# Conserved currents in higher-spin theories

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.

• 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? May 26, 2013 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. May 26, 2013 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, 2013 at 22:31
• Hi Debangshu, could you please mention the paper of Fronsdal that you were talking about in your comment? Thanks! Oct 14, 2015 at 8:42