We denote this curve byγ =(γ1,γ2):[0, 1] →H. So d(P,l1)=
1
0
γ1(t)2
+
γ2(t)21/2
dt/γ2(t)
≥ 1
0
γ1(t)dt/(a+r)≥ 1
0
γ1(t)dt/(a+r)≥(a−r)/(a+r),
since the difference in the x-coordinate is at least a−r. This bound is however independent of the pointPonl2, and hence
d(l1,l2)≥(a−r)/(a+r) >0.
5.7 Hyperboloid model of the hyperbolic plane
There is another basic model of the hyperbolic plane which should be mentioned, namely the hyperboloid model. This model is entirely analogous to our study of the sphere as an embedded surface in R3, but with a few signs in inner-products changed. One advantage of this model is that one can prove formulae which occur in hyperbolic trigonometry with a minimum of calculation, much as we did when proving the spherical versions. If one wants to prove these formulae without using the hyperboloid model, it will involve the manipulation of formulae — a good place to start would be the formula in Exercise5.7.
Let us take a Lorentzian inner-product, onR3, corresponding to the matrix
⎛
⎝1 0 0
0 1 0
0 0 −1
⎞
⎠.
Setq(x)= x,x =x2+y2−z2and letSbe the surfaceq(x)= −1, the hyperboloid of two sheetsx2+y2=z2−1. LetS+be the upper-sheet of the hyperboloid (i.e.
z>0). Now consider the stereographic projection mapπfrom the point(0, 0,−1), mappingS+onto the unit discD, given by
π(x,y,z)=x+iy 1+z. Then
u+iv= x+iy 1+z
=⇒ u2+v2= 1−z2
(1+z)2 = 1−z 1+z.
S+
D
C
(0,0,–1)
So ifr=u2+v2,
z=1+r2
1−r2 =⇒ 1+z= 2
1−r2. Therefore
x=(1+z)u= 2u
1−r2, y= 2v 1−r2. Now
∂x
∂u = ∂
∂u
2u
1−u2−v2 = 2
(1−r2)2(1+u2−v2),
∂y
∂u = 4uv (1−r2)2,
∂z
∂u = ∂(1+z)
∂u = 4u
(1−r2)2. Setting
σ(u,v)= 2u
1−r2, 2v
1−r2,1+r2 1−r2 ,
5.7 HYPERBOLOID MODEL OF THE HYPERBOLIC PLANE 109
we have
σu:=∂σ
∂u = 2
(1−r2)2
1+u2−v2, 2uv, 2u , σv :=∂σ
∂v = 2
(1−r2)2
2uv, 1+v2−u2, 2v , the second equation following from the first by symmetry.
Thetangent spacetoS+at a pointa ∈ S+ will by definition consist of vectors x such that, for t small, a+tx,a+tx = −1+O(t2), i.e. with a,x = 0.
However, withσas above, we haveσ (u,v),σ(u,v) = −1, and so, differentiating with respect touandv, we see that bothσuandσvare in the tangent space atσ (u,v).
A routine check verifies that they are always linearly independent foru2+v2<1, and soσu,σvform a basis for the tangent space toS+atσ(u,v). Moreover,, determines a symmetric bilinear form on this vector space, and hence viadσ (which identifiese1withσuande2withσv) corresponds to a symmetric bilinear form onR2. This bilinear form may be written asEdu2+2Fdu dv+Gdv2, where
E= σu,σu = 4 (1−r2)4
1+u2−v22
+4u2v2−4u2
= 4 (1−r2)4
1−
u2+v22
= 4 (1−r2)2, F = σu,σv =0 by inspection, and
G= σv,σv = 4
(1−r2)2 by symmetry.
This then is just the hyperbolic metric on the Poincaré disc model of the hyperbolic plane.
We now look for isometries of the hyperboloid model. LetO(2, 1)denote the group of 3×3 matrices which preserve the above Lorentzian inner-product; these are just the matricesPfor which
Pt
⎛
⎝1 0 0
0 1 0
0 0 −1
⎞
⎠P=
⎛
⎝1 0 0
0 1 0
0 0 −1
⎞
⎠.
In particular, the action ofO(2, 1)onR3preserves the hyperboloid. We however want the elements which send the upper-sheetS+of the hyperboloid to itself and do not switch the sheets. This further condition defines an index two subgroupO+(2, 1)of O(2, 1). Note that
⎛
⎝1 0 0
0 1 0
0 0 −1
⎞
⎠
is inO(2, 1)but not inO+(2, 1). IfPis inO(2, 1), then by taking determinants of the defining equation, we deduce that det(P)2=1, that is det(P)= ±1.
Since the inner-products on the tangent spaces are determined by the Loretzian inner-product on R3, we deduce furthermore that any element of O+(2, 1) will preserve the metric. It will moreover preserve the lengths of curves onS+in the given metric, where for a given curveγ (t)=(γ1(t),γ2(t),γ3(t))withγ :[0, 1] →S+, its length is defined to be
1
0 ˙γ (t),γ (t)˙ dt= 1
0 (γ˙12+ ˙γ22− ˙γ32)1/2dt.
There are two types of element ofO+(2, 1)which will be of particular interest to us.
The first are just the rotations about thez-axis, that is matrices of the following form:
⎛
⎝cosθ −sinθ 0 sinθ cosθ 0
0 0 1
⎞
⎠
for 0 ≤ θ <2π. This matrix is clearly inO(2, 1), and, as any such matrix may be joined to the identity matrixIby a continuous curve of such matrices inO(2, 1), the matrix is inO+(2, 1).
The second type of element which is of interest to us is those of the form:
⎛
⎝1 0 0 0 coshd −sinhd 0 −sinhd coshd
⎞
⎠
ford ≥ 0. Again, this matrix is readily checked to be inO(2, 1), and as any such matrix may be joined to the identity matrixIby a continuous curve of such matrices inO(2, 1), the matrix is inO+(2, 1). This matrixPhas the useful property that
P
⎛
⎝ 0 sinhd coshd
⎞
⎠ =
⎛
⎝0 0 1
⎞
⎠.
With the aid of elements of the above two types, we see that the action ofO+(2, 1) onS+is transitive. Given an arbitrary vector(x,y,z)t ∈S+, we can first rotate it to a vector of the form(0, sinhd, coshd)tby means of an element of the first type, and then send this to to(0, 0, 1)tby means of an element of the second type.
We note, for later use, that matrices of the above two types have determinant+1;
one can in fact show that elements of these two types generateO+(2, 1)(cf. the proof of Theorem2.19), and so det(P) =1 for all P ∈ O+(2, 1)(we shall not however need this latter fact).
Given two arbitrary points, with the aid of elements of the above two types, we may send the first point to(0, 0, 1), and the second point to(0, sinhd, coshd)for
5.7 HYPERBOLOID MODEL OF THE HYPERBOLIC PLANE 111 somed >0. Under the stereographic projection map, these correspond to the points 0 andisinhd/(1+coshd)=itanh(d/2)in the disc modelD. Since distances on S+ correspond to distances inD, we see that the two points are distance d apart.
Moreover, we note that the Lorentzian inner product of the two vectors is−coshd. We deduce therefore that for any two vectorsxandyrepresenting points onS+, we havex,y = −coshd, wheredis the hyperbolic distance between the two points.
We are now in a position to prove the hyperbolic cosine formula. It is possible to prove this in exactly the same way as we proved the spherical cosine formula, by defining an appropriate cross-product of vectors inR3associated to the Lorentzian inner-product defined above, but it is probably clearer to use the above isometries.
Proposition 5.27 (Hyperbolic cosine formula) Letbe a hyperbolic triangle, with anglesα,β,γ, and sides of length a,b,c (the side of length a being opposite the vertex with angleα, and similarly for b and c). Then
cosha=coshb coshc − sinhb sinhc cosα.
Proof We work on the hyperboloid model, where we may assume that the vertices of the triangle correspond to pointsA=(0, 0, 1),B=(0, sinhc, coshc), and
C=(sinαsinhb, cosαsinhb, coshb)
on S+. If the metric onS+is denoted byd, thend(A,B) = c,d(A,C) = b and d(B,C)=a. From the above discussion,−coshais just the Lorentzian inner-product of the position vectorsBandC, namely
cosα sinhb sinhc−coshb coshc.
Proposition 5.28 (Hyperbolic sine formula) With the notation as above,
sinha/sinα=sinhb/sinβ =sinhc/sinγ.
Proof Given pointsA,B,C, we can consider the matrixM(A,B,C)whose columns are the position vectorsA,BandC, and take its determinant. Clearly, this is invariant when we operate onR3by elements of the above two types, as the determinant of such matrices was seen to be one. We may therefore reduce down to the pointsA,B,Cbeing in the particularly simple form assumed in the proof of the previous result. With these three points, the determinant ofM(A,B,C)is checked to be−sinhcsinhbsinα, and this was therefore the number obtained from our original three points. The determinant of the matrix is however invariant under cyclic permutations of the points, from which we deduce
sinhc sinhb sinα=sinha sinhc sinβ =sinhb sinha sinγ.
Dividing through by sinha sinhb sinhcgives the result required.
Exercises
5.1 Verify that the Mửbius transformationζ →i(1+ζ )/(1−ζ)from the disc model of the hyperbolic plane to the upper half-plane model may be defined by an element of SU(2).
5.2 Suppose thatz1,z2 are points in the upper half-plane, and suppose the hyperbolic line throughz1 andz2meets the real axis at pointsz1∗andz∗2, wherez1 lies on the hyperbolic line segment[z1∗,z2], and where one ofz∗1andz∗2might be∞. Show that the hyperbolic distanceρ(z1,z2)=logr, whereris the cross-ratio of the four points z∗1,z1,z2,z2∗, taken in an appropriate order.
5.3 If a is a point of the upper half-plane, show that the Mửbius transformation g given by
g(z)= z−a z− ¯a
defines an isometry from the upper half-plane modelH to the disc modelDof the hyperbolic plane, sendinga to zero. Deduce that for pointsz1,z2in the upper half- plane, the hyperbolic distance is given byρ(z1,z2)=2 tanh−1z1−z2
z1−¯z2.
5.4 Letldenote the hyperbolic line inH given by a semicircle with centrea ∈ Rand radiusr>0. Show that the reflectionRlis given by the formula
Rl(z)=a+ r2
¯ z−a.
5.5 Given two hyperbolic lines meeting at a point, show that the locus of points equidistant from the two lines forms two further hyperbolic lines through the point. Show that in a hyperbolic triangle, none of whose vertices are at infinity, the angle bisectors are concurrent.
5.6 Show that any isometryg of the disc modelDfor the hyperbolic plane iseitherof the form (for somea∈Dand 0≤θ <2π)
g(z)=eiθ z−a 1− ¯az, orof the form
g(z)=eiθ z¯−a 1− ¯az¯. 5.7 For arbitrary pointsz,winC, prove the identity
|1− ¯zw|2= |z−w|2+(1− |z|2)(1− |w|2).
EXERCISES 113
Given pointsz,win the unit disc model of the hyperbolic plane, prove the identity sinh2(ρ(z,w)/2)= |z−w|2
(1− |z|2)(1− |w|2), whereρdenotes the hyperbolic distance.
5.8 LetAbe the sector in the unit disc model of the hyperbolic plane given by 0≤θ≤α for someα < π. Show thatAisconvex, i.e. for any pointsP,Q∈A, the hyperbolic line segmentPQlies entirely inA. IfBis a vertical strip in the upper half-plane model of the hyperbolic plane, show thatBis convex. Deduce that any hyperbolic triangle is the intersection of three suitable convex subsets of the hyperbolic plane, and is therefore itself a convex subset.
5.9 LetT be a hyperbolic triangle; show that the radius of any inscribed hyperbolic circle is less than cosh−1(3/2). Generalize this result to hyperbolic polygons.
5.10 Ifαandβare positive numbers withα+β < π, show that there exists a hyperbolic triangle (one vertex at infinity) with angles 0,αandβ. For any positive numbersα, βandγ, withα+β+γ < π, show that there exists a hyperbolic triangle with these angles. [Hint: For the second part, you may need a continuity argument.]
5.11 Show that two distinct hyperbolic lines have a common perpendicular if and only if they are ultraparallel, and that in this case the perpendicular is unique. By taking this common perpendicular to be the imaginary axis, find a simple proof for Proposition5.26.
5.12 Show that the composite of two reflections in distinct hyperbolic lines has finite order if and only if the lines meet at a point in (the interior of) the hyperbolic plane with angle which is a rational multiple of π. [Hint: For the ultraparallel case, use the argument from the previous exercise.]
5.13 Fix a pointPon the boundary ofD, the disc model of the hyperbolic plane. Give a description of the curves inDthat are orthogonal to every hyperbolic line that passes throughP.
5.14 If two hyperbolic triangles have the same side-lengths, including the degenerate cases of some sides being infinite, prove that the triangles are congruent, i.e. there is an isometry of the hyperbolic plane sending one onto the other. Using the argument from Exercise5.10, or otherwise, prove that the same holds if they have the same angles (also including the degenerate cases of some angles being zero).
5.15 IfABCis a hyperbolic triangle, with the angle atAat leastπ/2, show that the side BC has maximum length. Given pointsz1,z2in the hyperbolic plane, letwbe any point of the hyperbolic line segment joiningz1toz2, andwbe any point not on the hyperbolic line passing through the other three. Show that
ρ(w,w)≤max{ρ(w,z1),ρ(w,z2)}.
Deduce that the diameter of a hyperbolic triangle (that is, sup{ρ(P,Q): P,Q ∈ }) is equal to the length of its longest side. Show that the corresponding result holds for Euclidean triangles, but does not hold for spherical triangles.
5.16 Given two distinct hyperbolic circles of radiusρ, show that there is a hyperbolic circle of radius strictly less thanρwhich contains their intersection. (Hint: After applying an appropriate isometry, one may assume that the hyperbolic circles have centres
−a+ibanda+ibinH, fora,b>0.)
Deduce that, for any finite set of points in the hyperbolic plane, there is a unique hyperbolic circle of minimum radius which encloses them (some will of course lie on the circle). LetGbe a finite subgroup ofPSL(2,R); show that the action ofGon the upper half-plane has a fixed point, and deduce thatGis a cyclic group. Prove that any finite subgroup of Isom(H)is either cyclic or dihedral.
6 Smooth embedded surfaces
We now move on from the classical geometries, as described in previous chapters, to rather more general two-dimensional geometries. In this chapter, we study the concrete case of smooth surfaces embedded inR3; any such surface has a metric, defined in terms of infima of lengths of curves on the embedded surface, as was the case for the sphere. In Chapter8, we generalize further and introduce the concept of abstract smooth surfaces, equipped with a Riemannian metric. These will represent a common generalization of both the embedded surfaces (as studied in this chapter), and general Riemannian metrics on open subsets ofR2(as studied in Chapter4). In the remainder of this book, we shall develop the theory of curvature and geodesics in this more general context, thus providing an introduction to two central concepts from elementary differential geometry, and culminating in a proof of the Gauss–Bonnet theorem for arbitrary compact smooth surfaces.