Consider the variation of the EH action with respect to the metric
$$\delta S_{EH} = \int d^4x ~(\delta\sqrt{-g} R + \sqrt{-g}\delta g^{\mu\nu} R_{\mu\nu})$$ Now I make use of $$ \delta\sqrt{-g} = \frac{\sqrt{-g}}{2} g^{\mu\nu}\delta g_{\mu\nu} = -\frac{\sqrt{-g}}{2}g_{\mu\nu}\delta g^{\mu\nu}$$
$$\delta S_{EH} = \int d^4x ~\sqrt{-g}~\delta g^{\mu\nu}(-\frac{g_{\mu\nu}}{2} R + R_{\mu\nu}).$$
Now, since, this is the Einstein-Cartan formalism, it is not a priori clear if the connection is torsionless. Therefore, the Ricci tensor $R_{\mu\nu}$ may not be symmetric in $\mu \leftrightarrow \nu$. Therefore, the equation of motion that comes out should be
$$R_{(\mu, \nu)}-\frac{1}{2}g_{\mu\nu} R = 0$$ where $\mu, \nu$ is symmetrized. This is because to derive the equation of motion we must perform $\delta S/\delta g^{\alpha\beta}$ and use $$\frac{\delta g^{\mu\nu}(x)}{\delta g^{\alpha\beta}(x')} = \frac{1}{2}(\delta^{\mu}_{\alpha}\delta^{\nu}_{\beta}+\delta^{\mu}_{\beta}\delta^{\nu}_{\alpha})\delta^{(4)}(x-x')$$ which will lead to the symmetrization. Now, if there is a matter field present we can safely have
$$R_{(\mu, \nu)}-\frac{1}{2}g_{\mu\nu} R = T_{\mu\nu}$$
where $T_{\mu\nu}$ is the usual symmetric stress-energy tensor. But I believe this is in contradiction with what is known here. Can somebody explain where I am going wrong?
Best Answer
The Wikipedia page is wrong (or is misstating things$^1$). Varying with respect to the tetrad leads to the non-symmetric field equations (which can be seen as a nicer way of doing things). If you vary with respect to the metric, the (metric) field equations and stress-energy tensor must necessarily be symmetric (e.g. see here for an example of the metric variation). Varying w.r.t the metric does lead to $$ \sqrt{-g} \delta g^{ab} (G_{ab} - T_{ab}) =0 \\ \Rightarrow G_{(ab)} = T_{ab} $$ as you said. But remember that you must vary the action fully, i.e. with respect to the set that fully characterises that theory: for Einstein-Cartan that's either $\{g,\Gamma\}$ or $\{g,T\}$, where $\Gamma$ is the connection and $T$ is torsion. If working with the tetrad formalism the set would be $\{e,\omega\}$, the tetrad and Lorentz connection. You can find an example of the first-order formulation Here.
$^1$Note that on the Wiki they actually write that on the RHS of their field equations is the Belinfante–Rosenfeld stress–energy tensor, meaning the variation has been computed with respect to the Tetrad, not the metric as stated.