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?

Duplicate(s) in PhySE : 


*

*D1: How tensor fields transform under special conformal transformation?
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.
References:


*

*Francesco P., Mathieu P., Senechal D., Conformal field theory
 A: 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.
A: $\newcommand \act\vartriangleright$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.
