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It is common to refer to fixed points of the renormalization group as scale invariant theories. This statement can be formulated as $$ \beta(\mu) \Big |_{\mu^*} = 0 \; \; \Longrightarrow \; \; T^{\mu}_{\mu} = 0 .$$

However, I never saw a proof of this fact and I do not think it is trivial. How can I approach it?

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This is a hard problem, it is about the conditions which ensure that scale invariance implies conformal invariance. In two dimensions any unitary local scale invariant theory is conformally invariant. In four dimensions it is not yet known a set of necessary and sufficient conditions. Here are some references.

[1] J. Polchinski, Scale and Conformal Invariance in Quantum Field Theory, Nucl. Phys. B 303 (1988) 226.

[2] M. A. Luty, J. Polchinski and R. Rattazzi, The a-theorem and the Asymptotics of 4D Quantum Field Theory, JHEP 01 (2013) 152 1204.5221.

[3] Y. Nakayama, Scale invariance vs conformal invariance, Phys. Rept. 569 (2015) 1 1302.0884.

[4] A. Dymarsky, Z. Komargodski, A. Schwimmer and S. Theisen, On Scale and Conformal Invariance in Four Dimensions, JHEP 10 (2015) 171 1309.2921.

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  • $\begingroup$ Thanks for your answer! However I have one doubt left: it seems to me that you are assuming that the theory at the fixed point is a CFT. I know that it is a standard result, but I have never found a proof of that either, only some qualitative arguments. I asked about scale invariance because I feel that proving that a theory is scale invariant could be easier than proving it is invariant under the whole conformal group. However, if you also happen to have references for $\beta = 0 \Rightarrow $ CFT I would be equally satisfied! $\endgroup$
    – PPIP
    Jun 5, 2020 at 15:56
  • $\begingroup$ $T^\mu_\mu=0$ is the definition of a CFT. If you just assume scale invariance (namely $\beta = 0$) in principle one might have $T^\mu_\mu = \partial^\mu J$, $J$ being the virial current. $\endgroup$
    – MannyC
    Jun 5, 2020 at 16:00
  • $\begingroup$ Oh, you are asking why $\beta=0$ implies scale invariance? That's by definition, $\beta$ is the derivative with respect to the scale. $\endgroup$
    – MannyC
    Jun 5, 2020 at 16:01

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