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What is the purpose and meaning of taking the 't Hooft parameter to infinity?

Physics Asked on July 12, 2021

I am following Hong Liu’s MIT 8.821 String Theory and Holographic Duality lectures. He starts discussing the large-$N$ expansion in the context of a hermitian matrix model described by the Lagrangian

$$mathcal{L}=-frac{1}{g^2}text{Tr} left{frac{1}{2}partial_muPhipartial^muPhi+frac{1}{4}Phi^4right},$$

where $g$ is the coupling constant and $Phi$ is a $Ntimes N$ hermitian matrix. He shows that the large-$N$ limit of this model only makes sense when we consider the ‘t Hooft parameter

$$lambda_text{‘t Hooft}=g^2N,$$

and take the $Nrightarrowinfty$ limit while keeping $lambda_text{‘t Hooft}$ fixed.

At some point in the lecture, a student asks a question regarding whether $lambda_text{‘t Hooft}$ should be a small parameter or not, and Liu remarks that it does not matter, and later on in the course he will eventually take the limit $lambda_text{‘t Hooft}rightarrowinfty$.

My question is what is the point in considering the limit $lambda_text{‘t Hooft}rightarrowinfty$? What does it mean physically? Is it regarding some sort of a weak-strong coupling duality?

One Answer

As pointed out by @bolbteppa, in the lecture note Chapter3:DualityToolbox Liu answers this question. Taking the limit $Nrightarrow infty, lambda_{t'Hooft}rightarrowinfty$ corresponds to the semi-classical gravity limit. This means that strongly coupled $mathcal{N}=4$ SYM is dual to classical gravity!

Correct answer by gsuer on July 12, 2021

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