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I'm reading the book Covariant Loop quantum gravity by C.Rovelli where in 3.2 the action of a dirac fermion is presented in the tetrad formalism:

$$S= \int \bar{\psi} \gamma^{I} D\psi \wedge e^J \wedge e^{K} \wedge e^{L} \epsilon_{IJKL} $$

where $D = D_{\mu} dx^{\mu}$

I can't obtain from this action the typical action in terms of the coordinates. Could anyone give some guidance on how to proceed?

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Here's a standard formulation of the Dirac field lagrangian (where $\hbar = 1 = c$, and $\bar{\Psi} \equiv \Psi^{\dagger} \, \gamma^0$, as usual) : \begin{equation}\tag{1} \mathscr{L}_{\textsf{D}} = i \, \frac{1}{2} \big( \bar{\Psi} \: \Gamma^{\mu} \, (\, D_{\mu} \, \Psi \,) - (\, D{_{\mu}} \, \bar{\Psi} \,) \, \Gamma^{\mu} \, \Psi \, \big) - m \, \bar{\Psi} \, \Psi. \end{equation} Here, $D_{\mu}$ is the covariant derivative of the Dirac field, under local Lorentz transformations and under arbitrary coordinates transformations. Neglecting the electromagnetic and Yang-Mills fields for simplicity : \begin{equation}\tag{2} D_{\mu} \, \Psi \equiv \partial_{\mu} \, \Psi + \frac{1}{2} \: \omega_{\mu}^{\; ab} \: M_{ab} \, \Psi, \end{equation} where $M_{ab}$ is the set of the Lorentz group generators : \begin{equation}\tag{3} M_{ab} = i \, \frac{1}{2} ( \gamma_a \, \gamma_b - \gamma_b \, \gamma_a), \end{equation} and $\omega_{\mu}^{\; ab}$ are the spin-connection coefficients. Those can be defined from the tetrad field. Under some coordinates $x^{\mu}$, the tetrad is $\boldsymbol{e}^a \equiv e_{\mu}^a(x) \, \boldsymbol{d}x^{\mu}$ and can be found from the metric : \begin{equation}\tag{4} d\boldsymbol{s}^2 = g_{\mu \nu} \: \boldsymbol{d}x^{\mu} \otimes \boldsymbol{d}x^{\nu} \equiv \eta_{ab} \: e_{\mu}^a(x) \, e_{\nu}^b(x) \: \boldsymbol{d}x^{\mu} \otimes \boldsymbol{d}x^{\nu} \equiv \eta_{ab} \: \boldsymbol{e}^a \otimes \boldsymbol{e}^b. \end{equation} Then the "generalized" gamma matrices are \begin{equation}\tag{5} \Gamma^{\mu}(x) \equiv \gamma^a \: e_a^{\mu}(x), \end{equation} and the spin connection coefficients are given by this formula : \begin{equation}\tag{6} \omega_{\mu \; b}^{\; a}(x) = e_{\lambda}^a \: \nabla_{\mu} \, e_b^{\lambda}. \end{equation} Using the lagrangian density (1) above in the Euler-Lagrange equation gives the usual Dirac equation, in a gravitational field (i.e. curved spacetime). The calculations are laborious.

Is that what you are looking for ?

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  • $\begingroup$ Thank you for your answer but this is not what I was looking for exactly. What I need to know is why the action I showed in the question is equivalent with the action you havee given (meaning that they give the same equations of motion). More precisely my question is: How is it possible to obtain the dirac equation in curved space from the action I wrote in the question? $\endgroup$ – Pam Mar 18 '17 at 0:52
  • $\begingroup$ Where is the mass parameter in your action ? Is this for a massless Dirac field only ? $\endgroup$ – Cham Mar 18 '17 at 1:23
  • $\begingroup$ I was wondering the same when I saw the equation in the book I mentioned. I'm guessing the action is for a massless dirac field. $\endgroup$ – Pam Mar 18 '17 at 1:41
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After you have written $\gamma^I= e^I_\mu \gamma^\mu$ you can use ${\bf e}^I= e_\mu^I dx^\mu$ to write
$$ \epsilon_{IJKL} {\bf e}^I\wedge {\bf e}^J\wedge {\bf e}^K\wedge{\bf e}^L = \sqrt{g}d^4x $$ and so get the usual coordinate action. $$ S= \int \bar \psi \gamma^\mu \nabla_\mu \psi \sqrt{g}d^dx. $$

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