10Mobius group
IA Groups
10.1 M¨obius maps
We want to study maps
f
:
C → C
in the form
f
(
z
) =
az+b
cz+d
with
a, b, c, d ∈ C
and ad − bc = 0.
We impose
ad − bc
= 0 or else the map will be constant: for any
z, w ∈ C
,
f
(
z
)
− f
(
w
) =
(az+b)(cw+d)−(aw+b)(cz+d)
(cw+d)(cz+d)
=
(ad−bc)(z−w)
(cw+d)(cz+d)
. If
ad − bc
= 0, then
f
is constant and boring (more importantly, it will not be invertible).
If
c
= 0, then
f
(
−
d
c
) involves division by 0. So we add
∞
to
C
to form
the extended complex plane (Riemann sphere)
C ∪ {∞}
=
C
∞
(cf. Vectors and
Matrices). Then we define
f
(
−
d
c
) =
∞
. We call
C
∞
a one-point compactification
of
C
(because it adds one point to
C
to make it compact, cf. Metric and Topology).
Definition (M¨obius map). A M¨obius map is a map from
C
∞
→ C
∞
of the form
f(z) =
az + b
cz + d
,
where
a, b, c, d ∈ C
and
ad − bc
= 0, with
f
(
−
d
c
) =
∞
and
f
(
∞
) =
a
c
when
c
= 0.
(if c = 0, then f(∞) = ∞)
Lemma. The M¨obius maps are bijections C
∞
→ C
∞
.
Proof.
The inverse of
f
(
z
) =
az+b
cz+d
is
g
(
z
) =
dz−b
−cz+a
, which we can check by
composition both ways.
Proposition. The M¨obius maps form a group
M
under function composition.
(The M¨obius group)
Proof. The group axioms are shown as follows:
0.
If
f
1
(
z
) =
a
1
z+b
1
c
1
z+d
1
and
f
2
(
z
) =
a
2
z+b
2
c
2
z+d
2
, then
f
2
◦f
1
(
z
) =
a
2
a
1
z+b
1
c
1
z+d
1
+ b
2
c
2
a
1
z+b
1
c
1
z+d
1
+ d
2
=
(a
1
a
2
+ b
2
c
1
)z + (a
2
b
1
+ b
2
d
1
)
(c
2
a
1
+ d
2
c
1
)z + (c
2
b
1
+ d
1
d
2
)
. Now we have to check that
ad − bc
= 0:
we have (
a
1
a
2
+
b
2
c
1
)(
c
2
b
1
+
d
1
d
2
)
−
(
a
2
b
1
+
b
2
d
1
)(
c
2
a
1
+
d
2
c
1
) = (
a
1
d
1
−
b
1
c
1
)(a
2
d
2
− b
2
c
2
) = 0.
(This works for
z
=
∞, −
d
1
c
1
. We have to manually check the special cases,
which is simply yet more tedious algebra)
1. The identity function is 1(z) =
1z+0
0+1
which satisfies ad − bc = 0.
2.
We have shown above that
f
−1
(
z
) =
dz−b
−cz+a
with
da − bc
= 0, which are
also M¨obius maps
3. Composition of functions is always associative
M
is not abelian. e.g.
f
1
(
z
) = 2
z
and
f
2
(
z
) =
z
+ 1 are not commutative:
f
1
◦ f
2
(z) = 2z + 2 and f
2
◦ f
1
(z) = 2z + 1.
Note that the point at “infinity” is not special.
∞
is no different to any other
point of the Riemann sphere. However, from the way we write down the M¨obius
map, we have to check infinity specially. In this particular case, we can get quite
far with conventions such as
1
∞
= 0,
1
0
= ∞ and
a·∞
c·∞
=
a
c
.
Clearly
az+b
cz+d
=
λaz+λb
λcz+λd
for any
λ
= 0. So we do not have a unique represen-
tation of a map in terms of
a, b, c, d
. But
a, b, c, d
does uniquely determine a
M¨obius map.
Proposition. The map
θ
:
GL
2
(
C
)
→ M
sending
a b
c d
7→
az + b
cz + d
is a
surjective group homomorphism.
Proof.
Firstly, since the determinant
ad−bc
of any matrix in
GL
2
(
C
) is non-zero,
it does map to a M¨obius map. This also shows that θ is surjective.
We have previously calculated that
θ(A
2
) ◦ θ(A
1
) =
(a
1
a
2
+ b
2
c
1
)z + (a
2
b
1
+ b
2
d
1
)
(c
2
a
1
+ d
2
c
1
)z + (c
2
b
1
+ d
1
d
2
)
= θ(A
2
A
1
)
So it is a homomorphism.
The kernel of θ is
ker(θ) =
A ∈ GL
2
(C) : (∀z) z =
az + b
cz + d
We can try different values of
z
:
z
=
∞ ⇒ c
= 0;
z
= 0
⇒ b
= 0;
z
= 1
⇒ d
=
a
.
So
ker θ = Z = {λI : λ ∈ C, λ = 0},
where I is the identity matrix and Z is the centre of GL
2
(C).
By the isomorphism theorem, we have
M
∼
=
GL
2
(C)/Z
Definition (Projective general linear group
PGL
2
(
C
)). (Non-examinable) The
projective general linear group is
PGL
2
(C) = GL
2
(C)/Z.
Since
f
A
=
f
B
iff
B
=
λA
for some
λ
= 0 (where
A, B
are the corresponding
matrices of the maps), if we restrict
θ
to
SL
2
(
C
), we have
θ|
SL
2
(C)
:
SL
2
(
C
)
→ M
is also surjective. The kernel is now just {±I}. So
M
∼
=
SL
2
(C)/{±I} = PSL
2
(C)
Clearly PSL
2
(C)
∼
=
PGL
2
(C) since both are isomorphic to the M¨obius group.
Proposition. Every M¨obius map is a composite of maps of the following form:
(i) Dilation/rotation: f(z) = az, a = 0
(ii) Translation: f(z) = z + b
(iii) Inversion: f(z) =
1
z
Proof. Let
az+b
cz+d
∈ M.
If c = 0, i.e. g(∞) = ∞, then g(z) =
a
d
z +
b
d
, i.e.
z 7→
a
d
z 7→
a
d
z +
b
d
.
If
c
= 0, let
g
(
∞
) =
z
0
, Let
h
(
z
) =
1
z−z
0
. Then
hg
(
∞
) =
∞
is of the above form.
We have
h
−1
(
w
) =
1
w
+
z
0
being of type (iii) followed by (ii). So
g
=
h
−1
(
hg
) is
a composition of maps of the three forms listed above.
Alternatively, with sufficient magic, we have
z 7→ z +
d
c
7→
1
z +
d
c
7→ −
ad + bc
c
2
(z +
d
c
)
7→
a
c
−
ad + bc
c
2
(z +
d
c
)
=
az + b
cz + d
.
Note that the non-calculation method above can be transformed into another
(different) composition with the same end result. So the way we compose a
M¨obius map from the “elementary” maps are not unique.