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?


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.