 Solution

The connection is torsion-free, so on a scalar ${\nabla }_a{\nabla }_b\varphi ={\partial }_a{\partial }_b\varphi -{\Gamma }_{ab}^c{\partial }_c\varphi$ is symmetric in $a,b$. A covector also carries a connection acting on its own index, and commuting derivatives gives curvature, generally nonzero.

Use the curvature convention $[{\nabla }_a,{\nabla }_b]v_c=R^d{}_{cab}{v}_d$, consistent with the identity requested in the PDF. Some texts define the opposite Riemann sign, reversing both formulas. Expanding the prescribed derivative combination of the Killing equation givesThe three commutator terms are $[-R^d{}_{cab}+R^d{}_{bac}+R^d{}_{abc}]v_d=2R^d{}_{abc}{v}_d$, by antisymmetry in the last two indices and the first Bianchi identity. ThusIn flat Cartesian Minkowski coordinates this says ${\partial }_a{\partial }_b{v}_c=0$. Hence all Killing fields are affine:The constant term gives four translations and the antisymmetric term six Lorentz generators, rotations and boosts. Conversely every such field satisfies the Killing equation.

For the static block metric, time derivatives vanish and $g_{0i}=0$, $g^{00}=-1/f$. The Christoffel formula gives ${\Gamma }_{00}^0={\Gamma }_{ij}^0=0$ and ${\Gamma }_{0i}^0={\Gamma }_{i0}^0=({\partial }_{i}f)/(2f)$. The vector $v^a=(1,0,0,0)$ has covector components $v_0=-f$, $v_i=0$. Its only potentially nonzero derivatives are ${\nabla }_i{v}_0=-{\partial }_{i}f/2$ and ${\nabla }_0{v}_i={\partial }_{i}f/2$; spatial-spatial and time-time derivatives vanish. Their symmetric sums are zero, proving ${\partial }_t$ is a Killing vector field.