Killing Vector and Ricci curvature scalar

[Psi-String]

Homework Statement

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

Homework 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

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

 

[George Jones]

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

 

[Psi-String]

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

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 ??

Thanks for Reply!

 

[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 ??

Thanks for Reply!

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??

 

[George Jones]

Is the above correct??

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).

 

[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