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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 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 $z \rightarrow \omega(z)$.

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    $\begingroup$ 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. $\endgroup$
    – Theo Johnson-Freyd
    Commented Aug 26, 2012 at 4:19
  • $\begingroup$ thanks for the input. i fixed the typo in the author's name and made specific that h is not necessarily and integer. $\endgroup$
    – Yaniel Cabrera
    Commented Aug 26, 2012 at 18:45
  • $\begingroup$ 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. $\endgroup$
    – orbifold
    Commented Nov 28, 2012 at 21:37
  • $\begingroup$ Cross-posted from physics.stackexchange.com/q/34741/2451 $\endgroup$
    – Qmechanic
    Commented Sep 1, 2014 at 13:22

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