 Canonical momentum of the spatial metric

With the convention $K_{ij}=({\dot{h}}_{ij}-D_i{N}_j-D_j{N}_i)/(2N)$ for extrinsic curvature, differentiation of the gravitational Lagrangian gives ${\pi }^{ij}=\sqrt{h}(K^{ij}-K{h}^{ij})$ in units $16\pi G=1$. It is a weight-one tensor density. In three spatial dimensions, $\pi =-2\sqrt{h}K$, soThe lapse function and shift vector have no time derivatives in this action and consequently have vanishing conjugate momenta. Changing the sign convention for extrinsic curvature changes the corresponding momentum relation; it must not be silently combined with this formula.