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 6

= 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 6

=

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

0

w ¯w

+

¯

B

0

w

+

B

0

¯w

+

C

0

= 0

with A

0

, C

0

∈ 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.