Triangulating Hamiltonian Constraint in LQG

[AfonsoDeAlbuquerque]

TL;DR Summary

Transition from classical Euclidean Hamiltonian constraint to the triangulated one in LQG

Im trying to obtain regularized (and triangulated) version of Hamiltonian constraint in the LQG. However, one step remains unclear to me.
I am starting with the Euclidean Hamiltonian:$$H_E=\frac{2}{\kappa} \int_\Sigma d^3 x N(x)\epsilon^{abc} \text{Tr}(F_{ab},\{A_c,V\})
$$

Now i have to introduce the triangulation $T$ of $\Sigma$ into tetrahedra $\Delta$ with the following "setup":

i) $v(\Delta)$ denotes vertex of the $\Delta$,

ii)$s_I(\Delta)$ are three edges meeting in $v(\Delta)$ ($I=1,2,3$),

iii) $\alpha_{IJ}(\Delta)=s_I(\Delta) \circ a_{IJ}(\Delta)\circ s_J(\Delta)^{-1}$ denotes loop based at $v(\Delta)$

iv) $a_{IJ}(\Delta)$ is the 4th edge of $\Delta$, connecting endpoints of $s_I$ and $s_J$ distinct from $v(\Delta)$.

Using the above prescription, triangulated Hamiltonian takes form (https://arxiv.org/abs/gr-qc/9606089):

$$H_E^T= \sum_{\Delta \in T}H_E^\Delta;\;\;\;\; H_E^\Delta:= \frac{-2}{3}N_v \epsilon^{IJK} Tr(h_{\alpha_{IJ}} h_{s_K}\{h^{-1}_{s_K}(\Delta),V\}),$$

which upon shrinking to the point $\Delta \rightarrow v$ reproduces the classical expression

In order to obtain this, I have to use holonomies ($\dot{s}^a_I$ is vector tangent to the segment ):$$h_{s_I}=1-\epsilon \dot{s}^a_I A^i_a \tau_i + h.c,\;\;\;\;\; h_{\alpha_{IJ}}=1-\frac{\epsilon^2}{2} \dot{s}^a_I \dot{s}^b_J F^i_{ab} \tau_i + h.c.$$Here is my problem - i don't know how to relate $\epsilon ^{abs}$'s to the $\epsilon ^{IJK} \dot{s}^a_I \dot{s}^b_J \dot{s}^c_K$ and how to get that $-2/3$ factor in front of the triangulated expression - is this related to the limit when tetrahedra shrinks to the point?

I tried reverse route (following https://en.wikipedia.org/wiki/Hamiltonian_constraint_of_LQG), and plugging for holonomies expressions with $A$ and $F$ i get:

$$H^\Delta_E = -2 N(v)\epsilon^{IJK} Tr\Big((1-\frac{\epsilon^2}{2} \dot{s}^a_I \dot{s}^b_J F^i_{ab} \tau_i)(1-\epsilon \dot{s}^c_K A^j_c \tau_j) \{(1+\epsilon \dot{s}^c_I A^j_c \tau_j),V\}\Big)$$Since identity $1$ commutes with $V$, only term with $A$ will survive, then since $\epsilon \dot{s}^c_I A^j_c \tau_j$ is present, only identity is picked in the middle. Then, only term proportional to the $F_{ab}$ will matter, thus:

$$H^\Delta_E = -2 N(v)\epsilon^{IJK} Tr\Big(-\frac{\epsilon^2}{2} \dot{s}^a_I \dot{s}^b_J F^i_{ab} \tau_i)\{\epsilon \dot{s}^c_I A^j_c \tau_j,V\}\Big)$$.I know that i can use $Tr (\tau_i \tau_j)\sim \delta_{ij}$ (i don't know which renormalization is used in Thiemann's work) to get rid of the generators $\tau$ and $j \rightarrow i$.
However, I am stuck with the expression $\sim \epsilon ^{IJK} \dot{s}^a_I \dot{s}^b_J \dot{s}^c_K$ and i don't know how to relate this to the $\epsilon^{abc}$ and how to get that $-2/3$ in front of the triangulated expression.

How can i relate tangents of the segments to the classical nontriangulated expressions? How to generalize this to the more complex terms found in the literature (for example $\sim \int_\Sigma d^3 x N\{A_a,V\}\epsilon^{abc}Tr\Big(\{A_b(x),V^{3/4}\}\{A_c,V^{3/4}\}\Big)$?
In thesis https://arxiv.org/abs/1910.00469 (page 85-86) i found that "$\epsilon ^{IJK} \dot{s}^a_I \dot{s}^b_J \dot{s}^c_K$ is equal to the $6$ times the coordinate volume of the tetrahedron $\Delta$ ". What does that statement mean? Is it related to the $d^3 x$ in the integral and $\epsilon ^{abc}$ or $V$? .$V$ is present in the both "versions" of $H_E$, so it doesn't look like naive substitution will be correct.

I'm trying to follow article in https://arxiv.org/abs/gr-qc/9606089, with supplementary material (Thiemann's book "Modern Canonical Quantum General Relativity" and https://arxiv.org/abs/1007.0402).

 

[Ambitwistor]

I would appreciate any help or guidance in understanding how to relate the triangulated expression to the nontriangulated one and how to generalize it to more complex terms. Thank you.
Thank you for your question. The triangulated version of the Hamiltonian constraint in LQG can be a bit tricky to understand, but I will try my best to explain it to you.

First, let's look at the expression for the Euclidean Hamiltonian:

$$H_E=\frac{2}{\kappa} \int_\Sigma d^3 x N(x)\epsilon^{abc} \text{Tr}(F_{ab},\{A_c,V\})$$

This is the nontriangulated version, where we are integrating over the entire 3-dimensional space $\Sigma$. Now, to obtain the triangulated version, we need to introduce a triangulation $T$ of $\Sigma$ into tetrahedra $\Delta$. This means that we are dividing $\Sigma$ into smaller and smaller tetrahedra, until we have a mesh of tetrahedra covering the entire space.

Now, let's look at the setup for this triangulation. We have vertices $v(\Delta)$, which are the points where the tetrahedra meet. We also have three edges $s_I(\Delta)$ that meet at each vertex. These edges are labeled $I=1,2,3$. Finally, we have a fourth edge $a_{IJ}(\Delta)$ that connects the endpoints of the three edges $s_I(\Delta)$ and $s_J(\Delta)$. This edge is distinct from the vertex $v(\Delta)$.

Next, we need to define a loop based at each vertex, which is denoted by $\alpha_{IJ}(\Delta)$. This loop is constructed by starting at the vertex $v(\Delta)$, following the edge $s_I(\Delta)$, then following the edge $a_{IJ}(\Delta)$, and finally following the edge $s_J(\Delta)$ back to the vertex. This loop is important because it allows us to translate the nontriangulated expression into a triangulated one.

Now, let's look at the triangulated Hamiltonian:

$$H_E^T= \sum_{\Delta \in T}H_E^\Delta;\;\;\;\; H_E