 Solution

There are no ${\partial }_{t}N$ or ${\partial }_t{N}^i$ terms in the displayed Einstein-Hilbert action, so the lapse function and shift vector have zero canonical momenta. Their equations impose constraint equations in field theory rather than independent propagation; they are Lagrange multipliers in the Hamiltonian formulation of general relativity.

With the question's positive sign for $K_{ij}=({\dot{h}}_{ij}-D_i{N}_j-D_j{N}_i)/(2N)$, vary ${\dot{h}}_{ij}$ at fixed $h,N,N^i$. Since $\delta K_{ij}=\delta {\dot{h}}_{ij}/(2N)$,Thus the canonical momentum of the spatial metric is a weight-one tensor density. Write $\pi =h_{ij}{\pi }^{ij}$; in three spatial dimensions $\pi =-2\sqrt{h}K$, soThe action has suppressed its overall $1/(16\pi G)$, and this normalization is retained here. Reversing the sign convention for extrinsic curvature would reverse the momentum relation; the calculation uses exactly the convention supplied in this question.