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Work on a regular level surface, so $d\beta \ne 0$. A nonzero normal covector has the local form $n_a=f{\nabla }_a\beta$. Extend it as a normal to neighbouring level surfaces. Since the Levi-Civita connection is torsion-free,Antisymmetrizing the product with $n_c=f{\beta }_{;c}$ over three indices kills the symmetric second-derivative term and the term containing two copies of $d\beta$. HenceEquivalently $n\wedge dn=0$, the Frobenius theorem condition for hypersurface orthogonality.

For the second covariant derivative of a Killing vector, the curvature sign needs explicit care. Denote the curvature defined on the cover by $C^a{}_{bcd}$, and obtain its other index positions by ordinary raising and lowering with the metric tensor. With $D_{abc}=k_{a;bc}$, the Killing equation gives $D_{abc}=-D_{bac}$, while the supplied Ricci identity givesSuccessive use of these two facts givesSubstitute $D_{acb}=D_{abc}-C_a{}^d{}_{bc}{k}_d$. The curvature symmetries and first Bianchi identity then implyThus the identity consistent with the cover convention isThe plus-sign version is valid for the opposite curvature tensor $\tilde{R}=-C$. With ordinary metric index raising, the printed plus sign and the cover's curvature convention cannot both be used. This is the curvature-sign convention in Killing derivative identities, not a failure of the Killing equation.

An explicit sign check uses $d{s}^2=e^{2x}d{t}^2-d{x}^2-d{y}^2-d{z}^2$ and the timelike Killing vector $k={\partial }_t$. Here $k_t=e^{2x}$, ${\Gamma }^t{}_{tx}=1$ and ${\Gamma }^x{}_{tt}=e^{2x}$. Direct differentiation gives $k_{t;xx}=e^{2x}$, whereas the cover definition gives $C^t{}_{xxt}=-1$ and $C_{txx}{}^d{k}_d=-e^{2x}$. This also verifies the corrected sign without relying on a remembered curvature convention.

To prove the static Killing field is a Ricci eigenvector, use the permitted alternative of adapted coordinates. Hypersurface orthogonality supplies hypersurfaces orthogonal to $k$. Choose spatial coordinates on one such hypersurface and carry them along the flow of the Killing vector, using its flow parameter as $t$. The flow preserves both the metric tensor and $k$, so it preserves orthogonality to $k$. Consequently the mixed metric components vanish, and the Killing equation makes all coefficients independent of $t$:The only nonzero Christoffel symbols involving time can bein particular ${\Gamma }^0{}_{ij}={\Gamma }^i{}_{0j}={\Gamma }^0{}_{00}=0$. In $C_{0i}$, the derivative of the connection trace vanishes because ${\Gamma }^a{}_{0a}=0$; the other derivative is either a time derivative of a static coefficient or a derivative of ${\Gamma }^j{}_{0i}=0$. Every connection-product term contains a vanishing coefficient of these same types. Thus $C_{0i}=0$. The one-form $C_{ab}{k}^b=C_{a0}$ has only a time component, and is proportional to $k_a$. ThereforeChanging the overall curvature sign leaves this zero identity unchanged.

Finally a perfect fluid in signature $(+---)$ has $T_{ab}=(\rho +p)u_a{u}_b-p{g}_{ab}$ and $u^a{u}_a=1$. Contract the Einstein field equations with $k^b$, and antisymmetrize with $k_a$. Terms proportional to $g_{ab}$ disappear, as does the Ricci term just proved. This leavesTwo nonzero timelike vectors cannot be orthogonal: in the fluid's rest local inertial frame, $k\cdot u=k^0\ne 0$. The hypothesis $\rho +p>0$ therefore forces $u_{[a}{k}_{b]}=0$. Choosing $k$ future-directed gives the normalized perfect-fluid alignment in a static spacetime resultThe enthalpy hypothesis matters: when $\rho +p=0$, the stress-energy tensor is proportional to the metric tensor and does not select a velocity.