# Palatini action (General Relativity)

I'm reading this lecture notes (Lecture III: Ashtekar variables for general relativity) about Tetrad Formalism in General Relativity. In page 8-9 the Palatini action is defined (basically the Einstein-Hilbert action but in function of tetrads $$e$$ and the Lorentz action $$\omega$$);

$$$${\displaystyle S=\frac{1}{16\pi G}\int_M d^{4}x\;|e|\;e_{I}^{\mu }e_{J}^{\nu }{F _{\mu \nu }}^{IJ}(\omega)}$$$$

Thus, the autor say that $$\varepsilon^{\mu \nu c d} \epsilon_{IJKL}e^K_c e^L_d=2ee_I^{[\mu}e_J^{\nu]}$$ allows us to rewrite the Palatini action above in differential-form notation as follows: $$$${\displaystyle S=\frac{1}{64\pi G} \int_M \epsilon_{IJKL} e^K\land e^L \land {F }^{IJ}(\omega)}$$$$

But, even with this hint I don't see how to obtain this last expression using the first. Thanks in advance for any hint.

I am going to assume that $$F_{[ab]}=\tfrac12\left(F_{[ab]}-F_{[ba]}\right)$$. Hence,
$$e^a_Ie^b_JF^{IJ}_{ab}=e^{[a}_Ie^{b]}_JF^{IJ}_{ab}$$
$$d^4x|e|e^{[a}_Ie^{b]}_JF^{IJ}_{ab}=d^4x|e|\frac{1}{2e}\epsilon^{abcd}\epsilon_{IJKL}e^K_ce^L_dF^{IJ}_{ab}=\tfrac12\text{sgn}(e)d^4x\epsilon^{abcd}\epsilon_{IJKL}e^K_ce^L_dF^{IJ}_{ab}=\tfrac14\epsilon_{IJKL}\mathbf{e}^K\wedge\mathbf{e}^L\wedge\mathbf{F}^{IJ}$$ where I've used the definition of the wedge product and the one form. The result follows immediately.