10Mobius group
IA Groups
10.3 Permutation properties of M¨obius maps
We have seen that the M¨obius map with three fixed points is the identity. As a
corollary, we obtain the following.
Proposition. Given
f, g ∈ M
. If
∃z
1
, z
2
, z
3
∈ C
∞
such that
f
(
z
i
) =
g
(
z
i
), then
f = g. i.e. every M¨obius map is uniquely determined by three points.
Proof.
As M¨obius maps are invertible, write
f
(
z
i
) =
g
(
z
i
) as
g
−1
f
(
z
i
) =
z
i
. So
g
−1
f has three fixed points. So g
−1
f must be the identity. So f = g.
Definition (Three-transitive action). An action of
G
on
X
is called three-
transitive if the induced action on
{
(
x
1
, x
2
, x
3
)
∈ X
3
:
x
i
pairwise disjoint}
,
given by g(x
1
, x
2
, x
3
) = (g(x
1
), g(x
2
), g(x
3
)), is transitive.
This means that for any two triples
x
1
, x
2
, x
3
and
y
1
, y
2
, y
3
of distinct elements
of X, there exists g ∈ G such that g(x
i
) = y
i
.
If this g is always unique, then the action is called sharply three transitive
This is a really weird definition. The reason we raise it here is that the
M¨obius map satisfies this property.
Proposition. The M¨obius group M acts sharply three-transitively on C
∞
.
Proof.
We want to show that we can send any three points to any other three
points. However, it is easier to show that we can send any three points to 0
,
1
, ∞
.
Suppose we want to send
z
1
→ ∞, z
2
7→
0
, z
3
7→
1. Then the following works:
f(z) =
(z − z
2
)(z
3
− z
1
)
(z − z
1
)(z
3
− z
2
)
If any term
z
i
is
∞
, we simply remove the terms with
z
i
, e.g. if
z
1
=
∞
, we have
f(z) =
z−z
2
z
3
−z
2
.
So given also
w
1
, w
2
, w
3
distinct in
C
∞
and
g ∈ M
sending
w
1
7→ ∞, w
2
7→
0, w
3
7→ 1, then we have g
−1
f(z
i
) = w
i
.
The uniqueness of the map follows from the fact that a M¨obius map is
uniquely determined by 3 points.
3 points not only define a M¨obius map uniquely. They also uniquely define
a line or circle. Note that on the Riemann sphere, we can think of a line as a
circle through infinity, and it would be technically correct to refer to both of
them as “circles”. However, we would rather be clearer and say “line/circle”.
We will see how M¨obius maps relate to lines and circles. We will first recap
some knowledge about lines and circles in the complex plane.
Lemma. The general equation of a circle or straight line in C is
Az¯z +
¯
Bz + B ¯z + C = 0,
where A, C ∈ R and |B|
2
> AC.
A
= 0 gives a straight line. If
A
= 0
, B
= 0, we have a circle centered at the
origin. If C = 0, the circle passes through 0.
Proof.
This comes from noting that
|z − B|
=
r
for
r ∈ R >
0 is a circle;
|z − a|
=
|z − b|
with
a
=
b
is a line. The detailed proof can be found in Vectors
and Matrices.
Proposition. M¨obius maps send circles/straight lines to circles/straight lines.
Note that it can send circles to straight lines and vice versa.
Alternatively, M¨obius maps send circles on the Riemann sphere to circles on
the Riemann sphere.
Proof.
We can either calculate it directly using
w
=
az+b
cz+d
⇔ z
=
dw−b
−cw+a
and
substituting
z
into the circle equation, which gives
A
′
w ¯w
+
¯
B
′
w
+
B
′
¯w
+
C
′
= 0
with A
′
, C
′
∈ R.
Alternatively, we know that each M¨obius map is a composition of translation,
dilation/rotation and inversion. We can check for each of the three types. Clearly
dilation/rotation and translation maps a circle/line to a circle/line. So we simply
do inversion: if w = z
−1
Az¯z +
¯
Bz + B ¯z + C = 0
⇔ Cw ¯w + Bw +
¯
B ¯w + A = 0
Example. Consider
f
(
z
) =
z−i
z+i
. Where does the real line go? The real line
is simply a circle through 0
,
1
, ∞
.
f
maps this circle to the circle containing
f(∞) = 1, f(0) = −1 and f (1) = −i, which is the unit circle.
Where does the upper half plane go? We know that the M¨obius map is
smooth. So the upper-half plane either maps to the inside of the circle or the
outside of the circle. We try the point
i
, which maps to 0. So the upper half
plane is mapped to the inside of the circle.