martinbn
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I might be wrong, but this is not how I read Rovelli. For him the graphs are completely abstract combinatorial objects without an embedding in an apriori manifold. The space-time is a consequence. Even if you start with a manifold, representing space at a time or space-time, and consider graphs on them shouldn't you identify ones obtained after a diffeomorphisms? In other words for the state space the ##\gamma## and ##\phi(\gamma)## should be in the same equivalence class.Nullstein said:Yes, of course I'm thinking of non-trivial group actions. But more specifically, the action of the diffeomorphism group in this formalism is given by ##(U_\phi\psi_\gamma)(\vec g) =\psi_{\phi(\gamma)}(\vec g)##, where ##\phi## is a diffeomorphism. And unless ##\phi=\mathrm{id}##, this action has the peculiar feature that ##U_\phi\psi_\gamma## is orthogonal to ##\psi_\gamma##.
A spacetime is a manifold equipped with a metric. The graphs are embedded into the spacetime manifold. The metric becomes a quantum field instead of a classical field. (LQG gives meaning only to coarse functions of the metric, smeared along lower dimensional submanifolds.) The points that don't lie on the graph are still there in the spacetime manifold and diffeomorphisms still act on the whole manifold. Such points just don't carry any geometry due to the peculiar nature of the theory. Now the hope is to construct states such that e.g. ##\left<\hat g_{\mu\nu}(x)\right>## (or at least the smeared versions of it) resembles classical solutions to the EFE with some quantum corrections. You can then ask for example if there are diffeomorphisms ##\phi## such that something like ##\left<(\phi^*\hat g)_{\mu\nu}(x)\right> = \left<\hat g_{\mu\nu}(x)\right>## holds. (It is not clear what properties really are desirable, but this is one reasonable thing one could ask for). Then you could call ##\phi## a quantum spacetime isometry.