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A nowhere-zero differential one-form $n$ is hypersurface orthogonal if locally $n=Fdt$ with $F\ne 0$. Its kernel consists of tangent vectors to the level hypersurfaces of $t$. Frobenius theorem makes this equivalent to $n\wedge dn=0$, where $d$ is the exterior derivative. With a torsion-free Levi-Civita connection, this is $n_{[\alpha }{\nabla }_{\beta }{n}_{\gamma ]}=0$. For a timelike unit normal, projecting the derivative on both indices gives zero antisymmetric part, which explains why hypersurface orthogonality implies symmetric extrinsic curvature.

Differentiate the spatial projection tensor before making any contractions:For the spatial projector derivative identity, its first term vanishes after projection on $\mu$. The definition of the extrinsic curvature of a spatial hypersurface then gives the stronger tensor identityHere $\lambda$ remains a free index. For a normal to a genuine foliation, the extrinsic curvature is symmetric: projecting ${\nabla }_{[\nu }{n}_{\mu ]}$ gives zero because locally $n_{\mu }$ is a scalar multiple of a time gradient. This is hypersurface orthogonality implies symmetric extrinsic curvature.

Contract $\rho$ with $\lambda$ in the stronger identity to obtain exactly the contraction displayed in the PDF:The last equality follows from the transversality of the extrinsic curvature. Thus the literal printed identity is valid, although both its sides vanish; the uncontracted identity explains its geometric origin.

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.

Differentiating the spatial projection tensor and projecting its derivative and covariant indices leaves $-n^{\lambda }{K}_{\sigma \rho }$, where $K$ is the extrinsic curvature of a spatial hypersurface. Contracting $\lambda$ with $\rho$ gives zero because $K$ is transverse. This distinction separates the informative tensor identity, with a free normal index, from its vanishing trace. Hypersurface orthogonality implies symmetric extrinsic curvature.