Does the Riemman tensor and the covariant derivate commute?

[vertices]

A random question - Does the Riemman tensor and the covariant derivate commute?

a yes/no answer would suffice, but any explanation would be welcome:)

From the equations, it looks as though they do for flat metrics - but if we have other manifolds, it seems to me that the Christoffel symbols may be functions of different coordinates, and as such, might be affected by the covariant derivative.

 

[Stingray]

The Riemann tensor of a flat metric is always zero (by definition). The covariant derivative of any zero tensor is (unsurprisingly) zero.

But more generally, what you're asking is only true if the covariant derivative of the Riemann tensor vanishes. That is a very special case.

 

[vertices]

The covariant derivative of any zero tensor is (unsurprisingly) zero.

Many thanks for the reply Stingray.

I'm really going to show my ignorance here, but a zero tensor is a scalar no? From what I understand, the covariant derivative of scalar is just the partial derivative - why would this neccessarily be zero?

 

[Stingray]

You can have zero tensors of any rank. In this case, it is not a scalar.

Still, if all components of a tensor vanish in one coordinate system, they vanish in all others as well. You can easily check this using the standard transformation rules.

Regardless, the covariant derivative of something that is everywhere zero is zero. For example, you probably know that
$$\nabla_\mu A_\nu = \partial_\mu A_\nu - \Gamma_{\mu\nu}^\gamma A_\gamma .$$
If $A_\mu = 0$ over some finite region, it immediately follows that $\nabla_\mu A_\nu = \partial_\mu A_\nu = 0$ in that same region. This result is coordinate-independent.

 

[vertices]

You can have zero tensors of any rank. In this case, it is not a scalar.

Still, if all components of a tensor vanish in one coordinate system, they vanish in all others as well. You can easily check this using the standard transformation rules.

Regardless, the covariant derivative of something that is everywhere zero is zero. For example, you probably know that
$$\nabla_\mu A_\nu = \partial_\mu A_\nu - \Gamma_{\mu\nu}^\gamma A_\gamma .$$
If $A_\mu = 0$ over some finite region, it immediately follows that $\nabla_\mu A_\nu = \partial_\mu A_\nu = 0$ in that same region. This result is coordinate-independent.

I see.

Thanks again, Stingray:)

 

[zerotensor]

more generally, what you're asking is only true if the covariant derivative of the Riemann tensor vanishes. That is a very special case.

If the metric is not flat, then the covariant derivative does not commute. Generally speaking, in curved spaces the order in which the covariant derivatives are taken *does* matter (they do not commute). Perhaps I am mistaken, but my understanding is that the Riemann tensor actually gives us the "commutator" of covariant derivatives.

 

[Stingray]

If the metric is not flat, then the covariant derivative does not commute. Generally speaking, in curved spaces the order in which the covariant derivatives are taken *does* matter (they do not commute). Perhaps I am mistaken, but my understanding is that the Riemann tensor actually gives us the "commutator" of covariant derivatives.

My impression was that the OP was asking whether $\nabla_a R_{bcd}{}^{f} = R_{bcd}{}^{f} \nabla_a$. You're right that the Riemann tensor does give the commutator of two covariant derivatives.