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For a non-null Killing vector field $V$ of a Levi-Civita connection, hypersurface orthogonality implies that $\eta =V^{♭}/F$, with $F=g(V,V)$, is a closed differential form. The Killing equation and $V^{♭}\wedge d{V}^{♭}=0$ give $2F{\nabla }_{\mu }{V}_{\nu }=V_{\nu }{\partial }_{\mu }F-V_{\mu }{\partial }_{\nu }F$, whence $d\eta =0$. The Poincare lemma gives $\eta =d\varphi$ locally. For a timelike $V$, $\varphi$ supplies local static time coordinates. This does not assert global exactness.

Write $F=V_{\rho }{V}^{\rho }<0$ and $A_{\mu \nu }={\nabla }_{\mu }{V}_{\nu }$. The Killing equation makes $A_{\mu \nu }$ antisymmetric. Expanding hypersurface orthogonality and contracting with $V^{\alpha }$ givesHere metric compatibility gives ${\partial }_{\mu }F=2V^{\rho }{A}_{\mu \rho }$. Thereforeand the normalized Killing one-form satisfiesThus the required exponent is $\boxed{n=-2}$. The notation $|V{|}^2$ in this paper denotes the negative scalar $F$, rather than a positive squared norm; the answer uses only its real reciprocal. More generally, for a multiplier $h(F)$ the antisymmetric derivative is $(h/F+h^{\prime })(V_{\nu }{\partial }_{\mu }F-V_{\mu }{\partial }_{\nu }F)$, so $h=C/F$ is the universally valid choice. If $F$ is constant, other exponents can work in that particular metric.

Use an affine parameter $\lambda$ along a generator of a null hypersurface. Its null twist vanishes by hypersurface orthogonality. The screen metric is positive definite, so ${\hat{\sigma }}^{ab}{\hat{\sigma }}_{ab}\ge 0$. The Einstein field equations and null energy condition imply $R_{ab}{U}^a{U}^b\ge 0$, giving from the Null Raychaudhuri equationStarting with ${\theta }_0<0$, the null expansion stays negative as long as the congruence remains regular. Set $z=-1/\theta >0$. Then $z^{\prime }={\theta }^{\prime }/{\theta }^2\le -1/2$, soA finite negative null expansion cannot persist to $\lambda -{\lambda }_0=2/|{\theta }_0|$. Thus the null focusing theorem givesThis is a focal breakdown of the transverse congruence, provided the generator and regular geometry extend that far; it is not by itself a curvature singularity or a statement that the individual null geodesic terminates there.

Differentiate the normalization of the timelike unit normal with the metric-compatible covariant derivative:Thus the derivative of the normal is already orthogonal to the normal in its second index. Expanding the second spatial projection tensor in the definition of the extrinsic curvature of a spatial hypersurface givesTo prove symmetry using Frobenius theorem, set $F_{\alpha \beta }={\nabla }_{\alpha }{n}_{\beta }-{\nabla }_{\beta }{n}_{\alpha }$. Hypersurface orthogonality is equivalent toContract this with $n^{\alpha }{P}^{\beta }{}_{\mu }{P}^{\lambda }{}_{\nu }$. The last two terms vanish because a projected normal vanishes, while $n^{\alpha }{n}_{\alpha }=-1$ in the first term. HenceThe antisymmetric part of the doubly projected covariant derivative of $n$ is therefore zero. This proves that hypersurface orthogonality implies symmetric extrinsic curvature:No geodesic assumption for the normal congruence is needed; its normal acceleration may be nonzero.