# Translation and Dilation transformations within the conformal group

I am using Di Francesco's book P.39. The equation that the generators of the transformations satisfy is given by: $$iG_a \Phi = \frac{\delta x^{\mu}}{\delta w_a} \partial_{\mu} \Phi - \frac{\delta F}{\delta w_a},$$ where $\left\{w_a\right\}$ are a set of parameters for the transformation and $G_a$ is the corresponding generator. $F = F(\Phi(\mathbf{x}) )= \Phi'(\mathbf{x'})$ and $\Phi \equiv \Phi(\mathbf{x})$

He then considers some examples. For a translation, $\mathbf{x'} = \mathbf{x} + \mathbf{a}$ and $F = \Phi'(\mathbf{x+a}) = \Phi(\mathbf{x})$. I suppose the last equality there is a supposition (i.e we impose the condition that the field is invariant under translations in the coordinates). In this case, $F = \text{Id}$. I guess this warrants Francesco's statement that $\delta F/\delta w^v = 0$, but this is not general right? It is only in the case when the fields are not affected by the transformation?

Now consider a dilation. $\mathbf{x'} = \lambda \mathbf{x}$ and $F = \Phi'(\lambda \mathbf{x}) = \lambda^{-\Delta} \Phi(\mathbf{x})$, where $\Delta$ is the scaling dimension of the field. Could someone explain this last equality?

We can use the first equation above to find the generator of dilations. In this case, $w_a = \lambda$ and $x'^{\mu} - x^{\mu} = \lambda x^{\mu} - x^{\mu} \Rightarrow \delta x^{\mu}/\delta \lambda = x^{\mu}.$ I know that the generator is supposed to be $D = -ix^{\mu} \partial_{\mu}$ which seems to mean that $\delta F/\delta \lambda = 0$ But how so? By chain rule, $$\frac{\delta F}{\delta \lambda^{-\Delta}} \frac{\delta \lambda^{-\Delta}}{\delta \lambda} = -\Delta \lambda^{-1-\Delta} \Phi,$$

Many thanks.

The first equality for translations is true for any field in general. Since translations form an Abelian group, their irreducible representations are one-dimensional. Thus, under a translation, it is generally true for an irreducible field (for any spin) $\Phi'(x+a) = \Phi(x)$. Of course, the situation changes when the Lorentz group is considered. In this case, the field also transforms nontrivially and $F \neq Id$.
The second case with dilations is not generally true. It is only true for a class of fields called quasi-primary fields that transform as $\Phi'(\lambda x) = \lambda^{-\Delta} \Phi(x)$. However, in a CFT, it is a fact that the complete set of fields can be grouped into "conformal families", each with its representative primary field and its descendants. Transformation properties of the descendants can be derived from its corresponding primary field. Thus, it is often the case in CFTs, that one only discusses the transformation of primary fields. This is the reason Francesco chooses to discuss the example that he did rather than something else.
• @Prahar: How does irreducible representations of abelian group being 1D guarantee the scalar field will transform trivially? It could transform like $(1+a)\Phi(x)$ also right? – user7757 Jun 18 '14 at 5:34