= Hypersurface orthogonality implies symmetric extrinsic curvature
{title2=$K_{\mu\nu}=K_{\nu\mu}$}
Unit normalization gives $n^\nu\nabla_\mu n_\nu=0$. Hence $K_{\mu\nu}=-P^\alpha{}_\mu P^\beta{}_\nu\nabla_\alpha n_\beta=-P^\alpha{}_\mu\nabla_\alpha n_\nu$. Set $F_{\mu\nu}=\nabla_\mu n_\nu-\nabla_\nu n_\mu$. <Frobenius theorem> for a hypersurface normal gives $n_\alpha F_{\beta\gamma}+n_\beta F_{\gamma\alpha}+n_\gamma F_{\alpha\beta}=0$. Contracting with $n^\alpha P^\beta{}_\mu P^\gamma{}_\nu$ leaves $-P^\beta{}_\mu P^\gamma{}_\nu F_{\beta\gamma}=0$, precisely the vanishing antisymmetric part of the <extrinsic curvature of a spatial hypersurface>.
Back to article page
