Palatini identity

In general relativity and tensor calculus, the Palatini identity is

\delta R_{\sigma \nu }=\nabla _{\rho }\delta \Gamma _{\nu \sigma }^{\rho }-\nabla _{\nu }\delta \Gamma _{\rho \sigma }^{\rho },

where $\delta \Gamma _{\nu \sigma }^{\rho }$ denotes the variation of Christoffel symbols and $\nabla _{\rho }$ indicates covariant differentiation.^[1]

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 _{\nu \sigma }^{\rho })-\nabla _{\nu }({\mathcal {L}}_{\xi }\Gamma _{\rho \sigma }^{\rho }),

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 $\Gamma _{\mu \nu }^{\lambda }$ as

{R^{\rho }}_{\sigma \mu \nu }=\partial _{\mu }\Gamma _{\nu \sigma }^{\rho }-\partial _{\nu }\Gamma _{\mu \sigma }^{\rho }+\Gamma _{\mu \lambda }^{\rho }\Gamma _{\nu \sigma }^{\lambda }-\Gamma _{\nu \lambda }^{\rho }\Gamma _{\mu \sigma }^{\lambda }

.

Its variation is

\delta {R^{\rho }}_{\sigma \mu \nu }=\partial _{\mu }\delta \Gamma _{\nu \sigma }^{\rho }-\partial _{\nu }\delta \Gamma _{\mu \sigma }^{\rho }+\delta \Gamma _{\mu \lambda }^{\rho }\Gamma _{\nu \sigma }^{\lambda }+\Gamma _{\mu \lambda }^{\rho }\delta \Gamma _{\nu \sigma }^{\lambda }-\delta \Gamma _{\nu \lambda }^{\rho }\Gamma _{\mu \sigma }^{\lambda }-\Gamma _{\nu \lambda }^{\rho }\delta \Gamma _{\mu \sigma }^{\lambda }

.

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

\nabla _{\mu }\delta \Gamma _{\nu \sigma }^{\rho }=\partial _{\mu }\delta \Gamma _{\nu \sigma }^{\rho }+\Gamma _{\mu \lambda }^{\rho }\delta \Gamma _{\nu \sigma }^{\lambda }-\Gamma _{\mu \nu }^{\lambda }\delta \Gamma _{\lambda \sigma }^{\rho }-\Gamma _{\mu \sigma }^{\lambda }\delta \Gamma _{\nu \lambda }^{\rho }

.

Solving this equation for $\partial _{\mu }\delta \Gamma _{\nu \sigma }^{\rho }$ 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 _{\nu \sigma }^{\rho }-\nabla _{\nu }\delta \Gamma _{\mu \sigma }^{\rho }

.

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 _{\nu \sigma }^{\rho }-\nabla _{\nu }\delta \Gamma _{\rho \sigma }^{\rho }

.

Palatini identity

Proof

See also

Notes

References

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