10Mobius group

IA Groups

10.3 Permutation properties of obius maps
We have seen that the 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 obius map is uniquely determined by three points.
Proof.
As 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
obius map satisfies this property.
Proposition. The 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) =
zz
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 obius map is
uniquely determined by 3 points.
3 points not only define a 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 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. obius maps send circles/straight lines to circles/straight lines.
Note that it can send circles to straight lines and vice versa.
Alternatively, 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
=
dwb
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 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
) =
zi
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 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.