# Relation of conformal symmetry and traceless energy momentum tensor

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 .

• I suggest the book of Di Francesco et all. – 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 $T = 0$.