A problem from Sean Carroll's about Killing vectors

[shichao116]

I'm now stuck in the second part of problem 12 in Chapter 3. The problem is " Show that any Killing vector K^\mu satisfies the following relations:
\nabla_\mu\nabla_\sigma K^\rho = R^\rho_{\sigma\mu\nu}K^\nu
K^\lambda\nabla_\lambda R = 0

Where R is Riemann tensor.

I can prove the first one by using the definition of Riemann tensor, i.e. the commutator of two covariant derivatives, Killing equations associated with Killing vector, and Bianchi identity.

But for the second one, in the book it is said that we can prove it by contracting the first equation, i.e.
\nabla_\mu\nabla_\sigma K^\mu = R_{\sigma\nu}K^\nu
and the contracted Bianchi identity
\nabla_\mu(R^{\mu\nu}-1/2g^{\mu\nu}R)=0

What I do is multiplying Killing vector to the contracted Bianchi identity and then I get to where I stuck:
1/2K^\mu\nabla_\mu R = K_\nu\nabla_\mu R^{\mu\nu}

obviously the left hand side is what we need to prove to be zero. But I failed to show the right hand side to be zero after tried many ways.

Can anyone give me some clue how to do that ?

Thanks a lot

 

[WannabeNewton]

Hi there mate! This is what I used to do the problem. First note that you can use the first and second equations under the second paragraph of your post (the Bianchi identity and the one directly above it) to very easily write \triangledown _{\nu }\triangledown _{\mu }\triangledown ^{\nu }\xi ^{\mu } = \frac{1}{2}\xi ^{\nu }\triangledown _{\nu }R (can you indeed show this? It is rather trivial). Now utilize the formula (\triangledown _{a}\triangledown _{b} - \triangledown _{b}\triangledown _{a})T^{c_1c_2} = -R_{abe}^{c_1}T^{ec_2} - R_{abe}^{c_2}T^{c_1e} (which you can derive yourself - it may be a teensy bit tedious but it is straightforward)

EDIT: IMO it is worth noting the equation for the commutator of the covariant derivative for rank 2 tensors because I have needed it quite a few times (another notable time I had to use it was to show that the inhomogeneous curved space - time Maxwell equations, the first one in my signature, implied \triangledown ^{a}j_{a} = 0)

 

[shichao116]

Hey bro, thanks a lot. That clear things up. I used to get where you showed in the first equation in your reply but did not proceed because I never tried to use a Riemann tensor on a tensor of rank 2 or higher. Now I get some new experience. :)

 

[WannabeNewton]

No problem! Post again if you get stuck or something.