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I have a very basic question about conformal field theory: Does the partition function need a UV regulator, or is it finite even without? That is, does $\int D\phi \exp(-S)$ converge, or do we need to do the standard QFT procedure of only integrating over modes with frequency less than some $\Lambda$, and then using counterterms and taking $\Lambda\to\infty$?

On the one hand, the fact that all beta functions vanish for a CFT makes me think the partition function is finite, since the theory is invariant under increasing the cutoff $\Lambda$. Also, in e.g. this reference $^{{\text{[note: this is a direct link to downloading the pdf]}}}$ eq 114 seems to imply that the CFT partition function on the cylinder is finite.

On the other hand, there is (for example) the famous Cardy-Calabrese formula for the entropy of the 1+1D CFT vacuum restricted to a length $l$ interval: $S=\frac{c}{3}\log\left(\frac{l}{\epsilon}\right)$ + finite. Here, the short-distance cutoff $\epsilon$ is required to make the entropy finite. So, certain quantities in a CFT do require a regulator, which makes me think the partition function itself may well need a regulator.

Which point of view is right?

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$\newcommand{\cM}{\mathcal{M}}$Your theory is a CFT, therefore it's scale invariant. So you do not have a natural energy scale of the system, so you cannot choose a cutoff, or else, conformality is gone. The partition function had better be finite.

Oh, but wait! The conformal anomaly! Albeit classically your theory is scale-invariant, quantum mechanically this might not be the case.

If you put your CFT on a funky (i.e. non-flat) manifold it might not necessarily be scale-invariant. In particular, if $T_{\mu\nu}$ is the stress tensor of the CFT, on a manifold $\cM$ $$ \left<T^\mu_{\ \ \ \mu}\right> = \underset{\text{A-type anomaly}}{\underbrace{(-1)^{1+\frac{d}{2}} a\ \mathrm{Euler}(\cM)}} + \underset{\text{B-type anomaly}}{\underbrace{c_i \mathrm{Weyl}^i(\cM)}},$$ where $\mathrm{Euler}(\cM)$ is the Euler density, and $\mathrm{Weyl}^i(\cM)$ are the various Weyl invariants of the manifold. This classification was put forward by Deser & Schwimmer in hep-th/9302047. The A-type anomaly messes with Weyl invariance, but not with scale invariance, while the B-type has to introduce a scale, through regularisation, to make the path-integral well-defined. Roughly speaking, you should think of that scale as the curvature of the manifold.

This is a UV regulator. The path-integral measure will be fine at long scales, since the curvature of the manifold will be very small, but plagued by the conformal anomaly at short scales, since, then, the curvature of the manifold will really be singular. Thus, indeed, on a funky manifold, a CFT partition function requires a UV regulator.

The conformal anomaly lies at the heart of the divergence in the Cardy-Calabrese formula. The partition function of a 2d CFT has a universal logarithmic divergence due to the conformal anomaly, from points of divergent curvature. In the Cardy-Calabrese case, these are the conical singularities at the branch points, when they perform the replica calculation.

Finally, the above also answers why partition functions of CFTs on the plane, the cylinder, or the torus are finite without the need for regularising. Because all of these manifolds are flat, and thus anomaly-free!

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  • $\begingroup$ Great answer. I admit I find it quite surprising that a regulator shows up in the entropy formula even on flat space, where the partition function is finite. But I suppose there's no reason that finiteness of the partition function should imply finiteness of entanglement entropy. $\endgroup$ Oct 29, 2022 at 17:55
  • $\begingroup$ @nodumbquestions Well, entanglement entropy is QFT is (almost) always infinite. See e.g. this phys.SE question. In contrast, Rényi entropies are finite. In the case of a CFT finiteness of partition functions on flat space implies finiteness of Rényi entropies. However, when you take the limit $n\to 1$ in the replica trick, you're no longer on flat space but you have conical singularities coming from the collapsed replicas. The original Cardy-Calabrese paper is quite readable and explains that. $\endgroup$ Oct 30, 2022 at 12:59
  • $\begingroup$ No, the renyi entropies aren't finite. See Eq 3.7 in the original paper. You can see that $tr\rho^n$ is divergent as the cutoff $a\to 0$. $\endgroup$ Nov 4, 2022 at 0:02

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