4Hyperbolic geometry
IB Geometry
4.4 Geometry of the hyperbolic disk
So far, we have worked with the upper half-plane model. This is since the upper
half-plane model is more convenient for these calculations. However, sometimes
the disk model is more convenient. So we also want to understand that as well.
Recall that
ζ ∈ D 7→ z
=
i
1+ζ
1−ζ
∈ H
is an isometry, with an (isometric) inverse
z ∈ H 7→ ζ
=
z−i
z+i
∈ D
. Moreover, since these are M¨obius maps, circle-lines are
preserved, and angles between the lines are also preserved.
Hence, immediately from previous work on H, we know
(i) PSL(2, R)
∼
=
{M¨obius transformations sending D to itself} = G.
(ii)
Hyperbolic lines in
D
are circle segments meeting
|ζ|
= 1 orthogonally,
including diameters.
(iii) G
acts transitively on hyperbolic lines in
D
(and also on pairs consisting
of a line and a point on the line).
(iv)
The length-minimizing geodesics on
D
are a segments of hyperbolic lines
parametrized monotonically.
We write ρ for the (hyperbolic) distance in D.
Lemma. Let G be the set of isometries of the hyperbolic disk. Then
(i) Rotations z 7→ e
iθ
z (for θ ∈ R) are elements of G.
(ii) If a ∈ D, then g(z) =
z−a
1−¯az
is in G.
Proof.
(i)
This is clearly an isometry, since this is a linear map, preserves
|z|
and
|dz|, and hence also the metric
4|dz|
2
(1 − |z|
2
)
2
.
(ii)
First, we need to check this indeed maps
D
to itself. To do this, we first
make sure it sends {|z| = 1} to itself. If |z| = 1, then
|1 − ¯az| = |¯z(1 − ¯az)| = |¯z − ¯a| = |z −a|.
So
|g(z)| = 1.
Finally, it is easy to check
g
(
a
) = 0. By continuity,
G
must map
D
to itself.
We can then show it is an isometry by plugging it into the formula.
It is an exercise on the second example sheet to show all
g ∈ G
is of the form
g(z) = e
iθ
z − a
1 − ¯az
or
g(z) = e
iθ
¯z − a
1 − ¯a¯z
for some θ ∈ R and a ∈ D.
We shall now use the disk model to do something useful. We start by coming
up with an explicit formula for distances in the hyperbolic plane.
Proposition. If 0 ≤ r < 1, then
ρ(0, re
iθ
) = 2 tanh
−1
r.
In general, for z
1
, z
2
∈ D, we have
g(z
1
, z
2
) = 2 tanh
−1
z
1
− z
2
1 − ¯z
1
z
2
.
Proof.
By the lemma above, we can rotate the hyperbolic disk so that
re
iθ
is
rotated to r. So
ρ(0, re
iθ
) = ρ(0, r).
We can evaluate this by performing the integral
ρ(0, r) =
Z
r
0
2 dt
1 − t
2
= 2 tanh
−1
r.
For the general case, we apply the M¨obius transformation
g(z) =
z − z
1
1 − ¯z
1
z
.
Then we have
g(z
1
) = 0, g(z
2
) =
z
2
− z
1
1 − ¯z
1
z
2
=
z
1
− z
2
1 − ¯z
1
z
2
e
iθ
.
So
ρ(z
1
, z
2
) = ρ(g(z
1
), g(z
2
)) = 2 tanh
−1
z
1
− z
2
1 − ¯z
1
z
2
.
Again, we exploited the idea of performing the calculation in an easy case, and
then using isometries to move everything else to the easy case. In general, when
we have a “distinguished” point in the hyperbolic plane, it is often convenient to
use the disk model, move it to 0 by an isometry.
Proposition. For every point
P
and hyperbolic line
`
, with
P 6∈ `
, there is a
unique line
`
0
with
P ∈ `
0
such that
`
0
meets
`
orthogonally, say
` ∩ `
0
=
Q
, and
ρ(P, Q) ≤ ρ(P,
˜
Q) for all
˜
Q ∈ `.
This is a familiar fact from Euclidean geometry. To prove this, we again
apply the trick of letting P = 0.
Proof.
wlog, assume
P
= 0
∈ D
. Note that a line in
D
(that is not a diameter)
is a Euclidean circle. So it has a center, say C.
Since any line through
P
is a diameter, there is clearly only one line that
intersects
`
perpendicularly (recall angles in
D
is the same as the Euclidean
angle).
`
0
P
D
C
`
Q
It is also clear that
P Q
minimizes the Euclidean distance between
P
and
`
.
While this is not the same as the hyperbolic distance, since hyperbolic lines
through
P
are diameters, having a larger hyperbolic distance is equivalent to
having a higher Euclidean distance. So this indeed minimizes the distance.
How does reflection in hyperbolic lines work? This time, we work in the
upper half-plane model, since we have a favorite line L
+
.
Lemma (Hyperbolic reflection). Suppose
g
is an isometry of the hyperbolic
half-plane
H
and
g
fixes every point in
L
+
=
{iy
:
y ∈ R
+
}
. Then
G
is either
the identity or g(z) = −¯z, i.e. it is a reflection in the vertical axis L
+
.
Observe we have already proved a similar result in Euclidean geometry, and
the spherical version was proven in the first example sheet.
Proof.
For every
P ∈ H \ L
+
, there is a unique line
`
0
containing
P
such that
`
0
⊥ L
+
. Let Q = L
+
∩ `
0
.
L
+
`
0
P
Q
We see `
0
is a semicircle, and by definition of isometry, we must have
ρ(P, Q) = ρ(g(P ), Q).
Now note that
g
(
`
0
) is also a line meeting
L
+
perpendicularly at
Q
, since
g
fixes
L
+
and preserves angles. So we must have
g
(
`
0
) =
`
0
. Then in particular
g
(
P
)
∈ `
0
. So we must have
g
(
P
) =
P
or
g
(
P
) =
P
0
, where
P
0
is the image
under reflection in L
+
.
Now it suffices to prove that if
g
(
P
) =
P
for any one
P
, then
g
(
P
) must be
the identity (if g(P ) = P
0
for all P , then g must be given by g(z) = −¯z).
Now suppose g(P ) = P , and let A ∈ H
+
, where H
+
= {z ∈ H : Re z > 0}.
L
+
P
P
0
A
A
0
B
Now if g(A) 6= A, then g(A) = A
0
. Then ρ(A
0
, P ) = ρ(A, P ). But
ρ(A
0
, P ) = ρ(A
0
, B) + ρ(B, P ) = ρ(A, B) + ρ(B, P ) > ρ(A, P ),
by the triangle inequality, noting that
B 6∈
(
AP
). This is a contradiction. So
g
must fix everything.
Definition (Hyperbolic reflection). The map
R
:
z ∈ H 7→ −¯z ∈ H
is the
(hyperbolic) reflection in
L
+
. More generally, given any hyperbolic line
`
, let
T
be the isometry that sends ` to L
+
. Then the (hyperbolic) reflection in ` is
R
`
= T
−1
RT
Again, we already know how to reflect in
L
+
. So to reflect in another line
`
,
we move our plane such that ` becomes L
+
, do the reflection, and move back.
By the previous proposition,
R
`
is the unique isometry that is not identity
and fixes `.