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

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.

• You're going about the wrong way around. The CFT is by definition a QFT whose stress tensor is traceless which then implies that it is invariant under conformal transformations. You are trying to prove the inverse. Now, the inverse is likely true, but is a much more difficult thing to prove and requires additional assumptions of locality and unitarity. I don't believe there is even a full proof of the inverse statement in all dimensions. Jul 17, 2021 at 12:45
• The yellow book makes this extremely clear in the paragraph below (2.35) -- "The tracelessness of the energy-momentum tensor then implies the invariance of the action under conformal transformations. The converse is NOT true, since $\partial_\rho \epsilon^\rho$ is not an arbitrary function." Jul 17, 2021 at 12:47
• I am not trying to show the inverse, I am trying to show that $\delta S = \int \partial_\mu \epsilon_\nu T^{\mu \nu}$ for conformal transformations by simply upgrading the translation parameter to a spacetime-dependent parameter, which is done in the yellow book in Eq. (4.34). I do not understand why this is the case because upgrading the translation parameter doesn’t take into account the internal transformation of the fields too under a conformal transformation Jul 17, 2021 at 13:41
• Jul 17, 2021 at 14:26

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.
• Two questions: is your stress tensor here the Hilbert stress tensor? And why have you applied the formula for how the action transforms under a diffeomorphism to the conformal transformation? I’d expect all actions are trivially diffeomorphism invariant because they’re essentially a change of coordinates, so substituting a conformal transformation into your diffeomorphism formula will always give $\delta S = 0$, but a conformal transformation is clearly not a symmetry for all actions and isn’t just a diffeomorphism. Can you justify the use of that formula more please? Jul 18, 2021 at 11:53
• If the metric is held fixed when performing this diffeomorphism, how could the variation of the action possibly be $\delta S = \int \partial_\mu \epsilon_\nu \frac{\delta S}{\delta g_{\mu \nu}}$ which looks like you have varied the metric as $\delta g_{\mu \nu} = \partial_{( \mu} \epsilon_{\nu)}$. The essense of my question is the following: how can I show that $\delta S = \int \partial_{\mu} \epsilon_{\nu} T^{\mu \nu}$ using the Noether's theorem trick that the yellow book uses by simply upgrading $\epsilon \rightarrow \epsilon(x)$ without considering internal degres of freedom? Jul 19, 2021 at 14:23
• @Matt0410 - $x^\mu \to x^\mu + \epsilon^\mu$ is a symmetry of the action but $x^\mu \to x^\mu + \epsilon^\mu(x)$ is NOT. Therefore, if we look at the variation of the action under the latter transformation, the variation must depend on the derivative of $\epsilon^\mu(x)$ (since it is supposed to vanish when $\epsilon^\mu(x)$ is a constant. Defining the coefficient of $\partial_\mu \epsilon_\nu(x)$ to be $T^{\mu\nu}(x)$, we find the aforementioned result. Jul 19, 2021 at 14:26