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 . Hypersurface orthogonality is equivalent toContract this with . The last two terms vanish because a projected normal vanishes, while in the first term. HenceThe antisymmetric part of the doubly projected covariant derivative of 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.
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