2. Length and distance in hyperbolic geometry 2.1 The upper half-plane There are several different ways of constructing hyperbolic geometry. These different constructions are called models. In this lecture 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 R 2 with C, by noting that the point (x, y) R 2 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 imaginary 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 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 L z 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 2.1. 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. Before we can define distances in H we need to recall how to calculate path integrals in C (equivalently, in R 2 ). 1
i π(z) z R L z Figure 2.1: Stereographic projection. Notice how as z approaches i, the image π(z) gets large; this motivates defining π(i) =. 2.2 Path integrals By a path σ in the complex plane C, we mean the image of a (differentiable) function σ( ) : [a, b] C, where [a, b] R is an interval. 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 σ 1 : [0, 1] C by σ 1 (t) = t+it and define σ 2 : [0, 1] C by σ 2 (t) = t 2 + it 2. Then σ 1 and σ 2 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: b f = f(σ(t)) σ (t) dt; (2.1) σ a here denotes the usual modulus of a complex number, in this case, σ (t) = (Re σ (t)) 2 + (Im σ (t)) 2. Remark. To calculate the integral of f along the path σ we have to choose a parametrisation 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 parametrisations of a given path will always give the same answer. For this reason, we shall sometimes identify a path with its parametrisation. Exercise 2.1 Consider the two parametrisations σ 1 : [0, 2] H : t t + i, 2
σ 2 : [1, 2] H : t (t 2 t) + i. Verify that these two parametrisations define the same path σ. Let f(z) = 1/ Im(z). Calculate σ f using both of these parametrisations. The point of this exercise is to show that we can often simplify calculating the integral σ f of a function f along a path σ by choosing a good parametrisation. So far we have only defined how to integrate along differentiable paths, that is we have assumed that σ(t) is a differentiable function of t. 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 differentiable if σ is continuous and is differentiable except at finitely many points. (Roughly speaking this means that we allow the possibility that the curves has finitely many corners.) For example, the path σ(t) = (t, t ), 1 t 1 is piecewise differentiable: it is differentiable everywhere except at the origin, where it has a corner. To define σ f for a piecewise 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. 2.3 Distance in hyperbolic geometry We are now is a position to define the hyperbolic metric in the upper halfplane model of hyperbolic space. To do this, we first define the length of an arbitrary piecewise 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. 1 b Im(z) = σ (t) a Im(σ(t)) dt. Examples. σ 1. Consider the path σ(t) = a 1 +t(a 2 a 1 )+ib, 0 t 1 between a 1 +ib and a 2 + ib. Then σ (t) = a 2 a 1 and Im(σ(t)) = b. Hence 1 0 a 2 a 1 b dt = a 2 a 1. 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
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 { (2t 2) + i(1 + t), 0 t 1, σ(t) = (2t 2) + i(3 t), 1 t 2. Hence and Hence σ (t) = Im(σ(t)) = { 2 + i = 5, 0 t 1, 2 i = 5, 1 t 2, { 1 + t, 0 t 1, 3 t, 1 t 2. 1 5 2 5 0 1 + t dt + 1 3 t dt = 5 log(1 + t) 1 5 log(3 t) 0 2 1 = 2 5 log 2, which is about 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 2.2: The first path has hyperbolic length 4, the second path has hyperblic length 3.1 Exercise 2.2 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. 4
(ii) Show that log 1/a. (Notice that as a 0, we have that log 1/a. This motivates why we call R { } the circle at infinity.) 2.4 Hyperbolic distance We are now in a position to define the hyperbolic distance between two points on H. Definition. Let z, z H. We define the hyperbolic distance d(z, z ) between z and z to be d(z, z ) = inf{length H (σ) σ is a piecewise differentiable path with end-points z and z }. Remark. Thus we consider all piecewise 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 some path (a geodesic), and that this path is unique. Remark. Sometimes we write d H for d if we want to emphasise the dependence on H. Exercise 2.3 Show that d H satisfies the triangle inequality: d H (x, z) d H (x, y) + d H (y, z), x, y, z H. That is, the distance between two points is increased if one goes via a third point. 5