Consider the groupPSL(2,R)of Mửbius transformations z→ az+b
cz+d witha,b,c,d ∈ Rand deta b
c d
= 1. It is easy to check that these are precisely the Mửbius transformations ofC∞which sendR∪ {∞}toR∪ {∞}and sendH to H. A Mửbius transformation withrealcoefficients may always be represented by a realmatrix with determinant±1; the condition that the determinant is positive is just saying that the upper half-plane is sent to itself (and not to the lower half-plane).
Proposition 5.2 The elements of PSL(2,R) are isometries on H , and hence preserve the lengths of curves.
Proof Recall thatPSL(2,R)is generated by the elements
z→z+a (a∈R) (translations)
z→az (a∈R+) (dilations)
z→ −1 z
Claim Each of these preserves the metric|dz|2/y2.
The first two are clear — we check the third. Setw= −1/z.
Sincedw/dz=1/z2, the induced map onC=R2is given as multiplication by 1/z2. Under this linear map, the Euclidean metric|dw|2atwcorresponds to|dz|2/|z|4 atz. Note however that
Im(w)=Im(−1/z)= − 1
|z|2Im¯z= Imz
|z|2. Thus
|dw|2
Im(w)2 = |dz|2/|z|4
Im(z)2/|z|4 = |dz|2 (Imz)2,
as required, and the map is an isometry.
5.2 GEOMETRY OF THE UPPER HALF-PLANE MODEL H 93 We shall see later (Remark5.17) thatPSL(2,R)is in fact an index two subgroup of the full isometry group.
Remark 5.3 Since PSL(2,R)contains the Mửbius transformations of the form z→az+b, witha>0, it acts transitively onH, i.e. for any pointsz1,z2∈H, there existsg∈PSL(2,R)withg(z1)=z2.
Recall from Chapter2that any Mửbius transformation onCsends circles and straight lines to circles and straight lines. Since it is analytic, it also preserves angles, by the argument in Section4.1. So ifLis the imaginary axis, andg ∈PSL(2,R), we deduce thatg(L)is a circle or a straight line, orthogonal to the real axis, sincegsendsR∪{∞}
to itself.
Therefore, ifL+ := {it : t > 0}, theng(L+)is either a vertical half-line, or a semicircle (whose ends are on the real axis). We call these thehyperbolic linesinH. We note that two distinct hyperbolic lines meet in at most one point.
Lemma 5.4 Through any two points z1,z2∈ H , there exists a unique hyperbolic line l.
Proof This is clearly true if Rez1=Rez2. Suppose then Rez1=Rez2. We can locate the centre of the required semicircle by the construction as shown of a perpendicular bisector, and hencelis uniquely determined.
z1
z2
l
Lemma 5.5 PSL(2,R)acts transitively on the set of hyperbolic lines.
Proof We show that for any hyperbolic linel, there existsg ∈PSL(2,R)such that g(l)=L+, from which the general statement follows.
This is clear iflis a vertical line. Ifl is a semicircle with end-pointss<t ∈R, then we take
g(z)= z−t z−s
(noting that det1−t
1−s
=t−s>0).
s
l
t
We observe thatg(t)=0,g(s)= ∞and sog(l)=L+. Remark 5.6 If, in the proof of Lemma5.5, we composegwithz→ −1z, we obtain a Mửbius transformationhwithh(s)=0,h(t)= ∞. We can also ensure (scaling by a real number) that a given pointP ∈ lgoes toi, say. Thus PSL(2,R)in fact acts transitively on pairs(l,P)consisting of a hyperbolic lineland a pointP∈l.
Definition 5.7 We now consider the metric defined by the Riemannian metric onH, where the distance between any two points is the infimum of the lengths of piecewise smooth curves joining the points; we shall call this thehyperbolic distanceρ. So Proposition5.2implies thatPSL(2,R)preserves hyperbolic distance.
Given pointsz1,z2∈ H, there is a unique hyperbolic line throughz1andz2; we let z∗1,z2∗be as shown in the diagram (possibly withz2∗= ∞, if the hyperbolic line is a vertical half-line).
z2 z1
z2* z1*
As we argued above, there exists an elementh ∈ PSL(2,R)withh(z1∗) = 0 and h(z∗2) = ∞, which therefore sends the hyperbolic line through z1 and z2 to the positive imaginary axisL+. Therefore,h(z1)=iu,h(z2)=iv withu <v. Sinceh preserves distances,ρ(z1,z2)=ρ(iu,iv).
Let us therefore consider the case when the points are of the formz1=iu,z2 = iv, with u < v. Supposeτ : [0, 1] → H is a piecewise smooth curve such that τ(t)=i f(t)∈L+for allt, withτ(0)=iu,τ(1)=iv.
We say that a hyperbolic line segment γ : [0, 1] → H is monotonically parametrized ifρ(γ (0),γ (t))is a monotonic function oft. This property is clearly preserved when we take the imageh◦γofγ under an elementhofPSL(2,R). In the specific case under consideration,τ being monotonically parametrized is equivalent to the (piecewise smooth) functionf being monotonic increasing. By the Mean Value theorem, this is equivalent to the condition thatf(t)≥ 0, whenever the derivative exists and is continuous.
5.2 GEOMETRY OF THE UPPER HALF-PLANE MODEL H 95
Suppose now thatτ is monotonically parametrized; then
lengthτ = 1
0
|df/dt|
f dt
= 1
0
df/dt f dt
=logv u. We claim that this isρ(z1,z2).
Proposition 5.8 Suppose z1, z2are points of H , andγ :[0, 1] →H is a piecewise smooth curve from z1to z2. Thenlengthγ ≥ρ(z1,z2), with equality if and only ifγ is a monotonic parametrization of the hyperbolic line segment[z1,z2].
Proof As argued above, we may reduce to the case whenz1 = iu,z2 = iv, with u < v. Supposeγ = γ1+iγ2 : [0, 1] → H is a piecewise smooth curve with γ (0)=iu,γ (1)=iv. Then
lengthγ = 1
0
dγ1
dt
2
+ dγ2
dt
21/2
dt γ2(t)
≥ 1
0
dγ2
dt dt
γ2(t)
≥ 1
0
dγ2/dt γ2
dt
= [logγ2]10 = logv u,
with equality if and only if ddtγ1 = 0 and ddtγ2 ≥ 0 wherever they exist and are continuous, i.e. if and only ifγ1≡0 andγ2is monotonic.
Thus the hyperbolic distance between two pointsz1,z2 ∈ H is just the length of the unique hyperbolic line segment[z1,z2]between them. Moreover, if a hyperbolic line segmentγ : [0, 1] →H is monotonically parametrized, thenρ(γ (0),γ (t)) = lengthγ|[0,t]for allt∈ [0, 1].
Remark 5.9 For a general continuous curve γ : [0, 1] → H withγ (0)=z1, γ (1) = z2, we can define its length (when it exists) by taking dissections D : 0=t0 <t1< ã ã ã<tN =1; we setPi =γ (ti)and˜sD =N
i=1ρ(Pi−1,Pi), and we define lengthγ = supD˜sD. It is now a formal consequence of the triangle inequality (exactly as in the proof of Proposition 2.10) that lengthγ ≥ ρ(z1,z2), with equality if and only ifγ is a monotonic parametrization of the hyperbolic line segment[z1,z2].