Varying the Dirac action with differential forms

Question

The Dirac action in a curved spacetime can be written in terms of the vierbein $\{ e^a \}$ and spin connection $\{ \omega^{ab} \}$ differential forms. Let the spinor field $\psi$ be interpreted as a spinor-valued $0$-form and define the $1$-form $\gamma = \gamma_a e^a$, where $\{ \gamma_a \}$ are the gamma matrices. The spinor covariant derivative is given by

$$ D\psi = \mathrm{d} \psi + \frac{1}{4} \omega^{ab}[\gamma_a,\gamma_b] \psi $$

The standard Dirac action takes the form

$$ I = -i\int_M \bar{\psi} \gamma \wedge * D\psi$$

where $*$ is the Hodge dual operator. Now I would like to vary this action with respect to the veilbein. In the appendix of this book, equation (A.115) claims that the variation of the action with respect to the veilbein is given by

$$ \delta I = \int_M i \bar{\psi} \gamma_a * D \psi \wedge \delta e^a$$

but I cannot show this. Here is my attempt

$$ \delta I = -i \int_M \bar{\psi} \delta \gamma \wedge * D \psi + \bar{\psi} \gamma \wedge \delta (* D \psi) \\= -i \int_M \bar{\psi} \gamma_a \delta e^a \wedge * D \psi + \bar{\psi} \gamma\wedge \delta(* D \psi)$$

where I have assumed the gamma matrices $\{ \gamma_a \}$, despite having a veilbein index, are unaffected by the variation, hence $\delta \gamma = \delta (\gamma_a e^a) = \gamma_a \delta e^a$. The first term I can rearrange to

$$ i \int_M \bar{\psi} \gamma_a * D \psi \wedge \delta e^a $$

where I have just reversed the order of the wedge product, picking up a minus sign. This gives me the variation that the author claims, but I still have the second part to deal with. This contains the term $\delta (* D\psi)$. I know the Hodge dual should depend on the veilbein because the author has used this fact in equation (A.107) so this variation will not vanish in general. I am unsure how to evaluate it. I would ideally like it to vanish.

Any help would be greatly appreciated.

Answer 1

If you have vanishing torsion, then I think you can do the following trick: $$ \bar\psi \gamma\wedge \delta(\ast D\psi)=\bar\psi\gamma_aD\psi\wedge\delta\ast e^a=\bar\psi\gamma_aD\psi\wedge\delta e^b\wedge\ast(e^a\wedge e_b) $$ Since $D$ and $\delta$ commute (plus vanishing torsion condition) this term is a total derivative, because $D\delta e^a=0=D\ast(e^a\wedge e_b)$. Of course for this to make sense, the indices must be Lorentz indices (i.e. frame indices, not coordinate ones), so that the Hodge operator is related to the eta metric meaning that $D\epsilon_{abcd}=0$. I just exploited the symmetry of the Hodge product for this.