4Hyperbolic geometry
IB Geometry
4.5 Hyperbolic triangles
Definition (Hyperbolic triangle). A hyperbolic triangle
ABC
is the region
determined by three hyperbolic line segments
AB, BC
and
CA
, including extreme
cases where some vertices
A, B, C
are allowed to be “at infinity”. More precisely,
in the half-plane model, we allow them to lie in
R ∪{∞}
; in the disk model we
allow them to lie on the unit circle |z| = 1.
We see that if A is “at infinity”, then the angle at A must be zero.
Recall for a region R ⊆ H, we can compute the area of R as
area(R) =
ZZ
R
dx dy
y
2
.
Similar to the sphere, we have
Theorem (Gauss-Bonnet theorem for hyperbolic triangles). For each hyperbolic
triangle ∆, say,
ABC
, with angles
α, β, γ ≥
0 (note that zero angle is possible),
we have
area(∆) = π − (α + β + γ).
Proof.
First do the case where
γ
= 0, so
C
is “at infinity”. Recall that we like
to use the disk model if we have a distinguished point in the hyperbolic plane.
If we have a distinguished point at infinity, it is often advantageous to use the
upper half plane model, since ∞ is a distinguished point at infinity.
So we use the upper-half plane model, and wlog
C
=
∞
(apply
PSL
(2
, R
))
if necessary. Then
AC
and
BC
are vertical half-lines. So
AB
is the arc of a
semi-circle. So AB is an arc of a semicircle.
C C
B
A
β
π − α
α
β
We use the transformation
z 7→ z
+
a
(with
a ∈ R
) to center the semi-circle at 0.
We then apply
z 7→ bz
(with
b >
0) to make the circle have radius 1. Thus wlog
AB ⊆ {x
2
+ y
2
= 1}.
Now we have
area(T ) =
Z
cos β
cos(π−α)
Z
∞
√
1−x
2
1
y
2
dy dx
=
Z
cos β
cos(π−α)
1
√
1 − x
2
dx
= [−cos
−1
(x)]
cos β
cos(π−α)
= π − α − β,
as required.
In general, we use
H
again, and we can arrange
AC
in a vertical half-line.
Also, we can move
AB
to
x
2
+
y
2
= 1, noting that this transformation keeps
AC vertical.
B
A
α
C
γ
δ
β
We consider ∆
1
= AB∞ and ∆
2
= CB∞. Then we can immediately write
area(∆
1
) = π − α − (β + δ)
area(∆
2
) = π − δ −(π −γ) = γ − δ.
So we have
area(T ) = area(∆
2
) − area(∆
1
) = π − α − β − γ,
as required.
Similar to the spherical case, we have some hyperbolic sine and cosine rules.
For example, we have
Theorem (Hyperbolic cosine rule). In a triangle with sides
a, b, c
and angles
α, β, γ, we have
cosh c = cosh a cosh b − sinh a sinh b cos γ.
Proof. See example sheet 2.
Recall that in
S
2
, any two lines meet (in two points). In the Euclidean plane
R
2
, any two lines meet (in one point) iff they are not parallel. Before we move
on to the hyperbolic case, we first make a definition.
Definition (Parallel lines). We use the disk model of the hyperbolic plane. Two
hyperbolic lines are parallel iff they meet only at the boundary of the disk (at
|z| = 1).
Definition (Ultraparallel lines). Two hyperbolic lines are ultraparallel if they
don’t meet anywhere in {|z| ≤ 1}.
In the Euclidean plane, we have the parallel axiom: given a line
`
and
P 6∈ `
,
there exists a unique line
`
0
containing
P
with
` ∩ `
0
=
∅
. This fails in both
S
2
and the hyperbolic plane — but for very different reasons! In
S
2
, there are no
such parallel lines. In the hyperbolic plane, there are many parallel lines. There
is a more deep reason for why this is the case, which we will come to at the very
end of the course.