# Inferring that $\Gamma^{\mu}_{\alpha\beta}$ must be the Levi-Civita connection for $g_{\mu\nu}$ from the Einstein-Hilbert action

Consider a metric manifold $(M,g_{\mu\nu})$. Consider also a connection $\Gamma^{\mu}_{\alpha\beta}$ defined on it, in turn defining a covariant derivative $$\nabla_\mu e_{\nu} = \Gamma_{\mu\nu}^\rho e_\rho,$$ without making any assumption on the symmetry of its lower indices (i.e. without the hypothesis of vanishing torsion).

Can we infer from the variation of the Einstein-Hilbert action $$S_{EH}[g_{\mu\nu}, \Gamma^{\rho}_{\alpha\beta}] = \int d^D\!x \sqrt{-g} R_{\mu\nu} g^{\mu\nu},$$ with respect to $\Gamma^{\rho}_{\alpha\beta}$, that the connection should satisfy both $$\Gamma^{\rho}_{\mu\nu}=\Gamma^{\rho}_{\nu\mu}\qquad \text{and}\qquad \nabla_{\alpha}g_{\mu\nu}=0?$$ I tried using the Palatini formula $$\delta R_{\mu\nu} = \nabla_\nu \delta \Gamma^{\sigma}_{\sigma \mu}- \nabla_\sigma \delta\Gamma^{\sigma}_{\nu\mu} + 2 S^{\rho}_{\sigma\nu}\delta\Gamma^{\sigma}_{\rho \mu},$$ where $S^\alpha_{\mu\nu}=\Gamma^\alpha_{[\mu\nu]}$ is torsion, which gives $$\delta S_{EH} = \int d^D\!x \sqrt{-g} \left[ \nabla_\nu \delta \Gamma^{\sigma}_{\sigma \mu}- \nabla_\sigma \delta\Gamma^{\sigma}_{\nu\mu} + 2 S^{\rho}_{\sigma\nu}\delta\Gamma^{\sigma}_{\rho \mu} \right] g^{\mu\nu}$$ and integrating by parts $$\delta S_{EH} = \int d^D\!x \delta \Gamma^{\sigma}_{\nu\mu} \left[ \nabla_{\sigma} \left( \sqrt{-g} g^{\mu\nu} \right)- \delta^\nu_\sigma \nabla_\rho\left(\sqrt{-g}g^{\sigma\mu}\right) + 2S^{\nu}_{\alpha\sigma} g^{\alpha\mu} \right].$$ Now I can use $$\nabla\sqrt{-g} = -\frac{1}{2}\sqrt{-g} g_{\alpha\beta}\nabla_\sigma g^{\alpha\beta},$$ but I seem to get nowhere.

• – Qmechanic Nov 16 '16 at 16:47
• @Qmechanic My confusion arises on what to do when mixing the first order formulation with torsion. – Brightsun Nov 16 '16 at 16:52

In fact, Levi-Civita is not the only solution. If you assume that either the torsion or the non-metricity (covariant derivative of the metric) is zero, then Levi-Civita is your solution. But in the general case, you obtain the following connection (if the dimension is $D>2$):

$$\Gamma_{\mu \nu}{}^{\rho}= \Gamma^{LC}{}_{\mu \nu}{}^{\rho} + A_{\mu}\delta^{\rho}_{\nu}$$

that reflects the so-called "projective symmetry" of the Einstein-Hilbert-Palatini action.

You can calculate the torsion and the non-metricity in order to check what I said before:

$$T_{\mu \nu}{}^{\rho}= A_{\mu}\delta^{\rho}_{\nu} -A_{\nu}\delta^{\rho}_{\mu} \\ \nabla_{\mu} g_{\nu \rho}= -2A_{\mu}g_{\nu \rho}$$

If any of them vanishes then $A_\mu = 0$ and then the other one vanishes as well.

It is worth pointing out that, even for this general solution, the equation of the metric turns into the Einstein's one. And these and other details tell us that this $A_\mu$ field seems to be undetectable (there is no difference between the Einstein-Hilbert dynamics and the Einstein-Hilbert-Palatini dynamics).

Check the following reference:

A. N. Bernal et al. Physics Letters B 768 (2017), 280-287. https://arxiv.org/abs/1606.08756