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Can someone answer this question, I have chosen this from Bodendorfer's article on 'An Elementary Introduction to Loop Quantum Gravity' from section 3 General relativity in the connection formulation and quantum kinematics exercise 3.6.10

Ashtekar-Lewandowski Vaccum

A Cylindrical Function Ψ = 1, this state is called Ashtekar-Lewandowski Vacuum. Show that it corresponds to a maximally degenerate Spatial Geometry by evaluating the Vacuum expectation value of the Flux Operator

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Well, the answer is in the question – you need to evaluate the flux operator!

It's not that hard to see actually. Draw an arbitrary auxiliary graph, and note that the Ashtekar-Lewandowski vacuum state corresponds to a spin network on this graph with all spins equal to zero (because the spin-0 representation of $SU(2)$ is the trivial representation that assigns $1$ to each of its elements, which leads exactly to the state $\Psi[A] = 1$).

Now remember that area in LQG is proportional to $$ A \sim \sum \sqrt{j(j+1)}. $$

Substitute $j = 0$ and you will get $A = 0$ – all areas vanish for such a state. I believe that's what "maximally degenerate spatial geometry" means.

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