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From Lee's book, the differential Bianchi identity states that for the Riemann curvature tensor, $$R_{ijkl;m} + R_{ijlm;k} + R_{ijmk;l}=0$$ The the proof is, contract on $i, l$, and then on $j, k$ after raising an index on each pair. Then we obtain the "contracted Bianchi identity" $$2R_{ij;}^j = S_{;i}$$ where $R_{ij}$ is the Ricci tensor, $S$ is scalar curvature, and the indices after semicolon denoting the component of the covariant derivative with respect to that index.

First, how do we contract on $i, l$? Isn't it only correct to contract on the placement of the index (e.g. the first and the last index)?

I can use symmetry to make the second term $-R_{ijml;k}$, then contract on the first and fourth index. But then I don't know how to contract $i, l$ for the last term $R_{ijmk;l}$ since then I can't use symmetry of the curvature tensor.

user9812334
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1 Answers1

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Here's the idea: define a covariant $5$-tensor $T$ by $$T_{ijklm} = R_{ijkl;m} + R_{ijlm;k} + R_{ijmk;l}.$$ The differential Bianchi identity says that $T$ is the zero tensor. First contract $T$ on its first and fourth indices, yielding a $3$-tensor $U$ which is also zero: $$0 = U_{jkm}= T_{ijk}{}^i{}_m = R_{jk;m} - R_{jm;k} + R_{ijmk;}{}^{i}.$$ Then contract $U$ on its first two indices: $$0 = U_{j}{}^j{}_m = S_{;m} - R_{jm;}{}^{j} - R_{im;}{}^{i}.$$

Jack Lee
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