Reversing Indices in Contractions: Can it be Done?

[unscientific]

Suppose I have something like
\left( \nabla_\mu \nabla_\beta - \nabla_\beta \nabla_\mu \right) V^\mu = R_{\nu \beta} V^\nu
Can since all the terms involving $\mu$ on the left and $\nu$ on the right are contractions, can I simply do:
\left( \nabla^\mu \nabla_\beta - \nabla_\beta \nabla^\mu \right) V_{\mu} = R^\nu_{\beta} V_{\nu}?

 

[Orodruin]

As long as the connection is metric compatible.

 

[unscientific]

As long as the connection is metric compatible.

But my concern is that in the first term on the LHS, the gradient operator is acting on the gradient operator via $\nabla_\mu (\nabla_\beta) V^\mu$, and not on the vector $V^\mu$. I thought for contractions, these have to be acting on one another?

 

[Orodruin]

But my concern is that in the first term on the LHS, the gradient operator is acting on the gradient operator via $\nabla_\mu (\nabla_\beta) V^\mu$, and not on the vector $V^\mu$. I thought for contractions, these have to be acting on one another?

No, this is not how to interpret the LHS. The operators are each acting on $V^\mu$, one after the other.

 

[unscientific]

No, this is not how to interpret the LHS. The operators are each acting on $V^\mu$, one after the other.

Ok, now I'm confused. I thought the chain rule applies:
\nabla^\mu \left( \nabla_\mu A_\nu \right) = \nabla^\mu \left( \nabla_\mu\right) A_\nu + \nabla_\mu \nabla^\mu A_\nu

 

[unscientific]

No, this is not how to interpret the LHS. The operators are each acting on $V^\mu$, one after the other.

If I have something like $F_{\mu \nu} = \nabla_\mu A_\nu - \nabla_\nu A_\mu$, what will $\nabla^\mu F_{\mu \nu}$ look like?

 

[Orodruin]

What do you mean by $\nabla_\mu(\nabla^\mu)$? This is an operator applied to an operator. The only way to make sense of this is to apply the right operator to something first and then apply the left to the result.

 

[haushofer]

If I have something like $F_{\mu \nu} = \nabla_\mu A_\nu - \nabla_\nu A_\mu$, what will $\nabla^\mu F_{\mu \nu}$ look like?

You should write out this covariant derivative first in terms of F, and then in that expression write out F in terms of A. If your connection is symmetric, F can be expressed in terms of partial derivatives only.