 Solution

Write $q_{\alpha \beta }=g_{\alpha \beta }+n_{\alpha }{n}_{\beta }$ for the induced metric. The spatial covariant derivative projects its derivative index and both tensor indices:Four-dimensional metric compatibility removes the derivative of $g$. Each remaining term contains an undifferentiated normal on a projected index and hence vanishes. ThereforeThis establishes metric compatibility of the spatial covariant derivative directly from its projected definition.

Differentiating the normal's normalization givesso $n^{\mu }{\nabla }_{\nu }{n}_{\mu }=0$. Consequently the second projection in the extrinsic curvature definition is redundant:Finally differentiate $P^{\lambda }{}_{\nu }={\delta }^{\lambda }{}_{\nu }+n^{\lambda }{n}_{\nu }$ and project:The first term vanishes by $P^{\nu }{}_{\beta }{n}_{\nu }=0$, and the second is precisely the defining curvature. The positive sign follows from the specified positive-expansion convention, rather than from an independent projector convention.