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Write $h_{ij}={}^{(3)}{g}_{ij}$ for the induced metric and lower the shift vector with it, $N_i=h_{ij}{N}^j$. In the stated negative-shift convention the four-dimensional components and inverse components areMultiplying the given future unit normal by this metric tensor gives $n_{\mu }=(-N,0,0,0)$. In particular the tangential normal components vanish, but their covariant derivatives do not:The relevant Christoffel symbol isFor the second equality, the last parenthesis is $2h_{k\ell }{}^{(3)}{\Gamma }^{\ell }{}_{ij}$, so its contraction subtracts the two spatial connection terms in the symmetrized derivative of $N_i$. ThereforeThis is extrinsic curvature with a negative shift. Both the future-normal orientation and the definition $K=-\nabla n$ matter: replacing the threading by $d{x}^i+N^{i}dt$ reverses the shift terms. The spatial induced metric is generally a function of time as well as position; restricting it to time-independent data would simply make the displayed time derivative vanish.

The unimodular spatial metric separates volume from shape: writing $h_{ij}=a^2{\tilde{h}}_{ij}$ and $a^6=h=\det h_{ij}$ gives $\det \tilde{h}=1$. Contract the extrinsic curvature with a negative shift with $h^{ij}$. Metric compatibility and the Jacobi determinant derivative formula giveConsequentlyFor zero shift vector, $\dot{h}/h=6\dot{a}/a$, soThe infinitesimal spatial volume is $\sqrt{h}{d}^{3}x=a^3{d}^{3}x$; its logarithmic change per unit normal proper time is $3H$. Thus this $H$ is the local volume Hubble parameter, reducing to the usual Hubble parameter in a homogeneous universe. The determinant-defined $a$ refers to fixed spatial coordinates, while this normal volume expansion has a geometric interpretation.

Set $h=\det h_{ij}$. Contract the extrinsic curvature with a negative shift with $h^{ij}$. The Jacobi determinant derivative formula and metric compatibility giveConsequentlyThe local volume element is $\sqrt{h}{d}^{3}x$, so for zero shift vector the proper-time derivative along the normal is $d/ds=N^{-1}{\partial }_t$ andHere $\theta$ is the expansion scalar. Writing $a_{\mathrm{loc}}=h^{1/6}$ defines a local linear scale from this volume, whence the local volume Hubble parameter isThe factor of three converts volume expansion into linear expansion, and $N$ converts coordinate time into proper time. This agrees with the usual Hubble parameter in a homogeneous FLRW metric.

For a volume-shape decomposition the unimodular spatial metric is ${\gamma }_{ij}=a_{\mathrm{loc}}^{-2}{h}_{ij}$, with determinant one. If instead the printed positive power $a_{\mathrm{loc}}^2{h}_{ij}$ is taken literally, its determinant is $a_{\mathrm{loc}}^{12}$; it is a conformal rescaling but not the unimodular shape metric. The trace calculation does not need that rescaling.

Write $h_{ij}$ for the spatial metric tensor and lower the shift vector with $h_{ij}$. For the negative-shift convention, the four-dimensional metric and its inverse have componentsThe normal has covariant components $n_{\mu }=(-N,0,0,0)$, so $n_{\mu }{n}^{\mu }=-1$ and it annihilates every tangent vector to the spatial slice. Since $n_i=0$, its covariant derivative gives $n_{i;j}=N{\Gamma }^0{}_{ij}$ and therefore $K_{ij}=-N{\Gamma }^0{}_{ij}$.

Compute the required Christoffel symbol using the symmetric connection formula. Its time-index term and spatial-index term combine to giveThe spatial expression in parentheses is ${\dot{h}}_{ij}+D_i{N}_j+D_j{N}_i$, where $D$ is the covariant derivative of $h$. Thus the extrinsic curvature with a negative shift isThis derivation fixes both signs from the actual 3+1 decomposition of spacetime, rather than transferring a formula using the opposite shift vector convention. The connection formula used here has $g_{\lambda \kappa ,\nu }$ as its second differentiated term; the printed repeated-index version is a typographical error.