I'm trying to drill down the exact relation between conformal symmetry, Weyl symmetry, and tracelessness of the energy-momentum tensor. However, I'm getting quite confused because every book I can find seems to be treating this subject extremely sloppily.

First, following the exposition here, a conformal transformation is defined to be a diffeomorphism which satisfies $$g'_{\mu\nu}(x') = \Omega^{-2} g_{\mu\nu}(x)$$ followed by a Weyl transformation (i.e. a local rescaling of the metric and fields), so that the composite of the two maps transforms the metric as $$g_{\mu\nu}(x) \to g'_{\mu\nu}(x') = g_{\mu\nu}(x).$$ In fact, this is already a big source of confusion, because many sources call these diffeomorphisms conformal transformations in themselves, while other sources call Weyl transformations conformal transformations. But as far as I can tell, the "true" conformal transformations people actually use require both of these transformations. In other words, using the common nomenclature, a conformal transformation is a conformal transformation plus a conformal transformation. As far as I can tell, no source defines a conformal transformation explicitly, and most describe it as "a diffeormorphism that preserves angles", which is a completely vacuous statement.

In any case, both steps also affect the matter fields $\Phi$, so the variation of the action under a conformal transformation should have four terms, $$\delta S = \int_M d^4x \left(\frac{\delta S}{\delta \Phi} (\delta^d \Phi + \delta^w \Phi) + \frac{\delta S}{\delta g_{ab}} (\delta^d g_{ab} + \delta^w g_{ab}) \right)$$ where $\delta^d$ is the variation due to the diffeomorphism and $\delta^w$ is the variation due to the Weyl transformation. Let us number these contributions $\delta S_1$ through $\delta S_4$.

Polchinski performs the derivation in one line, blithely ignoring all terms except for $\delta S_4$. Meanwhile, di Francesco ignores all terms except for $\delta S_1$ (e.g. see Eq. 4.34). This is supposed to be analogous to an argument in chapter 2, which their own errata indicate are completely wrong, because they forgot to include $\delta S_3$. Unfortunately, they didn't correct chapter 4.

In any case, di Francesco claims that tracelessness of the energy-momentum tensor implies conformal invariance, which is the statement $\delta S = 0$. I've been unable to prove this. We know that $\delta S_1 + \delta S_3 = 0$ by diffeomorphism invariance, and $\delta S_4 = 0$ by tracelessness. But that doesn't take care of $\delta S_2$, which is the subject of this question. We cannot say it vanishes on-shell, because symmetries must hold off-shell.

I run into a similar problem trying to prove a converse. Suppose we have conformal invariance. Then $\delta S = 0$, and we know $\delta S_1 + \delta S_3 = 0$. At this point I can't make any further progress without assuming the matter is on-shell, $\delta S_2 = 0$. Then we know $\delta S_4 = 0$, but this does not prove the tracelessness of the energy-momentum tensor, because the Weyl transformation in a conformal transformation is not a general Weyl transformation, but rather is quite restricted.

In other words, I can't prove either direction, and I think all the proofs I've seen in books are faulty, forgetting about the majority of the terms in the variation. What is going on here?

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    $\begingroup$ My ideas are not clear enough to post a full answer. But let me remark this paper arxiv.org/abs/1510.08042 where in the intro they include $\delta S_2 + \delta S_4$. And also this paper arxiv.org/abs/1702.07079 where they use a different viewpoint. Namely they couple the theory to a background $g_{\mu\nu}$ and then study the variation of the partition function $\delta Z[g_{\mu\nu}]$, which doesn't depend on the fields since they are integrated over. $\endgroup$ – MannyC Dec 22 '18 at 20:29
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    $\begingroup$ Last remark: most people give diff invariance for granted because it always holds, so conformal transformations are thought of as a subset of Weyl transformations for which $\Omega^2 g_{\mu\nu}$ and $g_{\mu\nu}$ are diffeomorphic. That is, they always assume $\delta S_1+\delta S_3 = 0$. $\endgroup$ – MannyC Dec 22 '18 at 20:30
  • $\begingroup$ @Mane.andrea Agreed on your second comment, I just included the diffeomorphism for completeness. Regarding the first, I'm very surprised this is a matter of current debate! And these papers are far clearer than any books I've managed to find... $\endgroup$ – knzhou Dec 22 '18 at 22:08

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