Understanding why $\nabla_i\nabla^j\nabla_jA^i = \nabla^j\nabla_i\nabla_jA^i - R_i^s\nabla_sA^i +R_s^j\nabla_jA^s $

Question

I know that generaly we can define the Riemann curvature tensor as:

$$\nabla_a\nabla_dV^b -\nabla_d\nabla_aV^b = - {R_{adi}}^bV^i$$

Consider the equation $$\nabla^j\nabla_jA^i = \frac{4\pi}{c}J^i \space \space \space\space\space\space (1)$$

If we then apply $\nabla_i$ on (1) and apply the continuity equation $\nabla_iJ^i = 0$ as well as the lorenz gauge $\nabla_iA^i = 0$ we can deduce the identity

$$\nabla_i\nabla^j\nabla_jA^i = \nabla^j\nabla_i\nabla_jA^i - R_i^s\nabla_sA^i +R_s^j\nabla_jA^s $$

I am unsure how to deduce this expression though as when I try to apply the equations and identities above I can not get it. Any help would be appreciated.

Answer 1

Though defined using vector fields, the Riemann tensor also describes the antisymmetrized second derivatives of tensors of higher order: $$ (\nabla_i\nabla_j-\nabla_j\nabla_i)T^{kl}=R^k{}_{mij}T^{ml}+R^l{}_{mij}T^{km} $$ This can be shown a number of ways; one convenient path is to consder tensors of the form $T=U\otimes V$ for vector fields $U,V$ and extend to the rest by $C^\infty M$-linearity. (Similar formulae exist for tensors of higher order; they can be derived in much the same way.)

Once you have this identity, you can apply it to $(\nabla_i\nabla_j-\nabla_j\nabla_i)\nabla^jA^i$.