# Supergravity, torsion and diffeomorphism invariance

The action for $N=1$ supergravity in an $4$ 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 $a,b\ldots = 0\ldots 3$ 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.

I'm not expert, but it seems that you are mixing things here.

First, I'd point out that the usual normalization of the gamma matrices is $\{\gamma^a,\gamma^b\} = 2\eta^{ab}\boldsymbol{1}$, where $\boldsymbol{1}$ is the unit matrix in the spin-space. Here, the $a$ and $b$ indices are tangent space indices.

When you write the action

$$S= \int e\left( R + \overline{\psi}_a \gamma^{abc} D_b \psi_c \right),$$

the spacetime indices have been "translated" to tangent space indices through the vielbein.

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.

False! The connection is a general connection, neither symmetric nor anti-symmetric... but a sum of those!

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.

Actually, if you use the Maurer-Cartan structure equations to calculate the Riemann curvature, the spin-connection is anti-symmetric in those two indices, and you'd notice this is not an issue. The point is that from the Lorentz (or spin) connection you can obtain the Levi-Civita connection, but is not just a "translation" of indices.

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.

True!

In the second order formalism, the spin connection is not torsion free due to the presence of fermions.

Not true! Pure (Einstein-Cartan) gravity implies vanishing torsioin... the converse is not true! However, the inclusion of torsion is more general than its suppression.

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.

The contorsion IS present in the above action, you can split the spin connection into $\omega \to \bar{\omega} + \kappa$ (where $\bar{\omega}$ is the torsion-free part of the connection), in both... the Einstein-Cartan and the Rarita-Schwinger actions.

This does not spoil the invariance under diffeomorphisms.

You can notice that (of course) if the contorsion is independent... one could add invariant terms constructed with $\kappa$. These are non-minimal generalizations of "gravitational" theories with torsion.

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 agree with you!

Cheers!