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Index gymnastics-Maxwell's equations

WannabeNewton

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1. Homework Statement
Let ##X^{ab}## be an antisymmetric tensor such that ##\nabla^{(a}X^{b)c} = 0##. Show that if ##R_{ab} = 0##, then ##F_{ab} = R_{abcd}X^{cd}## satisfies Maxwell's equations ##\nabla^a F_{ab} = 0, \nabla_{[a}F_{bc]} = 0## where ##R_{abcd}## is the Riemann curvature tensor and ##R_{ab}## is the Ricci tensor.

3. The Attempt at a Solution
Well to start with, ##\nabla^a X^{bc} = -\nabla^b X^{ac} = \nabla^b X^{ca} = -\nabla^c X^{ba}## so ##\nabla^a X^{bc}## is antisymmetric in all its indices. Thus, ##\nabla^a F_{ab} = R_{abcd}\nabla^{a}X^{cd} + X^{cd}\nabla^a R_{abcd} = -R_{b[acd]}\nabla^{a}X^{cd} -2X^{cd}\nabla_{[b}R_{c]d} = 0## since ##R_{b[acd]} = 0## by the first Bianchi identity and ##R_{ab} = 0## by hypothesis.

Showing that ##\nabla_{[a}F_{bc]} = 0## should be just as easy but for some reason I can't seem to find a way to show it. We have ##\nabla_{a}F_{bc} =\nabla_{a}X^{de} R_{bcde} + X^{de}\nabla_{a}R_{bcde}## so ##\nabla_{[a}F_{bc]} = \nabla_{[a}X^{de}R_{bc]de}## since ##\nabla_{[a}R_{bc]de} = 0## by the second Bianchi identity. One can either try to show that ##\nabla_{[a}X^{de}R_{bc]de} = 0## directly or (equivalently) show that ##\epsilon^{abcf}\nabla_{a}X^{de}R_{bcde} = 0##. I know I'm missing something very trivial but I can't seem to proceed from here either way. Thanks in advance for any help!
 
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