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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.

For an arbitrary square matrix $A$, the derivative of the determinant in direction $H$ is the displayed expression. Differentiating the determinant's multilinear expansion gives the sum of entries of $H$ weighted by their cofactors, equal to the displayed trace. If $A$ is invertible, $\mathrm{adj}(A)=(\det A)A^{-1}$ gives $(\det A)\mathrm{tr}(A^{-1}H)$, and division by $\det A$ gives the Jacobi determinant derivative formula for the logarithmic derivative. The adjugate expression remains valid at singular matrices.

Let ${\gamma }_{ij}$ denote the spatial induced metric and $\gamma =\det {\gamma }_{ij}$. Its conformally rescaled version ${\tilde{\gamma }}_{ij}={\gamma }^{-1/3}{\gamma }_{ij}$ has determinant one. The Jacobi determinant derivative formula consequently givesThus the determinant measures the volume-changing part of the spatial metric, while the unit-determinant part contains shape changes. Metric compatibility for the spatial covariant derivative also gives ${\gamma }^{ij}{D}_i{N}_j=D_i{N}^i$.

There is a sign inconsistency in the PDF. Contracting its defining expression for $K_{ij}$, with its threading convention $d{x}^i-N^{i}dt$ and its future normal $n^{\mu }=N^{-1}(1,N^i)$, givesnot the displayed trace with a minus sign before the divergence. This is the trace of extrinsic curvature with a negative shift. For a concrete check, take ${\gamma }_{ij}={\delta }_{ij}$, $N=1$ and $N^i=b{x}^i$. The defining formula gives $K_{ij}=-b{\delta }_{ij}$ and $K=-3b$, whereas the printed trace would give $+3b$. Reversing the shift convention can produce the other trace formula, but also changes the shift terms in $K_{ij}$ and the normal; the two conventions cannot be combined.

For the requested expansion interpretation, set the shift vector to zero. The sign discrepancy then disappears. A small coordinate volume has proper volume proportional to $\sqrt{\gamma }$, and proper time along its normal is $ds=Ndt$. Its expansion rate isThereforeThis is the local volume Hubble parameter, the mean of the three local directional expansion rates. In a homogeneous isotropic background $\gamma =a^6$, it reduces to $\dot{a}/(Na)$. The interpretation does not require the local expansion to be isotropic: the trace-free part of the extrinsic curvature can still describe shear.

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.

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.

A Cartesian tensor of order $n$ is a multilinear geometric object whose components under an orthogonal change of axes obey the tensor component transformation law,An isotropic tensor has the same components under every proper rotation. For the Kronecker delta,Thus ${\delta }_{ij}$ is isotropic, even under improper orthogonal transformations.

To prove multiplicativity of the determinant from the given formula, first consider $C_{pqr}={ϵ}_{ijk}{A}_{ip}{A}_{jq}{A}_{kr}$. Exchanging any two of $p,q,r$ changes its sign, so it is proportional to the Levi-Civita symbol. Its $123$ component is the determinant, as follows by expanding the given double-epsilon expression. ThereforeSubstitute $(AB{)}_{il}=A_{ip}{B}_{pl}$ into that same determinant formula:HenceNo invertibility assumption was used, so this identity includes singular matrices.

Now differentiate the double-epsilon formula for a differentiable $A(t)$. Its three differentiated terms are equal after renaming the paired dummy indices: the simultaneous swaps in both epsilon symbols cancel their signs. Thuswhere $C_{il}$ is the cofactor of entry $A_{il}$. For an invertible matrix, the adjugate matrix identity gives $C_{il}=(\det A)(A^{-1}{)}_{li}$. Consequently the Jacobi determinant derivative formula isThe order of the factors inside the trace follows from the reversed cofactor indices; no commutativity of $A$ and $\dot{A}$ is assumed. The cofactor version is also Jacobi's formula and remains valid when $A$ is singular, although the inverse version then cannot be used.

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.

For the threading convention $d{x}^i-N^{i}dt$ and $K_{ij}=-({\dot{\gamma }}_{ij}+D_i{N}_j+D_j{N}_i)/(2N)$, contract with ${\gamma }^{ij}$. Metric compatibility gives ${\gamma }^{ij}{D}_i{N}_j=D_i{N}^i$, and the Jacobi determinant derivative formula gives ${\gamma }^{ij}{\dot{\gamma }}_{ij}={\partial }_t\log \det \gamma$. With zero shift, $-K/3=(3N{)}^{-1}{\partial }_t\log \sqrt{\gamma }$ is the local volume Hubble parameter. Reversing the shift convention reverses both the normal's spatial component and the shift terms; the two conventions must not be mixed.