Alternative definition of the Ricci Scalar

Question

I came across this definition of the Ricci Scalar on its Spanish Wikipedia page:

$$R=-g^{\mu\nu}\left(\Gamma_{\mu\nu}^{\lambda} \Gamma_{\lambda\sigma}^{\sigma} - \Gamma_{\mu\sigma}^{\lambda}\Gamma_{\nu\lambda}^{\sigma}\right) - \partial_{\nu}\left(g^{\mu\nu}\Gamma_{\mu \sigma}^{\sigma} - g^{\mu\sigma}\Gamma_{\mu \sigma}^{\nu}\right).$$

Does anyone know where it came from or how to demonstrate it?

I'll leave the link for the Wikipedia page here: https://es.wikipedia.org/wiki/Curvatura_escalar_de_Ricci

Answer 1

The indices in one of the terms in OP's expression might not be correct. The rest are. The following is a direct derivation.

The following two identities (which aren't immediately obvious) are needed: $$\partial_\mu g^{\sigma\lambda} = -g^{\sigma\rho}g^{\lambda\nu} \partial_\mu g_{\rho\nu} \\ g^{\mu\sigma}\Gamma^\nu_{\rho\sigma} - g^{\mu\sigma}g^{\nu\lambda}\partial_\rho g_{\sigma\lambda} = -g^{\nu\lambda}\Gamma^\mu_{\rho\lambda}.$$

The standard formula for the Ricci scalar obtained by contracting the Riemann tensor is $$R = g^{\mu\nu}\partial_\rho\Gamma^\rho_{\nu\mu} - g^{\mu\nu}\partial_\nu\Gamma^\rho_{\rho\mu} + g^{\mu\nu}\Gamma^\rho_{\rho\lambda}\Gamma^\lambda_{\mu\nu} - g^{\mu\nu}\Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\rho\nu}.$$ Integrating the first two terms by parts, $$R = \color{red}{\partial_\rho(g^{\mu\nu}\Gamma^\rho_{\nu\mu})} - \color{blue}{(\partial_\rho g^{\mu\nu})\Gamma^\rho_{\nu\mu}} - \color{red}{\partial_\nu(g^{\mu\nu}\Gamma^\rho_{\rho\mu})} + \color{green}{(\partial_\nu g^{\mu\nu})\Gamma^\rho_{\rho\mu}} + \color{green}{g^{\mu\nu}\Gamma^\rho_{\rho\lambda}\Gamma^\lambda_{\mu\nu}} - \color{blue}{g^{\mu\nu}\Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\rho\nu}} \\ = \color{red}{-\partial_\nu(g^{\nu\sigma}\Gamma^\rho_{\rho\sigma} - g^{\mu\sigma}\Gamma^\nu_{\mu\sigma})} + \color{green}{\Gamma^\rho_{\rho\mu}(\partial_\nu g^{\nu\mu}+g^{\lambda\nu}\Gamma^\mu_{\lambda\nu})} - \color{blue}{\Gamma^\rho_{\mu\nu}(\partial_\rho g^{\mu\nu}+g^{\mu\lambda}\Gamma^\nu_{\rho\lambda})}.$$ Applying the identities, $$\color{green}{\color{green}{\Gamma^\rho_{\rho\mu}(\partial_\nu g^{\nu\mu}+g^{\nu\lambda}\Gamma^\mu_{\nu\lambda})} = \Gamma^\rho_{\rho\mu}(-g^{\nu\sigma}g^{\mu\lambda}\partial_\nu g_{\sigma\lambda}+g^{\nu\sigma}\Gamma^\mu_{\nu\sigma}) = -\Gamma^\rho_{\rho\mu}g^{\mu\lambda}\Gamma^\nu_{\nu\lambda} = -g^{\mu\nu}\Gamma^\rho_{\rho\mu}\Gamma^\sigma_{\sigma\nu}} \\ \color{blue}{-\Gamma^\rho_{\mu\nu}(\partial_\rho g^{\mu\nu}+g^{\mu\lambda}\Gamma^\nu_{\rho\lambda}) = -\Gamma^\rho_{\mu\nu}(g^{\mu\lambda}\Gamma^\nu_{\rho\lambda} -g^{\mu\sigma}g^{\nu\lambda}\partial_\rho g_{\sigma\lambda}) = \Gamma^\rho_{\mu\nu}g^{\nu\lambda}\Gamma^\mu_{\rho\lambda} = g^{\mu\nu}\Gamma^\sigma_{\mu\rho}\Gamma^\rho_{\nu\sigma}}.$$ Putting it together, we finally get $$R = \color{red}{-\partial_\nu(g^{\nu\sigma}\Gamma^\rho_{\rho\sigma} - g^{\mu\sigma}\Gamma^\nu_{\mu\sigma})} - g^{\mu\nu}(\color{green}{\Gamma^\rho_{\rho\mu}\Gamma^\sigma_{\sigma\nu}}-\color{blue}{\Gamma^\sigma_{\mu\rho}\Gamma^\rho_{\nu\sigma}}).$$ Expanding this expression and applying the identities again, the standard formula is rederived.

The proofs of the identities are left as an exercise.

Answer 2

By performing a few integration by parts on the terms with derivatives of the connection, one can obtain this natural boundary term (total derivative). As a sketch, try using relations of the form $$ \partial ( g \Gamma) \propto \Gamma \partial g + g \partial \Gamma \ $$ with the original Ricci scalar definition, and you will obtain the definition above.

If you want more information, you can search for the 'Einstein' action or 'Gamma squared action', which refers to the quadratic connection term part. See also https://arxiv.org/abs/2103.15906, which uses this decomposition.