I'm doing an introductory studying on supergravity, and I'm studying in particular on the book Supergravity of D. Z. Freedman and A.V. Proeyen. In this book they describe supersymmetry using Majorana supercharges when possible. For instance in $D=4$ dimension and for one supersymmetry $\mathcal{N}=1$, the supercharges anticomutation rule is

\begin{equation} \left\{Q_\alpha,\bar{Q}^{\beta}\right\}= -\frac{1}{2}\left(\gamma^\mu\right)_\alpha{}^{\beta}P_\mu \end{equation} Moreover I will use the following notation

  • $D_\mu$ is the covariant derivative containing only the spin connection, hence for istance on spinor we have \begin{equation} D_\mu\psi = \left(\partial_\mu+\frac{1}{4}\omega_{\mu ab}\gamma^{ab}\right)\psi \end{equation}
  • $\nabla_\mu$ is the covariant derivative containing also the possible affine connection term, hence for instance on a spin 3/2 field we have \begin{equation} \nabla_\mu\psi_\nu= \left(\partial_\mu+\frac{1}{4}\omega_{\mu ab}\gamma^{ab}\right)\psi_\nu - \Gamma^\rho_{\mu\nu}\psi_\rho \end{equation}

$\mathcal{N}=1$ in $D=4$ dimension pure supergravity (i.e. no matter coupling) is described by an action containing only the vielbein field $e^\mu_a$ describing the graviton, and the Majorana Rarita - Schwinger field $\psi_\mu$, describing the gravitino. The action is

\begin{equation} S =\frac{1}{2k^2}\int d^4x\,e\left\{e^{a\mu}e^{b\nu}R_{\mu\nu ab}-\bar{\psi}_\mu\gamma^{\mu\nu\rho}D_{\nu}\psi_{\rho}\right\} \end{equation}

where $k^2 = 8\pi G$ ($G$ being the Newton constant) and, as ususal, $\gamma^{\mu\nu\rho}$ is the totally antisymmetrize product of three $\gamma$-matrices.

One can work in the first order formalism, and then consider the variation of the action with respect to the spin connection $\omega$. Thus one gets an equation for the spin connection that can be solved to obtain a connection with torsion.

Now my problem is, why in the action there is $D_\mu$ only and not $\nabla_\mu$? I mean:

  • If we have zero torsion then of course we don't have any affine connection, because this one would be symmetric in the lower indices which are contracted with the antisymmetric one of $\gamma^{\mu\nu\rho}$

  • When we have torsion though (and in the first order formalism we do find a non zero torsion) the above reasoning doesn't work anymore. However Freedman and Proeyen sais

We don't have any Christoffel connection in the action because this would be inconsistent with local supersymmetry

I really can't understand why. Can somebody please help me?

  • $\begingroup$ Which page in F&P? $\endgroup$ – Qmechanic Jun 1 '18 at 16:00
  • $\begingroup$ Chapter 9, page 191 $\endgroup$ – M. M. R. Jun 1 '18 at 21:09

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