10Mobius group

IA Groups



10.1 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 6= 0.
We impose
ad bc 6
= 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)
=
(adbc)(zw)
(cw+d)(cz+d)
. If
ad bc
= 0, then
f
is constant and boring (more importantly, it will not be invertible).
If
c 6
= 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 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 6
= 0, with
f
(
d
c
) =
and
f
(
) =
a
c
when
c 6
= 0.
(if c = 0, then f() = )
Lemma. The obius maps are bijections C
C
.
Proof.
The inverse of
f
(
z
) =
az+b
cz+d
is
g
(
z
) =
dzb
cz+a
, which we can check by
composition both ways.
Proposition.
The obius maps form a group
M
under function composition.
(The 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 6
= 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
) 6= 0.
(This works for
z 6
=
,
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 6= 0.
2.
We have shown above that
f
1
(
z
) =
dzb
cz+a
with
da bc 6
= 0, which are
also 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 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 6
= 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
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
adbc
of any matrix in
GL
2
(
C
) is non-zero,
it does map to a 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 6= 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 6
= 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 obius group.
Proposition.
Every obius map is a composite of maps of the following form:
(i) Dilation/rotation: f(z) = az, a 6= 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 6
= 0, let
g
(
) =
z
0
, Let
h
(
z
) =
1
zz
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
obius map from the “elementary” maps are not unique.