# Visualising the Hamiltonian constraint in inhomogeneous LQC

1. Jul 14, 2012

### julian

In this paper called "Stepping out of Homogeneity in Loop quantum Cosmology" - http://arxiv.org/pdf/0805.4585.pdf. On page 4 they say "where the sum is over the couples of distinct faces at each tetrahedron, $U_{ff'} = U_f U_{f_1} U_{f_2} \dots U^{-1}_{f'}$ where $l_{ff'} = \{ f , f_1; f_2; \dots f^{\prime -1} \}$ is the link of the oriented faces around the edge where $f$ and $f'$ join,".

Could someone clarify this for me? Possibly with diagrams?

Last edited: Jul 14, 2012
2. Jul 15, 2012

### francesca

Suppose you have a triangulated 3d-manifold and you want the curvature on this manifold. The triangulation is made by several tetrahedra: pick one of them, consider it on its own, and consider one of its bone (an edge). The curvature is obtained by performing the parallel transport (a Wilson loop) of the Ashtekar connection around the edge. So, if you start from the center of a tetrahedron, first you go from there to a face, you puncture it, and you come back to the starting point from another face by puncturing it. If you have a complicated triangulation, there would be some other tetrahedra attached to the same edge, sharing each one a face with the adjacent one: in this case, in order to come back, you would puncture more faces.
All the information about the curvature is stored in the faces. In principle, you can go from the center of the tetrahedron to the face, grasp the information about the curvature, and come back without puncturing the face: here we don't want to do this, so we take couple of distinct faces.
To get the full Hamiltonian constraint, you have to repeat this for each edge and sum everything.

I hope this could answer to your question. This construction follows the very first Hamiltonian constraint proposed by Rovelli/Smolin in the good old days.
If you need more details, maybe you can try this paper
http://arxiv.org/abs/1110.3020
that explain the same model with an up-to-date prospective and some developments.
Cheers,
Frances