Is boundary well defined if variation of metric don't vanish on the boundary? Suppose that you want to calculate the variation $\delta S$ of an action induced by some arbitrary variation $\delta g_{\mu \nu}$ of the spacetime metric :
\begin{equation}
S = \int_{\Omega} \mathscr{L} \, \sqrt{- g} \; d^D x.
\end{equation}
The domain of integration $\Omega$ is any finite part of spacetime, and $\partial \, \Omega$ is its boundary.  Usually, $\Omega$ and $\partial \, \Omega$ are fixed during the variation and we're asking that $\delta g_{\mu \nu} = 0$ on $\partial \, \Omega$.
If $\delta g_{\mu \nu} \ne 0$ on $\partial \, \Omega$, does it make sense to talk about a fixed boundary in General Relativity ?
If the metric isn't fixed, I'm not sure it make any sense to talk about a given region of spacetime.
I need opinions and/or references on this.
EDIT : Take note that the question is not about any boundary conditions imposed on the hypersurface $\partial \, \Omega$.  Whatever what are the boundary conditions, if you have $\delta g_{\mu \nu} \ne 0$ on $\partial \, \Omega$ (maybe it is the momentum that is fixed, or whatever else), can we still define the boundary ?
I have the feel that the coordinates and boundary are dependent on the metric, in general relativity, but I'm not sure. 
 A: The boundary of a subset of a topological space is abstractly defined as the set-theoretic difference between its closure and its interior. Since topological spaces in general have neither coordinates nor metrics, this notion is independent of the metric.
Since the spacetime manifold is a manifold, it is a topological space (locally homeomorphic to $\mathbb{R}^n$) even when not equipped with a metric (and a Lorentzian (pseudo-)metric tensor does not induce a metric topology, contrary to the Riemannian case). Therefore, the notation of a boundary of a subset of $M$ is completely independent of any choice of coordinates or metric, since it purely relies on the topology of the manifold.
Similarly, integration on the manifold does not actually rely on a choice of coordinates, it is abstractly defined by the integration of differential forms over chains. Those objects are almost always used in some particular coordinate system by physicists, but the differential geometric objects described by them are coordinate-independent: Every choice of coordinates will yield the same result for invariant quantities such as the value of an integral. 
In this specific case, the Lagrangian is just a function of fields, which are in turn functions of the manifold, and if it is properly coordinate-invariant (which it should be!), then multiplying it by the volume form provided by the metric and integrating the entire thing over a spacetime volume is coordinate-invariant, too.
A: There is no problem with saying that we have a region with boundary as far as the underlying manifold goes. The problem is that you don't know the metric on the boundary or how to integrate. You should rephrase your question accordingly and in that case yes you need to be careful what boundary conditions you take as already said.
A: We have the following action:
$$
\begin{equation}
S = \int_{\Omega} \mathscr{L} \, \sqrt{- g} \; d^D x
\end{equation}
$$
Let's apply a local transformation on $g_{ab}$ and $x^{\mu}$, only in the boundary $\partial \Omega$.
$$\delta g_{ab}=Y_{ab}(...)\epsilon (x)$$
$$\delta x^{\mu}=X^{\mu}(...)\epsilon (x)$$
Where $X^{\mu}$ and $Y_{ab}$ are functions of the fields, metric and space. With a little bit of work (I'm letting as an exercise) you get this variation on the boundary:
$$
\begin{equation}
\delta S = \int_{\partial \Omega} \left[\mathscr{L} \, \sqrt{- g} \;\delta x^{\mu}\;+\;\frac{\partial \left(\mathscr{L} \, \sqrt{- g} \right)}{\partial \left(\partial _{\mu} g_{ab}\right)}\delta_0 g_{ab} \right]  \; ds_{\mu}
\end{equation}
$$
Here $\delta _0$ means vertical variations: $\delta_{0}=\delta - \delta x^{\mu}\partial_{\mu}$. Note that $\delta x_{\mu}$ contain is variation over the manifold (here only in the boundary since we are integrating only in the boundary). Fixing the boundary is simply taking $X^{\mu}=0$. Then: 
$$
\begin{equation}
\delta S = \int_{\partial \Omega} \frac{\partial \left(\mathscr{L} \, \sqrt{- g} \right)}{\partial \left(\partial _{\mu} g_{ab}\right)}\delta g_{ab}\; ds_{\mu}
\end{equation}
$$
So, you have boundary terms of the variation of the metric only if the lagrangian has some $\partial_{\mu}g_{ab}$ term. Then, if you have some dynamical role for the metric you may have this boundary term that are related with things getting "in and out" of the region of space-time mapped by $\Omega$, or at least a topological counter one. Note that $\Omega$ here is simple a set of events and don't have any geometrical notion.
A: It does not make sense to have $\delta g_{\mu \nu}\neq 0$. Actually, for finite region, even if the variation of metric is zero at boundary, the derivative of metric is not zero, it will contribute a surface term to the variation of actio,  thus we have to add another surface term to cancel this contribution, which consists of exterior curvature of the boundary. The term originally comes from Hawking, Carters' paper on black hole thermodynamics. 
A: The question is why you would want to do this in the first place. The equations of motion that you obtain one a compact region still aren't the real equations of motion, since the compact region is a mathematical choice to simplify the formulation of the problem. It is implicitly understood that the true equations of motion are only obtained on the limit of this region becoming very large or infinite
