Can we always find a Hilbert space corresponding to a region of spacetime? In defining entanglement entropy in field theories we take a region A in spacetime. Now if $\mathcal{H}$ is the Hilbert space of the field theory we assume that we can decompose this Hilbert space into 
$$\mathcal{H}=\mathcal{H}_A\otimes \mathcal{H}_{A^{c}}$$
where $H_{A^c}$ refers to the Hilbert space corresponding to the compliment of region A. Now, if we have a pure state $\rho \in \mathcal{H}$ then, we can calculate the Von-Neumann Entanglement Entropy of $\rho_A \in \mathcal{H}_A$. But, it is not obvious to me that we can always decompose $\mathcal{H}$ in the above fashion for any arbitrary region A. Can someone please give some arguments as to why this is true? 
 A: First, to associate Hilbert spaces you should choose a proper time-like foliation of the space-time. Then each space-like hypersurface of the foliation will be described by quantum state in a given Hilbert space. In other words, Hilbert space are given for fixed time always. The time evolution will then act on observables or states of this Hilbert space. 
Now, in local QFTs we have the following condition that can be imposed in local observables: 
$$
\left[\mathcal{O}_i(x),\mathcal{O}_j(y)\right]=0
$$
for $x-y$ space-like. Remember that local observables are observables that behaves properly under Poincare transformation: 
$$
U(\Lambda, b)\mathcal{O}_i(x)U(\Lambda, b)^{\dagger}=R_i\,^j\,\mathcal{O}_j(\Lambda x+b)
$$
where $R$ is a representation of the Lorentz symmetry. This is all you need to construct Hilbert spaces for each point $x$ of the space-like hypersurface: For each point in space $x$ we construct the Hilbert space $\mathcal{H}_x$ associated to a complete set of local observables at point $x$, namely $\{\mathcal{O}_{i}(x),...\}$.
To construct a Hilbert space of a given region $A$ of space, just make the tensor product of the Hilbert spaces $\mathcal{H}_x$'s for $x\in A$:
$$
\mathcal{H}_{A}=\bigotimes_{x\in A}\mathcal{H}_x
$$
What makes local QFT so special is that, in principle, all the observables in that theory could be constructed out of local fields $\phi(x)$, $\psi_{\alpha}(x)$, $A_{\mu}(x)$ and etc. So this means that the Hilbert space of the theory is given by 
$$
\mathcal{H}=\bigotimes_{\forall x} \mathcal{H}_x
$$
This means that all the states of the theory is described by the superpositions you can make with theses copies $\mathcal{H}_x$. Eigenstates of non-local observables like the energy $H$ will be entangled states, e.g. the vacuum state $|0\rangle$. This is why accelerated observers sees the Unruh effect. Accelerated observer just have access to a region $A$ of the whole space $A\cup A^{c}$ due to the Rindler horizon, so you get a description:
$$
\rho_A= tr_{A^{c}} \{|0\rangle\langle 0|\}
$$
that is clearly a mixed state. Actually if you do all the computations right you get that for accelerated observer at acceleration $2\pi/\beta$ separate the vacuum state as: 
$$
|0\rangle=\sum_{i}e^{-(1/2)\beta E_i}|E_i\rangle _A \otimes|E_i\rangle _{A^c}
$$
where $E_i$ is the eigenvalue for the energy $H_{A/A^{c}}$ restricted just to the region $A$ or $A^{c}$. Doing the trace of the region behind the Rindler horizon you will get that the density matrix is:
$$
\rho_A=\sum_{i}e^{-\beta E_i}|E_i\rangle \langle E_i|_A
$$
a density matrix of a canonical ensemble at temperature $\beta^{-1}$.
Note that all this is based on the fact that things like $\bigotimes_{x}\mathcal{H}_x$ makes sense. To make sense of it we need to regularize and renormalize it. As David M point out, this may be not possible. A obvious way to see if this is possible is search for a lattice regulator, then we simple make the tensor product of each cell. 
A: Surprisingly, the answer is no, we can not always find such a decomposition. I'm not an expert on this subject, but I can point you to the following seminar given by Simeon Hellerman: http://pirsa.org/displayFlash.php?id=17110138.
A caveat is this seminar is based on work in progress, so this issue may not be completely settled, but it is an interesting question.
As he points out (and this is related to the other answers presented here), if the continuum quantum field theory has a lattice regularization, then this tensor factorization must exist. However, for 2d CFTs with unequal central charges $c_{L}\neq c_{R}$ a lattice regularization does not exist and its no longer obvious if entanglement entropy does as well. There are also gravitational anomalies in higher dimensions (see e.g. http://inspirehep.net/record/192309?ln=en) but I'm not aware if the implications of these anomalies on entanglement entropy has been similarly worked out.
Edit: For more details see the paper https://arxiv.org/abs/2101.03320
A: This actually follows from the axioms of mathematical QFT. Given two manifolds (possibly with boundaries), then the Hilbert space corresponding to the disjoint union of the two manifolds is given by the tensor product of the Hilbert spaces on the individual manifolds. This corresponds exactly to this decomposition.
