Diff(M) and requirements on GR observables This question is kind of inspired in this one:
Diff(M) as a gauge group and local observables in theories with gravity
The conundrum i'm trying to understand is how is derived the (quite) extraordinary statement that in GR there are no local observables. I just want to stress that this is indeed an extraordinarily counter-intuitive assertion (with extraordinarily dramatic consequences for any compatible theory of quantum reality), and it deserves an extraordinarily robust explanation. 
Among the statements i see are mostly involved in this argument are:


*

*It is usually argued that Diff(M) is a gauge transformation. This point i don't have any sort of issue; the atlas of reference frames we use to describe space-time are indeed a human convention and physical laws should not depend on such conventions

*A physical observable should be invariant under any gauge transformation. whaaa? i mean, this is clearly preposterous non-sense; actually i think this is more dangerous than simple non-sense, is just circular self-justification. This is actually saying that observables should be scalars?
I can provide you a proof that this is non-sense by saying that the momentum of a particle is an observable in my reference-frame, and in other Diff(M) gauge (that is, another reference frame) i will see a different momentum of the same observable. All eigenvalues and eigenvectors transform according to the vector representation of the Poincare group. You are now of course free of dismiss such 'proof' as non-sense because my assumption that the momentum of a local particle is an observable is non-sense, but then, how are you sustaining that part of the argument, without actually saying again that Diff(M) is a gauge transformation? how do you avoid the circularity in this argument?
U(1) gauge is a bad example because this is an internal symmetry; Diff(M) are space-time symmetries and i don't think that what is true in there (A eletromagnetic vector potential being unobservable, but B and E being observable, hence observables are invariant under U(1) )

*It is usually argued that observables in GR formally exists in the asymptotic boundary of the space-time. Is there an argument for this that is unrelated/independent to the previous point?
 A: Instead of re-inventing the wheel, here is a bunch of references:
Here is something I wrote back when I was blogging, the discussion was kind of fun. 
There is also a pretty good explanation in this paper by Giddings- Hartle-Marolf on how this statement doesn't contradict all kind of semi-intuitive ideas people have  (and started commenting on here already) about local observables. 
Finally, this paper by Arkani-Hamed and collaborators has very nice introduction explaining why dynamical gravity, and not just Diff invariance, is what prevents the existence of local observables.
A: Who says that there are no local observables?  There are two local degrees of freedom in General Relativity, related to the two polarizaiton modes of local gravitational waves.  And they are locally observable--hence the LIGO experiment.
Now, general relativity in 2+1 dimensions turns out to be topological--all the degrees of freedom of the theory are determined by the boundary conditions and the matter equation of state.  But that's certainly not true of the 3+1 dimensional world that we inhabit.
