= Solution
The <state–operator correspondence> maps $\mathcal O$ to the lowest one-particle state on $S^4\times\mathbb R$. On a sphere of radius $r$, cylinder energy equals scaling dimension divided by $r$. The primary has $\Delta=5$.
A translation generator raises the dimension by one and transforms as a vector of $SO(5)$. States with no angular momentum arise from scalar descendant pairs $P^2$, so the allowed descendants are
$$
(P^2)^n\mathcal O,\qquad n=0,1,2,\ldots.
$$
Their dimensions are $5+2n$, and hence the one-quantum, zero-angular-momentum energies are
$$
\boxed{E_n=\frac{5+2n}{r},\qquad n=0,1,2,\ldots}.
$$
These are the $l=0$ normal-mode energies of a massless scalar in global <Anti-de Sitter spacetime>.
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