Converting gravity action from differential forms to tensor component I deal with this action but i really do not know how differential forms can be expressed in terms of tensor components.
$$
S\stackrel{\text{def}}{=} M_{pl}^2\int_{M}\epsilon_{abcd}\left(e^a\wedge e^{b}\wedge\Omega^{cd}\right)=M_{pl}^2\int_{M}d^4x\, \epsilon^{\mu\nu\rho\sigma}\epsilon_{abcd}\,e^{a}_{\mu}\,e^{b}_{\nu}\, R^{cd}_{\rho\sigma}[\omega]
$$
Can anyone help me in this way? 
Or introduce me a book or paper related to my questions?
 A: I'm not sure of your conventions for the Riemann tensor or curvature two forms, but this should be fine. we have that $$\Omega^{cd} = R^{cd}{}_{\mu\nu} dx^\mu \wedge dx^\nu$$ in local coordinates (we must be careful really, since the coordinates may not be defined everywhere on the manifold.) Next, the tetrad/vierbein satisfies $e^a = e^a_\mu dx^\mu$. So the left hand side amounts to
$$M_{pl}^2 \int_M \epsilon_{abcd}(e^a_\mu dx^\mu)\wedge (e^b_\nu dx^\nu)\wedge R^{cd}{}_{\rho\sigma}dx^\rho \wedge dx^\sigma 
$$
The basis 1 forms are the only things that care about the wedge product/exterior product, but the functions that sit in front of them can be moved:
$$M_{pl}^2 \int_M \epsilon_{abcd}(e^a_\mu dx^\mu)\wedge (e^b_\nu dx^\nu)\wedge R^{cd}{}_{\rho\sigma}dx^\rho \wedge dx^\sigma = M_{pl}^2 \int_M \epsilon_{abcd}e^a_{\mu} e^b_\nu R^{cd}{}_{\rho\sigma}dx^\mu\wedge dx^\nu\wedge dx^\rho \wedge dx^\sigma
$$
Note however, that the $dx^\mu\wedge dx^\nu\wedge dx^\rho \wedge dx^\sigma$ is a permutation of the volume element $d^4x$ provided that no two indices are identical (otherwise it vanishes). Thus $dx^\mu\wedge dx^\nu\wedge dx^\rho \wedge dx^\sigma = \epsilon^{\mu\nu\rho\sigma} d^4 x$, and hence we have the result.
