Hilbert spaces in LQG

[meteor]

Well, you all know that LQG has different kinds of Hilbert spaces (4 in the sake of truth). You start with the kinematical Hilbert space that is the vector space of all possible quantum states of spacetime. However, all these spacetimes are not physically real, not all of them make sense. Then you have three constraints that "select" all the spacetimes that are physically possible. First you apply the Gauss constraint to the Kinematical Hilbert Space to obtain the gauge invariant Hilbert space, to it you apply the diffeomeorphism constraint to obtain the diffeomorphism invariant Hilbert space, and to it you apply the Hamiltonian constraint to obtain the physical Hilbert space, this is the "correct" space of quantum states of spacetime. Then you know that the kinematical Hilbert space is an infinite dimensional vector space, but my doubt is if the physical Hilbert space is also infinite dimensional. It should seem that not, since the number of quantum states is less than in the kinematical Hilbert space, but given my problems to understand the concept of infinity (you know the set of natural numbers has the same cardinality that the set of rational numbers even if it seems paradoxical) I'm not sure. So the question is, is the physical Hilbert space infinite dimensional, and if not, what's its dimension?

[marcus]

Meteor, you asked about the dimension of the physical Hilbert space of Loop Gravity. It is infinite.
I should be able to give you a page reference in Rovelli's book
tomorrow, or some more satisfactory response. Tonight I'm already falling asleep and cant think clearly.

[jeff]

Originally posted by meteor
Well, you all know that LQG has different kinds of Hilbert spaces (4 in the sake of truth). You start with the kinematical Hilbert space that is the vector space of all possible quantum states of spacetime. However, all these spacetimes are not physically real, not all of them make sense. Then you have three constraints that "select" all the spacetimes that are physically possible. First you apply the Gauss constraint to the Kinematical Hilbert Space to obtain the gauge invariant Hilbert space, to it you apply the diffeomeorphism constraint to obtain the diffeomorphism invariant Hilbert space, and to it you apply the Hamiltonian constraint to obtain the physical Hilbert space, this is the "correct" space of quantum states of spacetime.

Firstly, the kinematical space of states by definition consists of the states that satisfy both the gauss and diffeomorphism constraints, but not necessarily the hamiltonian constraint. Secondly, these aren't states of spacetime, but if LQG had turned out to be correct, they woud've been states of the gravitational field from which spacetime would've been recovered in the classical limit.

Originally posted by meteor
...is the physical Hilbert space infinite dimensional, and if not, what's its dimension?

Both the kinematical and physical space of states are infinite dimensional.

[meteor]

marcus and jeff, thanks for the response

Firstly, the kinematical space of states by definition consists of the states that satisfy both the gauss and diffeomorphism constraints, but not necessarily the hamiltonian constraint

But that's not what says this post of J.Baez. He says that the kinematical Hilbert space doesn't include the Gauss constraint and the diffeomorphism constraint
http://www.lns.cornell.edu/spr/1999-05/msg0016258.html

[jeff]

Originally posted by meteor
[B]...Baez...says that the kinematical Hilbert space doesn't include the Gauss constraint...

Consider the following excerpt from the link you gave:

...we start with a big "kinematical" Hilbert space...Then we impose the Gauss constraint and get the "gauge-invariant" Hilbert space...

By "gauge-invariant Hilbert space" baez meant the gauge-invariant part of the initially "big" kinematical Hilbert space, etc.

Now consider this excerpt from the same link:

And finally comes the Hamiltonian constraint...Solutions to this will form the "physical Hilbert space".