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How can I show that a given space-time is static, i.e. exists a time-like Killing vector $\xi = \partial_0$ that $\partial_0 g_{\mu \nu} = 0$ (Killing eq.) and $g_{0i}=0$, if and only if the relation $$\xi_{[\alpha}\nabla_{\mu}\xi_{\nu]} = 0$$ holds?

I have already proved that if $\partial_0 g_{\mu \nu}=0$ and $g_{i0}=0$, then $\xi_{[\alpha}\nabla_{\mu}\xi_{\nu]} = 0$. I have no idea how to prove the inverse.

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  • $\begingroup$ Nitpick: Killing's equation is usually written as $\nabla_{(\mu} \xi_{\nu)} = 0$, without any reference to a particular coordinate $x^0$. $\endgroup$ – Michael Seifert Jul 2 '19 at 6:32
  • $\begingroup$ Yes, you are right. But for this special case the equation $\nabla _{(\mu}\xi_{\nu)}=0$ takes the form $\partial_0 g_{\mu \nu} = 0$. $\endgroup$ – Daemonium Jul 2 '19 at 6:35
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    $\begingroup$ You need to use Frobenius' theorem. The condition on the Killing field implies that the integral line of the field are hypersurface orthogonal. $\endgroup$ – MBN Jul 2 '19 at 9:45
  • $\begingroup$ If this is homework, please add the homework-and-exercises tag. $\endgroup$ – Ben Crowell Jul 2 '19 at 12:01

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