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I had previously asked this question. This is kind of a continuation of that.

I recently found this expression which seems to be called the "Fierz-Pauli action" which is apparently the quadratic action for gravitons about $AdS_{d+1}$,

$S = \frac{c }{2 l_p ^{d-1} }\int d^{d+1}x \sqrt{-g} \left ( \frac{1}{4} \nabla_\mu h_{\rho \lambda}( \nabla^{\mu} h^{\rho \lambda} - 2\nabla^\rho h^{ \mu \lambda}) + \frac{1}{2}\nabla_\mu h^{\mu \nu}\nabla_\nu h - \frac{1}{4} \nabla_\mu h \nabla^\mu h - \frac{d(d-1) }{2L^2} ( h^{\mu \nu}h_{\mu \nu} - \frac{1}{2} h^2 ) + O(h^3) \right )$

  • Can someone help derive (or reference) this?

I am aware of general expressions for curvature invariants expanded to second order in metric fluctuation about some chosen background. But its not clear to me as to how does one covariantly impose that the background is some $AdS$...

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  • $\begingroup$ The only way to impose that the background is AdS is to say $R=\Lambda gg$, i.e. Riemann tensor is expressed in terms of metric. The question is not clear to me. Maybe you are asking how to impose this in Lagrangian, i.e. to impose restrictions on $\nabla$ as a part of the action. If so, then it is not needed anywhere, one can simply write what $\nabla$ is below the action. $\endgroup$ – John Jun 14 '14 at 17:09
  • $\begingroup$ @John If you expand $R$ in terms of $g_b + h$ then there will a lot of places where this background metric $g_b$ will occur. Now how does one say that this $g_b$ is $AdS$ and that too without breaking the covariant structure of the action? [...or if you know of a reference which explains this quoted equation of mine..] $\endgroup$ – user6818 Jun 14 '14 at 18:33
  • $\begingroup$ Still I cannot see the problem. You expand the action over some background, sometimes you have to use $[\nabla,\nabla]=gg$ or $Riemann=gg$ or $Ricci=g$ or $R=const$ but at no place you have to use explicit form of the metric tensor. For example, the mass-like term is thanks to $R=const$ and does not depend on a particular form of $g$. The quadratic lagrangian you wrote can be derived by using the relations above (i.e. $Riemann=gg$) and nothing else. $\endgroup$ – John Jun 14 '14 at 20:58
  • $\begingroup$ @John You can see equation 13 in this paper, arxiv.org/pdf/1010.2411v2.pdf which gives the expression for the quadratic fluctuation for the $R$. You can see that it clearly contains many terms dependent on the background metric $\bar{g}$. So one is faced with the problem of having to impose all of these to be that of the $AdS$ - how do you achieve that? $\endgroup$ – user6818 Jun 15 '14 at 11:54
  • $\begingroup$ OK, one can ask different questions about expansion over some fixed background. Nice questions will have nice answers. As I said, the action above can be derived just from $R=gg$ and nothing else needs to be used. Of course, one can simply start expanding everything over the background, the worst thing being Christoffels, this is what these guys did. In (13) one can use (8), which can be solved for $\Gamma$ in terms of fluctuations and background, so that eventually you will have to use $R=gg$ for the background if it is AdS and nothing else. I must say that the question is still pretty vague $\endgroup$ – John Jun 15 '14 at 13:44

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