In LQG we choose a two-complex (or other) on which to discretise GR before quantising. I'm a little uncertain about the nature of this choice. Is the two-complex an a priori property of the universe (i.e. on the same standing as the spacetime manifold)? Or is it simply a "lens" with which we view and regularise our theory (analogous to the scale in QFT renormalisation), independent of the overall dynamics?


In LQG we often talk about how space takes on a discrete character, but the truth is somewhat more subtle. Space in LQG is not a lattice, but the sum over all lattices which respect our boundary conditions (warning! see Edit 2 below). This is how we implement diffeomorphism invariance in the theory in a background independent way.

In many respects, a certain 'lattice' can be thought of much like a Feynman diagram. It is a picture that captures one configuration the system can evolve through but if you want to actually calculate anything you must sum over all the diagrams.

If your interested, in modern LQG we usually work in a basis where these lattices are described by something called spin networks which are oriented graphs where the links only meet at their end points which we call nodes. $$ $$ If you're wondering in what way the discreetness ends up being manifest in the theory, by looking at the flux through a surface one can define a well behaved operator which acts on a spin network state $|S\rangle$ to give it's area. This operator is simply called the area operator and has the following eigenvalue spectrum

$$\hat{A}_S|S\rangle = 8\pi \ell_p^2 \beta \sum_{p \in(\Gamma \cap S)} \sqrt{j_p(j_p+1)}|S\rangle$$

Where $\ell_p$ is the Plank length, $\beta$ is a free parameter in LQG, $\Gamma$ is the set of all of our spin networks and $j_p$ are spin values which are summed over. This is all highly analogous to the z-component of the angular momentum operator from regular quantum mechanics and indeed there is also a volume operator which is analogous to the length squared angular momentum operator from QM, though I won't go into any detail on it as I am not well acquainted with it.


If you would like a better explain action of spin networks here is the place to go: What is the role spin-networks play in Loop Quantum Gravity?

Edit 2

Reading over this again I see that I was a bit imprecise in my language and this probably caused the confusion. I claimed that space is the sum over all lattices but really, it's more precise for me to say that space is a lattice, but which lattice you are on is determined quantum mechanically. In the canonical theory, this is determined by which eigenvalues of the area and volume operators you find when you measure space.

In the covariant theory this is implemented by writing the transition amplitude as the sum over all geometries that agree with our initial and final state.

  • $\begingroup$ I am familiar with spin networks. Do you think you could expand further on the sum over lattices? I have been reading [1], and it makes no mention of such a sum; instead it seems to indicate that we fix the triangulation. $\endgroup$ Mar 9 '16 at 13:42
  • $\begingroup$ [1] Carlo Rovelli and Francesca Vidotto. Covariant loop quantum gravity : an elementary introduction to quantum gravity and spinfoam theory. Cambridge University Press, Cambridge, 2015. $\endgroup$ Mar 9 '16 at 13:42
  • $\begingroup$ Additionally it indicates that once we move to a Turaev-Viro theory, amplitudes are independent of the triangulation. I'm not sure what this might suggest -- perhaps that spacetime is something like an "equivalence class" of triangulations? $\endgroup$ Mar 9 '16 at 13:48
  • $\begingroup$ Hi Josh, Rovelli and Vidotto's book is a book on the covariant theory, not the canonical one which I assumed you were working in. In the covariant theory, the sum over geometries happens in the partition function $Z$ and the corresponding transition amplitudes $W$. Check out page 106 If your book to see this sum in action. $\endgroup$
    – Mason
    Mar 9 '16 at 17:05
  • $\begingroup$ Sorry, perhaps I should have specified. I assume you mean the expression for $W_\Delta$. It still seems to me that although we may be summing over all possible configurations (i.e. assignments of lengths to edges etc.) of a triangulation, the triangulation $\Delta$ itself is still a fixed quantity... $\endgroup$ Mar 9 '16 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.