# Ricci tensor along a Killing vector

1. Jan 5, 2009

### 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.

2. Jan 6, 2009

### schieghoven

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

3. Jan 7, 2009

### Stingray

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

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

The same thing works for the Riemann or Weyl tensors.

4. Jan 9, 2009

### student85

So it is zero then?

5. Jan 9, 2009

### 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).

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