Triangulation of the Hamiltonian constraint in Loop quantum gravity

Question

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.

Answer 1

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.