# Relation of Killing's vector field, Spacetime dimension, and Riemannian curvature tensor

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.

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 $$-[\nabla_a,\,\nabla_c]\xi^a=-g^{ad}[\nabla_a,\,\nabla_c]\xi_d=-g^{ad}R_{acd}^{\quad b}\xi_b=-R_c^{\:b}\xi_b.$$
• 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 $\nabla_{a} \xi^{a} = 0$? Apr 22, 2019 at 14:24