Superspace as the Hilbert Space for Quantum Gravity This is a question I've asked in physics.stackexchange, but have obtained no answers:
Let $\mathcal{A}$ be the Ashtekar connection. Since $^{(3)}g_{AB}=i\frac{\delta}{\delta\mathcal{A}^{AB}}$ (see R. Penrose, 2004: Road to Reality. Vintage Books, 1136 pp.), the Ashtekar connection, in a way, represents the graviton. Let $\mathcal{Riem}$ denote the configuration space of GR. 
If one managed to quantize the metric to yield the graviton, it would make sense to say that the $\mathcal{Riem}$ is also quantized, i.e., Wheeler's superspace becomes a Hilbert space with elements as gravitons. There is an complex inner product defined between the ``de-projections'' (the inverse projection mapping) of gravitons $\mathcal{A,A}_1$, defined as $\langle\mathcal{A}|\mathcal{A}_1\rangle$. An algebra of self-adjoint unbounded operators (of which the Hamiltonian is an element) on this quantum superspace corresponds to a set of all observables. The set of all possible states corresponds to the projective superspace.
Thus, is it possible to interpret the superspace as a Hilbert space (as mentioned above)and provide a rigorous mathematical formulation of quantum gravity?
 A: I started writing this as comments, but then ran out of space to give a satisfactory reply. (I rant too much to be confined to 400 characters!)
At any rate, the answer is: no.
But some clarification is necessary.
I'm old and forgetful, so I will note John Baez describes the basic algorithm to geometric quantization fairly well.
Examples Worth Considering
Harmonic Oscillator
A fascinating discussion worth examining, rich enough to actually do several different geometric quantizations. See, e.g., Adrian Lim's A Non-standard Geometric Quantization of the Harmonic Oscillator. The symplectic manifold $\mathcal{M}$ isn't used, but the square-integrable sections of the complex line bundle is a natural Hilbert space...so that's what's used.
Yang-Mills Theory
With Yang-Mills theory, we have the configuration space give us the Hilbert space structure quite naturally. This is a quirky feature unique to Yang-Mills theories, because of the differential geometric structure underpinning the space of connections has a "naturally obvious" Hilbert structure we may equip it. For review papers on this, see:


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*M. Asorey and P.K. Mitter, "Regularized, continuum Yang-Mills process and Feynman-Kac functional integral". Commun.Math.Phys. 80 43-58 (1981) eprint.

*M. Asorey and F. Falceto, "Geometric regularization of gauge theories". Nucl Phys. B327 427-460 (1989) eprint.


You say you are interested in quantum gravity, so let me warn you problems with this approach if you try it with gravity (e.g., the Ashtekar formalism):


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*There are relatively few interesting examples of gauge-invariant measures for diffeomorphism invariant theories, which is bad news for GR.

*For equipping the Hilbert structure atop the configuration space in Yang-Mills theories, you need to fix spacetime: this spells disaster for attempting the same recipe in GR.


Alas, there are too few free lunches (or even costly meals) in quantum gravity.
Remark. We should note that we took the space of connections, then equipped it with some extra structure. Rigorously speaking, this produces a new distinct object whose underlying "stuff" was the space we began with. Physicists blur this distinction for Yang-Mills, but really it's two different objects. (End of Remark)
Closing Thoughts
With regards to your motivations to the question, regarding ${}^{(3)}\!g_{AB}=i\delta/\delta A^{AB}$, try not to take Penrose as the gospel here. (Caveat: I only have his fourth printing from August 2005, so he may have revised his material on Ashtekar variables considerably.) Really, when he says "components of the [inverse] metric", he means the densitized triad.
(Oddly enough, wikipedia has a surprisingly decent article on the Ashtekar formalism)
Finally, I hope the peculiarity of electronic communication hasn't made me sound angry or negative --- something that happens all too easily. Frankly this is an intriguing question, and I hope I haven't inadvertently discouraged you any with my answer :)
