Why $N\to \infty$ limit implies $g_s \to 0$ in holographic QCD?

Question

One basic difficulty in QCD is that it does not contain a small dimensionless quantity that would allow for perturbative calculation of low-energy observables.

A remarkable feature of holographic dualities is that in many cases one description of the physics is strongly coupled, while the dual description is weakly coupled and amenable to perturbative analysis of observables.

However, at first blush, the gravitational side of the correspondence being a weakly coupled dual is not a simple realization at all. Following Hong Liu's MIT lectures "String Theory and Holographic Duality, Lec18", "Chapter 3: Duality Toolbox", I found the following relations between parameters of SYM and gravity \begin{equation} g^2_{YM} = 4\pi g_s \end{equation} \begin{equation} \lambda\equiv~g^2_{YM}N = \frac{R^4}{\alpha'^2} \end{equation} \begin{equation} \frac{\pi^4}{2N^2} = \frac{G_N}{R^8} \end{equation}

where the different parameters are defined as in the lecture notes. Moreover, he argues that treating gravity as classical background field and restricting length scales much larger than string length, the parameters take the limits $G_N/R^8 \to 0$ and $\alpha'/R^2 \to 0$ which corresponds to $N \to \infty$ and $\lambda \to \infty$, respectively. He concludes by saying that the strong coupling limit is described by classical gravity.

I would like to understand how one can realize the weakly coupled behaviour of the dual theory, specially, in terms of matching of parameters. In other words, why the previous parameter relations holds and why the $N\to \infty$ limit implies $g_s \to 0$?

Answer 1

The main point is that when Hong Liu writes on p.3 that the 't Hooft coupling $\lambda$ is arbitrary and $N\to \infty$ in the classical string limit (2nd case), he means that $\lambda$ is finite and $N\to \infty$, so that $g_s\to 0$.

Moreover, in the semi-classical gravity limit (1st case) on p.2, Hong Liu's sentence restrict length scales much larger than string length seems to imply the exact opposite of the relevant restriction, namely that the string length $\ell_s=\sqrt{\alpha^{\prime}}\ll R$ (in natural units).

Answer 2

You have to be careful with the order of limits. The parameters are related by $$ N^2 = \frac{\pi^4 L^8}{2G_{10}} = \frac{1}{16\pi^2 g_s^2} , \qquad \lambda = g_{YM}^2 N = \frac{L^4}{2\alpha'^2} . $$ The limits are taken in the following order:

1. First, we take $N \to \infty$ keeping $\lambda$ fixed. This is equivalent to taking $g_s \to 0$. On the string side, this limit removes all higher genus corrections so this is a classical limit (only the sphere contributions survive). On the QCD side, this is the planar limit.

2. After this, we take $\lambda \to \infty$. On the stringy side, this corresponds $\alpha' \to 0$ (in units of AdS radius $L$). This kills off any stringy corrections so only the lowest mass states contribute (namely the graviton, dilaton and KB field). On the QCD side, this is a strong coupling limit.