Gravitino Equation of motion in second-order formalism

Question

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.

Answer 1

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.

Then straight forward integration by parts gives the first part of the second line in (9.7). Finally we note that, due to the antisymmetry of the gamma matrices,

$\gamma^{\mu\nu\rho}D_{\mu}D_{\nu}\psi_\rho=\frac{1}{2}\gamma^{\mu\nu\rho}[D_\mu D_\nu-D_\nu D_\mu]\psi_\rho=\frac{1}{8}\gamma^{\mu\nu\rho}R_{\mu\nu ab}\gamma^{ab}\psi_\rho$,

which gives the final answer.