The Rarita-Schwinger action in curved $n$-dimensional spacetime is

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

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 $\gamma^{ab} = \gamma^{[a} \gamma^{b]}$.

In flat space, the covariant derivative becomes a normal derivative, and the action then has a symmetry $\psi_c \rightarrow \psi_c + \partial_c \phi $, with $\phi$ an arbitrary function. This freedom can be used to eliminate some of the degrees of freedom from the field $\psi_c$ which correspond to lower spin. However, in curved space there is no corresponding symmetry under $\psi_c \rightarrow \psi_c + D_c \phi$. For this reason, it is said that the Rarita-Schwinger action in curved spacetime is inconsistent. My question is, what goes wrong when you don't have this extra symmetry? And do the problems manifest at the classical level or only at the quantum level?

  • $\begingroup$ In order for the action to be invariant you have to rotate all objects according to the representation of the Lorentz algebra they take values in. In particular you need to transform spin connection, which means that the gravity part is affected. Note, that local Lorentz rotations affect vielbein-tetrad also but not the metric tensor. $\endgroup$ – John May 28 '14 at 18:05
  • $\begingroup$ I'm not sure I understand you. In my transformations above, the spin connection does transform, and this is taken account of in the transformation law of the covariant derivative $D_b \psi_c$. $\endgroup$ – Steven May 28 '14 at 18:28
  • $\begingroup$ Then, OK, RS is invariant under local Lorentz if you transform spin-connection as well, but then the spin-connection is a background field. So where you found that it is not invariant? $\endgroup$ – John May 28 '14 at 18:49
  • $\begingroup$ These notes for example: damtp.cam.ac.uk/research/gr/members/gibbons/… Bottom of page 21 and top of page 22. What puzzles me here is the form of the transformation for the RS field. More broadly speaking, others say things like the only consistent theory incorporating spin 3/2 fermions is supergravity. $\endgroup$ – Steven May 28 '14 at 19:38
  • $\begingroup$ What is discussed there is a different issue. The action is invariant under Lorentz transformations, but in order to describe correct number of degrees of freedom we should take care of the gauge symmetry that RS has on its own. The action is not gauge invariant since the commutator of two derivatives produces Rieman tensor (this is not a problem with spin-1, whose gauge parameter is a scalar).In order to compensate this non-invariance one can modify the graviton transformations as this variation also produces Rieman (in fact Ricci, but this is enough).Lorentz and RS symmetry are different $\endgroup$ – John May 28 '14 at 22:25

If you don't have the gauge gauge invariance, it's not necessarily inconsistent, it's just describing more degrees of freedom, i.e. the 4 degrees of a massive spin 3/2 rather than the 2 degrees of freedom of a massless spin 3/2. To have gauge invariance for the spin 3/2 you can do it on a background Einstein space with negative curvature, and you need to include a mass term with a mass fixed to a particular value relative to the AdS radius.

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