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I have a (hopefully) quick question: is it possible to have a null Killing field $\xi ^ \mu$ such that the twist 1-form $\omega_{\mu} = \epsilon_{\mu\nu\alpha\beta}\xi^\nu \nabla^\alpha \xi^\beta \neq 0$ but the exterior derivative $(d \omega)_{\mu\nu} = 2\nabla_{[\mu}\omega_{\nu]} = 0$? Or does $(d \omega)_{\mu\nu} = 0$ always imply $\omega_{\mu} = 0$ for a null Killing field?

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Following this reference formulae (8),(9),(10), (document)page 295, your twist 1-from is zero.

This can be done by looking at the square of $\omega_\mu V^\mu$ for an arbitrary vector $V$.

This brings different contractions for the Levi-Civita tensors, and they are all zero, due to the properties of the null Killing field.

Reference : Null-Killing vector dimensional reduction and Galilean geometrodynamics B.Julia H.Nicolai NUCLEAR PHYSICS B

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  • $\begingroup$ Minor comment to the answer (v1): Please consider providing reference info about the link in the answer, so that the link can be reconstruct in case of future link rot. $\endgroup$ – Qmechanic Sep 2 '13 at 8:32

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