Poisson brackets in curved spacetime The time evolution of any field $\phi$ is given in terms of the Poisson bracket with the Hamiltonian,
$$
\frac{\partial\phi}{\partial t} = \{\phi, H\}.
$$
How does this relation change in curved spacetime, say AdS? Is the definition of Poisson Bracket any different from what it is in flat spacetime?
 A: It's been some time since you asked this, but I'll give it a shot. 
The biggest problems in curved spacetime are defining exactly what you mean by a Hamiltonian, and what you mean by a Poisson bracket.
Let's say you're dealing with a real scalar field $\phi$ which minimizes some action
$$
W = \int_{\mathcal{M}}\mathcal{L}(\phi,\partial_\mu \phi,x) d^4 x
$$
which is the integration of a scalar density Lagrangian $\mathcal{L}$ over a manifold $\mathcal{M}$. From here you have a number of options: you can try to foliate the manifold into surfaces $\Sigma(t)$ and get a specific momentum, Hamiltonian, and Poisson bracket defined by quantities on these surfaces... or you can say you don't want to commit to a foliation yet. For any Cauchy surface $\Sigma$ we can define a momentum $\pi_\Sigma:\Sigma\rightarrow\mathbb{R}$ as a scalar density
$$
\pi_\Sigma = n_\mu\dfrac{\partial \mathcal{L}}{\partial\partial_\mu \phi}
$$
where $n_\mu$ is the future-pointing normal of $\Sigma$. We have the Euler-Lagrange equations
$$
\frac{\partial\mathcal{L}}{\partial\phi} = \partial_\mu \dfrac{\partial \mathcal{L}}{\partial\partial_\mu \phi}
$$
(note the $\partial_\mu$ is equivalent to using a $\nabla_\mu$ here, because the covariant divergence of a density is the same as the coordinate divergence).
The space of all fields which minimize $W$, call it $\mathscr{V}_W$, is an infinite-dimensional space, but it is not necessarily a Banach space - particularly if the equations of motion aren't linear in $\phi$ (that is, if the Lagrangian is not quadratic). We'll assume that it is quadratic for this simple example, but generally you'd have to view the space $\mathscr{V}_W$ as an infinite-dimensional manifold, and define things like the symplectic form on the tangent Banach spaces of this manifold.
If the manifold satisfies certain hyperbolicity requirements, then knowing the values of $\phi$ and its momentum $\pi_\Sigma$ on a Cauchy surface $\Sigma$ allows us to integrate the equations of motion and determine $\phi$ on the entire manifold. Thus, for a given surface $\Sigma$, $\phi$ and $\pi_\Sigma$ act as coordinates for our infinite manifold $\mathscr{V}_W$.
Now, to have a Poisson bracket, you need a symplectic form $\omega:\mathscr{V}_W\times\mathscr{V}_W\rightarrow\mathbb{R}$. This needs to be nondegenerate and antisymmetric. It turns out that, for any surface $\Sigma$,
$$
\omega[\phi_1,\phi_2] = \int_\Sigma \sqrt{-\frac{G_\Sigma}{g}}(\phi_1 \pi_{\Sigma,2} - \phi_2 \pi_{\Sigma,1})d^3 x
$$
where $G_\Sigma$ is the determinant of the 3-metric on $\Sigma$ and $g$ is the determinant of the 4-metric on $\mathcal{M}$, is such a form. It is independent of the surface $\Sigma$ of integration, which can be seen by using Stokes' Theorem and applying the equations of motion.
Thus, the value of the symplectic form $\omega(\phi_1,\phi_2)$ for two fields is independent of how we choose to foliate.
Now, let's back up to our earlier statement about $\phi$ and $\pi_\Sigma$ being coordinates. This means if I have some real functional $\mathcal{F}[\phi]$, I can define a derivative in terms of those two sets of coordinates. Specifically, at a given field $\phi\in\mathscr{V}_W$, there is a variational derivative $\delta \mathcal{F}_\phi:\mathscr{V}_W\rightarrow\mathbb{R}$ which is a linear map and can be written as
$$
\delta\mathcal{F}_\phi[\psi] = \int_\Sigma \sqrt{-\frac{G_\Sigma}{g}}\left(\frac{\delta \mathcal{F}}{\delta\phi(x)}\psi(x) + \frac{\delta\mathcal{F}}{\delta\pi_\Sigma(x)}\pi_{\Sigma,\psi}(x)\right) d^3 x
$$
In the same way that symplectic forms are combined with derivatives on finite symplectic manifolds to get a Poisson bracket (I won't repeat the whole process here; it's getting late and I need some sleep :-) ), we can now define the Poisson bracket of two functionals $\mathcal{F},\mathcal{G}$ as
$$
\{\mathcal{F},\mathcal{G}\} = \int_\Sigma \sqrt{-\frac{G_\Sigma}{g}}\left(\frac{\delta \mathcal{F}}{\delta\phi(x)}\frac{\delta \mathcal{G}}{\delta\pi_\Sigma(x)} - \frac{\delta\mathcal{F}}{\delta\pi_\Sigma(x)}\frac{\delta \mathcal{G}}{\delta\phi(x)}\right) d^3 x
$$
This, like $\omega$, is independent of the integration surface (which follows directly from $\omega$'s independence).
Okay. So we've done a lot so far without fixing any foliation or coordinates on the manifold. This is nice. To get a strict analog of the Hamiltonian formula, though, we need to choose some foliation. Instead, we can look at a more general example. Let $\xi^\mu$ be some vector field which represents a deformation of the manifold; that is, we are shifting coordinates and transforming the fields as
$$
x^\mu \mapsto x^\mu + \epsilon\xi^\mu
$$
$$
\phi \mapsto \phi - \epsilon\xi^\mu\partial_\mu\phi
$$
where $\epsilon$ is some very small number (the momentum transformation is more complicated, as $\xi^\mu$ might pull surfaces into other surfaces, which means we're transitioning between different momentum functions. Again I will be lazy for now and leave that out). We will furthermore restrict ourselves to fields $\xi^\mu$ that leave the action invariant. Now we can invoke Noether's Theorem!
Define the functional
$$
H_\xi[\phi] = \int_\Sigma \sqrt{-\frac{G_\Sigma}{g}}\left(\pi_\Sigma \delta_\xi \phi - n_\mu \xi^\mu \mathcal{L}\right)d^3 x
$$
where $\delta_\xi \phi = - \xi^\mu \partial_\mu \phi$. This functional is conserved (independent of $\Sigma$) by Noether's theorem. Then it is the case (again, sleepy, no proof) that
$$
\{\phi,H_\xi\} = \xi^\mu \partial_\mu \phi
$$
which is the curved-spacetime analog to your question.
