Usage of the symmetry of the Riemann tensor

Question

On the way to derive the energy-momentum pseudotensor the expression $(154)$ must be derived from $(153)$. $U^{ijkm}$ has the same symmetry properties as the Riemann tensor $R^{ijkm}$. What symmetry property is used to obtain $(154)$?

Answer 1

Putting $p=j$,$$U^{ijkm}_{\qquad,j}=2\Gamma^a_{aj}U^{ijkm}-\Gamma^i_{aj}U^{ajkm}-\Gamma^j_{aj}U^{iakm}-\Gamma^k_{aj}U^{ijam}-\Gamma^{m}_{aj}U^{ijka}.$$We wish to simplify the right-hand side to $\Gamma^a_{aj}U^{ijkm}-\Gamma^k_{aj}U^{ijam}-\Gamma^m_{aj}U^{ijka}$, i.e. to claim$$\Gamma^a_{aj}U^{ijkm}=\Gamma^i_{aj}U^{ajkm}+\Gamma^j_{aj}U^{iakm}.$$Since $U$ is antisymmetric in its first two indices while $\Gamma$ is symmetric in its lower indcies, $\Gamma^i_{aj}U^{ajkm}=0$, so the conjecture simplifies to $\Gamma^a_{aj}U^{ijkm}=\Gamma^j_{aj}U^{iakm}$. But since $a,\,j$ are dummy variables that don't survive contraction in either of these expressions, we can exchange them to rewrite the left-hand side as $\Gamma^j_{ja}U^{iakm}$. Since $\Gamma^j_{ja}=\Gamma^j_{aj}$, we're done.