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To make it simple, take the spin foam formalism of ($SU(2)$) 3D gravity. My question is about the choice of the data that will replace the (smoothly defined) fields $e$ (the triad) and $\omega$ (the connection) on the disretized version of space-time $\mathcal{M}$: the 2-complex $\Delta$: why choose to replace $e$ by the assignment of elemnts $e\in su(2)$ to each 1-cell of $\Delta$, and elements $g_{e}\in SU(2)$ to each edge in the dual 2-complex $\mathcal{J}_{\Delta}$? I mean, these are both $su(2)$-valued 1-forms, thus, roughly speaking, assigning elements of $su(2)$ to vectors of the tangent bundle $\mathcal{TM}$, in other terms assigning elements of $su(2)$ to infinitesimal displacements represented by the 1-cells of $\Delta$. I can understand the choise for $e$, but not for $\omega$, why this is so? Why it is not the inverse choice?

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Note: this question is cross-posted to MathOverflow,… – Pieter Naaijkens Dec 2 '11 at 10:57

Well, I'll take a shot at this - most of what I'm going to say comes from the text of Thiemann or Rovelli. The choice of $g_e\in SU(2)$ is connected to the holonomy, which we choose to use because it is a gauge-invariant. In fact, it's probably more pedagogical to say "the loops come from the holonomy $g_e\in SU(2)$, which we know can be made to be gauge-invariant Wilson loops" or something. Now as far as the choice for the 2-complex, we want something related to the triads so that the original Poisson algebra between the connection $A$ and fields $E$ is preserved. It actually turns out that $E$ is an $SU(2)$-valued vector density - dual to a 2-form represented by the 2-complex. In other words, we don't have two 1-forms, but rather a 1-form and an $SU(2)$-valued vector.

So that's a short answer coming mostly from Thiemann's excellent text. This issue is taken up in depth in the third of the Ashtekar series of papers:

Ashtekar, Corichi, Zapata (1998), Quantum Theory of Geometry: III. Non-commutativity of Riemannian structures, Class. Quan. Grav. 15, 2955-2972.

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