Degrees of freedom in General Relativity and well-posedness of the EFE I would like to understand what are the degrees of freedom in GR. I have read a few  previous posts already, but none of them really help me. Below, I will try to write down the entangled web of thoughts I have/know and hopefully someone we help me reach epiphany.
Let us consider the vacuum Einstein's field equation (EFE). It has 10 independent equations (naively I guess...). The metric itself has 10 independent components (naively?) so all seems good. But now, trouble starts...
1) First, due to general covariance, 4 components of the metric are really redundant. That is, if some $g_{\mu \nu}$ solve the EFE, then $g_{\mu \nu} \rightarrow g'_{\mu \nu}$ will also solve it. So, actually, the EFE under-determine the metric!? There is a first confusion here. To me, it is a bit like saying: here we have a system of 10 polynomial equations with 10 unknowns, but the system is nonetheless under-determined.
I mean, if we look at say some standard PDE's like the wave equation, for appropriate initial data, we have a unique solution.
Are we then saying that even given appropriator initial data, the EFE are still under-determined?? Even though the number of unknowns and equations match? Or are we strictly speaking about the physical "interpretation" of the solution? In the sense that, given a solution, say in cartesian coordinates, we could translate it into polar coordinates, which would not "really" change anything, only the individual numerical values, but the description of the solution is still the same? In this case, cant we say that the wave equation is then also under-determined since we could play that same game here as well?
2)Ok, now let us say I am fine with the above. We end up with 6 "really" independent components of the metric, for still 10 equations.
But, now, somehow, 4 equation of EFE are actually redundant because of the bianchi identity. I dont see how this makes sense. The Bianchi identity is,well, an identity. So it always holds, and therefore in my eyes it cannot provide with any additional constraints. More directly, if we start if $G_{\mu \nu} = 0$, which ones should we remove? and how do we recover me? (which we should be able to since they are supposedly redundant). For example, if we choose to "forget" about the first one, i.e. $G_{00}=0$, how can we recover it? From the Bianchi, we have: $\nabla_t G^{00} + \nabla_i G^{i0} = 0$, but that is not enough (even with all the other EFE) to deduce $G_{00} = 0$? In fact, with the EFE, we would have $\nabla_t G^{00} = 0$, but we can't get any further?
3)Now, even at this stage, I dont see how we get to 2 degree of freedom. I usually hear like diffeomorphism invariance removes 4 dof (as above), but Bianchi removes another 4?? However, Bianchi "at best" removes redundant EFE and does not "touch" the metric like diffeomorphism invariance does?? 
4) Finally, this was for the vaccum equation. Now, if we add a non-trivial stress-energy tensor, things get worse, since this looks like one is "adding" even more unknowns to the EFE. Even with Equations of state, one typically introduces more unknowns (for example for perfect fluid, pressure and 4-velocity). How does this work out? I hear that you introduce the continuity equation $\nabla_\mu T^{\mu \nu}=0$, but this is a direct consequence of the EFE, and therefore are not additional independent equations! So we still have only 10 equations!?
Thank you if you managed to read all this and thanks in advance for your responses.
 A: This is cleaner in the ADM formulation than it is directly from the Einstein equation.  
If you're not familiar with the ADM formulation, it is done by converting the Hilbert Lagrangian to a constrained (and vanishing) hamiltonian, and getting Hamilton's equations out of it.
https://en.wikipedia.org/wiki/ADM_formalism
If you do this, you will get four "gauge" degrees of freedom which derive from the coordinate system of the underlying spacetime.  Associated with these degrees of freedom, you will get four constraints.  What will be left over will be twelve first-order equations in the 3-metric and its first derivative.  Since we usually talk about degrees of freedom in terms of second-ordere equations, and you can always convert a system of equations of the form:
$$\begin{align}
{\dot q_{i}} &= f_{i}(q,p)\\
{\dot p_{i}} &= g_{i}(q,p)
\end{align}$$
into a set of equations of the form:
$${\ddot q_{i}} = h_{i}(q, {\dot q})$$
Then, given a fixed choice for your coordinate system (which is equivalent to choosing a fixed gauge in E&M), you have six second-order equations for the 3-metric, and four constraints, leaving you with two local degrees of freedom.
