Killing fields as eigenvectors of Ricci tensor

  • #1
WannabeNewton
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Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
WannabeNewton
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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)?
 
Last edited:
  • #3
WannabeNewton
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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##.
 
  • #4
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Is there a way to do it more directly, without having to introduce an additional arbitrary vector field?
 
  • #5
WannabeNewton
Science Advisor
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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##.
 

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