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 .

  • $\begingroup$ I suggest the book of Di Francesco et all. $\endgroup$ – Rexcirus Feb 11 '16 at 21:06

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 $T = 0$.

  • $\begingroup$ But we know that conformal invariance implies certain Ward identifies for the trace of the em tensor. These have the same form as the Ward identities of a theory that is Weyl inv. Therefore, can we conclude that conformal inv. implies Weyl inv.? If we can, then shouldn't that imply that the trace of the em tensor must be zero for conformally inv theories? $\endgroup$ – Sounak Sinha Sep 28 '20 at 17:40

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