# Spinor Representation of the Hamiltonian Constraint in 3D LQG with a Non-
zero Cosmological Constant

Valentin Bonzom<EMAIL_ADDRESS>LIPN, UMR CNRS 7030, Institut
Galilée, Université Paris 13, Sorbonne Paris Cité, 99, avenue Jean-Baptiste
Clément, 93430 Villetaneuse, France, EU  Maïté Dupuis
<EMAIL_ADDRESS>Perimeter Institute, 31 Caroline St North,
Waterloo N2L 2Y5, Ontario, Canada  Department of Applied Mathematics,
University of Waterloo, Waterloo, Ontario, Canada  Qiaoyin Pan
<EMAIL_ADDRESS>Perimeter Institute, 31 Caroline St North, Waterloo
N2L 2Y5, Ontario, Canada  Department of Applied Mathematics, University of
Waterloo, Waterloo, Ontario, Canada

###### Abstract

We develop in a companion article the kinematics of three-dimensional loop
quantum gravity in Euclidean signature and with a negative cosmological
constant, focusing in particular on the spinorial representation which is
well-known at zero cosmological constant. In this article, we put this
formalism to the test by quantizing the Hamiltonian constraint on the dual of
a triangulation. The Hamiltonian constraints are obtained by projecting the
flatness constraints onto spinors, as done in the flat case by the first
author and Livine. Quantization then relies on $q$-deformed spinors. The
quantum Hamiltonian constraint acts in the $q$-deformed spin network basis as
difference equations on physical states, which are thus the Wheeler-DeWitt
equations in this framework. Moreover, we study how physical states transform
under Pachner moves of the canonical surface. We find that those
transformations are in fact $q$-deformations of the transition amplitudes of
the flat case as found by Noui and Perez. Our quantum Hamiltonian constraints
therefore build a Turaev-Viro model at real $q$.

## Introduction

Three-dimensional gravity is often taken as a testing ground for new
approaches to quantum gravity as it is much better understood compared to the
four-dimensional case. Here we consider 3D gravity as a topological model
(meaning the triad can be degenerate) through the BF formulation. A criterion
for the validity of novel approaches to its quantization is that they
reproduce the results obtained via previous approaches, such as its
topological invariance. In the absence of a cosmological constant, non-
perturbative quantization in the canonical approach, i.e. the loop quantum
gravity (LQG) based on the BF formulation of gravity Rovelli:2007quantum ;
Thiemann:2007zz , and the covariant approach, based on spin foams and more
precisely the Ponzano-Regge model Ponzano:1969semi ; Baez:1997zt ; Baez:1999sr
; Oriti:2001qu ; Livine:2010zx , give consistent results. In particular, it
has been shown by Noui and Perez in Noui:2004iy  that the scalar products of
physical states defined in LQG are given by the Ponzano-Regge amplitudes.

A more recent approach has emerged which aims at quantizing the Hamiltonian
constraint instead of the flatness constraint derived from the BF formulation.
In the case of vanishing cosmological constant (which we will often refer to
as the flat case), the Hamiltonian constraint translates in the spin network
basis as _difference equations_ on the coefficients of the physical states.
These difference equations should really be seen as Wheeler-DeWitt equations
for 3D LQG. In particular, they are solved by spin network evaluations, as
expected from the Ponzano-Regge model Bonzom:2011hm ; Bonzom:2011nv .

When the cosmological constant is non-zero, the connection between LQG and
spin foams is less clear. On the spin foam side, the Turaev-Viro model
Turaev:1992hq  is known to provide the partition function of 3D gravity in
Euclidean signature with a positive cosmological constant Archer:1991rz . It
is a sum of states in $\mathrm{SU}_{q}(2)$ representation, with $q$ a root of
unity encoding the cosmological constant. It is thus a $q$-deformation of the
Ponzano-Regge model, further providing a regularization through a natural cut-
off on reprensations when replacing $\mathrm{SU}(2)$ with
$\mathrm{SU}_{q}(2)$. The large spin limit of the $q$-$6j$ symbol matches
Regge calculus for curved tetrahedra Taylor:2006j . The Turaev-Viro model thus
provides an example of the interplay between the cosmological constant, curved
geometries and the quantum group deformation of Lie groups.

On the LQG side, the Hamiltonian takes a more complicated form when the
cosmological constant is non-zero, so much that even how to discretize it has
been unclear and it seems to evade traditional LQG methods. It has
nevertheless been conjectured for a long time that the quantum theory ought to
be described by quantum groups, as expected from the spin foam model
Smolin:1994qb ; Smolin:1995vq ; Borissov:1996ge ; Major:1995yz ; Smolin:2002sz
. One (indirect) way to relate 3D LQG with a non-vanishing cosmological
constant to the Turaev-Viro model is to take the Chern-Simons formulation of
3D gravity and consider the Witten-Reshetikhin-Turaev path integral
${\mathcal{Z}}_{WRT}({\mathcal{M}})$ on a 3-manifold ${\mathcal{M}}$ with the
Chern-Simons action with opposite levels, say $k$ and $-k$. It has been well-
known that the Turaev-Viro state sum matches such path integral as
${\mathcal{Z}}_{TV}({\mathcal{M}})=|{\mathcal{Z}}_{WRT}({\mathcal{M}})|^{2}$
Walker:1991wi ; Turaev:2010tw .

A more direct approach for bridging the two quantum gravity approaches would
be to work on the BF formulation with a cosmological constant term itself.
Canonical analysis for the BF action written with the standard triad and
connection variables leads to a torsion equation independent of the
cosmological constant so that the kinematical Hilbert space upon quantization
is spanned by the $\mathrm{SU}(2)$ spin network as in the case with a zero
cosmological constant. In this setting, one can expect, through the connection
of cosmological constant and quantum group deformation, that the quantum group
structure would only appear at the level of the physical Hilbert space since
only the curvature equation depends on the cosmological constant. One proposal
to realize this and thereby connect LQG directly to the Turaev-Viro model was
given in Noui:2011im ; Noui:2011aa ; Pranzetti:2014xva . There, a new
curvature constraint was defined via a new (Poisson non-commutative)
connection, leading to a redefinition of physical scalar product and
recovering the Turaev-Viro amplitude.

From the geometrical point of view, the non-deformed kinematical structures
given by imposing the torsion-free, or Gauss, constraints represent discrete
flat 2D geometries. Then the deformed dynamical structures are expected to
describe the gluing of these flat 2D geometries to approximate the curved 3D
geometries, as the deformed version of the case with a zero cosmological
constant Bonzom:2011hm . Indeed, one can approximate a curved 3D geometry by
gluing flat 2D pieces and take the limit as the sizes of these pieces approach
zero. It was moreover argued in Dittrich:2008pw ; Bahr:2009ku ; Bahr:2009mc
that the symmetries survive at the discrete level when one uses curved 2D
pieces instead of flat ones.

This suggests us to rethink the definition of the kinematics and dynamics in
the BF formulation with a non-vanishing cosmological constant. The kinematical
phase space defined with the Gauss constraint can in fact be deformed so that
its quantization naturally leads to a quantum group deformation. In
particular, this kinematical phase space describes 2D _curved_ geometries
Bonzom:2014wva . Then the physical phase space, defined by imposing the
flatness constraint, describes the gluing of these 2D curved geometries into
3D curved geometries. The gap between the discrete, classical theory and the
continuous action was further filled recently in Dupuis:2020ndx .

This programme was carried out in Bonzom:2014wva ; Bonzom:2014bua  to a large
degree. There, the phase space is defined in terms of deformed fluxes and
holonomies and the Poisson structure is based on the Heisenberg double of
$\mathrm{SU}(2)$. In Bonzom:2014bua  we have investigated the quantization,
using the same techniques as in Bonzom:2011hm , i.e. by building a Hamiltonian
constraint out of the flatness constraints. It can be classically interpreted
as generating displacements of the vertices of the triangulation
BonzomDittrich . At the quantum level, the Hamiltonian constraints give rise
to _difference equations_ , which can therefore be considered as Wheeler-
DeWitt equations in the spin network basis. We considered in Bonzom:2014bua
the (simple) case of the boundary of the tetrahedron, and showed that the
solution to those difference equations is the $q$-$6j$ symbol.

Here we are interested in using the spinorial formalism for LQG instead of
holonomies and fluxes, and in further extracting all building blocks for the
transition amplitudes, i.e. go beyond the case of the tetrahedron from
Bonzom:2014bua . In a companion article Bonzom:2021rib , we revisit all
kinematical aspects of this deformed LQG model in more details, and in the
spinor representation (this was initiated in Dupuis:2014fya ). In particular,
the quantization of the deformed spinors can be performed in terms of
$q$-bosons. We then use those $q$-bosons to define the invariant operators
which are needed for the quantization of the Hamiltonian constraint in spinor
variables.

In this article, we describe the dynamics using the deformed spinors and
$q$-bosons of Bonzom:2021rib . The Hamiltonian constraint built from the
deformed spinors is a direct generalization of the non-deformed version given
in Bonzom:2011nv . At the quantum level, the Hamiltonian constraints also give
rise to difference equations which are direct $q$-deformed generalizations of
those of Bonzom:2011nv . Here we moreover go further by providing the
transformations of the physical states (in the spin network basis) under
Pachner moves of the canonical surface. This is equivalent to finding the
building blocks for spin foams, or for the transition amplitudes, as
emphasized by Noui and Perez Noui:2004iy . In particular, we find that those
building blocks are exactly those of the Turaev-Viro model in a version with
$q$ real (note that this version suffers from the same finiteness issues as
the Ponzano-Regge model).

This article is organized as follows. In Section I, we concisely recall the
discrete classical phase space in terms of the holonomies and (deformed)
fluxes introduced in Bonzom:2014wva , as well as the Gauss constraints (used
to defined the kinematical phase space) and the flatness constraints (used to
define the dynamical phase spaces). In Section II we move on to the
construction of the deformed spinors, following Bonzom:2021rib , and of the
Hamiltonian constraints. The quantization is performed in Section III, again
following the prescriptions of Bonzom:2021rib . This is where in particular we
find the difference equations encoding the Wheeler-DeWitt equations in the
spin network basis. Then in Section IV, we study how solutions to the
difference equations are related under Pachner moves, thereby providing the
building blocks for the transition amplitudes à la Noui-Perez.

##  I Classical phase space and constraints

We start by recalling the main ingredients of the classical phase space for 3D
loop gravity with a negative cosmological constant $\Lambda$ in Euclidean
signature. More details for the mathematical setup can be found in
Bonzom:2014wva ; Bonzom:2021rib . The phase space is associated to a graph
$\Gamma$ which is dual to a cellular decomposition of the canonical surface
$\Sigma$. It has $V$ vertices, $E$ edges and $F$ faces (the connected
components of $\Sigma\setminus\Gamma$).

Here and throughout the paper, we use $\kappa:=\frac{G\sqrt{-\Lambda}}{c}$,
for a cosmological constant $\Lambda<0$. It is a parameter which deforms the
Poisson structure with respect to the case of vanishing cosmological constant.

###  I.1 Phase space for a single edge

We first consider a single edge and associate to it a phase space: the
Heisenberg double $({\mathcal{D}}(\mathrm{SU}(2)),\pi_{H})$ of
$\mathrm{SU}(2)$. It is the group
${\mathcal{D}}(\mathrm{SU}(2))=\mathrm{SL}(2,{\mathbb{C}})\cong\mathrm{SU}(2)\bowtie\mathrm{AN}(2)$
with Poisson structure $\pi_{H}$ fully determined by a classical $r$-matrix
$r\in{\mathfrak{s}l}(2,{\mathbb{C}})\otimes{\mathfrak{s}l}(2,{\mathbb{C}})$.
The Poisson brackets can be compactly written as

$\\{d_{1},d_{2}\\}=-r_{21}d_{1}d_{2}+d_{1}d_{2}r=rd_{1}d_{2}-d_{1}d_{2}r_{21}\,,\quad\forall
d\in\mathrm{SL}(2,{\mathbb{C}})\,,$ (1)

where $d_{1}=d\otimes\mathbb{I}\,,d_{2}=\mathbb{I}\otimes d$. The $r$-matrix
is chosen as

$r=\frac{i\kappa}{4}\sum_{i=1}^{3}\sigma_{i}\otimes\rho_{i}=\frac{i\kappa}{4}\left(\begin{array}[]{cccc}1&0&0&0\\\
0&-1&4&0\\\ 0&0&-1&0\\\ 0&0&0&1\end{array}\right)\,,$ (2)

Here $\sigma_{1,2,3}$ are the Pauli matrices while
$\rho_{i}=\sigma_{i}+\tfrac{1}{2}[\sigma_{3},\sigma_{i}]$. Finally, $r_{21}$
is given by the permutation of the two vector space components of $r$ (in the
above $4\times 4$ representation, $r_{21}$ is simply the matrix transpose of
$r$). The equality of the two expressions in (1) is guaranteed by the property
that $r_{s}:=\frac{1}{2}(r+r_{21})$ is the Casimir thus
$[r_{s},d_{1}d_{2}]=0$.

It is important for loop gravity to split an $\mathrm{SL}(2,{\mathbb{C}})$
element via the Iwasawa decomposition into the product of an $\mathrm{AN}(2)$
element and an $\mathrm{SU}(2)$ element. One can write
$d\in\mathrm{SL}(2,{\mathbb{C}})$ in exactly two ways as

$d=\ell
u=\tilde{u}\tilde{\ell}\,,\quad\ell,\tilde{\ell}\in\mathrm{AN}(2)\,,\quad
u,\tilde{u}\in\mathrm{SU}(2)\,.$ (3)

This phase space can be seen as a deformation of the holonomy-flux phase space
at $\Lambda=0$. In that case, the phase space of an edge is $\mathrm{ISU}(2)$,
the holonomies are described by $\mathrm{SU}(2)$ and the fluxes are described
by ${\mathbb{R}}^{3}$. Here, in this deformed phase space, we keep an
$\mathrm{SU}(2)$ subgroup to describe holonomies while the (deformed) fluxes
correspond to an $\mathrm{AN}(2)$ subgroup. This splitting between holonomies
and fluxes is performed via the above Iwasawa decompositions, i.e. $u$ and
$\tilde{u}$ are holonomies, while $\ell$ and $\tilde{\ell}$ are fluxes. The
physical reason why there are two (equivalent) ways to perform this split will
become clear below.

We call the constraint,

${\mathcal{C}}=\mathbf{1}\in\mathrm{SL}(2,{\mathbb{C}})\quad\text{for}\quad{\mathcal{C}}:=\ell
u\tilde{\ell}^{-1}\tilde{u}^{-1}$ (4)

the _ribbon constraint_ , associated to every edge of $\Gamma$. It has six
real components and forms a set of second class constraints with respect to
the above Poisson brackets (meaning the brackets between the components do not
close).

The ribbon constraint has a natural graphical interpretation. Since the edges
of $\Gamma$ are embedded on a surface, there is a natural clockwise walk
around each of them. It goes $i)$ along the edge on one side, $ii)$ crosses it
at its end, $iii)$ goes back along the edge on its other side and $iv)$
finally crosses it again to close the walk. Equivalently, one thickens the
edge by taking a tubular neighborhood in $\Sigma$, as in Figure 1. The
boundary has four pieces which naturally correspond to the four parts of the
walk above.

If $e$ is an edge in $\Gamma$, then we denote $R(e)$ its thickening, called
_ribbon edge_. The boundary pieces parallel to $e$ will be called the _long
edges_ of $R(e)$ and the boundary pieces which cross $e$ at its ends will be
called the _short edges_ of $R(e)$. We can orient the long and short edges
clockwise around $R(e)$. The matrices
$u,\ell,\tilde{u}^{-1},\tilde{\ell}^{-1}$ are then assigned in this order and
as pictured in Figure 1. In particular, $u$ and $\tilde{u}$ are assigned to
the long edges of $R(e)$, while $\ell$ and $\tilde{\ell}$ are assigned to its
short edges. Equivalently, we can think of $u$ and $\tilde{u}$ as associated
to $e$ itself and $\ell$ and $\tilde{\ell}$ to each _half-edge_ , i.e. a pair
of an edge and an incident vertex.

To fix the position of the variables around the ribbon, one can use the
orientation on $\Gamma$ and decide for instance that $u$ is oriented opposite
to $e$. This is the convention we will use. The ribbon constraint
${\mathcal{C}}=\ell u\tilde{\ell}^{-1}\tilde{u}^{-1}\cong\mathbb{I}$ is then a
flatness constraints around $R(e)$.

The Poisson brackets (1) can be equivalently written as brackets between
holonomies and fluxes

$\begin{array}[]{llll}\\{\ell_{1},\ell_{2}\\}=-[r_{21},\ell_{1}\ell_{2}]\,,&\\{\ell_{1},u_{2}\\}=-\ell_{1}r_{21}u_{2}\,,&\\{u_{1},\ell_{2}\\}=\ell_{2}ru_{1}\,,&\\{u_{1},u_{2}\\}=-[r,u_{1}u_{2}]\,,\\\
\\{\tilde{\ell}_{1},\tilde{\ell}_{2}\\}=[r_{21},\tilde{\ell}_{1}\tilde{\ell}_{2}]\,,&\\{\tilde{\ell}_{1},\tilde{u}_{2}\\}=-\tilde{u}_{2}r_{21}\tilde{\ell}_{1}\,,&\\{\tilde{u}_{1},\tilde{\ell}_{2}\\}=\tilde{u}_{1}r\tilde{\ell}_{2}\,,&\\{\tilde{u}_{1},\tilde{u}_{2}\\}=[r,\tilde{u}_{1}\tilde{u}_{2}]\,.\end{array}$
(5)

All other Poisson brackets e.g. $\\{\ell_{1},\tilde{u}_{2}\\}$ can also be
obtained by combining (5) and (4).

Figure 1: A ribbon edge $R(e)$. The variables $\ell,u,\tilde{\ell},\tilde{u}$
are assigned to the four sides of the ribbon edge and they are subject to the
ribbon constraint $\ell u\tilde{\ell}^{-1}\tilde{u}^{-1}$ represented as the
trivialization of the loop around $R(e)$. The position of these variables are
fixed such that the direction of $u$ and $\tilde{u}$ are opposite to that of
the edge $e$ (in gray).

The $\mathrm{AN}(2)$ elements $\ell$ and $\tilde{\ell}$ can be parametrized as
$2\times 2$ lower triangular matrices

$\ell=\left(\begin{array}[]{cc}\lambda&0\\\
z&\lambda^{-1}\end{array}\right)\,,\quad\tilde{\ell}=\left(\begin{array}[]{cc}\tilde{\lambda}&0\\\
\tilde{z}&\tilde{\lambda}^{-1}\end{array}\right)\,,\quad\lambda,\tilde{\lambda}\in{\mathbb{R}}^{+}\,,\quad
z,\tilde{z}\in{\mathbb{C}}\,.$ (6)

By also writing the holonomies $u$ and $\tilde{u}$ in the fundamental
representation, one can write all Poisson brackets between the matrix elements
of $u,\tilde{u}$ and
$\lambda,\tilde{\lambda},z,\tilde{z},\bar{z},\bar{\tilde{z}}$ (See
Bonzom:2014wva ; Bonzom:2021rib  for details).

###  I.2 Ribbon graph phase space

We extend the phase space defined above for a single edge to the whole graph
$\Gamma$, by taking the product of $\mathrm{SL}(2,{\mathbb{C}})$ over the
edges. Graphically, one thickens every edge of $\Gamma$ as before. However,
this is not enough since there will be interactions between the group elements
of different edges which meet at vertices of $\Gamma$. An advantageous
graphical representation is to thicken $\Gamma$ itself. Each vertex $v$ of
valency $d$ is fattened to a _ribbon vertex_ $R(v)$, i.e. a $d$-gon whose
boundary edges correspond to the edges incident to $v$, and are glued to the
short edges of the ribbon edges (the boundary vertices of $R(v)$ correspond to
the corners at $v$ between adjacent edges). In other words, a ribbon vertex is
a face whose boundary sides are dressed with fluxes $\ell$s and
$\tilde{\ell}$s.

Overall, the holonomies $u_{e},\tilde{u}_{e}$ are labeled by the edges of
$\Gamma$ and assigned to the long edges of the ribbon edges. There are two
fluxes associated to every edge of $\Gamma$, denoted
$\ell_{e}\in\mathrm{AN}(2)$ if $e$ is outgoing at $v$,
$\tilde{\ell}_{e}\in\mathrm{AN}(2)$ if $e$ is incoming at $v$. Since each is
in fact assigned to a half-edge $(e,v)$ (equivalently a short edge of $R(e)$),
we will use the generic notation $\ell_{ev}$ for either one of them.

Since there are also two holonomies along $e$, it is tempting to distinguish
them in terms of half-edges. This is possible using orientations. We denote
$u_{ev}$ the $\mathrm{SU}(2)$ matrix which points towards $\ell_{ev}$ (so that
if $\ell_{ev}=\tilde{\ell}_{e}^{-1}$ then $u_{ev}=\tilde{u}_{e}^{-1}$ and else
$\ell_{ev}=\ell_{e}$ and $u_{ev}=u_{e}$).

###  I.3 Gauss and flatness constraints

The phase space for $\Gamma$ described above is constrained by the ribbon
constraints $\mathcal{C}_{e}=\mathbf{1}$ on every edge. Gravity further
imposes two additional sets of constraints, namely the Gauss and flatness
constraints. Gauss constraints are associated to vertices and impose that the
ordered products of the fluxes along the short edges of every $R(v)$ is
trivial. Flatness constraints are associated to the faces and impose that the
ordered products of the holonomies along the long edges which border every
face is trivial. Those two sets of constraints are first class.

To write the Gauss constraints explicitly, choose (randomly) one edge of
reference at each vertex of $\Gamma$ and call it $e_{1}$, and order the edges
from 1 to $n$, i.e. $e_{1},\dotsc,e_{n}$ by going counter-clockwise around
$v$. Notice that all the $\mathrm{AN}(2)$ matrices $\ell_{e_{i}v}$ on the
boundary of the ribbon vertex $R(v)$ are oriented counter-clockwise, as shown
in fig.2, for any choice of orientations of the edges incident to $v$. The
Gauss constraint is then simply the flatness around $R(v)$. It reads

$\ell_{e_{n}v}\dotsm\ell_{e_{1}v}=\mathbbm{1}\,.$ (7)

We repeat this construction on faces instead of vertices: choose a random edge
of reference around each face $f$ and denote it $e_{1}$, then
$e_{2},\dotsc,e_{d}$ are the edges encountered counter-clockwise around $f$.
For all possible orientations of the edges $e_{1},\dotsc,e_{d}$ on the
boundary of $f$ are, the $\mathrm{SU}(2)$ matrices $u_{e_{i}v}$ are all
counter-clockwise. The flatness constraint on $f$ reads

$u_{e_{d}v_{1}}\dotsm u_{e_{2}v_{3}}u_{e_{1}v_{2}}=\mathbbm{1}\,,$ (8)

as pictured in Figure 3.

Figure 2: A vertex $v$ on which edges meet becomes a ribbon vertex $R(v)$
incident to ribbon edges. The ribbon vertex is here depicted with a dashed
boundary. Due to the clockwise orientation of the short edges of every ribbon
edge, the matrix $\ell_{ev}$ around $R(v)$ are all oriented counter-clockwise.
Figure 3: When considering ribbon edges and ribbon vertices, the face on the
left bounded by edges $e_{1},\cdots,e_{6}$ becomes bounded by long edges on
the right. All matrices $u_{e_{i}v_{i+1}}\,,i=1,\cdots,6$ are oriented
counter-clockwise around the face.

Gauss constraints generate local $\mathrm{SU}(2)$ transformations through the
Poisson brackets Bonzom:2014wva ; Bonzom:2021rib . As usual in symplectic
geometry, first class constraints are not only imposed but one also needs to
quotient out the phase space by the orbits they generate. This is called the
symplectic quotient. Here, one obtains
${\mathcal{P}}_{\text{kin}}=\mathrm{SL}(2,{\mathbb{C}})^{E}//\mathrm{SU}(2)^{V}$
which is called the _kinematical phase space_ , where $E$ and $V$ denotes the
number of edges and vertices in $\Gamma$.

It was shown in Bonzom:2014wva  that the Gauss constraint for a trivalent
vertex geometrically represents the hyperbolic cosine law, implying that the
kinematical phase space describes hyperbolic geometry (at least hyperbolic
triangles in that case).

On the other hand, flatness constraints generate (deformed) translations
Bonzom:2014wva . The _physical phase space_ is then obtained via the
symplectic quotient of the kinematical phase space by the flatness
constraints,
${\mathcal{P}}_{\text{phys}}={\mathcal{P}}_{\text{kin}}//\mathrm{AN}(2)^{F}$
with $F$ the number of the faces in $\Gamma$.

In the $\Lambda\to 0$ limit, one recovers the Poincaré phase space structure
of first-order 3D gravity. In particular, the flatness constraints generate an
${\mathbb{R}}^{3}$ action, i.e. translations. Geometrically, those
translations simply move the vertices of the triangulation (dual to the faces
of $\Gamma$) around (three directions for the three components of the
constraints). The flatness constraints also enforce the dihedral angles to be
functions of the angles within triangles as in flat, Euclidean geometry
(recall that dihedral angles measure the extrinsic curvature at the discrete
level) Bonzom:2011hm . This geometric picture arises when the constraints are
written projected in the basis determined by the fluxes themselves. It is also
possible to describe them in a spinor basis Bonzom:2011nv .

##  II Spinorial representation

###  II.1 Deformed spinors

Here we describe the $\kappa$-deformed spinors which can be used to describe
the phase space and the constraints in place of the variables
$u_{ev},\ell_{ev}$. We will only give the main ingredients needed to construct
the Hamiltonian constraint. The fully detailed construction appears in
Bonzom:2021rib .

The building blocks are two independent pairs of $\kappa$-deformed spinor
variables
$(\zeta^{\kappa}_{0},\zeta^{\kappa}_{1}),(\tilde{\zeta}^{\kappa}_{0},\tilde{\zeta}^{\kappa}_{1})$
and their complex conjugates
$(\bar{\zeta}^{\kappa}_{0},\bar{\zeta}^{\kappa}_{1}),(\bar{\tilde{\zeta}}^{\kappa}_{0},\bar{\tilde{\zeta}}^{\kappa}_{1})$.
The norms of these $\kappa$-deformed spinor variables are
$\bar{\zeta}_{A}^{\kappa}\zeta_{A}^{\kappa}=\frac{2}{\kappa}\sinh(\frac{\kappa
N_{A}}{2})\,,\bar{\tilde{\zeta}}_{A}^{\kappa}\tilde{\zeta}_{A}^{\kappa}=\frac{2}{\kappa}\sinh(\frac{\kappa\tilde{N}_{A}}{2})\,,A=0,1$
where $N_{A}$ and $\tilde{N}_{A}$ are real functions of the $\kappa$-deformed
spinor variables 111 $N_{A}$ $(A=0,1)$ is the norm of the $\kappa$-spinors at
$\kappa\rightarrow 0$ thus the norm of the standard spinor variables and same
for $\tilde{N}_{A}$. . They satisfy the Poisson brackets

$\begin{array}[]{llll}\\{\zeta_{A}^{\kappa},\bar{\zeta}^{\kappa}_{B}\\}=-i\,\delta_{AB}\,\cosh(\frac{\kappa
N_{A}}{2}),&\quad\\{N_{A},\zeta_{B}^{\kappa}\\}=i\,\delta_{AB}\,\zeta^{\kappa}_{A},&\quad\\{N_{A},\bar{\zeta}_{B}^{\kappa}\\}=-i\,\delta_{AB}\,\bar{\zeta}^{\kappa}_{A}\,,&\\\\[8.5359pt]
\\{\tilde{\zeta}_{A}^{\kappa},\bar{\tilde{\zeta}}^{\kappa}_{B}\\}=-i\,\delta_{AB}\,\cosh(\frac{\kappa\tilde{N}_{A}}{2}),&\quad\\{\tilde{N}_{A},\tilde{\zeta}_{B}^{\kappa}\\}=i\,\delta_{AB}\,\tilde{\zeta}^{\kappa}_{A},&\quad\\{\tilde{N}_{A},\bar{\tilde{\zeta}}_{B}^{\kappa}\\}=-i\,\delta_{AB}\,\bar{\tilde{\zeta}}^{\kappa}_{A}\,,&\quad
A,B=0,1\,,\end{array}$ (9)

and all other Poisson brackets vanish.

Let $\epsilon=\left(\begin{smallmatrix}0&-1\\\ 1&0\end{smallmatrix}\right)$.
These $\kappa$-deformed spinor variables can be used to define _two_ types of
deformed spinors:

  * •

The $\mathrm{SU}(2)$-covariant spinors, transforming under $\mathrm{SU}(2)$
gauge transformations in a covariant way. We denote them $|t\rangle$ and
$|\tilde{t}\rangle$, and their duals $|t]$ and $|\tilde{t}]$,

$\begin{array}[]{ll}|t\rangle=\left(\begin{array}[]{c}t_{+}\\\
t_{-}\end{array}\right)=\left(\begin{array}[]{c}e^{\frac{\kappa
N_{1}}{4}}\zeta^{\kappa}_{0}\\\ e^{-\frac{\kappa
N_{0}}{4}}\zeta^{\kappa}_{1}\end{array}\right)\,,&\qquad|t]=\epsilon|\overline{t}\rangle=\left(\begin{array}[]{c}-\bar{t}_{-}\\\
\bar{t}_{+}\end{array}\right)=\left(\begin{array}[]{c}-e^{-\frac{\kappa
N_{0}}{4}}\bar{\zeta}^{\kappa}_{1}\\\ e^{\frac{\kappa
N_{1}}{4}}\bar{\zeta}^{\kappa}_{0}\end{array}\right)\,,\\\
|\tilde{t}\rangle=\left(\begin{array}[]{c}\tilde{t}_{+}\\\
\tilde{t}_{-}\end{array}\right)=\left(\begin{array}[]{c}e^{\frac{\kappa\tilde{N}_{1}}{4}}\tilde{\zeta}^{\kappa}_{0}\\\
e^{-\frac{\kappa\tilde{N}_{0}}{4}}\tilde{\zeta}^{\kappa}_{1}\end{array}\right)\,,&\qquad|\tilde{t}]=\epsilon|\bar{\tilde{t}}\rangle=\left(\begin{array}[]{c}-\bar{\tilde{t}}_{-}\\\
\bar{\tilde{t}}_{+}\end{array}\right)=\left(\begin{array}[]{c}-e^{-\frac{\kappa\tilde{N}_{0}}{4}}\bar{\tilde{\zeta}}^{\kappa}_{1}\\\
e^{\frac{\kappa\tilde{N}_{1}}{4}}\bar{\tilde{\zeta}}^{\kappa}_{0}\end{array}\right)\,.\end{array}$
(10)

  * •

The braided-covariant spinors, transforming in a braided-covariant way
Bonzom:2021rib . We denote them $|\tau\rangle$ and $|\tilde{\tau}\rangle$ and
their duals $|\tau]$ and $|\tilde{\tau}]$,

$\begin{array}[]{ll}|\tau\rangle=\left(\begin{array}[]{c}\tau_{+}\\\
\tau_{-}\end{array}\right)=\left(\begin{array}[]{c}e^{-\frac{\kappa
N_{1}}{4}}\zeta_{0}^{\kappa}\\\ e^{\frac{\kappa
N_{0}}{4}}\zeta^{\kappa}_{1}\end{array}\right)\,,&\qquad|\tau]=\epsilon|\bar{\tau}\rangle=\left(\begin{array}[]{c}-\bar{\tau}_{-}\\\
\bar{\tau}_{+}\end{array}\right)=\left(\begin{array}[]{c}-e^{\frac{\kappa
N_{0}}{4}}\bar{\zeta}^{\kappa}_{1}\\\ e^{-\frac{\kappa
N_{1}}{4}}\bar{\zeta}_{0}^{\kappa}\end{array}\right)\,,\\\
|\tilde{\tau}\rangle=\left(\begin{array}[]{c}\tilde{\tau}_{+}\\\
\tilde{\tau}_{-}\end{array}\right)=\left(\begin{array}[]{c}e^{-\frac{\kappa\tilde{N}_{1}}{4}}\tilde{\zeta}^{\kappa}_{0}\\\
e^{\frac{\kappa\tilde{N}_{0}}{4}}\tilde{\zeta}^{\kappa}_{1}\end{array}\right)\,,&\qquad|\tilde{\tau}]=\epsilon|\bar{\tilde{\tau}}\rangle=\left(\begin{array}[]{c}-\bar{\tilde{\tau}}_{-}\\\
\bar{\tilde{\tau}}_{+}\end{array}\right)=\left(\begin{array}[]{c}-e^{\frac{\kappa\tilde{N}_{0}}{4}}\bar{\tilde{\zeta}}^{\kappa}_{1}\\\
e^{-\frac{\kappa\tilde{N}_{1}}{4}}\bar{\tilde{\zeta}}^{\kappa}_{0}\end{array}\right)\,.\end{array}$
(11)

The norms are $\langle
t|t\rangle=\langle\tau|\tau\rangle=\tfrac{2}{\kappa}\sinh\tfrac{\kappa}{2}(N_{0}+N_{1})$,
and similarly
$\langle\tilde{t}|\tilde{t}\rangle=\langle\tilde{\tau}|\tilde{\tau}\rangle=\tfrac{2}{\kappa}\sinh\tfrac{\kappa}{2}(\tilde{N}_{0}+\tilde{N}_{1})$
(the norm of a dual is the same since $\epsilon^{\dagger}=\epsilon^{-1}$).
They match if the so-called norm matching condition holds, which is just
$N_{0}+N_{1}\cong\tilde{N}_{0}+\tilde{N}_{1}$.

Holonomies and fluxes can be reconstructed as follows,

$\begin{array}[]{ll}\ell=\left(\begin{array}[]{cc}\exp(\frac{\kappa}{4}(N_{1}-N_{0}))&0\\\
-\kappa\bar{\zeta}_{0}^{\kappa}\zeta_{1}^{\kappa}&\exp(\frac{\kappa}{4}(N_{0}-N_{1}))\end{array}\right)\,,&\qquad\tilde{\ell}=\left(\begin{array}[]{cc}\exp(\frac{\kappa}{4}(\tilde{N}_{0}-\tilde{N}_{1}))&0\\\
\kappa\bar{\tilde{\zeta}}^{\kappa}_{0}\tilde{\zeta}^{\kappa}_{1}&\exp(\frac{\kappa}{4}(\tilde{N}_{1}-\tilde{N}_{0}))\end{array}\right)\,,\\\\[17.07182pt]
u=\dfrac{|\tau\rangle[\tilde{t}|-|\tau]\langle\tilde{t}|}{\sqrt{\langle\tau|\tau\rangle\langle\tilde{t}|\tilde{t}\rangle}}\,,&\qquad\tilde{u}=\dfrac{|t\rangle[\tilde{\tau}|-|t]\langle\tilde{\tau}|}{\sqrt{\langle
t|t\rangle\langle\tilde{\tau}|\tilde{\tau}\rangle}}\,,\end{array}$ (12)

with $N_{0}+N_{1}=\tilde{N}_{0}+\tilde{N}_{1}$. It is straightforward to check
that the deformed spinors are related to one another by parallel transport via
fluxes and holonomies.

$|\tau\rangle=e^{-\frac{\kappa(N_{0}+N_{1})}{4}}\ell^{-1}|t\rangle\,,\quad|\tilde{\tau}]=e^{\frac{\kappa(\tilde{N}_{0}+\tilde{N}_{1})}{4}}\tilde{\ell}|\tilde{t}]\,,\quad
u|\tilde{t}]=|\tau\rangle\,,\quad\tilde{u}|\tilde{\tau}]=|t\rangle\,.$ (13)

Those relations have a natural graphical interpretation: the spinors can be
assigned to the _corners_ of the ribbon edge. For instance, $|\tilde{t}]$ is
at the source end of the long edge carrying $u$ and $|\tau\rangle$ is at its
target end. This is depicted in Figure 4 (we do not include the factors
$e^{-\frac{\kappa(N_{0}+N_{1})}{4}}$ and
$e^{\frac{\kappa(\tilde{N}_{0}+\tilde{N}_{1})}{4}}$ in the graphical
representation).

Figure 4: The ribbon edge with the holonomies on its long edges, fluxes on its
short edges and spinors on its corners.

It will be convenient to consider _non-normalized_ versions for $|\tau\rangle$
and $|\tilde{t}]$ instead. We denote them
$|\tau^{\text{nn}}\rangle=e^{\frac{\kappa(N_{0}+N_{1})}{4}}|\tau\rangle$ and
$|\tilde{t}^{\text{nn}}]=e^{\frac{\kappa(\tilde{N}_{0}+\tilde{N}_{1})}{4}}|\tilde{t}]$
respectively. The holonomy $\tilde{u}$ can also be written with the non-
normalized deformed spinors $\tau^{\text{nn}}$ and $\tilde{t}^{\text{nn}}$ as

$u=\frac{|\tau^{\text{nn}}\rangle[\tilde{t}^{\text{nn}}|-|\tau^{\text{nn}}]\langle\tilde{t}^{\text{nn}}|}{\sqrt{\langle\tau^{\text{nn}}|\tau^{\text{nn}}\rangle\langle\tilde{t}^{\text{nn}}|\tilde{t}^{\text{nn}}\rangle}}$
(14)

To avoid writing separate equations for $|t\rangle$ and $|t]$, we introduce
the following notations,

$\begin{aligned} t_{A}^{-}&:=(-1)^{\frac{1}{2}-A}t_{A},\\\ \
t_{A}^{+}&:=\bar{t}_{-A},\end{aligned}\qquad\text{and}\qquad\begin{aligned}
\tilde{t}_{A}^{-}&:=(-1)^{\frac{1}{2}-A}\tilde{t}_{A}\,,\\\
\tilde{t}_{A}^{+}&:=\bar{\tilde{t}}_{-A}\,.\end{aligned}$ (15)

for $A=\pm 1/2$. Similar but exchanged notations are used for $\tau$ and
$\tilde{\tau}$,

$\begin{aligned} \tau_{A}^{-}&:=\tau_{A}\,,\\\
\tau_{A}^{+}&:=(-1)^{\frac{1}{2}+A}\bar{\tau}_{-A}\,,\end{aligned}\qquad\text{and}\qquad\begin{aligned}
\tilde{\tau}_{A}^{-}&:=\tilde{\tau}_{A}\,,\\\
\tilde{\tau}_{A}^{+}&:=(-1)^{\frac{1}{2}+A}\bar{\tilde{\tau}}_{-A}\,.\end{aligned}$
(16)

For reference, we explicitly write the spinors and dual spinors with those
notations in a footnote222
$\begin{array}[]{llll}|t\rangle=\left(\begin{array}[]{c}t_{+}^{-}\\\
-t_{-}^{-}\end{array}\right)\,,&|t]=\left(\begin{array}[]{c}-t_{+}^{+}\\\
t_{-}^{+}\end{array}\right)\,,&|\tilde{t}\rangle=\left(\begin{array}[]{c}\tilde{t}_{+}^{-}\\\
-\tilde{t}_{-}^{-}\end{array}\right)\,,&|\tilde{t}]=\left(\begin{array}[]{c}-\tilde{t}_{+}^{+}\\\
\tilde{t}_{-}^{+}\end{array}\right)\,,\\\\[8.5359pt] \langle
t|=\left(\begin{array}[]{cc}t_{-}^{+},&t_{+}^{+}\end{array}\right)\,,&[t|=\left(\begin{array}[]{cc}t_{-}^{-},&t_{+}^{-}\end{array}\right)\,,&\langle\tilde{t}|=\left(\begin{array}[]{cc}\tilde{t}_{-}^{+},&\tilde{t}_{+}^{+}\end{array}\right)\,,&[\tilde{t}|=\left(\begin{array}[]{cc}\tilde{t}_{-}^{-},&\tilde{t}_{+}^{-}\end{array}\right)\,,\\\\[8.5359pt]
|\tau\rangle=\left(\begin{array}[]{c}\tau_{+}^{-}\\\
\tau_{-}^{-}\end{array}\right)\,,&|\tau]=\left(\begin{array}[]{c}\tau_{+}^{+}\\\
\tau_{-}^{+}\end{array}\right)\,,&|\tilde{\tau}\rangle=\left(\begin{array}[]{c}\tilde{\tau}_{+}^{-}\\\
\tilde{\tau}_{-}^{-}\end{array}\right)\,,&|\tilde{\tau}]=\left(\begin{array}[]{c}\tilde{\tau}_{+}^{+}\\\
\tilde{\tau}_{-}^{+}\end{array}\right)\,,\\\\[8.5359pt]
\langle\tau|=\left(\begin{array}[]{cc}\tau_{-}^{+},&-\tau_{+}^{+}\end{array}\right)\,,&[\tau|=\left(\begin{array}[]{cc}-\tau_{-}^{-},&\tau_{+}^{-}\end{array}\right)\,,&\langle\tilde{\tau}|=\left(\begin{array}[]{cc}\tilde{\tau}_{-}^{+},&-\tilde{\tau}_{+}^{+}\end{array}\right)\,,&[\tilde{\tau}|=\left(\begin{array}[]{cc}-\tilde{\tau}_{-}^{-},&\tilde{\tau}_{+}^{-}\end{array}\right)\,,\end{array}$
(17) where the subscripts $A=\pm\frac{1}{2}$ have been notated as $A=\pm$ for
simplicity.. The norms read

$\begin{array}[]{ll}\langle
t|t\rangle=\frac{1}{2}\sum_{\epsilon=\pm}\sum_{A=\pm\frac{1}{2}}\epsilon(-1)^{\frac{1}{2}-A}\,t_{A}^{\epsilon}t_{-A}^{-\epsilon}\,,&\langle\tilde{t}|\tilde{t}\rangle=\frac{1}{2}\sum_{\epsilon=\pm}\sum_{A=\pm\frac{1}{2}}\epsilon(-1)^{\frac{1}{2}-A}\,\tilde{t}_{A}^{\epsilon}\tilde{t}_{-A}^{-\epsilon}\,,\\\\[4.26773pt]
\langle\tau|\tau\rangle=\frac{1}{2}\sum_{\epsilon=\pm}\sum_{A=\pm\frac{1}{2}}\epsilon(-1)^{\frac{1}{2}-A}\,\tau_{A}^{\epsilon}\tau_{-A}^{-\epsilon}\,,&\langle\tilde{\tau}|\tilde{\tau}\rangle=\frac{1}{2}\sum_{\epsilon=\pm}\sum_{A=\pm\frac{1}{2}}\epsilon(-1)^{\frac{1}{2}-A}\,\tilde{\tau}_{A}^{\epsilon}\tilde{\tau}_{-A}^{-\epsilon}\,.\end{array}$
(18)

and the holonomies

$u_{AB}=-\frac{1}{\sqrt{\langle\tau|\tau\rangle\langle\tilde{t}|\tilde{t}\rangle}}\sum_{\epsilon=\pm}\epsilon\tau^{\epsilon}_{A}\tilde{t}^{\epsilon}_{-B}\qquad\tilde{u}^{-1}_{AB}=\frac{1}{\sqrt{\langle\tilde{\tau}|\tilde{\tau}\rangle\langle
t|t\rangle}}\sum_{\epsilon=\pm}\epsilon\
\tilde{\tau}^{\epsilon}_{A}t^{\epsilon}_{-B}$ (19)

###  II.2 Gauge invariant quantities

We have described the spinors associated to an edge. Consider now two edges
$e_{1},e_{2}$ meeting at a vertex $v$ and incident to the same corner of
$\Gamma$. The ribbon edges $R(e_{1})$ and $R(e_{2})$ share a corner where we
have a spinor of $e_{1}$ and a spinor of $e_{2}$. Their scalar product is
gauge-invariant. Given fixed orientations of the edges, there are four
possible products (each spinor and its dual). There are moreover four
configurations of orientations, shown in Figure 5. For instance, the four
scalar products for the bottom right configuration are $\langle
t_{2}|\tau_{1}\rangle$, $\langle t_{2}|\tau_{1}]$, $[t_{2}|\tau_{1}\rangle$
and $[t_{2}|\tau_{1}]$.

Figure 5:  Two edges meet at a vertex and share a corner. There are four
configurations of orientations and we indicate the spinors at the common
corner.

Using the notations (15) and (16), we can give a uniform expression for the
four scalar products at fixed orientations. For example, when both edges are
outgoing,

$E^{\epsilon_{2},\epsilon_{1}}_{e_{2}e_{1}}:=\sum_{A=\pm
1/2}\tau^{\epsilon_{2}}_{2,-A}t^{\epsilon_{1}}_{1,A}=\begin{cases}\langle
t_{2}|\tau_{1}]&\text{for $\epsilon_{1}=+,\epsilon_{2}=+$}\\\ \langle
t_{2}|\tau_{1}\rangle&\text{for $\epsilon_{1}=-,\epsilon_{2}=+$}\\\
[t_{2}|\tau_{1}]&\text{for $\epsilon_{1}=+,\epsilon_{2}=-$}\\\
[t_{2}|\tau_{1}\rangle&\text{for $\epsilon_{1}=-,\epsilon_{2}=-$}\end{cases}$
(20)

The other orientations are obtained by changing $\tau_{1}$ to
$\tilde{\tau}_{1}$ and $t_{2}$ to $\tilde{t}_{2}$ and the invariant is still
denoted $E^{\epsilon_{2},\epsilon_{1}}_{e_{2},e_{1}}$.

It will be convenient to encode all orientations and have a fully uniform way
of writing the invariant. We orient the corner between $e_{1}$ and $e_{2}$
counter-clockwise. We say that the orientation $o_{i}$ of $e_{i}$ for $i=1,2$
is positive if it matches that of the corner, and negative otherwise. We
denote the spinors meeting there as $t_{e_{1}v}$ and $t_{e_{2}v}$ according to

$\begin{array}[]{|l|l|}\hline\cr&t_{e_{1}v}\\\ \hline\cr
o_{1}=1&\tilde{\tau}_{1}\\\ \hline\cr o_{1}=-1&\tau_{1}\\\
\hline\cr\end{array}\hskip
56.9055pt\begin{array}[]{|l|l|}\hline\cr&t_{e_{2}v}\\\ \hline\cr
o_{2}=1&t_{2}\\\ \hline\cr o_{2}=-1&\tilde{t}_{2}\\\ \hline\cr\end{array}$
(21)

so that

$E^{\epsilon_{2},\epsilon_{1}}_{e_{2}e_{1}}=\begin{cases}\langle
t_{e_{2}v}|t_{e_{1}v}]&\text{for $\epsilon_{1}=+,\epsilon_{2}=+$}\\\ \langle
t_{e_{2}v}|t_{e_{1}v}\rangle&\text{for $\epsilon_{1}=-,\epsilon_{2}=+$}\\\
[t_{e_{2}v}|t_{e_{1}v}]&\text{for $\epsilon_{1}=+,\epsilon_{2}=-$}\\\
[t_{e_{2}v}|t_{e_{1}v}\rangle&\text{for
$\epsilon_{1}=-,\epsilon_{2}=-$}\end{cases}$ (22)

###  II.3 Hamiltonian constraint

By plugging $u$ and $\tilde{u}$ from (12) into the flatness constraint (8),
one obtains a spinorial expression of the constraint. By then taking the
matrix elements of the constraints between different spinors, we get some
scalar constraints which we call Hamiltonian constraints. They are the
$\kappa$-deformed versions of Bonzom:2011nv .

We first write the Hamiltonian constraints generally, i.e. on faces of
arbitrary lengths, then specialize them to the case of faces of length 3.

####  II.3.1 The Hamiltonian on a face of arbitrary degree

Let $f$ be a face of length $d$. We will introduce a constraint, derived from
the flatness constraint, for every pair of edges $(e,e^{\prime})$ around $f$.
Label the edges counter-clockwise around $f$ as $e_{1},\dotsc,e_{d}$. Without
loss of generality, we set the pair $(e,e^{\prime})$ which labels our function
to $(e_{1},e_{k})$ for $k\in\\{2,\dotsc,d\\}$. Label the vertices around $f$
as $v_{1},\dotsc,v_{d}$ counter-clockwise, such that $e_{i}$ is incident to
$v_{i}$ and $v_{i+1}$, for $i=1,\dotsc,d\mod d$, as shown in Figure 6. We
assume that $f$ visits each vertex and edge exactly once (as when $\Gamma$ is
dual to a simplicial complex), so that all $e_{i}$s and $v_{i}$s are distinct.

Figure 6: A sunny graph with edges $e_{1},\cdots,e_{d}$ counterclockwise
oriented around the face $f$. Each triple of edges
$(e_{i},e_{i-1},e^{\prime}_{i})$ are incident to a vertex $v_{i}$.

By convention, we denote the orientation $o_{i}=1$ if $e_{i}$ is counter-
clockwise and $o_{i}=-1$ elsewise (this is the relative orientation of the
edge with respect to the counter-clockwise orientation of $f$). With the
notation $u_{ev}$ introduced earlier, the flatness constraint reads
$u_{e_{d}v_{1}}\dotsm u_{e_{2}v_{3}}u_{e_{1}v_{2}}=\mathbbm{1}$ in
$\mathrm{SU}(2)$. In order to simplify the notations a bit, we will use

$u_{e_{i}f}:=u_{e_{i}v_{i+1}}=\begin{cases}\tilde{u}_{e_{i}}^{-1}\,,\qquad\text{if
$o_{1}=1$}\,,\\\ u_{e_{i}}\,,\qquad\text{if $o_{1}=-1$}\,.\end{cases}$ (23)

Furthermore we denote $t_{e_{i}v_{i}}$ the spinor along the long edge of
$R(e_{i})$ which is incident to both $f$ and $v_{i}$. It is determined by the
orientation of $e_{i}$,

$\displaystyle o_{i}=1$ $\displaystyle\quad\Rightarrow\quad
t_{e_{i}v_{i}}=t_{e_{i}}\quad\text{and}\quad
t_{e_{i}v_{i+1}}=\tilde{\tau}_{e_{i}}\,,$ (24) $\displaystyle o_{i}=-1$
$\displaystyle\quad\Rightarrow\quad
t_{e_{i}v_{i}}=\tilde{t}_{e_{i}}\quad\text{and}\quad
t_{e_{i}v_{i+1}}=\tau_{e_{i}}\,.$

Notice that we can combine the parallel transport relations (13) with the
notations (15), (16) to relate the spinors which are on both ends of the long
edge of $e_{i}$ incident to $f$,

$t^{\epsilon_{i}}_{e_{i}v_{i},-A}=-o_{i}\sum_{B=\pm
1/2}t_{e_{i}v_{i+1},-B}^{-\epsilon_{i}}(-1)^{\frac{1}{2}-B}\,u_{e_{i}f,BA}\,,\qquad
t_{e_{i}v_{i+1},A}^{\epsilon_{i}}=o_{i}\sum_{B=\pm
1/2}u_{e_{i}f,AB}\,(-1)^{\frac{1}{2}-B}t_{e_{i}v_{i},B}^{-\epsilon_{i}}\,.$
(25)

The flatness constraint on $f$ is thus $u_{e_{d}f}\dotsm
u_{e_{1}f}=\mathbbm{1}$. Assume momentarily that all edges are counter-
clockwise. Then,
$\tilde{u}_{e_{d}}^{-1}\dotsm\tilde{u}_{e_{1}}^{-1}=\mathbbm{1}$ implies for
all $k$

$[t_{e_{k}}|\tilde{u}_{e_{k-1}}^{-1}\dotsm\tilde{u}_{e_{2}}^{-1}|\tilde{\tau}_{e_{1}}\rangle=[t_{e_{k}}|\tilde{u}_{e_{k}}\tilde{u}_{e_{k+1}}\dotsm\tilde{u}_{e_{d}}\tilde{u}_{e_{1}}|\tilde{\tau}_{e_{1}}\rangle=\langle\tilde{\tau}_{e_{k}}|\tilde{u}_{e_{k+1}}\dotsm\tilde{u}_{e_{d}}|t_{e_{1}}]\,.$
(26)

In the first equality, we have used the constraint itself, while in the second
equality we have used the parallel transport relations on the edges $e_{1}$
and $e_{k}$. By then rewriting $\tilde{u}_{1_{1}},\dotsc,\tilde{u}_{e_{d}}$
with (12), one obtains the following result: a constraint written as a sum of
products of scalar invariants living on the corners around the face.
Obviously, one can change $[t_{e_{k}}|$ to $\langle t_{e_{k}}|$ and
$|\tilde{\tau}_{e_{1}}\rangle$ to $|\tilde{\tau}_{e_{1}}]$ without changing
that result (qualitatively). Similarly, one should be able to write this
function for arbitrarily chosen edge orientations. The notations we have
introduced will help us write it in the most generic way.

Going back to arbitrary edge orientations around $f$, we consider

$E_{e_{1}\to e_{k}}^{\epsilon_{1},\epsilon_{k}}=\sum_{A,B=\pm
1/2}t_{e_{k}v_{k},-A}^{\epsilon_{k}}\left(u_{e_{k-1}f}\cdots
u_{e_{2}f}\right)_{AB}t_{e_{1}v_{2},B}^{\epsilon_{1}}$ (27)

as the generalization of the left hand side of (26). Using the parallel
transport relations (25), it reads

$E_{e_{1}\to e_{k}}^{\epsilon_{1},\epsilon_{k}}=-o_{1}o_{k}\sum_{C,D=\pm
1/2}t_{e_{k}v_{k+1},-C}^{-\epsilon_{k}}(-1)^{\frac{1}{2}+C}\bigl{(}u_{e_{k}f}u_{e_{k-1}f}\dotsm
u_{e_{2}f}u_{e_{1}f}\bigr{)}_{CD}(-1)^{\frac{1}{2}+D}t_{e_{1}v_{1},D}^{-\epsilon_{1}}\,.$
(28)

If the flatness constraint holds, the holonomy going counter-clockwise from
$e_{1}$ to $e_{k}$ can then be replaced with the holonomy the other way around
$f$, i.e. clockwise. We thus define

$E_{e_{1}\leftarrow e_{k}}^{\epsilon_{1},\epsilon_{k}}=\sum_{A,B=\pm
1/2}t_{e_{1}v_{1},-A}^{-\epsilon_{1}}\bigl{(}u_{e_{d}f}\dotsm
u_{e_{k+1}f}\bigr{)}_{AB}t_{e_{k}v_{k+1},B}^{-\epsilon_{k}}\,.$ (29)

So if the flatness constraint holds, then

$E_{e_{1}\to e_{k}}^{\epsilon_{1},\epsilon_{k}}+o_{1}o_{k}E_{e_{1}\leftarrow
e_{k}}^{\epsilon_{1},\epsilon_{k}}=0.$ (30)

Indeed, using the flatness constraint in (28) we get

$E_{e_{1}\to e_{k}}^{\epsilon_{1},\epsilon_{k}}=-o_{1}o_{k}\sum_{C,D=\pm
1/2}t_{e_{k}v_{k+1},-C}^{-\epsilon_{k}}(-1)^{\frac{1}{2}+C}\bigl{(}u_{e_{d}f}\dotsm
u_{e_{k+1}f}\bigr{)}^{-1}_{CD}(-1)^{\frac{1}{2}+D}t_{e_{1}v_{1},D}^{-\epsilon_{1}}\,.$
(31)

For any matrix $g\in\mathrm{SU}(2)$, the matrix elements of the inverse can be
written $g^{-1}_{CD}=(-1)^{\frac{1}{2}-D}g_{-D-C}(-1)^{\frac{1}{2}-C}$. This
can be used to transform the above expression into
$o_{1}o_{k}E_{e_{1}\leftarrow e_{k}}^{\epsilon_{1},\epsilon_{k}}$. The last
step to define our Hamiltonian constraints is to rewrite $E_{e_{1}\to
e_{k}}^{\epsilon_{1},\epsilon_{k}}$ and $E_{e_{1}\leftarrow
e_{k}}^{\epsilon_{1},\epsilon_{k}}$ in terms of scalars like (22). The matrix
elements of the holonomies are indeed

$\displaystyle u_{e_{i}f,A_{i+1}A_{i}}$
$\displaystyle=o_{i}\frac{1}{N_{e_{i}}}\sum_{\epsilon_{i}=\pm}\epsilon_{i}\
t_{e_{i}v_{i+1},A_{i+1}}^{\epsilon_{i}}\
t_{e_{i}v_{i},-A_{i}}^{\epsilon_{i}}\,,$ (32) $\displaystyle\text{with}\quad
N_{e_{i}}$
$\displaystyle=\frac{1}{2}\sqrt{\sum_{\epsilon_{i},\epsilon^{\prime}_{i}=\pm}\sum_{A,B=\pm
1/2}\epsilon\epsilon^{\prime}(-1)^{\frac{1}{2}-A}(-1)^{\frac{1}{2}-B}\,t_{e_{i},v_{i+1},A}^{\epsilon_{i}}t_{e_{i}v_{i+1},-A}^{-\epsilon_{i}}\,t_{e_{i},v_{i},B}^{\epsilon^{\prime}_{i}}t_{e_{i}v_{i},-B}^{-\epsilon^{\prime}_{i}}}\,,$
(33)

so that one can re-organize the products over the vertices instead of edges,

$\displaystyle E_{e_{1}\to e_{k}}^{\epsilon_{1},\epsilon_{k}}$
$\displaystyle=\sum_{\epsilon_{2},\dotsc,\epsilon_{k-1}=\pm\atop
A_{2},\dotsc,A_{k}=\pm
1/2}\biggl{(}\prod_{i=2}^{k-1}\frac{o_{i}\epsilon_{i}}{N_{e_{i}}}\biggr{)}\biggl{(}\prod_{i=2}^{k}t^{\epsilon_{i}}_{e_{i}v_{i},-A_{i}}t^{\epsilon_{i-1}}_{e_{i-1}v_{i},A_{i}}\biggr{)}\,,$
(34) $\displaystyle E_{e_{1}\leftarrow e_{k}}^{\epsilon_{1},\epsilon_{k}}$
$\displaystyle=(-1)^{d-k}\sum_{\epsilon_{k+1},\dotsc,\epsilon_{d}=\pm\atop
A_{k+1},\dotsc,A_{d+1}=\pm
1/2}\biggl{(}\prod_{i=k+1}^{d}\frac{o_{i}\epsilon_{i}}{N_{e_{i}}}\biggr{)}\biggl{(}\prod_{i=k+1}^{d+1}t^{-\epsilon_{i}}_{e_{i}v_{i},-A_{i}}t^{-\epsilon_{i-1}}_{e_{i-1}v_{i},A_{i}}\biggr{)}\,.$
(35)

We can now use the quadratic invariants defined in Equation (22),
$E_{e_{i}e_{i-1}}^{\epsilon_{i},\epsilon_{i-1}}=\sum_{A=\pm
1/2}t_{e_{i}v_{i},-A}^{\epsilon_{i}}t_{e_{i-1}v_{i},A}^{\epsilon_{i-1}}$,
which encodes all four scalar products of the two spinors meeting at $v_{i}$,
i.e.

$E_{e_{i}e_{i-1}}^{\epsilon_{i},\epsilon_{i-1}}=\begin{cases}\langle
t_{e_{i}v_{i}}|t_{e_{i-1}v_{i}}]&\qquad\text{for
$\epsilon_{i}=+,\epsilon_{i-1}=+$}\\\ \langle
t_{e_{i}v_{i}}|t_{e_{i-1}v_{i}}\rangle&\qquad\text{for
$\epsilon_{i}=+,\epsilon_{i-1}=-$}\\\
[t_{e_{i}v_{i}}|t_{e_{i-1}v_{i}}]&\qquad\text{for
$\epsilon_{i}=-,\epsilon_{i-1}=+$}\\\
[t_{e_{i}v_{i}}|t_{e_{i-1}v_{i}}\rangle&\qquad\text{for
$\epsilon_{i}=-,\epsilon_{i-1}=-$},\end{cases}$ (36)

where the spinors $t_{e_{i}v_{i}}$ and $t_{e_{i-1}v_{i-1}}$ are given by the
rule (24) according to the orientations. This leads us to the following
definition of the Hamiltonian constraints.

######  Definition II.1.

Let $f$ be a face of degree $d$, with edges labeled by $e_{1},\dotsc,e_{d}$
counter-clockwise around $f$. A Hamiltonian is associated to $f$ and a pair of
edges along $f$ with a sign attached to each of them. Without loss of
generality, the pair can be chosen to be $(e_{1},e_{k})$ with signs
$(\epsilon_{1},\epsilon_{k})\in\\{+,-\\}^{2}$, for $k\in\\{2,\dotsc,d\\}$, and
the Hamiltonian is

$h_{f,e_{1},e_{k}}^{\epsilon_{1},\epsilon_{k}}=\sum_{\epsilon_{2},\dotsc,\epsilon_{k-1}=\pm}\biggl{(}\prod_{i=2}^{k}\frac{o_{i}\epsilon_{i}}{N_{e_{i}}}E_{e_{i}e_{i-1}}^{\epsilon_{i},\epsilon_{i-1}}\biggr{)}+(-1)^{d-k}\epsilon_{1}\epsilon_{k}\frac{N_{e_{1}}}{N_{e_{k}}}\sum_{\epsilon_{k+1},\dotsc,\epsilon_{d}=\pm}\biggl{(}\prod_{i=k+1}^{d+1}\frac{o_{i}\epsilon_{i}}{N_{e_{i}}}E_{e_{i}e_{i-1}}^{-\epsilon_{i},-\epsilon_{i-1}}\biggr{)}\,.$
(37)

The Hamiltonian constraint (37) captures the flatness constraint completely
with all choices of pairs $(e_{1},e_{k})$ and of signs
$(\epsilon_{1},\epsilon_{k})$. The proof is the same as in the vector case at
$\kappa=0$, see Bonzom:2011hm .

####  II.3.2 Application to faces of degree three

Figure 7:  On the left, a triangular face with its adjacent edges. On the
right, we have shown the ribbon graph it gives rise to.

Let us discuss more explicitly the case of triangular faces. We use the
notations and orientations of Figure 7 as an example. In particular
$o_{1}=o_{6}=-1$ and $o_{2}=1$. Here there are three choices of pairs of edges
(which label the Hamiltonians (37)), which correspond to the three corners of
the face.

On a corner, say between the edges $e_{2}$ and $e_{6}$, there are four
invariant quantities quadratic in the spinors, $\langle
t_{2}|\tau_{6}\rangle,\langle
t_{2}|\tau_{6}],[t_{2}|\tau_{6}\rangle,[t_{2}|\tau_{6}]$ which are encoded in
the scalar product (22),

$E_{26}^{\epsilon_{2},\epsilon_{6}}=\sum_{A=\pm
1/2}t_{2,-A}^{\epsilon_{2}}\tau_{6,A}^{\epsilon_{6}}=\begin{cases}\langle
t_{2}|t_{6}]&\text{for $\epsilon_{2}=\epsilon_{6}=+$},\\\ \langle
t_{2}|t_{6}\rangle&\text{for $\epsilon_{2}=-\epsilon_{6}=+$},\\\
[t_{2}|t_{6}]&\text{for $\epsilon_{2}=-\epsilon_{6}=-$},\\\
[t_{2}|t_{6}\rangle&\text{for $\epsilon_{2}=\epsilon_{6}=-$}\,.\end{cases}$
(38)

Similarly at the corners between $e_{1},e_{2}$ and $e_{6},e_{1}$,

$E_{12}^{\epsilon_{1},\epsilon_{2}}=\sum_{A=\pm
1/2}\tilde{t}_{1,-A}^{\epsilon_{1}}\tilde{\tau}_{2,A}^{\epsilon_{2}},\qquad
E_{61}^{\epsilon_{6},\epsilon_{1}}=\sum_{A=\pm
1/2}\tilde{t}_{6,-A}^{\epsilon_{6}}\tau_{1,A}^{\epsilon_{1}}\,.$ (39)

The flatness constraint $\tilde{u}_{2}u_{1}^{-1}u_{6}^{-1}=\mathbb{I}$ implies
that if $|\tau_{6}\rangle$ and $|\tau_{6}]$ are transported around the face
via $\tilde{u}_{2}u_{1}^{-1}u_{6}^{-1}$, the above quadratic quantities are
left unchanged, that is

$\displaystyle\langle
t_{2}|\tilde{u}_{2}u_{1}^{-1}u_{6}^{-1}|\tau_{6}\rangle=\langle
t_{2}|\tau_{6}\rangle\,,$ $\displaystyle\langle
t_{2}|\tilde{u}_{2}u_{1}^{-1}u_{6}^{-1}|\tau_{6}]=\langle t_{2}|\tau_{6}]\,,$
(40)
$\displaystyle[t_{2}|\tilde{u}_{2}u_{1}^{-1}u_{6}^{-1}|\tau_{6}\rangle=[t_{2}|\tau_{6}\rangle\,,$
$\displaystyle[t_{2}|\tilde{u}_{2}u_{1}^{-1}u_{6}^{-1}|\tau_{6}]=[t_{2}|\tau_{6}]\,.$

Similarly at the corners where $e_{1},e_{2}$ and $e_{6},e_{1}$ meet
respectively,

$\displaystyle\langle\tilde{t}_{1}|u_{1}^{-1}u_{6}^{-1}\tilde{u}_{2}|\tilde{\tau}_{2}\rangle=\langle\tilde{t}_{1}|\tilde{\tau}_{2}\rangle\,,$
$\displaystyle\langle\tilde{t}_{6}|u_{6}^{-1}\tilde{u}_{2}u_{1}^{-1}|\tau_{1}\rangle=\langle\tilde{t}_{6}|\tau_{1}\rangle\,,$
(41)
$\displaystyle\langle\tilde{t}_{1}|u_{1}^{-1}u_{6}^{-1}\tilde{u}_{2}|\tilde{\tau}_{2}]=\langle\tilde{t}_{1}|\tilde{\tau}_{2}]\,,$
$\displaystyle\langle\tilde{t}_{6}|u_{6}^{-1}\tilde{u}_{2}u_{1}^{-1}|\tau_{1}]=\langle\tilde{t}_{6}|\tau_{1}]\,,$
$\displaystyle[\tilde{t}_{1}|u_{1}^{-1}u_{6}^{-1}\tilde{u}_{2}|\tilde{\tau}_{2}\rangle=[\tilde{t}_{1}|\tilde{\tau}_{2}\rangle\,,$
$\displaystyle[\tilde{t}_{6}|u_{6}^{-1}\tilde{u}_{2}u_{1}^{-1}|\tau_{1}\rangle=[\tilde{t}_{6}|\tau_{1}\rangle\,,$
$\displaystyle[\tilde{t}_{1}|u_{1}^{-1}u_{6}^{-1}\tilde{u}_{2}|\tilde{\tau}_{2}]=[\tilde{t}_{1}|\tilde{\tau}_{2}]\,,$
$\displaystyle[\tilde{t}_{6}|u_{6}^{-1}\tilde{u}_{2}u_{1}^{-1}|\tau_{1}]=[\tilde{t}_{6}|\tau_{1}]\,.$

In fact, this set of constraints simply amounts to rewriting the constraint
$\tilde{u}_{2}u_{1}^{-1}u_{6}^{-1}=\mathbb{I}$ in the coherent state basis.
Therefore, as long as those vectors are generic (hence linearly independent),
this whole set is equivalent to
$\tilde{u}_{2}u_{1}^{-1}u_{6}^{-1}=\mathbb{I}$.

Let us consider the constraint $\langle
t_{2}|\tilde{u}_{2}u_{1}^{-1}u_{6}^{-1}|\tau_{6}]-\langle t_{2}|\tau_{6}]$ and
rewrite it like in (37). Use the parallel transport relations
$u_{6}^{-1}|\tau_{6}]=-|\tilde{t}_{6}\rangle$ and $\langle
t_{2}|\tilde{u}_{2}=[\tilde{\tau}_{2}|$ which gives $\langle
t_{2}|\tilde{u}_{2}u_{1}^{-1}u_{6}^{-1}|\tau_{6}]=-[\tilde{\tau}_{2}|u_{1}^{-1}|\tilde{t}_{6}\rangle=[\tilde{t}_{6}|u_{1}|\tilde{\tau}_{2}\rangle$.
Then use
$u_{1}=\frac{1}{N_{e_{1}}}(|\tau_{1}\rangle[\tilde{t}_{1}|-|\tau_{1}]\langle\tilde{t}_{1}|)$
so that the constraint becomes

$\displaystyle\langle
t_{2}|\tilde{u}_{2}u_{1}^{-1}u_{6}^{-1}|\tau_{6}]-\langle t_{2}|\tau_{6}]$
$\displaystyle=\frac{1}{N_{e_{1}}}\bigl{(}[\tilde{t}_{6}|\tau_{1}\rangle[\tilde{t}_{1}|\tilde{\tau}_{2}\rangle-[\tilde{t}_{6}|\tau_{1}]\langle\tilde{t}_{1}|\tilde{\tau}_{2}\rangle\bigr{)}-\langle
t_{2}|\tau_{6}]$ (42)
$\displaystyle=-\frac{1}{N_{e_{1}}}\sum_{\epsilon=\pm}\epsilon\,E^{-,\epsilon}_{61}E^{\epsilon,-}_{12}-E_{26}^{+,+}$

which is exactly the specialization of (37) to $d=3$, $k=2$,
$o_{1}=o_{6}=-o_{2}=-1$ and $\epsilon_{2}=\epsilon_{6}=+$,

$h_{e_{2}e_{6}}^{\epsilon_{2},\epsilon_{6}}=E_{26}^{\epsilon_{2},\epsilon_{6}}+\frac{1}{N_{e_{1}}}\sum_{\epsilon_{1}=\pm}\epsilon_{1}E_{12}^{\epsilon_{1},-\epsilon_{2}}E_{61}^{-\epsilon_{6},\epsilon_{1}}\,.$
(43)

where we recall that
$N_{e_{1}}=\sqrt{\langle\tau_{1}|\tau_{1}\rangle\langle\tilde{t}_{1}|\tilde{t}_{1}\rangle}$.

This way, the Hamiltonian constraint does not involve holonomy variables
anymore like in (40), but only the quadratic invariants of spinors.

##  III Quantum hamiltonian constraint

We now proceed to the quantization of the system. The aim is to quantize the
Hamiltonian constraints and solve them at the quantum level. This requires
quantizing the quadratic invariant
$E^{\epsilon_{i},\epsilon_{i-1}}_{e_{i}e_{i-1}}$. It has been constructed
consistently with the quantization of holonomies and fluxes in the companion
paper Bonzom:2021rib . We therefore start with recalling the main ingredients
before proceeding to the construction of the quantum Hamiltonian.

###  III.1 Quantum deformed spinors

Let $q:=e^{\hbar\kappa}$ and denote
$[n]:=\frac{q^{\frac{n}{2}}-q^{-\frac{n}{2}}}{q^{\frac{1}{2}}-q^{-\frac{1}{2}}}$
the $q$-numbers. The $\kappa$-deformed spinors are quantized as $q$-bosons
(the same way they are at $\kappa=0$ Livine:2012 ). Consider two independent
pairs of $q$-boson operators $(a,a^{\dagger})$ and $(b,b^{\dagger})$
satisfying the relations

$\begin{array}[]{llll}aa^{\dagger}-q^{\mp\frac{1}{2}}a^{\dagger}a=q^{\pm\frac{N_{a}}{2}}\,,&a^{\dagger}a-q^{\pm\frac{1}{2}}aa^{\dagger}=-q^{\pm\frac{N_{a}+1}{2}}\,,&[N_{a},a^{\dagger}]=a^{\dagger}\,,&[N_{a},a]=-a\,,\\\
bb^{\dagger}-q^{\mp\frac{1}{2}}b^{\dagger}b=q^{\pm\frac{N_{b}}{2}}\,,&b^{\dagger}b-q^{\pm\frac{1}{2}}bb^{\dagger}=-q^{\pm\frac{N_{b}+1}{2}}\,,&[N_{b},b^{\dagger}]=b^{\dagger}\,,&[N_{b},b]=-b\,.\end{array}$
(44)

with $a^{\dagger}a=[N_{a}]$, $aa^{\dagger}=[N_{a}+1]$, $b^{\dagger}b=[N_{b}]$,
$bb^{\dagger}=[N_{b}+1]$. We furthermore introduce two other independent pairs
of $q$-bosons denoted $(\tilde{a},\tilde{a}^{\dagger})$ and
$(\tilde{b},\tilde{b}^{\dagger})$, and satisfying the same relations as above.
The tilde and non-tilde operators are chosen to commute with each other.

We will then use the following quantization map

$\begin{array}[]{lll}(\zeta^{\kappa}_{0},\zeta^{\kappa}_{1})\rightarrow(a,b)\,,&(\bar{\zeta}^{\kappa}_{0},\bar{\zeta}^{\kappa}_{1})\rightarrow(a^{\dagger},b^{\dagger})\,,&(N_{0},N_{1})\rightarrow(N_{a},N_{b})\,,\\\
(\tilde{\zeta}^{\kappa}_{0},\tilde{\zeta}^{\kappa}_{1})\rightarrow(\tilde{a},\tilde{b})\,,&(\bar{\tilde{\zeta}}^{\kappa}_{0},\bar{\tilde{\zeta}}^{\kappa}_{1})\rightarrow(\tilde{a}^{\dagger},\tilde{b}^{\dagger})\,,&(\tilde{N}_{0},\tilde{N}_{1})\rightarrow(\tilde{N}_{a},\tilde{N}_{b})\,.\end{array}$
(45)

In previous works by the first and second author, the fluxes $\ell$ and
$\tilde{\ell}$ had been quantized in terms of the quantum algebra
${{\mathcal{U}}_{q}({\mathfrak{s}}{\mathfrak{u}}(2))}$. This can be replicated
in a manner which is consistent with the $q$-bosons. Indeed, the Jordan map
builds generators of ${{\mathcal{U}}_{q}({\mathfrak{s}}{\mathfrak{u}}(2))}$
out of the above $q$-bosons,

$\begin{array}[]{lll}J_{+}=a^{\dagger}b\,,&J_{-}=ab^{\dagger}\,,&K=q^{\frac{J_{z}}{2}}=q^{\frac{N_{a}-N_{b}}{4}}\,,\\\
\tilde{J}_{+}=\tilde{a}^{\dagger}\tilde{b}\,,&\tilde{J}_{-}=\tilde{a}\tilde{b}^{\dagger}\,,&\tilde{K}=q^{\frac{\tilde{J}_{z}}{2}}=q^{\frac{\tilde{N}_{a}-\tilde{N}_{b}}{4}}\,,\end{array}$
(46)

where $J_{\pm}\,,K\equiv q^{\frac{J_{z}}{2}}$ and
$\tilde{J}_{\pm}\,,\tilde{K}\equiv q^{\frac{\tilde{J}_{z}}{2}}$ are two
independent copies of the
${{\mathcal{U}}_{q}({\mathfrak{s}}{\mathfrak{u}}(2))}$ generators satisfying
the relations

$KJ_{\pm}K^{{-1}}=q^{\pm\frac{1}{2}}J_{\pm}\,,\quad[J_{+},J_{-}]=[2J_{z}]\,,\quad\tilde{K}\tilde{J}_{\pm}\tilde{K}^{{-1}}=q^{\pm\frac{1}{2}}\tilde{J}_{\pm}\,,\quad[\tilde{J}_{+},\tilde{J}_{-}]=[2\tilde{J}_{z}]\,,$
(47)

and others vanish. Performing the quantization as follows,

$\begin{array}[]{lcl}\ell=\left(\begin{array}[]{cc}\exp(\frac{\kappa}{4}(N_{1}-N_{0}))&0\\\
-\kappa\bar{\zeta}_{0}^{\kappa}\zeta_{1}^{\kappa}&\exp(\frac{\kappa}{4}(N_{0}-N_{1}))\end{array}\right)&\quad\longrightarrow&L=\left(\begin{array}[]{cc}K^{-1}&0\\\
-q^{\frac{1}{4}}(q^{\frac{1}{2}}-q^{-\frac{1}{2}})J_{+}&K\end{array}\right)\,,\\\
\tilde{\ell}=\left(\begin{array}[]{cc}\exp(\frac{\kappa}{4}(\tilde{N}_{0}-\tilde{N}_{1}))&0\\\
\kappa\bar{\tilde{\zeta}}^{\kappa}_{0}\tilde{\zeta}^{\kappa}_{1}&\exp(\frac{\kappa}{4}(\tilde{N}_{1}-\tilde{N}_{0}))\end{array}\right)&\quad\longrightarrow&\widetilde{L}=\left(\begin{array}[]{cc}\tilde{K}&0\\\
q^{-\frac{1}{4}}(q^{\frac{1}{2}}-q^{-\frac{1}{2}})\tilde{J}_{+}&\tilde{K}^{-1}\end{array}\right)\,,\end{array}$
(48)

one finds, as in Bonzom:2021rib , that $L\in
F_{q^{-1}}(\mathrm{AN}(2))\cong{{\mathcal{U}}_{q^{-1}}({\mathfrak{s}}{\mathfrak{u}}(2))}$
and $\widetilde{L}\in
F_{q}(\mathrm{AN}(2))\cong{{\mathcal{U}}_{q}({\mathfrak{s}}{\mathfrak{u}}(2))}$
Semenov:1994po .

The map (45) quantizes the $\kappa$-deformed spinors (11) as follows333If we
do not distinguish the two copies of the $q$-bosons, one has
${\bf{t}}^{\epsilon}=\tilde{{\bf{t}}}^{\epsilon}$ and
$\bm{\uptau}^{\epsilon}={\widetilde{\bm{\uptau}}}^{\epsilon}$, which could
lead to a wrong quantization of the holonomies. However, it will not affect
the results when we consider the quantum Hamiltonian constraint as it is
defined by the product of scalar operators, which are given by the tensor
product of quantum spinors. Different quantum spinors in the expression of a
scalar operator (see (55)) act on different vector space thus do not
communicate. Nevertheless, distinguishing the tilde and non-tilde $q$-bosons
makes the quantum holonomies in terms of the quantum spinors well-defined as
$\mathrm{SU}_{q}(2)$ or $\mathrm{SU}_{q^{-1}}(2)$ elements (see Bonzom:2021rib
).

$\begin{array}[]{lllllll}|t\rangle=\left(\begin{array}[]{c}e^{\frac{\kappa
N_{1}}{4}}\zeta^{\kappa}_{0}\\\ e^{-\frac{\kappa
N_{0}}{4}}\zeta^{\kappa}_{1}\end{array}\right)&\longrightarrow&{\bf{t}}^{-}=\left(\begin{array}[]{c}{\bf{t}}^{-}_{+}\\\
{\bf{t}}^{-}_{-}\end{array}\right)=\left(\begin{array}[]{c}q^{\frac{N_{b}}{4}}a\\\
q^{-\frac{N_{a}}{4}}b\end{array}\right)\,,&&|t]=\left(\begin{array}[]{c}-e^{-\frac{\kappa
N_{0}}{4}}\bar{\zeta}^{\kappa}_{1}\\\ e^{\frac{\kappa
N_{1}}{4}}\bar{\zeta}^{\kappa}_{0}\end{array}\right)&\longrightarrow&{\bf{t}}^{+}=\left(\begin{array}[]{c}{\bf{t}}^{+}_{+}\\\
{\bf{t}}^{+}_{-}\end{array}\right)=\left(\begin{array}[]{c}-b^{\dagger}q^{-\frac{N_{a}+1}{4}}\\\
a^{\dagger}q^{\frac{N_{b}+1}{4}}\end{array}\right)\,,\\\\[11.38092pt]
|\tau\rangle=\left(\begin{array}[]{c}e^{-\frac{\kappa
N_{1}}{4}}\zeta_{0}^{\kappa}\\\ e^{\frac{\kappa
N_{0}}{4}}\zeta^{\kappa}_{1}\end{array}\right)&\longrightarrow&\bm{\uptau}^{-}=\left(\begin{array}[]{c}\bm{\uptau}^{-}_{+}\\\
\bm{\uptau}^{-}_{-}\end{array}\right)=\left(\begin{array}[]{c}q^{-\frac{N_{b}}{4}}a\\\
q^{\frac{N_{a}}{4}}b\end{array}\right)\,,&&|\tau]=\left(\begin{array}[]{c}-e^{\frac{\kappa
N_{0}}{4}}\bar{\zeta}^{\kappa}_{1}\\\ e^{-\frac{\kappa
N_{1}}{4}}\bar{\zeta}_{0}^{\kappa}\end{array}\right)&\longrightarrow&\bm{\uptau}^{+}=\left(\begin{array}[]{c}\bm{\uptau}^{+}_{+}\\\
\bm{\uptau}^{+}_{-}\end{array}\right)=\left(\begin{array}[]{c}-b^{\dagger}q^{\frac{N_{a}+1}{4}}\\\
a^{\dagger}q^{-\frac{N_{b}+1}{4}}\end{array}\right)\,,\\\\[11.38092pt]
|\tilde{t}\rangle=\left(\begin{array}[]{c}e^{\frac{\kappa\tilde{N}_{1}}{4}}\tilde{\zeta}^{\kappa}_{0}\\\
e^{-\frac{\kappa\tilde{N}_{0}}{4}}\tilde{\zeta}^{\kappa}_{1}\end{array}\right)&\longrightarrow&\tilde{{\bf{t}}}^{-}=\left(\begin{array}[]{c}\tilde{{\bf{t}}}^{-}_{+}\\\
\tilde{{\bf{t}}}^{-}_{-}\end{array}\right)=\left(\begin{array}[]{c}q^{\frac{\tilde{N}_{b}}{4}}\tilde{a}\\\
q^{-\frac{\tilde{N}_{a}}{4}}\tilde{b}\end{array}\right)\,,&&|\tilde{t}]=\left(\begin{array}[]{c}-e^{-\frac{\kappa\tilde{N}_{0}}{4}}\bar{\tilde{\zeta}}^{\kappa}_{1}\\\
e^{\frac{\kappa\tilde{N}_{1}}{4}}\bar{\tilde{\zeta}}^{\kappa}_{0}\end{array}\right)&\longrightarrow&\tilde{{\bf{t}}}^{+}=\left(\begin{array}[]{c}\tilde{{\bf{t}}}^{+}_{+}\\\
\tilde{{\bf{t}}}^{+}_{-}\end{array}\right)=\left(\begin{array}[]{c}-\tilde{b}^{\dagger}q^{-\frac{\tilde{N}_{a}+1}{4}}\\\
\tilde{a}^{\dagger}q^{\frac{\tilde{N}_{b}+1}{4}}\end{array}\right)\,,\\\\[11.38092pt]
|\tilde{\tau}\rangle=\left(\begin{array}[]{c}e^{-\frac{\kappa\tilde{N}_{1}}{4}}\tilde{\zeta}^{\kappa}_{0}\\\
e^{\frac{\kappa\tilde{N}_{0}}{4}}\tilde{\zeta}^{\kappa}_{1}\end{array}\right)&\longrightarrow&{\widetilde{\bm{\uptau}}}^{-}=\left(\begin{array}[]{c}{\widetilde{\bm{\uptau}}}^{-}_{+}\\\
{\widetilde{\bm{\uptau}}}^{-}_{-}\end{array}\right)=\left(\begin{array}[]{c}q^{-\frac{\tilde{N}_{b}}{4}}\tilde{a}\\\
q^{\frac{\tilde{N}_{a}}{4}}\tilde{b}\end{array}\right)\,,&&|\tilde{\tau}]=\left(\begin{array}[]{c}-e^{\frac{\kappa\tilde{N}_{0}}{4}}\bar{\tilde{\zeta}}^{\kappa}_{1}\\\
e^{-\frac{\kappa\tilde{N}_{1}}{4}}\bar{\tilde{\zeta}}^{\kappa}_{0}\end{array}\right)&\longrightarrow&{\widetilde{\bm{\uptau}}}^{+}=\left(\begin{array}[]{c}{\widetilde{\bm{\uptau}}}^{+}_{+}\\\
{\widetilde{\bm{\uptau}}}^{+}_{-}\end{array}\right)=\left(\begin{array}[]{c}-\tilde{b}^{\dagger}q^{\frac{\tilde{N}_{a}+1}{4}}\\\
\tilde{a}^{\dagger}q^{-\frac{\tilde{N}_{b}+1}{4}}\end{array}\right)\,.\end{array}$
(49)

Those objects are in fact really spinors for some
${{\mathcal{U}}_{q}({\mathfrak{s}}{\mathfrak{u}}(2))}$ actions:
${\bf{t}}^{\epsilon}$ and $\tilde{{\bf{t}}}^{\epsilon}$ are spinors under the
${{\mathcal{U}}_{q}({\mathfrak{s}}{\mathfrak{u}}(2))}$ adjoint right action,
while $\bm{\uptau}^{\epsilon}$ and ${\widetilde{\bm{\uptau}}}^{\epsilon}$ are
spinors under the ${{\mathcal{U}}_{q^{-1}}({\mathfrak{s}}{\mathfrak{u}}(2))}$
adjoint right action Bonzom:2021rib .

Notice that the map $q\to q^{-1}$ exchanges ${\bf{t}}^{\epsilon}$ with
$\bm{\uptau}^{\epsilon}$ and $\tilde{{\bf{t}}}^{\epsilon}$ with
${\widetilde{\bm{\uptau}}}^{\epsilon}$ (the operators
$a,b,\tilde{a},\tilde{b}$ are invariant under $q\to q^{-1}$, (44)).

###  III.2 Kinematical Hilbert space

The kinematical state space was defined in Bonzom:2021rib . We here describe
the corresponding spin network basis. Each edge $e$ of $\Gamma$ carries an
irreducible representation $V_{j_{e}}$ of
${{\mathcal{U}}_{q}({\mathfrak{s}}{\mathfrak{u}}(2))}$, characterized by its
spin $j_{e}\in{\mathbb{N}}/2$. The Gauss constraint then projects the tensor
product of the incident representations at each vertex onto the invariant
subspace.

We consider the usual magnetic basis $\\{|j,m\rangle\\}_{m=-j,\dotsc,j}$ on
each $V_{j}$, which diagonalizes $K$, i.e.
$K|j,m\rangle=q^{\frac{m}{2}}|j,m\rangle$. The $q$-bosons act on $V_{j}$ as

$\begin{array}[]{ll}a^{\dagger}|j,m\rangle=\sqrt{[j+m+1]}\,|j+\frac{1}{2},m+\frac{1}{2}\rangle\,,&a|j,m\rangle=\sqrt{[j+m]}\,|j-\frac{1}{2},m-\frac{1}{2}\rangle\,,\\\\[4.26773pt]
b^{\dagger}|j,m\rangle=\sqrt{[j-m+1]}\,|j+\frac{1}{2},m-\frac{1}{2}\rangle\,,&b|j,m\rangle=\sqrt{[j-m]}\,|j-\frac{1}{2},m+\frac{1}{2}\rangle\,,\\\\[4.26773pt]
N_{a}|j,m\rangle=(j+m)\,|j,m\rangle\,,&N_{b}|j,m\rangle=(j-m)\,|j,m\rangle\,.\end{array}$
(50)

In particular $a^{\dagger},b^{\dagger}$ ($a,b$) map $V_{j}$ to $V_{j+1/2}$ (to
$V_{j-1/2}$). The tilde $q$-bosons
$\tilde{a},\tilde{a}^{\dagger},\tilde{b},\tilde{b}^{\dagger}$ act on these
basis in the same way by definition. It leads to the Wigner-Eckart theorem for
the quantum spinors (49):

$\displaystyle\langle
j_{1},m_{1}|{\bf{t}}^{\epsilon}_{m}|j_{2},m_{2}\rangle=\delta_{j_{1},j_{2}+\epsilon/2}\,\sqrt{[d_{j_{1}}]}\,_{q}C^{j_{1}\,\,\,\frac{1}{2}\,\,\,j_{2}}_{m_{1}\,m\,m_{2}}\,,$
(51a) $\displaystyle\langle
j_{1},m_{1}|\bm{\uptau}^{\epsilon}_{m}|j_{2},m_{2}\rangle=\delta_{j_{1},j_{2}+\epsilon/2}\,\sqrt{[d_{j_{1}}]}\,_{q^{-1}}C^{j_{1}\,\,\,\frac{1}{2}\,\,\,j_{2}}_{m_{1}\,m\,m_{2}}\,,$
(51b) $\displaystyle\langle
j_{1},m_{1}|\tilde{{\bf{t}}}^{\epsilon}_{m}|j_{2},m_{2}\rangle=\delta_{j_{1},j_{2}+\epsilon/2}\,\sqrt{[d_{j_{1}}]}\,_{q}C^{j_{1}\,\,\,\frac{1}{2}\,\,\,j_{2}}_{m_{1}\,m\,m_{2}}\,,$
(51c) $\displaystyle\langle
j_{1},m_{1}|{\widetilde{\bm{\uptau}}}^{\epsilon}_{m}|j_{2},m_{2}\rangle=\delta_{j_{1},j_{2}+\epsilon/2}\,\sqrt{[d_{j_{1}}]}\,_{q^{-1}}C^{j_{1}\,\,\,\frac{1}{2}\,\,\,j_{2}}_{m_{1}\,m\,m_{2}}\,,$
(51d)

where ${}_{q}C^{j_{1}}_{m_{1}}{}^{j_{2}}_{m_{2}}{}^{j_{3}}_{m_{3}}$ is the
Clebsch-Gordan coefficient for
${{\mathcal{U}}_{q}({\mathfrak{s}}{\mathfrak{u}}(2))}$.

Before enforcing the Gauss constraints, the space of states is
$\bigoplus_{\\{j_{e}\\}}\bigotimes_{e}V_{j_{e}}\otimes V_{j_{e}}^{*}$, where
$V_{j_{e}}$ is associated to the target end of $e$ and $V^{*}_{j_{e}}$ to its
source. At each vertex $v$, the Gauss constraint enforces a projection of the
tensor product of the vectors meeting at $v$ onto the trivial representation.
If the edges meeting at $v$ are denoted $e_{1},\dotsc,e_{n}$, we further
denote $\operatorname{Inv}(j_{e_{1}v}\otimes\dotsb\otimes j_{e_{n}v})$ the
space of _intertwiners_ , i.e. the invariant subspace of the tensor product
$V_{j_{e_{1}}}\otimes\dotsb\otimes V_{j_{e_{n}}}$ if all $e_{i}$s are incoming
at $v$, and we dualize to $V_{j_{e_{i}}}^{*}$ if $e_{i}$ is outgoing at $v$.
Therefore, the kinematical Hilbert space is given by

$\mathcal{H}_{\text{kin}}=\bigoplus_{\\{j_{e}\in{\mathbb{N}}/2\\}}\bigotimes_{v}\operatorname{Inv}(j_{e_{1}v}\otimes\dotsb\otimes
j_{e_{n}v})\,.$ (52)

A basis is obtained at fixed spins $\\{j_{e}\\}$ by specifying a basis of
$\operatorname{Inv}(j_{e_{1}v}\otimes\dotsb\otimes j_{e_{n}v})$ for all $v$.
We denote an element of this space as $i_{j_{e_{1}v}\dotsb j_{e_{n}(v)v}}$
(the letter $i$ refering to intertwiner). A kinematical state $|\psi\rangle$
thus admits the expansion

$|\psi\rangle=\sum_{\\{j_{e}\\}}\sum_{\\{i_{v}\\}}\psi(\\{j_{e},i_{v}\\})\
|\\{j_{e},i_{v}\\}\rangle,$ (53)

with

$|\\{j_{e},i_{v}\\}\rangle=\bigotimes_{v}i_{j_{e_{1}v}\dotsb j_{e_{n}(v)v}}$
(54)

where the sum over each $i_{v}$ runs over a basis of the invariant space at
$v$. The state $|\\{j_{e},i_{v}\\}\rangle$ is called a _spin network state_.
They form a basis of $\mathcal{H}_{\text{kin}}$. In the case of trivalent
vertices, the invariant space $\operatorname{Inv}(j_{e_{1}v}\otimes
j_{e_{2}v}\otimes j_{e_{3}v})$ is one-dimensional.

###  III.3 Scalar operators

We proceed to the quantization of the quadratic invariant (22),
$E^{\epsilon_{2},\epsilon_{1}}_{e_{2}e_{1}}$. The quantization of the spinors
$t_{e_{1}v},t_{e_{2}v}$ themselves is given by (49). As can be seen from (51),
the operators
${\bf{t}}^{\epsilon},\bm{\uptau}^{\epsilon},\tilde{{\bf{t}}}^{\epsilon},{\widetilde{\bm{\uptau}}}^{\epsilon}$
transform as spinors under
${{\mathcal{U}}_{q}({\mathfrak{s}}{\mathfrak{u}}(2))}$ or
${{\mathcal{U}}_{q^{-1}}({\mathfrak{s}}{\mathfrak{u}}(2))}$. Therefore to
ensure that $E^{\epsilon_{2},\epsilon_{1}}_{e_{2}e_{1}}$ is quantized as a
quantum group invariant, one needs to contract the two spinor operators via
some $q$-Clebsh-Gordan coefficients,
${}_{q}C^{\frac{1}{2}\,\,\,\,\frac{1}{2}\,\,0}_{A\,B\,0}=(-1)^{\frac{1}{2}-A}q^{\frac{A}{2}}\delta_{B,-A}$
or
${}_{q^{-1}}C^{\frac{1}{2}\,\,\,\,\frac{1}{2}\,\,0}_{A\,B\,0}=(-1)^{\frac{1}{2}-A}q^{-\frac{A}{2}}\delta_{B,-A}$.

Since changing the orientation of an edge exchanges ${\bf{t}}^{\epsilon}$ with
$\tilde{{\bf{t}}}^{\epsilon}$, and $\bm{\uptau}^{\epsilon}$ with
${\widetilde{\bm{\uptau}}}^{\epsilon}$, and since ${\bf{t}}^{\epsilon}$ and
$\tilde{{\bf{t}}}^{\epsilon}$ are in fact the same operator (and also
$\bm{\uptau}^{\epsilon}$ and ${\widetilde{\bm{\uptau}}}^{\epsilon}$), one
would expect the quantum operator for
$E^{\epsilon_{2},\epsilon_{1}}_{e_{2}e_{1}}$ to be independent of the
orientations of $e_{1}$ and $e_{2}$. This is entirely possible to proceed this
way.

We will however not do so. Our motivation is that while
$E^{\epsilon_{2},\epsilon_{1}}_{e_{2}e_{1}}$ would be independent of
orientations, the vector space on which it acts does depend on orientations
($V_{j}^{*}$ versus $V_{j}$). Therefore the action on an intertwiner would in
fact depend explictly on the orientations. Instead, we decide to perform the
quantization so that its action on intertwiners is independent of
orientations.

This requires changing the spinor operator to its $q^{-1}$ version when
flipping the orientation. Obviously, this exchanges the $\mathbf{t}$s with the
$\bm{\uptau}$s. However, we prefer to keep the same letter for the spinor
operator because we think exchanging $\mathbf{t}$s with $\bm{\uptau}$s could
be confusing in the ribbon picture. We therefore define
$\overline{{{\bf{t}}}^{\epsilon}}:=\bm{\uptau}^{\epsilon}$ and same with the
tildes, and eventually444 Note that the definition of $o_{1}$ is opposite to
that in Bonzom:2021rib  which leads to a slight difference for the definition
of the scalar operator (55) compared to that in Bonzom:2021rib . This is
because, in Bonzom:2021rib , $o_{1}$ and $o_{2}$ are considered to be the
orientation of edges relative to vertex $v$ and $+1$ (resp. $-1$) denotes
outgoing (resp. incoming). Here, on contrast, $o_{1}$ and $o_{2}$ are
considered to be the orientation of edges relative to the orientation of the
face $f$. As an example, when edges $e_{2}$ and $e_{1}$ are both outgoing to
$v$, $e_{2}$ is counter-clockwise while $e_{1}$ is clockwise relative to $f$.

${\bf
E}^{\epsilon_{2},\epsilon_{1}}_{e_{2}e_{1}}=-o_{1}\sqrt{[2]}(-1)^{\frac{1-o_{1}}{2}\frac{1+\epsilon_{1}}{2}}(-1)^{\frac{1-o_{2}}{2}\frac{1+\epsilon_{2}}{2}}\sum_{A=\pm\frac{1}{2}}\,{}_{q^{o_{1}}}C^{\frac{1}{2}\,\,\,\,\frac{1}{2}\,\,0}_{A\,-A\,0}\,{\bf
T}_{e_{1}v,A}^{o_{1}\epsilon_{1}}\otimes{\bf
T}_{e_{2}v,-A}^{-o_{2}\epsilon_{2}}\\\
=\begin{cases}\sum_{A=\pm\frac{1}{2}}(-1)^{\frac{1}{2}-A}q^{\frac{A}{2}}\epsilon_{2}\,\tilde{{\bf{t}}}_{A}^{\epsilon_{1}}\otimes{\widetilde{\bm{\uptau}}}_{-A}^{\epsilon_{2}}&\text{
for }\,-o_{1}=o_{2}=-1\\\\[5.69046pt]
\sum_{A=\pm\frac{1}{2}}(-1)^{\frac{1}{2}+A}q^{\frac{A}{2}}\tilde{{\bf{t}}}_{A}^{\epsilon_{1}}\otimes\overline{\bm{\uptau}_{-A}^{-\epsilon_{2}}}&\text{
for }\,o_{1}=o_{2}=1\\\\[5.69046pt]
\sum_{A=\pm\frac{1}{2}}(-1)^{\frac{1}{2}-A}q^{-\frac{A}{2}}\epsilon_{1}\epsilon_{2}\,\overline{{\bf{t}}_{A}^{-\epsilon_{1}}}\otimes{\widetilde{\bm{\uptau}}}_{-A}^{\epsilon_{2}}&\text{
for }\,-o_{1}=-o_{2}=1\\\\[5.69046pt]
\sum_{A=\pm\frac{1}{2}}(-1)^{\frac{1}{2}+A}q^{-\frac{A}{2}}\epsilon_{1}\,\overline{{\bf{t}}_{A}^{-\epsilon_{1}}}\otimes\overline{\bm{\uptau}_{-A}^{-\epsilon_{2}}}&\text{
for }\,-o_{1}=o_{2}=1\end{cases}\,,$ (55)

where ${\bf
T}_{e_{2}v,A}^{-o_{2}\epsilon_{2}}={{\bf{t}}}_{e_{2}v,A}^{\epsilon_{2}}$ if
$o_{2}=-1$ while ${\bf
T}_{e_{2}v,A}^{-o_{2}\epsilon_{2}}=\overline{{{\bf{t}}}_{e_{2}v,A}^{-\epsilon_{2}}}$
if $o_{2}=1$, and similarly for ${\bf T}_{e_{1}v,A}^{o_{1}\epsilon_{1}}$. We
then extend this definition to the space
$\operatorname{Inv}(j_{e_{1}v}\otimes\dotsb\otimes j_{e_{n}v})$ of invariant
vectors at $v$ by tensoring with the identity as necessary. It comes

${\bf
E}^{\epsilon_{2},\epsilon_{1}}_{e_{2}e_{1}}\,i_{j_{1}j_{2}k}=\sqrt{[d_{j_{1}}][d_{j_{2}}][d_{l_{1}}][d_{l_{2}}]}\delta_{l_{1},j_{1}+\frac{\epsilon_{1}}{2}}\delta_{l_{2},j_{2}+\frac{\epsilon_{2}}{2}}\left\\{\begin{array}[]{ccc}l_{1}&j_{1}&\frac{1}{2}\\\
j_{2}&l_{2}&k\end{array}\right\\}_{q}(-1)^{l_{1}+l_{2}+k}i_{l_{1}l_{2}k}\,,$
(56)

It thus maps the interwtiner space $\operatorname{Inv}(j_{1}\otimes
j_{2}\otimes k)$ to $\operatorname{Inv}(l_{1}\otimes l_{2}\otimes k)$.

This definition also works for two edges $e_{i},e_{i+1}$ sharing a corner in
$\operatorname{Inv}(j_{e_{1}v}\otimes\dotsb\otimes j_{e_{n}v})$, for
$i=1,\dotsc,n-1$. In the trivalent case, this gives
$E_{e_{3}e_{2}}^{\epsilon_{3},\epsilon_{2}}i_{j_{1}j_{2}j_{3}}$ exactly as in
(56) with $e_{1}\to e_{2},e_{2}\to e_{3},e_{3}\to e_{1}$

${\bf
E}^{\epsilon_{3},\epsilon_{2}}_{e_{3}e_{2}}\,i_{j_{1}j_{2}j_{3}}=\sqrt{[d_{j_{2}}][d_{j_{3}}][d_{l_{2}}][d_{l_{3}}]}\delta_{l_{2},j_{2}+\frac{\epsilon_{2}}{2}}\delta_{l_{3},j_{3}+\frac{\epsilon_{3}}{2}}\left\\{\begin{array}[]{ccc}l_{2}&j_{2}&\frac{1}{2}\\\
j_{3}&l_{3}&j_{1}\end{array}\right\\}_{q}(-1)^{l_{2}+l_{3}+j_{1}}i_{j_{1}l_{2}l_{3}}.$
(57)

For the case $i=n$, i.e. $E^{\epsilon_{1},\epsilon_{n}}_{e_{1}e_{n}}$, the
definition has to be amended to obtain an invariant operator Bonzom:2021rib
and eventually one finds the same expression for
$E^{\epsilon_{1},\epsilon_{3}}_{e_{1}e_{3}}i_{j_{1}j_{2}j_{3}}$ as (56) with
the appropriate permutation of the indices, i.e.

${\bf
E}^{\epsilon_{1},\epsilon_{3}}_{e_{1}e_{3}}\,i_{j_{1}j_{2}j_{3}}=\sqrt{[d_{j_{3}}][d_{j_{1}}][d_{l_{3}}][d_{l_{1}}]}\delta_{l_{3},j_{3}+\frac{\epsilon_{3}}{2}}\delta_{l_{1},j_{1}+\frac{\epsilon_{1}}{2}}\left\\{\begin{array}[]{ccc}l_{3}&j_{3}&\frac{1}{2}\\\
j_{1}&l_{1}&j_{2}\end{array}\right\\}_{q}(-1)^{l_{3}+l_{1}+j_{2}}i_{l_{1}j_{2}l_{3}}.$
(58)

###  III.4 Quantum Hamiltonian constraint

We now need to quantize the classical Hamiltonian (37) as a well-defined
operator on $\mathcal{H}_{\text{kin}}$ (defined in (52)). The first step is
obviously to use the quantization map described in the previous section
turning the observables $E_{e_{i}e_{i-1}}^{\epsilon_{i},\epsilon_{i-1}}$ into
operators ${\bf E}_{e_{i}e_{i-1}}^{\epsilon_{i},\epsilon_{i-1}}$. The second
step is concerned with quantization ambiguities. Indeed, factors $N_{e_{i}}$
appear in (37) and they are expected to be diagonal on the spin network basis,
as a function of $j_{e_{i}}$ only in fact. Notice however that the operators
$E^{\epsilon_{2},\epsilon_{1}}_{e_{2}e_{1}}$ change the spins of the edges
$e_{1},e_{2}$ by $\epsilon_{1}/2$ and $\epsilon_{2}/2$. There are therefore
ordering ambiguities, the result differing according to whether $N_{e_{i}}$ is
before or after some operators $E$ which changes $j_{e_{i}}$. We found an
ordering, see below, which ultimately leads to a topological model, which
would presumably not be true for other orderings.

Let us introduce

$\mathbf{h}_{f,e_{1},e_{p}}^{\epsilon_{1},\epsilon_{p}}=\frac{1}{\mathbf{N}_{e_{1}v_{2}}}\,\left(\sum_{\epsilon_{2},\dotsc,\epsilon_{p-1}=\pm}\prod_{i=2}^{p}{\bf
E}_{e_{i}e_{i-1}}^{\epsilon_{i},\epsilon_{i-1}}\frac{o_{i}\epsilon_{i}}{\mathbf{N}_{e_{i}v_{i}}}\right)+(-1)^{d-p}\epsilon_{1}\epsilon_{p}\frac{1}{\mathbf{N}_{e_{p}v_{p+1}}}\,\left(\sum_{\epsilon_{p+1},\dotsc,\epsilon_{d}=\pm}\prod_{i=p+1}^{d+1}{\bf
E}_{e_{i}e_{i-1}}^{-\epsilon_{i},-\epsilon_{i-1}}\frac{o_{i}\epsilon_{i}}{\mathbf{N}_{e_{i}v_{i}}}\right)\,,$
(59)

where $\mathbf{N}_{e_{i}v_{i}}$ is diagonal on $V_{j_{e_{i}}}$ (or its dual),
$\mathbf{N}_{e_{i}v_{i}}|j_{e_{i}},m_{e_{i}}\rangle=[d_{j_{e_{i}}}]|j_{e_{i}},m_{e_{i}}\rangle$.
We include the vertex $v_{i}$ in the notation because here
$\mathbf{N}_{e_{i}v_{i}}$ only acts on the space of intertwiners at $v_{i}$,
where $e_{i}$ and $e_{i-1}$ meet. As already discussed, the ordering is
important because
$[\mathbf{N}_{e_{i}v_{i}},E^{\epsilon_{i-1},\epsilon_{i}}_{e_{i-1}e_{i}}]\neq
0$. However
$[\mathbf{N}_{e_{i}v_{i}},E^{\epsilon_{i},\epsilon_{i+1}}_{e_{i}e_{i+1}}]=0$
by definition, so that the operators ${\bf
E}_{e_{i}e_{i-1}}^{\epsilon_{i},\epsilon_{i-1}}\frac{o_{i}\epsilon_{i}}{\mathbf{N}_{e_{i}v_{i}}}$
which act on the space of intertwiners at $v_{i}$ commute with one another.
Here $\mathbf{N}_{e_{i}v_{i}}$ is placed to the right of ${\bf
E}_{e_{i}e_{i-1}}^{\epsilon_{i},\epsilon_{i-1}}$, which is also the case if
one reconstructs the quantum holonomies from the quantum spinors
Bonzom:2021rib .

However, the operator $\mathbf{h}_{f,e_{1},e_{p}}^{\epsilon_{1},\epsilon_{p}}$
as such is not defined on $\mathcal{H}_{\text{kin}}$. Indeed, a state in
$\mathcal{H}_{\text{kin}}$ is a superposition of spin network states which
assigns a spin to each edge along with the space $V_{j}$ to the target end and
$V^{*}_{j}$ to the source end. Say the edge $e_{1}$ gets the spin $j_{1}$.
Then the first term of the above operator acts on $e_{1}$ with ${\bf
E}_{e_{2}e_{1}}^{\epsilon_{2},\epsilon_{1}}$ which shifts the spin $j_{1}$ to
$j_{1}+\epsilon_{1}/2$, on the intertwiner which sits at the vertex where
$e_{1}$ and $e_{2}$ meet. It thus maps $V_{j_{1}}$ to
$V_{j_{1}+\epsilon_{1}/2}$, or $V_{j_{1}}^{*}$ to
$V_{j_{1}+\epsilon_{1}/2}^{*}$ depending on orientations, but not both, i.e.
it does not shift $j_{1}$ at the vertex where $e_{d}$ and $e_{1}$ meet.
Therefore the operator brings the state out of $\mathcal{H}_{\text{kin}}$.

Similarly, the second term of
$\mathbf{h}_{f,e_{1},e_{p}}^{\epsilon_{1},\epsilon_{p}}$ acts on $e_{1}$
through ${\bf E}_{e_{1}e_{d}}^{-\epsilon_{1},-\epsilon_{d}}$. This shifts
$j_{1}$ to $j_{1}-\epsilon_{1}/2$ at the vertex where $e_{d},e_{1}$ meet. If
${\bf E}_{e_{2}e_{1}}^{\epsilon_{2},\epsilon_{1}}$ in the first term acted on
$V_{j_{1}}$, then this operator acts on $V^{*}_{j_{1}}$ (or the other way
around).

We thus turn $\mathbf{h}_{f,e_{1},e_{p}}^{\epsilon_{1},\epsilon_{p}}$ into a
well-defined operator on $\mathcal{H}_{\text{kin}}$ by multiplying it by a
product of operators ${\bf E}_{e_{i}e_{i-1}}^{\epsilon_{i},\epsilon_{i-1}}$ so
that the intertwiners of both ends of the same edge have the same spin. Notice
that the first term in (59) only contains the shift operators for
$i=2,\cdots,p$, one can add ${\bf
E}_{e_{i}e_{i-1}}^{\epsilon_{i},\epsilon_{i-1}}$ for all the remaining
vertices, i.e. $i=p+1,\cdots,d+1$, so that the spin changes for both ends of
each edge are the same. For the second term in (59), adding these shift
operators also shift all the spins $j_{i}-\epsilon_{i}/2$ to $j_{i}$ thus
drags the state back in ${\mathcal{H}}_{\text{kin}}$. This is the method which
was already used in Bonzom:2011nv  for constructing the quantum Hamiltonian in
the spinor representation in the flat case.

######  Definition III.1.

We define the quantum Hamiltonian on the face $f$, labeled by the pair of
edges $(e_{1},e_{p})$, to be

${\bf
H}_{f,e_{1},e_{p}}^{\epsilon_{1},\epsilon_{p},\epsilon_{p+1},\dotsc,\epsilon_{d}}=\biggl{[}\prod_{i=p+1}^{d+1}{\bf
E}_{e_{i}e_{i-1}}^{\epsilon_{i},\epsilon_{i-1}}\biggr{]}\
\mathbf{h}_{f,e_{1},e_{p}}^{\epsilon_{1},\epsilon_{p}}.$ (60)

Compared to the operator (59), the quantum Hamiltonian defined as such not
only depends on $\epsilon_{1}$ and $\epsilon_{p}$, but also
$\epsilon_{p+1},\cdots,\epsilon_{d+1}$. The physical Hilbert space is spanned
by the physical states which are the solutions to the quantum Hamiltonian. In
the spin representation, the coefficients of these physical spin network
states satisfy a set of difference equations, which is stated in the following
theorem.

######  Theorem III.2.

The constraint

$\forall k_{e}\qquad\langle\\{k_{e}\\}|\ {\bf
H}_{f,e_{1},e_{p}}^{\epsilon_{1},\epsilon_{p},\epsilon_{p+1},\dotsc,\epsilon_{d}}\
|\psi\rangle=0,$ (61)

is equivalent to the following set of _difference equations_ on the spin
network coefficients
$\psi(k_{1},k_{2},\cdots,k_{d},\\{k_{e}\\}_{e\notin\partial f})$ of
$|\psi\rangle$,

$\sum_{\tilde{\epsilon}_{2},\dotsc,\tilde{\epsilon}_{p-1}=\pm}\biggl{(}\prod_{i=2}^{p}A^{\tilde{\epsilon}_{i},\tilde{\epsilon}_{i-1}}_{o_{i}}(k_{i},k_{i-1},l_{i})\biggr{)}\
\psi(k_{1}-\frac{\epsilon_{1}}{2},k_{2}-\frac{\tilde{\epsilon}_{2}}{2},\dotsc,k_{p-1}-\frac{\tilde{\epsilon}_{p-1}}{2},k_{p}-\frac{\epsilon_{p}}{2},\dotsc,k_{d}-\frac{\epsilon_{d}}{2},\\{k_{e}\\}_{e\not\in\partial
f})\\\
+(-1)^{d-p}\alpha^{\epsilon_{1},\epsilon_{p}}(k_{1},k_{p})\sum_{\tilde{\epsilon}_{p+1},\dotsc,\tilde{\epsilon}_{d}=\pm}\biggl{(}\prod_{i=p+1}^{d+1}B^{\tilde{\epsilon}_{i},\tilde{\epsilon}_{i-1}}_{o_{i}}(k_{i}-\frac{\epsilon_{i}}{2},k_{i-1}-\frac{\epsilon_{i-1}}{2},l_{i})\biggr{)}\\\
\psi(k_{1},\dotsc,k_{p},k_{p+1}-\frac{\epsilon_{p+1}}{2}+\frac{\tilde{\epsilon}_{p+1}}{2},\dotsc,k_{d}-\frac{\epsilon_{d}}{2}+\frac{\tilde{\epsilon}_{d}}{2},\\{k_{e}\\}_{e\not\in\partial
f})=0.$ (62)

Here

  * •

$l_{1},\dotsc,l_{d}$ are the spins carried by the edges
$e^{\prime}_{1},\dotsc,e^{\prime}_{d}$ incident to $f$, see Figure 6.

  * •

By definition,
$\tilde{\epsilon}_{1}=\epsilon_{1},\tilde{\epsilon}_{p}=\epsilon_{p}$, while
$\epsilon_{p+1},\dotsc,\epsilon_{d}$ are fixed.

  * •

The coefficients are

$\displaystyle
A^{\tilde{\epsilon}_{i},\tilde{\epsilon}_{i-1}}_{o_{i}}(k_{i},k_{i-1},l_{i})$
$\displaystyle=o_{i}\tilde{\epsilon}_{i}[d_{k_{i}}](-1)^{k_{i}+k_{i-1}+l_{i}}\begin{Bmatrix}k_{i}&k_{i}-\frac{\tilde{\epsilon}_{i}}{2}&\frac{1}{2}\\\
k_{i-1}-\frac{\tilde{\epsilon}_{i-1}}{2}&k_{i-1}&l_{i}\end{Bmatrix}_{q}\,,$
(63) $\displaystyle
B^{\tilde{\epsilon}_{i},\tilde{\epsilon}_{i-1}}_{o_{i}}(k_{i},k_{i-1},l_{i})$
$\displaystyle=o_{i}\tilde{\epsilon}_{i}[d_{k_{i}}](-1)^{k_{i}+k_{i-1}+l_{i}}\begin{Bmatrix}k_{i}&k_{i}+\frac{\tilde{\epsilon}_{i}}{2}&\frac{1}{2}\\\
k_{i-1}+\frac{\tilde{\epsilon}_{i-1}}{2}&k_{i-1}&l_{i}\end{Bmatrix}_{q}\,,$
(64) $\displaystyle\alpha^{\epsilon_{1},\epsilon_{p}}(k_{1},k_{p})$
$\displaystyle=\epsilon_{1}\epsilon_{p}\frac{[d_{k_{p}}]}{[d_{k_{1}-\frac{\epsilon_{1}}{2}}]}\,.$
(65)

Those constraints are recursions on the physical states. They generalize the
one found in Bonzom:2011nv  for a triangular face. Improving on Bonzom:2011hm
, the differences are shifts of the spins by $1/2$ instead of $1$. Moreover,
edge orientations are kept arbitrary.

Those constraints have two types of contributions: the _$A$ -terms_ and the
_$B$ -terms_. Notice that
$\mathbf{h}_{f,e_{1},e_{p}}^{\epsilon_{1},\epsilon_{p}}$ contains all the
operators ${\bf
E}_{e_{i}e_{i-1}}^{\tilde{\epsilon}_{i},\tilde{\epsilon}_{i-1}}$ exactly once,
for $i=1,\dotsc,d$. Whether an operator ${\bf
E}_{e_{i}e_{i-1}}^{\tilde{\epsilon}_{i},\tilde{\epsilon}_{i-1}}$ gives rise to
an $A$-term or a $B$-term depends on the choice of the reference edges $e_{1}$
and $e_{p}$ around $f$. It is important that the coefficients
$A^{\tilde{\epsilon}_{i},\tilde{\epsilon}_{i-1}}_{o_{i}}(k_{i},k_{i-1},l_{i})$
and
$B^{\tilde{\epsilon}_{i},\tilde{\epsilon}_{i-1}}_{o_{i}}(k_{i},k_{i-1},l_{i})$
are _local_ : they only depend on the spins incident to the vertex and are
determined by the choice of a corner on that vertex. As a consequence, for
example, if one considers another constraint on the same face with $e_{q}$,
$q<p$, chosing $e_{1}$ as reference edge, then the coefficients
$A^{\tilde{\epsilon}_{i},\tilde{\epsilon}_{i-1}}_{o_{i}}(k_{i},k_{i-1},l_{i})$
for $i=1,\dotsc,q$ would be the same as those appearing above, and similarly
for the $B$-terms. The structure of the constraint is schematically pictured
in Figure 8.

Figure 8:  The schematic representation of the quantum constraint (62) with
its $A$-terms and $B$-terms associated to the corners around the face.

By exchanging the role of $e_{1}$ with $e_{p}$, the $A$-terms become the
$B$-terms and vice versa. The constraint obtained this way is equivalent to
(62), as we now show. First, evaluate (62) on
$k_{1}+\epsilon_{1}/2,\dotsc,k_{d}+\epsilon_{d}/2$, and then flip the signs of
all the $\epsilon_{i}$ and $\tilde{\epsilon}_{i}$. That gives the constraint

$\sum_{\tilde{\epsilon}_{2},\dotsc,\tilde{\epsilon}_{p-1}=\pm}\biggl{(}\prod_{i=2}^{p}A^{-\tilde{\epsilon}_{i},-\tilde{\epsilon}_{i-1}}_{o_{i}}(k_{i}-\frac{\epsilon_{i}}{2},k_{i-1}-\frac{\epsilon_{i-1}}{2},l_{i})\biggr{)}\
\psi\Bigl{(}k_{1},k_{2}-\frac{\epsilon_{2}}{2}+\frac{\tilde{\epsilon}_{2}}{2},\dotsc,k_{p-1}-\frac{\epsilon_{p-1}}{2}+\frac{\tilde{\epsilon}_{p-1}}{2},k_{p},\dotsc,k_{d},\\{k_{e}\\}_{e\not\in\partial
f}\Bigr{)}\\\
+(-1)^{d-p}\alpha^{\epsilon_{1},\epsilon_{p}}(k_{1}-\frac{\epsilon_{1}}{2},k_{p}-\frac{\epsilon_{p}}{2})\sum_{\tilde{\epsilon}_{p+1},\dotsc,\tilde{\epsilon}_{d}=\pm}\biggl{(}\prod_{i=p+1}^{d+1}B^{-\tilde{\epsilon}_{i},-\tilde{\epsilon}_{i-1}}_{o_{i}}(k_{i},k_{i-1},l_{i})\biggr{)}\\\
\psi\Bigl{(}k_{1}-\frac{\epsilon_{1}}{2},\dotsc,k_{p}-\frac{\epsilon_{p}}{2},k_{p+1}-\frac{\tilde{\epsilon}_{p+1}}{2},\dotsc,k_{d}-\frac{\tilde{\epsilon}_{d}}{2},\\{k_{e}\\}_{e\not\in\partial
f}\Bigr{)}=0\,.$ (66)

We then use the key relation between the coefficients $A$ and $B$,

$B^{\tilde{\epsilon}_{i},\tilde{\epsilon}_{i-1}}_{o_{i}}(k_{i},k_{i-1},l_{i})=-A_{o_{i}}^{-\tilde{\epsilon}_{i},-\tilde{\epsilon}_{i-1}}(k_{i},k_{i-1},l_{i})$
(67)

to get

$(-1)^{p}\sum_{\tilde{\epsilon}_{2},\dotsc,\tilde{\epsilon}_{p-1}=\pm}\biggl{(}\prod_{i=2}^{p}B^{\tilde{\epsilon}_{i},\tilde{\epsilon}_{i-1}}_{o_{i}}(k_{i}-\frac{\epsilon_{i}}{2},k_{i-1}-\frac{\epsilon_{i-1}}{2},l_{i})\biggr{)}\
\psi\Bigl{(}k_{1},k_{2}-\frac{\epsilon_{2}}{2}+\frac{\tilde{\epsilon}_{2}}{2},\dotsc,k_{p-1}-\frac{\epsilon_{p-1}}{2}+\frac{\tilde{\epsilon}_{p-1}}{2},k_{p},\dotsc,k_{d},\\{k_{e}\\}_{e\not\in\partial
f}\Bigr{)}\\\
+\alpha^{\epsilon_{1},\epsilon_{p}}(k_{1}-\frac{\epsilon_{1}}{2},k_{p}-\frac{\epsilon_{p}}{2})\sum_{\tilde{\epsilon}_{p+1},\dotsc,\tilde{\epsilon}_{d}=\pm}\biggl{(}\prod_{i=p+1}^{d+1}A^{\tilde{\epsilon}_{i},\tilde{\epsilon}_{i-1}}_{o_{i}}(k_{i},k_{i-1},l_{i})\biggr{)}\\\
\psi\Bigl{(}k_{1}-\frac{\epsilon_{1}}{2},\dotsc,k_{p}-\frac{\epsilon_{p}}{2},k_{p+1}-\frac{\tilde{\epsilon}_{p+1}}{2},\dotsc,k_{d}-\frac{\tilde{\epsilon}_{d}}{2},\\{k_{e}\\}_{e\not\in\partial
f}\Bigr{)}=0,$ (68)

where we recognize the matrix element $\langle\\{k_{e}\\}|{\bf
H}_{f,e_{p},e_{1}}^{\epsilon_{1},\epsilon_{2},\dotsc,\epsilon_{p}}|\psi\rangle$
and have shown the equivalence

$\langle\\{k_{e}\\}|{\bf
H}_{f,e_{p},e_{1}}^{\epsilon_{p},\epsilon_{1},\epsilon_{2},\dotsc,\epsilon_{p-1}}|\psi\rangle=0\quad\Leftrightarrow\quad\langle\\{k_{e}\\}|\
{\bf
H}_{f,e_{1},e_{p}}^{\epsilon_{1},\epsilon_{p},\epsilon_{p+1},\dotsc,\epsilon_{d}}\
|\psi\rangle=0.$ (69)

Proof of Theorem III.2. There are two types of terms in (60), whose action on
spin network states is now presented. First,

$\prod_{i=p+1}^{d+1}{\bf
E}_{e_{i}e_{i-1}}^{\epsilon_{i},\epsilon_{i-1}}\frac{1}{\mathbf{N}_{e_{1}v_{2}}}\left(\prod_{i=2}^{p}{\bf
E}_{e_{i}e_{i-1}}^{\epsilon_{i},\epsilon_{i-1}}\frac{o_{i}\epsilon_{i}}{\mathbf{N}_{e_{i}v_{i}}}\right)|\\{j_{e}\\}\rangle\\\
=\frac{1}{[d_{k_{1}}]}\prod_{i=2}^{p}\frac{o_{i}\epsilon_{i}}{[d_{j_{i}}]}\prod_{i=2}^{d+1}\delta_{k_{i},j_{i}+\frac{\epsilon_{i}}{2}}[d_{k_{i}}][d_{j_{i}}](-1)^{k_{i}+k_{i-1}+l_{i}}\begin{Bmatrix}k_{i}&k_{i}-\frac{\epsilon_{i}}{2}&\frac{1}{2}\\\
k_{i-1}-\frac{\epsilon_{i-1}}{2}&k_{i-1}&l_{i}\end{Bmatrix}_{q}|\\{k_{i}\\}_{i=1,\dotsc,d},\\{j_{e}\\}_{e\not\in\partial
f}\rangle\,,$ (70)

where we have applied the action (56) of ${\bf
E}_{e_{i}e_{i-1}}^{\epsilon_{i},\epsilon_{i-1}}$ on the intertwiner
$i_{j_{i-1}j_{i}l_{i}}$ at the vertex where $e_{i-1}\,,e_{i}$ and
$e^{\prime}_{i}$ meet for all $i=1,\cdots,d$. Each operator
$1/\mathbf{N}_{e_{i}v_{i}}$ acts before the shift operator ${\bf
E}_{e_{i}e_{i-1}}^{\epsilon_{i},\epsilon_{i-1}}$ thus the result picks up a
factor $1/[d_{j_{i}}]$. For $i=1,\dotsc,d$, the spin $j_{i}$ is shifted to
$j_{i}+\frac{\epsilon_{i}}{2}$ after the action of ${\bf
E}_{e_{i}e_{i-1}}^{\epsilon_{i},\epsilon_{i-1}}$. The spins of of edges not on
the boundary of the face $f$ remain unchanged. In addition,
$\frac{1}{\mathbf{N}_{e_{1}v_{2}}}$ acts after ${\bf
E}_{e_{2}e_{1}}^{\epsilon_{2},\epsilon_{1}}$ thus the result picks up the
factor $1/[d_{k_{1}}]$. As each edge is incident to two vertices, the assigned
spin shows up in two intertwiners thus the term
$\sqrt{[d_{k_{i}}][d_{j_{i}}]}$ appears twice in the result, which gives the
factor $[d_{k_{i}}][d_{j_{i}}]$. The $q$-$6j$ symbols and the sign factors
naturally follows from (56).

Secondly,

$\prod_{i=p+1}^{d+1}{\bf
E}_{e_{i}e_{i-1}}^{\epsilon_{i},\epsilon_{i-1}}\frac{1}{\mathbf{N}_{e_{p}v_{p+1}}}\left(\prod_{i=p+1}^{d+1}{\bf
E}_{e_{i}e_{i-1}}^{-\tilde{\epsilon}_{i},-\tilde{\epsilon}_{i-1}}\frac{o_{i}\epsilon_{i}}{\mathbf{N}_{e_{i}v_{i}}}\right)|\\{j_{e}\\}\rangle\\\
=\frac{1}{[d_{j_{p}-\frac{\tilde{\epsilon}_{p}}{2}}]}\prod_{i=p+1}^{d+1}\frac{o_{i}\epsilon_{i}}{[d_{j_{i}}]}\delta_{k_{i},j_{i}-\frac{\tilde{\epsilon}_{i}}{2}+\frac{\epsilon_{i}}{2}}(-1)^{k_{i}-\frac{\epsilon_{i}}{2}+k_{i-1}-\frac{\epsilon_{i-1}}{2}+l_{i}}(-1)^{k_{i}+k_{i-1}+l_{i}}[d_{j_{i}-\frac{\tilde{\epsilon}_{i}}{2}}][d_{j_{i-1}-\frac{\tilde{\epsilon}_{i-1}}{2}}]\\\
\sqrt{[d_{j_{i}}][d_{k_{i}}][d_{j_{i-1}}][d_{k_{i-1}}]}\begin{Bmatrix}k_{i}-\frac{\epsilon_{i}}{2}&j_{i}&\frac{1}{2}\\\
j_{i-1}&k_{i-1}-\frac{\epsilon_{i-1}}{2}&l_{i}\end{Bmatrix}_{q}\begin{Bmatrix}k_{i}-\frac{\epsilon_{i}}{2}&k_{i}&\frac{1}{2}\\\
k_{i-1}&k_{i-1}-\frac{\epsilon_{i-1}}{2}&l_{i}\end{Bmatrix}_{q}\
|\\{j_{i}\\}_{i=1,\dotsc,p},\\{k_{i}\\}_{i=p+1,\dotsc,d},\\{j_{e}\\}_{e\not\in\partial
f}\rangle\\\
=\frac{[d_{k_{p}}]}{[d_{k_{1}}][d_{k_{1}-\frac{\epsilon_{1}}{2}}]}\prod_{i=p+1}^{d+1}o_{i}\epsilon_{i}\delta_{k_{i},j_{i}-\frac{\tilde{\epsilon}_{i}}{2}+\frac{\epsilon_{i}}{2}}(-1)^{k_{i}-\frac{\epsilon_{i}}{2}+k_{i-1}-\frac{\epsilon_{i-1}}{2}+l_{i}}(-1)^{k_{i}+k_{i-1}+l_{i}}[d_{k_{i}-\frac{\epsilon_{i}}{2}}]^{2}[d_{k_{i}}]\\\
\begin{Bmatrix}k_{i}-\frac{\epsilon_{i}}{2}&j_{i}&\frac{1}{2}\\\
j_{i-1}&k_{i-1}-\frac{\epsilon_{i-1}}{2}&l_{i}\end{Bmatrix}_{q}\begin{Bmatrix}k_{i}-\frac{\epsilon_{i}}{2}&k_{i}&\frac{1}{2}\\\
k_{i-1}&k_{i-1}-\frac{\epsilon_{i-1}}{2}&l_{i}\end{Bmatrix}_{q}\
|\\{j_{i}\\}_{i=1,\dotsc,p},\\{k_{i}\\}_{i=p+1,\dotsc,d},\\{j_{e}\\}_{e\not\in\partial
f}\rangle\,.$ (71)

Here, two shift operators act on each site for $i=p+1,\cdots,d+1$ and we
denote $k_{i}=j_{i}-\frac{\tilde{\epsilon}_{i}}{2}+\frac{\epsilon_{i}}{2}$.
The first shift operator ${\bf
E}_{e_{i}e_{i-1}}^{-\tilde{\epsilon}_{i},-\tilde{\epsilon}_{i-1}}$ (in the
bracket) acts on the spin network state and shifts $j_{i}$ and $j_{i-1}$ to
$j_{i}-\tilde{\epsilon}_{i}/2$ and $j_{i-1}-\tilde{\epsilon}_{i-1}/2$
respectively. It also gives the first $q$-$6j$ symbol in the third line and
the term
$(-1)^{k_{i}-\frac{\epsilon_{i}}{2}+k_{i-1}-\frac{\epsilon_{i-1}}{2}+l_{i}}\sqrt{[d_{j_{i}}][d_{j_{i-1}}][d_{j_{i}-\frac{\tilde{\epsilon}_{i}}{2}}][d_{j_{i-1}-\frac{\tilde{\epsilon}_{i-1}}{2}}]}$.
The result picks up a factor $1/[d_{j_{i}}]$ by the action of
$1/\mathbf{N}_{e_{i}v_{i}}$ before the shift operator. In addition,
$1/\mathbf{N}_{e_{p}v_{p+1}}$ acts on the spin network state after ${\bf
E}_{e_{p+1}e_{p}}^{-\tilde{\epsilon}_{p+1},-\tilde{\epsilon}_{p}}$ and thus
brings a factor $1/[d_{j_{p}-\frac{\tilde{\epsilon}_{p}}{2}}]$. The action of
the second shift operator ${\bf
E}_{e_{i}e_{i-1}}^{\epsilon_{i},\epsilon_{i-1}}$ shifts the spins
$j_{i}-\tilde{\epsilon}_{i}/2$ and $j_{i-1}-\tilde{\epsilon}_{i-1}/2$ to
$k_{i}$ and $k_{i-1}$ respectively and brings the second $q$-$6j$ symbol in
the third line as well as the term
$(-1)^{k_{i}+k_{i-1}+l_{i}}\sqrt{[d_{k_{i}}][d_{k_{i-1}}][d_{k_{i}-\frac{\epsilon_{i}}{2}}][d_{k_{i-1}-\frac{\epsilon_{i-1}}{2}}]}$.
Note that the spin $j_{1}=k_{1}$ and $j_{p}=k_{p}$ are kept unchanged in the
result as $\epsilon_{1}=\tilde{\epsilon}_{1}$ and
$\epsilon_{p}=\tilde{\epsilon}_{p}$. The last equality is the rearrangement of
the result.

Putting them together, using the orthogonality of the spin network states,
$\langle\\{k_{e}\\}|\\{j_{e}\\}\rangle\propto\prod_{e}\delta_{k_{e},j_{e}}$
and eliminating the common terms
$\frac{1}{[d_{k_{1}}]}\prod_{i=p+1}^{d+1}[d_{k_{i}-\frac{\epsilon_{i}}{2}}][d_{k_{i}}](-1)^{k_{i}+k_{i-1}+l_{i}}\begin{Bmatrix}k_{i}&k_{i}-\frac{\epsilon_{i}}{2}&\frac{1}{2}\\\
k_{i-1}-\frac{\epsilon_{i-1}}{2}&k_{i-1}&l_{i}\end{Bmatrix}_{q}$ leads to the
expected difference equations. ∎

The dependence of $\psi$ on the orientations is given by the following lemma.

######  Lemma III.3.

If $|\psi\rangle$, with spin network coefficients $\psi(\\{j_{e}\\})$,
satisfies all the constraints (62) for given edge orientations $\\{o_{e}\\}$,
then $(-1)^{2j^{*}}\psi(\\{j_{e}\\})$ satisfies all the constraints on the
same graph with reversed orientation $-o_{e^{*}}$ on the edge $e^{*}$.

Proof. Consider the constraint (62) on the fixed face $f$. If
$e^{*}\not\in\partial f$, multiplication by $(-1)^{2j^{*}}$ does not change
anything. If $e^{*}\equiv e_{s}\in\\{e_{2},\dotsc,e_{p-1}\\}$, then the
coefficient
$A^{\tilde{\epsilon}_{s},\tilde{\epsilon}_{s-1}}_{o_{s}}(k_{s},k_{s-1},l_{s})$
changes sign. Moreover, it is the only one that depends on $o_{s}$. The state
coefficient on the first line of (62) changes from
$\psi(k_{1}-\epsilon_{1},\dotsc,k_{e_{p}}-\tilde{\epsilon}_{p},\dotsc)$ to
$(-1)^{2k_{s}+1}\psi(k_{1}-\epsilon_{1},\dotsc,k_{p}-\tilde{\epsilon}_{p},\dotsc)$
since $(-1)^{\tilde{\epsilon}_{s}}=-1$. Moreover, the coefficients $B$’s are
independent of the orientation $o_{s}$ and the state coefficient on the second
line changes from $\psi(k_{1},\dotsc,k_{p},\dotsc)$ to
$(-1)^{2k_{s}}\psi(k_{1},\dotsc,k_{p},\dotsc)$. Factorizing $(-1)^{2k_{s}}$
from the equation reveals that only the first line is modified, by
$-o_{s}\times(-1)=o_{s}$. The constraint therefore still holds. If
$e^{*}\equiv e_{s}\in\\{e_{p+1},\cdots,e_{d+1}\\}$, the coefficient
$B^{\tilde{\epsilon}_{s},\tilde{\epsilon}_{s-1}}_{o_{s}}(k_{s}-\frac{\epsilon_{s}}{2},k_{s-1}-\frac{\epsilon_{s-1}}{2},l_{s})$
changes sign while the coefficients $A$’s remain unchanged. The same analysis
leads to the same conclusion.

The argument is the same for all edges in the boundary of $f$, since the
orientation of any of those edges appears in a single coefficient of the
equation. ∎

##  IV Pachner moves

We now show how to relate physical states on triangulations which are related
by Pachner moves. This is an extension of Noui:2004iy  to $q$ real (using
Hamiltonian constraints instead of projection on flat connections). In two
dimensions, there are two types of Pachner moves, the $3-1$ moves and the
$2-2$ moves (as well as their inverses).

###  IV.1 2-2 Pachner move

The 2-2 Pachner move changes a portion of the graph into another one as
follows,

$\begin{array}[]{c}\includegraphics{2to2.pdf}\end{array}$ (72)

We denote the initial graph which contains the left hand side as $\Gamma_{i}$,
and the final graph which contains the right hand side as $\Gamma_{f}$. The
orientation of all edges are left arbitrary.

######  Theorem IV.1.

Let $|\psi_{f}\rangle$ on $\Gamma_{f}$ be defined in the spin network basis by

$\psi_{f}(j_{1},j_{2},j_{3},j_{4},j_{0},\dotsc)=(-1)^{j_{1}+j_{2}+j_{3}+j_{4}}[d_{j_{0}}]\sum_{j_{5}}(-1)^{(1-o_{5})j_{5}+(1-o_{0})j_{0}}\begin{Bmatrix}j_{1}&j_{2}&j_{0}\\\
j_{3}&j_{4}&j_{5}\end{Bmatrix}_{q}\,\psi_{i}(j_{1},j_{2},j_{3},j_{4},j_{5},\dotsc)\,.$
(73)

where the ellipses denote spins which are the same on both sides (for edges
which are not affected by the move). Then $|\psi_{i}\rangle$ is a state which
satisfies all the constraints on $\Gamma_{i}$ if and only if
$|\psi_{f}\rangle$ satisfies all the constraints on $\Gamma_{f}$.

Since the 2-2 move is its own inverse, there is a symmetry between both sides
of the move. This must translate into a symmetry which exchanges the role of
$|\psi_{i}\rangle$ and $|\psi_{f}\rangle$ in (73). This is indeed true thanks
to the orthonormality of the $q$-$6j$ symbols,

$\sum_{j_{5}}[d_{j_{5}}][d_{j_{0}}]\begin{Bmatrix}j_{1}&j_{2}&j_{0}\\\
j_{3}&j_{4}&j_{5}\end{Bmatrix}_{q}\begin{Bmatrix}j_{1}&j_{2}&j^{\prime}_{0}\\\
j_{3}&j_{4}&j_{5}\end{Bmatrix}_{q}=\delta_{j_{0},j^{\prime}_{0}}\,,$ (74)

which transforms (73) into

$\psi_{i}(j_{1},j_{2},j_{3},j_{4},j_{5},\dotsc)=(-1)^{j_{1}+j_{2}+j_{3}+j_{4}}[d_{j_{5}}]\sum_{j^{\prime}_{0}}(-1)^{(1-o_{5})j_{5}+(1-o_{0})j^{\prime}_{0}}\begin{Bmatrix}j_{1}&j_{2}&j^{\prime}_{0}\\\
j_{3}&j_{4}&j_{5}\end{Bmatrix}_{q}\,\psi_{f}(j_{1},j_{2},j_{3},j_{4},j^{\prime}_{0},\dotsc)\,.$
(75)

Proof of Theorem IV.1. There are four faces involved in the move on each side.
Clearly, $|\psi_{i}\rangle$ and $|\psi_{f}\rangle$ satisfy the same
constraints associated to faces which are not among those four. Therefore, we
can focus on the four faces involved in the move, and for symmetry reasons, we
can simply look at the constraints on two faces: the face $f_{12}$ which has
$e_{1},e_{2}$ in its boundary, and the face $f_{14}$ which has $e_{1},e_{4}$
in its boundary.

##### Face $f_{12}$.

It has a different boundary on $\Gamma_{f}$ and $\Gamma_{i}$, due to the
disappearance of $e_{5}$. On $\Gamma_{i}$, there are constraints where ${\bf
E}^{\epsilon_{5},\epsilon_{1}}_{e_{5}e_{1}}$ and ${\bf
E}^{\epsilon_{2},\epsilon_{5}}_{e_{2}e_{5}}$ are both among the $A$-terms of
the constraint (62). Let us denote the two reference edges ($e_{1}$ and
$e_{k}$ in (62)) $e$ and $e^{\prime}$, which may be $e_{1}$ and/or $e_{2}$.
Then the difference equations (62) read

$\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e\underset{\text{c.c.}}{\to}e^{\prime}}A)\sum_{\tilde{\epsilon}_{5}=\pm}A_{o_{5}}^{\tilde{\epsilon}_{5},\tilde{\epsilon}_{1}}(k_{5},k_{1},k_{4})A_{o_{2}}^{\tilde{\epsilon}_{2},\tilde{\epsilon}_{5}}(k_{2},k_{5},k_{3})\psi_{i}\Bigl{(}k_{1}-\frac{\tilde{\epsilon}_{1}}{2},k_{2}-\frac{\tilde{\epsilon}_{2}}{2},k_{3},k_{4},k_{5}-\frac{\tilde{\epsilon}_{5}}{2},\dotsc\Bigr{)}\\\
+(-1)^{d_{12,i}-d_{ee^{\prime}}}\alpha^{\epsilon_{e},\epsilon_{e^{\prime}}}(k_{e},k_{e^{\prime}})\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e^{\prime}\underset{\text{c.c.}}{\to}e}B)\psi_{i}(k_{1},k_{2},k_{3},k_{4},k_{5},\dotsc)=0,$
(76)

where $d_{12,i}$ denotes the number of boundary edges of $f_{12}$ in
$\Gamma_{i}$ and $d_{ee^{\prime}}$ the number of edges from $e$ to
$e^{\prime}$ counter-clockwise. Notice that $\tilde{\epsilon}_{1}$
(respectively $\tilde{\epsilon}_{2}$) is fixed if $e=e_{1}$ (respectively if
$e^{\prime}=e_{2}$) and summed over otherwise. We have indicated in $\psi_{i}$
only the spins which are involved in the move.

We have written
$\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e\underset{\text{c.c.}}{\to}e^{\prime}}A)$
and
$\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e^{\prime}\underset{\text{c.c.}}{\to}e}B)$
schematically the coefficients of the equation which are associated to corners
_not_ involved in the move. Here
$\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e\underset{\text{c.c.}}{\to}e^{\prime}}A)$
is the product of the $A$-terms over the corners from $e$ to $e^{\prime}$
going counter-clockwise, except for the two corners with $e_{5}$, whose
$A$-terms are distinguished. Then
$\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e^{\prime}\underset{\text{c.c.}}{\to}e}B)$
is the product of the $B$-terms over the corners from $e^{\prime}$ to $e$
counter-clockwise. This is depicted in the Figure 9.

Figure 9:  A graphical representation of Equations (76) on the LHS and (77) on
the RHS.

On the other hand, a state on $\Gamma_{f}$ must also satisfy a constraint
along the face $f_{12}$ with the two reference edges $e$ and $e^{\prime}$. It
reads

$\langle\\{k_{e}\\}_{\Gamma_{f}}|{\bf
H}_{f_{12},e,e^{\prime}}|\psi_{f}\rangle\propto\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e\underset{\text{c.c.}}{\to}e^{\prime}}A)A_{o_{2}}^{\epsilon_{2},\epsilon_{1}}(k_{2},k_{1},k_{0})\psi_{f}\Bigl{(}k_{1}-\frac{\tilde{\epsilon}_{1}}{2},k_{2}-\frac{\tilde{\epsilon}_{2}}{2},k_{3},k_{4},k_{0},\dotsc\Bigr{)}\\\
+(-1)^{d_{12,i}-d_{ee^{\prime}}+1}\alpha^{\epsilon_{e},\epsilon_{e^{\prime}}}(k_{e},k_{e^{\prime}})\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e^{\prime}\underset{\text{c.c.}}{\to}e}B)\psi_{f}(k_{1},k_{2},k_{3},k_{4},k_{0},\dotsc).$
(77)

There is also a constraint where
$\mathbf{E}_{e_{1}e_{2}}^{\tilde{\epsilon_{1}},\tilde{\epsilon}_{2}}$ gives
rise to a $B$-term, but as we have shown this is equivalent to the above
constraint. Here it is important that the products of the $A$-terms and
$B$-terms over all corners except the one where $e_{1}$ and $e_{2}$ meet are
the same as in (76). The reason is obviously that those terms are local and
the 2-2 move does not involve their corners. As in (76),
$\tilde{\epsilon}_{1}$ and $\tilde{\epsilon}_{2}$ may be fixed or summed over.

We now plug (73) into (77) to check that it vanishes, provided the constraint
(76) holds. First compute, with
$j_{1,2}:=k_{1,2}-\frac{\tilde{\epsilon}_{1,2}}{2}$,

$A_{o_{2}}^{\tilde{\epsilon}_{2},\tilde{\epsilon}_{1}}(k_{2},k_{1},k_{0})\psi_{f}(j_{1},j_{2},k_{3},k_{4},k_{0},\dotsc)=o_{2}\tilde{\epsilon}_{2}[d_{k_{2}}][d_{k_{0}}]\sum_{j_{5}}(-1)^{(1-o_{5})j_{5}+(1-o_{0})k_{0}}(-1)^{j_{1}+j_{2}+k_{3}+k_{4}}(-1)^{k_{0}+k_{1}+k_{2}}\\\
\begin{Bmatrix}k_{1}&j_{1}&\frac{1}{2}\\\
j_{2}&k_{2}&k_{0}\end{Bmatrix}_{q}\begin{Bmatrix}j_{1}&j_{2}&k_{0}\\\
k_{3}&k_{4}&j_{5}\end{Bmatrix}_{q}\psi_{i}(j_{1},j_{2},k_{3},k_{4},j_{5},\dotsc)\,.$
(78)

The Biedenharn-Elliott identity on $q$-$6j$ symbols gives precisely

$(-1)^{j_{1}+j_{2}+k_{3}+k_{4}}(-1)^{k_{0}+k_{1}+k_{2}}\begin{Bmatrix}k_{1}&j_{1}&\frac{1}{2}\\\
j_{2}&k_{2}&k_{0}\end{Bmatrix}_{q}\begin{Bmatrix}j_{1}&j_{2}&k_{0}\\\
k_{3}&k_{4}&j_{5}\end{Bmatrix}_{q}\\\
=\sum_{k_{5}}[d_{k_{5}}](-1)^{k_{5}+j_{5}+\frac{1}{2}}\begin{Bmatrix}k_{1}&k_{2}&k_{0}\\\
k_{3}&k_{4}&k_{5}\end{Bmatrix}_{q}\begin{Bmatrix}k_{1}&j_{1}&\frac{1}{2}\\\
j_{5}&k_{5}&k_{4}\end{Bmatrix}_{q}\begin{Bmatrix}k_{5}&j_{5}&\frac{1}{2}\\\
j_{2}&k_{2}&k_{3}\end{Bmatrix}_{q}\,.$ (79)

Setting $j_{5}=k_{5}-\frac{\tilde{\epsilon}_{5}}{2}$ to change the summation
over $j_{5}$ to one over $\tilde{\epsilon}_{5}$ (there are no other values of
$j_{5}$ allowed by the triangular inequalities on the $q$-$6j$ symbol), we get

$A_{o_{2}}^{\epsilon_{2},\epsilon_{1}}(k_{2},k_{1},k_{0})\psi_{f}(j_{1},j_{2},k_{3},k_{4},k_{0},\dotsc)=o_{2}\epsilon_{2}[d_{k_{2}}][d_{k_{0}}]\sum_{k_{5},\tilde{\epsilon}_{5}}(-1)^{(1-o_{5})j_{5}+(1-o_{0})k_{0}}(-1)^{k_{5}+j_{5}+\frac{1}{2}}\\\
[d_{k_{5}}]\begin{Bmatrix}k_{1}&k_{2}&k_{0}\\\
k_{3}&k_{4}&k_{5}\end{Bmatrix}_{q}\begin{Bmatrix}k_{1}&j_{1}&\frac{1}{2}\\\
j_{5}&k_{5}&k_{4}\end{Bmatrix}_{q}\begin{Bmatrix}k_{5}&j_{5}&\frac{1}{2}\\\
j_{2}&k_{2}&k_{3}\end{Bmatrix}_{q}\psi_{i}(j_{1},j_{2},k_{3},k_{4},j_{5},\dotsc)\,.$
(80)

Using $\epsilon_{5}=(-1)^{\frac{1}{2}+j_{5}-k_{5}}$, we find
$(-1)^{k_{5}+j_{5}+\frac{1}{2}}=\epsilon_{5}(-1)^{2k_{5}}$. We also use
$(-1)^{(1-o_{5})j_{5}}=o_{5}(-1)^{(1-o_{5})k_{5}}$ and notice that a $q$-$6j$
symbol can be factored. Thus,

$\langle\\{k_{e}\\}_{\Gamma_{f}}|{\bf
H}_{f_{12},e,e^{\prime}}|\psi_{f}\rangle\propto\sum_{k_{5},\tilde{\epsilon}_{5}}(-1)^{(1-o_{5})k_{5}+(1-o_{0})k_{0}}(-1)^{k_{1}+k_{2}+k_{3}+k_{4}}\begin{Bmatrix}k_{1}&k_{2}&k_{0}\\\
k_{3}&k_{4}&k_{5}\end{Bmatrix}_{q}[d_{k_{0}}]\\\
\Biggl{(}\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e\underset{\text{c.c.}}{\to}e^{\prime}}A)o_{5}\epsilon_{5}[d_{k_{5}}](-1)^{2k_{5}}(-1)^{k_{1}+k_{2}+k_{3}+k_{4}}o_{2}\epsilon_{2}[d_{k_{2}}]\begin{Bmatrix}k_{1}&j_{1}&\frac{1}{2}\\\
j_{5}&k_{5}&k_{4}\end{Bmatrix}_{q}\begin{Bmatrix}k_{5}&j_{5}&\frac{1}{2}\\\
j_{2}&k_{2}&k_{3}\end{Bmatrix}_{q}\psi_{i}(j_{1},j_{2},k_{3},k_{4},j_{5},\dotsc)\\\
+(-1)^{d_{12,i}-d_{ee^{\prime}}}\alpha^{\epsilon_{e},\epsilon_{e^{\prime}}}(k_{e},k_{e^{\prime}})\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e^{\prime}\underset{\text{c.c.}}{\to}e}B)\psi_{i}(k_{1},k_{2},k_{3},k_{4},k_{5},\dotsc)\Biggr{)}\,.$
(81)

We now recognize the coefficients in (76),

$\langle\\{k_{e}\\}_{\Gamma_{f}}|{\bf
H}_{f_{12},e,e^{\prime}}|\psi_{f}\rangle\propto\sum_{k_{5}}(-1)^{(1-o_{5})k_{5}+(1-o_{0})k_{0}}(-1)^{k_{1}+k_{2}+k_{3}+k_{4}}\begin{Bmatrix}k_{1}&k_{2}&k_{0}\\\
k_{3}&k_{4}&k_{5}\end{Bmatrix}_{q}[d_{k_{0}}]\\\
\Biggl{(}\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e\underset{\text{c.c.}}{\to}e^{\prime}}A)\sum_{\tilde{\epsilon}_{5}=\pm}A_{o_{5}}^{\tilde{\epsilon}_{5},\tilde{\epsilon}_{1}}(k_{5},k_{1},k_{4})A_{o_{2}}^{\tilde{\epsilon}_{2},\tilde{\epsilon}_{5}}(k_{2},k_{5},k_{3})\psi_{i}(j_{1},j_{2},k_{3},k_{4},j_{5},\dotsc)\\\
+(-1)^{d_{12,i}-d_{ee^{\prime}}}\alpha^{\epsilon_{e},\epsilon_{e^{\prime}}}(k_{e},k_{e^{\prime}})\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e^{\prime}\underset{\text{c.c.}}{\to}e}B)\psi_{i}(k_{1},k_{2},k_{3},k_{4},k_{5},\dotsc)\Biggr{)},$
(82)

and conclude that (77) vanished provided (76) and (73).

##### Face $f_{14}$.

We now perform the same analysis on the constraints which act on the face
$f_{14}$. We use the same notation as for the face $f_{12}$, i.e. let $e$ and
$e^{\prime}$ be two reference edges around $f_{14}$ and consider the
Hamiltonian constraints associated to them on $\Gamma_{i}$ and $\Gamma_{f}$.
On $\Gamma_{i}$, the Hamiltonians contain the operator ${\bf
E}_{e_{1}e_{4}}^{\epsilon_{1},\epsilon_{4}}$, which without loss of generality
can be considered to give rise to an $A$-term. The constraints on the spin
network coefficients of $|\psi_{i}\rangle$ read

$\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e\underset{\text{c.c.}}{\to}e^{\prime}}A)A_{o_{1}}^{\tilde{\epsilon}_{1},\tilde{\epsilon}_{4}}(k_{1},k_{4},k_{5})\psi_{i}(j_{1},k_{2},k_{3},j_{4},k_{5},\dotsc)\\\
+(-1)^{d_{14,i}-d_{ee^{\prime}}}\alpha^{\epsilon_{e},\epsilon_{e^{\prime}}}(k_{e},k_{e^{\prime}})\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e^{\prime}\underset{\text{c.c.}}{\to}e}B)\psi_{i}(k_{1},k_{2},k_{3},k_{4},k_{5},\dotsc)=0,$
(83)

with $j_{1}=k_{1}-\tilde{\epsilon_{1}}/2,j_{4}=k_{4}-\tilde{\epsilon}_{4}/2$.
The sign $\tilde{\epsilon}_{1}$ (respectively $\tilde{\epsilon}_{4}$) is fixed
if $e=e_{1}$ (respectively if $e^{\prime}=e_{4}$) and summed over otherwise.
Here, $\prod_{e\underset{\text{c.c.}}{\to}e^{\prime}}A$ is the product of the
$A$-terms from $e$ to $e^{\prime}$ counter-clockwise, except for the one on
the corner of $e_{1},e_{4}$ which has been singled out. As for
$\prod_{e^{\prime}\underset{\text{c.c.}}{\to}e}B$, it is the product of the
$B$-terms going counter-clockwise from $e^{\prime}$ to $e$.

On $\Gamma_{f}$, we need to look at two types of constraints. Either the
operators ${\bf E}^{\tilde{\epsilon}_{0},\tilde{\epsilon}_{4}}_{e_{0}e_{4}}$
and ${\bf E}^{\tilde{\epsilon}_{1},\tilde{\epsilon}_{0}}_{e_{1}e_{0}}$ which
enter ${\bf H}_{f_{14},e,e^{\prime}}$ on $\Gamma_{f}$ both contribute to
$A$-terms of the constraint (or both to $B$-terms but this is the same), or
one gives rise to an $A$-term and the other one to a $B$-term.

In case they both give rise to $A$-terms, we are in the same situation as in
our previous analysis on the face $f_{12}$, with the role of $\Gamma_{i}$ and
$\Gamma_{f}$ exchanged. Since the relation (73) between $|\psi_{f}\rangle$ and
$|\psi_{i}\rangle$ can be inverted with the same form, we have nothing to
prove.

If ${\bf E}^{\tilde{\epsilon}_{0},\tilde{\epsilon}_{4}}_{e_{0}e_{4}}$
contributes to a $B$-term, and ${\bf
E}^{\tilde{\epsilon}_{1},\tilde{\epsilon}_{0}}_{e_{1}e_{0}}$ contributes to an
$A$-term, this means that $e_{0}=e$ is a reference edge chosen for the
constraint. The Hamiltonians of this type on $\Gamma_{f}$ are ${\bf
H}_{f_{14},e_{0},e^{\prime}}^{\epsilon_{0},\epsilon_{e^{\prime}},\dotsc,\epsilon_{4}}$
and they are labeled by signs for all the edges from $e^{\prime}$ to $e_{0}$
counter-clockwise. The matrix elements read

$\langle\\{k_{e}\\}_{\Gamma_{f}}|{\bf
H}^{\epsilon_{0},\epsilon_{e^{\prime}},\dotsc,\epsilon_{4}}_{f_{14},e_{0},e^{\prime}}|\psi_{f}\rangle\propto\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e_{1}\underset{\text{c.c.}}{\to}e^{\prime}}A)A_{o_{1}}^{\tilde{\epsilon}_{1},\epsilon_{0}}(k_{1},k_{0},k_{2})\psi_{f}(j_{1},k_{2},k_{3},j_{4},j_{0},\dotsc)\\\
+(-1)^{d_{14,f}-d_{e_{0}e^{\prime}}}\alpha^{\epsilon_{0},\epsilon_{e^{\prime}}}(k_{0},k_{e^{\prime}})\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e^{\prime}\underset{\text{c.c.}}{\to}e_{4}}B)B_{o_{0}}^{\epsilon_{0},\tilde{\epsilon}_{4}}(j_{0},j_{4},k_{3})\psi_{f}(k_{1},k_{2},k_{3},l_{4},k_{0},\dotsc)\,,$
(84)

where $\epsilon_{0}$ is fixed (but $\tilde{\epsilon}_{4}$ only is if
$e^{\prime}=e_{4}$) and $j_{0,4}=k_{0,4}-\epsilon_{0,4}/2$, and
$l_{4}=j_{4}+\tilde{\epsilon}_{4}/2$.

We now plug (73) into the above matrix elements. We first look at the
$A$-term,

$A_{o_{1}}^{\epsilon_{1},\epsilon_{0}}(k_{1},k_{0},k_{2})\psi_{f}(j_{1},k_{2},k_{3},j_{4},j_{0},\dotsc)=[d_{j_{0}}]\sum_{k_{5}}o_{1}\epsilon_{1}[d_{k_{1}}](-1)^{k_{0}+k_{1}+k_{2}}(-1)^{(1-o_{5})k_{5}+(1-o_{0})j_{0}}(-1)^{j_{1}+k_{2}+k_{3}+j_{4}}\\\
\begin{Bmatrix}j_{1}&k_{2}&j_{0}\\\
k_{3}&j_{4}&k_{5}\end{Bmatrix}_{q}\begin{Bmatrix}k_{1}&j_{1}&\frac{1}{2}\\\
j_{0}&k_{0}&k_{2}\end{Bmatrix}_{q}\psi_{i}(j_{1},k_{2},k_{3},j_{4},k_{5},\dotsc).$
(85)

The relevant Biedenharn-Elliott identity is

$(-1)^{k_{0}+k_{1}+k_{2}+j_{0}+j_{1}+k_{3}+k_{5}}\begin{Bmatrix}j_{1}&k_{2}&j_{0}\\\
k_{3}&j_{4}&k_{5}\end{Bmatrix}_{q}\begin{Bmatrix}k_{1}&j_{1}&\frac{1}{2}\\\
j_{0}&k_{0}&k_{2}\end{Bmatrix}_{q}=\sum_{l_{4}}[d_{l_{4}}](-1)^{j_{4}+l_{4}+\frac{1}{2}}\begin{Bmatrix}j_{4}&l_{4}&\frac{1}{2}\\\
k_{1}&j_{1}&k_{5}\end{Bmatrix}_{q}\begin{Bmatrix}j_{4}&l_{4}&\frac{1}{2}\\\
k_{0}&j_{0}&k_{3}\end{Bmatrix}_{q}\begin{Bmatrix}k_{1}&k_{2}&k_{0}\\\
k_{3}&l_{4}&k_{5}\end{Bmatrix}_{q}\,.$ (86)

As for the $B$-term,

$\sum_{l_{4}}B_{o_{0}}^{\epsilon_{0},\tilde{\epsilon}_{4}}(j_{0},j_{4},k_{3})\psi_{f}(k_{1},k_{2},k_{3},l_{4},k_{0},\dotsc)=\sum_{k_{5},l_{4}}o_{0}\epsilon_{0}[d_{j_{0}}][d_{k_{0}}](-1)^{j_{0}+k_{3}+j_{4}}(-1)^{(1-o_{5})k_{5}+(1-o_{0})k_{0}}(-1)^{k_{1}+k_{2}+k_{3}+l_{4}}\\\
\begin{Bmatrix}j_{4}&l_{4}&\frac{1}{2}\\\
k_{0}&j_{0}&k_{3}\end{Bmatrix}_{q}\begin{Bmatrix}k_{1}&k_{2}&k_{0}\\\
k_{3}&l_{4}&k_{5}\end{Bmatrix}_{q}\psi_{i}(k_{1},k_{2},k_{3},l_{4},k_{5},\dotsc).$
(87)

We recognize the two same $q$-$6j$ symbols as in the Biendenharn-Elliott
identity above. We can thus factor them out, so that the matrix elements of
the Hamiltonian are proportional to

$\langle\\{k_{e}\\}_{\Gamma_{f}}|{\bf
H}^{\epsilon_{0},\epsilon_{e^{\prime}},\dotsc,\epsilon_{4}}_{f_{14},e_{0},e^{\prime}}|\psi_{f}\rangle\propto\sum_{k_{5},l_{4}}(-1)^{j_{0}+k_{3}+j_{4}}(-1)^{(1-o_{5})k_{5}+(1-o_{0})k_{0}}(-1)^{k_{1}+k_{2}+k_{3}+l_{4}}\begin{Bmatrix}j_{4}&l_{4}&\frac{1}{2}\\\
k_{0}&j_{0}&k_{3}\end{Bmatrix}_{q}\begin{Bmatrix}k_{1}&k_{2}&k_{0}\\\
k_{3}&l_{4}&k_{5}\end{Bmatrix}_{q}[d_{j_{0}}]\\\
\Biggl{(}\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e_{1}\underset{\text{c.c.}}{\to}e^{\prime}}A)o_{0}o_{1}\epsilon_{1}[d_{k_{1}}][d_{l_{4}}](-1)^{k_{2}+j_{4}-j_{0}-k_{5}}(-1)^{j_{4}+l_{4}+\frac{1}{2}}(-1)^{j_{0}+k_{3}+j_{4}}(-1)^{k_{1}+k_{2}+k_{3}+l_{4}}\begin{Bmatrix}j_{4}&l_{4}&\frac{1}{2}\\\
k_{1}&j_{1}&k_{5}\end{Bmatrix}_{q}\\\
\times\psi_{i}(j_{1},k_{2},k_{3},j_{4},k_{5},\dotsc)+(-1)^{d_{14,f}-d_{e_{0}e^{\prime}}}\alpha^{\epsilon_{0},\epsilon_{e^{\prime}}}(k_{0},k_{e^{\prime}})\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e^{\prime}\underset{\text{c.c.}}{\to}e_{4}}B)o_{0}\epsilon_{0}[d_{k_{0}}]\psi_{i}(k_{1},k_{2},k_{3},l_{4},k_{5},\dotsc)\Biggr{)}$
(88)

It now suffices to show that the expression into brackets vanishes thanks to
(83). Let us take care of the signs:
$(-1)^{k_{2}+j_{4}-j_{0}-k_{5}}(-1)^{j_{4}+l_{4}+\frac{1}{2}}(-1)^{j_{0}+k_{3}+j_{4}}(-1)^{k_{1}+k_{2}+k_{3}+l_{4}}=(-1)^{2(k_{2}-k_{5}+k_{3})}(-1)^{4j_{4}}(-1)^{\frac{1+\tilde{\epsilon}_{4}}{2}}(-1)^{k_{1}+l_{4}+k_{5}}=-\tilde{\epsilon}_{4}(-1)^{k_{1}+l_{4}+k_{5}}$.
Replace the sign factor in the bracket, we get

$-\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e_{1}\underset{\text{c.c.}}{\to}e^{\prime}}A)o_{0}o_{1}\epsilon_{1}[d_{k_{1}}][d_{l_{4}}](-1)^{j_{1}-k_{1}+j_{4}-l_{4}}(-1)^{j_{1}+l_{4}+k_{5}+\frac{1}{2}}\begin{Bmatrix}j_{4}&l_{4}&\frac{1}{2}\\\
k_{1}&j_{1}&k_{5}\end{Bmatrix}_{q}\psi_{i}(j_{1},k_{2},k_{3},j_{4},k_{5},\dotsc)\\\
+(-1)^{d_{14,f}-d_{e_{0}e^{\prime}}}\alpha^{\epsilon_{0},\epsilon_{e^{\prime}}}(k_{0},k_{e^{\prime}})\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e^{\prime}\underset{\text{c.c.}}{\to}e_{4}}B)o_{0}\epsilon_{0}[d_{k_{0}}]\psi_{i}(k_{1},k_{2},k_{3},l_{4},k_{5},\dotsc)\\\
=-o_{0}\tilde{\epsilon}_{4}[d_{l_{4}}]\Bigl{(}\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e_{1}\underset{\text{c.c.}}{\to}e^{\prime}}A)A_{o_{1}}^{\epsilon_{1},\tilde{\epsilon}_{4}}(k_{1},l_{4},k_{5})\psi_{i}(j_{1},k_{2},k_{3},j_{4},k_{5},\dotsc)\\\
+(-1)^{d_{14,i}-d_{e_{4}e^{\prime}}}\alpha^{\tilde{\epsilon}_{4},\epsilon_{e^{\prime}}}(l_{4},k_{e^{\prime}})\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e^{\prime}\underset{\text{c.c.}}{\to}e_{4}}B)\psi_{i}(k_{1},k_{2},k_{3},l_{4},k_{5},\dotsc)\Bigr{)}.$
(89)

The expression into brackets on the RHS is exactly the constraint (83) on
$\Gamma_{i}$ with the choice $e=e_{4}$ of reference edge and arbitrary
$\tilde{\epsilon}_{4}$ fixed. ∎

###  IV.2 Removing an edge

Consider two adjacent faces $F$ and $f$, separated by an edge $e_{0}$. We
consider the move which consists in removing $e_{0}$ (as well as its two end
vertexs). By performing a series of 2-2 Pachner moves (described in Section
IV.1), we can always assume that $f$ is triangular,

$\begin{array}[]{c}\includegraphics{3to1.pdf}\end{array}$ (90)

If $|\psi_{i}\rangle$ is a state which satisfies all the constraints before
the edge removal, we want to describe how it transforms through the move.

######  Theorem IV.2.

$|\psi_{f}\rangle$ with spin network coefficients

$\psi_{f}(j_{1},j_{2},\dotsc)=(-1)^{(1+o_{1})j_{1}+(1+o_{2})j_{2}}\sqrt{[d_{j_{1}}]\,[d_{j_{2}}]}\,\psi_{i}(0,j_{1},j_{2},j_{2},j_{1},\dotsc)$
(91)

is a solution of the constraints on the graph after the edge removal. Here
$o_{1},o_{2}$ are the orientations of the edges $e_{1},e_{2}$ with respect to
$f$ (counter-clockwise oriented) and
$\psi(j_{0},j_{1},j_{2},j_{3},j_{4},\dotsc)$ is the spin network coefficient
of $|\psi_{i}\rangle$.

In other words, $|\psi_{i}\rangle$ gives rise to a solution of the constraints
on $\Gamma_{f}$, obtained by keeping only its $j_{0}=0$ components. We will
use this relation to study the 3-1 Pachner move.

Proof. Consider two reference edges $e,e^{\prime}$ in $F\cup f$, and the
associated constraint such that
$\mathbf{E}_{e_{1}e_{2}}^{\tilde{\epsilon}_{1},\tilde{\epsilon}_{2}}$ is an
$A$-term (without loss of generality since $A$\- and $B$-terms can be
exchanged). Its matrix elements $\langle\\{k_{e}\\}|{\bf H}_{F\cup
f,e,e^{\prime}}|\psi_{f}\rangle$ read

$\langle\\{k_{e}\\}|{\bf H}_{F\cup
f,e,e^{\prime}}|\psi_{f}\rangle\propto\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e\underset{\text{c.c.}}{\to}e^{\prime}}A)A_{o_{2}}^{\tilde{\epsilon}_{2},\tilde{\epsilon}_{1}}(k_{2},k_{1},l_{2})\psi_{f}(j_{1},j_{2},\dotsc)+(-1)^{d_{f}-d_{ee^{\prime}}}\alpha^{\epsilon_{e},\epsilon_{e^{\prime}}}(k_{e},k_{e^{\prime}})\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e^{\prime}\underset{\text{c.c.}}{\to}e}B)\psi_{f}(k_{1},k_{2},\dotsc),$
(92)

with $j_{1,2}=k_{1,2}-\tilde{\epsilon}_{1,2}/2$, and $d_{f}$ denotes the
number of boundary edges surrounding $F\cup f$. Here,
$\prod_{e\underset{\text{c.c.}}{\to}e^{\prime}}A$ is the product of the
$A$-terms from $e$ to $e^{\prime}$ counter-clockwise, except for the one on
the corner of $e_{1},e_{2}$ which has been singled out. We will show those
matrix elements vanish as soon as the constraints on $f$ and on $F$ are both
satisfied on $|\psi_{i}\rangle$, given (91).

On $f$, we have the constraint, for fixed $\epsilon_{1},\epsilon_{2}$, and
$j_{1,2}=k_{1,2}-\epsilon_{1,2}/2$,

$B_{o_{2}}^{-\epsilon_{2},-\epsilon_{1}}(k_{2},k_{1},l_{2})\psi_{i}(k_{0},j_{1},j_{2},k_{3},k_{4},\dotsc)\\\
=\alpha^{\epsilon_{1},\epsilon_{2}}(k_{1},k_{2})\sum_{\epsilon_{0}=\pm}A_{o_{0,f}}^{\epsilon_{0},-\epsilon_{2}}(k_{0},j_{2},k_{3})A_{o_{1}}^{-\epsilon_{1},\epsilon_{0}}(j_{1},k_{0},k_{4})\psi_{i}(k_{0}-\frac{\epsilon_{0}}{2},k_{1},k_{2},k_{3},k_{4},\dotsc),$
(93)

where $o_{0,f}$ is the orientation of $e_{0}$ relative to $f$. On $F$ there is
a constraint similar to (92), from the Hamiltonians
$\mathbf{H}_{F,e,e^{\prime}}$ with the same signs $\epsilon$s. It reads,

$\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e\underset{\text{c.c.}}{\to}e^{\prime}}A)\sum_{\tilde{\epsilon}_{0}=\pm}A_{o_{0,F}}^{\tilde{\epsilon}_{0},\tilde{\epsilon}_{4}}(k_{0},k_{4}+\frac{\tilde{\epsilon}_{4}}{2},k_{1})A_{o_{3}}^{\tilde{\epsilon}_{3},\tilde{\epsilon}_{0}}(k_{3}+\frac{\tilde{\epsilon}_{3}}{2},k_{0},k_{2})\psi_{i}(k_{0}-\frac{\tilde{\epsilon}_{0}}{2},k_{1},k_{2},k_{3}-\frac{\tilde{\epsilon}_{3}}{2},k_{4}-\frac{\tilde{\epsilon}_{4}}{2},\dotsc)\\\
+(-1)^{d_{f}-d_{ee^{\prime}}}\alpha^{\epsilon_{e},\epsilon_{e^{\prime}}}(k_{e},k_{e^{\prime}})\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e^{\prime}\underset{\text{c.c.}}{\to}e}B)\psi_{i}(k_{0},k_{1},k_{2},k_{3},k_{4},\dotsc)=0\,,$
(94)

where $o_{0,F}$ is the orientation of $e_{0}$ as the boundary of $F$, which is
opposite to $o_{0,f}$. Here $\tilde{\epsilon}_{4}$ (respectively
$\tilde{\epsilon}_{3}$) is fixed if $e=e_{4}$ (respectively if
$e^{\prime}=e_{3}$) and summed over otherwise.

We now specialise (93) and (94) to $k_{0}=0$, where they simplify a lot.
First, that enforces $\epsilon_{0}=-$ in (93) and $\tilde{\epsilon}_{0}=-$ in
(94), so that those sums reduce to a single term. In (94) we further take
$k_{3}=k_{2}$ and $k_{4}=k_{1}$. All $q$-$6j$-symbols with a spin equal to 0
can be evaluated as $\\{\begin{smallmatrix}j_{1}&k_{1}&\frac{1}{2}\\\
\frac{1}{2}&0&k_{4}\end{smallmatrix}\\}_{q}=\delta_{j_{1},k_{4}}(-1)^{j_{1}+k_{1}+\frac{1}{2}}/\sqrt{[2][d_{j_{1}}]}$.

As a consequence, (93) gives

$B_{o_{2}}^{-\epsilon_{2},-\epsilon_{1}}(k_{2},k_{1},l_{2})\psi_{i}(0,j_{1},j_{2},j_{2},j_{1},\dotsc)=-\alpha^{\epsilon_{1},\epsilon_{2}}(k_{1},k_{2})o_{0,f}o_{1}\epsilon_{2}\frac{[d_{j_{1}}]}{[2]\sqrt{[d_{j_{1}}][d_{j_{2}}]}}\psi_{i}(\frac{1}{2},k_{1},k_{2},j_{2},j_{1},\dotsc),$
(95)

where $k_{3}=j_{2}$ and $k_{4}=j_{1}$ on the last term is enforced by the
special evaluations of the $q$-$6j$-symbols with a spin 0. Equation (94) gives

$\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e\underset{\text{c.c.}}{\to}e^{\prime}}A)o_{0,f}o_{3}\epsilon_{1}\frac{[d_{k_{2}}]}{[2]\sqrt{[d_{k_{1}}][d_{k_{2}}]}}\psi_{i}(\frac{1}{2},k_{1},k_{2},j_{2},j_{1},\dotsc)\\\
+(-1)^{d_{f}-d_{ee^{\prime}}}\alpha^{\epsilon_{e},\epsilon_{e^{\prime}}}(k_{e},k_{e^{\prime}})\sum_{\\{\tilde{\epsilon}\\}}(\prod_{e^{\prime}\underset{\text{c.c.}}{\to}e}B)\psi_{i}(0,k_{1},k_{2},k_{2},k_{1},\dotsc)=0\,.$
(96)

The term $\psi_{i}(\frac{1}{2},k_{1},k_{2},j_{2},j_{1},\dotsc)$ can be
eliminated using (95). Moreover we turn the $B$-coefficient of this equation
into an $A$-coefficient using
$-A_{o_{2}}^{\epsilon_{2},\epsilon_{1}}(k_{2},k_{1},l_{2})=B_{o_{2}}^{-\epsilon_{2},-\epsilon_{1}}(k_{2},k_{1},l_{2})$.
It is then enough to recognize $\psi_{f}$ as given in (91) to obtain that (92)
vanishes.

∎

###  IV.3 3-1 Pachner move

The 3-1 Pachner move removes a triangular face from the graph and replaces it
with a vertex. The edges incident to the face become incident to the vertex,

$\begin{array}[]{c}\includegraphics{Move3to1.pdf}\end{array}$ (97)

The orientation of all the edges are left arbitrary.

######  Theorem IV.3.

If $|\psi_{i}\rangle$ is a state on the initial graph $\Gamma_{i}$ which
satisfy all the constraints, then its spin network coefficients can be written

$\psi_{i}(j_{1},j_{2},j_{3},j_{4},j_{5},j_{6},\dotsc)=(-1)^{(1+o_{1})j_{1}+(1+o_{2})j_{2}+(1+o_{6})j_{6}}(-1)^{j_{3}+j_{4}+j_{5}}\begin{Bmatrix}j_{1}&j_{2}&j_{3}\\\
j_{4}&j_{5}&j_{6}\end{Bmatrix}_{q}\psi_{f}(j_{3},j_{4},j_{5},\dotsc)\,.$ (98)

where $\psi_{f}(j_{3},j_{4},j_{5},\dotsc)$ are the spin network coefficients
of a state $|\psi_{f}\rangle$ which satisfies all the constraints on the final
graph $\Gamma_{f}$.

Proof. Let us write the constraints on the triangular face. There is one
constraint for each pair of edges of the boundary. For the pair
$(e_{2},e_{6})$, for instance, one gets

$\sum_{\epsilon_{1}=\pm}A_{o_{1}}^{\epsilon_{1},\epsilon_{2}}(k_{1},k_{2},k_{3})A_{o_{6}}^{\epsilon_{6},\epsilon_{1}}(k_{6},k_{1},k_{5})\psi_{i}(j_{1},j_{2},k_{3},k_{4},k_{5},j_{6},\dotsc)\\\
+\alpha^{\epsilon_{2},\epsilon_{6}}(k_{2},k_{6})B_{o_{2}}^{\epsilon_{2},\epsilon_{6}}(j_{2},j_{6},k_{4})\psi_{i}(k_{1},k_{2},k_{3},k_{4},k_{5},k_{6},\dotsc)=0\,.$
(99)

Here $j_{i}=k_{i}-\epsilon_{i}/2$, for $i=1,2,6$. The coefficients are

$\displaystyle
A_{o_{1}}^{\epsilon_{1},\epsilon_{2}}(k_{1},k_{2},k_{3})=o_{1}\epsilon_{1}[d_{k_{1}}](-1)^{k_{1}+k_{2}+k_{3}}\begin{Bmatrix}k_{1}&j_{1}&\frac{1}{2}\\\
j_{2}&k_{2}&k_{3}\end{Bmatrix}_{q},$ (100) $\displaystyle
A_{o_{6}}^{\epsilon_{6},\epsilon_{1}}(k_{6},k_{1},k_{5})=o_{6}\epsilon_{6}[d_{k_{6}}](-1)^{k_{1}+k_{5}+k_{6}}\begin{Bmatrix}k_{6}&j_{6}&\frac{1}{2}\\\
j_{1}&k_{1}&k_{5}\end{Bmatrix}_{q},$ $\displaystyle
B_{o_{2}}^{\epsilon_{2},\epsilon_{6}}(j_{2},j_{6},k_{4})=o_{2}\epsilon_{2}[d_{j_{2}}](-1)^{j_{2}+k_{4}+j_{6}}\begin{Bmatrix}k_{6}&j_{6}&\frac{1}{2}\\\
j_{2}&k_{2}&k_{4}\end{Bmatrix}_{q}\,.$

We thus have the recursion

$\sum_{\epsilon_{1}=\pm}o_{1}o_{2}o_{6}[d_{k_{1}}](-1)^{2k_{1}+k_{2}+k_{3}+k_{5}+k_{6}+\frac{1-\epsilon_{1}}{2}+\frac{\epsilon_{2}}{2}+\frac{\epsilon_{6}}{2}}\begin{Bmatrix}k_{1}&j_{1}&\frac{1}{2}\\\
j_{2}&k_{2}&k_{3}\end{Bmatrix}_{q}\begin{Bmatrix}k_{6}&j_{6}&\frac{1}{2}\\\
j_{1}&k_{1}&k_{5}\end{Bmatrix}_{q}\psi_{i}(j_{1},j_{2},k_{3},k_{4},k_{5},j_{6},\dotsc)\\\
+(-1)^{k_{2}+k_{4}+k_{6}}\begin{Bmatrix}k_{6}&j_{6}&\frac{1}{2}\\\
j_{2}&k_{2}&k_{4}\end{Bmatrix}_{q}\psi_{i}(k_{1},k_{2},k_{3},k_{4},k_{5},k_{6},\dotsc)=0,$
(101)

and similarly for the pairs $(e_{1},e_{2})$, $(e_{6},e_{1})$. A similar result
for the flat case was found in Bonzom:2011nv , where $q$ is set to 1. Those
recursions determine the dependence of $\psi_{i}$ on $j_{1},j_{2},j_{6}$ up to
a single initial condition. As the recursion involve three terms, it may seem
like several initial conditions are required. However, at $k_{1}=0$, only two
terms are left in the recursion, as shown in Equation (95). This means that
from the initial condition $\psi_{i}(0,k_{3},k_{3},k_{4},k_{5},k_{5})$, one
gets
$\psi_{i}(\frac{1}{2},k_{3}-\epsilon_{2}/2,k_{3},k_{4},k_{5},k_{5}-\epsilon_{6}/2)$.
Then this determines $\psi_{i}$ for arbitrary $k_{1},k_{2},k_{6}$. The result
is known to be

$\psi_{i}(k_{1},k_{2},k_{3},k_{4},k_{5},k_{6},\dotsc)=\begin{Bmatrix}k_{1}&k_{2}&k_{3}\\\
k_{4}&k_{5}&k_{6}\end{Bmatrix}_{q}\,\phi(k_{3},k_{4},k_{5},\dotsc),$ (102)

where $\phi(k_{3},k_{4},k_{5},\dotsc)$ is independent of $k_{1},k_{2},k_{6}$.
To determine $\phi$, we set $k_{1}=0$,

$\phi(k_{3},k_{4},k_{5},\dotsc)=\begin{Bmatrix}0&k_{3}&k_{3}\\\
k_{4}&k_{5}&k_{5}\end{Bmatrix}_{q}^{-1}\psi_{i}(0,k_{3},k_{3},k_{4},k_{5},k_{5},\dotsc)=(-1)^{k_{3}+k_{4}+k_{5}}\sqrt{[d_{k_{3}}][d_{k_{5}}]}\psi_{i}(0,k_{3},k_{3},k_{4},k_{5},k_{5},\dotsc).$
(103)

We conclude with Theorem IV.2. ∎

## Conclusion

In this article, we have given a realization of the interplay between the
cosmological constant, curved geometries and quantum group structure in the 3D
loop quantum gravity framework in Euclidean signature with a negative
cosmological constant, which we call the deformed LQG model. In particular,
the deformed constraints at the classical level represent discrete hyperbolic
geometries, as shown in Bonzom:2014wva . Upon the standard quantization
procedure, these deformed constraints become quantum constraints with a
quantum group structure.

We have focused on the Hamiltonian constraints, obtained from the flatness
constraints. We have written them with the deformed spinors and performed the
quantization following the companion article Bonzom:2021rib . The result is a
generalization of the quantum Hamiltonian constraints derived in Bonzom:2011nv
for flat space. By studying the way the solutions to the quantum constraints
change under Pachner moves, we provide a generalization of the Noui-Perez
transition amplitudes Noui:2004iy  to $q\neq 1$ real: the transition
amplitudes are the coefficients relating the physical states in the spin
network basis under Pachner moves. Here, they clearly lead to a Turaev-Viro
model for $q$ real. It is a topological model (with the same finiteness issues
as the $q=1$ version, the Ponzano-Regge model).

Our method is radically different from Noui:2004iy  however and maybe more in
the spirit of LQG. On its way to linking deformed LQG to spin foams, our
method derives Wheeler-DeWitt equations as difference equations on the spin
network coefficients of the states, see Equation (62). In the flat case, the
Hamiltonian constraint can be interpreted as displacements of the vertices of
the triangulation BonzomDittrich . Our difference equations (62) are quantum
implementations of those symmetries.

Although our constraints are in fact derived from the flatness constraints, we
believe that this approach is promising to study both how to incorporate the
cosmological constant in 4D and how to write an interesting dynamics for
curved 4D geometries. A first step on the continuous theory has been initiated
in Girelli:2021pol .

## Acknowledgements

The authors would like to thank Florian Girelli and Etera Livine for early
participation of this work. This research was supported in part by Perimeter
Institute for Theoretical Physics. Research at Perimeter Institute is
supported by the Government of Canada through the Department of Innovation,
Science and Economic Development Canada and by the Province of Ontario through
the Ministry of Research, Innovation and Science. QP is supported by a NSERC
Discovery grant awarded to MD. VB is partially supported by the
ANR-20-CE48-0018 ”3DMaps” grant. The University of Waterloo and the Perimeter
Institute for Theoretical Physics are located in the traditional territory of
the Neutral, Anishnawbe and Haudenosaunee peoples. We thank them for allowing
us to conduct this research on their land.

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