2Spherical geometry
IB Geometry
2.1 Triangles on a sphere
One main object of study is spherical triangles – they are defined just like
Euclidean triangles, with
AB, AC, BC
line segments on
S
of length
< π
. The
restriction of length is just merely for convenience.
A
B C
c
b
a
α
β
γ
We will take advantage of the fact that the sphere sits in R
3
. We set
n
1
=
C × B
sin a
n
2
=
A × C
sin b
n
3
=
B × A
sin c
.
These are unit normals to the planes
OBC, OAC
and
OAB
respectively. They
are pointing out of the solid OABC.
The angles
α, β, γ
are the angles between the planes for the respective sides.
Then 0
< α, β, γ < π
. Note that the angle between n
2
and n
3
is
π
+
α
(not
α
itself — if α = 0, then the angle between the two normals is π). So
n
2
· n
3
= −cos α
n
3
· n
1
= −cos β
n
1
· n
2
= −cos γ.
We have the following theorem:
Theorem (Spherical cosine rule).
sin a sin b cos γ = cos c − cos a cos b.
Proof. We use the fact from IA Vectors and Matrices that
(C × B) · (A × C) = (A · C)(B · C) − (C · C)(B · A),
which follows easily from the double-epsilon identity
ε
ijk
ε
imn
= δ
jm
δ
kn
− δ
jn
δ
km
.
In our case, since C · C = 1, the right hand side is
(A · C)(B · C) − (B · A).
Thus we have
−cos γ = n
1
· n
2
=
C × B
sin a
·
A × C
sin b
=
(A · C)(B · C) − (B · A)
sin a sin b
=
cos b cos a − cos c
sin a sin b
.
Corollary (Pythagoras theorem). If γ =
π
2
, then
cos c = cos a cos b.
Analogously, we have a spherical sine rule.
Theorem (Spherical sine rule).
sin a
sin α
=
sin b
sin β
=
sin c
sin γ
.
Proof. We use the fact that
(A × C) × (C × B) = (C · (B × A))C,
which we again are not bothered to prove again. The left hand side is
−(n
1
× n
2
) sin a sin b
Since the angle between n
1
and n
2
is
π
+
γ
, we know n
1
×
n
2
= C
sin γ
. Thus
the left hand side is
−C sin a sin b sin γ.
Thus we know
C · (A × B) = sin a sin b sin γ.
However, since the scalar triple product is cyclic, we know
C · (A × B) = A · (B × C).
In other words, we have
sin a sin b sin γ = sin b sin c sin α.
Thus we have
sin γ
sin c
=
sin α
sin a
.
Similarly, we know this is equal to
sin β
sin b
.
Recall that for small a, b, c, we know
sin a = a + O(a
3
).
Similarly,
cos a = 1 −
a
2
2
+ O(a
4
).
As we take the limit
a, b, c →
0, the spherical sine and cosine rules become the
usual Euclidean versions. For example, the cosine rule becomes
ab cos γ = 1 −
c
2
2
−
1 −
a
2
2
1 −
b
2
2
+ O(k(a, b, c)k
3
).
Rearranging gives
c
2
= a
2
+ b
2
− 2ab cos γ + O(k(a, b, c, )k
3
).
The sine rule transforms similarly as well. This is what we would expect, since
making
a, b, c
small is equivalent to zooming into the surface of the sphere, and
it looks more and more like flat space.
Note that if
γ
=
π
, it then follows that
C
is in the line segment given by
AB
.
So c = a + b. Otherwise, we get
cos c > cos a cos b − sin a sin b = cos(a + b),
since cos γ < 1. Since cos is decreasing on [0, π], we know
c < a + b.
Corollary (Triangle inequality). For any P, Q, R ∈ S
2
, we have
d(P, Q) + d(Q, R) ≥ d(P, R),
with equality if and only if Q lies is in the line segment P R of shortest length.
Proof.
The only case left to check is if
d
(
P, R
) =
π
, since we do not allow our
triangles to have side length π. But in this case they are antipodal points, and
any Q lies in a line through P R, and equality holds.
Thus, we find that (S
2
, d) is a metric space.
On
R
n
, straight lines are curves that minimize distance. Since we are calling
spherical lines lines, we would expect them to minimize distance as well. This is
in fact true.
Proposition. Given a curve Γ on
S
2
⊆ R
3
from
P
to
Q
, we have
`
=
length
(Γ)
≥
d
(
P, Q
). Moreover, if
`
=
d
(
P, Q
), then the image of Γ is a spherical line segment
P Q.
Proof. Let Γ : [0, 1] → S and ` = length(Γ). Then for any dissection D of [0, 1],
say 0 = t
0
< ··· < t
N
= 1, write P
i
= Γ(t
i
). We define
˜
S
D
=
X
i
d(P
i−1
, P
i
) > S
D
=
X
i
|
−−−−→
P
i−1
P
i
|,
where the length in the right hand expression is the distance in Euclidean 3-space.
Now suppose
` < d
(
P, Q
). Then there is some
ε >
0 such that
`
(1 +
ε
)
<
d(P, Q).
Recall from basic trigonometric that if θ > 0, then sin θ < θ. Also,
sin θ
θ
→ 1 as θ → 0.
Thus we have
θ ≤ (1 + ε) sin θ.
for small θ. What we really want is the double of this:
2θ ≤ (1 + ε)2 sin θ.
This is useful since these lengths appear in the following diagram:
2 sin θ
2θ
2θ
This means for P, Q sufficiently close, we have d(P, Q) ≤ (1 + ε)|
−−→
P Q|.
From Analysis II, we know Γ is uniformly continuous on [0
,
1]. So we can
choose D such that
d(P
i−1
, P
i
) ≤ (1 + ε)|
−−−−→
P
i−1
P
i
|
for all i. So we know that for sufficiently fine D,
˜
S
D
≤ (1 + ε)S
D
< d(P, Q),
since
S
D
→ `
. However, by the triangle inequality
˜
S
D
≥ d
(
P, Q
). This is a
contradiction. Hence we must have ` ≥ d(P, Q).
Suppose now
`
=
d
(
P, Q
) for some Γ : [0
,
1]
→ S
,
`
=
length
(Γ). Then for
every t ∈ [0, 1], we have
d(P, Q) = ` = length Γ|
[0,t]
+ length Γ|
[t,1]
≥ d(P, Γ(t)) + d(Γ(t), Q)
≥ d(P, Q).
Hence we must have equality all along the way, i.e.
d(P, Q) = d(P, Γ(t)) + d(Γ(t), Q)
for all Γ(t).
However, this is possible only if Γ(
t
) lies on the shorter spherical line segment
P Q, as we have previously proved. So done.
Note that if Γ is a curve of minimal length from
P
to
Q
, then Γ is a
spherical line segment. Further, from the proof of this proposition, we know
length
Γ
|
[0,t]
=
d
(
P,
Γ(
t
)) for all
t
. So the parametrisation of Γ is monotonic.
Such a Γ is called a minimizing geodesic.
Finally, we get to an important theorem whose prove involves complicated
pictures. This is known as the Gauss-Bonnet theorem. The Gauss-Bonnet
theorem is in fact a much more general theorem. However, here we will specialize
in the special case of the sphere. Later, when doing hyperbolic geometry, we
will prove the hyperbolic version of the Gauss-Bonnet theorem. Near the end of
the course, when we have developed sufficient machinery, we would be able to
state the Gauss-Bonnet theorem in its full glory. However, we will not be able
to prove the general version.
Proposition (Gauss-Bonnet theorem for
S
2
). If ∆ is a spherical triangle with
angles α, β, γ, then
area(∆) = (α + β + γ) − π.
Proof.
We start with the concept of a double lune. A double lune with angle
0
< α < π
is two regions
S
cut out by two planes through a pair of antipodal
points, where α is the angle between the two planes.
A
A
0
It is not hard to show that the area of a double lune is 4
α
, since the area of the
sphere is 4π.
Now note that our triangle ∆ =
ABC
is the intersection of 3 single lunes,
with each of
A, B, C
as the pole (in fact we only need two, but it is more
convenient to talk about 3).
A
A
0
B
B
0
C
C
0
Therefore ∆ together with its antipodal partner ∆
0
is a subset of each of the 3
double lunes with areas 4
α,
4
β,
4
γ
. Also, the union of all the double lunes cover
the whole sphere, and overlap at exactly ∆ and ∆
0
. Thus
4(α + β + γ) = 4π + 2(area(∆) + area(∆
0
)) = 4π + 4 area(∆).
This is easily generalized to arbitrary convex
n
-gons on
S
2
(with
n ≥
3).
Suppose
M
is such a convex
n
-gon with interior angles
α
1
, ··· , α
n
. Then we
have
area(M) =
n
X
1
α
i
− (n − 2)π.
This follows directly from cutting the polygon up into the constituent triangles.
This is very unlike Euclidean space. On
R
2
, we always have
α
+
β
+
γ
=
π
.
Not only is this false on
S
2
, but by measuring the difference, we can tell the
area of the triangle. In fact, we can identify triangles up to congruence just by
knowing the three angles.