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?
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.
