Palatini identity

In general relativity and tensor calculus, the Palatini identity is


 * $$ \delta R_{\sigma\nu} = \nabla_\rho \delta \Gamma^\rho_{\nu\sigma} - \nabla_\nu \delta \Gamma^\rho_{\rho\sigma},$$

where $$\delta \Gamma^\rho_{\nu\sigma}$$ denotes the variation of Christoffel symbols and $$\nabla_\rho$$ indicates covariant differentiation.

The "same" identity holds for the Lie derivative $$\mathcal{L}_{\xi} R_{\sigma\nu}$$. In fact, one has


 * $$ \mathcal{L}_{\xi} R_{\sigma\nu} = \nabla_\rho (\mathcal{L}_{\xi} \Gamma^\rho_{\nu\sigma}) - \nabla_\nu (\mathcal{L}_{\xi} \Gamma^\rho_{\rho\sigma}),$$

where $$\xi = \xi^{\rho}\partial_{\rho}$$ denotes any vector field on the spacetime manifold $$M$$.

Proof
The Riemann curvature tensor is defined in terms of the Levi-Civita connection $$\Gamma^\lambda_{\mu\nu}$$ as


 * $${R^\rho}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma} + \Gamma^\rho_{\mu\lambda} \Gamma^\lambda_{\nu\sigma} - \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}$$.

Its variation is



\delta{R^\rho}_{\sigma\mu\nu} = \partial_\mu \delta\Gamma^\rho_{\nu\sigma} - \partial_\nu \delta\Gamma^\rho_{\mu\sigma} + \delta\Gamma^\rho_{\mu\lambda} \Gamma^\lambda_{\nu\sigma} + \Gamma^\rho_{\mu\lambda} \delta\Gamma^\lambda_{\nu\sigma} - \delta\Gamma^\rho_{\nu\lambda} \Gamma^\lambda_{\mu\sigma} - \Gamma^\rho_{\nu\lambda} \delta\Gamma^\lambda_{\mu\sigma}$$.

While the connection $$\Gamma^\rho_{\nu\sigma}$$ is not a tensor, the difference $$\delta\Gamma^\rho_{\nu\sigma}$$ between two connections is, so we can take its covariant derivative



\nabla_\mu \delta \Gamma^\rho_{\nu\sigma} = \partial_\mu \delta \Gamma^\rho_{\nu\sigma} + \Gamma^\rho_{\mu\lambda} \delta \Gamma^\lambda_{\nu\sigma} - \Gamma^\lambda_{\mu\nu} \delta \Gamma^\rho_{\lambda\sigma} - \Gamma^\lambda_{\mu\sigma} \delta \Gamma^\rho_{\nu\lambda}$$.

Solving this equation for $$\partial_\mu \delta \Gamma^\rho_{\nu\sigma}$$ and substituting the result in $$\delta{R^\rho}_{\sigma\mu\nu}$$, all the $$\Gamma \delta \Gamma$$-like terms cancel, leaving only



\delta{R^\rho}_{\sigma\mu\nu} = \nabla_\mu \delta\Gamma^\rho_{\nu\sigma} - \nabla_\nu \delta\Gamma^\rho_{\mu\sigma}$$.

Finally, the variation of the Ricci curvature tensor follows by contracting two indices, proving the identity



\delta R_{\sigma\nu} = \delta {R^\rho}_{\sigma\rho\nu} = \nabla_\rho \delta \Gamma^\rho_{\nu\sigma} - \nabla_\nu \delta \Gamma^\rho_{\rho\sigma}$$.