# Killing vectors in an static space-time

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.

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