# Conformal Gravity

Lubos, in his comment to a question, says that (https://physics.stackexchange.com/q/61281)

First of all, one can't gauge a symmetry without modifying (enriching) the field contents. Gauging a symmetry means to add a gauge field and the appropriate interactions (e.g. by covariantize all terms with derivatives, in the case of both Yang-Mills and diffeomorphism symmetries).

Recently, I have seen papers on "Conformal Gravity" i.e. ( http://arxiv.org/pdf/1306.5220.pdf ), and they say that the following action is invariant under local conformal transformations

$S = \int d^4 x \sqrt{-g} \Big( \frac{1}{12} \phi^2 R + \frac{1}{2} \partial_\mu \phi \partial^\mu \phi -\frac{1}{4} \lambda \phi^4 \Big)$

where the local conformal transformations are given by

$\widetilde{g}_{\mu\nu} = e^{-2\sigma(x)} g_{\mu\nu} \,, \qquad \tilde{\phi} = e^{\sigma(x)} \phi$

Now, if this action is locally conformal invariant so that the conformal symmetry is gauged, should not one enrich the field content by adding gauge fields and the appropriate interactions, and covariantize all terms with derivatives?

• Note that the action you wrote is equivalent to the action $$S=\frac{1}{2}\int\text{d}^4x\sqrt{-g}\nabla^{a}\phi\nabla_{a}\phi$$ upon degauging. Here $\nabla_{a}$ is the conformally covariant derivative $$\nabla_{a}=e_{a}{}^{\mu}\partial_{\mu}-\frac{1}{2}\omega_{a}{}^{bc}M_{bc}-\mathfrak{f}_{a}{}^{b}K_b-\mathfrak{b}_a\mathbb{D}~,$$ where extra gauge fields and generators associated with the conformal algebra have been included. See e.g. arxiv.org/abs/0906.4399 for more details. Feb 21 at 6:24

The conformal transformation $g'_{\mu\nu} = e^{-2\sigma}g_{\mu\nu}$, $sigma = sigma(x)$ leads to the transformation of the Ricci scalar $$R' = e^{2\sigma}R - 12e^{2\sigma}(2\sigma_{,\mu}^{,\nu} - 2\sigma_{,\mu}\sigma^{,\mu}$$ Since $\phi' = e^{\sigma}$ then $$\frac{1}{12}\phi'^2R' = \frac{1}{12}\phi^2 R - \phi^2(2\square\sigma - 2\sigma_{,\mu}\sigma^{,\mu})$$ The Ricci tensor and scalar are not conformal invariant. The scalar field Lagrangian $\frac{1}{2}\phi_{,\mu}\phi^{,\mu} - \frac{1}{4}\lambda\phi^4$ transforms with the $\phi \rightarrow e^\sigma\phi$ and with some work it is easy to see that $\phi^2(2\square\sigma – 2\sigma_{,\mu}\sigma^{,\mu})$ in the Ricci scalar transformation is subtracted out and the total Lagrangian is invariant. This is why there is that $1/6$ factor that keeps showing up in conformal invariant Lagrangians.
This occurs without the requirement of a gauge covariant operator. The conformal transformation is a sort of rescaling of the metric. It has a relationship with conformal transformations in complex variables that preserve angle measures. There are the Cauchy Riemann equations that can be computed for a function $z = x + iy$ $\rightarrow u = v + iw$. If you take a derivative $du/dz$ separate out the real from imaginary parts you get these equations. If you do a similar thing with quaternions you can get the equations that define the field tensors of gauge theory. As a result the conformal transformations and gauge transformations are different in their structure.
• It is called local because $\sigma = \sigma(x)$ May 25, 2016 at 23:08
• Of course I understand that, but even if $\sigma$ is a constant global parameter, the action would be the same, if no covariantization is needed. So why call it local if I cannot distinguish whether $\sigma$ is constant of $\sigma = \sigma(x)$. May 26, 2016 at 6:19