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In the paper I am reading, there is a statement that seems to use the fact that the scaling dimension of stress tensor is $\Delta=d$, and I would like to show that this is correct. I found a similar question, but it only gave a rough description and I could not find a complete proof. So I'm trying to prove it, and I'm confused about the basics.

First, I will give my proof and I would like you to check if it is correct or not. After that, I feel that a naive idea would contradict $\Delta=d$. So I would like to know what is wrong with this. Finally, I would like to know the proof using $T_{\mu\nu}=-\frac{2}{\sqrt{g}}\frac{\delta S}{\delta g^{\mu\nu}}$, which is suggested in the previous question.

  1. My proof: I understand that the scaling dimension $\Delta_{\mathcal{O}}$ of a field $\mathcal{O}$ is defined by the behavior under scale transformation (e.g. Di Fransesco's textbook eq(2.121)), \begin{align} x'&=\lambda x\\ \mathcal{O}'(\lambda x)&=\lambda^{-\Delta_{\mathcal{O}}}\mathcal{O}(x). \end{align} We know that canonical stress tensor is given by \begin{align} T_{\mu\nu} =-\eta^{\mu\nu}\mathcal{L} +\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\Phi)}\partial^{\nu}\Phi. \end{align} Of course, we need to modify this tensor by adding some term so that it is symmetric and traceless. However, for my purposes this is not important and we will ignore it. If the theory we consider have conformal symmetry, i.e. it is CFT, action $S[\Phi]=\int d^{d}x\ \mathcal{L}$ is invariant under scale transformation, $S[\Phi']=\int d^{d}x'\ \mathcal{L}'(x')=S[\Phi]$, which implies that $\mathcal{L}'(x')=\lambda^{-d}\mathcal{L}(x)$. Therefore stress tensor transforms as \begin{align} T'_{\mu\nu}(x')=-\eta^{\mu\nu}\mathcal{L}'(x') + \frac{\partial \mathcal{L}'}{\partial(\partial'_{\mu}\Phi')}\partial'^{\nu}\Phi'=\lambda^{-d}T_{\mu\nu}(x), \end{align} which means that scaling dimension of $T_{\mu\nu}$ is $\Delta=d$ as we desire. My first question is whether this proof is correct.

  2. Naive idea: Stress tensor $T_{\mu\nu}$ is $(0,2)$ tensor. We know that under general coordinate transformation, $x\rightarrow x'$, $(0,2)$ tensor behaves \begin{align} T'_{\mu\nu}(x')=\frac{\partial x^{\rho}}{\partial x'^{\mu}}\frac{\partial x^{\sigma}}{\partial x'^{\nu}}T_{\rho\sigma}(x). \end{align} In special case, where it is scale transformation, we get $T'_{\mu\nu}(x')=\lambda^{-2}T_{\mu\nu}(x)$, which indicates $\Delta=2$. This seems to contradict the previous proof. Where was the mistake?

  3. Final question: Answer to this question suggested that $\Delta=d$ can be shown using $T_{\mu\nu}=-\frac{2}{\sqrt{g}}\frac{\delta S}{\delta g^{\mu\nu}}$. I would like to know how we proof it along this line?

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  • $\begingroup$ This specialization to Lagrangian theories certainly works. But the conformal algebra shows that a conserved primary operator of spin-2 must have dimension $d$. Also, your $\Delta = 2$ confusion is explained at physics.stackexchange.com/questions/727332/… $\endgroup$ Commented Aug 5, 2023 at 14:22
  • $\begingroup$ Thank you for your comment. I understand my confusion a little better now. The mistake I made was to apply usual tensor transformation law to stress tensor $T_{\mu\nu}$, and since stress tensor is a (quasi-)primary field, I should have actually applied $T'_{\mu\nu}(x')=\Omega^{\Delta-2}\frac{\partial x^{\rho}}{\partial x'^{\mu}} \frac{\partial x^{\sigma}}{\partial x'^{\nu}}T_{\rho\sigma}(x)$, where $\Omega=\lambda^{-1}$, is that correct? $\endgroup$
    – sakata
    Commented Aug 6, 2023 at 5:26
  • $\begingroup$ The exact equation you wrote would say that conformal primaries also have a prescribed transformation law under something which isn't a conformal transformation. What you should do here is write that a conformal transformation $x'(x)$ is given by $x''(x)$ times a local rescaling where $x''(x)$ is a rotation. Then changing $x'$ to $x''$ and $\Omega^{\Delta - 2}$ to $\Omega^\Delta$ will make your equation correct. $\endgroup$ Commented Aug 6, 2023 at 10:43

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