Relation of conformal symmetry and traceless energy momentum tensor

Question

In usual string theory, or conformal field theory textbook, they states traceless energy momentum tensor $T_{a}^{\phantom{a}a}=0$ implies (Here energy momentum tensor is usual one which is symmetric and follows conservation law) conformal theory. (i.e, see page 3 )

I wonder how they are related to each other.

I found similar question Why does Weyl invariance imply a traceless energy-momentum tensor? and get some idea about weyl invariance.

and get some another useful information from Conformal transformation/ Weyl scaling are they two different things? Confused! which dictates that conformal transformation and weyl transformation is totally different things .

Answer 1

Note that under an infinitesimal change in the metric of the form $g \to g + \delta g$ the action changes to $$ \delta S = \int T^{ab} \delta g_{ab} $$ Now, under Weyl transformations we have $$ g_{ab} \to e^{2\omega} g_{ab} \qquad \implies \qquad \delta g_{ab} = 2 \omega g_{ab} $$ For Weyl transformations $\omega$ is completely arbitrary. If we consider a conformal transformation then, the metric also transforms as above except that $\omega = \frac{1}{d} \nabla_a \xi^a$ where $\xi^a$ is a conformal killing vector, i.e. $\omega$ takes a specific functional form.

Either way, for both conformal or Weyl transformations $\delta g_{ab} =2\omega g_{ab}$. Thus, for either of these transformations, the variation in the metric is $$ \delta S = 2 \int \omega T $$ Thus, if the trace of energy momentum tensor vanishes, $T = 0$, then $$ \delta S = 0 $$ and we have a symmetry of our theory!

OK. So we have shown that if $T = 0$, then the theory is invariant under Weyl and conformal transformations. What about the inverse statement? Can we infer from Weyl and conformal invariance that $T = 0$? The latter is a more subtle question.

Weyl or conformal invariance implies $$ \int \omega T = 0 $$ Now, when talking about Weyl invariance, the above is true for arbitrary $\omega$. In this case, we can most certainly conclude that $T = 0$ (for instance take $\omega \propto \delta^4(x)$ or some smoothed out version thereof and we immediately reach this conclusion.

When talking about conformal invariance, $\omega$ is not arbitrary and we cannot conclude that $T$ must vanish. For instance, in a flat background, $\omega$ takes the form $\lambda + a_\mu x^\mu$ where $\lambda$ and $a_\mu$ are arbitrary constants. Thus, all we can conclude is that we must have $$ \int T = 0 ~, \qquad \int x^\mu T = 0 $$ These two conditions no longer imply that $T = 0$. Thus, as per this argument the inverse statement is not necessarily true in conformal field theories. I'm not sure if there is any other argument that can be used to justify that $T$ must vanish in CFTs, but so far, all the CFTs we study have $T = 0$.

Answer 2

To complement the answer above, here is another point of view:

Suppose we have a theory of a field $\phi$ that has been written in a coordinate invariant form by coupling to a background metric $g$. An infinitesimal conformal transformation, $x\mapsto\tilde{x}=x+\epsilon\xi$ acts on the theory by $$x\mapsto\tilde{x}$$ $$\phi(x)\mapsto \Omega^{-\Delta}\tilde{\phi}(\tilde{x}).$$ Alternatively, one can understand this as the composition of transformations $$x\mapsto\tilde{x}\mapsto\tilde{x}$$ $$\phi(x)\mapsto\tilde{\phi}(\tilde{x})\mapsto\Omega^{-\Delta}\phi(x)$$ $$g(x)\mapsto\tilde{g}(\tilde{x})=\Omega^{-2}g(\tilde{x})\mapsto g(\tilde{x}) $$

Of course, the first trivially produces no transformation in the action. Thus, we can equivalently compute the change in the action by only doing the latter $$\phi(x)\mapsto\Omega^{-\Delta}\bar{\phi}(x)$$ $$g_{\mu\nu}(x)\mapsto\Omega^2g_{\mu\nu}(x).$$ The three of these give the same variation of the action.

Using the later, the variation of the action, which is a total derivative by the assumption that our theory has conformal symmetry, is $$\int\text{d}^Dx\partial_\mu (\cdots)^\mu=\Delta S=2\int\text{d}^Dx\frac{\delta S}{\delta g_{\mu\nu}}g_{\mu\nu}\omega-\Delta\int\text{d}^Dx\frac{\delta S}{\delta\phi}\omega\phi=-\int\text{d}^Dx T\omega+\Delta\int\text{d}^Dx\partial_\mu\left(\frac{\partial L}{\partial\partial_\mu\phi}\omega\phi\right).$$ In here we have set $\Omega=1+\omega$ and used the equations of motion to rewrite the last term as a total derivative (actually, here there might be some subtleties if the action depends on derivatives of the metric, since then the energy momentum tensor only coincides with the functional derivative for transformations of compact support; due to this, this procedure may only apply to theories whose initial coupling to gravity was minimal). From this we conclude that $T\omega$ is a total derivative. For scale transformations, $\omega$ is constant so that this implies $T=\partial_\mu K^\mu$ is a total derivative. For special conformal transformations, $\omega$ is linear in $x$. Since $T$ is a total derivative, the statement that $\omega T$ is a total derivative implies that $T$ is a double derivative $$\partial_\mu(\cdots)^\mu=b\cdot x\partial_\mu K^{\mu}=-b_\mu\cdot K^\mu+\partial_{\mu}(b\cdot x K^\mu)\Rightarrow K^\mu=\partial_{\nu}(x^\mu K^\nu-(\cdots)^\nu),$$ i.e. $$T=\partial_\mu\partial_\nu L^{\mu\nu}$$

This brings forth the possibility of modifying the action by adding terms propotional to $R_{\mu\nu}L^{\mu\nu}$ and $RL$, with $L=L^\mu_\mu$ According to Weyl versus conformal invariance in quantum field theory by Kara Farnsworth and Markus A. Luty and Valentina Prilepina, found in https://arxiv.org/abs/1702.07079, one can choose the constants of proportionality so that the resulting Einstein-Hilbert energy-momentum tensor is traceless on-shell. Note that this traceless-ness is only on shell, so that one cannot in general conclude from it that the theory has invariance under arbitrary rescalings of the metric.