Why are the ADM momentum and angular momentum such defined? In Sec. 4.3 of the book "A relativist's toolkit" by Eric Poisson, it
is explained how to extract ADM angular momentum from the ADM
Hamiltonian. He says that it suffices to replace the lapse $N$ and the shift $N^a$ with the components of the vector field associated to the quantity you are looking for. For example, the angular momentum is recovered by replacing $N=0$ and 
$$
N^a=\phi^a
$$
where $\phi^a$ is the generator of rotations at asymptotic infinity.
(Similarly, many references say that the ADM momentum $P_b$ is recovered for $N=0$ and $$ N^a=\delta^a_b $$. )
I can see that this "trick" gives you back the correct momenta, but I don't understand what is the rationale behind it.
I understand the procedure for the ADM energy, in which you replace $N=1$ and $N^a=0$, because of course the Hamiltonian is associated with the energy. But I don't understand the connection of the same Hamiltonian with the other momenta.
 A: In ADM form of General Relativity, we must apply 3+1 decomposition to the entire space-time and how to evolute time-slice is arbitrary. This arbitrariness is reflected on the arbitrariness of  the vector $(N,N^a)$. 
Suppose that the "space coordinate" of an particle is fixed and denote the tangent vector of its world line as $t^{\alpha}$, we have (equation 4.36 of your textbook)
$$t^{\alpha} = N n^{\alpha} + N^a e_a^{\alpha}$$
And the Hamiltonian is the generator of the evolution in the direction $t^{\alpha}$. 
In asymptotic flat space-time, we have asymptotic Minkowski coordinates $(\bar{t},\bar{x},\bar{y},\bar{z})$. Roughly speaking, for a rest observer at infinity, his watch is $\bar{t}$ and his ruler is $\bar{x},\bar{y},\bar{z}$. From now on, we will denote this coordinate as $x^{\alpha}$.
Now, let us choose the initial time slice as $\Sigma_{\bar{t}}$. In this time slice, we choose $y^a$ the same as $\bar{x},\bar{y},\bar{z}$. So at infinity, which is flat, we have
$$n^{\alpha} = \delta^{\alpha}_{0}, e_a^{\alpha} = \delta^{\alpha}_a.$$
The evolution vector now become
$$t^{\alpha} = N\delta^{\alpha}_0 + N^a\delta^{\alpha}_a.$$
If $N = 1$ and $N^a = 0$, we have
$$t^\alpha = (1,0,0,0)$$
So, the Hamiltonian is the generator of the evolution in direction $\bar{t}$. It is just the energy of the entire space-time, at least for the observer at infinity. 
If $N = 0$ and $N^a = \delta^a_b$, we have
$$t^\alpha = \delta^{\alpha}_b$$
So, the Hamiltonian is the generator of the evolution in direction $x^b$. It is just the momentum in direction $x^b$ of the entire space-time, at least for the observer at infinity.
If $N = 0$ and $N^a = \phi^a = \frac{\partial y^{a}}{\partial \phi}$, we have
$$t^\alpha = \frac{\partial x^{\alpha}}{\partial \phi}$$
For example, for rotation in $\bar{z}$ direction, we have
$$\bar{x} = r\sin\theta\cos\phi, \; \bar{y} = r\sin\theta\sin\phi, \; \bar{z} = r\cos\theta$$
we then have $$t^{\alpha} = r\sin\theta(0,-\sin\phi, \cos\phi,0)$$
So, the Hamiltonian is the generator of the evolution rotation around $\bar{z}$. It is just the angular momentum in direction $\bar{z}$ of the entire space-time, at least for the observer at infinity.
