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Contracted Bianchi identities

In general relativity and tensor calculus, the contracted Bianchi identities are:

Proof
Start with the Bianchi identity : R_{abmn;\ell} + R_{ab\ell m;n} + R_{abn\ell;m} = 0. Contract both sides of the above equation with a pair of metric tensors: : g^{bn} g^{am} (R_{abmn;\ell} + R_{ab\ell m;n} + R_{abn\ell;m}) = 0, : g^{bn} (R^m {}_{bmn;\ell} - R^m {}_{bm\ell;n} + R^m {}_{bn\ell;m}) = 0, : g^{bn} (R_{bn;\ell} - R_{b\ell;n} - R_b {}^m {}_{n\ell;m}) = 0, : R^n {}_{n;\ell} - R^n {}_{\ell;n} - R^{nm} {}_{n\ell;m} = 0. The first term on the left contracts to yield a Ricci scalar, while the third term contracts to yield a mixed Ricci tensor, : R_{;\ell} - R^n {}_{\ell;n} - R^m {}_{\ell;m} = 0. The last two terms are the same (changing dummy index n to m) and can be combined into a single term which shall be moved to the right, : R_{;\ell} = 2 R^m {}_{\ell;m}, which is the same as : \nabla_m R^m {}_\ell = {1 \over 2} \nabla_\ell R. Swapping the index labels l and m on the left side yields : \nabla_\ell R^\ell {}_m = {1 \over 2} \nabla_m R. ==See also==
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