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Supergravity, torsion and diffeomorphism invariance

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The action for $N=1$ supergravity in an $n$ spacetime dimenions is

$$ S= \int e\left( R + \overline{\psi}_a \gamma^{abc} D_b \psi_c \right) $$ Here $R$ is the scalar curvature, $e=\det(e_{a\mu})$, and $e_{a\mu}$ is the frame field. $\psi = \psi_{\mu} dx^{\mu}$ is a spinor valued one-form. The indices $a,b\ldots = 0\ldots n-1 $ are internal indices that transform under the Lorentz group. The frame field $e_{a\mu}$ can be used to 'convert' spacetime indices to internal indices, and vice-versa. The gamma matrices obey $\gamma^a \gamma^b +\gamma^b \gamma^a = \eta^{ab} $, with $\eta^{ab}$ the 'internal metric'. $\gamma^{ab\ldots z} = \gamma^{[a} \gamma^b \ldots \gamma^{z]} $ denotes an antisymmetrised product of gamma matrices. The covariant derivative is

$$ D \psi = d \psi + \frac{1}{2} \omega_{ab} \gamma^{ab} $$

My question is the following: The RS field $\psi_{\mu}$ has a spacetime index and a spinor index, yet in the above action there is no affine connection part in the covariant derivative. The contribution from the torsion-free part of the affine connection vanishes because it is symmetric in two indices which get contracted with the antisymmetrised product of gamma matrices. But that still leaves the contorsion part. In the first order formalism, the spin connection and frame field are taken to be independent variables, so in general the spin connection may have torsion. In the second order formalism, the spin connection is not torsion free due to the presence of fermions. So in either case the contorsion tensor is non-zero. This leads me to believe that not having it in the above action will mean that the action is not invariant under diffeomorphisms.

Secondly, because the torsion is not in general zero, it seems to me that the RS action should actually be split into two pieces

$$ e\left( \overline{\psi}_a \gamma^{abc} D_b \psi_c - (D_b \overline{\psi}_a) \gamma^{abc} \psi_c \right) $$

This is because a complex conjugation should send each of those terms to each other, so that the action is real. However, if you only have one of those terms and the torsion is non-zero, when you complex conjugate you have to use integration by parts to 'move the covariant deriative to the other side', whereupon you pick up torsion tensor contributions from the covariant derivative acting on $e$, and the action is not real.

I'd be grateful if anyone could shed some light on either of these issues.

This post imported from StackExchange Physics at 2014-06-01 12:06 (UCT), posted by SE-user Steven
asked Jun 1, 2014 in Theoretical Physics by Steven (45 points) [ no revision ]

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