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?

Duplicate(s) in PhySE :

D1 is unanswered.

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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.


  1. Francesco P., Mathieu P., Senechal D., Conformal field theory
  • $\begingroup$ Can you explain why (4.31) in Di Francesco doesn't answer your question? Maybe you want the finite transformations? $\endgroup$
    – d_b
    Dec 16, 2019 at 19:33
  • $\begingroup$ 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? $\endgroup$
    – Adam
    Dec 16, 2019 at 22:03
  • $\begingroup$ @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. $\endgroup$
    – Ajay Mohan
    Dec 17, 2019 at 5:53
  • $\begingroup$ @d_b Yes, as you said, I want to know this: $\Phi'(x)=(?)\Phi((g^{-1}(b))x)$. How does one find this using: $K_{\mu} \Phi(x)$? $\endgroup$
    – Ajay Mohan
    Dec 17, 2019 at 6:01
  • $\begingroup$ @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). $\endgroup$
    – Ajay Mohan
    Dec 17, 2019 at 6:18

2 Answers 2


The fields transform under finite conformal transformations as$^1$ $$ \Phi^a(x') \mapsto {\Phi^a}'(x) = \Omega(x')^\Delta\,D(R(x'))^{\phantom{b}a}_b \,\Phi^b(x')\,. \tag{1}\label{main} $$ as given in equation $(55)$ of $[1]$. Let's break it down:

  • $\Delta$ is the conformal dimension of $\Phi$.
  • $\Omega$ is the conformal factor of the transformation.
  • $D$ is the spin representation of $\Phi$.
  • $R$ is the rotation Jacobian of the transformation

So let's compute these things. The spin and the conformal dimension $\Delta$ are given. The first thing we have to look at is the Jacobian. $$ \frac{\partial x^{\prime \mu}}{\partial x^\nu} = \Omega(x') R^\mu_{\phantom{\mu}\nu}(x')\,. $$ This implicitly defines both $\Omega$ and $R$ and it is not ambiguous because we require $R \in \mathrm{SO}(d)$, namely $$ R^{\mu}_{\phantom{\mu}\nu} \,\eta^{\nu\rho}\,R^{\lambda}_{\phantom{\lambda}\rho} \,\eta_{\lambda\kappa}= R^{\mu}_{\phantom{\mu}\kappa}\,. $$ You can immediately see that for the Poincaré subgroup of the conformal group $\Omega(x')= 1$, whereas for dilatations $\Omega(x') = \lambda$ and for special conformal transformations $$ \Omega(x') = \frac{1}{1+2(b\cdot x') + b^2 {x'}^2}\,.\tag{2}\label{omega} $$ This can be proven with a bit of algebra. Using your definition of SCT $$ {x'}^\mu = \frac{x^\mu - b^\mu x^2}{1+-2(b\cdot x) + b^2 {x}^2}\,, $$ one can check $$ \frac{\partial x^{\prime \mu}}{\partial x^\rho} \eta_{\mu\nu} \frac{\partial x^{\prime \mu}}{\partial x^\lambda} = \frac{\eta_{\rho\lambda}}{(1-2(b\cdot x) + b^2 x^2)^2}\,. $$ That means that the Jacobian is an orthogonal matrix up to a factor, which is the square root of whatever multiplies $\eta_{\rho\lambda}$. Then we have to re-express that as a function of $x'$. After some algebra again one finds that it suffices to change the sign to the term linear in $b$.

Finally, how does one compute $R$? Well, it's just the Jacobian divided by $\Omega$. In the case of special conformal transformations one has (there might be mistakes, redo it for safety) $$ R^{\mu}_{\phantom{\mu}\nu} = \delta^\mu_\nu + \frac{2 b_\nu x^\mu - 2 b^\mu (b_\nu x^2+ x_\nu - 2 (b\cdot x) x_\nu) -2b^2 x^\mu x_\nu }{1-2b\cdot x +b^2 x^2}\,, $$ which, as before, needs to be expressed in terms of $x'$.

If you are interested in $\Phi$ scalar then $D(R) = 1$ and you can just plug \eqref{omega} into \eqref{main} to obtain the transformation. If you want to consider also spinning $\Phi$ then it's not much harder.

For spin $\ell=1$ the $D$ is just the identity, namely $$ D(R)^{\phantom{\nu}\mu}_\nu = R^{\phantom{\nu}\mu}_\nu\,. $$ For higher spins one just has to take the product $$ D(R)^{\phantom{\nu_1\cdots \nu_\ell}\mu_1\cdots \mu_\ell}_{\nu_1\cdots \nu_\ell} = R^{\phantom{\nu_1}\mu_1}_{\nu_1}\cdots R^{\phantom{\nu_\ell}\mu_\ell}_{\nu_\ell}\,. $$ Again, by plugging these definitions in \eqref{main} you obtain the desired result.

$\;[1]\;\;$TASI Lectures on the Conformal Bootstrap, David Simmons-Duffin, 1602.07982

$\;{}^1\;\;$The way the transformations are written in the lecture notes linked above differs a bit from Di Francesco Mathieu Sénéchal. The difference is that Di Francesco et al. make an active transformation $x \to x'$ with $$ \Phi(x) \mapsto \Phi'(x') = \mathcal{F}(\Phi(x))\,, $$ while David Simmons Duffin makes essentially the inverse transformation $x' \to x$ $$ \Phi(x') \mapsto \Phi'(x) = \mathcal{F}^{-1}(\Phi(x'))\,. $$ That is why in the above discussion the indices of $R^\mu_{\phantom{\mu}\nu}$ get swapped when passed inside $D$ as $D(R) = R^{\phantom{\nu}\mu}_{\nu}$. And that's also why we get a factor $\lambda^\Delta$ rather than $\lambda^{-\Delta}$ as Di Francesco et al. This is all consistent as long as it is clear what one is doing.


$\newcommand \act\vartriangleright$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.

  • $\begingroup$ 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 ? $\endgroup$ Dec 17, 2019 at 3:13
  • $\begingroup$ 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. $\endgroup$
    – Gonenc
    Dec 17, 2019 at 3:19
  • $\begingroup$ @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? $\endgroup$
    – Ajay Mohan
    Dec 17, 2019 at 6:43
  • $\begingroup$ @Gonenc If that's the case, then why hasn't Di Francesco defined special conformal transformations in a similar manner? $\endgroup$
    – Ajay Mohan
    Dec 17, 2019 at 6:46

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