If $xi_mu$ is a Killing vector is it true that its contravariant representation $xi^mu$ also is a Killing vector?

Question

Question
I have worked on an exercise and proved that
$$
nabla_{mu} xi_{nu}+nabla_{nu} xi_{mu}=0
$$
i.e. that $xi_mu$ is a Killing vector. Now I wonder if it therefore follows that the contravariant repsentation $xi^mu$ also is a Killing vector?
My own attempt trying to prove that it is true
$$
begin{aligned}
nabla_{mu} xi_{nu}+nabla_{nu} xi_{mu}
&=
nabla_{mu}(g_{nu nu} xi^{nu}) +nabla_{nu}( g_{mu mu} xi^{mu})

&= g_{nu nu} nabla_{mu}( xi^{nu}) +g_{mu mu} nabla_{nu}(  xi^{mu}) 
end{aligned}
$$
but here I believe I've already severely abused the notation by having more than two indices of the same kind and furthermore do not know where to go.

Eletie · Accepted Answer

You can raise the indices as follows,
$$  g^{mu lambda} g^{nu rho} ( nabla_{mu} xi_{nu} + nabla_{nu} xi_{mu}) = 0 
nabla^{lambda}xi^{rho} + nabla^{rho} xi^{lambda} = 0   ,
$$
which is equivalent. This is just requiring the Lie derivative of the inverse metric to vanish, which is the correct way to think about the killing vector equation with components $xi^{mu}$.
This seems like what you're looking for?

AFG · Answer

You can't lower the index like that, use $xi_nu=g_{nualpha}xi^alpha$ instead, and you will see that $nabla_muxi^nu+nabla_nuxi^mu=0$ does not hold. That equation is also wrong by construction, as you can't have the same index up and down if it is not summed.

Answered by AFG on June 13, 2021

If $xi_mu$ is a Killing vector is it true that its contravariant representation $xi^mu$ also is a Killing vector?

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