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For a nondegenerate metric tensor and its Levi-Civita connection, contracting the Christoffel symbol formula gives ${\Gamma }^{\rho }{}_{\rho \mu }=\frac{1}{2}{g}^{ab}{\partial }_{\mu }{g}_{ab}$. The Jacobi determinant derivative formula identifies this with ${\partial }_{\mu }\log \sqrt{|\det g|}$. The volume density, rather than the determinant without an absolute value, works in every fixed metric signature.

For a nondegenerate metric tensor, differentiating $g_{ac}{g}^{cb}={\delta }_a{}^b$ gives $\delta g_{ab}=-g_{ac}{g}_{bd}\delta g^{cd}$. Combining this with the Jacobi determinant derivative formula proves the displayed volume-density variation. It is the determinant contribution when defining a stress-energy tensor by varying the inverse metric tensor.

Vary the inverse metric tensor while holding the matter fields fixed. Differentiating $g_{a\mu }{g}^{\mu b}={\delta }_a{}^b$ gives $\delta g_{ab}=-g_{a\mu }{g}_{b\nu }\delta g^{\mu \nu }$. The Jacobi determinant derivative formula consequently implies the inverse metric volume variationSince the matter Lagrangian in this question contains no derivatives of the metric tensor, its metric variation is algebraic. The product rule givesMultiplying by $-2/\sqrt{-g}$ proves the equivalence of the stress-energy definitions:The variations of the symmetric metric tensor are understood symmetrically. If the matter action contained metric derivatives, the definition would instead require a functional derivative and the corresponding integration by parts.

Contract the Christoffel symbol formula for the Levi-Civita connection. The first and third derivative terms cancel after relabelling their summed indices:The Jacobi determinant derivative formula gives ${\partial }_{\mu }g=g{g}^{\rho \sigma }{\partial }_{\mu }{g}_{\rho \sigma }$. In a Lorentzian chart $g<0$, and hence ${\partial }_{\mu }\sqrt{-g}=\frac{1}{2}\sqrt{-g}{g}^{\rho \sigma }{\partial }_{\mu }{g}_{\rho \sigma }$. Thus the trace is the logarithmic derivative of the volume density:In particular, the covariant divergence formula for a vector field is ${\nabla }_{\mu }{A}^{\mu }=(-g{)}^{-1/2}{\partial }_{\mu }(\sqrt{-g}{A}^{\mu })$.

First, the inverse metric tensor printed in the PDF has a missing factor: its lower-left entry must be $g^{i0}={\alpha }^{-2}{\beta }^i$, symmetric with $g^{0i}$. This correction follows either from symmetry or direct block multiplication and does not change the evolution equation used here.

Let $D_t={\partial }_t-{\beta }^k{\partial }_k$. Contract the spatial metric tensor evolution equation with ${\gamma }^{ij}$. Both shift-gradient terms contribute ${\partial }_k{\beta }^k$, givingThe Jacobi determinant derivative formula applies to every derivative of $\gamma =\det ({\gamma }_{ij})$ and gives $D_t\gamma =\gamma {\gamma }^{ij}{D}_t{\gamma }_{ij}$. Therefore the spatial determinant evolves asEquivalently, the spatial volume evolution identity is $D_t\sqrt{\gamma }=\sqrt{\gamma }({\partial }_k{\beta }^k-\alpha K)$, with $K$ the trace of the extrinsic curvature of a spatial hypersurface.

For $S_t(z)=\Phi (t,0,z)$, differentiability with respect to initial data gives the variational equationThe Jacobi determinant derivative formula givesThe mixed derivatives in the Hamiltonian vector field cancel:Consequently $J^{\prime }(t)=0$, and $J(0)=1$ gives preservation of phase space volume:The same argument applies to $\Phi (t,s)$ for every starting time $s$. This is the Liouville theorem in Hamiltonian mechanics. Explicit time dependence of $H$ does not affect the cancellation.

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.

In a 3+1 decomposition of spacetime with $K_{ij}=-({\partial }_t{\gamma }_{ij}-{\mathcal{L}}_{\beta }{\gamma }_{ij})/(2\alpha )$, contract the metric evolution with ${\gamma }^{ij}$ and use the Jacobi determinant derivative formula. Here $\gamma =\det {\gamma }_{ij}$, $K={\gamma }^{ij}{K}_{ij}$, $\alpha$ is the lapse function, and $\beta$ is the shift vector. The identity separates spatial transport and coordinate compression from the geometric volume change caused by extrinsic curvature.