Killing fields as eigenvectors of Ricci tensor

[WannabeNewton]

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.

 

[WannabeNewton]

Ok if $\xi^a$ is a null killing vector field, then $R_{ab}\xi^a \xi^b = 2\omega^2$ where $\omega^2$ is the norm of the twist $\omega_a$. If $\xi^a$ is an eigenvector of $R_{ab}$ then $\omega^2 = 0$ and since this is the twist, this implies $\omega_a = 0$.

So for a null killing vector field $\xi^a$, either (1) $\nabla_{[a}\omega_{b]} = 0$ implies $\omega_a = 0$, in which case $\xi^a = \alpha\nabla^a \beta$ hence $\nabla^{a}\xi^{b} = \nabla^{[a}\xi^{b]} = \nabla^{[a}\alpha \nabla^{b]}\beta$ thus $R^{a}{}{}_b \xi^b = \nabla_b \nabla^a \xi^b = \nabla_b (\nabla^{[a}\alpha \nabla^{b]}\beta) = 0$,
or (2) there exists a space-time with some null killing field $\xi^a$ such that $\omega_a \neq 0$ but $\nabla_{[a}\omega_{b]} = 0$ for $\xi^a$, which would mean that the problem statement is incorrect as given and should specify that the killing field is non-null. Does anyone know if (1) is true or have an example of (2)?

 

[WannabeNewton]

If anyone is interested, it just so happens that for any null killing field $\xi^a$, $\nabla_{[a}\omega_{b]} = 0$ implies $\omega_a = 0$. As noted in post #2, this then implies that $\xi^a$ is an eigenvector of $R_{ab}$.

To see this, first note that since $\xi^a \xi_a = 0$ we have $\xi^a \nabla_b \xi_a = 0 = -\xi^a \nabla_a \xi_b$. Also, from the calculations in post #1 we have that $\nabla_{[a}\omega_{b]} = 0\Rightarrow \xi^b \nabla_a \nabla^c \xi^a = \xi^c \nabla_a \nabla^b \xi^a$ hence $\xi^b (\xi_c \nabla_a \nabla^c \xi^a) = 0$ but $\xi^a$ is an arbitrary null killing field so it must be that $\xi_c \nabla_a \nabla^c \xi^a = 0$ thus $\nabla_a \xi_b \nabla^a \xi^b = 0$.

Now let $\nu ^a$ be an arbitrary vector field and consider $(\omega _a \nu ^a)^2\\ = (\epsilon_{abcd}\nu^a \xi^b \nabla^c \xi^d)(\epsilon^{efgh}\nu_e \xi_f \nabla_g \xi_h)\\ = -4!(\nu^a \xi^b \nabla^c \xi^d )(\nu_{[a}\xi_{b}\nabla_{c}\xi_{d]})$
Using $\xi^a \xi_a = \xi^a \nabla_b \xi_a = \xi^a \nabla_a \xi_b = \nabla_a \xi_b \nabla^a \xi^b = 0$, it is easy to see that $(\omega _a \nu ^a)^2 = 0$ hence $\omega_a = 0$.

 

[R136a1]

Is there a way to do it more directly, without having to introduce an additional arbitrary vector field?

 

[WannabeNewton]

Uh well the argument would be extremely similar. First note that $\omega_a$ is null, $\omega^a \omega_a = \epsilon^{abcd}\epsilon_{aefg}(\xi_b \nabla_c \xi_d )(\xi^e \nabla^f \xi^g)\\ = -6(\xi_b \nabla_c \xi_d )(\xi^{[b} \nabla^c \xi^{d]} )\\ = -2\{(\xi_b \xi^b) \nabla_c \xi_d \nabla^c \xi^d - (\xi_b \nabla^b \xi^d )\xi^c \nabla_c \xi_d + (\xi_b \nabla^b \xi^c )\xi^d \nabla_c \xi_d\} = 0$

Also note that $\xi^a \omega_a = \epsilon_{[ab]cd}\xi^{(a}\xi^{b)}\nabla^c \xi^d = 0$. Hence $\omega^a = \alpha \xi^a$, where $\alpha$ is a scalar field, because two orthogonal null vector fields must be parallel. Now $\nabla_b \omega_{c} = \xi_c \nabla_b \alpha + \alpha \nabla_b \xi_c$ therefore $\xi_{[a}\nabla_b \omega_{c]} = \xi_{[a}\xi_c \nabla_{b]} \alpha + \alpha \xi_{[a}\nabla_b \xi_{c]}$. But $\xi_{[a}\xi_{c]} = 0$ so we are left with $\xi_{[a}\nabla_b \omega_{c]} = \alpha \xi_{[a}\nabla_b \xi_{c]}$.

Thus if $\nabla_{[a}\omega_{b]} = 0$ then $\alpha \xi_{[a}\nabla_b \xi_{c]} = 0$ which implies $\alpha = 0$, directly yielding $\omega^a = 0$, or $\xi_{[a}\nabla_b \xi_{c]} = 0$ implying $\epsilon_{eabc}\omega^{e} \propto\xi_{[a}\nabla_b \xi_{c]} = 0$ hence $\omega^{d}\propto \epsilon^{dabc}\epsilon_{eabc}\omega^{e} = 0$.