Defining an inversion transformation in coordinates as

$$ x^\mu\rightarrow \mathcal{I}x^\mu = \frac{x^\mu}{x^2}, \tag1 $$

if we want to study these transformations on tensor operators $\mathcal{O}$ we could make use of the usual transformation rule, this is

$$ \mathcal{O}^\prime_{\nu_1\nu_2\cdots\nu_n}(\mathcal{I}x) = \mathcal{O}_{\mu_1\mu_2\cdots\mu_n}(x) \frac{\partial x^{\mu_1}}{\partial \mathcal{I}x^{\nu_1}} \frac{\partial x^{\mu_2}}{\partial \mathcal{I}x^{\nu_2}} \cdots \frac{\partial x^{\mu_n}}{\partial \mathcal{I}x^{\nu_n}} \tag2 $$

Because of (1), $$ \frac{\partial(\mathcal{I}x^\mu)}{\partial x^\nu} = \frac{1}{x^2} \mathcal{I}^\mu_{\nu}(x) \Rightarrow \frac{\partial x^\mu}{\partial \mathcal{I}x^\nu} = x^2 \mathcal{I}^\mu_{\nu}(x) \tag3 $$ where $\mathcal{I}^\mu_{\nu}(x) = \delta^\mu_\nu - 2\frac{x^\mu x_\nu}{x^2}$ is the so-called inversion tensor.

Therefore, introducing (3) into (2) we should get $$ \mathcal{O}^\prime_{\nu_1\nu_2\cdots\nu_n}(\mathcal{I}x) = (x^2)^n \mathcal{I}^{\mu_1}_{\nu_1}\mathcal{I}^{\mu_2}_{\nu_2}\cdots\mathcal{I}^{\mu_n}_{\nu_n} \mathcal{O}_{\mu_1\mu_2\cdots\mu_n}(x) \tag4 $$

Nevertheless, if you check these notes, section 2.2. Conformal transformations of fields, it is claimed that the power of $x^2$ should be the scaling dimension of the operator, $\Delta$, instead of $n$ as it seems to be from the tensor transformation rule.

Why is $\Delta$ (and not $n$) the correct power? How to actually prove that the inversion transformation of the tensor has the form of (4) but with $(x^2)^n$ exchanged by $(x^2)^\Delta$ (eq (2.46) in aforementioned notes)?

  • $\begingroup$ You are taking a transformation law from a theory that has more symmetry (general covariance) and hoping it works in a theory that only has conformal symmetry. To see that (2) is wrong and should be replaced by Osborn's (2.45), you don't need inversions. A global rescaling of co-ordinates would also fail to capture $\Delta$. $\endgroup$ Commented Sep 12, 2022 at 13:55
  • $\begingroup$ @ConnorBehan then how does one deduce Osborn's (2.45)? And why (2) is wrong? I don't understand it, please elaborate $\endgroup$
    – Vicky
    Commented Sep 12, 2022 at 22:11
  • $\begingroup$ Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Commented Oct 31, 2022 at 12:10

1 Answer 1


You are considering a very special type of field, a conformal primary field. It can be defined as follows:

In $d>2$ spacetime dimensions, a conformal primary field $\hat{\phi}^M_{\rho}(x)$, in the $\rho$ representation of the Lorentz group and with conformal dimension $\Delta$, transforms under a conformal transformation $\eta_{\mu \nu}\mapsto \Omega^2(x)\eta_{\mu \nu}$ as \begin{equation}\label{eq:def2} \hat{\phi'}^M_{\rho}(x')=\Omega^{\Delta}(x)\mathcal{D}{\left[R(x)\right]^M}_{N}\hat{\phi}^N_{\rho}(x) \end{equation} where ${R^{\mu}}_{\nu}(x)=\Omega^{-1}(x)\frac{\partial x^{\mu}}{\partial x'^{\nu}}$ and $\mathcal{D}{\left[R(x)\right]^M}_{N}$ implements the action of $R$ in the $SO(d-1,1)$ representation of $\hat{\phi}^{M}_{\rho}(x)$.

What is $\Omega$ for an inversion $x'{_\mu}=x_{\mu}/x^2$? Use the definition of conformal transformation:

\begin{equation} \frac{\partial x^{\rho}}{\partial {x'}^{\mu}}\frac{\partial x^{\sigma}}{\partial x'^{\nu}}\eta_{\rho \sigma}=\Omega^2(x)\eta_{\mu \nu}, \end{equation}

from which you get $\Omega(x)=x^2$.

That's why a conformal primary transforms with a factor of $x^{2\Delta}$ under inversions. Fur further reading see for example this paper.

  • $\begingroup$ But you're starting by defining or assuming that conformal fields transform as shown in your 1st equation. I´d say that my question is precisely related to the actual prove of such a formula $\endgroup$
    – Vicky
    Commented Nov 30, 2022 at 22:59
  • $\begingroup$ Yes, I mean, that is the definition of conformal primary. There exists actually another, maybe more intuitive, definition in terms of the generators of the conformal group. For a proof of the equivalence of both definitions and for further discussions, see e.g this paper $\endgroup$ Commented Dec 1, 2022 at 12:53

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