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After the proposal of Maldacena (AdS/CFT), there have been numerous attempts to find out gravity duals of various kinds of CFT. Klebanov and Polyakov gave one such correspondence here. The claim is this: The singlet sector of the critical $O(N)$ model with the $(\phi^{a}\phi^{a})^{2}$ interaction is dual to the minimal bosonic theory in $AdS_4$. Of course, we have to take the large $N$ limit. Simplest such $O(N)$ invariant theory (free, with no interactions) is: \begin{equation*} S= \frac{1}{2}\int d^3 x \sum_{a=1}^{N}(\partial_{\mu}\phi^{a})^{2} \end{equation*} How do I derive the conserved currents in this theory? $\phi^{a}$ are $N$-component vector fields and NOT $N \times N$ matrix fields.

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O.k. I figured it out myself. It seems this formalism was developed by Fronsdal long time back. However, another question has popped up. Spin-$s$ massless fields are given by totally symmetric tensor $\phi_{n_1 n_2..n_s}$. There is a condition though on these fields. They have to be "double traceless" i.e I contract them with $g_{\mu_1 \mu_2}g_{\mu_3 \mu_4}$. Why do I need to impose this condition? – Debangshu May 26 '13 at 14:54
Can you post your results? I don't know anything about this Fronsdal stuff, but the c.c.'s are not so difficult to write down, namely $J_\mu = \phi^a \partial_\mu \phi^a,$ $J_{\mu \nu} = \phi^a \partial_\mu \partial_\nu \phi^a$ etc. The derivatives need to act on both fields, so they are conserved (but don't know the TeX to do this). They are traceless because $\partial^2 \phi^a = 0.$ If $\phi^a$ is real the odd spins vanish identically. – Vibert May 26 '13 at 15:44
Vibert, one can write the currents as follows, $j_{\mu_1...\mu_s}=\bar{\phi}(x) (\overleftarrow{\partial_\mu}-\overrightarrow{\partial_\mu})^s\phi(x)$. These are not traceless, the trace is given by superpotential. One can easily make them traceless, but the formulas are no so nice – John Jul 10 '13 at 22:31
Hi Debangshu, could you please mention the paper of Fronsdal that you were talking about in your comment? Thanks! – Orbifold Oct 14 '15 at 8:42

That looks weird but you have to impose the double trace constraint in order to have the right number of degrees of freedom. This is easy to see by going into light-cone and analyzing the equations of motion.

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