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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.

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Perhaps you could include a little more detail about what you have already done and where you are stuck? – joshphysics Jun 25 '13 at 20:04
Well, $S_{EH}(e)=\int \frac{1}{2}\epsilon_{IJKL}~e^I\wedge e^J\wedge F^{KL}(\omega(e))~=~\int \frac{1}{2}\epsilon_{IJKL} e_{\alpha}^I e_{\beta}^J F_{\gamma\delta}^{K L} dx^{\alpha}\wedge dx^{\beta} \wedge dx^{\gamma} \wedge dx^{\delta}~=~\int 4!\frac{1}{2}\epsilon_{IJKL} e_{\alpha}^I e_{\beta}^J F_{\gamma\delta}^{K L}\epsilon^{\alpha\beta\gamma\delta}d^{4}x$. From the other side I know $S_{EH}(g)~=~\int~d^4x\sqrt{-g}R~=~\int d^4x\sqrt{-g} R_{\mu\nu} g^{\mu\nu}~=~\int d^4x e e^{\mu}_I e^{\nu~I}R_{\mu\nu\rho\sigma}e^{\rho}_J e^{\sigma~J}~=~\int d^4x e e^{\mu}_I e^{\rho}_J F^{IJ}_{\mu\rho}$. – Hamurabi Jun 25 '13 at 20:19
With which trick can I tie both ends together? – Hamurabi Jun 25 '13 at 20:22
What is a relationship between the spin-connection and the Riemann tensor? – Jerry Schirmer Jun 25 '13 at 21:01
The curvature is $F^{IJ}=d\omega^{I J}+ \omega^{I}_{K}\wedge \omega^{K J}$, where the spin and Levi-Civita connections are related as $\omega^{I}_{\mu~J}=e^{I}_{\nu}\nabla_{\mu}e^{\nu}_{J}$. Cartan's first structure equation is $d_{\omega}e^I=0$, where $d_{\omega}$ is the $\omega$-twisted differential. Cartan's second structure equation is $F^{I J}_{\mu\nu}=e^{I\rho}e^{J\sigma}R_{\mu\nu\rho\sigma}$. – Hamurabi Jun 25 '13 at 21:07
up vote 4 down vote accepted

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}


\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}

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Thanks Trimok. I have done the computation but am ending up with a factor 4 to much. I have used together with the identity that you wrote down. It could be that the 4! that I wrote in the question is wrong. – Hamurabi Jun 27 '13 at 9:29
Are you ending up with a factor $4$ too much, or a factor $4!$ too much ? – Trimok Jun 27 '13 at 9:42
@Hamurabi : I think your $4!$ term is not correct. The correct equation is $ dx^{\alpha}\wedge dx^{\beta} \wedge dx^{\gamma} \wedge dx^{\delta} = \epsilon^{\alpha\beta\gamma\delta} d^4x$ – Trimok Jun 27 '13 at 10:26
@Hamurabi : I made an EDIT in the answer, with the complete calculus – Trimok Jun 27 '13 at 11:15
No, it is correct. The notation $$\large \delta_{i_{k+1}....i_n}^{j_{k+1}....j_n}$$ means that all the antisymmetric possibilities of the $\delta$ functions are considered, each with the weight $1$. The total weight is $(n-k)!$ because there are $(n-k)!$ terms. But the notation $$\large \delta_{[i_{k+1}}^{j_{k+1}}...\delta_{i_{n}]}^{j_{n}}$$ means that all the antisymmetric possibilities of the $\delta$ functions are considered, but each with the weight $\frac{1}{(n-k)!}$, so that the total weight is $1$. – Trimok Jun 27 '13 at 16:49

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