4Hyperbolic geometry
IB Geometry
4.3 Two models for the hyperbolic plane
That’s enough preparation. We can start talking about hyperbolic plane. We
will in fact provide two models of the hyperbolic plane. Each model has its own
strengths, and often proving something is significantly easier in one model than
the other.
We start with the disk model.
Definition (Poincar´e disk model). The (Poincar´e) disk model for the hyperbolic
plane is given by the unit disk
D ⊆ C
∼
=
R
2
, D = {ζ ∈ C : |ζ| < 1},
and a Riemannian metric on this disk given by
4(du
2
+ dv
2
)
(1 − u
2
− v
2
)
2
=
4|dζ|
2
(1 − |ζ|
2
)
2
, (∗)
where ζ = u + iv.
Note that this is similar to our previous metric for the sphere, but we have
1 − u
2
− v
2
instead of 1 + u
2
+ v
2
.
To interpret the term
|
d
ζ|
2
, we can either formally set
|
d
ζ|
2
= d
u
2
+ d
v
2
, or
interpret it as the derivative dζ = du + idv : C → C.
We see that (
∗
) is a scaling of the standard Riemannian metric by a factor
depending on the polar radius
r
=
|ζ|
2
. The distances are scaled by
2
1−r
2
, while
the areas are scaled by
4
(1−r
2
)
2
. Note, however, that the angles in the hyperbolic
disk are the same as that in
R
2
. This is in general true for metrics that are just
scaled versions of the Euclidean metric (exercise).
Alternatively, we can define the hyperbolic plane with the upper-half plane.
Definition (Upper half-plane). The upper half-plane is
H = {z ∈ C : Im(z) > 0}.
What is the appropriate Riemannian metric to put on the upper half plane?
We know D bijects to H via the M¨obius transformation
ϕ : ζ ∈ D 7→ i
1 + ζ
1 − ζ
∈ H.
This bijection is in fact a conformal equivalence, as defined in IB Complex
Analysis/Methods. The idea is to pick a metric on
H
such that this map is
an isometry. Then
H
together with this Riemannian metric will be the upper
half-plane model for the hyperbolic plane.
To avoid confusion, we reserve the letter
z
for points
z ∈ H
, with
z
=
x
+
iy
,
while we use ζ for points ζ ∈ D, and write ζ = u + iv. Then we have
z = i
1 + ζ
1 − ζ
, ζ =
z − i
z + i
.
Instead of trying to convert the Riemannian metric on
D
to
H
, which would be
a complete algebraic horror, we first try converting the Euclidean metric. The
Euclidean metric on R
2
= C is given by
hw
1
, w
2
i = Re(w
1
w
2
) =
1
2
(w
1
¯w
2
+ ¯w
1
w
2
).
So if
h·, ·i
eucl
is the Euclidean metric at
ζ
, then at
z
such that
ζ
=
z−i
z+i
, we
require (by definition of isometry)
hw, vi
z
=
dζ
dz
w,
dζ
dz
v
eucl
=
dζ
dz
2
Re(w¯v) =
dζ
dz
2
(w
1
v
1
+ w
2
v
2
),
where w = w
1
+ iw
2
, v = v
1
+ iv
2
.
Hence, on H, we obtain the Riemannian metric
dζ
dz
2
(dx
2
+ dy
2
).
We can compute
dζ
dz
=
1
z + i
−
z − i
(z + i)
2
=
2i
(z + i)
2
.
This is what we get if we started with a Euclidean metric. If we start with the
hyperbolic metric on
D
, we get an additional scaling factor. We can do some
computations to get
1 − |ζ|
2
= 1 −
|z − i|
2
|z + i|
2
,
and hence
1
1 − |ζ|
2
=
|z + i|
2
|z + i|
2
− |z − i|
2
=
|z + i|
2
4 Im z
.
Putting all these together, metric corresponding to
4|dζ|
2
(1−|ζ|
2
)
2
on D is
4 ·
4
|z + i|
4
·
|z + i|
2
4 Im z
2
· |dz|
2
=
|dz|
2
(Im z)
2
=
dx
2
+ dy
2
y
2
.
We now use all these ingredients to define the upper half-plane model.
Definition (Upper half-plane model). The upper half-plane model of the hyper-
bolic plane is the upper half-plane H with the Riemannian metric
dx
2
+ dy
2
y
2
.
The lengths on
H
are scaled (from the Euclidean one) by
1
y
, while the areas
are scaled by
1
y
2
. Again, the angles are the same.
Note that we did not have to go through so much mess in order to define
the sphere. This is since we can easily “embed” the surface of the sphere in
R
3
.
However, there is no easy surface in
R
3
that gives us the hyperbolic plane. As
we don’t have an actual prototype, we need to rely on the more abstract data of
a Riemannian metric in order to work with hyperbolic geometry.
We are next going to study the geometry of
H
, We claim that the following
group of M¨obius maps are isometries of H:
PSL(2, R) =
z 7→
az + b
cz + d
: a, b, c, d ∈ R, ad − bc = 1
.
Note that the coefficients have to be real, not complex.
Proposition. The elements of
PSL
(2
, R
) are isometries of
H
, and this preserves
the lengths of curves.
Proof. It is easy to check that PSL(2, R) is generated by
(i) Translations z 7→ z + a for a ∈ R
(ii) Dilations z 7→ az for a > 0
(iii) The single map z 7→ −
1
z
.
So it suffices to show each of these preserves the metric
|dz|
2
y
2
, where
z
=
x
+
iy
.
The first two are straightforward to see, by plugging it into formula and notice
the metric does not change.
We now look at the last one, given by z 7→ −
1
z
. The derivative at z is
f
0
(z) =
1
z
2
.
So we get
dz 7→ d
−
1
z
=
dz
z
2
.
So
d
−
1
z
2
=
|dz|
2
|z|
4
.
We also have
Im
−
1
z
= −
1
|z|
2
Im ¯z =
Im z
|z|
2
.
So
|d(−1/z)|
2
Im(−1/z)
2
=
|dz|
2
|z
4
|
(Im z)
2
|z|
4
=
|dz|
2
(Im z)
2
.
So this is an isometry, as required.
Note that each
z 7→ az
+
b
with
a >
0
, b ∈ R
is in
PSL
(2
, R
). Also, we can
use maps of this form to send any point to any other point. So
PSL
(2
, R
) acts
transitively on H. Moreover, everything in PSL(2, R) fixes R ∪ {∞}.
Recall also that each M¨obius transformation preserves circles and lines in the
complex plane, as well as angles between circles/lines. In particular, consider the
line
L
=
iR
, which meets
R
perpendicularly, and let
g ∈ PSL
(2
, R
). Then the
image is either a circle centered at a point in
R
, or a straight line perpendicular
to R.
We let
L
+
=
L ∩ H
=
{it
:
t >
0
}
. Then
g
(
L
+
) is either a vertical half-line
or a semi-circle that ends in R.
Definition (Hyperbolic lines). Hyperbolic lines in
H
are vertical half-lines or
semicircles ending in R.
We will now prove some lemmas to justify why we call these hyperbolic lines.
Lemma. Given any two distinct points
z
1
, z
2
∈ H
, there exists a unique
hyperbolic line through z
1
and z
2
.
Proof.
This is clear if
Re z
1
=
Re z
2
— we just pick the vertical half-line through
them, and it is clear this is the only possible choice.
Otherwise, if Re z
1
6= Re z
2
, then we can find the desired circle as follows:
R
z
1
z
2
It is also clear this is the only possible choice.
Lemma. PSL(2, R) acts transitively on the set of hyperbolic lines in H.
Proof.
It suffices to show that for each hyperbolic line
`
, there is some
g ∈
PSL
(2
, R
) such that
g
(
`
) =
L
+
. This is clear when
`
is a vertical half-line, since
we can just apply a horizontal translation.
If it is a semicircle, suppose it has end-points s < t ∈ R. Then consider
g(z) =
z − t
z − s
.
This has determinant
−s
+
t >
0. So
g ∈ PSL
(2
, R
). Then
g
(
t
) = 0 and
g
(
s
) =
∞
.
Then we must have
g
(
`
) =
L
+
, since
g
(
`
) is a hyperbolic line, and the only
hyperbolic lines passing through ∞ are the vertical half-lines. So done.
Moreover, we can achieve
g
(
s
) = 0 and
g
(
t
) =
∞
by composing with
−
1
z
.
Also, for any
P ∈ `
not on the endpoints, we can construct a
g
such that
g
(
P
) =
i ∈ L
+
, by composing with
z 7→ az
. So the isometries act transitively on
pairs (`, P ), where ` is a hyperbolic line and P ∈ `.
Definition (Hyperbolic distance). For points
z
1
, z
2
∈ H
, the hyperbolic distance
ρ
(
z
1
, z
2
) is the length of the segment [
z
1
, z
2
]
⊆ `
of the hyperbolic line through
z
1
, z
2
(parametrized monotonically).
Thus
PSL
(2
, R
) preserves hyperbolic distances. Similar to Euclidean space
and the sphere, we show these lines minimize distance.
Proposition. If
γ
: [0
,
1]
→ H
is a piecewise
C
1
-smooth curve with
γ
(0) =
z
1
, γ
(1) =
z
2
, then
length
(
γ
)
≥ ρ
(
z
1
, z
2
), with equality iff
γ
is a monotonic
parametrisation of [
z
1
, z
2
]
⊆ `
, where
`
is the hyperbolic line through
z
1
and
z
2
.
Proof.
We pick an isometry
g ∈ PSL
(2
, R
) so that
g
(
`
) =
L
+
. So without loss
of generality, we assume z
1
= iu and z
2
= iv, with u < v ∈ R.
We decompose the path as γ(t) = x(t) + iy(t). Then we have
length(γ) =
Z
1
0
1
y
p
˙x
2
+ ˙y
2
dt
≥
Z
1
0
|˙y|
y
dz
≥
Z
1
0
˙y
y
dt
= [log y(t)]
1
0
= log
v
u
This calculation also tells us that
ρ
(
z
1
, z
2
) =
log
v
u
. so
length
(
γ
)
≥ ρ
(
z
1
, z
2
)
with equality if and only if
x
(
t
) = 0 (hence
γ ⊆ L
+
) and
˙y ≥
0 (hence monotonic).
Corollary (Triangle inequality). Given three points z
1
, z
2
, z
3
∈ H, we have
ρ(z
1
, z
3
) ≤ ρ(z
1
, z
2
) + ρ(z
2
, z
3
),
with equality if and only if z
2
lies between z
1
and z
2
.
Hence, (H, ρ) is a metric space.