# Why holographic renormalization?

Why is there a need to perform holographic renormalization for the normal $AdS_5\times S^5$/CFT$_4$ correspondence if the brane theory is conformal? Since the flow along the AdS direction $r$ is related to the renormalization scale does this not explicitly introduce an "energy" parameter that breaks conformal invariance in the SYM side of the duality?

## 1 Answer

The fact that the boundary theory is conformal means that renormalization does not induce running of the coupling. However, there are divergences which have to be regularized and renormalized. The regularization requires the introduction of an arbitrary scale, which is not Weyl invariant and leads to a conformal anomaly (in even dimensions).

Correspondingly, also the bulk theory has to be regularized by introducing a cutoff $\epsilon$ on the radial coordinate. The supergravity fields have to be expanded close to the horizon and local counterterms have to be introduced to subtract the divergences when taking the limit of $\epsilon\to 0$. For the metric, the regularization procedure requires picking a reference metric $g_{(0)}$ from the conformal structure on the boundary. For $d$ (boundary dimension) even, the dependence of the counterterm on the chosen reference metric leads to a renormalized Lagrangian, that is not Weyl invariant. One picks up exactly the expected Weyl anomaly.

This is a very neat example of a connection of boundary UV physics (the cutoff) and bulk IR physics (divergences close to the boundary) which lead to the same Weyl anomaly.

For details, see the paper by Henningson and Skenderis. There are also these very instructive lecture notes on holographic renormalization with the example of renormalization of the action of a massive bulk scalar.

Addendum: Example of why CFT correlators need regularization/renormalization
It is well-known that conformal invariance greatly restricts the form of CFT correlation functions. For example, two-point functions of a scalar operator $\mathcal{O}$ are restricted to $$\langle\mathcal{O}(x)\mathcal{O}(0)\rangle=\frac{C}{x^{2\Delta}}$$ where $\Delta$ is the scaling dimension of $\mathcal{O}$ and $C$ is a normalization constant. It is much less known though, that this is only a bare correlator and it is not valid at $x^{2}=0$. A correlator should be a well-defined distribution and have well-defined Fourier transforms: $$G(p)=\int d^dx\, e^{-ipx}\frac{C}{x^{2\Delta}}=\frac{C\pi^{d/2}2^{d-2\Delta}\Gamma\left(\frac{d-2\Delta}{2}\right)}{\Gamma{\Delta}}p^{2\Delta-d}.$$ Since the $\Gamma$-function is undefined for negative integer arguments, we can see that regularization is necessary when $\Delta=\frac{d}{2}+n$, where $n$ is a positive integer. This can be done, for example, using dimensional regularization. After addition of a counterterm in the action the correlator becomes $$G(p)=p^{2\Delta-d}\left(C_1 \log\frac{p^2}{\mu^2}+C_2\right),$$ which clearly is a scale dependent expression. Scale invariance is an anomalous symmetry in the full quantum theory. However, in $\mathcal{N}=4$ super Yang-Mills the coupling is protected from running by supersymmetry. So it is not conformal symmetry that leads to a vanishing $\beta$-function, but it is SUSY. The vanishing of the $\beta$-function for this particular theory is discussed here including some references.

• I still am confused. The correlators of a CFT do not need any regularization, right? I mean, in a general CFT. Then, why if this CFT has a holographic dual needs to get renormalized (even at the stress energy tensor level)? – Marion Feb 5 '15 at 11:42
• CFT correlators do need a regularization in general. However, the beta function vanishes, so you do not get a running coupling. However, the regularization does affect the path integral where is breaks conformal invariance and you get anomalies. – physicus Feb 5 '15 at 13:15
• What is the reason of this regularization? – Marion Feb 5 '15 at 13:53
• I am not an expert in conformal perturbation theory, but there must be divergent integrals when you compute correlators just as in any QFT. You regularize and then demand that the bare coupling is independent of the arbitrary scale you introduced. Usually, you find that the renormalized coupling runs consequently, but in a CFT the beta function vanishes. Therefore, the perturbative computation is valid at all energy scales as it should be in a conformal theory. However, even in the conformal case you have to get rid of divergences. – physicus Feb 5 '15 at 18:18