# What sets AdS radius of the Vasiliev dual to the O(N) vector model?

In $\mathrm{AdS}_5$/$\mathrm{CFT}_4$ the AdS radius $R$ is determined in terms of the string length by the gauge theory t'Hooft parameter as follows $$\frac{R}{l_{\rm s}} \sim \lambda^{1/4}$$ Consequently large t'Hooft parameter corresponds to dropping derivative corrections to the spacetime effective action.

In the free O(N) model there is no such parameter. What sets the AdS radius in this case?

There's also another scale in the game in "ordinary" AdS/CFT: while $\lambda$ sets string length, $N$ sets Planck length. Large $N$ suppresses quantum effects, while large $\lambda$ suppresses stringy effects. Stringy (higher derivative) effects have no obvious parallel in Vasiliev, and $\lambda$ has no obvious parallel in the $O(N)$ model. On the other hand, Quantum effects seem like a natural thing to have in the bulk, and $N$ looks rather like the $N$ in SYM. So perhaps the answer is that it's $N$ setting the radius of AdS relative to the Planck length.
Maybe the view of Vasiliev being something like a tensionless, $l_s\to\infty$ limit of strings, with the first Regge trajectory becoming massless, would be a helpful point of view here? I'd need to think a little more to make this feeling any more precise...
• That makes some sense. In the free O(N) model we effectively have $\lambda = 0$. Then the equation $R/l_s = 0$ can be satisfied for any value of $R$ since $l_s \to \infty$ in the tensionless limit as you say. So I think the answer is that all values of $R$ are equivalent for the O(N) model. – user11881 Dec 19 '13 at 3:18
• @user11881 I think that (= all values of $R$ being admissible) is true even for "ordinary" AdS/CFT, where it is always the ratio $\frac{R}{l_s}$ that appears in observables. In Vasiliev theories, this ratio $\rightarrow$ zero – GuSuku Jan 18 '14 at 3:53