Extrinsic curvature with a negative shift ($K_{ij}=-({\dot{h}}_{ij}+D_i{N}_j+D_j{N}_i)/(2N)$)

For $d{s}^2=-N^{2}d{t}^2+h_{ij}(d{x}^i-N^{i}dt)(d{x}^j-N^{j}dt)$ and the future unit normal, the convention $K_{ij}=-n_{i;j}$ gives $K_{ij}=-({\dot{h}}_{ij}+D_i{N}_j+D_j{N}_i)/(2N)$. Lower the shift vector with $h$. The sign of the shift terms changes when the opposite threading convention is used.

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.