Let the Einstein-Hilbert action be rewritten as a functional of the tetrad $e$ (units shall be set to $1$) such that

$S_{EH}(e)=\int \frac{1}{2}\epsilon_{IJKL}~e^I\wedge e^J\wedge F^{KL}(\omega(e))$, where $\epsilon$ is the Levi-Civta-symbol as usual, $F$ is the curvature of the spin connection $\omega$ and $I,J,K,L$ denote internal indices, indicating that the object carries a rep. of the Lorentz group.

How do I get back to the usual action $S_{EH}(g)=\int~d^4x\sqrt{-g}R$? I am trying to figure it out using identities but can't finish till the end. Could anybody give a little more detailed account, please? Thank you.

# [Physics] Einstein action as a functional of the tetrad (first order formulation of gravity)

differential-geometrygeneral-relativitygravityhomework-and-exercisesquantum-gravity

## Best Answer

The Levi-Civita symbol in curved space-time, $\epsilon^{\alpha\beta\gamma\delta}$ ,is defined as :

$$\epsilon^{\alpha\beta\gamma\delta} = e~ \epsilon^{I'J'K'L'} e_{I'}^\alpha e_{J'}^\beta e_{K'}^\gamma e_{L'}^\delta,$$

where $\epsilon^{I'J'K'L'}$ is the standard Levi-Civita symbol (flat spacetime).

The presence of the $e$ is because $\epsilon^{\alpha\beta\gamma\delta}$ has to transform as a tensor.

You could then use the properties of the standard Levi-Civita symbol. For instance, the product $\epsilon^{I'J'K'L'} \epsilon_{IJKL}$ could be obtained as a determinant of the matrix of Kronecker delta symbols $\delta^M_N $. The product equals $1$ if $I'J'K'L'$ is a even permutation of $IJKL$, and $-1$ if $I'J'K'L'$ is a odd permutation of $IJKL$, $0$ in other cases.

And you are going to use the fact that the matrix $e_I^\alpha$ and $e_\beta^J$, are inverse matrix.

[EDIT] Here is the complete calculus:

\begin{align} S_{EH}(g)&=\int\sqrt{-g}R\,\mathrm{d}^4x\\ &=\int \sqrt{-g} R_{\mu\nu} g^{\mu\nu}\,\mathrm{d}^4x\\ &=\int e\, e^{\mu}_I e^{\nu~I}R_{\mu\nu\rho\sigma}e^{\rho}_J e^{\sigma~J}\,\mathrm{d}^4x\\ &=\int e\, e^{\mu}_I e^{\rho}_J F^{IJ}_{\mu\rho}\,\mathrm{d}^4x \end{align}

and

\begin{align} S_{EH}(e)&=\int \frac{1}{2}\epsilon_{IJKL}~e^I\wedge e^J\wedge F^{KL}(\omega(e))\\ &=~\int \frac{1}{2}~\epsilon_{IJKL}~ e_{\alpha}^I \,\mathrm{d}x^{\alpha} \wedge e_{\beta}^J \,\mathrm{d}x^{\beta} \wedge (\frac{1}{2} F_{\gamma\delta}^{K L} \wedge \,\mathrm{d}x^{\gamma} \wedge \,\mathrm{d}x^{\delta})\\ &=~\int \frac{1}{4}~\epsilon_{IJKL}~ e_{\alpha}^I~ e_{\beta}^J ~F_{\gamma\delta}^{K L} \,\mathrm{d}x^{\alpha}\wedge \,\mathrm{d}x^{\beta} \wedge \,\mathrm{d}x^{\gamma} \wedge \,\mathrm{d}x^{\delta}\\ &=~\int \frac{1}{4}\epsilon_{IJKL}~ \epsilon^{\alpha\beta\gamma\delta}~e_{\alpha}^I ~e_{\beta}^J ~F_{\gamma\delta}^{K L}\,\mathrm{d}^{4}x\\ &=~\int e\frac{1}{4}~\epsilon_{IJKL} ~\epsilon^{I'J'K'L'} ~e_{I'}^\alpha ~e_{J'}^\beta ~e_{K'}^\gamma ~e_{L'}^\delta ~e_{\alpha}^I ~e_{\beta}^J ~F_{\gamma\delta}^{K L}\,\mathrm{d}^{4}x\\ &=~\int e \frac{1}{4}~\epsilon_{IJKL} ~\epsilon^{I'J'K'L'} ~\delta_{I'}^I ~\delta_{J'}^J ~e_{K'}^\gamma ~e_{L'}^\delta ~F_{\gamma\delta}^{K L}\,\mathrm{d}^{4}x\\ &=~\int e \frac{1}{4}~\epsilon_{IJKL} ~\epsilon^{IJK'L'} ~e_{K'}^\gamma ~e_{L'}^\delta ~F_{\gamma\delta}^{K L}\,\mathrm{d}^{4}x\\ &=~\int e \frac{1}{4} ~2 ~(\delta_{K}^L \delta_{K'}^{L'} - \delta_{K}^{L'} \delta_{K'}^L)~ e_{K'}^\gamma ~e_{L'}^\delta ~F_{\gamma\delta}^{K L}\,\mathrm{d}^{4}x\\ &=~\int ~e ~\frac{1}{4} 2*2~(\delta_{K}^L \delta_{K'}^{L'})~ e_{K'}^\gamma ~e_{L'}^\delta ~F_{\gamma\delta}^{K L}\,\mathrm{d}^{4}x\\ &=~\int ~e ~e_{K}^\gamma ~e_{L}^\delta ~F_{\gamma\delta}^{K L}\,\mathrm{d}^{4}x \end{align}