 Solution

Write ${\gamma }_{ij}={}^{(3)}{g}_{ij}$, $\gamma =\det {\gamma }_{ij}$ and $D_i$ for the spatial covariant derivative. With the negative-shift convention, the extrinsic curvature of a spatial hypersurface is $K_{ij}=-({\dot{\gamma }}_{ij}+D_i{N}_j+D_j{N}_i)/(2N)$. Contracting with ${\gamma }^{ij}$ and using metric compatibility givesThe logarithmic determinant identity makes ${\gamma }^{ij}{\dot{\gamma }}_{ij}=\dot{\gamma }/\gamma$, henceThis is also the fractional volume expansion, with a minus sign from the extrinsic-curvature convention.

Define $a={\gamma }^{1/6}$ so that $\sqrt{\gamma }=a^3$. The useful unit-determinant conformal decomposition is ${\gamma }_{ij}=a^2{\tilde{\gamma }}_{ij}$ with $\det \tilde{\gamma }=1$, equivalently ${\tilde{\gamma }}_{ij}=a^{-2}{\gamma }_{ij}$. Multiplying by $a^2$ instead would be a different conformal rescaling, not this unit-determinant one; no conformal decomposition is needed for the trace identity itself.

With zero shift vector, $\dot{\gamma }/\gamma =6\dot{a}/a$ and the local volume Hubble parameter isAlong a normal trajectory proper time satisfies $ds=Ndt$, so $H=a^{-1}da/ds$ is the local physical volume-expansion rate. The printed equality with $\dot{a}/a$ requires the proper-time gauge $N=1$; zero shift alone does not set the lapse to unity. In a homogeneous Friedmann-Lemaître-Robertson-Walker metric it reduces to the ordinary Hubble parameter, while in an inhomogeneous spacetime it can vary from one normal trajectory to another.