 Small determinant does not imply a small metric perturbation

For $h_{ij}=\mathrm{diag}(L,0,0)$, the determinant is zero for every $L$, even when one perturbation component is arbitrarily large. Linearized metric theory instead requires the eigenvalues of $h$ relative to the background spatial metric to be small, equivalently all its components in a background orthonormal frame to be small. The determinant of $1+h$ then expands as $1+\mathrm{tr}h+O(h^2)$, and the volume element as $1+\mathrm{tr}h/2+O(h^2)$.