In classical gravity, a state may be described by the combination of the 3 elements:(adsbygoogle = window.adsbygoogle || []).push({});

(1) An n-dimensional manifold M

(2) The specification of which subbundle F of the linear frame bundle LM is the orthonormal frame bundle with signature (n-1) +, (1) -.

(3) A GL(n) connection A.

One can then prove that for a fixed F, the connection A reduces to a SO(n-1,1) connection W.

In the quantization of a classical system (i.e., in a quantum systerm that has said classical system as its h-bar -> 0 limit), the state space may always be taken as the same, but now with the addition of "transition probabilities". The limit, itself, may be described as the limit of the transition probability T(z1,z2) to the Kroenecker delta delta(z1,z2) for any 2 states z1, z2, as h-bar -> 0. This mode of quantization -- where the state space is quantized rather than the operator algebra -- leads to what are called coherent states. Thus, if z is a classical state then W_z is the corresponding coherent state, and Tr(W_{z1} W_{z2}) = T(z1,z2).

This mode of description also transcends the distinction between classical vs. quantum. Both pure classical and pure quantum systems can be described under this same umbrella, as well as everything in between.

A sector is then any subset of states that have 0 transition probability with the rest of the state space. The total state space decomposes into an orthogonal sum of sectors, called "coherent subspaces". The extreme cases are

* pure quantum system -- only one coherent subspace

* pure classical system -- one sector, essentially, for each state (i.e. every 2 states have 0 transition probability between them)

* hybrid -- each sector describing a "superselection mode" or "coherent subspace". The parameters that index the sectors then comprise the classical variables of the system.

The question, then, naturally avails itself, no matter what formalism is used to attempt to quantize gravity: what are the coherent states? More precisely, what is the coherent state W_{M,F,A} corresponding to the classical state (M,F,A)?

How many sectors?

In particular, how would one describe W_{M,F,A} and W_{M,F',A'} for 2 DISTINCT frame subbundles F and F' of TM? Going by the Unruh-Davies effect, one has that the state spaces of two frames lie in distinct coherent subspaces when the frames are mutually accelerating. Here, that would mean that F parameterizes between DIFFERENT sectors.

That is: F must be a CLASSICAL variable.

The same goes for the M parameter.

This brings the issue of background back to the foreground, big time (pun intended). It's one thing to make F part of the "dynamic" background (i.e., to make F, itself, subject to the overall system's dynamics), but it's an entirely different thing to make it part of the *non-quantum* background (i.e., an external or classical mode). However, there is a marked tendency in LQG and amongst those who work in LQG to confuse the two.

Just because you're quantizing the connection does not mean that either F or M are brought under the umbrella of the whole endeavor. There is nothing that says that M ought to be anything but an ordinary manifold. It means two completely different things to quantize a geometry (as in the manifold M, itself), versus to quantize structures (like A or even F) that are sitting ON TOP of the classical geometry M.

But trying to bring M (or even F) under the programme, you're biting off more than you can chew. The issue with the coherent states W_{M, F, A}, W_{M, F', A'} when F and F' are different, already shows that.

Rovelli has had a tendency to think he could evade these issues by going into denial about M. But that simply doesn't cut it. No matter how the theory is formulated, no matter whether it be classical, quantum or a combination of the two, there will somewhere down the line be SOME definition of the coherent states W_{M, F, A}. And it's at this point that the issue of the classical geometry (along with all the issues raised here) returns.

So, there is no denying M. Even if it's "not there", M still cannot be evaded.

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