- #1
etotheipi
- Homework Statement
- To prove ##{\nabla_a R_{bcd}}^a + \nabla_b R_{cd} - \nabla_c R_{bd}##
- Relevant Equations
- N/A
Start with the Bianchi identity,$${\nabla_{[a} R_{bc]d}}^e = 0$$$${\nabla_{a} R_{bcd}}^e - {\nabla_{a} R_{cbd}}^e + {\nabla_{b} R_{cad}}^e - {\nabla_{b} R_{acd}}^e + {\nabla_{c} R_{abd}}^e - {\nabla_{c} R_{bad}}^e = 0$$Use definition of Ricci tensor$$
\left[ {\nabla_{a} R_{bcd}}^e + \nabla_b R_{cd} - \nabla_c R_{bd} \right] + \left[ {\nabla_c R_{abd}}^e - {\nabla_a R_{cbd}}^e - {\nabla_b R_{acd}}^e \right] = 0
$$How to proceed from here? Thanks
\left[ {\nabla_{a} R_{bcd}}^e + \nabla_b R_{cd} - \nabla_c R_{bd} \right] + \left[ {\nabla_c R_{abd}}^e - {\nabla_a R_{cbd}}^e - {\nabla_b R_{acd}}^e \right] = 0
$$How to proceed from here? Thanks