 Spatial projector derivative identity

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.