# Killing Vector and Ricci curvature scalar

1. Nov 30, 2008

### Psi-String

1. The problem statement, all variables and given/known data

I'm currently self-studying Carroll's GR book and get stuck by proving
the following identity:

$$K^\lambda \nabla _\lambda R = 0$$

where K is Killing vector and R is the Ricci Scalar

2. Relevant equations

Mr.Carroll said that it is suffice to show this by knowing:

$$\nabla _\mu \nabla _\sigma K^\mu = R_{\sigma \nu}K^\nu$$

Bianchi identity $$\nabla ^ \mu R_{\rho \mu} = \frac{1}{2} \nabla _\rho R$$

and Killing equation $$\nabla _\mu K_\nu + \nabla _\nu K_\mu = 0$$

3. The attempt at a solution

The work I done so far :

$$K^\lambda \nabla _\lambda R = 2 K^\lambda \nabla ^\mu R_{\mu \lambda} = 2 \left( \nabla ^\mu R_{\mu \lambda} K^\lambda -R_{\mu \lambda} \nabla ^\mu K^\lambda \right) = 2 \nabla ^\mu \nabla _\sigma \nabla _\mu K^\sigma$$

Note that $$R_{\mu \lambda} \nabla ^\mu K^\lambda =0$$, since

$$R_{\mu \lambda} \nabla ^\mu K^\lambda = - R_{\mu \lambda} \nabla^\lambda K^\mu = -R_{\lambda\mu} \nabla^\lambda K^\mu = -R_{\mu \lambda} \nabla ^\mu K^\lambda$$

And I can't get any further

Could someone help?? Thanks in advace

2. Nov 30, 2008

### George Jones

Staff Emeritus
$\nabla ^\mu \nabla _\sigma \nabla _\mu K^\sigma = 0$ in the same way that $R_{\mu \lambda} \nabla ^\mu K^\lambda =0$.

3. Nov 30, 2008

### Psi-String

Hi~

$R_{\mu \lambda} \nabla ^\mu K^\lambda =0$
is due to the Killing Equation and the symmetry of Ricci tensor $$R_{\mu \lambda} = R_{\lambda \mu}$$

But it seems like
$$\nabla ^\mu \nabla _\sigma \nabla _\mu K^\sigma \neq \nabla _\sigma \nabla ^\mu \nabla _\mu K^\sigma$$ ??

Last edited: Nov 30, 2008
4. Nov 30, 2008

### Psi-String

I think I figure it out!

$$\nabla ^\mu \nabla _\sigma \nabla _\mu K^\sigma$$

$$= (g^{\mu \sigma} \nabla_\sigma) (g_{\sigma \mu}\nabla^\mu)(\nabla_\mu K^\sigma)$$

$$= g^{\mu \sigma}g_{\sigma \mu} (\nabla _\sigma \nabla^\mu \nabla_\mu K^\sigma) + g^{\mu \sigma}(\nabla^\mu \nabla_\mu K^\sigma)(\nabla_\sigma g_{\sigma \mu})$$

$$= \nabla_\sigma \nabla^\mu \nabla_\mu K^\sigma$$

the second term of the third line vanishes due to metric compatibility.

Is the above correct??:uhh:

Last edited: Nov 30, 2008
5. Nov 30, 2008

### George Jones

Staff Emeritus
Yes, metric compatibility is used, but be careful with the indices. You have used $\mu$ and $\sigma$ four times each in the product $(g^{\mu \sigma} \nabla_\sigma) (g_{\sigma \mu}\nabla^\mu)(\nabla_\mu K^\sigma)$.

6. Nov 30, 2008

### Psi-String

I noticed that. But I can't derive it out if I use different indices...

I don't know how to turn

$$g^{\mu \sigma}g_{\lambda \rho}(\nabla_\sigma \nabla^\rho \nabla _\mu K^\lambda)$$

into

$$\nabla _\lambda \nabla^\mu \nabla_\mu K^\lambda$$