# Twist of null Killing fields

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?

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