I am a undergraduate student who studying general relativity. While I am reading numerical relativity written by Masaru Shibata, I do not understand how to derive the equation $\nabla^{a} \nabla_{a} \xi_{c} + \frac{D-2}{D} \nabla_{c} \nabla_{a} \xi^{a} = -R^{d}_{c} \xi_{d}$ from following steps. Thank you.

These steps are come from Masaru Shibata's book:

The Killing's vector (field) $\xi^{a}$ would satisfy the Killing's equation $ \nabla_{a} \xi_{b} + \nabla_{b} \xi_{a}= 0 $, where $\nabla$ is the covariant derivative.

Then we can combine the Killing's equation and Riemannian curvature tensor $\nabla_{a} \nabla_{b} \xi_{c} - \nabla_{b} \nabla_{a} \xi_{c} = R^{d}_{abc} \xi_{d}$,

We get $\nabla_{a} \nabla_{b} \xi_{c} + \nabla_{b} \nabla_{c} \xi_{a} = R^{d}_{abc} \xi_{d}$. By cyclic permutation of above equation, $\nabla_{a} \nabla_{b} \xi_{c} = -R^{d}_{bac} \xi_{d}$.

By contracting $\nabla_{a} \nabla_{b} \xi_{c} = -R^{d}_{bac} \xi_{d}$, we get $\nabla^{a} \nabla_{b} \xi_{c} = -R^{d}_{c} \xi_{d}$

The final step uses Killing's equation($\nabla_{a} \xi^{a}= 0$) and rewrite $\nabla^{a} \nabla_{b} \xi_{c} = -R^{d}_{c} \xi_{d}$ to following:

$\nabla^{a} \nabla_{a} \xi_{c} + \frac{D-2}{D} \nabla_{c} \nabla_{a} \xi^{a} = -R^{d}_{c} \xi_{d}$, where D is the spacetime dimension.


1 Answer 1


OK, let's spell this out a bit more. Firstly, any vector field $\xi_c$ satisfies $\nabla_a\nabla_b\xi_c-\nabla_b\nabla_a\xi_c=R_{abc}^{\quad d}\xi_d$, not just Killing fields. (This can be taken as a definition of the Riemann tensor.) Then the cyclic permutation step is this: $$\nabla_a\nabla_b\xi_c+\nabla_b\nabla_c\xi_a+\nabla_c\nabla_a\xi_b=\frac12(R_{abc}^{\quad d}+R_{bca}^{\quad d}+R_{cab}^{\quad d})\xi^d=0\\\implies\nabla_a\nabla_b\xi_c=-(\nabla_b\nabla_c\xi_a+\nabla_c\nabla_a\xi_b)=-R_{bca}^d\xi_d.$$Contracting $a$ with $b$ gives $\square\xi_c=-R_c^{\:d}\xi_d$, so$$\nabla_a\xi^a=0\implies-R_c^{\:d}\xi_d=\square\xi_c=\square\xi_c+\frac{D-2}{D}\nabla_c\nabla_a\xi^a.$$I don't have access to Shibata's book, but it's weird he'd set out to derive this equation for Killing vector fields specifically, since the coefficient $\frac{D-2}{D}$ could be anything; $\nabla_a\xi^a=0$, after all. (Are you sure it's not meant to apply to arbitrary vector fields?) Weirder still, there's a much quicker proof of its applicability to Killing fields: $$\square\xi_c+\frac{D-2}{D}\nabla_c\nabla_a\xi^a=\nabla^a\nabla_a\xi_c=-\nabla^a\nabla_c\xi_a=-[\nabla^a,\,\nabla_c]\xi_a=-[\nabla_a,\,\nabla_c]\xi^a,$$where square brackets denote a commutator (obtained by using $\nabla_a\xi^a=0$ again). But $$-[\nabla_a,\,\nabla_c]\xi^a=-g^{ad}[\nabla_a,\,\nabla_c]\xi_d=-g^{ad}R_{acd}^{\quad b}\xi_b=-R_c^{\:b}\xi_b.$$

  • $\begingroup$ Thank you for your answer :D. The chapter that I read is just the appendix and the whole derivation is based on killing vector $\xi^{a}$. Therefore, I am not sure whether it can be applied to arbitrary vector fields. Besides, it is true that we can have freedom to add $\frac{D-2}{D} \nabla_{a} \xi^{a}$ when $\nabla_{a} \xi^{a} = 0$? $\endgroup$
    – Ricky Pang
    Apr 22, 2019 at 14:24

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