Bianchi and Ricci Identities: Understanding and Applying in Tensor Calculus

[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

 

[TSny]

$${\nabla_{a} R_{bcd}}^e - {\nabla_{a} R_{cbd}}^e + {\nabla_{b} R_{cad}}^e - {\nabla_{b} R_{acd}}^e + {\nabla_{c} R_{abd}}^e - {\nabla_{c} R_{bad}}^e = 0$$

You should be able to reduce the left side to just 3 terms by using the "skew symmetry" of the curvature tensor.

$$\left[ {\nabla_{a} R_{bcd}}^e + \nabla_b R_{cd} - \nabla_c R_{bd} \right] + \left[ {\nabla_c R_{abd}}^e - {\nabla_a R_{cbd}}^e - {\nabla_b R_{acd}}^e \right] = 0$$

You can see something's wrong here. Some of the terms have an index $e$ while other terms don't.

 

[etotheipi]

Thanks! What about,$${\nabla_{a} R_{bcd}}^e - {\nabla_{a} R_{cbd}}^e + {\nabla_{b} R_{cad}}^e - {\nabla_{b} R_{acd}}^e + {\nabla_{c} R_{abd}}^e - {\nabla_{c} R_{bad}}^e = 0$$ $${\nabla_{a} R_{bcd}}^e + {\nabla_{b} R_{cad}}^e - {\nabla_{c} R_{bad}}^e = 0
$$Since $a$ and $e$ are free indices, we can choose to let $a = e$, and sum over all $a$ [I'll write in the sum explicitly, and not use the convention here]$$\sum_a \left[ {\nabla_{a} R_{bcd}}^a + {\nabla_{b} R_{cad}}^a - {\nabla_{c} R_{bad}}^a \right] = 0$$Those final two terms we can contract over $a$,$$\sum_a \left[ {\nabla_{a} R_{bcd}}^a \right] + \nabla_b R_{cd} - \nabla_c R_{bd} = 0$$

 

[wrobel]

for such problems it turns to be convenient to take into account that in a neighborhood of any point $x'$ there are local coordinates such that $\Gamma_{ij}^k(x')=0$. This holds provided $\Gamma_{ij}^k=\Gamma_{ji}^k$

 

[TSny]

Thanks! What about,$${\nabla_{a} R_{bcd}}^e - {\nabla_{a} R_{cbd}}^e + {\nabla_{b} R_{cad}}^e - {\nabla_{b} R_{acd}}^e + {\nabla_{c} R_{abd}}^e - {\nabla_{c} R_{bad}}^e = 0$$ $${\nabla_{a} R_{bcd}}^e + {\nabla_{b} R_{cad}}^e - {\nabla_{c} R_{bad}}^e = 0
$$Since $a$ and $e$ are free indices, we can choose to let $a = e$, and sum over all $a$ [I'll write in the sum explicitly, and not use the convention here]$$\sum_a \left[ {\nabla_{a} R_{bcd}}^a + {\nabla_{b} R_{cad}}^a - {\nabla_{c} R_{bad}}^a \right] = 0$$Those final two terms we can contract over $a$,$$\sum_a \left[ {\nabla_{a} R_{bcd}}^a \right] + \nabla_b R_{cd} - \nabla_c R_{bd} = 0$$

Looks good to me