Understanding Higher Order Poles in Conformal Transformations

[BenTheMan]

I suppose this is the proper place for this question:)

I am learning about conformal field theories and have a question about poles of order > 1.

If a conformal transformation acts as

$$z \rightarrow f(z)$$,

f(z) must be both invertable and well-defined globally. I want to show that f(z) must have the form

$$f(z) = \frac{az+b}{cz+d}$$.

Here is what I have so far. A general complex function f(z) can have branch cuts, essential singularities and multiple poles of varying order.

If f(z) has a branch cut then it is not well-defined in the vicinity of the branch point---i.e. it is multi-valued.

If f(z) has an essential singularity, then the transformaiton is not invertable.

So f(z) must look like

$$f(z) = \frac{P(z)}{Q(z)}$$,

where P(z) and Q(z) are power series in z. If f(z) has multiple poles (of any order), then it is not well-defined at those points. That is, if f(z) has poles at z0, z1, ..., then it maps all of these points to 0.

The case I am having trouble with is when f(z) has higher order poles, i.e. poles of order n > 1. The textbook I am using ("Conformal Field Theories" by Di Francesco, Mathieu, and Senechal) says "the image of a small neighborhood of z0 (the location of the pole) is wrapped n times around z0, and thus is not invertable". Can someone explain this statement to me? In what sense is there "wrapping" about the higher order poles?

 

[mathwonk]

the way to prove this result is to first show all automorphisms of C arelinear (using weierstrass and fund thm of algebra).

then to find all automorphisms of the sphere it suffices to find a subgroup of them that acts transitively and contains all automorphisms fixing one point.

but the linear fractional transformations do act transitively, and they contain the stabilizer of infinity, namely the linear transformations, by the previous result. QED.

or see a good complex book like cartan.

 

[quasar987]

If a conformal transformation acts as

$$z \rightarrow f(z)$$,

f(z) must be both invertable and well-defined globally. I want to show that f(z) must have the form

$$f(z) = \frac{az+b}{cz+d}$$.

There's something I don't get.

A transformation of the form you wrote is called a homographic map (or a Möbius transformation). A homographic map is just a composition of 4 conformal maps so it is conformal. But unless I'm mistaken, you're asked to show the converse of that, namely that every conformal map is a homographic.

This doesn't sound right. Can someone enlighten me?

 

[mathwonk]

he is aparently trying to prove every holomorphic automorphism of the riemann sphere has that form. i have said how to prove it.

 

[BenTheMan]

he is aparently trying to prove every holomorphic automorphism of the riemann sphere has that form. i have said how to prove it.

Yes, and thank you:)

 

[quasar987]

Ok, so if we acknowledge the extended complex plane as a field, then all holomorphic conformal field automorphisms are homographic.

 

[BenTheMan]

What does homographic mean?

 

[quasar987]

$$f(z) = \frac{az+b}{cz+d}$$.

A transformation of this form is called a homographic map or a Möbius transformation.

 

[BenTheMan]

Is there any deep reason why it's called a Mobius transformation? Was he just the first one to fool with them?