# How do fields transform under special conformal transformations?

## A Question in Classical Field Theory

$$\underline{\text{Assumption 1}}$$: The definition of a transformation specifies how both the coordinates and the fields transform: They are namely $$(1$$-$$1)$$ and $$(1$$-$$2)$$. $$x'^{\mu}=x'^{\mu}(x) \tag{1-1}$$ $${\Phi'}(x')=\mathcal{F}(\Phi(x)) \tag{1-2}$$

I've seen the definitions for translations, Lorentz transformations, and dilatations. For example, the definition for translation is given as follows: $$x'^{\mu} = x^{\mu}+a^{\mu} \tag{i-1}$$ $$\Phi'(x)=\Phi(x-a) \tag{i-2}$$

However, after considerable searches on various online sources, I have only been able to find $$(1$$-$$1)$$ for special conformal transformations and not $$(1$$-$$2)$$.

$${x'}^{\mu}=\frac{x^{\mu}-b^{\mu}x^2}{1-2b\cdot x + b^2 x^2} \tag{ii-1}$$

Question: How do fields transform under special conformal transformations?

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## My Thoughts

I have read Francesco's book [1] on this topic. But I'm unable to follow his arguments in sections $$4.1$$ and $$4.2$$ of his book.

In section $$4.1$$, he finds the generators of the conformal group assuming that the fields transform trivially ($$\Phi'(x')=\Phi(x)$$)$$^\mathbf{*}$$, which are provided in $$(4.18)$$ of the book. He uses these generators to determine the conformal algebra in $$(4.19)$$.

$$^\mathbf{*}$$ This goes against Assumption 1. For example, if we consider Lorentz transformations, we know that $$\Phi'(x') = \Phi(x)$$ is true only for scalar fields.

Any help would be greatly appreciated. Just keep in mind that for the purposes of this question, the relevant field of interest is classical field theory and not quantum field theory.

References:

1. Francesco P., Mathieu P., Senechal D., Conformal field theory
• Can you explain why (4.31) in Di Francesco doesn't answer your question? Maybe you want the finite transformations? – d_b Dec 16 '19 at 19:33
• You have to choose how the field transforms. Clearly, they assume that the field transform trivially in this book. This does not go against your assumption 1, it is one possibility. Could it be the only allowed one for conformal transformations? – Adam Dec 16 '19 at 22:03
• @d_b Let's see what he does for Poincare transformations (4.28). The way I follow his arguments: (1) Assumes fields transform trivially to get the conformal algebra (4.19) $\to$ (2) Shows how the generators act on $\Phi(x)$, by translating the generator $L_{\mu \nu}$ using the conformal algebra in (4.19) (which was derived assuming that fields transform trivially). I don't understand this step. – Ajay Mohan Dec 17 '19 at 5:53
• @d_b Yes, as you said, I want to know this: $\Phi'(x)=(?)\Phi((g^{-1}(b))x)$. How does one find this using: $K_{\mu} \Phi(x)$? – Ajay Mohan Dec 17 '19 at 6:01
• @Adam In $(2.121)$ for example,he defines dilatations by showing both how the coordinates and the fields transform. The fields transform like this: $\Phi'(x) = \lambda^{-\Delta}\Phi(\lambda^{-1}x)$. To me, defines is a strong word: It means that, under dilatations, there is only one way for the fields to transform that is accepted by all humans (I agree that the choice is made arbitrarily by humans). – Ajay Mohan Dec 17 '19 at 6:18

$$\newcommand \act\vartriangleright$$I think you are in general confused about how symmetries act on fields. If we have a symmetry group $$G$$ and coordinates transform as $$x \to g \act x$$ for $$g \in G$$, then fields should transform as

$$\phi(x) \to R(g)\phi(g \act x)$$

where $$R(g)$$ stands for a particular representation of $$G$$. You can think of representations as "transformation rules". They satisfy $$R(gh) = R(g)R(h)$$. Note that any group has a trivial representation given by $$R(g) = \text{Id}$$ i.e. the identity for any $$g \in G$$. Let's take the example of Lorentz group. Then our fields will transform as

$$\phi(x) \to R(\Lambda) \phi (\Lambda^\mu_{\;\;\nu}\, x^\nu)$$

of course the question is what $$R(\Lambda)$$ is. If you chose a scalar particle, which mean by definition that the representation you chose is the trivial representation $$R(\Lambda)=\text{Id}$$. Then the field transforms as

$$\phi(x) \to \phi (\Lambda^\mu_{\;\;\nu}\, x^\nu)$$

If you have a vector field, which by definition means that you chose the fundamental (or defining) representation (think about the usual Lorentz transformation matrices), the fields transform as

$$A^\rho(x) \to \Big( R(\Lambda) A(\Lambda x) \Big)^\rho =\Lambda^\rho_{\;\;\delta}\, A^\delta(\Lambda x)$$

If you have a two-tensor (for example $$F_{\mu\nu}$$ of electrodynamics or $$g_{\mu\nu}$$ the metric, then

$$F^{\mu\nu}(x) \to \Big( R(\Lambda) F(\Lambda x) \Big)^{\mu\nu} =\Lambda^\mu_{\;\;\rho}\,\Lambda^\nu_{\;\;\delta}\, F^{\rho\delta}(\Lambda x)$$

Note that Di Francesco "cheats" a bit. He is actually talking about the transformation of spacetime coordinate, which is in the defining representation of the conformal group to derive the conformal algebra. So he uses the transformation

$$x^\mu \to\big( R_{\text{defining}} (g) x\big)^\mu$$

which are given in (4.15) to derive the infinitesimal version of this algebra. Since the algebra is defined by the defining representation, the result follows. I think he says consider a field that doesn't transform to simplify the algebra that he has to do.

If you are for example interested in how a spin 1 field concretely transforms under (any) conformal transformation, you can check Rychkov's lecture notes in particular chapter 2.

• These lectures that I was following youtube.com/playlist?list=PLDfPUNusx1Ep5g1jIKXqpNX_t_Zz-kAlQ mention that the fields discussed in CFTs are projective representations. I suppose it means R(g) in your is a projective rep ? Is that correct ? Could you please comment on that since it is closely related to your answer ? – symanzik138 Dec 17 '19 at 3:13
• Projective representation means that $R(g h) = c(g,h) R(g)R(h)$ for some constant $c(g,h)$. I didn't want to get into the nitty gritty of representation theory in my answer though. – Gonenc Dec 17 '19 at 3:19
• @Gonenc You say the way a field transforms is determined by the representation $R(g)$. For example: In $(2.121)$, Di Francesco defines dilatations by the two equations (one for how the coordinates transform and one for how the fields transform). Defines is a strong word: With your understanding, I assume it means that humans all over the world have decided that $R(\lambda)=\lambda^{-\Delta_{\Phi}}$ for dilatations (where $\Delta_{\Phi}$ is a number assigned to operator $\Phi$ called the scaling dimension). Am I correct in making this statement? – Ajay Mohan Dec 17 '19 at 6:43
• @Gonenc If that's the case, then why hasn't Di Francesco defined special conformal transformations in a similar manner? – Ajay Mohan Dec 17 '19 at 6:46