There are no or 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 , vary at fixed . Since ,
Thus the canonical momentum of the spatial metric is a weight-one tensor density. Write ; in three spatial dimensions , so
The action has suppressed its overall , 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.
Perform the Legendre transform in mechanics, using . The trace contraction gives
To avoid differentiating a tensor density as though it were an ordinary tensor, set . Integration by parts with the Levi-Civita connection of gives
after dropping the surface term. Also
Therefore the requested Hamiltonian is
Varying the lapse function and shift vector imposes the Hamiltonian constraint and momentum constraint . The total canonical generator also contains multipliers for the vanishing lapse and shift canonical momenta. With an asymptotic boundary, surface terms must be restored to define the Arnowitt-Deser-Misner energy.
For a closed universe, the spatial Cauchy hypersurface is compact without boundary. There is consequently no asymptotic surface term in the gravitational Hamiltonian. On solutions, the Hamiltonian constraint and momentum constraint give
This is the vanishing canonical Hamiltonian of a closed universe: evolution generated by lapse and shift is a gauge transformation, and there is no asymptotic time-translation charge. It does not mean the local stress-energy tensor or Riemann curvature tensor is zero. In an asymptotically flat universe the same bulk constraint equations in field theory hold, but a boundary Arnowitt-Deser-Misner energy can be nonzero.

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