Killing equation (${\nabla }_{(\mu }{K}_{\nu )}=0$)

The Killing equation tests whether a candidate vector field generates a local isometry, without needing to solve for its flow of a vector field. For the Levi-Civita connection of a metric tensor $g$, it isParentheses mean averaging over the two index orders. The equation says the symmetric part of the covariant derivative vanishes, not that the entire covariant derivative vanishes. The Lie derivative of the metric tensor gives the coordinate formThe last two terms account for how tangent vectors move with the field; checking only $K^{\rho }{\partial }_{\rho }{g}_{\mu \nu }$ misses this. See Carroll, section 5, equations (5.33) and (5.41)-(5.42).

For the Euclidean Killing vector field $K=-y{\partial }_x+x{\partial }_y$, the Euclidean metric in Cartesian coordinates makes covariant derivatives ordinary partial derivatives. Its diagonal equations give ${\partial }_x{K}_x={\partial }_y{K}_y=0$, and its mixed equation gives ${\partial }_x{K}_y+{\partial }_y{K}_x=1-1=0$. Thus a varying vector field can preserve a constant metric tensor. In contrast, $K=x{\partial }_x+y{\partial }_y$ gives $({\mathcal{L}}_{K}g{)}_{xx}=2$ and fails the Killing equation, despite the constant metric components.

If $K={\partial }_z$ is a coordinate basis vector, its components are constant and the coordinate form reduces to ${\partial }_z{g}_{\mu \nu }=0$. Hence independence of the metric components from $z$ supplies a Killing vector field. This shortcut applies to that coordinate generator, not to an arbitrary varying vector field.

For a Killing vector field, set $F_{ab}={\nabla }_a{K}_b$ and $a^s=K^t{\nabla }_t{K}^s$. The Killing equation makes $F$ antisymmetric, and expansion of normalized three-index antisymmetrization gives the displayed identity. On a Killing horizon, hypersurface orthogonality removes its first right-hand term and relates $F^{ab}{F}_{ab}$ to surface gravity.

For a Killing vector field and the curvature convention $[{\nabla }_{\mu },{\nabla }_{\nu }]V^{\alpha }=R^{\alpha }{}_{\beta \mu \nu }{V}^{\beta }$,The identity follows by combining the Ricci identity, Killing equation, and first Bianchi identity.

Let $C^a{}_{bcd}={\partial }_d{\Gamma }_{bc}^a-{\partial }_c{\Gamma }_{bd}^a-{\Gamma }_{ec}^a{\Gamma }_{bd}^e+{\Gamma }_{ed}^a{\Gamma }_{bc}^e$, with all other index positions obtained by the metric tensor. This convention is the negative of the convention $[{\nabla }_c,{\nabla }_d]V^a=R^a{}_{bcd}{V}^b$. The Ricci identity is $k_{a;bc}-k_{a;cb}=C_a{}^d{}_{bc}{k}_d$, with $k_{a;bc}={\nabla }_c{\nabla }_b{k}_a$. For a Killing vector, the Killing equation makes this second derivative antisymmetric in $a,b$. Applying the commutator three times gives $2k_{a;bc}=(C_a{}^d{}_{bc}+C_c{}^d{}_{ba}-C_b{}^d{}_{ac})k_d=-2C_{abc}{}^d{k}_d$, where the last step uses the first Bianchi identity. Switching the sign of the Riemann curvature tensor switches the sign in this identity; moving a raised index by ordinary metric contraction does not switch it.

Every Killing vector field $K$ preserves the scalar curvature: ${\mathcal{L}}_{K}R=K^{\alpha }{\nabla }_{\alpha }R=0$. One index proof rewrites this derivative as a doubly contracted curvature commutator acting on the antisymmetric tensor ${\nabla }_{[\mu }{K}_{\lambda ]}$; both terms vanish by symmetry of the Ricci tensor.