10Mobius group

IA Groups

10.2 Fixed points of M¨obius maps

Definition (Fixed point). A fixed point of f is a z such that f(z) = z.

We know that any M¨obius map with

c

= 0 fixes

∞

. We also know that

z → z

+

b

for any

b 6

= 0 fixes

∞

only, where as

z 7→ az

for

a 6

= 0

,

1 fixes 0 and

∞

. It turns out that you cannot have more than two fixed points, unless you

are the identity.

Proposition.

Any M¨obius map with at least 3 fixed points must be the identity.

Proof.

Consider

f

(

z

) =

az+b

cz+d

. This has fixed points at those

z

which satisfy

az+b

cz+d

=

z ⇔ cz

2

+ (

d − a

)

z − b

= 0. A quadratic has at most two roots, unless

c = b = 0 and d = a, in which the equation just says 0 = 0.

However, if c = b = 0 and d = a, then f is just the identity.

Proposition.

Any M¨obius map is conjugate to

f

(

z

) =

νz

for some

ν 6

= 0 or to

f(z) = z + 1.

Proof.

We have the surjective group homomorphism

θ

:

GL

2

(

C

)

→ M

. The

conjugacy classes of GL

2

(C) are of types

λ 0

0 µ

7→ g(z) =

λz + 0

0z + µ

=

λ

µ

z

λ 0

0 λ

7→ g(z) =

λz + 0

0z + λ

= 1z

λ 1

0 λ

7→ g(z) =

λz + 1

λ

= z +

1

λ

But the last one is not in the form

z

+ 1. We know that the last

g

(

z

) can

also be represented by

1

1

λ

0 1

, which is conjugate to

1 1

0 1

(since that’s its

Jordan-normal form). So z +

1

λ

is also conjugate to z + 1.

Now we see easily that (for

ν 6

= 0

,

1),

νz

has 0 and

∞

as fixed points,

z

+ 1

only has ∞. Does this transfer to their conjugates?

Proposition. Every non-identity has exactly 1 or 2 fixed points.

Proof.

Given

f ∈ M

and

f 6

=

id

. So

∃h ∈ M

such that

hfh

−1

(

z

) =

νz

. Now

f

(

w

) =

w ⇔ hf

(

w

) =

h

(

w

)

⇔ hfh

−1

(

h

(

w

)) =

h

(

w

). So

h

(

w

) is a fixed point

of

hfh

−1

. Since

h

is a bijection,

f

and

hfh

−1

have the same number of fixed

points.

So

f

has exactly 2 fixed points if

f

is conjugate to

νz

, and exactly 1 fixed

point if f is conjugate to z + 1.

Intuitively, we can show that conjugation preserves fixed points because if we

conjugate by

h

, we first move the Riemann sphere around by

h

, apply

f

(that

fixes the fixed points) then restore the Riemann sphere to its original orientation.

So we have simply moved the fixed point around by h.