Hi guys! I need help on a problem from one of my GR texts. Suppose that ##\xi^a## is a killing vector field and consider its twist ##\omega_a = \epsilon_{abcd}\xi^b \nabla^c \xi^d##. I must show that ##\omega_a = \nabla_a \omega## for some scalar field ##\omega##, which is equivalent to showing ##(d\omega)_{ab} = \nabla_{[a}\omega_{b]} = 0##, if and only if ##\xi^a## is an eigenvector of the Ricci tensor i.e. ##R^{a}{}{}_{b}\xi^b = \lambda \xi^{a}## for some scalar field ##\lambda##.
First note that ##\nabla_{[a}\omega_{b]} = 0## if and only if ##\nabla_{a}\omega^{abc} = 0## where ##\omega^{abc} = \epsilon^{abcd}\omega_{d}## is the dual of the twist; inserting the expression for ##\omega_a## we find ##\omega^{abc} = -6\xi^{[a}\nabla^{b}\xi^{c]}## (see the formulas for ##\epsilon_{abcd}## in section B.2 of Wald, particularly page 433). This is easy to see as ##\nabla_{[a}\omega_{b]} = 0 \Rightarrow \epsilon^{efgh}\epsilon_{abgh}\nabla_{e}\omega_{f} = 0 \Rightarrow \epsilon_{abcd}\nabla_{e}(\xi^{[e}\nabla^{c}\xi^{d]}) = 0 \Rightarrow \nabla_{e}(\xi^{[e}\nabla^{a}\xi^{b]}) = \nabla_{e}\omega^{ebc} = 0##
because ##\nabla^{[a}\xi^{b]} = \nabla^a \xi^b## on account of ##\xi^a## being a killing vector field. For the converse, ##\epsilon^{abcd}\nabla_{c}\omega_{d} = \epsilon^{dcba}\epsilon_{defg}\nabla_{c}(\xi^{e}\nabla^{f}\xi^{g}) = -6\nabla_{c}(\xi^{[c}\nabla^{b}\xi^{a]}) = \nabla_{c}\omega^{cba} = 0## thus ##\nabla_{[a}\omega_{b]} = 0##.
Now on to the problem itself, if ##R^{a}{}{}_{b}\xi^b = \lambda \xi^{a}## then
##\nabla_a \omega^{abc} = -6\nabla_{a}(\xi^{[a}\nabla^{b}\xi^{c]} ) = -2(\xi^b \nabla_a \nabla^c \xi^a -\xi^c \nabla_a \nabla^b \xi^a + R^{cb}{}{}_{ad}\xi^{a}\xi^{d})\\ = -2(R^{c}{}{}_{d}\xi^{d}\xi^{b} - R^{b}{}{}_{d}\xi^{d}\xi^{c}) = -2(\lambda\xi^{c}\xi^{b} - \lambda \xi^{b}\xi^{c}) = 0.##
It's the converse I'm stuck on mainly. If ##\nabla_{[a}\omega_{b]} = 0## then, using the above, ##R^{c}{}{}_{d}\xi^{d}\xi^{b} = R^{b}{}{}_{d}\xi^{d}\xi^{c}##. If ##\xi^a## is non-null (##\xi^a \xi_a \neq 0##), then ##R^{c}{}{}_{d}\xi^{d} = \frac{R_{bd}\xi^b\xi^{d}}{\xi^b \xi_b}\xi^{c} = \lambda \xi^c ## as desired. However I don't get what to do when ##\xi^a## is null; I don't see how to show the desired result.
First note that ##\nabla_{[a}\omega_{b]} = 0## if and only if ##\nabla_{a}\omega^{abc} = 0## where ##\omega^{abc} = \epsilon^{abcd}\omega_{d}## is the dual of the twist; inserting the expression for ##\omega_a## we find ##\omega^{abc} = -6\xi^{[a}\nabla^{b}\xi^{c]}## (see the formulas for ##\epsilon_{abcd}## in section B.2 of Wald, particularly page 433). This is easy to see as ##\nabla_{[a}\omega_{b]} = 0 \Rightarrow \epsilon^{efgh}\epsilon_{abgh}\nabla_{e}\omega_{f} = 0 \Rightarrow \epsilon_{abcd}\nabla_{e}(\xi^{[e}\nabla^{c}\xi^{d]}) = 0 \Rightarrow \nabla_{e}(\xi^{[e}\nabla^{a}\xi^{b]}) = \nabla_{e}\omega^{ebc} = 0##
because ##\nabla^{[a}\xi^{b]} = \nabla^a \xi^b## on account of ##\xi^a## being a killing vector field. For the converse, ##\epsilon^{abcd}\nabla_{c}\omega_{d} = \epsilon^{dcba}\epsilon_{defg}\nabla_{c}(\xi^{e}\nabla^{f}\xi^{g}) = -6\nabla_{c}(\xi^{[c}\nabla^{b}\xi^{a]}) = \nabla_{c}\omega^{cba} = 0## thus ##\nabla_{[a}\omega_{b]} = 0##.
Now on to the problem itself, if ##R^{a}{}{}_{b}\xi^b = \lambda \xi^{a}## then
##\nabla_a \omega^{abc} = -6\nabla_{a}(\xi^{[a}\nabla^{b}\xi^{c]} ) = -2(\xi^b \nabla_a \nabla^c \xi^a -\xi^c \nabla_a \nabla^b \xi^a + R^{cb}{}{}_{ad}\xi^{a}\xi^{d})\\ = -2(R^{c}{}{}_{d}\xi^{d}\xi^{b} - R^{b}{}{}_{d}\xi^{d}\xi^{c}) = -2(\lambda\xi^{c}\xi^{b} - \lambda \xi^{b}\xi^{c}) = 0.##
It's the converse I'm stuck on mainly. If ##\nabla_{[a}\omega_{b]} = 0## then, using the above, ##R^{c}{}{}_{d}\xi^{d}\xi^{b} = R^{b}{}{}_{d}\xi^{d}\xi^{c}##. If ##\xi^a## is non-null (##\xi^a \xi_a \neq 0##), then ##R^{c}{}{}_{d}\xi^{d} = \frac{R_{bd}\xi^b\xi^{d}}{\xi^b \xi_b}\xi^{c} = \lambda \xi^c ## as desired. However I don't get what to do when ##\xi^a## is null; I don't see how to show the desired result.
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