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Conserved currents in higher-spin theories

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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.

This post imported from StackExchange Physics at 2014-07-28 11:12 (UCT), posted by SE-user Debangshu
asked May 26, 2013 in Theoretical Physics by DebangshuMukherjee (155 points) [ no revision ]
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?

This post imported from StackExchange Physics at 2014-07-28 11:12 (UCT), posted by SE-user Debangshu
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.

This post imported from StackExchange Physics at 2014-07-28 11:12 (UCT), posted by SE-user Vibert
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

This post imported from StackExchange Physics at 2014-07-28 11:12 (UCT), posted by SE-user John

1 Answer

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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.

This post imported from StackExchange Physics at 2014-07-28 11:12 (UCT), posted by SE-user John
answered Jul 10, 2013 by John (25 points) [ no revision ]

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