This is in part a reiteration of this old phys.SE question, which did not receive much attention.

It is usually stated (see e.g. Ref. 1, §4.2.2) that a traceless energy momentum tensor $\Theta {^\mu} {^\nu}$ implies the invariance of an action $S[\Phi]$ under conformal transformations (here and in what follows the metric $g{_\mu} {_\nu}$ is non-dynamical).

As pointed out in phys.SE/53003, in the usual textbook treatment of the issue, it appears that the underlying assumption that fields $\Phi$ satisfy the EOM, $\frac{\delta S}{\delta \Phi}=0$, is made. Indeed, on the one hand, the total variation of the action should include also a term proportional to $\frac{\delta S}{\delta \Phi}$, if fields carry a non-trivial representation of the conformal group. Moreover, the traceless condition $\Theta ^\mu _\mu =0$ is usually satisfied only on-shell (cf., for instance, the construction of $\Theta {^\mu} {^\nu}$ in §4.2.2 of Ref. 1, where the tracelessnes conditions follows from the conservation of the dilatation current $\partial _\mu j^\mu _D=0$).

This brings me to my question. Consider the three propositions:

  1. The theory is conformally invariant.
  2. It is possible to define an energy momentum tensor $\Theta ^\mu _\nu$ with identically vanishing trace: $\Theta ^\mu _\mu =0$.
  3. It is possible to define an energy momentum tensor $\Theta ^\mu _\nu$ whose trace vanishes identically along the equations of motion: $\Theta ^\mu _\mu \approx 0$.

What are the relations between these three propositions?

Ref. 1: Conformal field theory, P. Di Francesco, P. Mathieu, D. Senechal.

  • $\begingroup$ From Francesco-Mathieu-Senechal (page 101): "Under an arbitrary coordinate transformation, $\delta S\sim \int T\cdot\mathrm{d}\xi$. This is valid even if the equations of motion are not satisfied. If $\xi$ denotes a conformal transformation, $\delta S\sim \int \mathrm{tr}(T)\cdot\mathrm{div}\xi$, that is, tracelessness of the energy-momentum tensor implies invariance of $S$ under conformal transformations. The converse is not true, since $\mathrm{div}\xi$ is not arbitrary." $\endgroup$ – AccidentalFourierTransform Jan 31 '18 at 17:55
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    $\begingroup$ @AccidentalFourierTransform I find di Francesco's remarks quite confusing, which was the motivation of my original question. First of all, I don't get why he (apparently) treats the conformal transformation on coordinates and fields as if it was merely a coordinate transformation, which is not. Second, I believe that the equation $\text {tr}T=0$ is valid, in general, only on-shell; see for example the construction that he gives for the improved EM tensor, whose trace is equal to $\partial _\mu j_D ^\mu$. So, even if $\delta S \sim \intop \text {Tr} T$ [...] $\endgroup$ – pppqqq Jan 31 '18 at 19:10
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    $\begingroup$ [...] (which I believe to be wrong, see my answer below) it appears from this argument that one has in general $\delta S \neq 0$ for off-shell configurations (on the other hand, $\delta S =0$ for on-shell configurations is trivial). $\endgroup$ – pppqqq Jan 31 '18 at 19:12
  • $\begingroup$ General tip: Let's not have posts look like revision histories! $\endgroup$ – Qmechanic Dec 23 '18 at 13:12
  1. The conformal invariance of the classical action is not an on-shell statement. It is the same as saying that the Lorentz invariance was an on-shell statement. The conservation of the energy-momentum tensor in that case is an on-shell statement, that comes from Noëther theorem. In the case of conformal invariance, the tracelessness of the energy-momentum tensor is an on-shell statement.

Quantum mechanically, Noëther theorem should be changed by Ward Identities. Thus, the trace of the energy-momentum tensor should not be zero as an operator, but its expected value in the vacuum should be zero. This is assuming that the whole quantum theory is conformally invariant, which is the result of the classical action being invariant plus the path integral measure being invariant. If the path integral measure breaks conformal symmetry, then we have a conformal anomaly, and the trace of the energy momentum has a non-zero expected value in the vacuum. If this happens, all the stuff about 2-point and 3-poiny functions is not valid because there you assumed the path integral measure to be invariant.

  1. I think that from the on-shell tracelessness condition you could obtain the equations of motion, and from then you could obtain the action and see if it is invariant. Or other way, you could define the variation of the fields to be such that it cancelled the trace of the energy momentum tensor for off-shell fields. This way you would end up discovering the conformal weights of each field as well as that the whole transformation you defined is a conformal transformation (because by definition such a transformation changes the metric that way).
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    $\begingroup$ Dear Sasieta, thanks for your answer. I agree with you in (1), I believe as well that the action has to be invariant. I also understand your remarks on the quantum mechanical side. However, I'm not really sure about your point (3): first, I'm assuming that the action, and a fortiori the EOMs, are given, so I don't need to obtain them from the tracelessness condition (is this even possible?). Second, it isn't obvious at all that, given tracelessness on-shell I can define the variaton of the fields in such a way to have $\delta S=0$ (it is exactly what I'm trying to understand)... or is it? $\endgroup$ – pppqqq Jan 22 '18 at 0:21
  • $\begingroup$ Okay, what I meant is the following. Imagine the simplest case of a CFT, a (massless) free scalar in $d$ dimensions. Beginning with the usual energy-momentum tensor $$T_{\mu\nu} = \partial_\mu \phi \partial_\nu \phi - \frac{1}{2}\eta_{\mu \nu}\partial^\sigma \phi \partial_\sigma \phi $$ it can be made traceless for on-shell configurations by adding a total derivative: $$\Theta_{\mu\nu} = T_{\mu\nu} + \left(\dfrac{d-2}{2}\right) \partial_\mu (\phi \partial_\nu\phi)$$ which is as well symmetric and conserved on-shell ... $\endgroup$ – M. Sasieta Jan 22 '18 at 10:43
  • $\begingroup$ Suppose you are given this $\Theta_{\mu\nu}$ and told it is traceless for on-shell configurations. Then you can easily obtain the equations of motion for $\phi$, and from then, the action $S[\phi]$. Once you have the action, it is not an on-shell thing, so you can play with it and transform it with a conformal transformation to see if it is invariant. You can see that indeed it will be invariant under such a transformation. This is because if you define $\phi$ to transform as $$\phi(x) \rightarrow \lambda^{\left(\frac{d-2}{2}\right)}\phi(\lambda^{-1}x)$$ you can compute $\delta S[\phi] = 0$ $\endgroup$ – M. Sasieta Jan 22 '18 at 10:53

In Ref. 1 of the OP it is proved that $2\implies 1$ (I'm referring to the numbered propositions of the OP). What perhaps should be emphasized is that in such a theory (with $\Theta ^\mu _\mu$ exactly vanishing) conformal invariance holds in a special way: in such a way that all fields are defined to have vanishing conformal weight.

In this answer, conversely, I will prove that $1 \implies 3$, under some technical assumptions which can be found in Ref. 1 of the OP. I will actually prove the statement for a simply scale-invariant theory (under the same technical assumptions, such a theory turns out to be conformally invariant). If all fields have vanishing scaling dimension, the same proof shows that $1\implies 2$.

When the theory is scale invariant, there exists a conserved dilatation current: $$\partial _\mu j^\mu _D \approx 0,$$ where $\approx$ denotes equality on-shell.

In OP's Ref. 1 it is proved that, given some certain technical assumptions, in a scale invariant theory one can always define an energy-momentum tensor which satisfies: $$\Theta ^{\mu} _{\,\mu} = \partial _\mu j_D^\mu.$$

In particular, this implies:

  1. That $\Theta ^{\mu} _{\,\mu} \approx 0$ (trivial).
  2. That the theory is actually conformally invariant (not trivial).

Let me sketch the proof of the second point. The variation of a generic matter field $\Phi$ under an infinitesimal conformal transformation $x\to x+\xi(x)$ is given by: $$\delta _c \Phi =\delta _d \Phi + \delta _s \Phi,$$ where $\delta _d\Phi$ is the variation of $\Phi$ as a consequence of the diffeomorphism $x\to x+\xi (x)$, while $\delta _s\Phi$ is a local scale transformation: $$\delta _s \Phi = -\frac{\Delta}{d} (\partial _{\mu} \xi ^{\mu}) \Phi, $$where $\Delta$ is the conformal weight of $\Phi$. Correspondingly, the variation of the action is:$$\delta _c S =\intop \text {d}^d x \frac{\delta S}{\delta \Phi (x)}[\delta _d \Phi (x) + \delta _s \Phi (x)].$$

Now, the two crucial points:

  1. The first term is expressed in terms of $\Theta ^{\mu} _{\,\nu}$ by coupling the theory to gravity: $$S[\Phi]\to S[\Phi,g]$$ in such a way that the new action is invariant under diffeomorphisms. In this way, one has: $$\intop \text {d}^d x \frac{\delta S}{\delta \Phi (x)}\delta _d \Phi (x)=-\intop \text {d}^d x \frac{\delta S}{\delta g _{\mu \nu} (x)}\delta g _{\mu \nu} (x)=\intop \text {d}^d x T ^{\mu} _{\,\mu} \partial _{\nu} \xi ^\nu.$$ Here we have defined: $$T ^{\mu \nu}(x)\equiv \frac{\delta S}{\delta g _{\mu \nu} (x)}.$$
  2. The second term is, by definition of $j^\mu _D$: $$\intop \text {d}^d x \frac{\delta S}{\delta \Phi (x)}\delta _s \Phi (x)=\intop \text {d}^d x (\partial _\mu j_D ^\mu) (\partial _\nu \xi^\nu) .$$

Now, if we had $T^{\mu \nu}=\Theta ^{\mu \nu}$, then from the condition $\partial _\mu j^\mu _D = \Theta ^\mu _\mu$, summing up the two variations, we would immediately conclude that $\delta _c S=0$, i.e. the theory is conformally invariant. Of course, in general $T\neq \Theta$, but reconsidering the construction of $\Theta$ in Ref. 1 (in particular, Eqs. (4.42) and (4.43) of §4.2), one can see that the substitution $T\to \Theta$ under the integral sign is unconsequential. Conformal invariance then follows.

  • $\begingroup$ I'm having a similar confusion but I'm not understanding the last half of this answer. How exactly does conformal invariance follow? Are you defining a conformal transformation to be the composite of a diffeomorphism (which is conformal) plus a Weyl transformation? Why don't you include the variation of the metric in $\delta_c S$ to begin with? $\endgroup$ – knzhou Dec 22 '18 at 18:07
  • $\begingroup$ Dear @knzhou, thanks for asking, it actually took me a while to re-understand my own answer. I updated it, try to give a look. Let me add a few comments (i) I want to emphasize that I'm dealing with conformal transformations of a spacetime with a fixed Minkowski metric. These transformations act on regions of spacetime $\Omega\to \Omega '$ and fields $\Phi\to \Phi'$, while leaving the Minkowski metric $\eta\to \eta$ untouched; the statement of conformal invariance is: $S[\Omega ', \Phi ', \eta]=S[\Omega , \Phi , \eta]$. This should answer your last question. [...] $\endgroup$ – pppqqq Dec 22 '18 at 20:17
  • $\begingroup$ [...] (ii) I'm defining conformal transformations as coordinate transformations, and I'm also allowing the various fields to carry a non-trivial representation under the conformal group (i.e. a non-zero conformal weight). The conformal transformation of a field is thus composed by two parts: a coordinate transformation (the $\delta _d$ part) and a linear transformation (the $\delta _s$ part). I think it becomes clearer when you restrict to the case $\xi ^\mu (x)=\epsilon x^\mu$, which corresponds to scale transformation. $\endgroup$ – pppqqq Dec 22 '18 at 20:27
  • $\begingroup$ Looking at this more carefully, I'm unsure how the step numbered (2) above works. For example, consider the special case $\xi^\mu = \epsilon x^\mu$. The variation of $\Phi$ under the corresponding scale transformation is like $\delta \Phi = - \Delta \Phi + x^\mu \partial_\mu \Phi$. But the variation $\delta_s \Phi$ only contains the first term. That is, your local scale transformation only scales the local field value, without moving the field at all. Why can the extra term be ignored? $\endgroup$ – knzhou Dec 23 '18 at 13:00
  • $\begingroup$ My notation might be confusing you. You are right, the total variation of $\Phi$ under a scale transformation would be $\delta \Phi = \delta _d \Phi + \delta _s \Phi$, with $\delta _d \Phi = \epsilon x^\mu \partial _\mu \Phi$ and $\delta _s \Phi = -\epsilon \Delta \Phi$. Here the $\delta _d$ part is taking into account that the coordinate system is changing (if $\Phi$ was a general tensor field, I would also have to take into account the rotation of the local coordinate frame). The crucial point here is that the fields $\Phi$ and $\Phi + \delta _d \Phi$ are just the same field [...] $\endgroup$ – pppqqq Dec 25 '18 at 16:42

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