Physics Asked on April 3, 2021
Einsteins field equations can be derived from the Einstein-Hilbert action which only involves the scalar curvature $R$ of the spacetime manifold. This is simply
$$S = int_M R.$$
The volume form or measure here is left implicit as is quite common when the integral is written over a manifold.
Is there a similar Lagrangian for super-gravity and does it similarly involve a kind of scalar curvature suitably interpreted?
Consider the simple case of $mathcal{N}=1$ on-shell supergravity in 4 dimensions with no matter content. The full action is made up of the Einstein-Hilbert term and the Rarita-Schwinger action for the gravitino, with $$ S_{EH}=frac{1}{2kappa^2}intmathrm d^4 xdet e e^mu_ae^nu_bR^{ab}_{munu}(omega) $$ where $R_{munu}^{ab}$ is the field strength of the spin connection $omega_mu^{ab}$ and $det e$ takes the place of the volume element.
Since this SUGRA theory can formally be defined as a gauge theory with gauge group $text{SO}(1, d-1)$, the commutator of the supercovariant derivatives defines the Riemann tensor above: $$ [D_mu, D_nu]=frac14R_{munu}^{ab}gamma_{ab} $$
So $e^mu_ae^nu_bR^{ab}_{munu}(omega)$ is essentially the Ricci scalar in the spin-connection-vielbien formalism. The spin connection here is actually a function of the vielbien, and on introducing supersymmetry, also of the fermionic fields (which you can see when you find the EOM for $omega$ in the first-order formalism).
The presence of the scalar curvature in the action is a feature of all supergravity theories. There are of course Lagrangian descriptions for things like type IIA SUGRA as well, obtained as the low-energy effective action of type IIA string theory: $$ S = frac{1}{2kappa^2}intmathrm d^{10}xsqrt{-g}e^{-2phi}left(R+4partial_muPhipartial^muPhi-frac1{12}|H_3|^2right)-frac{1}{4kappa^2}intmathrm d^{10}xsqrt{-g}left(|F_2|^2+|tilde F_4|^2right)-frac{1}{4kappa^2}intmathrm d^{10}xsqrt{-g} B_2wedge F_4wedge F_4 $$ with the three terms corresponding respectively to the bosonic part, the RR part and the Chern-Simons part. You can remove the ugly $e^{-2phi}$ in the bosonic action by going to the Einstein frame through a conformal rescaling, whereupon you pick up the familiar Einstein-Hilbert term $intmathrm d^n x sqrt{-g} R$.
Correct answer by Nihar Karve on April 3, 2021
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