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So, under conformal transformations $$x\mapsto x'\\ \phi\mapsto\phi'(x')=\Omega^{(2-D)/2}\phi(x),$$ where $$\eta_{\mu\nu}\frac{\partial x^\mu}{\partial x^{'\alpha}}\frac{\partial x^\nu}{\partial x^{'\beta}}=\Omega^{-2}(x)\eta_{\alpha\beta},$$ the action transforms like $$S\mapsto\int d^Dx\,\Omega^{D-2}\partial_\mu(\Omega^{(2-D)/2}\phi)\partial^\mu(\Omega^{(2-D)/2}\phi)$$ (A quick way to get to this equations is by instead considering the associated Weyl transformation as described in an answer in Simple conceptual question conformal field theory). It is then obvious that the action is invariant under scale transformations, i.e. when $\Omega$ is constant. However, why is it invariant when $\Omega$ is not constant? What does one do with the terms involving derivatives of $\Omega$?

Ivan Burbano
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    The only non-constant $\Omega$ for which this will still be invariant is the one corresponding to special conformal transformations. – Connor Behan Jun 23 '21 at 23:32
  • So this theory is not Weyl invariant? In any case, let us assume that $\Omega$ is restricted by the equation relating the metrics in the different coordinate systems. This should be enough to proof the invariance of the action since that restriction is already enough to show that $\Omega$ comes from dilations or special conformal transformations (or compositions thereof). – Ivan Burbano Jun 23 '21 at 23:40
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    I think this is because $\phi$ is not a primary field. Rather, $\partial\phi$ is a primary field, and the action is invariant under conformal transform. This is true in 2d. For general dimensions, I am not sure. – Youran Jun 25 '21 at 12:28
  • Oh yeah, for sure. So the general action is Weyl invariant (so the $\Omega$ doesn't have to come from a conformal transformation) if we scale $\partial\phi$ instead of $\phi$. I remember something similar happening in electrodynamics (maybe in a book by Wald?). – Ivan Burbano Jun 25 '21 at 14:47
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    $\partial_\mu \phi$ is not primary in $d > 2$. – Connor Behan Jun 28 '21 at 16:10
  • @ConnorBehan I don't think see why it should work for $\Omega$ corresponding to special conformal transformations. The approach I tried was to see if it worked for inversions. The action is not invariant for these, even up to total derivatives because one ends up with an annoying $(D-2)^2x^{-2}\phi^2$ factor – Ivan Burbano Nov 05 '21 at 14:37
  • I can't remember if it's supposed to work for a single inversion. But it looks to fast to say that each $\partial_\mu$ simply picks up the Jacobian to the power of $-\frac{1}{D}$. – Connor Behan Nov 05 '21 at 17:09
  • @ConnorBehan It is not that each derivative picks up a factor of $\Omega^{-1}$. It is just that, after contraction with the inverse metric in the kinetic term this is what pops out. In any case, thanks for this conversation. It turns out that everything works out (I should also thank Bruno de Souza Leão Torres for conversations on this). I will try to type it as an answer in case anyone finds it useful in the future. – Ivan Burbano Nov 18 '21 at 13:49

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As mentioned by @ConnorBehan, the result is only true for Weyl factors coming from conformal transformations. Bruno de Souza Leão Torres stressed to me the importance that said conformal transformations are of the flat metric. Indeed, this proof will only show conformal invariance for the theory coupled to the flat metric.

The conformal transformations of the flat metric are well known: they are generated by translations, rotations, dilations, and special conformal transformations. Invariance under the first three can be checked. This answer will focus on invariance under the last. The key for this is noting that every special conformal transformation is a translation conjugated by inversions $I:x^\mu\mapsto x^\mu/x^2$. Thus, if we check the theory is invariant under inversions then it turns out that the theory will be invariant under special conformal transformations.

Inversions have the conformal factor $\Omega=x^{-2}$. Then, we have $$\partial_\mu(\Omega^{-\frac{D-2}{2}}\phi)=(D-2)(x^{2})^{\frac{D-4}{2}}x_\mu\phi+(x^2)^{\frac{D-2}{2}}\partial_\mu\phi.$$ We now need to compute the square of this vector and multiply by the remaining factor of $\Omega^{D-2}=(x^2)^{2-D}.$ The resulting term quadratic in derivatives is then clearly the original action. The remaining terms are $$2(D-2)(x^2)^{-1}x^\mu\phi\partial_\mu\phi+(D-2)^2(x^{2})^{-1}\phi^2.$$ On the first term we can use $2\phi\partial_\mu\phi=\partial_\mu(\phi^2)$. Moreover, note that $$\partial_\mu((x^2)^{-1}x^\mu)=(x^2)^{-1}(D-2).$$ Then these remaining terms are a total derivative $$(D-2)\partial_\mu((x^2)^{-1}x^\mu\phi^2).$$

As a final remark, as explained by Bruno in Simple conceptual question conformal field theory, one can modify the scalar field theory so that invariance under any Weyl transformation, even those that can't be undone by diffeomorphisms, is achieved.

Ivan Burbano
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