Torsion-free Einstein-Cartan action and Hilbert-Einstein action equations of motion equivalence

Question

Let $e^\alpha_{\ \mu}$ be the tetrad i.e.

\begin{equation} g_{\mu\nu} = \eta_{\alpha\beta}e^{\alpha}_{\ \mu}e^{\beta}_{\ \nu} \end{equation}

I'm denoting the internal indices using greek letters $\alpha,\beta,\gamma,\dots$ as in Weinberg. I will use a coordinate free description from now on so $e^\alpha$ are 1-forms on spacetime such that $e^\alpha(u) = e^\alpha_{\ \mu}u^\mu$. The spin connection is a $\mathfrak{so}(1, 3)$-valued connection 1-form $\omega^\alpha_{\ \beta}$ and its curvature is given by the second structural equation \begin{equation} R^\alpha_{\ \ \beta} = d\omega^\alpha_{\ \ \beta} + \omega^\alpha_{\ \ \gamma}\wedge \omega^\gamma_{\ \ \beta} \end{equation}

I am considering the torsion-free spin connection which directly relates to the Levi-Civita connection on space time. This directly relates to the curvature for the Levi-Civita connection on spacetime via the equation \begin{equation} e^\alpha_{\ \ \mu}e_{\beta}^{\ \ \nu}R^\mu_{\ \ \nu} = R^\alpha_{\ \ \beta} \end{equation}

The Einstein-Cartan action is (without the cosmological constant) \begin{equation} S[e,\omega] = \frac{1}{16\pi G} \int \epsilon_{\alpha\beta\gamma\delta} e^\alpha \wedge e^\beta \wedge R^{\gamma\delta} \end{equation} (see Krosnov, Formulations of General Relativity: Gravity, Spinors and Differential forms equation 3.60) which gives the equation of motion \begin{equation} \epsilon_{\alpha\beta\gamma\delta}e^\beta \wedge R^{\gamma\delta} = 0 \end{equation} when varied wrt the tetrad and gives the zero torsion condition when varied wrt the spin connection. My question is: how do I show that this equation implies the Einstein equation $R_{\mu\nu} = 0$?

Answer 1

I am assuming that since you are interested in the differential-forms formulation you are already familiar with the Hodge star operation. I would suggest you do the following.

1. Wedge from the left with a coframe in order to create the 4-form $$ \epsilon_{abcd}e^k\wedge e^b\wedge R_{[2]}^{cd}=0 $$ I use the $[2]$ subscript to distinguish the 2-form from the Ricci tensor.
2. Expand the curvature 2-form to get $$ \tfrac{1}{2}\epsilon_{abcd}R^{cd}{}_{mn}e^k\wedge e^b\wedge e^m\wedge e^n=0 $$
3. Operate on the latter with the Hodge star to get $$ \tfrac{1}{2}\epsilon_{abcd}R^{cd}{}_{mn}\ast (e^k\wedge e^b\wedge e^m\wedge e^n)=\tfrac{1}{2}\epsilon_{abcd}\epsilon^{kbmn}R^{cd}{}_{mn}=-\tfrac{1}{2}\delta^{kmn}_{acd}R^{cd}{}_{mn}. $$
4. Expand the generalized delta, and you should end up with $$ 2G^k_a=0, $$ where $G^k_a=e^k_\mu e_a^\nu G^\mu_\nu$ with $G_{(\mu\nu)}=R_{(\mu\nu)}-\tfrac{1}{2}Rg_{\mu\nu}$ being the Einstein tensor. Note that when you have torsion the Ricci tensor is in general asymmetric if I recall correctly, hence you also get an equation $G_{[\mu\nu]}=R_{[\mu\nu]}=0$. However, since your connection field equations imply the vanishing of torsion, $R_{[\mu\nu]}$ vanishes identically, and you recover the Einstein equations in vacuum.