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Using the lapse function $N$, shift vector $N^i$ and spatial metric $h_{ij}$, the Legendre transform in mechanics of the Einstein-Hilbert action gives, in units $16\pi G=1$,Here $\pi =h_{ij}{\pi }^{ij}$ and ${\pi }^{ij}$ is the canonical momentum of the spatial metric. Varying lapse and shift imposes the Hamiltonian constraint and momentum constraint. An asymptotic time translation requires an Arnowitt-Deser-Misner energy boundary term. The bulk constraints generate gauge evolution; the nonzero asymptotic energy comes from the boundary.

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.

In the paper's convention, ${\dot{h}}_{ij}=2N{K}_{ij}+D_i{N}_j+D_j{N}_i$. The Legendre transform in mechanics therefore givesIndeed ${\pi }^{ij}{K}_{ij}=\sqrt{h}(K_{ij}{K}^{ij}-K^2)$, and inversion of the canonical momentum of the spatial metric givesTo integrate the shift term without ambiguity about tensor densities, put $p^{ij}={\pi }^{ij}/\sqrt{h}$. Metric compatibility of the spatial covariant derivative gives, up to a boundary term,Thus the Hamiltonian formulation of general relativity hasVariation of the lapse function imposes the Hamiltonian constraint $\mathcal{H}=0$, while variation of the shift vector imposes the momentum constraint ${\mathcal{H}}_i=0$.

The action contains neither $\dot{N}$ nor ${\dot{N}}^i$, so their conjugate momentum vanishes. They impose constraints rather than independent evolution equations: after the Legendre transform in mechanics, the lapse function and shift vector are Lagrange multipliers in the Hamiltonian formulation of general relativity.

Use precisely the paper's sign convention for extrinsic curvature, $K_{ij}=({\dot{h}}_{ij}-D_i{N}_j-D_j{N}_i)/(2N)$. At fixed $h_{ij},N,N^i$,Consequently the canonical momentum of the spatial metric isThis is a weight-one tensor density, with indices raised and lowered using $h_{ij}$. Its trace and inverse relation areThe sign of $K_{ij}$ used here is opposite to the convention in extrinsic curvature of a spatial hypersurface; mixing the two conventions would reverse the momentum formula.