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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 Francesco CFT: A field $f(z)$ is primary if it transforms as $f(z) \rightarrow g(\omega)=\left( \frac{d\omega}{dz}\right)^{-h}f(z)$ under an infinitesimal conformal transformation $z \rightarrow \omega(z)$.

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

Can someone show me how they are indeed the same?

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Cross-posted to mathoverflow.net/questions/105489 –  Qmechanic Mar 25 '13 at 22:16
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1 Answer

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.

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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. –  yca Aug 23 '12 at 17:42
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@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. –  Ron Maimon Aug 24 '12 at 3:09
    
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. –  yca Aug 24 '12 at 13:43
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@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. –  Ron Maimon Aug 24 '12 at 14:23
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@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. –  Ron Maimon Aug 25 '12 at 20:16
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