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

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

$$$$g_{\mu\nu} = \eta_{\alpha\beta}e^{\alpha}_{\ \mu}e^{\beta}_{\ \nu}$$$$

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 $$$$R^\alpha_{\ \ \beta} = d\omega^\alpha_{\ \ \beta} + \omega^\alpha_{\ \ \gamma}\wedge \omega^\gamma_{\ \ \beta}$$$$

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 $$$$e^\alpha_{\ \ \mu}e_{\beta}^{\ \ \nu}R^\mu_{\ \ \nu} = R^\alpha_{\ \ \beta}$$$$

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

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.