Ricci tensor along a Killing vector

[La Guinee]

In Carrol's text, he shows that the covariant derivative of the Ricci scalar is zero along a Killing vector. He then goes on to say something about how this intuitively justifies our notion of geometry not changing along a Killing vector. This same informal reasoning would seem to imply that the Ricci tensor (and Riemann tensor for that matter) is covariantly constant along a Killing vector. However, Carroll has no discussion of this, nor can I find it in any other source (which leads me to think it's probably not true). My question is:
Is the covariant derivative of the Ricci tensor zero along a Killing vector? If so, how does one show this? If not, is there a conceptual way of understanding this and/or what is a counterexample? Thanks.

 

[schieghoven]

...the Ricci tensor (and Riemann tensor for that matter) is covariantly constant along a Killing vector.

I expect so, but I couldn't prove it myself. But, after all, the Killing vector generates an isometry of the manifold, and the Ricci tensor is defined entirely in terms of the metric. Good luck!

Dave

 

[Stingray]

You need to use Lie derivatives for it to work out. For example,

<br /> 0 = \mathcal{L}_\xi R_{ab} = \xi^c \nabla_c R_{ab} + 2 R_{c(a} \nabla_{b)} \xi^c<br />

The same thing works for the Riemann or Weyl tensors.

 

[student85]

You need to use Lie derivatives for it to work out. For example,

<br /> 0 = \mathcal{L}_\xi R_{ab} = \xi^c \nabla_c R_{ab} + 2 R_{c(a} \nabla_{b)} \xi^c<br />

The same thing works for the Riemann or Weyl tensors.

So it is zero then?

 

[Stingray]

Yes, Lie derivatives of curvature tensors with respect to Killing fields are always zero. The directional covariant derivative is not zero unless you're talking about a scalar quantity (like the Ricci scalar).