I want to consider conformal maps on suitable compactifications of $\mathbb{R}^{n}$. I know that a special conformal transformation: $$x_i\mapsto\frac{x_i-x^{2}b_i}{1-2b\cdot x+b^{2}x^{2}}$$ can be written as a composition of a spherical inversion, a translation (by $b$) and another inversion about the same circle. Since the conformal group is generated by special conformal transformations, translations, dilations and rotations, it should be possible to write the spherical inversion: $$x_i\mapsto\frac{x_i}{x^{2}}$$ as a composition of these maps. I can't, however, think of such a representation. How can it be obtained?
[Physics] Spherical inversion in terms of special conformal transformation
conformal-field-theory
Related Solutions
Fist, let me just say that this book will answer (almost) all of your questions with beautiful precision and wonderful detail; I highly highly recommend that you read the chapter on the conformal group from which I am stealing the following information.
Let me repeat one of the definitions that directly answers your question in the case of the pseudo-Euclidean manifolds $\mathbb R^{p,q}$;
Definition. The conformal group $\mathrm{Conf}(\mathbb R^{p,q})$ is the connected component containing the identity of the group of conformal diffeomorphisms of the conformal compactification of $\mathbb R^{p,q}$.
With this definition, one can show that for $p+q>2$, the group $\mathrm{Conf}(\mathbb R^{p,q})$ is isomorphic to either $\mathrm{SO}(p+1, q+1)$ or $\mathrm{O}(p+1, q+1)/\{\pm 1\}$ if $-1$ is in the connected component of $\mathrm O(p+1, q+1)$ containing $1$.
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.
Best Answer
The special conformal transformations as well as the translations, dilations and rotations are all continuously connected to the identity. This means that they contain parameters such that at some particular value the trasformation becomes trivial. For example, for $b=0$ the special conformal trasformation you write is simply $x_i\mapsto x_i$. The inversion map $$x_i \mapsto \frac{x_i}{x^2},$$ on the other hand, is not connected to the identity.
We can also note that an inversion changes the orientation of the space, while the other conformal transformations preserve the orientation. In $D$ (Euclidean) dimensions the special conformal transformations, translations, dilations and rotations together form the Lie group $SO(D+1,1)$. This is what, at least in physics, is normally referred to as "the conformal group". If we also allow for inversions this group is extended to $O(D+1,1)$.