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The Rarita-Schwinger action is

[tex]

\int \sqrt{g} \overline{\psi}_a \gamma^{abc} D_b \psi_c

[/tex]

Here ##g = \det(g_{\mu \nu})##, and the indices ##a, b \dots ## are 'internal' indices that transform under e.g. ##\mathrm{SO} (3,1) ## in ##3+1## dimensions. ##\gamma^{abc} = \gamma^{[a} \gamma^{b} \gamma^{c]}## with the gamma matrices obeying ##\gamma^a \gamma^b + \gamma^b \gamma^a = 2 \eta^{ab} ## and ##\eta^{ab}=\mathrm{diag}(1,1 \ldots 1,-1,-1 \ldots -1)## is the 'internal metric'. ##\psi_{\mu} = \psi_{c} e^{c}_{\mu} ## is a spinor-valued one form. Spacetime indices ##\mu, \nu## can be 'converted' to internal indices using the frame field ##e_a^{\mu}##, and vice versa. The covariant derivative is ##D_{\mu} \psi_{\nu} =\partial_{\mu} \psi_{\nu} + \frac{1}{4} \omega_{\mu}^{ab} \gamma_{ab} \psi_{\nu} ##. Here ##\omega## is taken to be the torsion free spin connection, and ##\gamma_{ab} = \gamma^{[a} \gamma^{b]}##.

My question is as follows; from my reading, it seems that the Rarita-Schwinger action above is not invariant under local Lorentz transformations, but the Rarita-Schwinger action plus gravity is invariant under a combined supersymmetry and Lorentz transform. However, I can't see how this action fails to be invariant. As far as I know, the transformation is

[tex] \psi_{c} \rightarrow \Lambda_c^b S \psi_b \\

D_b \psi_c \rightarrow \Lambda_b^e \Lambda_c^f S D_e \psi_f \\

\gamma^{abc} \rightarrow \Lambda_d^a \Lambda_e^b \Lambda_f^c S \gamma^{def} S^{-1}

[/tex]

Here ##S## is an element of the relevant spin group, and ##\Lambda## is a local Lorentz transformation.

To me, it looks like the action is invariant. I'm not sure how this reasoning breaks down.

Any help greatly appreciated. Thanks.

[tex]

\int \sqrt{g} \overline{\psi}_a \gamma^{abc} D_b \psi_c

[/tex]

Here ##g = \det(g_{\mu \nu})##, and the indices ##a, b \dots ## are 'internal' indices that transform under e.g. ##\mathrm{SO} (3,1) ## in ##3+1## dimensions. ##\gamma^{abc} = \gamma^{[a} \gamma^{b} \gamma^{c]}## with the gamma matrices obeying ##\gamma^a \gamma^b + \gamma^b \gamma^a = 2 \eta^{ab} ## and ##\eta^{ab}=\mathrm{diag}(1,1 \ldots 1,-1,-1 \ldots -1)## is the 'internal metric'. ##\psi_{\mu} = \psi_{c} e^{c}_{\mu} ## is a spinor-valued one form. Spacetime indices ##\mu, \nu## can be 'converted' to internal indices using the frame field ##e_a^{\mu}##, and vice versa. The covariant derivative is ##D_{\mu} \psi_{\nu} =\partial_{\mu} \psi_{\nu} + \frac{1}{4} \omega_{\mu}^{ab} \gamma_{ab} \psi_{\nu} ##. Here ##\omega## is taken to be the torsion free spin connection, and ##\gamma_{ab} = \gamma^{[a} \gamma^{b]}##.

My question is as follows; from my reading, it seems that the Rarita-Schwinger action above is not invariant under local Lorentz transformations, but the Rarita-Schwinger action plus gravity is invariant under a combined supersymmetry and Lorentz transform. However, I can't see how this action fails to be invariant. As far as I know, the transformation is

[tex] \psi_{c} \rightarrow \Lambda_c^b S \psi_b \\

D_b \psi_c \rightarrow \Lambda_b^e \Lambda_c^f S D_e \psi_f \\

\gamma^{abc} \rightarrow \Lambda_d^a \Lambda_e^b \Lambda_f^c S \gamma^{def} S^{-1}

[/tex]

Here ##S## is an element of the relevant spin group, and ##\Lambda## is a local Lorentz transformation.

To me, it looks like the action is invariant. I'm not sure how this reasoning breaks down.

Any help greatly appreciated. Thanks.

Last edited: