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Suppose one has some supergravity Lagrangian $\mathcal{L}_\text{sugra}$ in flat (Minkowski) space. Moreover, assume that this Lagrangian is on-shell, in the sense that the auxiliary fields have been integrated out. For the purposes of this question, I only wish to consider the bosonic part of the supergravity action, since this is the part that affects the classical geometry.

If I want to put this supergravity theory in an asymptotically AdS space, am I allowed to just add a cosmological constant term $-2\sqrt{-g}\Lambda$ to $\mathcal{L}_\text{sugra}$, or do I have to introduce additional terms to keep it supersymmetric?

My intuition is that for one supercharge (e.g., $\mathcal{N}=1$ SUGRA in 4D), then there's only one superpartner and it's fermionic, so can be ignored when looking at classical supergravity solutions. But what about more supercharges? Do I need to introduce additional terms then?

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A cosmological constant $\Lambda\det e$ can be added consistently to the supergravity action with any legal $\mathcal N$ and $D$, as long as:

  1. We add a mass term for the gravitino into the Lagrangian, $\sim \bar\psi_\mu\gamma^{\mu\nu}\psi_\nu\det e$
  2. The SUSY transformation law for the vielbein is $\delta_\epsilon e_\mu{}^a=\frac12\bar\epsilon\gamma^a\psi_\mu$
  3. The SUSY transformation law for the gravitino is $\delta_\epsilon\psi_\mu=D_\mu \epsilon+g\gamma_\mu\epsilon$

Here $\epsilon$ is the infinitesimal supersymmetry parameter, $D_\mu$ is the spin covariant derivative constructed out of the spin connection $\omega_\mu^{ab}$ and $g$ is the parameter that will determine the cosmological constant. Consistency then demands that the prefactor of the gravitino mass term is $\frac12g(D-2)$, while the cosmological constant itself will be $2g^2(D-1)(D-2)$. $g$ will be real due to the Majorana condition on $g\gamma_\mu\epsilon$, and so the cosmological constant term is of definite sign at least in any vacuum that preserves SUSY.

Also note that gauging supergravity (e.g. minimally coupling the gravitino(s) to the abelian gauge field in the case of $\{\mathcal N=2, D=4\}$, and similar for higher SUSY), automatically leads to a cosmological negative constant in the same way, at least when there are no scalars in the SUGRA multiplet. This is because the gravitino must transform as $D_\mu\epsilon+g\gamma_\mu\epsilon+gA_\mu\epsilon+...$, by supersymmetric invariance. If there are no scalar fields to absorb the transformation, a cosmological constant must appear, as above, with $g$ doubling up as the gauge coupling constant.

This construction of AdS SUGRA appears naturally while compactifying string theory on $\mathrm{AdS}_5\times S^5$ and M-theory on $\mathrm{AdS}_7\times S^4$ and similarly KK reductions of 11D $\mathcal N = 1$ SUGRA on some non-trivial spaces.

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  • $\begingroup$ Is there an easy way to see why gauging the supergravity automatically leads to a cosmological constant? $\endgroup$
    – arow257
    Jul 13 '21 at 21:04
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    $\begingroup$ @arow257 well it's just a consequence of supersymmetric invariance again; in the simplest case, the gravitino transformation will be something like $\delta_\epsilon\psi_\mu=D_\mu\epsilon+g\gamma_\mu\epsilon+gA_\mu\epsilon+...$, so $g$ also plays the role of the gauge coupling constant here. $\endgroup$ Jul 14 '21 at 4:17

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