# Equivalent Definitions of Primary Fields in CFT

+ 2 like - 1 dislike
5799 views

I have come across two similar definitions of primary fields in conformal field theory. Depending on what I am doing each definition has its own usefulness. I expect both definitions to be compatible but I can't seem to be able to show it. By compatible I mean definition 1 $\iff$ definition 2. I will write both definitions in the two-dimensional case and restricting to holomorphic transformations.

Def #1 from D'Francesco et al's CFT: A field $f(z)$ if it transforms as $f(z) \rightarrow g(\omega)=\left( \frac{d\omega}{dz}\right)^{-h}f(z), h\in\mathbb{R}$ under an infinitesimal conformal transformation $z \rightarrow \omega(z)$.

Def #2 from Blumenhagen et al's Intro to CFT: A field $f(z)$ is primary if it transforms as $f(z) \rightarrow g(z)=\left( \frac{d\omega}{dz}\right)^{h}f(\omega), h\in\mathbb{R}$ under an infinitesimal conformal transformation $z \rightarrow \omega(z)$.

This post imported from StackExchange MathOverflow at 2014-08-31 09:09 (UCT), posted by SE-user Daniel

Closed as per community consensus as the post is Not graduate-level upwards
closed Sep 4, 2014
Dear Daniel, What is $h$ (a standard part of CFT, but you should include it for completeness to improve the question)? And why aren't these manifestly equivalent? Note that $\bigl(\frac{dw}{dz}\bigr)^{-1} = \frac{dz}{dw}$. To see this hands-on, think about primary fields with $h=1$, which should be precisely one-forms on your curve (or maybe that's $h=-1$. As an aside, if you have been reading the notes I'm thinking of, then it's "Di Francesco", not "Francesco", I believe.

This post imported from StackExchange MathOverflow at 2014-08-31 09:09 (UCT), posted by SE-user Theo Johnson-Freyd
thanks for the input. i fixed the typo in the author's name and made specific that h is not necessarily and integer.

This post imported from StackExchange MathOverflow at 2014-08-31 09:09 (UCT), posted by SE-user Daniel
Primary fields are "operator valued" sections of the canonical bundle $K_S$ of your surface $S$ raised to some power $h$. The transformation rule simply expresses that fact. This should be a comment, but I do not have enough reputation to comment.

This post imported from StackExchange MathOverflow at 2014-08-31 09:09 (UCT), posted by SE-user orbifold
Cross-posted to mathoverflow.net/q/105489

This post imported from StackExchange Physics at 2014-09-01 15:35 (UCT), posted by SE-user Qmechanic

This is an obvious notation confusion that I answered already, just reposted on mathoverflow. It's probably already imported, but it's not grad level anyway, just replace letters to get the second transformation law from the first.

+ 4 like - 0 dislike

In the second definition, switch the two coordinate names "z" and "w" with each other, and remember that

$${dz\over dw} = ({dw\over dz})^{-1}$$

and then you see it's the same as the first.

answered Aug 23, 2012 by (7,720 points)
I don't think that is all b/c then the transformation is $f(\omega) \rightarrow g(\omega) = ...$ which is different from the first definition.

This post imported from StackExchange Physics at 2014-09-01 15:35 (UCT), posted by SE-user yca

@yca: What I wrote is the complete and correct answer. I didn't use "f" and "g", it doesn't matter what name you give to the relation between w and z.

The first definition is a relation between the new field $g$ evaluated at the new point and the old field $f$ at the the old point. The second definition is a relation between new field at the old point and the old field at the new point. Simply changing the names of the variables in one of the definitions does not give me the other.

This post imported from StackExchange Physics at 2014-09-01 15:35 (UCT), posted by SE-user yca

@yca: Yes, it is a simple swap of names. You just change the names of the variables. That's it. The points are arbitrary, so which one is "old" and which one is "new" doesn't matter. The transformation from z to w is inverse to the one from w to z. Please think about it more, there is nothing further one can say. This is the complete answer.

Thanks. I will wait for some other answer. If nothing comes up I will move these question to MathOverflow.

This post imported from StackExchange Physics at 2014-09-01 15:35 (UCT), posted by SE-user yca

@yca: there is no other answer! This is a complete and correct answer. Math overflow people will laugh at you and say the same thing. I have absolutely no confusion, and you should read it, learn it, understand it, and move on.