2Spherical geometry
IB Geometry
2.2 M¨obius geometry
It turns out it is convenient to identify the sphere
S
2
withe the extended complex
plane
C
∞
=
C ∪{∞}
. Then isometries of
S
2
will translate to nice class of maps
of C
∞
.
We first find a way to identify
S
2
with
C
∞
. We use coordinates
ζ ∈ C
∞
. We
define the stereographic projection π : S
2
→ C
∞
by
N
P
π(P )
π(P ) = (line P N) ∩ {z = 0},
which is well defined except where P = N, in which case we define π(N) = ∞.
To give an explicit formula for this, consider the cross-section through the
plane ONP .
O
N
r
P
z
π(P )
R
If
P
has coordinates (
x, y
), then we see that
π
(
P
) will be a scalar multiple of
x
+
iy
. To find this factor, we notice that we have two similar triangles, and
hence obtain
r
R
=
1 − z
1
.
Then we obtain the formula
π(x, y, z) =
x + iy
1 − z
.
If we do the projection from the South pole instead, we get a related formula.
Lemma. If
π
0
:
S
2
→ C
∞
denotes the stereographic projection from the South
Pole instead, then
π
0
(P ) =
1
π(P )
.
Proof. Let P (x, y, z). Then
π(x, y, z) =
x + iy
1 − z
.
Then we have
π
0
(x, y, z) =
x + iy
1 + z
,
since we have just flipped the z axis around. So we have
π(P )π
0
(P ) =
x
2
+ y
2
1 − z
2
= 1,
noting that we have x
2
+ y
2
+ z
2
= 1 since we are on the unit sphere.
We can use this to infer that
π
0
◦ π
−1
:
C
∞
→ C
∞
takes
ζ 7→
1
/
¯
ζ
, which is
the inversion in the unit circle |ζ| = 1.
From IA Groups, we know M¨obius transformations act on
C
∞
and form a
group G by composition. For each matrix
A =
a b
c d
∈ GL(2, C),
we get a M¨obius map C
∞
→ C
∞
by
ζ 7→
aζ + b
cζ + d
.
Moreover, composition of M¨obius map is the same multiplication of matrices.
This is not exactly a bijective map between
G
and
GL
(2
, C
). For any
λ ∈ C
∗
=
C \ {
0
}
, we know
λA
defines the same map M¨obius map as
A
.
Conversely, if
A
1
and
A
2
gives the same M¨obius map, then there is some
λ
1
6
= 0
such that A
1
= λA
2
.
Hence, we have
G
∼
=
PGL(2, C) = GL(2, C)/C
∗
,
where
C
∗
∼
=
{λI : λ ∈ C
∗
}.
Instead of taking the whole
GL
(2
, C
) and quotienting out multiples of the
identities, we can instead start with
SL
(2
, C
). Again,
A
1
, A
2
∈ SL
(2
, C
) define
the same map if and only if
A
1
=
λA
2
for some
λ
. What are the possible values
of λ? By definition of the special linear group, we must have
1 = det(λA) = λ
2
det A = λ
2
.
So
λ
2
=
±
1. So each M¨obius map is represented by two matrices,
A
and
−A
,
and we get
G
∼
=
PSL(2, C) = SL(2, C)/{±1}.
Now let’s think about the sphere. On
S
2
, the rotations
SO
(3) act as isometries.
In fact, the full isometry group of
S
2
is O(3) (the proof is on the first example
sheet). What we would like to show that rotations of
S
2
correspond to M¨obius
transformations coming from the subgroup SU(2) ≤ GL(2, C).
Theorem. Via the stereographic projection, every rotation of
S
2
induces a
M¨obius map defined by a matrix in SU(2) ⊆ GL(2, C), where
SU(2) =
a −b
¯
b ¯a
: |a|
2
+ |b|
2
= 1
.
Proof.
(i)
Consider the
r
(
ˆ
z, θ
), the rotations about the
z
axis by
θ
. These corresponds
to the M¨obius map ζ 7→ e
iθ
ζ, which is given by the unitary matrix
e
iθ/2
0
0 e
−iθ/2
.
(ii) Consider the rotation r(
ˆ
y,
π
2
). This has the matrix
0 0 1
0 1 0
−1 0 0
x
y
z
=
z
y
−x
.
This corresponds to the map
ζ =
x + iy
1 − z
7→ ζ
0
=
z + iy
1 + x
We want to show this is a M¨obius map. To do so, we guess what the
M¨obius map should be, and check it works. We can manually compute
that −1 7→ ∞, 1 7→ 0, i 7→ i.
1
−1
i
0
∞
The only M¨obius map that does this is
ζ
0
=
ζ − 1
ζ + 1
.
We now check:
ζ − 1
ζ + 1
=
x + iy − 1 + z
x + iy + 1 − z
=
x − 1 + z + iy
x + 1 − (z − iy)
=
(z + iy)(x − 1 + z + iy)
(x + 1)(z + iy) − (z
2
+ y
2
)
=
(z + iy)(x − 1 + z + iy)
(x + 1)(z + iy) + (x
2
− 1)
=
(z + iy)(x − 1 + z + iy)
(x + 1)(z + iy + x −1)
=
z + iy
x + 1
.
So done. We finally have to write this in the form of an SU(2) matrix:
1
√
2
1 −1
1 1
.
(iii) We claim that SO(3) is generated by r
ˆ
y,
π
2
and r(
ˆ
z, θ) for 0 ≤ θ < 2π.
To show this, we observe that
r
(
ˆ
x, ϕ
) =
r
(
ˆ
y,
π
2
)
r
(
ˆ
z, ϕ
)
r
(
ˆ
y, −
π
2
). Note that
we read the composition from right to left. You can convince yourself this
is true by taking a physical sphere and try rotating. To prove it formally,
we can just multiply the matrices out.
Next, observe that for v
∈ S
2
⊆ R
3
, there are some angles
ϕ, ψ
such that
g
=
r
(
ˆ
z, ψ
)
r
(
ˆ
x, ϕ
) maps v to
ˆ
x
. We can do so by first picking
r
(
ˆ
x, ϕ
) to
rotate v into the (x, y)-plane. Then we rotate about the z-axis to send it
to
ˆ
x.
Then for any
θ
, we have
r
(v
, θ
) =
g
−1
r
(
ˆ
x, θ
)
g
, and our claim follows by
composition.
(iv)
Thus, via the stereographic projection, every rotation of
S
2
corresponds
to products of M¨obius transformations of
C
∞
with matrices in
SU
(2).
The key of the proof is step (iii). Apart from enabling us to perform the
proof, it exemplifies a useful technique in geometry — we know how to rotate
arbitrary things in the
z
axis. When we want to rotate things about the
x
axis
instead, we first rotate the sphere to move the
x
axis to where the
z
axis used
to be, do those rotations, and then rotate it back. In general, we can use some
isometries or rotations to move what we want to do to a convenient location.
Theorem. The group of rotations
SO
(3) acting on
S
2
corresponds precisely
with the subgroup
PSU
(2) =
SU
(2)
/ ±
1 of M¨obius transformations acting on
C
∞
.
What this is in some sense a converse of the previous theorem. We are saying
that for each M¨obius map from
SU
(2), we can find some rotation of
S
2
that
induces that M¨obius map, and there is exactly one.
Proof. Let g ∈ PSU(2) be a M¨obius transformation
g(z) =
az + b
¯
bz + ¯a
.
Suppose first that
g
(0) = 0. So
b
= 0. So
a¯a
= 1. Hence
a
=
e
iθ/2
. Then
g
corresponds to r(
ˆ
z, θ), as we have previously seen.
In general, let
g
(0) =
w ∈ C
∞
. Let
Q ∈ S
2
be such that
π
(
Q
) =
w
. Choose a
rotation
A ∈ SO
(3) such that
A
(
Q
) =
−
ˆ
z
. Since
A
is a rotation, let
α ∈ PSU
(2)
be the corresponding M¨obius transformation. By construction we have
α
(
w
) = 0.
Then the composition
α ◦ g
fixes zero. So it corresponds to some
B
=
r
(
z, θ
).
We then see that g corresponds to A
−1
B ∈ SO(3). So done.
Again, we solved an easy case, where
g
(0) = 0, and then performed some
rotations to transform any other case into this simple case.
We have now produced a 2-to-1 map
SU(2) → PSU(2)
∼
=
SO(3).
If we treat these groups as topological spaces, this map does something funny.
Suppose we start with a (non-closed) path from
I
to
−I
in
SU
(2). Applying
the map, we get a closed loop from I to I in SO(3).
Hence, in
SO
(3), loops are behave slightly weirdly. If we go around this loop
in
SO
(3), we didn’t really get back to the same place. Instead, we have actually
moved from
I
to
−I
in
SU
(2). It takes two full loops to actually get back to
I
.
In physics, this corresponds to the idea of spinors.
We can also understand this topologically as follows: since
SU
(2) is defined by
two complex points
a, b ∈ C
such that
|a|
2
+
|b|
2
, we can view it as as three-sphere
S
3
in SO(3).
A nice property of
S
3
is it is simply connected, as in any loop in
S
3
can be
shrunk to a point. In other words, given any loop in
S
3
, we can pull and stretch
it (continuously) until becomes the “loop” that just stays at a single point.
On the other hand,
SO
(3) is not simply connected. We have just constructed
a loop in
SO
(3) by mapping the path from
I
to
−I
in
SU
(2). We cannot deform
this loop until it just sits at a single point, since if we lift it back up to
SU
(2), it
still has to move from I to −I.
The neat thing is that in some sense,
S
3
∼
=
SU
(2) is just “two copies” of
SO
(3). By duplicating
SO
(3), we have produced
SU
(2), a simply connected
space. Thus we say SU(2) is a universal cover of SO(3).
We’ve just been waffling about spaces and loops, and throwing around terms
we haven’t defined properly. These vague notions will be made precise in the
IID Algebraic Topology course, and we will then (maybe) see that
SU
(2) is a
universal cover of SO(3).