Palatini Action

Question

I'm studying the Palatini's action in terms of tetrads and the spin connection. In the Rovelli's book "Quantum Gravity", the expression for the action of General Relativity is written as: $$S[e,\omega]=\int_M\epsilon_{IJKL}(e^I\wedge e^J\wedge R^{KL}[\omega])$$

but, in the Baez and Muniain's book "Gauge Fields, Knots and Gravity", the action is written in the form:

$$S[e,\omega]=\int_M e_{I}^{\alpha}e_{J}^{\beta}R_{\alpha\beta}^{IJ}\text{vol}$$ where $vol$ is the volume form in four dimensions.

I have tried to pass from one expression to the other, but I couldn't. I have tried to use that $e^{I}=e^{I}_{\alpha}dx^{\alpha}$, $R^{KL}=R^{KL}_{\mu\nu}dx^{\mu}\wedge dx^{\nu}$ and the expression: $$dx^0\wedge dx^1\wedge dx^2\wedge dx^3=\frac{1}{4}\epsilon_{\alpha\beta\mu\nu}dx^{\alpha}\wedge dx^{\beta}\wedge dx^{\mu} \wedge dx^{\nu}$$ to put the first expression in the form of the second one, but I couldn't do it. I'm sure there is something missing in my calculations, but I cannot see what. Could anybody help me?

P.S I have successfully completed the exercise of writte the standard Einstein-Hilbert action:

$$S[g,\Gamma]=\int_{M}R\text{vol}$$

in the form of the first expression.

Answer 1

Your question revolves around proving that you can always use the vielbein isomorphism to convert coordinate tensor indices to frame tensor indices, and vice versa with the use of the inverse. In particular, $R$ is a coordinate scalar and thus, $R=R^{\mu\nu}{}_{\mu\nu}=R^{IJ}{}_{IJ}$. If someone gave you the relation $R^\lambda{}_{\rho\mu\nu}=e^\lambda_I e^J_\rho e^K_\mu e^L_\nu R^I{}_{JKL}$ as granted then you would be surely able to trivially show that $R^{IJ}{}_{IJ}=R^{IJ}{}_{\mu\nu}e^\mu_I e^\nu_J$. This means that all you have to do is prove this ``granted'' relation. Since this is a fundamental task in the tetradic approach I will just say that you need to do the proof yourself. You will have to use the duality relations $e^I_\mu e^\mu_J=\delta^I_J$, $e^I_\mu e^\nu_I=\delta^\nu_\mu$, the definition $$ R^{IJ}{}_{\mu\nu}=2\partial_{[\mu}\omega^{IJ}{}_{\nu]}+2\omega^{I}{}_{K[\mu}\omega^{KJ}{}_{\nu]}, $$ which follows from $R^{IJ}=\mathrm{d}\omega^{IJ}+\omega^{I}{}_K\wedge \omega^{KJ}$, and at some point you will also need the so-called vielbein postulate $$ \nabla_\mu e^I_\nu=0=\partial_\mu e^I_\nu+\omega^I{}_{J\mu}e^J_\nu-\Gamma^\lambda_{\mu\nu}e^I_\lambda, $$ to relate the components of the spin connection with the Christoffel symbols. It is also a standard exercise to show that especially the Levi-Civita spin connection can be written completely in terms of vielbeins and their partial derivatives but for that you will also need the zero-torsion condition $\mathrm{d}e^I=-\omega^I{}_J\wedge e^J$.

All you have to do is manipulate the expressions in a pretty straightforward way which unfortunately is a bit tedious. But it of course suffices to do it only once for purposes of understanding.

Answer 2

Starting from the "Rovelli" form $$S[e,\omega]=\int_M\epsilon_{IJKL}(e^I\wedge e^J\wedge R^{KL}[\omega]),$$ the integrand is a top form and must be expressable as $$ \epsilon_{IJKL}e^I\wedge e^J\wedge R^{KL}[\omega] = \ell \frac{1}{4!}\epsilon_{\mu\nu\rho\sigma}dx^\mu\wedge dx^\nu \wedge dx^\rho\wedge dx^\sigma=\ell\, \text{vol}, $$ for some scalar function $\ell$. To extract the scalar $\ell$, one simply contracts the components of the integrand with the Levi-Civita tensor (remember to account for overcounting and overall sign).

In doing this contraction, the non-trivial step is to evaluate $$ \epsilon^{\mu\nu\rho\sigma}\epsilon_{IJKL}e^I_\mu e^J_\nu. $$ This can be accomplished using the identity $$\epsilon^{\rho_1\ldots\rho_p\mu_{p+1}\ldots\mu_n}\epsilon_{\rho_1\ldots\rho_p\nu_{p+1}\ldots\nu_n}=(-1)^sp!(n-p)!\delta^{[\mu_{p+1}}_{\nu_{p+1}}\ldots\delta^{\mu_n]}_{\nu_n},$$ where $n$ is the spacetime dimension, $s$ the signature of the metric ($s=1$ for Lorentzian metrics), and $[\ldots]$ denotes anti-symmetrization. In our derivation, this works out as $$ \epsilon^{\mu\nu\rho\sigma}\epsilon_{IJKL}e^I_\mu e^J_\nu =\epsilon^{\mu\nu\rho\sigma}\epsilon_{\mu\nu\alpha\beta} e^\alpha_Ke^\beta_L = -(2!)^2 e^{[\rho}_Ke^{\sigma]}_L, $$ from which the second form of the Palatini action readily follows.