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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.