# Lapse and shift inside or outside the Poisson bracket?

For general relativity in the 3+1 ADM formulation, one has $$H=\int dx [N{\cal H}+N^a{\cal H}_a]$$ with $$N$$ and $$N^a$$ the lapse and shift which are undetermined Lagrange multipliers. The dynamical equations for the 3-metric and conjugate momentum are $$\dot{g_{ab}}=\{g_{ab},H\}$$, $$\dot{\pi^{ab}}=\{g^{ab},H\}$$, where the Poisson bracket can be taken to only be in terms of $$g_{ab}$$ and $$\pi^{ab}$$, ie. $$\dot{g_{ab}}(x)=\frac{\delta H}{\delta \pi^{ab}(x)}$$. One also have the constraints $${\cal H}\approx 0$$, $${\cal H}_a\approx0$$.

I've seen two general procedures -- (i) we solve for $$g_{ab}$$, $$\pi^{ab}$$ in terms of some unspecified $$N$$ and $$N^a$$ and then set these to be whatever we want (usually some function of $$g_{ab}$$, $$\pi^{ab}$$ but I'm not sure what the restrictions are here, it feels like there are none except that the dynamical equations are satisfied). Or, (ii) one makes some convenient gauge choice for $$g_{ab}$$ and $$\pi^{ab}$$ such that they satisfy the constraints, and then solve the dynamical equations for $$N$$ and $$N^a$$ in terms of $$g_{ab}$$ and $$\pi^{ab}$$.

In both cases, it is tempting to write $$\dot{g_{ab}}=N\{g_{ab},{\cal H}\}+N^a\{{\cal H}_a,g_{ab}\}$$, but I have not seen this. Instead I see $$\dot{g_{ab}}=\{g_{ab},N{\cal H}\}+\{g_{ab},N^a{\cal H}_a\}$$. Because the conjugate momenta to $$N$$ and $$N^a$$ don't appear in the Hamiltonian, the derivatives with respect to the lapse and shift don't contribute to the Poisson bracket, so that suggests I can take $$N$$ and $$N^a$$ out of the Poisson bracket. But if I set them to be equal to a function of $$g_{ab}$$, and $$\pi^{ab}$$ before doing this, then this suggests I should leave it inside the Poisson bracket.

Which is correct?