= Solution
{c}
A <Killing vector field> $K$ generates local <isometries>: its <flow of a vector field> $\Phi_s$ satisfies $\Phi_s^*g=g$. Equivalently, the <Lie derivative> of the <metric tensor> vanishes. For the <Levi-Civita connection>, <metric compatibility> converts the coordinate expression into the <Killing equation>:
$$
\boxed{\mathcal L_Kg=0,\qquad \nabla_\mu K_\nu+\nabla_\nu K_\mu=0.}
$$
\b[The <Levi-Civita connection> is an implicit extra assumption in this question.] The printed assumption of vanishing <torsion tensor> alone does not imply <metric compatibility>. For example, on <Minkowski spacetime>, take the <affine connection> whose only nonzero <Christoffel symbol> is $\Gamma^0{}_{00}=1$. The <vector field> $K=\partial_0$ preserves the <Minkowski metric>, but $K_0=-1$ gives $\nabla_0K_0=1$, contradicting the <Killing equation>. The following calculation, and the covariant formulas in part (b), use the <Levi-Civita connection>; this is the role of <metric compatibility in Killing identities>.
For the stated <curvature sign convention>, the <Ricci identity> on a <covector> has a minus sign:
$$
[\nabla_\mu,\nabla_\nu]K_\rho=-R^\sigma{}_{\rho\mu\nu}K_\sigma.
$$
Write $A_{\mu\nu\rho}=\nabla_\mu\nabla_\nu K_\rho$. Differentiating the <Killing equation> gives $A_{\mu\nu\rho}=-A_{\mu\rho\nu}$. Alternately using this relation and the <Ricci identity> gives
$$
\begin{aligned}
A_{\mu\nu\rho}
&=-A_{\mu\rho\nu}
=-A_{\rho\mu\nu}+R^\sigma{}_{\nu\mu\rho}K_\sigma\\
&=A_{\rho\nu\mu}+R^\sigma{}_{\nu\mu\rho}K_\sigma\\
&=A_{\nu\rho\mu}-R^\sigma{}_{\mu\rho\nu}K_\sigma+R^\sigma{}_{\nu\mu\rho}K_\sigma\\
&=-A_{\mu\nu\rho}-R^\sigma{}_{\rho\mu\nu}K_\sigma-R^\sigma{}_{\mu\rho\nu}K_\sigma+R^\sigma{}_{\nu\mu\rho}K_\sigma.
\end{aligned}
$$
The <first Bianchi identity> and antisymmetry of the last two indices of the <Riemann curvature tensor> reduce the three curvature terms to $2R^\sigma{}_{\mu\nu\rho}K_\sigma$. Consequently the <second covariant derivative of a Killing vector> is
$$
\boxed{\nabla_\mu\nabla_\nu K_\rho=R^\sigma{}_{\mu\nu\rho}K_\sigma.}
$$
Back to article page
