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Lorentz invariance of Rarita-Schwinger action

  1. May 27, 2014 #1
    The Rarita-Schwinger action 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]}##.

    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}

    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: May 27, 2014
  2. jcsd
  3. May 29, 2014 #2


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    The action is indeed invariant under LLT's :) Which text do you use and what does it say explicitly? You only need the grav. part if you consider Susy, as the gravitino is the superpartner of the graviton. But every term on it self should be invariant under gct's and LLT's.
  4. May 30, 2014 #3
    Yes, I had a misunderstanding. The RS action is invariant under local lorentz transformations. What it is not invariant under (at least in curved space, or when it is minimally coupled to something) is a gauge transformation which removes unphysical degrees of freedom from the action. One way to get around that problem is to add it to the gravity action in a supersymmetric combination.

    I do have some more questions about supergravity though, if youll humour me. And in particular, torsion.

    In the first order formalism, the spin connection and frame field are taken to be independent variables. In that case, there is nothing constraining the spin connection and therefore in general it can have torsion. This also means that the affine connection ##\Gamma_{\mu \nu}^{\sigma}## may in general have torsion - that is, a part which is antisymmetric under ##\mu \leftrightarrow \nu##. In the RS action, the covariant derivative of ##\psi_{\mu}## should include the affine connection since it has a spacetime index. The torsion free part of this connection doesn't contribute because it is symmetric and gets contracted with the antisymmetrised product of gamma matrices. But the contribution from the antisymmetric part (contorsion tensor) shouldn't vanish. However, in my reading I have found that in supergravity actions this contribution from the affine connection is not there. And, for example, in the book by Freedman, in the paragraph under 9.2 it says inclusion of these terms would be inconsistent with local supersymmetry. On the other hand, if you don't include them, as far as I can see the theory is not diffeomorphism invariant. What is going on here?

    Furthermore, if there is indeed torsion, as far as Im aware you need to split the RS action into two pieces as follows:

    \int \sqrt{g} \left( \overline{\psi}_a \gamma^{abc} D_b \psi_c - ( D \overline{\psi}_a) \gamma^{abc} \psi_c \right)

    The reason is that otherwise the action is not real. Complex conjugation sends each of these two terms to the other, so if you have both in the action then it is real. However, if you only have one term of these terms, in order to show it is real you have to use integration by parts to 'take the covariant derivative to the other side'. In doing so, you pick up an extra ##D e## term, which only vanishes if the torsion is zero, which isn't generally true in the first order formalism.

    Finally, the second order formalism seems to suffer from the same problems, since you get to the second order formalism by substituting the equation of motion given by the variation of the action wrt ##\omega##. But because there are fermions present, the spin connection you get from this is not torsion free, and I have all the same objections.
  5. Jun 2, 2014 #4


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    Hi bob, i can't properly go into your question now due to illness. If i can, you ll notice ;)
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