Triangulation of the Hamiltonian constraint in Loop quantum gravity 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:
$H_E^T= \sum_{\Delta \in T}H_E^\Delta;\;\;\;\; H_E^\Delta:= \frac{-2}{3}N_v \epsilon^{IJK} Tr(h_{\alpha_{IJ}} h_{s_K}\{h^{-1}_{s_K}(\Delta),V\}),$
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:
$H^\Delta_E = -2 N(v)\epsilon^{IJK} Tr\Big(-\frac{\epsilon^2}{2} \dot{s}^a_I \dot{s}^b_J F^i_{ab} \tau_i)\{\epsilon \dot{s}^c_I A^j_c \tau_j,V\}\Big)$.
Edit: Number $2$ in the front of the above expression will cancel with the $\sim \frac
{1}{
2} F$ so i need the factor $-2/3$
to compensate that.
I know that i can use $Tr (\tau_i \tau_j)\sim \delta_{ij}$ (i dont know which renormalization is used in Thiemann's work) to get rid of the generators $\tau$ and $j \rightarrow i$.
However, im 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 generalise 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)? $
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).
Edit: 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.
 A: I'm working on this regularization too.
The triangulated Hamiltonian $H_{E}^{\Delta}$ you wrote it's such that in the limit it tends to (here I take $k = 1$)
\begin{equation}
2 \int_{\Delta} \mathrm{d}^{3}x \, N(x) \epsilon^{abc} \mathrm{Tr}(F_{ab} \, \{ A_{c}, V \}),
\end{equation}
namely in this infinitesimal limit ($\epsilon \rightarrow 0$) you should expect something like
\begin{equation}
V_{\Delta} (2 N \epsilon^{abc} \mathrm{Tr}(F_{ab} \, \{ A_{c}, V \}))
\end{equation}
where the parenthesis is evaluated in the vertex $v$. Then using what you wrote, you obtain:
\begin{equation}
\begin{split}
H_{E}^{\Delta}
& =-\dfrac{2}{3} N(v) \epsilon^{IJK} 
\mathrm{Tr}(h_{\alpha_{IJ}} h_{s_{K}}^{-1} \{ h_{s_{K}}, V \})=\\
& = -\dfrac{2}{3} N(v) \epsilon^{IJK}
\mathrm{Tr}(-\dfrac{\epsilon^{2}}{2} \dot{s}^{a}_{I} \dot{s}^{b}_{J} F_{ab} \, \{\epsilon \dot{s}^{c}_{K} A_{c}, V \}) = \\
& = \dfrac{1}{3} N(v) \epsilon^{IJK}
\mathrm{Tr}(\epsilon^{2} \dot{s}^{a}_{I} \dot{s}^{b}_{J} F_{ab} \, \{\epsilon \dot{s}^{c}_{K} A_{c}, V \})
\end{split}
\end{equation}
Using the result in the comment
\begin{equation}
\epsilon^{abc}
\propto
\epsilon^{IJK} \dot{s}^{a}_{I} \dot{s}^{b}_{J} \dot{s}^{c}_{K}
\end{equation}
and since in our case the volume of a tetrahedron is given by
\begin{equation}
V_{\Delta}
= \dfrac{1}{6} \boldsymbol{a} \cdot (\boldsymbol{b} \times \boldsymbol{c})
= \dfrac{1}{6} \epsilon_{ijk} a^{i} b^{j} c^{k}
\end{equation}
where $\boldsymbol{a}$, $\boldsymbol{b}$ and $\boldsymbol{c}$ are the vectors identifying the 3 vertices of our tetrahedron in which the fourth one is in the origin of the coordinate system, we have
\begin{equation}
6 V_{\Delta} \epsilon^{abc}
= \epsilon^{IJK} \epsilon \dot{s}^{a}_{I} \epsilon \dot{s}^{b}_{J} \epsilon \dot{s}^{c}_{K}
\end{equation}
Putting all together, we obtain
\begin{equation}
\begin{split}
& V_{\Delta} (\dfrac{1}{3} N(v) 6 \epsilon^{abc}
\mathrm{Tr}(F_{ab} \, \{A_{c}, V \})
=\\
& = V_{\Delta} (2 N \epsilon^{abc} \mathrm{Tr}(F_{ab} \, \{ A_{c}, V \}))
\end{split}
\end{equation}
as expected. I think that you have to use the Tr of the $\tau$ only if you are interested in obtaining the limit written without the trace.
For any doubt contact me.
