# Can you take the cutoff to infinity at a conformal fixed point?

A conformal fixed point is defined by

$$\beta(g)=0$$

We hence know that couplings, masses and dimensions of operators do not flow in the effective Lagrangian when we change the renormalization scale $\mu$.

My question then is, can we take $\mu$ to infinity and retain finite results for all separated correlation functions at this point? Intuitively that seems sensible to me, but perhaps there is a technical obstruction.

If no, could you provide me with an example where separated correlation functions diverge in a CFT?

This question is related, but doesn't seem to answer my specific question.

• @Edwardhughes Maybe its is time to write that answer, I don't want the bounty going to waste. Very good question nevertheless. – Prathyush May 29 '15 at 11:29
• Yes - I shall do over the weekend. Thanks for the bounty btw! – Edward Hughes May 29 '15 at 14:46
• @Prathyush - find my answer below. Hopefully it's a lucid collection of thoughts! Many thanks once again for bringing more community attention to this via your bounty. It's nice to know there are people checking my reasoning! – Edward Hughes May 31 '15 at 16:51
• I did not use this site for a while, I found this questions after some searching. I do like your answer, as far as I can see it seems accurate. But I am only learning the subject, so it will take some time to fully understand it. It only take 7 hours for questions to disappear on the main page now, Perhaps that is a reason for a lack of attention. Do check out physicsoverflow, maybe you will find a larger communitybase, interested in your questions. – Prathyush Jun 1 '15 at 3:34

For clarity, I shall always assume that the theory we are considering is renormalizable, conformal at the quantum level, and well-defined at least at some finite energy scale.

The answer is yes, I believe. Here's my reasoning.

1. Correlation functions cannot have direct dependence on the renormalization scale $\mu$, by definition.

2. Therefore they can only depend on $\mu$ through physical parameters like the coupling, masses and dimensions.

3. In a conformal theory $\beta(g)=0$ implies that all physical parameters are fixed independent of $\mu$.

4. Therefore we may extend our well-defined theory at finite $\mu$ to $\infty$ by assuming a trivial group flow in the opposite direction.

So why was I confused? Well, people still talk about UV divergences in conformal field theories. But these are the divergences in the unrenormalized theory. Obviously many of these must still cancel out to ensure $\beta =0$. But not all of them!

In particular, the wavefunction renormalization $Z$ can (and often does) absorb a UV divergent factor when correcting the bare fields to interacting ones. Intuitively this makes sense to me, at least. If you want an example, check out this paper in which a certain form factor require field strength renormalization in $\mathcal{N}=4$ super-Yang-Mills theory.

To quote from that paper

The UV divergences, on the other hand, require renormalisation. In $\mathcal{N} = 4$ SYM theory, the appropriate combinations of the self-energies of the elementary fields and the one-particle-irreducible (1PI) corrections to the elementary vertices are UV finite, ensuring the vanishing of the $\beta$-function. The only sources for UV divergences are the insertions of composite operators as external states, which hence need to be renormalised.

So to conclude, you can take $\mu$ to $\infty$ in conformal field theories, provided that you go about field strength renormalization correctly when doing it!