What is meant with the fact that Super Yang-Mills flows to a conformal field theory in the infrared? Also, is this a general fact or does this depend on the fact of considering a certain class of theories (such as $N=4$, $d=3$ for example)?

I have seen this statement numerous times in various articles such as



Every time this fact is stated by some author it is given as a known thing, and it is not explained. With an online research I could not make sense of this.

Could someone please explain?


There is a more general statement: all 4D Lorentz invariant field theories flow to CFTs in the UV and the IR. A proof was given last year by Luty, Polchinski, and Ratazzi in http://arxiv.org/pdf/1204.5221.pdf. Their argument has some assumptions but they are fairly weak.

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    $\begingroup$ (To clarify, it is the assumptions that are weak and not the authors.) $\endgroup$ – Matthew Nov 6 '13 at 23:35
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    $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/1204.5221 $\endgroup$ – Qmechanic Nov 6 '13 at 23:36
  • $\begingroup$ The assumptions seem very strong to me. It has to be 4D, Lorentz invariant and weak coupling. I guess the theory also has to be massless. For some, I guess that might be the only interesting QFT, but there are other cases where for sure this statement is not true. $\endgroup$ – Adam Nov 6 '13 at 23:39
  • $\begingroup$ You're right about the dimensionality and Lorentz invariance, I edited this in. It doesn't have to be weak coupling, as long as you assume that scale invariance implies conformal invariance. It doesn't have to be massless, and it's almost always not. $\endgroup$ – Matthew Nov 6 '13 at 23:42
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    $\begingroup$ @Matthew Yeah I should've been clearer, but the point is theories flow to scale invariant field theories in the IR or UV and so they must also be conformal. This assumes that theories flow to scale invariant theories, which seems to be true (either you integrate out all scales or all scales become meaningless) but I'm not sure if this has been proven. $\endgroup$ – David M Nov 7 '13 at 0:51

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