4
$\begingroup$

Consider a field theory coupled with gravity described by the action:

$S=\int d^Dx \sqrt{-g} \left( \mathcal{R}-\Lambda+\mathcal{L}_m[\phi] \right)$,

with the requirement that g must be asymptotically AdS.

Usually in AdS/CFT correspondence the action evaluated on-shell reduces to a purely boundary term.

Is there a general formula to express this boundary term?

$\endgroup$
  • 2
    $\begingroup$ If you're fixing the boundary to be AdS, then what is there left to vary at the boundary? Or are you asking if there is a way to express the above action as $\oint d^{D-1}x \sqrt{|g_{ind}|}(stuff) + \int d^{D}x\sqrt{-g}\mathcal{L}_{m}$? $\endgroup$ – Jerry Schirmer Nov 1 '13 at 23:51
  • $\begingroup$ Yes, usually in holography you make an ansatz on the metric and on the fields and then you evaluate the action on the equations of motion. Typically the action reduces to a purely surface term, since the bulk term vanishes due to the equations of motion. I'm asking if there are general rule to express this boundary term without compute case by case the on shell action $\endgroup$ – Andy Bale Nov 2 '13 at 9:03
  • $\begingroup$ This happens somewhat generically in general relativity, due to the diffeomorphism invariance, though the relativity community typically will discuss this in terms of constraints and evolution equations. Any good discussion on a 3+1 formalism will show that the Hilbert action is equivalent to a sum of constraints that are satisfied on the boundary of your spacetime evolution. $\endgroup$ – Jerry Schirmer Nov 2 '13 at 16:46
  • $\begingroup$ Where could I find some material about this argument? $\endgroup$ – Andy Bale Nov 2 '13 at 20:40
  • 1
    $\begingroup$ See, amongst other places, "a relativist's tooklit" by Eric Poisson, or the original ADM paper, which is available on the arxiv. $\endgroup$ – Jerry Schirmer Nov 3 '13 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.