Obtaining curved space Dirac equation from action (tetrad formalism) 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?
 A: 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.
$$
S= \int  \bar \psi \gamma^\mu \nabla_\mu \psi \sqrt{g}d^dx. 
$$
A: 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 ?
