Gravitino Equation of motion in second-order formalism

In Freedman and Proeyen's text on supergravity they derive the equation of motion for the gravitino using the second order formalism. However, I'm not exactly clear as to how they use partial integration and the identities given to manipulate it into the final form. They give the gravitino action as (9.1),

$$S_{3/2}=-\frac{1}{2\kappa^2}\int d^Dx~e\bar{\psi}_\mu\gamma^{\mu\nu\rho}D_\nu\psi_\rho,\tag{9.1}$$

and go on to say on p. 189,

"In the second order formalism, partial integration is valid, so it is sufficient to vary $$\delta\bar{\psi}_\mu$$ and multiply by 2 obtaining, $$\delta S_{3/2}=-\frac{1}{\kappa^2}\int d^Dx~e(D_\mu\bar{\epsilon})\gamma^{\mu\nu\rho}D_\nu\psi_\rho$$ $$=\frac{1}{\kappa^2}\int d^Dx~e\bar{\epsilon}\gamma^{\mu\nu\rho}D_\mu D_\nu\psi_\rho=\frac{1}{8\kappa^2}\int d^Dx~e\bar{\epsilon}\gamma^{\mu\nu\rho}R_{\mu\nu ab}\gamma^{ab}\psi_{\rho}.\tag{9.7}$$ We integrated by parts and used (8.37) to move to the second line$$^4$$ and then used the Ricci Identity (7.123) to obtain the last expression."

The relevant equation they mention in the text are:

$$\nabla_\mu\gamma_\nu=0,\tag{8.37}$$

and

$$[D_\mu,D_\nu]\Phi=\frac{1}{2}R_{\mu\nu ab}M^{ab}\Phi,$$

$$[D_\mu,D_\nu]V^a =R_{\mu\nu cb}\eta^{ac}V^b,$$

$$[D_\mu,D_\nu]\Psi=\frac{1}{4}R_{\mu\nu ab}\gamma^{ab}\Psi,\tag{7.123}$$

where the last equation is for a spinor. This is all in section 9.1 in the textbook.

• You are saying you don't understand what you are reading. Where is the snag? What is it you don't get? – Cosmas Zachos Jul 12 at 16:29
• what is the question exactly? What equation is not clear? – Kosm Jul 13 at 7:19
• My problem is the manipulation from the unvaried action to the final form, ie. the three steps between S_{3/2} and the varied action with the curvature tensor. – huntercallum Jul 13 at 14:31

We do not vary with respect to $$e$$ as this would introduce cubic terms in $$\psi$$ and these terms are not universal as they are cancelled by various other terms in a full SUGRA theory.
The reality condition ensures that a variation by $$\bar{\psi}$$ is the same as a variation by $$\psi$$ hence the factor of two.
$$\gamma^{\mu\nu\rho}D_{\mu}D_{\nu}\psi_\rho=\frac{1}{2}\gamma^{\mu\nu\rho}[D_\mu D_\nu-D_\nu D_\mu]\psi_\rho=\frac{1}{8}\gamma^{\mu\nu\rho}R_{\mu\nu ab}\gamma^{ab}\psi_\rho$$,