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It has been proved that the $\phi^4$ theory is trivial in spacetime dimensions $d>4$. By trivial I mean that the field $\phi$ is a generalized free field, or in other words, it's only nonzero connected correlator is the two point correlator. This is a nonperturbative result, which manages to get around the fact that $\phi^4$ is nonrenormalizeable in dimensions $d>4$.

Here is the paper which proves this result: https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.47.1

Have there been any results similar to this for the Yang-Mills theory? Yang-Mills is nonrenormalizeable in dimensions $d>4$ as well, so I imagine that if there were a similar result, $d=4$ should also be the critical dimension.

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Good question. I am not aware of similar results for YM. The $\phi^4$ case uses correlation inequalities for ferromagnetic spin systems. Unfortunately, not many of those are known for gauge theories. YM is an example of model with non-Abelian group symmetry like $SU(N)$. Even for much simpler models with $O(N)$ symmetry like $N$-component $\phi^4$ or spherical spins, not much is known as far as correlation inequalities when $N\ge 3$.

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Note that in Yang-Mills theory you can always scale out the coupling constant, $$ {\cal L}= \frac{1}{g^2} {\rm Tr}\left( F_{\mu\nu}F^{\mu\nu}\right) \hspace{1cm} F_{\mu\nu}= \partial_\mu A_\nu-\partial_\nu A_\mu +[A_\mu,A_\nu] $$ so there is no limit in which non-abelian YM theory reduces to a free field theory.

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  • $\begingroup$ What about $\mathcal{L}={\rm Tr}(F_{\mu\nu}F^{\mu\nu})$ with $F_{\mu\nu}=\partial_{\mu} A_{\nu}-\partial_{\nu}A_{\mu}$ ? $\endgroup$ – Abdelmalek Abdesselam Feb 24 at 15:09
  • $\begingroup$ That's QED without matter ($U(1)$ Yang-Mills theory), which is a non-interacting theory for any value of the coupling, and in any number of dimensions. $\endgroup$ – Thomas Feb 24 at 15:46
  • $\begingroup$ I guess that is why I made sure to say that for the $\phi^4$ theory, the statement is that it is a generalized free field. So the n-point correlation functions decompose into a product of two-point functions. In the case of Yang-Mills, this happens in the large N limit, therefore this property can occur! $\endgroup$ – LucashWindowWasher Feb 24 at 17:14
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    $\begingroup$ I would still argue that there is no such thing as a non-interacting non-abelian Yang Mills field. You could try to Higgs the theory to $U(1)^N$, but this is a different theory. Large N is also an unusual case, usually argued to be an infinite number of free fields, possibly more acurately desribed as as a string theory. $\endgroup$ – Thomas Feb 24 at 17:42
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    $\begingroup$ @Thomas: What I was trying to explain in my previous comment is that by replacing $A_{\mu}$ by $g A_{\mu}$ you get $\mathcal{L}={\rm Tr}(F_{\mu\nu} F^{\mu\nu})$ with $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}+g[A_{\mu},A_{\nu}]$. So there is a "limit in which non-abelian YM reduces to a free field theory", namely, the $g\rightarrow 0$ limit. The limit is Abelian. So what. $\endgroup$ – Abdelmalek Abdesselam Feb 24 at 18:59

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