As k2 6= 1 the third equation implies that bd = 0. 5 MATH32051 Solutions for Week 5 Case 1: b = 0. If b = 0 then the second equation implies that adk1 = adk2. So either k1 = k2 or ad = 0. If ad = 0 then, as b = 0, we have ad − bc = 0 so γ is not a Mo¨bius transformation of H. Hence k1 = k2. Case 2: d = 0. If d = 0 then bc = bck1k2. So either k1k2 = 1 or bc = 0. If bc = 0 then, as d = 0, we have ad− bc = 0 so γ is not a Mo¨bius transformation of H. Hence k1k2 = 1. Here is a sketch of an alternative method. If γ1(z) = k1z and γ2(z) = k2z are conjugate then they have the same trace. The trace of γ1 is seen in Exercise 18.5 below to be ( √ k1 + 1/ √ k1) 2, and the trace of γ2 is ( √ k2 + 1/ √ k2) 2. Equating these shows (after some manipulation) that k1 = k2 or k1 = 1/k2. Solution 18.5 Let γ be hyperbolic. Then γ is conjugate to a dilation z 7→ kz. Writing this dilation in a normalised form z 7→ k√ k z 1√ k we see that τ(γ) = (√ k + 1√ k )2 . Solution 18.6 Let γ be an elliptic Mo¨bius transformation. Then γ is conjugate (as a Mo¨bius transfor- mation of D) to the rotation of D by θ, i.e. γ is conjugate to z 7→ eiθz. Writing this transformation in a normalised form we have z 7→ e iθ/2z e−iθ/2 , which has trace (eiθ/2 + e−iθ/2)2 = 4cos2(θ/2). Hence τ(γ) = 4 cos2(θ/2). 6 MATH32051 Solutions for Week 6 Solutions for Week 6 Solution 20.1 Our intuition is that, as b → 0, we should have that γb → id, the identity transformation in Mo¨b(H). To check this rigorously using our definition, notice that γb and id have matrices( 1 b 0 1 ) , ( 1 0 0 1 ) , respectively and that these are both normalised. Hence dMo¨b(H)(γb, id) = min{‖(1, b, 0, 1)−(1, 0, 0, 1)‖, ‖(1, b, 0, 1)−(−1, 0, 0,−1)‖} = ‖(0, b, 0, 0)‖ = |b| (here, ‖(a, b, c, d) = √ a2 + b2 + c2 + d2). Hence dMo¨b(H)(γb, id) = |b| → 0 as b→ 0. Solution 20.2 Fix q > 0 and let Γq = { γ(z) = az + b cz + d | a, b, c, d ∈ Z, b, c are divisible by q } . First note that id ∈ Γq (take a = d = 1, b = c = 0). Let γ1 = (a1z + b1)/(c1z + d2), γ2 = (a2z + b2)/(c2z + d2) ∈ Γq. Then γ1γ2(z) = (a1a2 + b1c2)z + (a1b2 + b1d2) (c1a2 + d1c2)z + (c1b2 + d1d2) . Now q divides b1, b2, c1, c2. Hence q divides a1b2 + b1d2 and c1a2 + d1c2. Hence γ1γ2 ∈ Γq. If γ(z) = (az + b)/(cz + d) ∈ Γq then γ−1(z) = (dz − b)/(−cz + a). Hence γ−1 ∈ Γq. Hence Γq is a subgroup of Mo¨b(H). Solution 20.3 The group generated by γ1(z) = z + 1 and γ2(z) = kz (k 6= 1) is not a Fuchsian group. Consider the orbit Γ(i) of i. First assume that k > 1. Then observe that γ−n2 γ m 1 γ n 2 (i) = γ −n 2 γ m 1 (k ni) = γ−n2 (k ni+m) = i+m/kn. By choosing n arbitrarily large we see that m/kn is arbitrarily close to, but not equal to, 0. Hence i is not an isolated point of the orbit Γ(i). Hence Γ(i) is not discrete. By Proposition 21.2.1, Γ is not a Fuchsian group. The case where 0 < k < 1 is similar, but with γ−n2 γ m 1 γ n 2 replaced by γ n 2 γ m 1 γ −n 2 Solution 21.1 There are two cases: z = 0 and z 6= 0. We have Γ(0) = {0}. When z 6= 0, Γ(z) consists of 4 corners of a square. Solution 23.1 See Figure 60.1. 1 MATH32051 Solutions for Week 6 Figure 60.1: Solution to Exercise 21.1. 2 MATH32051 Solutions for Week 7 Solutions for Week 7 Solution 24.1 (Not examinable—included for interest only!) Recall that a subset C ⊂ H is convex if: ∀z, w ∈ C, [z, w] ⊂ C; that is, the geodesic segment between any two points of C lies inside C. Let us first show that a half-plane is convex. We first show that the half-plane H0 = {z ∈ H | Re(z) > 0} is convex; in fact this is obvious by drawing a picture. Now let H be any half-plane; we have to show that H is convex. Recall that H is defined by a geodesic ℓ of H and that the group of Mo¨bius transformations of H acts transitively on geodesics. Hence we can find a Mo¨bius transformation γ of H that maps the imaginary axis to ℓ. Hence γ maps either H0 or {z ∈ H | Re(z) < 0} to H. In the latter case we can first apply the isometry z 7→ −z¯ so that H0 is mapped by an isometry to H. As isometries map geodesic segments to geodesic segments, we see that H is convex. Finally, let D = ∩Hi be an intersection of half-planes. Let z, w ∈ D. Then z, w ∈ Hi for each i. As Hi is convex, the geodesic segment [z, w] ⊂ Hi for each i. Hence [z, w] ⊂ D so that D is convex. Solution 25.1 (i) By Proposition 25.2.1, z ∈ H is on the perpendicular bisector of [z1, z2] if and only if dH(z, z1) = dH(z, z2). Note that dH(z, z1) = dH(z, z2) ⇔ cosh dH(z, z1) = cosh dH(z, z2) ⇔ 1 + |z − z1| 2 2y1 Im(z) = 1 + |z − z2|2 2y2 Im(z) ⇔ y2|z − z1|2 = y1|z − z2|2. (ii) Let z = x + iy. Then z is on the perpendicular bisector of 1 + 2i and (6 + 8i)/5 precisely when 8 5 |(x+ iy)− (1 + 2i)|2 = 2|(x+ iy)− ( 6 5 + 8i 5 ) |2, i.e. 4 ( (x− 1)2 + (y − 2)2) = 5 (( x− 6 5 )2 + ( y − 8 5 )2) . Expanding this out and collecting like terms gives x2 − 4x+ y2 = 0 and completing the square gives (x− 2)2 + y2 = 4 = 22. Hence the perpendicular bisector is the semi-circle with centre (2, 0) and radius 2. 1 MATH32051 Solutions for Week 7 Solution 25.2 The result in Proposition 25.2.1 continues to hold in Euclidean geometry. Indeed, let x = (x1, y1), y = (x2, y2) ∈ R2. Then the perpendicular bisector of the (Euclidean) straight line between x and y is {z = (x3, y3) | d(x, z) = d(z, y)}. (Here d denotes the Euclidean distance.) To see this, apply an isometry of R2 so that, without loss of generality, x = (0, 0) is the origin and y = (x2, 0) lies along the x-axis. Then the perpendicular bisector of the straight line between x and y is the vertical straight line with x co-ordinate equal to x2/2. Let z = (x3, y3) be a point that is equidistant from x = (0, 0) and y = (x2, 0). Then d(x, z) = d(z, y) is equivalent to (0− x3)2 + (0− y3)2 = (x3 − x2)2 + (0− y3)2. Expanding this out gives that x3 = x2/2, i.e. the set of points that are equidistant from x and y is the vertical straight line with x co-ordinate equal to x2/2. Solution 26.1 Let Γ = {γn | γn(z) = 2nz}. Let p = i and note that γn(p) = 2ni 6= p unless n = 0. For each n, [p, γn(p)] is the arc of imaginary axis from i to 2 ni. Suppose first that n > 0. Recalling that for a < b we have dH(ai, bi) = log b/a it is easy to see that the midpoint of [i, 2ni] is at 2n/2i. Hence Lp(γn) is the semicircle of radius 2 n/2 centred at the origin and Hp(γn) = {z ∈ H | |z| < 2n/2}. For n < 0 one sees that Hp(γn) = {z ∈ H | |z| > 2n/2}. Hence D(p) = ⋂ γn∈Γ\{Id} Hp(γn) = {z ∈ H | 1/ √ 2 < |z| < √ 2}. 2 MATH32051 Solutions for Week 8 Solutions for Week 8 Solution 27.1 Let γ(z) = (az + b)/(cz + d) ∈ PSL(2,Z) \ {id}. Suppose that γ(p) = p. Then aki + b = −ck2 + dki. Comparing real and imaginary parts shows that b = −ck2 and ak = dk. As k > 1, we have a = d. Hence we have 1 = ad− bc = a2+ c2k2. There are several ways from this to see that this implies that a = d = ±1, b = c = 0. Here’s one way. Note that 1 = a2 + c2k2 = |a + ick|2. Hence a + ick is a point on the unit circle in C. The only points on the unit circle in C where the real part is an integer occur at ±1 and at ±i. When a± 1 we have a = d = ±1, b = c = 0; here γ is the identity. When a + ick = ±i we must have that c = 0 (as otherwise there is no solution as c ≥ 1, k > 1) and a = 0, contradicting ad− bc = 1. Here’s another way. Note that a2 ≤ a2 + c2k2 = 1. Now a2 ≤ a2 + c2k2 = 1 and a2 is a non-negative integer. This is only possible if either (i) a = ±1 and c = 0 (and in this case γ is the identity) or (ii) a = 0 and c2k2 = 1. In case (ii), as k > 1 and c is an integer, this is only possible if c = 0; but then ad− bc = 0, a contradiction. Solution 28.1 Let p = i and let γn(z) = 2 nz. There are two sides: s1 = {z ∈ C | |z| = 1/ √ 2}, s2 = {z ∈ C | |z| = √ 2}. The side s1 is the perpendicular bisector of [p, γ−1(p)]. Hence γs1 , the side-pairing trans- formation associated to the side s1, is γs1(z) = (γ−1) −1(z) = 2z and pairs side s1 to side s2. Hence γs2(z) = γ −1 s1 (z) = z/2. Solution 30.1 (i) Suppose that γ1 has order m. Suppose that γ2 = g −1γ1g is conjugate to γ1. Then γm2 = (g −1γ1g)m = (g−1γ1g)(g−1γ1g) · · · (g−1γ1g) = g−1γm1 g = g−1g = id. Hence γ2 has finite order and the order of γ2 divides m. Suppose γ2 has order r < m. Then the same reasoning as above shows that id = γr2 = g −1γr1g. Hence γ1 has order r, a contradiction. Hence γ2 has order m. (ii) Suppose that γ has order m. Then (γ−1)m = (γm)−1 = id. Hence γ−1 has finite order and the order of γ−1 divides m. Suppose γ−1 has order r < m. Then id = (γ−1)r = (γr)−1 so that γ has order r, a contradiction. Hence γ−1 has order m. 1 MATH32051 Solutions for Week 9 Solutions for Week 9 Solution 32.1 Let Γ = 〈a, b | a4 = b2 = (ab)2 = e〉. First note that e = a4 = a3a and e = b2 = bb so that a−1 = a3 and b−1 = b. Now e = (ab)2 = abab and multiplying on the left first by a−1 and then b−1 gives that ab = ba3. (Note that one cannot write (ab)2 6= a2b2.) From this it follows that a2b = a(ab) = a(ba3) = (ab)a3 = ba3a3 = ba6 = ba2a4 = ba2 and similarly a3b = a(a2b) = a(ba2) = (ab)a3 = ba2a3 = ba5 = ba. Now let w ∈ Γ be a finite word in Γ. Then w = an1bm1 · · · ankbmk for suitable integers nj ,mj. Using the relations a 4 = b2 = e we can assume that nj ∈ {0, 1, 2, 3} and mj ∈ {0, 1}. Using the relations we deduced above that ab = ba3, a2b = ba2 and a3b = ba, we can move all of the as to the left and all of the bs to the right to see that we can write w = anbm for suitable integers n,m. Again, as a4 = b2 = e we can assume that n ∈ {0, 1, 2, 3} and m ∈ {0, 1}. Hence there are exactly 8 elements in Γ. Solution 33.1 Label the sides and vertices of the quadrilateral as in Figure 63.1. Then γ2 γ1 A B CD s4 s1 s3 s2 Figure 63.1: A hyperbolic quadrilateral. ( A s1 ) γ2→ ( D s3 ) ∗→ ( D s4 ) γ1→ ( C s2 ) ∗→ ( C s3 ) γ−1 2→ ( B s1 ) ∗→ ( B s2 ) γ−1 1→ ( A s4 ) ∗→ ( A s1 ) . 1 MATH32051 Solutions for Week 9 Hence the elliptic cycle is A→ D → C → B and the corresponding elliptic cycle transfor- mation is γ−11 γ −1 2 γ1γ2. If we let ∠A denote the internal angle at A, with similar notation for the other vertices, then the angle sum is sum(A) = ∠A+ ∠B + ∠C +∠D. By Poincare´’s Theorem, γ1, γ2 generate a Fuchsian group if and only if m(∠A+ ∠B + ∠C + ∠D) = 2π for some integer m ≥ 1. 2 MATH32051 Solutions for Week 10 Solutions for Week 10 Solution 35.1 Label the sides as in Figure 64.1. Then γ2 γ1 −1 0 1 s3 s2s1 s4 Figure 64.1: A fundamental domain for the free group on 2 generators. ( −1 s1 ) γ2→ ( 1 s2 ) ∗→ ( 1 s4 ) γ−1 1→ ( −1 s3 ) ∗→ ( −1 s1 ) , and ( ∞ s3 ) γ1→ ( ∞ s4 ) ∗→ ( ∞ s3 ) , and ( 0 s1 ) γ2→ ( 0 s2 ) ∗→ ( 0 s1 ) . Hence there are 3 vertex cycles: −1 → 1, ∞ and 0. The corresponding parabolic cycles are: γ−11 γ2, γ1 and γ2, respectively. −1→ 1 with corresponding parabolic cycle transformation γ−11 γ2, ∞ with corresponding parabolic cycle transformation γ1, 0 with corresponding parabolic cycle transformation γ2. Clearly γ1 is parabolic (it is a translation and so has a single fixed point at ∞). The map γ2 is parabolic; it is normalised and has trace τ(γ2) = (1 + 1) 2 = 4. Finally, the map γ−11 γ2 is given by: γ−11 γ2(z) = γ −1 1 ( z 2z + 1 ) = z 2z + 1 − 2 = −3z − 2 2z + 1 1 MATH32051 Solutions for Week 10 which is normalised; hence τ(γ−11 γ2) = (−3 + 1)2 = 4 so that γ−11 γ2 is parabolic. By Poincare´’s Theorem, as all parabolic cycle transformations are parabolic (and there are no elliptic cycles), the group Γ generated by γ1, γ2 is a Fuchsian group. As there are no elliptic cycles, there are no relations. Hence the group is isomorphic to 〈a, b〉 (take a = γ1, b = γ2), which is the free group on 2 generators. Solution 35.2 (i) The side-pairing transformation γ1 is a translation that clearly maps the side Re(z) = −(1+√2/2) to the side Re(z) = 1+√2/2. Hence γ1 is a side-pairing transformation. Recall that through any two points of H∪∂H there exists a unique geodesic. The map γ2 maps the point i √ 2/2 to itself and the point −(1 +√2/2) to 1 +√2/2. Hence γ2 maps the arc of geodesic [A,B] to [C,B]. Hence γ2 is a side-pairing transformation. (ii) Let s1 denote the side [B,A], s2 denote the side [B,C], s3 denote the side [C,∞] and s4 denote the side [A,∞]. Now ( B s1 ) γ2→ ( B s2 ) ∗→ ( B s1 ) . Hence we have an elliptic cycle E = B with elliptic cycle transformation γ2 and corresponding angle sum sum(E) = ∠B = π/2. As 4π/2 = 2π, the elliptic cycle condition holds with mE = 4. Now consider the following parabolic cycle:( ∞ s4 ) γ1→ ( ∞ s3 ) ∗→ ( ∞ s4 ) . Hence we have a parabolic cycle P1 =∞ with parabolic cycle transformation γ1. As γ1 is a translation, it must be parabolic (recall that all parabolic Mo¨bius transformations of H are conjugate to a translation). Hence the parabolic cycle condition holds. Finally, we have the parabolic cycle:( A s4 ) γ1→ ( C s3 ) ∗→ ( C s2 ) γ−1 2→ ( A s1 ) ∗→ ( A s4 ) . Hence we have a parabolic cycle P2 = A → C with parabolic cycle transformation: γ−12 γ1. Now γ −1 2 γ1 has the matrix( √ 2/2 1/2 −1 √2/2 )( 1 2 + √ 2 0 1 ) = ( √ 2/2 √ 2 + 1 −1 −2−√2/2 ) which is normalised. Hence the trace of γ−12 γ1 is(√ 2 2 − 2− √ 2 2 )2 = 4. 2 MATH32051 Solutions for Week 10 Using the fact that a Mo¨bius transformation is parabolic if and only if it has trace 4, we see that γ−12 γ1 is parabolic. Hence the parabolic cycle condition holds. By Poincare´’s Theorem, γ1 and γ2 generate a Fuchsian group. In terms of generators and relations, it is given by 〈a, b | b4 = e〉. (Here we take a = γ1, b = γ2. The relation b 4 comes from the fact that the elliptic cycle E = B has elliptic cycle transformation γE = γ2 with angle sum π/2. Hence mE = 4. The relation γ mE E is then b 4.) Solution 36.1 (i) Label the sides, vertices and side-pairing transformations as illustrated. [PLACE-HOLDER] Then we have: ( 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 ) γ3→ ( v8 s7 ) ∗→ ( v8 s8 ) γ4→ ( v7 s6 ) ∗→ ( v7 s7 ) γ−1 3→ ( v6 s5 ) ∗→ ( v6 s6 ) γ−1 4→ ( v9 s8 ) ∗→ ( v9 s9 ) γ−1 4→ ( v1 s10 ) ∗→ ( v1 s1 ) . This gives elliptic cycle E1 : v1 → v4 → v3 → v2 → v5 → v8 → v7 → v6 → v9 and elliptic cycle transformation γ5γ −1 4 γ −1 3 γ4γ3γ −1 2 γ −1 1 γ2γ1. The angle sum is sum(E1) = 9π/9 = π. Hence the elliptic cycle condition holds with m1 = 2. We also have: ( v10 s9 ) γ5→ ( v10 s10 ) ∗→ ( v10 s9 ) . Hence we have elliptic cycle E2 : v10 with corresponding elliptic cycle transformation γ5. The angle sum is sum(E2) = π/9. Hence the elliptic cycle condition holds with m2 = 18. 3 MATH32051 Solutions for Week 10 (ii) By Poincare´’s Theorem, γ1, . . . , γ5 generate a Fuchsian group Γ. Let aj = γj , j = 1, . . . , 5 denote abstract symbols. (Note that here we choose not to use a, b, c, d, e as the abstract symbols due to the possibility of confusion for e to mean the identity.) Then Γ has presentation 〈a1, a2, a3, a4, a5 | (a5a−14 a−13 a4a3a−12 a−11 a2a2)2 = a185 = e〉. Solution 38.1 The decagon D gives a triangulation of H/Γ with V = 2 vertices (the number of elliptic cycles), E = 5 edges (the number of pairs of paired sides), and F = 1 faces (the number of polygons). Hence χ(H/Γ) = V −E + F = 2− 5 + 1 = −2. The genus is then 2− 2g = −2, so that g = 2. (This is also clear geometrically by looking at how the sides are paired.) There are two marked points on H/Γ, one of order 2 and one of order 18, corresponding to the two elliptic cycles constructed above. 4 MATH32051 Solutions for Week 11 Solutions for Week 11 Solution 39.1 (i) First note that one side of the polygon is paired with itself. Introduce a new vertex at the mid-point of this side, introducing two new sides each of which is paired with the other. Label the polygon as in Figure 65.1. 2π/7 θ2 2π/3 θ1θ3 A s1 B s2 s3 D E s4s5 s6 γ1 γ2 γ3 F C Figure 65.1: Labelling the hyperbolic polygon, remembering to add an extra vertex to the side that is paired with itself. Then ( B s1 ) γ1→ ( B s2 ) ∗→ ( B s1 ) . This gives an elliptic cycle E1 = B with elliptic cycle transformation γ1 and angle sum sum(E1) = π. Hence the elliptic cycle condition holds with m1 = 2. We also have ( D s3 ) γ2→ ( D s4 ) ∗→ ( D s3 ) . This gives an elliptic cycle E2 = D with elliptic cycle transformation γ2 and angle sum sum(E2) = 2π/3. Hence the elliptic cycle condition holds with m1 = 3. Also ( F s5 ) γ3→ ( F s6 ) ∗→ ( F s5 ) . This gives an elliptic cycle E3 = F with elliptic cycle transformation γ3 and angle sum sum(E3) = 2π/7. Hence the elliptic cycle condition holds with m1 = 7. 1 MATH32051 Solutions for Week 11 Finally ( A s1 ) γ1→ ( C s2 ) ∗→ ( C s3 ) γ2→ ( E s4 ) ∗→ ( E s5 ) γ3→ ( A s6 ) ∗→ ( A s1 ) . This gives an elliptic cycle E4 = A → C → E with elliptic cycle transformation γ3γ2γ1. The angle sum is sum(E3) = θ1 + θ2 + θ3 = 2π. Hence the elliptic cycle condition holds with m4 = 1. Hence E4 is an accidental cycle. (ii) By Poincare´’s Theorem, γ1, γ2, γ3 generate a Fuchsian group Γ. In terms of generators and relations we can write Γ = 〈a, b, c | a2 = b3 = c7 = abc = e〉. (iii) To calculate the genus of H/Γ we use Euler’s formula 2−2g = V −E+F . Recall that each elliptic cycle on the polygon glues together to give one vertex on a triangulation of H/Γ. As there are 4 elliptic cycles we have V = 4. Each pair of paired sides in the polygon glue together to give one edge on a triangulation of H/Γ. As there are 6 sides in the polygon, there are E = 6/2 = 3 edges in the trinagulation of H/Γ. As we are only using 1 polygon, there is F = 1 face of the triangulation of H/Γ. Hence 2− 2g = V −E + F = 4− 3 + 1 = 2, so that g = 0. As the orders of the non-accidental elliptic cycles are 2, 3, 7, we see that sig(Γ) = (0; 2, 3, 7). Solution 39.2 From Exercise 14.1, we know that there exists a regular hyperbolic n-gon with internal angle θ provided (n− 2)π − 8θ > 0. When n = 8, this rearranges to θ ∈ [0, 3π/4). Label the vertices of the octagon as indicated in Figure 65.2. We have ( 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 ) γ3→ ( v8 s7 ) ∗→ ( v8 s8 ) γ4→ ( v7 s6 ) ∗→ ( v7 s7 ) γ−1 3→ ( v6 s5 ) ∗→ ( v6 s6 ) 2 MATH32051 Solutions for Week 11 s1s8 v7 v8 v1 v2 v3 v4 v5 v6 s4 s3 s2s7 s6 s5 γ1 γ4 γ3 γ2 θ Figure 65.2: See the solution to Exercise 39.2. 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 θ, the angle sum is 8θ Hence the elliptic cycle condition holds whenever there exists an integer m = mE such that 8mθ = 2π, i.e. whenever θ = π/4m for some integer m. When m = 1 this is an accidental cycle. Let θ be such that θ = π/4m for some integer m. Then by Poincare´’s Theorem, the group Γπ/4m 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: Γπ/4m = 〈γ1, γ2, γ3, γ4 | (γ−14 γ−13 γ4γ3γ−12 γ−11 γ2γ1)m = e〉. The quotient space H/Γπ/4m is a torus of genus 2. When m = 1, sig(Γπ/4) = (2,−) and H/Γπ/4 has no marked points. When m ≥ 2 then sig(Γπ/4) = (2,m) and H/Γπ/4m has one marked point of order m. Solution 40.1 (i) Consider the Dirichlet polygon and side-pairing transformations for the modular group that we constructed in Section 27. See Figure 65.3. The sides s1 and s2 are paired. This gives one cusp at the point ∞. There are two elliptic cycles: A→ B (which has an angle sum of 2π/3), and i (which has an angle sum of π). Hence when we glue together the vertices A and B we get a marked point of order 3, and the vertex i gives a marked point of order 2. We do not get any ‘holes’ when we glue together the sides. Hence we have genus 0. Thus the modular group has signature (0; 2, 3; 1). 3 MATH32051 Solutions for Week 11 A B s2s1 s3 s4 i Figure 65.3: A fundamental domain and side-pairing transformations for the modular group. (ii) By Proposition 23.2.1 it is sufficient to prove that the formula holds for a Dirichlet polygon D. Suppose that D has n vertices (hence n sides). We use the Gauss-Bonnet Theorem (Theorem 13.2.1). By Proposition 30.3.1, the angle sum along the jth non-accidental elliptic cycle Ej is sum(Ej) = 2π mj . Hence the sum of the interior angles of vertices on non-accidental elliptic cycles is r∑ j=1 2π mj . Suppose that there are s accidental cycles. (Recall that a 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. Suppose that there are c parabolic cycles. The angle sum along a parabolic cycle must be zero (the vertices must be on the boundary, and the angle between two geodesics that intersect on the boundary must be zero). As each vertex belongs to either a non-accidental elliptic cycle, to an accidental cycle or to a parabolic cycle, the sum of all the internal angles of D is given by 2π r∑ j=1 1 mj + s . By the Gauss-Bonnet Theorem, we have AreaH(D) = (n− 2)π − 2π r∑ j=1 1 mj + s . (0.1) 4 MATH32051 Solutions for Week 11 Consider now the space H/Γ. This is formed by taking D and glueing together paired sides. The vertices along each elliptic cycle, accidental cycle and parabolic cycle are glued together to form a vertex in H/Γ. Hence the number of vertices in H/Γ is equal to the number of cycles (elliptic, accidental and parabolic); hence D corresponds to a triangulation of H/Γ with V = r+ s+ c vertices. As paired sides are glued together, there are E = n/2 edges. Finally, as we only need the single polygon D, there is only F = 1 face. Hence 2− 2g = χ(H/Γ) = r + s+ c− n 2 + 1 which rearranges to give n− 2 = 2((r + s+ c)− (2− 2g)). (0.2) Substituting (0.2) into (0.1) we see that AreaH(D) = 2π r + s+ c− (2− 2g) − r∑ j=1 1 mj − s = 2π (2g − 2) + r∑ j=1 ( 1− 1 mj ) + c . (iii) We must show that (2g − 2) + r∑ j=1 ( 1− 1 mj ) + c ≥ 1 6 . (0.3) We assume that c ≥ 1. If g ≥ 1 then 2g − 2 + c ≥ 1 > 1/6, so that (0.3) holds. So it remains to check the cases when g = 0. If g = 0 and c ≥ 2 then 2g − 2 + c ≥ 0. As 1 − 1/mj ≥ 1/2, it follows that the left-hand side of (0.3) is at least 1/2. Hence (0.3) holds. So it remains to check the cases when g = 0 and c = 1. If g = 0 and c = 1 then 2g − 2 + c = −1. As mj ≥ 2, we see that 1 − 1/mj ≥ 1/2. Hence if r ≥ 3 then the left-hand side of (0.3) is at least 1/2. Hence (0.3) holds. It remains to check that case when g = 0, c = 1 and r = 2. In this case, it remains to check that s(k, l) = 1− 1 k − 1 l ≥ 1 6 (letting k = m1, l = m2). We may assume that k ≤ l. Now s(3, 3) = 1/3 > 1/6 and s(3, l) ≥ 1/3 for l ≥ 3. Hence we may assume that k = 2. Then s(2, 2) = 0, s(2, 3) = 1/6 and s(2, l) > 1/6. Hence the minimum is achieved for k = 2, l = 3. Hence the minimum is achieved for a Fuchsian group with signature (0; 2, 3; 1). By part (i), this is the signature of the modular group. Solution 42.1 We have κ(x) = lim r→0 12 π ( πr2 −Area(B(x, r)) r4 ) 5 MATH32051 Solutions for Week 11 where B(x, r) = {y ∈ S | d(x, y) < r}. In the Euclidean case Area(B(x, r)) = πr2. Clearly κ(x) = 0 for all x ∈ R2. In the hyperbolic case, we saw in Exercise 12.5 that Area(B(x, r)) = 4π sinh2 r/2. Recall that sinh2 r has Taylor series r + r3/3 + r5/5 + · · ·. Hence 4π sinh2 r/2 has Taylor series 4π ( r 2 + r3 233! + r5 255! + · · · )( r 2 + r3 233! + r5 255! + · · · ) = 4π ( r2 4 + r4 233! +O(r6) ) = πr2 + π r4 2 · 3! +O(r 6). Hence κ(x) = lim r→0 12 π ( πr2 − πr2 − π ( r4 2·3! ) +O(r6) ) r4 = lim r→0 −r4 +O(r6) r4 = −1. 6 欢迎咨询51作业君