10Mobius group

IA Groups

10.4 Cross-ratios

Finally, we’ll look at an important concept known as cross-ratios. Roughly

speaking, this is a quantity that is preserved by M¨obius transforms.

Definition

(Cross-ratios)

.

Given four distinct points

z

1

, z

2

, z

3

, z

4

∈ C

∞

,

their

cross-ratio is [

z

1

, z

2

, z

3

, z

4

] =

g

(

z

4

), with

g

being the unique M¨obius map that

maps z

1

7→ ∞, z

2

7→ 0, z

3

7→ 1. So [∞, 0, 1, λ] = λ for any λ 6= ∞, 0, 1. We have

[z

1

, z

2

, z

3

, z

4

] =

z

4

− z

2

z

4

− z

1

·

z

3

− z

1

z

3

− z

2

(with special cases as above).

We know that this exists and is uniquely defined because

M

acts sharply

three-transitively on C

∞

.

Note that different authors use different permutations of 1

,

2

,

3

,

4, but they

all lead to the same result as long as you are consistent.

Lemma. For z

1

, z

2

, z

3

, z

4

∈ C

∞

all distinct, then

[z

1

, z

2

, z

3

, z

4

] = [z

2

, z

1

, z

4

, z

3

] = [z

3

, z

4

, z

1

, z

2

] = [z

4

, z

3

, z

2

, z

1

]

i.e. if we perform a double transposition on the entries, the cross-ratio is retained.

Proof. By inspection of the formula.

Proposition. If f ∈ M, then [z

1

, z

2

, z

3

, z

4

] = [f(z

1

), f(z

2

), f(z

3

), f(z

4

)].

Proof.

Use our original definition of the cross ratio (instead of the formula). Let

g be the unique M¨obius map such that [z

1

, z

2

, z

3

, z

4

] = g(z

4

) = λ, i.e.

z

1

g

7−→ ∞

z

2

7→ 0

z

3

7→ 1

z

4

7→ λ

We know that gf

−1

sends

f(z

1

)

f

−1

7−−→ z

1

g

7−→ ∞

f(z

2

)

f

−1

7−−→ z

2

g

7−→ 0

f(z

3

)

f

−1

7−−→ z

3

g

7−→ 1

f(z

4

)

f

−1

7−−→ z

4

g

7−→ λ

So [f (z

1

), f(z

2

), f(z

3

), f(z

4

)] = gf

−1

f(z

4

) = g(z

4

) = λ.

In fact, we can see from this proof that: given

z

1

, z

2

, z

3

, z

4

all distinct and

w

1

, w

2

, w

3

, w

4

distinct in

C

∞

, then

∃f ∈ M

with

f

(

z

i

) =

w

i

iff [

z

1

, z

2

, z

3

, z

4

] =

[w

1

, w

2

, w

3

, w

4

].

Corollary. z

1

, z

2

, z

3

, z

4

lie on some circle/straight line iff [z

1

, z

2

, z

3

, z

4

] ∈ R.

Proof.

Let

C

be the circle/line through

z

1

, z

2

, z

3

. Let

g

be the unique M¨obius

map with

g

(

z

1

) =

∞

,

g

(

z

2

) = 0,

g

(

z

3

) = 1. Then

g

(

z

4

) = [

z

1

, z

2

, z

3

, z

4

] by

definition.

Since we know that M¨obius maps preserve circle/lines,

z

4

∈ C ⇔ g

(

z

4

) is on

the line through ∞, 0, 1, i.e. g(z

4

) ∈ R.