In it's form as a covariant 4 tensor field the symmetries of the curvature tensor are given by

$$R_{ijkl}=-R_{jikl}$$ $$R_{ijkl}=-R_{ijlk}$$ $$R_{ijkl}=R_{klij}$$ $$R_{ijkl}+R_{jkil}+R_{kijl}=0$$

Can we recast these in terms of the definition of the curvature as a (3,1) tensor? It would seem to me that the first symmetry implies


Is this correct? What about recasting the remaining 3? Is it possible and if so how?


Note that $R_{ijk}^{\;\;\; l} = R_{ijkm}g^{ml}$. So we have

$$R_{ijk}^{\;\;\; l} = R_{ijkm}g^{ml} = -R_{jikm}g^{ml} = -R_{jik}^{\;\;\;l}.$$

For the second identity, we have

$$R_{ijk}^{\;\;\; l} = R_{ijkm}g^{ml} = - R_{ijmk}g^{ml} = -R_{ij\; k}^{\;\; l}.$$

For the third, we have

$$R_{ijk}^{\;\;\; l} = R_{ijkm}g^{ml} = R_{kmij}g^{ml} = R_{k\; ij}^{\; l}.$$

Finally, we have

\begin{align*} R_{ijk}^{\;\;\; l} + R_{jki}^{\;\;\; l} + R_{kij}^{\;\;\; l} &= R_{ijkm}g^{ml} + R_{jkim}g^{ml} + R_{kijm}g^{ml}\\ &= (R_{ijkm} + R_{jkim} + R_{kijm})g^{ml}\\ &= 0g^{ml}\\ &= 0. \end{align*}

The take away is that if the identity involves permuting the indicies $i$, $j$, and $k$ only, then it holds for the $(3, 1)$-tensor too. If the identity involves permuting $l$ as well, then the position of the raised index changes.


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