Proving the energy-momentum tensor for a conformal field theory is traceless

Question

A trick to derive Noether currents that is frequently used in conformal field theory literature is the following: suppose we have an action $S[\phi]$ which has the infinitesimal symmetry $\phi(x) \rightarrow \phi'(x) = \phi(x) + \epsilon \Delta \phi(x)$, where $\epsilon$ is some small parameter of the transformation, then if we upgrade $\epsilon \rightarrow \epsilon(x)$ the action changes by $$ \delta S = \int \mathrm{d}^d x (\partial_\mu \epsilon) j^\mu + (\text{boundary terms}) \tag{1}$$

where $j^\mu$ is the conserved current for the original rigid symmetry transformation when $\epsilon \neq \epsilon(x)$.

Translations

For the example of translations, we have

$$ \phi(x) \rightarrow \phi'(x) = \phi(x-a) = \phi(x) - \epsilon^\mu \partial_\mu \phi(x) + O(\epsilon^2), \tag{2}$$

so $\Delta \phi(x) = - \partial_\mu \phi(x)$ in this example. Therefore, if I was to upgrade $\epsilon^\mu \rightarrow \epsilon^\mu(x)$ to now give us some sort of space-dependent translation, from $(1)$ we would find

$$ \delta S = \int \mathrm{d}^d x (\partial_\mu \epsilon_\nu) T^{\mu \nu} + (\text{boundary terms}) \tag{3}$$

where $T^{\mu \nu}$ is the energy-momentum tensor.

Conformal transformations

It is at this point that many texts prove that a theory is conformally symmetric if its energy-momentum tensor is traceless, see Eq. (4.34) of CFT by Di Francesco et al (which also uses the Noether's theorem trick Eq. (2.191) and Eq. (2.142) outlined above). First we assume $T^{\mu \nu}$ is symmetric, so we can write

$$ \delta S = \int \mathrm{d}^d x (\partial_\mu \epsilon_\nu) T^{\mu \nu} = \frac{1}{2} \int \mathrm{d}^d x (\partial_\mu \epsilon_\nu + \partial_\nu \epsilon_\mu) T^{\mu \nu} $$

Now we use the fact that for an infinitesimal conformal transformation $\partial_{(\mu} \epsilon_{\nu)} \propto \partial_\alpha \epsilon^\alpha g_{\mu \nu}$ and therefore

$$ \delta S \propto \int \mathrm{d}^d x \partial_\alpha \epsilon^\alpha T^\mu_{\ \mu} \tag{4}$$

so if $T^\mu_{\ \mu} = 0$ then $\delta S = 0$ up to boundary terms and we have conformal symmetry. I am extremely uncomfortable with this proof as it seems to be implying that by simply upgrading $\epsilon \rightarrow \epsilon(x)$, where $\epsilon(x)$ obeys the conditions of a conformal transformation of the manifold, we have the action of a conformal transformation on our fields, yet it does not take into account how the internal degrees of freedom of the field transform. More precisely, we are using the result $(1)$, then upgrading $(2)$ to a space-depdent translation with $\epsilon \rightarrow \epsilon(x)$ in order to derive the result $(3)$. Therefore, referring to $(2)$, replacing $\epsilon \rightarrow \epsilon(x)$ means we have changed our variation of the fields as

$$ \delta \phi(x) =- \epsilon^\mu \partial_\mu \phi(x) \quad \rightarrow \quad \delta \phi(x)= - \epsilon^\mu(x) \partial_\mu \phi(x) $$

This is quite clearly not a conformal transformation of the fields because it is assuming they transform as a scalar field under the conformal group. For example, Lorentz transformations are conformal and they transform via

$$ \phi(x) \rightarrow \phi'(x) = R(\Lambda) \phi(\Lambda^{-1} x)$$

where $\Lambda$ is our Lorentz transformation and $R$ is some spin representation, so we get an internal transformation too. Similarly, scale transformations go as

$$ \phi(x) \rightarrow \phi'(x) = \lambda^{-\Delta} \phi\left( \frac{x}{\lambda} \right) $$

where $\Delta$ is the scaling dimension of the fields. So simply defining the transformation on the coordinates is not enough, we need to know how the internal degrees of freedom transform too, which is why I am uncomfortable interpreting the space-dependent translation obtained from $\epsilon \rightarrow \epsilon(x)$ as a conformal transformation.

My question

Simply upgrading our rigid translation paramter $\epsilon^\mu$ to a function $\epsilon^\mu(x)$ alone is not enough to turn the translation into a conformal transformation as we need the additional information of the group action on the fields, so how is this proof above (from the book CFT by Di Francesco et al) justified?

I am fully aware of the definition of the energy-momentum tensor from variation of the metric, e.g. from this, but I would like justification for the method taken in this book, where they do indeed use the Noether's theorem trick and not definining it via the Hilbert tensor.

Answer 1

For any quantum field theory, under an infinitesimal diffeomorphism, the action transforms as $$ S \stackrel{\text{diff}}{\to} S + \int d^d x \partial_\mu \epsilon_\nu(x) T^{\mu\nu}(x) + {\cal O}(\epsilon^2) $$ This is true for any vector field $\epsilon^\mu(x)$. This equation defines the stress-tensor of the theory. We now simplify this for the special case of conformal Killing vectors which satisfy $$ \partial_\mu \epsilon_\nu(x) + \partial_\nu \epsilon_\mu(x) = \frac{2}{d} \eta_{\mu\nu} \partial_\rho \epsilon^\rho(x). $$ Thus, in this case, it follows that \begin{align} S \stackrel{\text{conf}}{\to} & S + \int d^d x \partial_\mu \epsilon_\nu(x) T^{\mu\nu}(x) + {\cal O}(\epsilon^2) \\ =& S + \frac{1}{2} \int d^d x [ \partial_\mu \epsilon_\nu(x) + \partial_\nu \epsilon_\mu(x) ] T^{\mu\nu}(x) + {\cal O}(\epsilon^2) \\ =& S + \frac{1}{d} \int d^d x \partial_\rho \epsilon^\rho(x) T^\mu{}_\mu(x) + {\cal O}(\epsilon^2) \end{align} The last line holds ONLY for conformal Killing vectors, not for all vector fields.