Suppose we have a theory with conformal invariance that has been extended to a diffeomorphism invariant theory in a way that the resulting energy-momentum tensor is traceless on-shell (which can always be done, as e.g. explained in my answer in Relation of conformal symmetry and traceless energy momentum tensor). In general we have that $\xi^\mu T_{\mu\nu}$ is the conserved current associated to any infinitesimal isometry $x\mapsto x+\xi$. For a traceless energy momentum tensor, this current is also conserved whenever $\xi$ induces an infinitesimal conformal transformation as well. However, this doesn't make it clear to me that this current gives the charges that generate conformal transformations. In particular, I don't know how to obtain this current from Noether's theorem. More physically, the current $\xi^\mu T_{\mu\nu}$ only seems to know about the change in coordinates. Why would it know about the scaling of the fields $\phi(x)\mapsto\Omega^{-\Delta}\tilde{\phi}(\tilde{x})$ in a conformal transformation?

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    $\begingroup$ I think the best way to understand what is going on with these transformations is to explicitly calculate the conserved current for an infinitesimal scaling in a toy model. $\endgroup$ Commented Mar 27, 2023 at 12:31
  • $\begingroup$ Currents for a symmetry never tell you which representation of this symmetry a certain matter field will furnish. $\endgroup$ Commented Mar 27, 2023 at 14:07
  • $\begingroup$ Maybe there is a caveat to that statement? If one writes the current in the Hamiltonian formalism then one can compute the Poisson bracket to see how it acts. For example, one can compute that if $\vec{L}=\vec{r}\times\vec{p}$, then $\vec{r}$ transforms like a vector under rotations since $\{L_i,x_j\}=\epsilon_{ijk}x_k$ $\endgroup$ Commented Mar 27, 2023 at 14:15
  • $\begingroup$ Ok sure. If $\xi^\mu T_{\mu\nu}$ knows about more than a change in coordinates because it's actually a composite operator built out of the matter fields in a known way, then it can be used to derive the transformation laws. $\endgroup$ Commented Mar 27, 2023 at 15:47
  • $\begingroup$ Right, I guess one could see that in specific examples, say with the energy momentum tensor here alves-nickolas.github.io/pdf/…. I guess I was just wondering if there was a cleaner argument that connects $\xi^\mu T_{\mu\nu}$ to the full conformal transformation of a theory in general. $\endgroup$ Commented Mar 27, 2023 at 15:52

1 Answer 1


Here is how to understand the association between a conformal transformation generated by a conformal Killing vector field $\xi^\mu$ and the conserved current $J_\nu=\xi^\mu T_{\mu\nu}$ using Noether's theorem.

First, replace $\xi\mapsto \xi'=\epsilon\xi$ for an $\epsilon$ vanishing at the boundaries of spacetime. In this replacing, keep the same conformal factor for the fields $\Omega^2=1+2\omega$ and let $\tilde{\phi}(\tilde{x})$ be the field transformed by the diffeomorphism $\epsilon\xi$. Consider then the transformation $$\begin{pmatrix} x \\ \phi(x) \\ g_{\mu\nu}(x) \end{pmatrix}\mapsto\begin{pmatrix} \tilde{x} \\ \tilde{\Omega}^{-\Delta}(\tilde{x})\tilde{\phi}(\tilde{x}) \\ g_{\mu\nu}(\tilde{x}) \end{pmatrix},$$ which reduces to a conformal transformation for $\epsilon=1$. This transformation can be broken down into the following composition $$\begin{pmatrix} x \\ \phi(x) \\ g_{\mu\nu}(x) \end{pmatrix}\mapsto\begin{pmatrix} x \\ \Omega^{-\Delta}(x)\phi(x) \\ \Omega^{2}(x)g_{\mu\nu}(x) \end{pmatrix}\mapsto\begin{pmatrix} \tilde{x} \\ \tilde{\Omega}^{-\Delta}(\tilde{x})\tilde{\phi}(\tilde{x}) \\ \tilde{\Omega}^{2}(\tilde{x})\tilde{g}_{\mu\nu}(\tilde{x}) \end{pmatrix}\mapsto\begin{pmatrix} \tilde{x} \\ \tilde{\Omega}^{-\Delta}(\tilde{x})\tilde{\phi}(\tilde{x}) \\ g_{\mu\nu}(\tilde{x}) \end{pmatrix}$$

The first is a Weyl transformation while the second is a diffeomorphism. Therefore, in a Weyl and diffeomorphism invariant theory, the variation of the action is only due to the last transformation $$\begin{pmatrix} x \\ \phi(x) \\ g_{\mu\nu}(x) \end{pmatrix}\mapsto\begin{pmatrix} x \\ \phi(x) \\ \Omega^{-2}(x)\bar{g}_{\mu\nu}(x) \end{pmatrix}$$ where $x\mapsto\bar{x}$ is the inverse transformation to $x\mapsto\tilde{x}$. In the latter we have $$\delta g_{\mu\nu}=-2\omega g_{\mu\nu}+\nabla_\mu\xi'_\nu+\nabla_\nu\xi'_\mu$$ and the variation of the action is $$\delta S=\int\text{d}^D{x}\frac{\delta S}{\delta g_{\mu\nu}}\delta g_{\mu\nu}\propto\int\text{d}^D{x}(-\omega T+T^{\mu\nu}\nabla_\mu\xi'_\nu).$$ In a Weyl invariant theory $T\equiv T^\mu_\mu=0$. Thus, on-shell, using the fact that $\epsilon$ vanishes at the boundaries to integrate by parts, and using the definition of a conformal Killing vector field, we have $$0=\int\text{d}^Dx (J^\mu\nabla_\mu\epsilon+\epsilon T^{\mu\nu}\nabla_\mu\xi_\nu)=\int\text{d}^Dx (J^\mu\nabla_\mu\epsilon+\frac{1}{D}\epsilon T\nabla_\alpha\xi^\alpha)=-\int\text{d}^Dx \nabla_\mu J^\mu\epsilon.$$ Thus $J^\mu$ is conserved.


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