#  Fermion coupling to loop quantum gravity: canonical formulation.

Jerzy Lewandowski 111Jerzy.Lewandowski(AT)fuw.edu.pl Cong Zhang222
czhang(AT)fuw.edu.cn Faculty of Physics, University of Warsaw, Pasteura 5,
02-093 Warsaw, Poland

###### Abstract

In the model of a fermion field coupled to loop quantum gravity, we consider
the Gauss and the Hamiltonian constraints. According to the explicit solutions
to the Gauss constraint, the fermion spins and the gravitational spin networks
intertwine with each other so that the fermion spins contribute to the volume
of the spin network vertices. For the Hamiltonian constraint, the
regularization and quantization procedures are presented in detail. By
introducing an adapted vertex Hilbert space to remove the regulator, we
propose a diffeomorphism covariant graph-changing Hamiltonian constraint
operator of the fermion field. This operator shows how fermions move in the
loop quantum gravity spacetime and simultaneously influences the background
quantum geometry. Moreover, as an innovation of our work, introducing the
vertex Hilbert space also fixes issues so that a densely defined symmetric
Hamiltonian constraint operator can be obtained.

##  I introduction

The real physical world consists of spacetime and matter. According to general
relativity (GR), “spacetime tells matter how to move; matter tells spacetime
how to curve"-John Archibald Wheeler, which should also be carried out in the
quantum theory. Loop quantum gravity (LQG) ashtekar1991lectures ;
ashtekar2004back ; rovelli2005quantum ; thiemann2007modern ;
han2007fundamental , as a background-independent and non-perturbative quantum
gravity theory, sets a stage for incorporating matters into quantum spacetime.
In rovelli1994physical ; brown1995dust ; giesel2010algebraic ;
domagala2010gravity ; lewandowski2011dynamics , the Brown-Kuchar model of
gravity coupled to dust as well as the Rovelli-Smolin model of gravity coupled
to massless Klein-Gordon field is quantized. In han2013spin ;
bianchi2013spinfoam , a minimal coupling of fermions and Yang-Mills fields to
covariant LQG dynamics is proposed. The quantum theory of spinor fields
coupled to LQG is well understood morales1994fermions ; morales1995loop ;
baez1998quantization ; thiemann1998kinematical ; bojowald2008fermions ;
barnett2015fermion ; gambini2015no . In thiemann1998quantumV , a systematic
procedure to couple standard model to the canonical LQG is proposed and
further developed in thiemann1998kinematical ; mansuroglu2021fermion ;
mansuroglu2021kinematics . With the present paper, by employing the procedure
in thiemann1998quantumV , we investigate the Gauss and the Hamiltonian
constraint in the model of fermion field coupled to LQG. In particular, the
Gauss constraint is solved explicitly and, the Hamiltonian constraint is
regularized and quantized by introducing the so-called vertex Hilbert space to
remove the regulator.

In the classical model of gravity coupled to the fermion field. The gravity
action $S_{G}$ can be formulated optionally with the first-order formulation
(see, e.g., ashtekar1991lectures ; holst1996barbero  for the Palatini-Host
action) or the second-order formulation (see, e.g., arnowitt1959dynamical ;
thiemann2007modern  for Hilbert-Einstein action). In the pure gravity case,
these two formulations are equivalent to each other up to boundary terms,
while for the case with the fermion field coupled, the equivalency is no
longer valid. For the first-order formulation, $S_{G}$ is a functional of an
SL($2,\mathbb{C}$) connection, and the fermion field will be coupled to it
directly. As a consequence, the fermion field will result in an on-shell
torsion term in the connection. However, for the second-order formulation, the
fermion field will be coupled to the torsion-free spin connection compatible
with the tetrad. Thus, there is no torsion involved in this formulation. In
the current paper, even though we adopt the second-order formulation for
discussion, the results for the first-order formulation can be obtained
analogously. Moreover, since no extra field is introduced for
deparametrization in our model, the dynamics will be governed by the
Hamiltonian constraint $H[N]$ with lapse functions $N$ rather than the
physical Hamiltonian. Then a problem arises that the Hamiltonian constraint
operator cannot be defined in the diffeomorphism invariant space. This problem
will be solved by, for instance, the master constraint framework
thiemann2006phoenix ; han2006master  or the deparametrization framework
domagala2010gravity ; lewandowski2011dynamics . In these frameworks, one
finally needs to consider the Hamiltonian constraint operator with a constant
lapse function or some dynamical lapse function. These operators can be
constructed directly with the Hamiltonian constraint operators
$\widehat{H[N]}$. Thus, one can only focus on how to define well the
Hamiltonian constraint operators $\widehat{H[N]}$ without loss of generality,
which is a main task of the current work.

The phase space of the fermion field coupled to gravity is composed of fields
$(A_{a}^{i},E^{b}_{j},\Psi,\Pi)$ on the spatial manifold $\Sigma$, where
$A_{a}^{i}$ is an SU(2) connection, $E^{b}_{j}$, the canonical conjugate to
$A_{a}^{i}$, is a densitized triad field, $\Psi$ denotes the fermion field,
and $\Pi$ is the canonical momentum conjugate to $\Psi$ ashtekar1991lectures ;
barbero1995real ; ashtekar1989new . With the variables
$(A_{a}^{i},E^{b}_{j})$, the gravitational Hilbert space is constructed by
using the spin networks ashtekar1994representation ;
lewandowski1994topological ; ASHTEKAR1995191 . Therein, the Hamiltonian
constraint of pure gravity as well as variables representing the geometric
observables are regularized and promoted to operators, and properties of the
operators are welll-studied rovelli1995discreteness ; ashtekar1997quantum ;
ashtekar1997quantumII ; thiemann1998quantum ; thiemann1998quantumII ;
ma2000qhat ; yang2016new ; thiemann2006phoenix ; han2006master ;
lewandowski2014symmetric ; yang2015new ; zhang2020first ; zhang2021firstII .
The pure-gravity Hamiltonian constraint comprises of the curvature of the
connection $A$. The curvature, as an operator, will attach loops on graphs of
the spin network states. In some regularization strategies, the final operator
is the limit as the loops shrink to a point. Then, the vertex Hilbert space is
necessarily introduced to define the limit alesci2010regularization ;
alesci2015hamiltonian ; assanioussi2015new ; zhang2018towards ;
zhang2019bouncing . In general, a vertex Hilbert space is a Hilbert space
averaged with diffeomorphisms preserving some particular vertices. Thus,
elements in a vertex Hilbert space are partially diffeomorphism invariant. In
the current work, the Hamiltonian constraint of fermion is also regularized by
introducing some regulator so that the fermion Hamiltonian constraint is the
limit of the regularized version as the regulator approaches $0$. In order to
define this limit, we also need to introduce an adapted vertex Hilbert space.
Besides, the vertex Hilbert space can do more than just defining limit in this
work. As shown in Sec. IV.2, the regularized fermion Hamiltonian constraint is
split into $H_{\mathcal{C}_{\epsilon}}$ plussing its complex conjugate. Then,
$H_{\mathcal{C}_{\epsilon}}$ can be quantized into an operator
$\widehat{H(\delta)}$ which is cylindrical consistent and diffeomorphism
covariant. However, the adjoint of $\widehat{H(\delta)}$, denoted by
$\widehat{H(\delta)}{}^{\dagger}$, is not densely defined. Actually, as shown
in Sec. IV.2, acting on some particular states associated to a graph $\gamma$,
the operator $\widehat{H(\delta)}$ can change $\gamma$ by erasing a segment
$e(\delta)$ of an edge $e\subset\gamma$. Using $\gamma^{\prime}$ to denote the
graph of $\gamma$ with $e(\delta)$ erased by $\widehat{H(\delta)}$,
$\widehat{H(\delta)}{}^{\dagger}$, acting on a state $\Psi_{\gamma^{\prime}}$
associated to $\gamma^{\prime}$, will add an edge $\tilde{e}(\delta)$ to
$\gamma^{\prime}$, where $\tilde{e}(\delta)$ is arbitrary edge connecting the
end points of $e(\delta)$. Thus, there will be infinitely many ways to add
$\tilde{e}(\delta)$ to $\gamma^{\prime}$, and the resulting states are
orthogonal to each other. Consequently, the result of
$\widehat{H(\delta)}{}^{\dagger}\Psi_{\gamma^{\prime}}$ is not normalizable,
i.e., $\Psi_{\gamma^{\prime}}$ is not in the domain of
$\widehat{H(\delta)}{}^{\dagger}$. This problem will be fixed by introducing
the vertex Hilbert space to define limit. In the vertex Hilbert space,
$\lim_{\delta\to 0}\widehat{H(\delta)}$ will be defined properly as an
operator $\hat{A}_{F}^{\dagger}$. The action of $\hat{A}_{F}^{\dagger}$
performs in a way that a projection is left multiplied in
$\hat{A}_{F}^{\dagger}$ as a factor. Then, if $\hat{A}_{F}^{\dagger}$, acting
on these particular states, erases segments, the projection will annihilate
the resulting states, so that the adjoint of $\hat{A}_{F}^{\dagger}$ is
densely defined. Actually, the same issue also exists in defining the
gravitational Hamiltonian constraint operator with certain regularization
strategy, and can be fixed with the same mechanism yang2015new .

The usual Hamiltonian analysis tells $\Pi^{\dagger}=\sqrt{q}\,\Psi$ with
$q\equiv|\det(E)|$. In the quantum theory, this equation is expected to be
realized in an appropriate form. In other words, one might require that the
adjoint of the operator $\hat{\Pi}$ is related to the operator $\hat{\Psi}$
via $\hat{\Pi}^{\dagger}=\widehat{\sqrt{q}}\hat{\Psi}$. Then, contradiction
appears because in our quantum theory $\sqrt{q}$ will become an operator
rather than some background c-number. This can be seen as follows. Considering
a non-trivial real-valued function $f(A)$ of the connection $A_{a}^{i}$, one
has $0=[\hat{\Pi},\widehat{f(A)}]$ on one hand, but, on the other hand,
$[\hat{\Pi},\widehat{f(A)}]^{\dagger}=[\widehat{f(A)},\hat{\Pi}^{\dagger}]=[\widehat{f(A)},\widehat{\sqrt{q}}\hat{\Psi}]\neq
0$. The two results do not coincide. To overcome this inconsistency, the
author in thiemann1998quantumV  introduces the Grassman-valued half-densities
$\tilde{\Psi}:=\sqrt[4]{q}\Psi$ and $\tilde{\Psi}^{\dagger}$ to define the
phase space of fermion. Moreover, in order to do quantization, the smeared
version of $\tilde{\Psi}$ by $\sqrt{\delta(x,y)}$, i.e.,
$\int\differential^{3}y\sqrt{\delta(x,y)}\tilde{\Psi}(y)$, is also introduced
by thiemann1998quantumV . With the smeared variables, the fermion sector is
quantized and the resulting quantum theory carries out the dieffeomorphsm-
invariance feature.

It will be useful to compare our work with the known models of fermions in
LQG. Among these models, a typical one is to introduce the path observables
(see, e.g., morales1994fermions ; morales1995loop ; baez1998quantization )
each of which combines the holonomy along a path $\alpha$ and the fermions at
the ends of $\alpha$. These path variables form a closed algebra under the
Poisson bracket so that the quantum theory is obtained by quantizing this
algebra. Another typical model is the one employed in the current work (see,
e.g., thiemann1998kinematical ). As aforementioned, this model considers the
algebra formed by $\int\differential^{3}y\sqrt{\delta(x,y)}\tilde{\Psi}(y)$,
which leads to a fermionic Fock space as the Hilbert space for fermions. Our
work develops this model from the following aspects. At first, our work
studies the Gauss constraint in details. We not only give the exact solutions
to the Gauss constraint but also study the physical pictures resulting from
these solutions. Second, for the Hamiltonian constraint of the fermion field,
we propose a different regularization strategy than that in
thiemann1998quantumV . Additionally, the vertex Hilbert space is introduced,
not only for removing the regulator in the regularized Hamiltonian constraint
but also for fixing the issues on the adjoint of $\widehat{H(\delta)}$, so
that a densely defined symmetric Hamiltonian constraint operator can be
obtained.

This paper is organized as follows. In Sec. II the classical theory of gravity
coupled to fermion field is introduced briefly. In Sec. III we introduce some
basic notions of the kinematical Hilbert space of pure gravity and revisit the
construction of the fermion kinematical Hilbert space. In Sec. IV, the Gauss
constraint and the Hamiltonian constraint are regularized and quantized, where
the adapted vertex Hilbert space is introduced and some physical results are
discussed. Finally, in V, we summarize the remarkable results and propose some
outlooks for further works.

##  II Classical theory of gravity coupled to Fermion

Let $\mathcal{M}$ denote the spacetime manifold, which is homeomorphism to
$\mathbb{R}\times\Sigma$ with $\Sigma$ being the spatial manifold. Given a
4-dimensional vector space $V$, let $\eta_{IJ}$ be the Minkowski metric on it.
A tetrad field $e_{\mu}^{I}$ gives the metric
$g_{\mu\nu}=\eta_{IJ}e^{I}_{\mu}e^{J}_{\nu}$ on $\mathcal{M}$. The curvature
of $g_{\mu\nu}$ defines the Einstein-Hilbert action in terms of the tetrad
fields,

$S_{H}[e]=\frac{1}{2\kappa}\int_{\mathcal{M}}\differential^{4}xR[e].$ (1)

where $\kappa=8\pi G$. Let $\Gamma^{I}_{\mu J}$ denote the spin connection
compatible with the tetrad so that

$\differential e^{I}+\Gamma^{I}{}_{J}\wedge e^{J}=0.$ (2)

The model of gravity coupled to the fermion field is described by the action

$\displaystyle
S[e,\Psi]=S_{H}[e]-\frac{i}{2}\int_{\mathcal{M}}\differential^{4}xe\,\left(\overline{\Psi}\gamma^{I}e^{\mu}_{I}\nabla_{\mu}\Psi-c.c\right)$
(3)

where $\gamma^{I}$ denotes the gamma matrices satisfying
$\gamma^{I}\gamma^{J}+\gamma^{J}\gamma^{I}=2\eta^{IJ}\mathbbm{1},$ and the
covariant derivative of $\Psi$ is

$\nabla_{\mu}\Psi=\partial_{\mu}\Psi-\frac{1}{4}\Gamma^{I}_{\mu
J}\gamma^{I}\gamma^{J}\Psi.$ (4)

Performing the 3+1-decomposition and doing canonical transformation
thiemann2007modern , we get the gravitational canonical pair
$(A_{a}^{i},E^{a}_{i})$. The Poisson brackets between them are

$\\{A_{a}^{i}(x),E^{b}_{j}(y)\\}=\kappa\beta\delta_{a}^{b}\delta^{i}_{j}\delta(x,y),$
(5)

where $\beta$ is the Barbero-Immirzi parameter. For the fermion field, we
split the Dirac fermion $\Psi$ into its chiral components and follow the
argument in thiemann1998quantumV  to introduce the half densities on $\Sigma$,

$\xi:=\sqrt[4]{q}\Psi_{-},\ \nu=\sqrt[4]{q}\Psi_{+}$ (6)

with $\Psi_{\pm}=\frac{1\pm\gamma^{5}}{2}\Psi$. Detailed Hamiltonian analysis
(see Appendix A) tells that the conjugate moments to $\xi$ and $\nu$ are their
complex conjugates, and the anti-Poisson brackets are

$\displaystyle\\{\xi_{A}(x),\xi_{B}^{\dagger}(y)\\}_{+}$
$\displaystyle=-i\delta_{AB}\delta(x,y),$ (7)
$\displaystyle\\{\nu_{A}(x),\nu_{B}^{\dagger}(y)\\}_{+}$
$\displaystyle=-i\delta_{AB}\delta(x,y).$

for all $A,B=\pm 1/2$.

The dynamics of this model is encoded in the Gauss constraint $G_{m}$, the
diffeomorphism constraint $H_{a}$ and the Hamiltonian constraint $H$, which
are

$\displaystyle G_{m}=$
$\displaystyle\frac{1}{\kappa\beta}D_{a}E^{a}_{l}+\frac{1}{2}(\xi^{\dagger}\sigma_{m}\xi+\nu^{\dagger}\sigma_{m}\nu),$
(8) $\displaystyle H_{a}=$
$\displaystyle\frac{1}{\kappa\beta}E^{b}_{i}F^{i}_{ab}+\frac{i}{2}\Big{\\{}\xi^{\dagger}D_{a}\xi-(D_{a}\xi)^{\dagger}\xi+\nu^{\dagger}D_{a}\nu-(D_{a}\nu)^{\dagger}\nu\Big{\\}}+\beta
K_{a}^{m}G_{m},$ $\displaystyle H=$ $\displaystyle
H_{G}+\frac{1}{2\sqrt{q}}\Big{[}i(\xi^{\dagger}E_{i}^{a}\sigma^{i}D_{a}\xi-(D_{a}\xi)^{\dagger}E_{i}^{a}\sigma^{i}\xi)-\beta
E^{a}_{i}K_{a}^{i}\xi^{\dagger}\xi-\frac{1}{\beta}(1+\beta^{2})D_{a}E^{a}_{i}\xi^{\dagger}\sigma^{i}\xi-\beta
E^{a}_{i}D_{a}\Big{(}\xi^{\dagger}\sigma^{i}\xi\Big{)}$
$\displaystyle-i(\nu^{\dagger}E_{i}^{a}\sigma^{i}D_{a}\nu-(D_{a}\nu)^{\dagger}E_{i}^{a}\sigma^{i}\nu)+\beta
E^{a}_{i}K_{a}^{i}\nu^{\dagger}\nu-\frac{1}{\beta}(1+\beta^{2})D_{a}E^{a}_{i}\nu^{\dagger}\sigma^{i}\nu-\beta\frac{1}{\sqrt{q}}E^{a}_{i}D_{a}\Big{(}\nu^{\dagger}\sigma^{i}\nu\Big{)}\Big{]}.$

Here, $H_{G}$ denotes the scalar constraint of pure gravity,

$H_{G}=\frac{1}{2\kappa\sqrt{q}}E^{a}_{i}E^{b}_{j}\left(F_{ab}^{m}\epsilon_{m}{}^{ij}-2(1+\beta^{2})K_{[a}^{i}K_{b]}^{j}\right).$
(9)

##  III Loop quantization of the theory: kinematics

###  III.1 the kinematical Hilbert space of pure gravity

In LQG, besides a fixed differentiability class $C^{m}$ with $m\geq 1$, a
semianalytic structure on $\Sigma$ is also necessary lewandowski2005uniqueness
. Then all local maps, differmorphisms, submanifold and function thereon are
assumed to be $C^{m}$ and semianalytic. Particularly, an edge is a
semianalytic curve embedded in $\Sigma$. A graph is a collection of edges
$\\{e_{1},\cdots,e_{n}\\}$ where these $e_{k}$ intersect each other at most at
the ending points. Given a graph $\gamma\subset\Sigma$, let $E(\gamma)$ denote
the set of its edges and $V(\gamma)$, its vertices. The number of elements in
$E(\gamma)$ ($V(\gamma)$ respectively) is denoted by $|E(\gamma)|$
($|V(\gamma)|$ respectively). A cylindrical function $\Psi$ of the Ashtekar
connection $A$ is a function that can be written in the form

$\Psi(A)=\psi_{\gamma}(h_{e_{1}}(A),\cdots,h_{e_{n}}(A))$ (10)

where $\psi_{\gamma}:{\mathrm{SU(2)}}^{|E(\gamma)|}\to\mathbb{C}$ is a complex
function on ${\mathrm{SU(2)}}^{|E(\gamma)|}$ and $h_{e}(A)\in{\mathrm{SU(2)}}$
is the parallel transport along an edge $e$ with respect to a given connection
$A$,

$h_{e}(A)=\mathcal{P}\exp\left(-\int_{e}A\right)=1+\sum_{n=1}^{\infty}(-1)^{n}\int_{0}^{1}\differential
t_{n}\int_{0}^{t_{n}}\differential t_{n-1}\cdots\int_{0}^{t_{2}}\differential
t_{1}A(t_{1})\cdots A(t_{n}).$ (11)

Given a cylindrical function $\Psi$ with respect to a graph $\gamma$, it can
always be rewritten via another graph $\gamma^{\prime}\supset\gamma$.
Therefore, for two cylindrical functions $\Psi^{(1)}$ and $\Psi^{(2)}$ with
respect to graphs $\gamma_{1}$ and $\gamma_{2}$ respectively, one can always
find another graph $\gamma_{3}$ with
$\gamma_{3}\supset\gamma_{1},\gamma_{3}\supset\gamma_{2}$, and rewrite
$\Psi^{(1)}$ and $\Psi^{(2)}$ by some functions $\psi_{\gamma_{3}}^{(1)}$ and
$\psi_{\gamma_{3}}^{(2)}$ respectively on
${\mathrm{SU(2)}}^{|E(\gamma_{3})|}$. Then the inner product of $\Psi^{(1)}$
and $\Psi^{(2)}$ is

$\langle\Psi^{(1)}|\Psi^{(2)}\rangle=\int_{{\mathrm{SU(2)}}^{n}}\differential\mu_{H}(g)\overline{\psi_{\gamma_{3}}^{(1)}(g_{1},\cdots,g_{n})}\psi_{\gamma_{3}}^{(2)}(g_{1},\cdots,g_{n})$
(12)

where $n=|E(\gamma_{3})|$ and $\differential\mu_{H}$ is the Haar measure on
${\mathrm{SU(2)}}^{n}$. Eq. (12) defines a measure $\differential\mu_{\rm AL}$
on the quantum configuration space $\mathcal{A}$. Thus (12) is always
rewritten as

$\langle\Psi^{(1)}|\Psi^{(2)}\rangle=\int_{\mathcal{A}}\differential\mu_{\rm
AL}(A)\,\overline{\Psi^{(1)}(A)}\Psi^{(2)}(A)$ (13)

The space of cylindrical function is denoted by ${\mathbf{Cyl}}$. The Hilbert
space $\mathcal{H}_{G}$ of the pure gravity is the completion of
${\mathbf{Cyl}}$ with the inner product define in (13).

Given a graph $\gamma$, the space of cylindrical functions with respect to
$\gamma$ is denoted by ${\mathbf{Cyl}}_{\gamma}$. The Cauchy completion of
${\mathbf{Cyl}}_{\gamma}$ with respect to (12) is denoted by
$\mathcal{H}_{G,\gamma}$. As shown in ashtekar2004back ,
$\mathcal{H}_{G,\gamma}$ admits the spin network decomposition

$\mathcal{H}_{G,\gamma}={\bigoplus_{\vec{j},\vec{l}}}\mathcal{H}_{G,\gamma}^{(\vec{j},\vec{l})}$
(14)

where $\vec{j}=\\{j_{1},j_{2},\cdots,j_{|E(\gamma)|}\\}$ assigns to each edge
of $\gamma$ an irreducible representation of ${\mathrm{SU(2)}}$, and
$\vec{l}=\\{l_{1},\cdots,l_{|V(\gamma)|}\\}$, to each vertex of $\gamma$ an
irreducible representation. Consider the assignments $\vec{j}^{\prime}$ such
that each representation is non-trivial. Besides, let $\vec{l}^{\prime}$
denote assignments of irreducible representations to vertices of $\gamma$
which are non-trivial at each spurious vertex of $\gamma$, where a vertex $v$
is spurious if it is bivalent, and if the edges $e_{i}$ and $e_{i+1}$ which
meet at $v$ are such that $e_{i}\circ e_{i+1}$ is itself an semianalytic
(i.e., $v$ just serves to split and edge). Then, we define
$\mathcal{H}_{G,\gamma}^{\prime},$ as

$\mathcal{H}_{G,\gamma}^{\prime}={\bigoplus_{\vec{j}^{\prime},\vec{l}^{\prime}}}\mathcal{H}_{G,\gamma}^{(\vec{j}^{\prime},\vec{l}^{\prime})}.$
(15)

Thanks to $\mathcal{H}_{G,\gamma}^{\prime}$, the space ${\mathbf{Cyl}}$ can be
decomposed as333For elements in an infinite direct sum, we require that all
but finitely many components are zero.

${\mathbf{Cyl}}=\bigoplus_{\gamma}\mathcal{H}_{G,\gamma}^{\prime}\oplus\mathbb{C}.$
(16)

The multiplication operator $D^{\iota}_{ab}(h_{e})$ acts on a cylindrical
function $\Psi(A)=\psi_{\gamma}(h_{e_{1}}(A),\cdots,h_{e_{n}}(A))$ as

$(D^{\iota}_{ab}(h_{e})\Psi)(A)=D^{\iota}_{ab}(h_{e}(A))\psi_{\gamma}(h_{e_{1}}(A),\cdots,h_{e_{n}}(A)),$
(17)

where $D^{\iota}_{ab}(h_{e}(A))$ denotes the Wigner-D matrix of
$h_{e}(A)\in{\mathrm{SU(2)}}$. Moreover, the derivative operators
$\hat{J}_{i}^{v,e}$, for $i=1,2,3$ and $v$ being source point $s_{e}$ of $e$
or the target point $t_{e}$ of $e$, act on $\Psi$ as

$(\hat{J}_{j}^{v,e}\Psi)(A)=\left\\{\begin{aligned}
&i\left.\frac{\differential}{\differential
t}\right|_{t=0}\psi_{\gamma}(h_{e_{1}},\cdots,h_{e}e^{t\tau_{j}},\cdots,h_{e_{n}}),\
v=s_{e},\\\ &i\left.\frac{\differential}{\differential
t}\right|_{t=0}\psi_{\gamma}(h_{e_{1}},\cdots,e^{-t\tau_{j}}h_{e},\cdots,h_{e_{n}}),\
v=t_{e},\end{aligned}\right.$ (18)

where $\tau_{j}=-i\sigma_{j}/2$ with $\sigma_{j}$ being the Pauli matrices.
With the operator $\hat{J}_{j}^{v,e}$, we can define an operator
$\hat{J}_{j}^{x,[e]}$ such that

$\hat{J}_{j}^{x,[e]}\Psi=\sum_{e^{\prime}\in[e]}\hat{J}^{x,e^{\prime}}_{j}\Psi,\forall\Psi\in{\mathbf{Cyl}},$
(19)

where $[e]$ is a maximal family of curves beginning at $x\in\Sigma$ such that
each two curves overlap on a connected initial segment containing $x$.

###  III.2 loop quantization of the fermion field

From now on, we will only focus on the single Weyl component $\xi$. However,
everything works similarly for the other chiral component $\nu$. To quantize
the fermion field, we follow thiemann1998quantumV  to use the modified
symplectic structure

$\\{\theta_{A}(x),\theta_{B}^{\dagger}(y)\\}_{+}=-i\delta_{AB}\delta_{x,y},\
A,B=\pm\frac{1}{2}$ (20)

where, in comparison with (7), the Dirac delta $\delta(x,y)$ is changed to the
Kronecker delta $\delta_{x,y}$. This change results from the following
canonical transformation from $\xi$ to $\theta$:

$\displaystyle\theta(x)$
$\displaystyle=\int_{\Sigma}\differential^{3}y\sqrt{\delta(x,y)}\xi(y),$ (21)
$\displaystyle\xi(x)$
$\displaystyle=\sum_{y\in\Sigma}\sqrt{\delta(x,y)}\theta(y).$

To prove the relation (21), one used the function
$f_{\epsilon}(x,y):=\chi_{\epsilon}(x,y)/\epsilon^{3}$ to regularize the Dirac
delta function thiemann1998quantumV , where $\chi_{\epsilon}$ is

$\chi_{\epsilon}(x,y):=\left\\{\begin{aligned} 1,\
&\sum_{a=1}^{3}(x^{a})^{2}+(y^{a})^{2}\leq\left(\frac{\epsilon}{2}\right)^{2},\\\
0,\ &\text{otherwise}.\end{aligned}\right.$

According to (21), $\xi(x)$ will be singular for regular $\theta(x)$. This
scenario contradicts that $\xi(x)$ is smooth as a classical field. To have a
consistent understanding of this formulation, one interprets the singular
fields $\xi(x)$ as defining the quantum configuration space of the fermion
field so that $\theta(x)$ is a regular-field coordinate of this quantum
configuration space.

As in pure LQG, the quantization starts by introducing the cylindrical
functions. Before doing so, we will first introduce a convenient field
$\zeta_{A}$ as

$\zeta_{A}(x)=\frac{1}{\sqrt{\hbar}}\theta_{A}(x),\ A=\pm\frac{1}{2}.$ (22)

####  III.2.1 the cylindrical functions of the fermion field

A fermionic graph $\gamma_{F}$ is a finite subset of $\Sigma$ with
$|\gamma_{F}|$ elements. Elements in $\gamma_{F}$ are called fermionic
vertices. An orientation of $\gamma_{F}$ is a surjection $n\mapsto
v_{F}^{(n)}\in\gamma_{F}$ with $1\leq n\leq|\gamma_{F}|$. The surjection
endows the elements in $\gamma_{F}$ with an order. Given an oriented graph
$\gamma_{F}$, we have a family of Grassmann numbers
$\\{\zeta^{\dagger}_{\pm\frac{1}{2}}(v_{F})\\}_{v_{F}\in\gamma_{F}}$, which
will be renamed to $\rho^{\dagger}_{n}$ with $1\leq n\leq 2|\gamma_{F}|$ such
that

$\displaystyle\rho_{2i-1}^{\dagger}\equiv\zeta_{\frac{1}{2}}^{\dagger}(v_{F}^{(i)}),\
\rho_{2i}^{\dagger}\equiv\zeta_{-\frac{1}{2}}^{\dagger}(v_{F}^{(i)}),\ 1\leq
i\leq|\gamma_{F}|.$ (23)

Then a function $\Psi$ of $\zeta^{\dagger}$ with respect to $\gamma_{F}$ takes
the general form

$\Psi(\zeta^{\dagger})=f_{0}+\sum_{n=1}^{2|\gamma_{F}|}\sum_{1\leq
i_{1}<i_{2}<\cdots<i_{n}\leq{2|\gamma_{F}|}}f_{i_{1}\cdots
i_{n}}\rho^{\dagger}_{i_{1}}\rho^{\dagger}_{i_{2}}\cdots\rho^{\dagger}_{i_{n}},$
(24)

where $f_{0}$ and $f_{i_{1}\cdots i_{n}}$ are complex numbers. Functions
taking the form (24) are called the (fermionic) cylindrical functions of
$\zeta^{\dagger}$. The space of cylindrical functions of $\zeta^{\dagger}$
will be denoted by ${\mathbf{Cyl_{F}}}$.

According to (24), each cylindrical function with respect to $\gamma_{F}$ can
be identical with a vector

$f_{\gamma_{F}}=\left(f_{0},\\{f_{i_{1}\cdots i_{n}}\\}_{1\leq
i_{1}<i_{2}<\cdots<i_{n}\leq{2|\gamma_{F}|}}\right)\in\mathbb{C}^{2^{2|\gamma_{F}|}}.$

Moreover, for a cylindrical function $\Psi$ expressed via
$f_{\gamma_{F}}\in\mathbb{C}^{2^{2|\gamma_{F}|}}$, there always exists a lager
graph $\gamma_{F}^{\prime}\supset\gamma_{F}$ such that $\Psi$ is rewritten via
some
$f^{\prime}_{\gamma_{F}^{\prime}}\in\mathbb{C}^{2^{2|\gamma_{F}^{\prime}|}}$.
Thus, given two functions $\Psi_{1},\Psi_{2}\in{\mathbf{Cyl_{F}}}$ on
$\gamma_{F}{}_{1}$ and $\gamma_{F}{}_{2}$ respectively, we can find another
graph $\gamma_{F}^{\prime}$ containing both $\gamma_{F}{}_{1}$ and
$\gamma_{F}{}_{2}$ to rewrite $\Psi_{1}$ and $\Psi_{2}$ with
$f_{\gamma_{F}^{\prime}}^{(1)},f_{\gamma_{F}^{\prime}}^{(2)}\in\mathbb{C}^{2^{2|\gamma_{F}^{\prime}|}}$.
Then the inner product of $\Psi_{1}$ and $\Psi_{2}$ is given by

$\langle\Psi_{1},\Psi_{2}\rangle=\int\differential\mu(\rho_{1}\rho_{1}^{\dagger})\differential\mu(\rho_{2}\rho_{2}^{\dagger})\cdots\differential\mu(\rho_{2|\gamma_{F}^{\prime}|}\rho_{2|\gamma_{F}|^{\prime}}^{\dagger})\Psi_{1}^{\dagger}\Psi_{2}$
(25)

with
$\differential\mu(\rho_{n}\rho_{n}^{\dagger})=\differential\rho_{n}^{\dagger}\differential\rho_{n}e^{\rho_{n}\rho_{n}^{\dagger}}$.
By applying (25), one can verify

$\langle\Psi_{1},\Psi_{2}\rangle=(f_{\gamma_{F}^{\prime}}^{(1)})^{\dagger}f_{\gamma_{F}^{\prime}}^{(2)}.$
(26)

Moreover, even though the graphs $\gamma_{F}^{\prime}$ containing both
$\gamma_{F}{}_{1}$ and $\gamma_{F}{}_{2}$ are not unique, Eq. (25) is
independent of the choice of $\gamma_{F}^{\prime}$ since
$\int\differential\mu(\rho_{n}\rho_{n}^{\dagger})=1$. Indeed, Eq. (25) defines
a measure $\differential\mu_{H}(\zeta^{\dagger}\zeta)$ on
${\mathbf{Cyl_{F}}}^{\dagger}\otimes{\mathbf{Cyl_{F}}}$, and will be rewritten
as

$\langle\Psi_{1},\Psi_{2}\rangle=\int\differential\mu_{H}(\zeta\zeta^{\dagger})\Psi_{1}^{\dagger}\Psi_{2}.$
(27)

The fermion Hilbert space $\mathcal{H}_{F}$ is the completion of
${\mathbf{Cyl_{F}}}$ with the inner product defined by (27), i.e.,

$\mathcal{H}_{F}=\overline{{\mathbf{Cyl_{F}}}}.$ (28)

On ${\mathbf{Cyl_{F}}}$, a type of operators are the multiplication operators
$\widehat{\zeta_{v_{F},A}^{\dagger}}$ given by

$(\widehat{\zeta_{v_{F},A}^{\dagger}}\Psi)(\zeta^{\dagger})=\zeta_{A}^{\dagger}(v_{F})\Psi(\zeta^{\dagger}),\
\forall\Psi\in{\mathbf{Cyl_{F}}}.$ (29)

Another type of operators are the derivative operators
$\hat{\zeta}_{v_{F},A}$, acting on $\Psi\in{\mathbf{Cyl_{F}}}$ as

$(\hat{\zeta}_{v_{F},A}\Psi)(\zeta^{\dagger})=\left(\frac{\partial}{\partial\zeta_{A}^{\dagger}(v_{F})}\Psi\right)(\zeta^{\dagger}),\
\forall\Psi\in{\mathbf{Cyl_{F}}}.$ (30)

It is easy to verify that $\widehat{\zeta_{v_{F},A}^{\dagger}}$ and
$\hat{\zeta}_{v_{F},A}$ are adjoint to each other, i.e.,

$\widehat{\zeta_{v_{F},A}^{\dagger}}=\hat{\zeta}_{v_{F},A}^{\dagger},$ (31)

which realizes the real condition. Moreover, one has

$[{\hat{\zeta}}_{v_{F},A},{\hat{\zeta}}^{\dagger}_{v_{F}^{\prime},B}]_{+}={\hat{\zeta}}_{v_{F},A},{\hat{\zeta}}^{\dagger}_{v_{F}^{\prime},B}+{\hat{\zeta}}^{\dagger}_{v_{F}^{\prime},B}{\hat{\zeta}}_{v_{F},A}=\delta_{AB}\delta_{v_{F}v_{F}^{\prime}},$
(32)

which implements the Poisson brackets (20) by defining

$\hat{\theta}_{A}(v_{F})=\sqrt{\hbar}\,\hat{\zeta}_{v_{F},A}.$ (33)

####  III.2.2 the spin network states of fermion field

Given a graph $\gamma_{F}$, the space of the fermionic cylindrical functions
with respect to $\gamma_{F}$ is a finite-dimensional Hilbert space, denoted by
$\mathcal{H}_{{F,}\gamma_{F}}$. Considering a graph $\gamma_{F}=\\{v_{F}\\}$
which is compronent of a single vertex, one has the the space
$\mathcal{H}_{{F,}\\{v_{F}\\}}\equiv\mathcal{H}_{v_{F}}$ consisting of
functions

$\Psi(\zeta^{\dagger})=a_{00}+a_{10}\zeta_{\frac{1}{2}}^{\dagger}(v_{F})+a_{01}\zeta_{-\frac{1}{2}}^{\dagger}(v_{F})+a_{11}\zeta_{\frac{1}{2}}^{\dagger}(v_{F})\zeta_{-\frac{1}{2}}^{\dagger}(v_{F})$
(34)

The inner product of
$\Psi^{(i)}(\zeta^{\dagger})=a_{00}^{(i)}+a_{10}^{(i)}\zeta_{\frac{1}{2}}^{\dagger}(v_{F})+a_{01}^{(i)}\zeta_{-\frac{1}{2}}^{\dagger}(v_{F})+a_{11}^{(i)}\zeta_{\frac{1}{2}}^{\dagger}(v_{F})\zeta_{-\frac{1}{2}}^{\dagger}(v_{F})$
with $i=1,2$ is

$\langle\Psi^{(1)},\Psi^{(2)}\rangle=\sum_{i,j\in\\{0,1\\}}(a_{ij}^{(1)})^{*}a_{ij}^{(2)}$
(35)

For convenience, we introduce the Dirac bra-ket notation $|i,j\rangle_{v_{F}}$
($i,j=0,1$) to denote the state $\Psi_{ij}$, where $\Psi_{ij}$ are the states
given by $\Psi_{00}(\zeta^{\dagger})=1$,
$\Psi_{10}(\zeta^{\dagger})=\zeta_{\frac{1}{2}}^{\dagger}(v_{F})$,
$\Psi_{01}(\zeta^{\dagger})=\zeta_{-\frac{1}{2}}^{\dagger}(v_{F})$ and
$\Psi_{11}(\zeta^{\dagger})=\zeta_{\frac{1}{2}}^{\dagger}(v_{F})\zeta_{-\frac{1}{2}}^{\dagger}(v_{F})$.
Then the states $|i,j\rangle_{v_{F}}$ form an orthonormal basis of
$\mathcal{H}_{v_{F}}$, i.e.,

${}_{v_{F}}\langle
i_{1},j_{1}|i_{2},j_{2}\rangle_{v_{F}}=\delta_{i_{1}i_{2}}\delta_{j_{1}j_{2}}.$
(36)

The action of $\hat{\zeta}_{v_{F},A}$ and $\hat{\zeta}_{v_{F},A}^{\dagger}$
for $A=\pm\frac{1}{2}$ on $\mathcal{H}_{v_{F}}$ reads

$\begin{array}[]{lll}\hat{\zeta}_{v_{F},\frac{1}{2}}^{\dagger}|0,i_{2}\rangle_{v_{F}}=|1,i_{2}\rangle_{v_{F}},&\hat{\zeta}_{v_{F},\frac{1}{2}}^{\dagger}|1,i_{2}\rangle_{v_{F}}=0,&\forall
i_{2}=0,1,\\\
\hat{\zeta}_{v_{F},\frac{1}{2}}|0,i_{2}\rangle_{v_{F}}=0,&\hat{\zeta}_{v_{F},\frac{1}{2}}|1,i_{2}\rangle_{v_{F}}=|0,i_{2}\rangle_{v_{F}},&\forall
i_{2}=0,1,\\\
\hat{\zeta}_{v_{F},-\frac{1}{2}}^{\dagger}|i_{1},0\rangle_{v_{F}}=(-1)^{i_{1}}|i_{1},1\rangle_{v_{F}},&\hat{\zeta}_{v_{F},-\frac{1}{2}}^{\dagger}|i_{1},1\rangle_{v_{F}}=0,&\forall
i_{1}=0,1,\\\
\hat{\zeta}_{v_{F},-\frac{1}{2}}|i_{1},0\rangle_{v_{F}}=0,&\hat{\zeta}_{v_{F},-\frac{1}{2}}|i_{1},1\rangle_{v_{F}}=(-1)^{i_{1}}|i_{1},0\rangle_{v_{F}},&\forall
i_{1}=0,1.\end{array}$ (37)

For the general case where $\gamma_{F}\subset\Sigma$ consists of more than one
vertices, we first associate to each $v_{F}\in\gamma_{F}$ the Hilbert space
$\mathcal{H}_{v_{F}}$. Then, fixing an orientation of $\gamma_{F}$, one has
the tensor product space

$\mathcal{H}_{F,\gamma_{F}}=\mathcal{H}_{v_{F}^{(1)}}\otimes\mathcal{H}_{v_{F}^{(2)}}\otimes\cdots\otimes\mathcal{H}_{v_{F}^{|\gamma_{F}|}}.$
(38)

An orthonormal basis of $\mathcal{H}_{F,\gamma_{F}}$ is composed of the
vectors

$|i_{1},i_{2},\cdots,i_{2|\gamma_{F}|}\rangle:=|i_{1},i_{2}\rangle_{v_{F}^{(1)}}\otimes|i_{3},i_{4}\rangle_{v_{F}^{(2)}}\otimes\cdots\otimes|i_{2|\gamma_{F}|-1},i_{2|\gamma_{F}|}\rangle_{v_{F}^{(|\gamma_{F}|)}}$
(39)

with $i_{k}\in\\{0,1\\}$ for all $1\leq k\leq 2|\gamma_{F}|$. Indeed, the
vector $|i_{1},i_{2},\cdots,i_{2|\gamma_{F}|}\rangle$ refers to the
cylindrical function
$\Psi_{\vec{i}}=\rho_{i_{1}}^{\dagger}\rho_{i_{2}}^{\dagger}\cdots\rho_{i_{n}}^{\dagger}$
with respect to $\gamma_{F}$. It is worth noting that the definition of
$\mathcal{H}_{F,\gamma_{F}}$ depends on the orientation of $\gamma_{F}$, since
$\zeta_{A}^{{\dagger}}(v_{F})$ are Grassmann numbers. This fact can be
illustrated more explicitly with the following examples. Consider another
orientation $n\to\tilde{v}_{F}^{(n)}$ of $\gamma_{F}$ such that

$\tilde{v}_{F}^{(1)}=v_{F}^{(2)},\tilde{v}_{F}^{(2)}=v_{F}^{(1)},\tilde{v}_{F}^{(k)}=v_{F}^{(k)},\
\forall k\geq 3.$ (40)

Then, under the new orientation, we have the tensor-product Hilbert space
$\tilde{\mathcal{H}}_{F,\gamma_{F}}$ possessing the basis

$\widetilde{|i_{1},i_{2},\cdots,i_{2|\gamma_{F}|}\rangle}:=|i_{1},i_{2}\rangle_{\tilde{v}_{F}^{(1)}}\otimes|i_{3},i_{4}\rangle_{\tilde{v}_{F}^{(2)}}\otimes\cdots\otimes|i_{2|\gamma_{F}|-1},i_{2|\gamma_{F}|}\rangle_{\tilde{v}_{F}^{(\gamma_{F})}}.$
(41)

By definition, $\widetilde{|i_{1},i_{2},\cdots,i_{2|\gamma_{F}|}\rangle}$
referes to the cylindrical funciton

$\tilde{\Psi}_{\vec{i}}=\rho_{i_{3}}^{\dagger}\rho_{i_{4}}^{\dagger}\rho_{i_{1}}^{\dagger}\rho_{i_{2}}^{\dagger}\cdots\rho_{i_{n}}^{\dagger}=(-1)^{(i_{1}+i_{2})(i_{3}+i_{4})}\Psi_{\vec{i}^{\prime}},$
(42)

where
$\vec{i}^{\prime}=\\{i_{3},i_{4},i_{1},i_{2},\cdots,i_{2|\gamma_{F}|}\\}$. Eq.
(42) gives the equivalence relation between $\mathcal{H}_{F,\gamma_{F}}$ and
$\tilde{\mathcal{H}}_{F,\gamma_{F}}$

$\widetilde{|i_{1},i_{2},\cdots,i_{2|\gamma_{F}|}\rangle}=(-1)^{(i_{1}+i_{2})(i_{3}+i_{4})}|i_{3},i_{4},i_{1},i_{2},\cdots,i_{2|\gamma_{F}|}\rangle.$
(43)

This equivalence relation can be analogously defined between the tensor
product spaces with different orientations. The fermionic Hilbert space with
respect to $\gamma_{F}$ is actually the space of equivalence classes
associated with this equivalence relation.

Indeed, the extra sign in (43) can be systematically obtained by introducing
the notion of graded objects. One can refer to Appendix B and the reference
therein for more details on the this notion. In our work, the Hilbert spaces
$\mathcal{H}_{v_{F}}$ are graded. The degree ${\mathfrak{d}}(i_{1},i_{2})$ of
each $|i_{1},i_{2}\rangle_{v_{F}}$ is

${\mathfrak{d}}(i_{1},i_{2})=i_{1}+i_{2}\mod 2.$ (44)

The operator algebra on $\mathcal{H}_{v_{F}}$ is also graded. By definition,
the degrees of the operators $\hat{\zeta}_{v_{F},A}$ and
$\hat{\zeta}_{v_{F},A}^{\dagger}$ are

${\mathfrak{d}}(\hat{\zeta}_{v_{F},A})=1={\mathfrak{d}}(\hat{\zeta}_{v_{F},A}^{\dagger}).$
(45)

A general principle to deal with these graded objects is that, whenever we
swap two items, an additional sign appears by the rule
$xy=(-1)^{{\mathfrak{d}}(x){\mathfrak{d}}(y)}yx$. Following this rule, we can
obtain the identity (43) manifestly.

Given a vertex $v_{F}$, let $\mathcal{H}_{v_{F}}^{\rm irr}$ be the subspace of
$\mathcal{H}_{v_{F}}$ spanned by $|0,1\rangle_{v_{F}}$, $|1,0\rangle_{v_{F}}$
and $|1,1\rangle_{v_{F}}$. Then the Hilbert space
$\mathcal{H}_{F,\gamma_{F}}^{\rm irr}$ with respect to $\gamma_{F}$ is defined
as

$\mathcal{H}_{F,\gamma_{F}}^{\rm
irr}=\bigotimes_{v_{F}\in\gamma_{F}}\mathcal{H}_{v_{F}}^{\rm irr},$ (46)

such that ${\mathbf{Cyl}}_{F}$ can be decomposed as

${\mathbf{Cyl}}_{F}=\bigoplus_{\gamma_{F}}\mathcal{H}_{F,\gamma_{F}}^{\rm
irr}\oplus\mathbb{C}.$ (47)

The kinematical Hilbert space $\mathcal{H}$ of the entire system is the tensor
product of $\mathcal{H}_{G}$ and $\mathcal{H}_{F}$, i.e.,

$\mathcal{H}=\overline{\mathcal{H}_{G}\otimes\mathcal{H}_{F}}.$ (48)

A densely subspace ${\mathbf{Cyl_{tot}}}$ of $\mathcal{H}$ is

${\mathbf{Cyl_{tot}}}={\mathbf{Cyl}}\otimes{\mathbf{Cyl_{F}}}.$ (49)

The states in ${\mathbf{Cyl_{tot}}}$ will be called the cylindrical states. To
obtain a cylindrical state, one needs a graph
$\gamma=\gamma_{G}\cup\gamma_{F}$, where the gravitational graph $\gamma_{G}$
is constituted of edges and their ending points as vertices, and the fermionic
graph $\gamma_{F}$ contains only vertices . To define a state with respect to
$\gamma$, besides the data for a (gauge variant) LQG spin network state, one
also needs to assign to each fermionic vertex $v_{F}$ a state
$|i_{1}(v_{F}),i_{2}(v_{F})\rangle_{v_{F}}$ with
$i_{1}(v_{F}),i_{2}(v_{F})\in\\{0,1\\}$. In principle, a fermionic vertex
$v_{F}$ can be located anywhere, regardless of the given gravitational graph
$\gamma_{G}$. However, if $v_{F}$ is chosen as point in $e\in E(\gamma_{G})$
but $v_{F}\notin V(\gamma_{G})$, we can always split $e$ at $v_{F}$ to define
a new graph $\tilde{\gamma}_{G}$. Then $v_{F}$ becomes a vertex of
$\tilde{\gamma}_{G}$. Moreover, because of
$\gamma_{G}\subset\tilde{\gamma}_{G}$, every cylindrical function with respect
to $\gamma_{G}$ can be rewritten by using $\tilde{\gamma}_{G}$. Thus, it is
sufficient to consider those graphs $\gamma=\gamma_{G}\cup\gamma_{F}$ where
each fermionic vertex $v_{F}$ satisfies either $v_{F}\in V(\gamma_{G})$ or
$v_{F}\notin\gamma_{G}$. Then, in $V(\gamma_{G})$, there could be bivalent
vertices which is a fermionic vertex. These vertices serve to split edge.
Thus, by (15), they are just spurious vertices. However, in contrast to (15),
we can put trivial SU(2) representation at these vertices to decompose the
Hilbert space with respect to graphs. Moreover precisely, let $\Gamma_{o}$ be
the set of graphs $\gamma=\gamma_{G}\cup\gamma_{F}$ where each fermionic
vertex $v_{F}$ satisfies either $v_{F}\in V(\gamma_{G})$ or
$v_{F}\notin\gamma_{G}$. Consider the assignments $\vec{j}^{\prime}$ to
$E(\gamma)$ such that each representation is nontrivial. Besides, let
$\vec{l}^{\prime}$ denote assignments of irreducible representations to
vertices of $\gamma$ which are non-trivial at each fake vertex of $\gamma_{G}$
where a vertex $v\in V(\gamma_{G})$ is fake if $v\notin\gamma_{F}$ and it is
spurious as defined in (15). Then, we define

$\mathcal{H}_{G,\gamma_{G}}^{\rm
irr}:={\bigoplus_{\vec{j}^{\prime},\vec{l}^{\prime}}}\mathcal{H}_{G,\gamma_{G}}^{(\vec{j}^{\prime},\vec{l}^{\prime})},$
(50)

which gives us the Hilbert space $\mathcal{H}_{\gamma}^{\rm irr}$ as

$\mathcal{H}_{\gamma}^{\rm irr}=\mathcal{H}_{G,\gamma_{G}}^{\rm
irr}\otimes\mathcal{H}_{F,\gamma_{F}}^{\rm irr}.$ (51)

Then we have the decomposition

${\mathbf{Cyl_{tot}}}=\bigoplus_{\gamma\in\Gamma_{o}}\mathcal{H}_{\gamma}^{\rm
irr}\oplus\mathbb{C}.$ (52)

##  IV The constraint operators for gravity coupled to fermion field

###  IV.1 the Gauss constraint

Classically, the Gauss constraint $G[\lambda]$ reads

$G[\lambda]=\int_{\Sigma}\differential^{3}x\lambda^{m}\left(\frac{1}{\kappa\beta}D_{a}E^{a}_{m}+\frac{1}{2}\xi^{\dagger}\sigma_{m}\xi\right).$
(53)

It is straightfoward to quantize it as the operator

$\widehat{G[\lambda]}=\sum_{v}\lambda^{m}(v)\hat{G}_{v,m}$ (54)

with

$\hat{G}_{v,m}=\hbar\sum_{[e]}\hat{J}_{m}^{v,[e]}+\hbar\,\hat{\zeta}_{v,A}^{\dagger}\frac{(\sigma_{m})_{AB}}{2}\hat{\zeta}_{v,B}$
(55)

Let us use $\hat{\mathcal{J}}_{v,m}$ to denote the second term in (55), namely

$\hat{\mathcal{J}}_{v,m}=\hat{\zeta}_{v,A}^{\dagger}\frac{(\sigma_{m})_{AB}}{2}\hat{\zeta}_{v,B}.$
(56)

On the fermionic Hilbert space $\mathcal{H}_{v}$ at vertex $v$, the action of
$\hat{\mathcal{J}}_{v,m}$ reads

$\displaystyle\hat{\mathcal{J}}_{v,m}|0,0\rangle_{v}$ $\displaystyle=0,\
\hat{\mathcal{J}}_{v,m}|1,1\rangle_{v}=0$ (57)
$\displaystyle\hat{\mathcal{J}}_{v,m}\left(|1,0\rangle_{v},|0,1\rangle_{v}\right)$
$\displaystyle=\left(|1,0\rangle_{v},|0,1\rangle_{v}\right)\frac{\sigma_{m}}{2}.$

According to (57), the operators $\hat{\mathcal{J}}_{v,m}$ for all $m=1,2,3$
behave as the angular moment operators. Thus, the operator
$\hat{\mathcal{J}}_{v,m}$ generates an SU(2) action on $\mathcal{H}_{v}$ as

$u\triangleright|\phi\rangle_{v}=\begin{pmatrix}|1,0\rangle_{v},|0,1\rangle_{v}\end{pmatrix}u\begin{pmatrix}\phi_{10}\\\
\phi_{01}\end{pmatrix}+\phi_{00}|0,0\rangle_{v}+\phi_{11}|1,1\rangle_{v}.$
(58)

where $|\phi\rangle_{v}=\sum_{ij}\phi_{ij}|i,j\rangle_{v}$ and
$u\in{\mathrm{SU(2)}}$. Therefore, $\mathcal{H}_{v}$ becomes a reducible
representation space of SU(2). The 1-dimensional space spanned by either
$|0,0\rangle_{v}$ or $|1,1\rangle_{v}$ is the trivial representation space,
and the 2-dimensional space spanned by $|0,1\rangle_{v}$ and $|1,0\rangle_{v}$
is the $1/2$-representation space where $|0,1\rangle_{v}$ and
$|1,0\rangle_{v}$ serve as the standard basis according to (57). This fact
leads to the decomposition

$\mathcal{H}_{v}\equiv\mathcal{H}_{0}\oplus\mathcal{H}_{0}\oplus\mathcal{H}_{1/2},$
(59)

where $\mathcal{H}_{j}$ denote the $j$-representation space of SU(2).

For a graph $\gamma=\gamma_{G}\cup\gamma_{F}$, a spin $j_{e}$ is assigned to
the edge $e\subset\gamma_{G}$. Then at each vertex $v\in V(\gamma)$ there is
the Hilbert space

$\mathcal{H}_{v}^{\rm tot}=\bigotimes_{e\text{ starts from
}v}\mathcal{H}_{j_{e}}\otimes\bigotimes_{e^{\prime}\text{ targets
}v}\mathcal{H}_{j_{e^{\prime}}}^{*}\otimes\mathcal{H}_{v}$ (60)

where $\mathcal{H}_{j}^{*}$ denotes the dual space of $\mathcal{H}_{j}$. On
$\mathcal{H}_{v}^{\rm tot}$, the infinitesimal SU(2) action gives the Gauss
constraint. Thus, the solution space to the Gauss constraint is

$\mathcal{H}^{\rm Gau}=\bigotimes_{x\in V(\gamma)}{\rm
Inv}\left(\mathcal{H}_{v}^{\rm tot}\right),$ (61)

where ${\rm Inv}\left(\mathcal{H}_{v}^{\rm
tot}\right)\subset\mathcal{H}_{v}^{\rm tot}$ is the SU(2)-invariant subspace.
To see ${\rm Inv}(\mathcal{H}_{v}^{\rm tot})$ more precisely, let us assume
all edges at $v$ are outgoing without loss of generality. Then we have

$\mathcal{H}_{v}^{\rm tot}=\bigotimes_{e\text{ at
}v}\mathcal{H}_{j_{e}}\otimes\mathcal{H}_{v}.$ (62)

Given an order of the edges at $v$, one can choose an orthonormal basis of
$\bigotimes\limits_{e\text{ at }v}\mathcal{H}_{j_{e}}$ composed of vectors
$|k_{2},k_{3},\cdots,k_{n},M\rangle$ satisfying

$\displaystyle\sum_{i=1}^{3}(L_{i}^{(l)})^{2}|k_{2},k_{3},\cdots,k_{n},M\rangle=$
$\displaystyle k_{l}(k_{l}+1)|k_{2},k_{3},\cdots,k_{n},M\rangle,\ \forall
l=2,\cdots,n$ (63) $\displaystyle
L_{3}^{(n)}|k_{2},k_{3},\cdots,k_{n},M\rangle=$ $\displaystyle
M|k_{2},k_{3},\cdots,k_{n},M\rangle,$ $\displaystyle
L_{1}^{(n)}|k_{2},k_{3},\cdots,k_{n},M\rangle=$ $\displaystyle\sum_{s=\pm
1}\frac{1}{2}\sqrt{(k_{n}-sM)(k_{n}+sM+1)}|k_{2},k_{3},\cdots,k_{n},M+s\rangle$
$\displaystyle L_{2}^{(n)}|k_{2},k_{3},\cdots,k_{n},M\rangle=$
$\displaystyle\sum_{s=\pm
1}\frac{-is}{2}\sqrt{(k_{n}-sM)(k_{n}+sM+1)}|k_{2},k_{3},\cdots,k_{n},M+s\rangle$

with $\hat{L}_{i}^{(l)}:=\sum_{k=1}^{l}\hat{J}_{i}^{v,e_{k}}.$ Let us define
$\mathrm{Inv}\left(\mathcal{H}_{G}^{(v)}\right)\subset\bigotimes\limits_{e\text{
at }v}\mathcal{H}_{j_{e}}$ as the subspace spanned by
$|k_{2},k_{3},\cdots,k_{n-1},0,0\rangle$ for all possible
$k_{2},k_{3},\cdots,k_{n-1}$. One has

$L_{i}^{(n)}|k_{2},k_{3},\cdots,k_{n-1},0,0\rangle=0,\ \forall i=1,2,3.$ (64)

Moreover, with the vectors $|k_{2},k_{3},\cdots,k_{n-1},1/2,M\rangle$, we
define

$|k_{2},k_{3},\cdots,k_{n-1}\rangle_{\rm
tot}={\frac{1}{\sqrt{2}}}\left|k_{2},k_{3},\cdots,k_{n-1},\frac{1}{2},\frac{1}{2}\right\rangle\otimes|0,1\rangle_{v}-{\frac{1}{\sqrt{2}}}\left|k_{2},k_{3},\cdots,k_{n-1},\frac{1}{2},-\frac{1}{2}\right\rangle\otimes|1,0\rangle_{v}.$
(65)

Then one has

$(L_{i}^{(n)}+\hat{\mathcal{J}}_{v,i})|k_{2},k_{3},\cdots,k_{n-1}\rangle_{\rm
tot}=0,\ \forall i=1,2,3.$ (66)

Let $\mathcal{H}_{\rm inv}\subset\mathcal{H}_{v}^{\rm tot}$ denote the
subspace spanned by $|k_{2},k_{3},\cdots,k_{n-1}\rangle_{\rm tot}$ for all
possible $k_{2},k_{3},\cdots,k_{n-1}$. Then ${\rm Inv}(\mathcal{H}_{v}^{\rm
tot})$ can be decomposed as

${\rm Inv}(\mathcal{H}_{v}^{\rm
tot})=\left(\mathrm{Inv}\left(\mathcal{H}_{G}^{(v)}\right)\otimes|0,0\rangle_{v}\right)\oplus\left(\mathrm{Inv}\left(\mathcal{H}_{G}^{(v)}\right)\otimes|1,1\rangle_{v}\right)\oplus\mathcal{H}_{\rm
inv},$ (67)

where
$\mathrm{Inv}\left(\mathcal{H}_{G}^{(v)}\right)\otimes|i_{1},i_{2}\rangle_{v}$
is the space composed of vectors $|\psi\rangle\otimes|i_{1},i_{2}\rangle$ for
all $|\psi\rangle\in\mathrm{Inv}\left(\mathcal{H}_{G}^{(v)}\right)$.

Let $v$ be a $n$-valence gauge invariant fermionic vertex, where the $i$th
edge $e_{i}$ is assigned to a spin $j_{i}$. According to the decomposition
(67), the gauge invariant Hilbert space ${\rm Inv}(\mathcal{H}_{v}^{\rm tot})$
contains the subspace $\mathcal{H}_{\rm inv}$, isometric to the gauge
invariant Hilbert space of a $(n+1)$-valence pure-gravity vertex where the
$i$th edge for $1\leq i\leq n$ is assigned to spin $j_{i}$, and the $(n+1)$th,
the spin $1/2$. Then once we consider the volume operator at $v$, this extra
spin $1/2$ will also have contribution. Since the extra spin $1/2$ originates
from the fermion filed, one gets an intuitive picture that fermion field
contributes to the volume of a vertex. Moreover, a $n$-valence vertex in pure-
LQG is always regarded as a polyhedron whose faces are dual to the edges. The
flux operators $\hat{J}^{v,e}_{i}$ associated to each edge $e$ have the
geometric interpretation of the area vector of the dual face. Then the pure-
LQG Gauss constraint is just the closure condition
$\sum_{e}\hat{J}^{v,e}_{i}=0$ ensuring that the faces can form a closed
polyhedron. Now, the fermion field is involved. Then the Gauss constraint (55)
implies

$\sum_{e}\hat{J}^{v,e}_{i}=-\hat{\mathcal{J}}_{v,i}$ (68)

where the right hand side does not vanish in general. Thus, the faces dual to
the edges could not give a closed polyhedron for states in $\mathcal{H}_{\rm
inv}$. By (68), the area defect of this unclosed polyhedron is filled by
$\hat{\mathcal{J}}_{v,i}$, i.e. the fermion spin at the vertex (see
mansuroglu2021fermion  for more details on the fermion spin). A direct
consequence of the above discussion is that the volume of a 3-valence vertex
with fermion does not vanish any more for states in $\mathcal{H}_{\rm inv}$.
Let $j_{i}$ with $i=1,2,3$ be the spins on the edges. Then the states in
$\mathcal{H}_{\rm inv}$ are spanned by $|k\rangle_{\rm tot}\equiv|k\rangle$
with $k=j_{3}\pm 1/2$. The action of the operator $\hat{q}_{123}$, the
operator proportional to the square of the volume operator ashtekar2004back ,
on $|k\rangle$ reads

$\displaystyle\langle k|\hat{q}_{123}|k+1\rangle=$
$\displaystyle\frac{-i}{4\sqrt{(2k+1)(2k+3)}}\sqrt{(j_{1}-j_{2}+k+1)(-j_{1}+j_{2}+k+1)(j_{1}+j_{2}-k)(j_{1}+j_{2}+k+2)}$
(69)
$\displaystyle\sqrt{(j_{3}-\frac{1}{2}+k+1)(-j_{3}+\frac{1}{2}+k+1)(j_{3}+\frac{1}{2}-k)(j_{3}+\frac{1}{2}+k+2)}.$

Then we have

$\displaystyle\left\langle
j_{3}-\frac{1}{2}\middle|\hat{q}_{123}\middle|j_{3}+\frac{1}{2}\right\rangle=$
$\displaystyle
i\frac{1}{16}\sqrt{(2j_{1}+2j_{2}-2j_{3}+1)(2j_{1}-2j_{2}+2j_{3}+1)}$ (70)
$\displaystyle\sqrt{(-2j_{1}+2j_{2}+2j_{3}+1)(2j_{1}+2j_{2}+2j_{3}+3)}.$

Since the associated Hilbert space is 2-dimensional, the whole Hilbert space
is the eigenspace of the volume operator
$\frac{\kappa_{0}{\beta}\ell_{p}^{3/2}}{{2\sqrt{2}}}\sqrt{|\hat{q}_{123}|}$
with eigenvalue

$V_{v}=\frac{\kappa_{0}{\beta}\ell_{p}^{3/2}}{{2\sqrt{2}}}\sqrt{\left|\left\langle
j_{3}-\frac{1}{2}\middle|\hat{q}_{123}\middle|j_{3}+\frac{1}{2}\right\rangle\right|}.$
(71)

###  IV.2 The Hamiltonian constraint

As discussed in thiemann1998quantumV ; thiemann1998kinematical , the smeared
Hamiltonian constraint in terms of $\theta_{A}(x)$ is

$H_{F}[N]:=\sum_{x\in\Sigma}N(x)H_{F}(x)$ (72)

where $H_{F}(x)$ is given by

$\displaystyle H_{F}=$ $\displaystyle
i\frac{1}{2\sqrt{q}}(\theta^{\dagger}E_{i}^{a}\sigma^{i}D_{a}\theta-(D_{a}\theta)^{\dagger}E_{i}^{a}\sigma^{i}\theta)-\beta\frac{1}{2\sqrt{q}}E^{a}_{i}K_{a}^{i}\theta^{\dagger}\theta-\frac{1+\beta^{2}}{\beta}\frac{1}{2\sqrt{q}}D_{a}E^{a}_{i}\theta^{\dagger}\sigma^{i}\theta-\beta\frac{1}{2\sqrt{q}}E^{a}_{i}D_{a}\Big{(}\theta^{\dagger}\sigma^{i}\theta\Big{)}.$
(73)

Fix a coordinate system $x^{a}$ on $\Sigma$ and a positive number $\epsilon$.
Divide $\Sigma$ into a family $\mathcal{C}_{\epsilon}$ of cells such that each
cell $C\in\mathcal{C}_{\epsilon}$ is cubic with the coordinate volume less
than $\epsilon^{3}$, and that different cells can only share points on their
boundaries. Given a graph $\gamma=\gamma_{G}\cup\gamma_{F}$, for each cell
$C\in\mathcal{C}_{\epsilon}$, let $\gamma_{C}$ denote $\gamma\cap C$. Since
the limit $\epsilon\to 0$ will be considered eventually, we will assume that
$\epsilon$ is small enough such that $\gamma_{C}\neq\emptyset$ is one of the
following types (see Fig. 1):

  * (i)

the type-A graph: $\gamma_{C}$ is composed of a single edge;

  * (ii)

the type-B graph: $\gamma_{C}$ is composed of a single fermionic vertex
without connecting any edges;

  * (iii)

the type-C graph: $\gamma_{C}$ is composed of edges intersecting a single
vertex.

Figure 1: Different types of graphs where in the lower left panel is a type-C
graph with a fermionic vertex and, the lower right panel, a type-C graph with
a gravitational vertex.

For each cell $C\in\mathcal{C}_{\epsilon}$, let us define

$\displaystyle H_{C}^{(1)}=$
$\displaystyle\int_{C}\differential^{3}xN(x)(D_{a}\theta)^{\dagger}(x)E^{a}_{i}(x)\sigma^{i}\theta(x),$
(74) $\displaystyle H_{C}^{(2)}=$
$\displaystyle\int_{C}\differential^{3}xN(x)E^{a}_{i}(x)K_{a}^{i}(x)\theta^{\dagger}(x)\theta(x),$
$\displaystyle H_{C}^{(3)}=$
$\displaystyle\int_{C}\differential^{3}xN(x)D_{a}E^{a}_{i}(x)\theta^{\dagger}(x)\sigma^{i}\theta(x),$

and introduce

$H_{\mathcal{C}_{\epsilon}}=\sum_{C\in\mathcal{C}_{\epsilon}}\frac{1}{V_{C}}\left(-iH_{C}^{(1)}-\frac{\beta}{2}H_{C}^{(2)}-\frac{1+\beta^{2}}{2\beta}H_{C}^{(3)}-\beta
H_{C}^{(1)}\right),$ (75)

where the volume $V_{C}$ of $C$ is
$V_{C}=\int_{C}\differential^{3}x\sqrt{|\det(E)|}.$ Then $H_{F}[N]$ is the
limit of $\frac{1}{2}H_{\mathcal{C}_{\epsilon}}$ plussing its complex
conjugate as $\epsilon\to 0$, i.e.

$H_{F}[N]=\frac{1}{2}\lim_{\epsilon\to 0}(H_{\mathcal{C}_{\epsilon}}+c.c.).$
(76)

To quantize $H_{\mathcal{C}_{\epsilon}}$, we need to promote

$\tilde{H}_{C}^{(i)}=\frac{1}{V_{C}}H_{C}^{(i)},\ i=1,2,3,\
C\in\mathcal{C}_{\epsilon},$ (77)

to an operator. As known in LQG, the volume operator is not invertible. We
thus need to regularize $V_{C}^{-1}$ as in yang2016new  to be
$V_{C}^{-1}=\left(\sqrt{V_{C}^{-1}}\right)^{2}$ with

$\displaystyle\sqrt{V_{C}^{-1}}=\kappa_{0}^{\prime}\frac{4\times
8}{6}\left(\frac{2}{\kappa\beta}\right)^{3}\sum_{e,e^{\prime},e^{\prime\prime}\text{
at
}v}\epsilon(e,e^{\prime},e^{\prime\prime})\tr\left(h_{e}^{-1}\\{h_{e},V_{C}\\}h_{e^{\prime}}^{-1}\\{h_{e^{\prime}},V_{C}\\}h_{e^{\prime\prime}}^{-1}\\{h_{e^{\prime\prime}},V_{C}\\}\right).$
(78)

where $\epsilon(e,e^{\prime},e^{\prime\prime})=0,\pm 1$ depending on the
orientation of $e\wedge e^{\prime}\wedge e^{\prime\prime}$, and
$\kappa_{0}^{\prime}$ is a constant to remove the dependence of partition.
Then, the operator $\widehat{\tilde{H}_{C}^{(i)}}$ is given by

$\widehat{\tilde{H}_{C}^{(i)}}=\widehat{\sqrt{V_{C}^{-1}}}\hat{H}_{C}^{(i)}\widehat{\sqrt{V_{C}^{-1}}},$
(79)

where $\hat{H}_{C}^{(i)}$ will be discussed latter, and the operator
$\widehat{\sqrt{V_{C}^{-1}}}$ reads

$\displaystyle\widehat{\sqrt{V_{C}^{-1}}}=\kappa_{0}^{\prime}\frac{4\times
8}{6}\left(\frac{2}{i\hbar\kappa\beta}\right)^{3}\sum_{v\text{ in
}C}\sum_{e,e^{\prime},e^{\prime\prime}\text{ at
}v}\epsilon(e,e^{\prime},e^{\prime\prime})\times$ (80)
$\displaystyle{\tr}\left(h_{e}^{-1}[h_{e},\hat{V}_{C}]h_{e^{\prime}}^{-1}[h_{e^{\prime}},\hat{V}_{C}]h_{e^{\prime\prime}}^{-1}[h_{e^{\prime\prime}},\hat{V}_{C}]\right).$

Here there are several issues on $\widehat{\sqrt{V_{C}^{-1}}}$. At first,
$\widehat{\sqrt{V_{C}^{-1}}}$ actually acts on states $\psi_{\gamma_{C}}$ with
respect to the graph $\gamma_{C}=\gamma\cap C$. Given a state
$\psi_{\gamma_{C}}$, according to (52), there is a unique graph
$\mathbf{P}(\gamma_{C})\subset\gamma_{C}$ such that
$\psi_{\gamma_{C}}\in\mathcal{H}_{\mathbf{P}(\gamma_{C})}^{\rm irr}$. The
summation in (80) over $v\in C$ actually means the summation over $v\in
V(\mathbf{P}(\gamma_{C}))$, and the summation over
$e,e^{\prime},e^{\prime\prime}$ means the summation over
$e,e^{\prime},e^{\prime\prime}\in E(\mathbf{P}(\gamma_{C}))$. As a
consequence, for $\gamma_{C}$ being a type-A or type-B graph, one has
$\widehat{\sqrt{V_{C}^{-1}}}\psi_{\gamma_{C}}=0$, and for $\gamma_{C}$ being a
type-C graph with the vertex $v_{C}$, one has

$\displaystyle\widehat{\sqrt{V_{C}^{-1}}}\psi_{\gamma_{C}}=\kappa_{0}^{\prime}\frac{4\times
8}{6}\left(\frac{2}{i\hbar\kappa\beta}\right)^{3}\sum_{e,e^{\prime},e^{\prime\prime}\in
E(\mathbf{P}(\gamma_{C}))}\epsilon(e,e^{\prime},e^{\prime\prime})\tr\left(h_{e}^{-1}[h_{e},\hat{V}_{v_{C}}]h_{e^{\prime}}^{-1}[h_{e^{\prime}},\hat{V}_{v_{C}}]h_{e^{\prime\prime}}^{-1}[h_{e^{\prime\prime}},\hat{V}_{v_{C}}]\right)\psi_{\gamma_{C}},$
(81)

where $\hat{V}_{v_{C}}$ is the volume operator introduced in
ashtekar1997quantumII . According to this result, the operator
$\widehat{\sqrt{V_{C}^{-1}}}$ at the most right of
$\widehat{\tilde{H}_{C}^{(i)}}$ will annihilate the states with respect to the
type-A and type-B graphs. Hence, only the states on type-C graphs is needed to
be considered. From now on, $\gamma_{C}$ will be referred to as the type-C
graphs until otherwise stated. The vertex for the edges in $\gamma_{C}$
intersecting will be denoted by $v_{C}$.

Let us begin with the operator $\widehat{\tilde{H}_{C}^{(1)}}$. Replacing
$E_{i}^{a}(x)$ by $-i\kappa\hbar\beta\delta/\delta A_{a}^{i}(x)$ in
$H_{C}^{(1)}$ in (74), we can quantize $H_{C}^{(1)}$ as

$\hat{H}_{C}^{(1)}=-i\kappa\hbar\beta\int_{C}\differential^{3}xN(x)\widehat{(D_{a}\theta)}^{\dagger}(x)\sigma^{i}\hat{\theta}(x)\frac{\delta}{\delta
A_{a}^{i}(x)}.$ (82)

Note that we put the derivative $\frac{\delta}{\delta A_{a}^{i}(x)}$ at the
most right so that the resulting operator is cylindrical consistent, i.e., the
results of states acted by the resulting operator do not depend on the edges
taking trivial representation. Given an edge $e:[0,\delta]\to C$ of
$\gamma_{C}$ with $e(0)=v_{C}$, let $U_{e}(t,0,A)$ denote the parallel
transport from $e(0)$ to $e(t)$ along $e$. Thus, $U_{e}(t,0,A)$ satisfies

$\frac{\differential}{\differential
t}U_{e}(t,0;A)=-A_{a}(e(t))\dot{e}^{a}(t)U(t,0;A),\ \text{ and }\ U(0,0;A)=I.$
(83)

Defining $h_{e}:=U({\delta,0};A)$, one has, for any $f_{a}^{i}(x)$,

$\displaystyle\int\differential^{3}xf_{a}^{i}(x)\frac{\delta}{\delta
A_{a}^{i}(x)}h_{e}=-\int_{0}^{{\delta}}\differential\tau
f_{a}^{i}(e(\tau))\dot{e}^{a}(\tau)U_{e}({\delta},\tau,A)\tau_{i}U_{e}(\tau,0,A).$
(84)

Thus, we get

$\displaystyle-i\kappa\hbar\beta\int_{C}\differential^{3}x\,N(x)\widehat{D_{a}\theta}^{\dagger}(x)\sigma^{i}\hat{\theta}(x)\frac{\delta}{\delta
A_{a}^{i}(x)}h_{e}=\kappa\hbar\beta
N(s_{e})\left(\hat{\theta}^{\dagger}(t_{e})h_{e}-\hat{\theta}^{\dagger}(s_{e})\right)\sigma^{i}\hat{\theta}(s_{e})\hat{J}^{v,e}_{i}h_{e}$
(85)

where we used

$\delta\times\dot{e}^{a}(t_{1})(D_{a}\theta)^{\dagger}(e(t_{1}))=\theta^{\dagger}(e(t_{1}+\delta))U_{e}(t_{1}+\delta,t_{1};A)-\theta^{\dagger}(e(t_{1}))+O(\delta^{2}).$
(86)

Then, for a state $\Psi_{C}$ with respect to $\gamma_{C}$, one has

$\displaystyle\hat{H}_{C}^{(1)}\Psi_{C}=\kappa\hbar\beta
N(v_{C})\sum_{e\in\gamma_{C}}\left(\hat{\theta}^{\dagger}(t_{e})h_{e}-\hat{\theta}^{\dagger}(v_{C})\right)\sigma^{i}\hat{\theta}(v_{C})\hat{J}^{v_{C},e}_{i}\Psi_{C},$
(87)

which gives $\widehat{\tilde{H}^{(1)}_{C}}$ acting on $\Psi_{C}$ as

$\displaystyle\widehat{\tilde{H}^{(1)}_{C}}\Psi_{C}=\kappa\hbar\beta
N(v_{C})\sum_{e\in\gamma_{C}}\widehat{\sqrt{V_{C}^{-1}}}\left(\hat{\theta}^{\dagger}(t_{e})h_{e}-\hat{\theta}^{\dagger}(v_{C})\right)\sigma^{i}\hat{\theta}(v_{C})\hat{J}^{v_{C},e}_{i}\widehat{\sqrt{V_{C}^{-1}}}\Psi_{C}.$
(88)

For the second term $\widehat{\tilde{H}_{C}^{(2)}}$, one has

$\hat{H}_{C}^{(2)}=\int_{C}\differential^{3}xN(x)E^{a}_{i}(x)K_{a}^{i}(x)\theta^{\dagger}(x)\theta(x).$
(89)

Taking advantage of the Thiemann’s trick to quantize pure-gravity Hamiltonian
constraint in LQG thiemann2007modern , one has

$\int_{C}\differential^{3}xf(x)E^{a}_{i}(x)K_{a}^{i}(x)=\frac{1}{2\kappa\beta^{2}}\left\\{\int_{C}\differential^{3}xf(x)H_{E}(x),V_{C}\right\\},$
(90)

where $H_{E}(x)$ is the Euclidean part of the pure-gravity Hamiltonian
constraint, i.e.

$H_{E}(x)=\frac{\epsilon_{ijk}F_{ab}^{i}(x)E^{a}_{j}(x)E^{b}_{k}(x)}{\sqrt{q(x)}}.$

Thus, $H_{C}^{(2)}$ is quantized as

$\displaystyle\hat{H}_{C}^{(2)}=\frac{1}{2i\kappa\hbar\beta^{2}}\left[N(v_{C})\hat{H}_{E,v_{C}}\hat{\theta}^{\dagger}(v_{C})\hat{\theta}(v_{C}),\hat{V}_{v_{C}}\right]$
(91)

where $\hat{H}_{E,v_{C}}$ denotes the Euclidean part of the pure-gravity
Hamiltonian constraint operator at $v_{C}$, and $\hat{V}_{v_{C}}$, the volume
operator at $v_{C}$. Eq. (91) leads to the operator
$\widehat{\tilde{H}^{(2)}_{C}}$ given by

${\displaystyle\widehat{\tilde{H}^{(2)}_{C}}=\frac{1}{2i\kappa\hbar\beta^{2}}}N(v_{C})\sqrt{\widehat{V_{v_{C}}^{-1}}}\left(\hat{H}_{E,v_{C}}\hat{V}_{v_{C}}-\hat{V}_{v_{C}}\hat{H}_{E,v_{C}}\right)\hat{\theta}^{\dagger}(v_{C})\hat{\theta}(v_{C})\sqrt{\widehat{V_{v_{C}}^{-1}}}$
(92)

Finally, for the third term $\widehat{\tilde{H}_{C}^{(3)}}$, we have

$\displaystyle\hat{H}_{C}^{(3)}\Psi_{C}=\kappa\hbar\beta
N(v_{C})\left(\sum_{e\text{ at
}v_{C}}\hat{J}_{i}^{v_{C},e}\right)\hat{\theta}^{\dagger}(v_{C})\sigma^{i}\hat{\theta}(v_{C})\Psi_{C},$
(93)

by taking advantage of the Gaussian constraint operator of the pure gravity.
Thus $\widehat{\tilde{H}_{C}^{(3)}}$ reads

$\displaystyle\widehat{\tilde{H}_{C}^{(3)}}\Psi_{C}=\kappa\hbar\beta
N(v_{C})\widehat{\sqrt{V_{C}^{-1}}}\left(\sum_{e\text{ at
}v_{C}}\hat{J}_{i}^{v_{C},e}\right)\hat{\theta}^{\dagger}(v_{C})\sigma^{i}\hat{\theta}(v_{C})\widehat{\sqrt{V_{C}^{-1}}}\Psi_{C}$
(94)

With (88), (92) and (94), the operator $\widehat{H_{\mathcal{C}_{\epsilon}}}$
is

$\widehat{H_{\mathcal{C}_{\epsilon}}}=\sum_{C\in\mathcal{C}_{\epsilon}^{(3)}}-i\widehat{\tilde{H}_{C}^{(1)}}-\frac{\beta}{2}\widehat{\tilde{H}_{C}^{(2)}}-\frac{1+\beta^{2}}{2\beta}\widehat{\tilde{H}_{C}^{(3)}}-\beta\widehat{\tilde{H}_{C}^{(1)}}$
(95)

where

$\mathcal{C}_{\epsilon}^{(3)}=\\{C\in\mathcal{C}_{\epsilon},\gamma\cap C\text{
is type-C}\\}.$

Even though $\widehat{H_{\mathcal{C}_{\epsilon}}}$ in (95) is defined with a
partition structure $\mathcal{C}_{\epsilon}$ on $\Sigma$, this partition
structure is indeed not necessary. One can define an operator equivalent to
$\widehat{H_{\mathcal{C}_{\epsilon}}}$ with $\Sigma$ endowed with another
structure, which is more convenient for the further study. To this end, let us
first introduce the following notion.

######  Definition IV.1 (removable vertex).

A vertex $v$ of a graph $\gamma$ is removable if it satisfies the following
conditions.

  * (i)

$v$ is a bivalence vertex connecting $e_{1}$ and $e_{2}$;

  * (ii)

The composition of $e_{1}$ and $e_{2}$ as a curve is $C^{m}$ and semianalytic.

Given a graph $\gamma$, one can obtain another graph $\ker(\gamma)$ by
removing all of its removable vertices. $\ker(\gamma)$ will be called the
kernel of $\gamma$. Let $\Gamma_{\ker}$ be the collection of the kernels of
all graphs in $\Gamma_{o}$. Fix once and for all a parametrization for each
$\gamma\in\Gamma_{\ker}$, where a parametrization of a graph is an assignment
to each edge $e\in E(\gamma)$ a parametrization $[0,1]\ni t\to e(t)\in\Sigma$.
For an edge $e$ of a kernel in $\Gamma_{\ker}$ taking $v$ as an endpoint, we
can define $e(v,\delta)\subset e$ as the segment starting from $v$ and ending
at either $e(\delta)$ for $v=e(0)$ or $e(1-\delta)$ for $v=e(1)$. Given a
graph $\gamma=\gamma_{G}\cup\gamma_{F}$, for each $v\in V(\gamma_{G})$,
mimicking the operator $\widehat{\tilde{H}_{C}^{(i)}}$ in (88), (92) and (94),
we define

$\widehat{\tilde{H}^{(i)}(v;\vec{\delta}_{v})}=\widehat{\sqrt{V_{v}^{-1}}}\widehat{H^{(i)}(v;\vec{\delta}_{v})}\widehat{\sqrt{V_{v}^{-1}}},\
i=1,2,3$ (96)

with

$\displaystyle\widehat{H^{(1)}(v;\vec{\delta}_{v})}=$
$\displaystyle\kappa\hbar\beta N(v)\sum_{e\text{ at
}v}\left(\hat{\theta}^{\dagger}(t_{e(v,\delta_{v,e})})h_{e(v,\delta_{v,e})}\sigma^{i}\hat{\theta}(v)-\hat{\theta}^{\dagger}(v)\sigma^{i}\hat{\theta}(v)\right)\hat{J}^{v,e}_{i},$
(97) $\displaystyle\widehat{H^{(2)}(v;\vec{\delta}_{v})}=$
$\displaystyle\frac{1}{2i\kappa\hbar\beta^{2}}N(v)\left(\hat{H}_{E,v}\hat{V}_{v}-\hat{V}_{v}\hat{H}_{E,v}\right)\hat{\theta}^{\dagger}(v)\hat{\theta}(v),$
$\displaystyle\widehat{H^{(3)}(v;\vec{\delta}_{v})}=$
$\displaystyle\kappa\hbar\beta N(v)\left(\sum_{e\text{ at
}v}\hat{J}_{i}^{v,e}\right)\hat{\theta}^{\dagger}(v)\sigma^{i}\hat{\theta}(v).$

where $\vec{\delta}_{v}=\\{\delta_{v,e}\\}_{e\text{ at }v}$ is a vector of
real numbers for each vertex. Taking advantage of these operators, we define

$\displaystyle\widehat{H(\delta)}:=$ $\displaystyle\sum_{v\in
V(\gamma_{G})}-i\widehat{\tilde{H}^{(1)}(v;\vec{\delta}_{v})}-\frac{\beta}{2}\widehat{\tilde{H}^{(2)}(v;\vec{\delta}_{v})}-\frac{1+\beta^{2}}{2\beta}\widehat{\tilde{H}^{(3)}(v;\vec{\delta}_{v})}-\beta\widehat{\tilde{H}^{(1)}(v;\vec{\delta}_{v})}$
(98) $\displaystyle=$ $\displaystyle\sum_{v\in
V(\gamma)}-i\widehat{\tilde{H}^{(1)}(v;\vec{\delta}_{v})}-\frac{\beta}{2}\widehat{\tilde{H}^{(2)}(v;\vec{\delta}_{v})}-\frac{1+\beta^{2}}{2\beta}\widehat{\tilde{H}^{(3)}(v;\vec{\delta}_{v})}-\beta\widehat{\tilde{H}^{(1)}(v;\vec{\delta}_{v})}$

where $\delta\equiv\\{\vec{\delta}_{v}\\}_{v\in V(\gamma)}$ and the second
step is a consequence of the operator $\sqrt{\widehat{V_{v}^{-1}}}$ in
$\widehat{\tilde{H}^{(i)}(v;\vec{\delta}_{v})}$. Actually, due to the operator
$\sqrt{\widehat{V_{v}^{-1}}}$ in
$\widehat{\tilde{H}^{(i)}(v;\vec{\delta}_{v})}$,
$\widehat{\tilde{H}^{(i)}(v;\vec{\delta}_{v})}$ with $v\notin\gamma_{G}$
vanish for all $i=1,2,3$. Comparing the operators $\widehat{H(\delta)}$ and
$\widehat{H_{\mathcal{C}_{\epsilon}}}$, one can verify easily that
$\widehat{H(\delta)}=\widehat{H_{\mathcal{C}_{\epsilon}}}$ with a suitably
chosen $\delta$. Hence, we can use $\widehat{H(\delta)}$ instead of
$\widehat{H_{\mathcal{C}_{\epsilon}}}$ for our further study. In the rest of
the paper, we will assume that $\delta_{v,e}$ is a constant for all $v$ and
$e$ for convenience. The discussion for non-constant $\delta$ can be discussed
similarly.

When the limit of $\widehat{H(\delta)}$ as $\delta\to 0$ is taken, the only
nontrivial term in (97) one needs to consider is

$\hat{\mathfrak{h}}_{1}(v,e;\delta):=\hat{\theta}^{\dagger}(t_{e(v,\delta)})h_{e(v,\delta)}\sigma^{i}\hat{\theta}(v)\hat{J}_{i}^{v,e}.$

By definition, we have

$\displaystyle\hat{\mathfrak{h}}_{1}(v,e;\delta)\left(D^{j}_{mn}(h_{e(v,\delta)})\otimes|k_{1},k_{2}\rangle_{v}\right)$
(99) $\displaystyle=$
$\displaystyle-2W_{j}W_{\frac{1}{2}}\sum_{J=j\pm\frac{1}{2}}\sum_{A,C}(-1)^{m+A-n-C}\begin{bmatrix}\frac{1}{2}&\frac{1}{2}&1\\\
j&j&J\end{bmatrix}\begin{pmatrix}j&\frac{1}{2}&J\\\
m&A&-(m+A)\end{pmatrix}\begin{pmatrix}J&\frac{1}{2}&j\\\
-(n+C)&C&n\end{pmatrix}\times$ $\displaystyle
D^{J}_{m+A,n+C}(h_{e(v,\delta)})\otimes\hat{\theta}_{C}(v)|k_{1},k_{2}\rangle_{v}\otimes\hat{\theta}_{A}^{\dagger}(t_{e(v,\delta)})|0,0\rangle_{t_{e(v,\delta)}},$

where $W_{j}=\sqrt{j(j+1)(2j+1)}$, $\begin{bmatrix}J&j&\frac{1}{2}\\\
\frac{1}{2}&1&J\end{bmatrix}$ is the $6j$ symbol and
$\begin{pmatrix}j_{1}&j_{2}&j_{3}\\\ m_{1}&m_{2}&m_{3}\end{pmatrix}$ denotes
the $3j$ symbol. Eq. (99) gives us the picture that the operator
$\hat{\mathfrak{h}}_{1}(v,e;\delta)$ moves the fermionic vertex at $v$ to
$t_{e(v,\delta)}\in e$, and simultaneously change the spin on $e(v,\delta)$.
To illustrate this statement, let us use a disk
$\mathord{\vbox{\hbox{\raisebox{0.0pt}{\includegraphics[width=6.50403pt]{fermionvertex}}}}}$
to represent a fermionic vertex and a solid line to represent the edge $e$.
Then, by omitting the explicit coefficients in the right hand of (99), one has

  * (1)

for $k_{1}=1=k_{2}$,

$\displaystyle\hat{\mathfrak{h}}_{1}(v,e;\delta)\mathord{\vbox{\hbox{\raisebox{0.0pt}{\includegraphics[width=65.04034pt]{hamiltonianaction1}}}}}\to\sum_{(i_{1},i_{2})\in\\{(1,0),(0,1)\\}}\mathord{\vbox{\hbox{\raisebox{0.0pt}{\includegraphics[width=65.04034pt]{hamiltonianaction2}}}}}+\sum_{(i_{1},i_{2})\in\\{(1,0),(0,1)\\}}\mathord{\vbox{\hbox{\raisebox{0.0pt}{\includegraphics[width=65.04034pt]{hamiltonianaction3}}}}}$
(100)

  * (2)

for $k_{1}=1$ and $k_{2}=0$

$\displaystyle\hat{\mathfrak{h}}_{1}(v,e;\delta)\mathord{\vbox{\hbox{\raisebox{0.0pt}{\includegraphics[width=65.04034pt]{hamiltonianaction4}}}}}\to$
$\displaystyle\sum_{(i_{1},i_{2})\in\\{(1,0),(0,1)\\}}\mathord{\vbox{\hbox{\raisebox{0.0pt}{\includegraphics[width=65.04034pt]{hamiltonianaction5}}}}}$
(101)

  * (3)

for $k_{1}=0$ and $k_{2}=1$

$\displaystyle\hat{\mathfrak{h}}_{1}(v,e;\delta)\mathord{\vbox{\hbox{\raisebox{0.0pt}{\includegraphics[width=65.04034pt]{hamiltonianaction6}}}}}\to$
$\displaystyle\sum_{(i_{1},i_{2})\in\\{(1,0),(0,1)\\}}\mathord{\vbox{\hbox{\raisebox{0.0pt}{\includegraphics[width=65.04034pt]{hamiltonianaction5}}}}}$
(102)

According to (100), (101) and (102), one gets
$\langle\hat{\mathfrak{h}}_{1}(v,e;\delta)\psi|\hat{\mathfrak{h}}_{1}(v,e;\delta^{\prime})\psi\rangle=0$
for $\delta\neq\delta^{\prime}$, which implies that the limit of
$\hat{\mathfrak{h}}_{1}(v,e;\delta)$ as $\delta\to 0$ does not exist. As a
consequence, the limit of $\widehat{H(\delta)}$ as $\delta\to 0$ does not
exist, too. However, as final operator should be define as $\lim_{\delta\to
0}\widehat{H(\delta)}$, we need to introduce the vertex Hilbert space
$\mathcal{H}_{\rm vtx}$ which is in the dual space of ${\mathbf{Cyl_{tot}}}$.
Once $\mathcal{H}_{\rm vtx}$ is introduced, $\widehat{H(\delta)}$ can be
promoted to an operator $\widehat{H(\delta)}{}^{*}$ in $\mathcal{H}_{\rm vtx}$
by the duality such that
$\widehat{H(\delta)}{}^{*}=\widehat{H(\delta^{\prime})}{}^{*}$ for
$\delta\neq\delta^{\prime}$. Then, the limit of $\widehat{H(\delta)}{}^{*}$ as
$\delta\to 0$ exists. It will be seen below that $\mathcal{H}_{\rm vtx}$ is
defined to be the kinematical Hilbert space averaged by diffeomorphisms
preserving some particular vertices.

Another issue motivating us to introduce the vertex Hilbert space is that the
adjoint operator to $\widehat{H(\delta)}$ is not densely defined. To see this,
let us investigate the adjoint of $\widehat{H^{(1)}(v;\delta)}$, a term of
$\widehat{H(\delta)}$. By definition, for a state
$\Psi_{\gamma}\in\mathcal{H}_{\gamma}^{\rm irr}$ with $\gamma$ taking $v$ as a
vertex, the adjoint $\widehat{H^{(1)}(v;\delta)}{}^{\dagger}$ acts on it such
that for all $\Phi_{\gamma^{\prime}}\in\mathcal{H}$,

$\displaystyle\langle\widehat{H^{(1)}(v;\delta)}{}^{\dagger}\Psi_{\gamma}|\Phi_{\gamma^{\prime}}\rangle=\langle\Psi_{\gamma}|\widehat{H^{(1)}(v;\delta)}\Phi_{\gamma^{\prime}}\rangle$
(103) $\displaystyle=$ $\displaystyle\sum_{\begin{subarray}{c}e\in
E(\gamma^{\prime})\\\ e\text{ at }v\end{subarray}}\kappa\hbar\beta
N(v)\left\langle(\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}-\hat{\theta}(v)\sigma^{i}\hat{\theta}(v)\hat{J}_{i}^{v,e})\Psi_{\gamma}\middle|\Phi_{\gamma^{\prime}}\right\rangle$

with
$\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}=\hat{J}_{i}^{v,e}\hat{\theta}^{\dagger}(v)\sigma^{i}h_{e(v,\delta)}^{-1}\hat{\theta}(t_{e(v,\delta)})$.
The subtlety arises due to the summation over edges of $\gamma^{\prime}$
rather than $\gamma$. Let us consider a state $\Psi_{\gamma}$ with $\gamma$
shown in Fig. 2. Moreover, the fermion state of $\Psi_{\gamma}$ at the
fermionic vertex $t_{e(v,\delta)}$ is chosen to be
$a|0,1\rangle_{t_{e(v,\delta)}}+b|1,0\rangle_{t_{e(v,\delta)}}$ for some
$a,b\in\mathbb{C}$. Then, for each graph $\gamma^{\prime}$ obtained from
$\gamma$ by using an edge $e(v,\delta)$ to connect the vertex
$t_{e(v,\delta)}$ and $v$, dropping the fermionic vertex $t_{e(v,\delta)}$ and
adding a fermionic vertex at $v$ (see the two examples in Fig. 2), one can
find a state $\Phi_{\gamma^{\prime}}\in\mathcal{H}_{\gamma^{\prime}}^{\rm
irr}$ such that
$\left\langle(\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}-\hat{\theta}(v)\sigma^{i}\hat{\theta}(v)\hat{J}_{i}^{v,e})\Psi_{\gamma}\middle|\Phi_{\gamma^{\prime}}\right\rangle\neq
0$, where $e\in\gamma^{\prime}$ is the edge containing $e(v,\delta)$ as a
segment. Because there exist uncountably infinitely many such graphs
$\gamma^{\prime}$ and the states $\Phi_{\gamma^{\prime}}$ associated to
different graphs are orthogonal to each other, $\Psi_{\gamma}$ is not in the
domain of $\widehat{H^{(1)}(v;\delta)}{}^{\dagger}$. Similar argument can be
done for all of states in $\mathcal{H}_{\gamma}^{\rm irr}$ (including those
taking $|1,1\rangle_{t_{e(v,\delta)}}$ at $t_{e(v,\delta)}$), which implies
that the entire Hilbert space $\mathcal{H}_{\gamma}^{\rm irr}$ is not
contained in the domain of $\widehat{H^{(1)}(v;\delta)}{}^{\dagger}$. Besides,
it is manifest that there are infinitely many such Hilbert spaces that are
excluded from the domain of $\widehat{H^{(1)}(v;\delta)}{}^{\dagger}$.
Actually, according to (99), the operator $\hat{\mathfrak{h}}_{1}(v,e;\delta)$
will change the state taking spin $j=\frac{1}{2}$ on $e(v,\delta)$ to a
superposition of states taking spins $j=0$ and $1$ on $e(v,\delta)$. In other
words, the operator $\hat{\mathfrak{h}}_{1}(v,e;\delta)$ can erase the segment
$e(v,\delta)\subset e$, which is indeed the essential reason for the problem
of defining $\widehat{H^{(1)}(v;\delta)}{}^{\dagger}$. It will be seen from
blow that this problem can also be fixed by introducing the vertex Hilbert
space to define the limit. After taking the limit as $\delta\to 0$ in the
vertex Hilbert space, we will get a projection $\chi_{+}(|\hat{J}^{v,e}|^{2})$
(see (136)). This projection is left multiplied to the operator corresponding
to ${\hat{\mathfrak{h}}_{1}(v,e;\delta)}$ in $\mathcal{H}_{\rm vtx}$, and kill
the state with $j=0$ on $e(v,\delta)$.

Figure 2: An example of a graph $\gamma$ (top panel) such that
$\mathcal{H}_{\gamma}^{\rm irr}$ is not in the domain of
$\widehat{H^{(1)}(v;\delta)}{}^{\dagger}$, and examples of graphs
$\gamma^{\prime}$ (bottom panel) such that
$\langle\widehat{H^{(1)}(v;\delta)}{}^{\dagger}\Psi_{\gamma}|\Phi_{\gamma^{\prime}}\rangle\neq
0$. The black disks represent fermionic vertices.

Finally, let us discuss why we choose $H_{\mathcal{C}_{\epsilon}}$ rather than
its complex conjugate for quantization. Suppose that one choose the complex
conjugate of $H_{\mathcal{C}_{\epsilon}}$ to do the above quantization. Then,
in the resulting operator, one will get a term involving
$\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}=\hat{J}_{i}^{v,e}\hat{\theta}^{\dagger}(v)\sigma^{i}h_{e(v,\delta)}^{-1}\hat{\theta}(t_{e(v,\delta)})$.
In contrast to $\hat{\mathfrak{h}}_{1}(v,e;\delta)$,
$\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}$ moves the fermionic vertex at
$t_{e(v,\delta)}\in e$ to $v$. Now let us consider the states
$\hat{\mathfrak{h}}_{1}(v,e;\delta)|\psi\rangle$ and
$\hat{\mathfrak{h}}_{1}(v,e;\delta^{\prime})|\psi\rangle$ with
$\delta\neq\delta^{\prime}$ for $|\psi\rangle$, says, being
$D^{j}_{mn}(h_{e})\otimes|k_{1},k_{2}\rangle_{v}$. Then, one gets
$\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}(\hat{\mathfrak{h}}_{1}(v,e;\delta^{\prime})|\psi\rangle)=0$
but
$\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}(\hat{\mathfrak{h}}_{1}(v,e;\delta)|\psi\rangle)\neq
0$. This result means that $\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}$ is
not diffeomorphism covariant. Namely, diffeomorphism equivalent states, i.e.,
the states $\hat{\mathfrak{h}}_{1}(v,e;\delta)|\psi\rangle$ and
$\hat{\mathfrak{h}}_{1}(v,e;\delta^{\prime})|\psi\rangle$, could be no longer
diffeomorphism equivalent after acted by
$\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}$. This facts leads to a problem
that $\sum_{e\text{ at }v}\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}$ cannot
be promoted to a well-defined operator in the diffeomorphism invariant Hilbert
space.

####  IV.2.1 the vertex Hilbert space $\mathcal{H}_{\rm vtx}$

Given a graph $\gamma$, we will consider the subgroup of $C^{m}$ semianalytic
diffeomorphisms which act on $V(\ker(\gamma))$ trivially, i.e. the
diffeomorphisms preserving every vertex of $\ker(\gamma)$. This subgroup will
be denoted by ${\rm Diff}_{V(\ker(\gamma))}$. There are two subgroups of ${\rm
Diff}_{V(\ker(\gamma))}$. The first one, denoted by ${\rm Diff}_{\gamma}$,
preserves $\gamma$. The other one, denoted by ${\rm Diff}^{0}_{\gamma}$,
preserves every edge of $\gamma$. Hence ${\rm Diff}^{0}_{\gamma}$ is a
subgroup of ${\rm Diff}_{\gamma}$. The quotient

${\rm GS}_{\gamma}={\rm Diff}_{\gamma}/{\rm Diff}^{0}_{\gamma}$ (104)

is the group of graph symmetries of $\gamma$. ${\rm GS}_{\gamma}$ is a finite
group with order $|{\rm GS}_{\gamma}|$. Given a state $\Psi_{\gamma}$ with
respect to $\gamma$ and a diffeomorphism $\phi$, $\phi\star\Psi_{\gamma}$
denotes the pullback of $\Psi$ under $\phi$. The averaging with respect to
${\rm GS}_{\gamma}$ defines a projection $P_{\gamma}:\mathcal{H}_{\gamma}^{\rm
irr}\to\mathcal{H}_{\gamma}^{\rm irr}$ ,

$P_{\gamma}:\Psi_{\gamma}\mapsto\frac{1}{|{\rm GS}_{\gamma}|}\sum_{\phi\in{\rm
GS}_{\gamma}}\phi\star\Psi_{\gamma}.$ (105)

The averaging with respect to the remaining diffeomorphisms ${\rm
Diff}_{V(\ker(\gamma))}/{\rm Diff}_{\gamma}$ defines

$\eta(\Psi_{\gamma}):=\sum_{\phi\in{\rm Diff}_{V(\ker(\gamma))}/{\rm
Diff}_{\gamma}}\phi\star P_{\gamma}(\Psi_{\gamma})=\frac{1}{|{\rm
GS}_{\gamma}|}\sum_{\phi\in{\rm Diff}_{V(\ker(\gamma))}/{\rm
Diff}_{\gamma}^{0}}\phi\star\Psi_{\gamma}.$ (106)

Obviously, $\eta(\Psi_{\gamma})$ belongs to the algebraic dual space
${\mathbf{Cyl_{tot}}}^{*}$ of ${\mathbf{Cyl_{tot}}}$. Taking advantage the
decomposition (52), one can extend $\eta$ to a well-defined operation on
${\mathbf{Cyl_{tot}}}$. Let
$\eta({\mathbf{Cyl_{tot}}})\subset{\mathbf{Cyl_{tot}}}^{*}$ denote the image
of ${\mathbf{Cyl_{tot}}}$ under $\eta$. The group averaging procedure
naturally endows $\eta({\mathbf{Cyl_{tot}}})$ with an inner product

$\left(\eta(\Psi_{\gamma}),\eta(\Psi_{\gamma^{\prime}}^{\prime})\right)=(\eta(\Psi_{\gamma})|\Psi_{\gamma^{\prime}}^{\prime}\rangle,$
(107)

where $(\eta(\Psi_{\gamma})|\Psi_{\gamma^{\prime}}^{\prime}\rangle$, the
action of $\eta(\Psi_{\gamma})\in{\mathbf{Cyl_{tot}}}^{*}$ on
$\Psi_{\gamma^{\prime}}$, is given by

$(\eta(\Psi_{\gamma})|\Psi_{\gamma^{\prime}}^{\prime}\rangle=\sum_{\phi\in{\rm
Diff}_{V(\ker(\gamma))}/{\rm
Diff}_{\gamma}^{0}}\langle\phi\star\Psi_{\gamma},\Psi_{\gamma^{\prime}}^{\prime}\rangle.$
(108)

Here, $\langle\cdot,\cdot\rangle$ is the inner product in $\mathcal{H}$. Then
the vertex Hilbert space $\mathcal{H}_{\rm vtx}$ is the completion of
$\eta({\mathbf{Cyl_{tot}}})$, i.e,

$\mathcal{H}_{\rm vtx}=\overline{\eta({\mathbf{Cyl_{tot}}})}.$ (109)

Given a finite subset $W\subset\Sigma$, let $\Gamma_{\ker}(W)$ be the
collection of graphs $\gamma$ satisfying $V(\ker(\gamma))=W$. Two graphs
$\gamma,\gamma^{\prime}\in\Gamma_{\ker}(W)$ are said to be equivalent, denoted
by $\gamma\sim_{d}\gamma^{\prime}$, if there exists a diffeomorphism
$\phi\in{\rm Diff}_{W}$, i.e., a diffeomorphism preserving $W$, such that
$\phi(\gamma)=\gamma^{\prime}$. Let $[\Gamma_{\ker}(W)]$ be the quotient space
$\Gamma_{\ker}(W)/\sim_{d}$, $[\gamma]\in[\Gamma_{\ker}(W)]$ be the
equivalence class of $\gamma$, and $\mathcal{S}_{\gamma}^{\rm irr}$ be the
image of $P_{\gamma}$. Then $\eta:\mathcal{S}_{\gamma}^{\rm
irr}\to{\mathcal{H}}_{\rm vtx}$ is isometric. By
$\eta(\mathcal{S}_{\gamma}^{\rm irr})$ denoting the image of
$\mathcal{S}_{\gamma}^{\rm irr}$ under $\eta$, one has

$\eta(\mathcal{S}_{\gamma}^{\rm irr})=\eta(\mathcal{S}_{\gamma^{\prime}}^{\rm
irr}),\ \forall\gamma,\gamma^{\prime}\in[\gamma].$ (110)

Thus, we can define

$\eta(\mathcal{S}_{[\gamma]}^{\rm
irr}):=\eta(\mathcal{S}_{\gamma^{\prime}}^{\rm irr})$ (111)

by choosing arbitrary $\gamma^{\prime}\in[\gamma]$. Let ${\rm FS}(\Sigma)$ be
the set of finite subsets of $\Sigma$. Then we have

$\eta({\mathbf{Cyl_{tot}}})=\bigoplus_{W\in{\rm
FS}(\Sigma)}\bigoplus_{[\gamma]\in{[\Gamma_{\ker}(W)]}}\eta({\mathcal{S}}_{[\gamma]}^{\rm
irr})\oplus\mathbb{C}.$ (112)

The factor $1/|{\rm GS}_{\gamma}|$ in (107) ensures that $\eta$ is an
isometric between $\mathcal{H}_{\gamma}^{\rm irr}$ and
$\eta(\mathcal{H}_{\gamma}^{\rm irr})$, i.e.,

$\eta({\mathbf{Cyl_{tot}}})\cong\bigoplus_{W\in{\rm
FS}(\Sigma)}\bigoplus_{[\gamma]\in{[\Gamma_{\ker}(W)]}}{\mathcal{S}}_{\sigma([\gamma])}^{\rm
irr}\oplus\mathbb{C}.$ (113)

where $\sigma([\gamma])\in[\gamma]$ is a representative of $[\gamma]$ fixed
once and for all.

####  IV.2.2 the Hamiltonian operator on $\mathcal{H}_{\rm vtx}$

Given $\Psi_{\gamma}\in\mathcal{S}_{\gamma}^{\rm irr}$, let
$\widehat{H(\delta)}{}^{*}$ be the dual of $\widehat{H(\delta)}$ acting on
$\eta(\Psi_{\gamma})$, i.e.,

$\left(\widehat{H(\delta)}{}^{*}\eta(\Psi_{\gamma})\middle|\Phi_{\gamma^{\prime}}\right\rangle=\left(\eta(\Psi_{\gamma})\middle|\widehat{H(\delta)}\Phi_{\gamma^{\prime}}\right\rangle,\forall\gamma^{\prime}\in\Gamma_{o},\
\Phi_{\gamma^{\prime}}\in\mathcal{H}_{\gamma^{\prime}}^{\rm irr}.$ (114)

The operator $\widehat{H(\delta)}$ takes the form
$\widehat{H(\delta)}=\sum_{v}N(v)\widehat{\sqrt{V_{v}^{-1}}}\hat{O}(v)\widehat{\sqrt{V_{v}^{-1}}}$
with $\hat{O}(v)$ representing some operator. Due to the inverse volume
operators, we have

$\displaystyle\lim_{\delta\to
0}\left(\widehat{H(\delta)}{}^{*}\eta(\Psi_{\gamma})\middle|\Phi_{\gamma^{\prime}}\right\rangle=\lim_{\delta\to
0}\sum_{v\in
V(\ker(\gamma))}N(v)\left(\eta(\Psi_{\gamma})\middle|\widehat{\sqrt{V_{v}^{-1}}}\hat{O}(v)\widehat{\sqrt{V_{v}^{-1}}}\Phi_{\gamma^{\prime}}\right\rangle.$
(115)

Recalling (98), we will consider the right hand term by term.

Let us begin with the operator $\widehat{H^{(1)}(v;\delta)}$ with $v\in
V(\ker(\gamma))$. By (115), we need to investigate

$I_{v}(\gamma^{\prime},\delta)=\left(\eta(\Psi_{\gamma})\middle|\widehat{H^{(1)}(v;\delta)}\Phi_{\gamma^{\prime}}\right\rangle$
(116)

for all graphs $\gamma^{\prime}$ taking $v$ as a vertex. By definition, we can
split $\widehat{H^{(1)}(v;\delta)}$ into two parts such that

$\widehat{H^{(1)}(v;\delta)}\Phi_{\gamma^{\prime}}=\kappa\hbar\beta
N(v)\hat{\mathfrak{h}}_{1}(v;\delta)\Phi_{\gamma^{\prime}}-\kappa\hbar\beta
N(v)\hat{\mathfrak{h}}_{2}(v)\Phi_{\gamma^{\prime}}$ (117)

with $\hat{\mathfrak{h}}_{1}(v;\delta)\Phi_{\gamma^{\prime}}=\sum_{e\text{ at
}v}\hat{\mathfrak{h}}_{1}(v,e;\delta)\Phi_{\gamma^{\prime}}$ and
$\hat{\mathfrak{h}}_{2}(v)=\sum_{[e]\text{ at
}v}\hat{\theta}^{\dagger}(v)\sigma^{i}\hat{\theta}(v)\hat{J}_{i}^{v,[e]}$.
Substituting (117) into (116), we will be concerned about the term

$I_{v}^{(1)}(\gamma^{\prime},\delta)=\left(\eta(\Psi_{\gamma})\middle|\hat{\mathfrak{h}}_{1}(v;\delta)\Phi_{\gamma^{\prime}}\right\rangle=\sum_{\begin{subarray}{c}e\in
E(\gamma^{\prime})\\\ e\text{ at
}v\end{subarray}}\left(\eta(\Psi_{\gamma})\middle|\hat{\mathfrak{h}}_{1}(v,e;\delta)\Phi_{\gamma^{\prime}}\right\rangle.$
(118)

Substituting (106) into (118), we get

$\displaystyle I_{v}^{(1)}(\gamma^{\prime},\delta)=$
$\displaystyle\sum_{\begin{subarray}{c}e\in E(\gamma^{\prime})\\\ e\text{ at
}v\end{subarray}}\sum_{\phi\in{\rm
Diff}^{\prime}}\left\langle\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}(\phi\star\Psi_{\gamma})\middle|\Phi_{\gamma^{\prime}}\right\rangle$
(119)

with ${\rm Diff}^{\prime}:={\rm Diff}_{V(\ker(\gamma))}/{\rm Diff}_{\gamma}$
for abbreviation, where we employ
$\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}=\hat{J}_{i}^{v,e}\hat{\theta}^{\dagger}(v)\sigma^{i}h_{e(v,\delta)}^{-1}\hat{\theta}(t_{e(v,\delta)})$.
Note that the summation in the right hand side of (119) is over edges of
$\gamma^{\prime}$, while the operator
$\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}$ acts on states associated to
graphs diffeomorphism to $\gamma$. Thus, we cannot get
$I^{(1)}_{v}(\gamma^{\prime},\delta)=\left(\hat{\mathfrak{h}}_{1}(v;\delta)^{\dagger}\eta(\Psi_{\gamma})|\Phi_{\gamma^{\prime}}\right\rangle$
naively from (119).

Consider the action of $\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}$ on
$D^{j_{e}}_{m_{e}n_{e}}(h_{e(v,\delta)})\otimes
D^{j_{e}^{\prime}}_{m_{e}^{\prime}n_{e}^{\prime}}(h_{e\setminus
e(v,\delta)})\otimes|k_{1},k_{2}\rangle_{t_{e(v,\delta)}}$, a general state
associated to the graph $\\{e\\}\cup\\{t_{e(v,\delta)}\\}$, where $e\setminus
e(v,\delta)$ is the other segment of $e$ complementing $e(v,\delta)$. Here and
in the rest, by a state associated to a graph $\alpha$, we refer to a state in
$\mathcal{H}_{\alpha}^{\rm irr}$. Acting on the state, the holonomy
$h_{e(v,\delta)}^{-1}$ in $\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}$
changes the spin $j_{e}$ on $e(v,\delta)$ to $j_{e}\pm\frac{1}{2}$. Then, for
$j_{e}=\frac{1}{2}$, the spin on $e(v,\delta)$ is changed to $0$, i.e., the
segment $e(v,\delta)$ is erased. Thus, for $j_{e}=\frac{1}{2}$, acted by the
holonomy, the state becomes a combination of one state associated $\\{e\\}$
and the other state associated to $\\{e\setminus e(v,\delta)\\}$. However, the
operator $\hat{J}_{i}^{v,e}$ in $\hat{\mathfrak{h}}_{1}(v;\delta)^{\dagger}$
will annihilate this one associated to $\\{e\setminus e(v,\delta)\\}$.
Consequently, acted by $\hat{\mathfrak{h}}_{1}(v;\delta)^{\dagger}$, the
resulting state is still a state associated to $\\{e\\}$, regardless of the
change of the fermionic vertices by $\hat{\theta}^{\dagger}(v)$ and
$\hat{\theta}(t_{e(v,\delta)})$.

$\phi_{o}\star\Psi_{\gamma}$ for some diffeomorphism $\phi_{o}$ is a state
associated to the graph $\phi_{o}(\gamma)$. Let $\tilde{\gamma}_{G}$ and
$\tilde{\gamma}_{F}$ denote the gravitational sector and the fermionic sector
of $\phi_{o}(\gamma)$, i.e.,
$\phi_{o}(\gamma)=\tilde{\gamma}_{G}\cup\tilde{\gamma}_{F}$. Then, the
discussion in the last paragraph tells that
$\hat{\mathfrak{h}}_{1}(v;\delta)^{\dagger}\phi_{o}\star\Psi_{\gamma}$ is a
state associated to some graph whose gravitational sector is
$\tilde{\gamma}_{G}\cup e(v,\delta)$. Thus,
$\left\langle\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}(\phi_{o}\star\Psi_{\gamma})\middle|\Phi_{\gamma^{\prime}}\right\rangle\neq
0$ for $\Phi_{\gamma^{\prime}}\in\mathcal{H}_{\gamma^{\prime}}^{\rm irr}$
implies $e\subset\tilde{\gamma}_{G}\cup e(v,\delta)$, which means either
$e\subset\phi_{o}(\gamma)$, or $e(v,\delta)\not\subset\phi_{o}(\gamma)$ but
$e\setminus e(v,\delta)\subset\phi_{o}(\gamma)$.

Let us first consider the case where $e(v,\delta)\not\subset\phi_{o}(\gamma)$
but $e\setminus e(v,\delta)\subset\phi_{o}(\gamma)$ such that
$\left\langle\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}(\phi_{o}\star\Psi_{\gamma})\middle|\Phi_{\gamma^{\prime}}\right\rangle\neq
0$. The hypothesis
$\left\langle\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}(\phi_{o}\star\Psi_{\gamma})\middle|\Phi_{\gamma^{\prime}}\right\rangle\neq
0$ implies $\tilde{\gamma}_{G}\cup e(v,\delta)=\gamma^{\prime}_{G}$ where
$\gamma^{\prime}_{G}$ is the gravitational sector of $\gamma^{\prime}$.
Together with $e(v,\delta)\not\subset\phi_{o}(\gamma)$, one gets
$\tilde{\gamma}_{G}=\gamma^{\prime}_{G}\setminus e(v,\delta)$. Since
$e(v,\delta)$ is a segment of an edge $e\subset\gamma^{\prime}_{G}$,
$t_{e(v,\delta)}$ is a 1-valence vertex in $\tilde{\gamma}_{G}$. Furthermore,
$t_{e(v,\delta)}$ is a 1-valence fermionic vertex because of
$\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}(\phi_{o}\star\Psi_{\gamma})\neq
0$. Thus, $t_{e(v,\delta)}$ is not a removable vertex, i.e., all of the
diffeomorphisms in ${\rm Diff}^{\prime}$ preserve this vertex. Now, let us
come to $\delta^{\prime}<\delta$. If $\sum_{\phi\in{\rm
Diff}^{\prime}}\left\langle\hat{\mathfrak{h}}_{1}(v,e;\delta^{\prime})^{\dagger}(\phi\star\Psi_{\gamma})\middle|\Phi_{\gamma^{\prime}}\right\rangle=\sum_{\phi\in{\rm
Diff}^{\prime}}\left\langle\hat{\mathfrak{h}}_{1}(v,e;\delta^{\prime})^{\dagger}(\phi\star\phi_{o}\star\Psi_{\gamma})\middle|\Phi_{\gamma^{\prime}}\right\rangle$
still does not vanish, there have to exist another diffeomorphism
$\phi_{o}^{\prime}$ which can move a fermionic vertex in $\phi_{o}(\gamma)$ to
$t_{e(v,\delta^{\prime})}$ such that
$\hat{\mathfrak{h}}_{1}(v,e;\delta^{\prime})^{\dagger}(\phi_{o}^{\prime}\star\phi_{o}\star\Psi_{\gamma})\neq
0$, and simultaneously preserve all of the other vertices but not necessarily
$t_{e(v,\delta)}$ in $\phi_{o}(\gamma)$ such that
$\left\langle\hat{\mathfrak{h}}_{1}(v,e;\delta^{\prime})^{\dagger}(\phi_{o}^{\prime}\star\phi_{o}\star\Psi_{\gamma})\middle|\Phi_{\gamma^{\prime}}\right\rangle\neq
0$. This can be done only if $\phi_{o}^{\prime}$ can move $t_{e(v,\delta)}$ to
$t_{e(v,\delta^{\prime})}$ which is impossible because $t_{e(v,\delta)}$ is
preserved by all diffeomorphisms in ${\rm Diff}^{\prime}$. Therefore,
$\sum_{\phi\in{\rm
Diff}^{\prime}}\left\langle\hat{\mathfrak{h}}_{1}(v,e;\delta^{\prime})^{\dagger}(\phi\star\Psi_{\gamma})\middle|\Phi_{\gamma^{\prime}}\right\rangle=0$
for all $\delta^{\prime}<\delta$. This discussion tells that for sufficiently
small $\delta$, the case with $e(v,\delta)\not\subset\phi_{o}(\gamma)$ but
$e\setminus e(v,\delta)\subset\phi_{o}(\gamma)$ can be excluded. Because we
finally need to consider the limit as $\delta\to 0$, choosing a sufficiently
small $\delta$ can be done without loss of generality.

For the case with $e\subset\phi_{o}(\gamma)$,
$\left\langle\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}(\phi_{o}\star\Psi_{\gamma})\middle|\Phi_{\gamma^{\prime}}\right\rangle\neq
0$ requires that the graph of
$\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}(\phi_{o}\star\Psi_{\gamma})$ is
equal to $\gamma^{\prime}$. Since $\delta$ is chosen to be sufficiently small,
in the graph $\gamma^{\prime}$ there is no fermionic vertex on the segment
$e(v,\delta)\subset e\subset\gamma^{\prime}$, and
$t_{e(v,\delta)}\in\gamma^{\prime}$ is also not a fermionic vertex. Therefore,
for
$\left\langle\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}(\phi_{o}\star\Psi_{\gamma})\middle|\Phi_{\gamma^{\prime}}\right\rangle\neq
0$, there cannot exist any fermionic vertices in
$e(v,\delta)\cup\\{t_{e(v,\delta)}\\}$, where $e(v,\delta)$ and
$t_{e(v,\delta)}$ now are thought of as a segment and a vertex in the graph of
$\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}(\phi_{o}\star\Psi_{\gamma})$.
This conclusion has two meanings. At first, $\phi_{o}\in{\rm Diff}^{\prime}$
must be the diffeomorphism which moves the closest fermionic vertex to $v$ in
$e\in\gamma$ to $t_{e(v,\delta)}$ so that $\hat{\theta}(t_{e(v,\delta)})$ can
kill it. The closest fermionic vertex to $v$ in $e\in\gamma$ will be denoted
by $v_{F}^{(v,e)}$, and Fig. 3 gives an illustration of $v_{F}^{(v,e)}$.
Second, it means that the fermion state at $v_{F}^{(v,e)}$ cannot be
$|1,1\rangle_{v_{F}^{(v,e)}}$ because it cannot be annihilated completely by
$\hat{\theta}(t_{v,\delta})$. For the second point, we introduce a projection
$\hat{\mathbb{P}}_{v_{F}}$ on the fermion Hilbert space $\mathcal{H}_{v_{F}}$
such that

$\hat{\mathbb{P}}_{v_{F}}|i_{1},i_{2}\rangle=\left\\{\begin{array}[]{cc}0,&i_{1}=1=i_{2},\\\
|i_{1},i_{2}\rangle,&\text{ otherwise. }\end{array}\right.$ (120)

The first point makes us decompose $\phi_{o}$ as follows. Let $\gamma_{\star}$
be the graph of $\gamma^{\prime}$ with $v_{F}^{(v,e)}$ dropped, and
$v\in\gamma$ promoted to a new fermionic vertex if it is not. Then, we find a
diffeomorphism $\phi_{1}\in{\rm Diff}_{V(\ker(\gamma))}/{\rm
Diff}_{\gamma_{\star}}={\rm Diff}_{V(\ker(\gamma_{\star}))}/{\rm
Diff}_{\gamma_{\star}}$ such that $\phi_{1}(\gamma_{\star})=\gamma^{\prime}$.
With $\phi_{1}$, we find $\phi_{2}\in{\rm
Diff}_{\phi_{1}(\gamma_{\star})}/{\rm Diff}_{\gamma}$ which moves
$v_{F}^{(v,e)}$ to $t_{e(v,\delta)}$. One thus gets
$\phi_{o}=\phi_{2}\circ\phi_{1}$. Introducing $\widehat{\mathfrak{H}}_{v}(e)$
as

$\widehat{\mathfrak{H}}_{v}(e):=\left\\{\begin{array}[]{cl}\hat{J}_{i}^{v,e}\hat{\theta}^{\dagger}(v)\sigma^{i}h_{e[v_{F}^{(v,e)},v]}^{-1}\hat{\theta}(v_{F}^{(v,e)})\hat{\mathbb{P}}_{v_{F}^{(v,e)}},&\text{
if }v_{F}^{(v,e)}\text{ exists}.\\\ 0&\text{\ otherwise}\end{array}\right.$
(121)

where $e[v_{F}^{(v,e)},v]$ denotes the edge from $v$ to $v_{F}^{(v,e)}$. We
thus get

$\left\langle\hat{\mathfrak{h}}_{1}(v,e;\delta)^{\dagger}(\phi_{o}\star\Psi_{\gamma})\middle|\Phi_{\gamma^{\prime}}\right\rangle=\left\langle\phi_{1}\star(\widehat{\mathfrak{H}}_{v}(e)\Psi_{\gamma})\middle|\Phi_{\gamma^{\prime}}\right\rangle,$
(122)

where the right hand side is independent of $\delta$, and
$\phi_{2}(\gamma^{\prime})=\gamma^{\prime}$ is used. Using (122), we finally
get

$\lim_{\delta\to 0}I^{(1)}_{v}(\gamma^{\prime},\delta)=\sum_{\phi\in{\rm
Diff}_{V(\gamma)}/{\rm
Diff}_{\gamma_{\star}}}\left\langle\phi\star\widehat{\mathfrak{H}}_{v}\Psi_{\gamma}\middle|\Phi_{\gamma^{\prime}}\right\rangle$
(123)

with

$\widehat{\mathfrak{H}}_{v}\Psi_{\gamma}=\sum_{e\text{ at
}v}\widehat{\mathfrak{H}}_{v}(e)\Psi_{\gamma},\forall\Psi_{\gamma}\in\mathcal{S}_{\gamma}^{\rm
irr}.$ (124)

Eq. (123) leads to

$\lim_{\delta\to
0}\left(\hat{\mathfrak{h}}_{1}(v;\delta)^{*}\eta(\Psi_{\gamma})\right|=\left(\eta(\hat{\mathfrak{H}}_{v}\Psi_{\gamma})\right|$
(125)

Finally, taking advantage of $\lim_{\delta\to
0}\hat{\mathfrak{h}}_{2}^{*}\eta(\Psi_{\gamma})=\eta(\hat{\mathfrak{h}}_{2}\Psi_{\gamma})$,
we have

$\lim_{\delta\to
0}(\widehat{H^{(1)}(v;\delta)}{})^{*}\eta(\Psi_{\gamma})=\kappa\hbar\beta
N(v)\eta\left[\left((\hat{\mathfrak{H}}_{v}-\sum_{e\text{ at
}v}\hat{J}_{i}^{v,e}\hat{\theta}^{\dagger}(v)\sigma^{i}\hat{\theta}(v)\right)\Psi_{\gamma}\right].$
(126)

Figure 3: An example of graphs $\gamma$ with $v_{F}^{(v,e)}$ being the closest
fermionic vertex to $v$ in $e\in E(\ker(\gamma))$, and
$e[v_{F}^{(v,e)},v]\subset e$ is the edge of $\gamma$ from $v$ to
$v_{F}^{(v,e)}$.

Now let us consider the operator $\widehat{H^{(2)}(v;\delta)}$. An issue on
defining $\lim\limits_{\delta\to 0}\widehat{H^{(2)}(v;\delta)}$ is the
operator $\hat{H}_{E,v}$ comprising $\widehat{H^{(2)}(v;\delta)}$. In this
paper, we will employ the work alesci2015hamiltonian  to define
$\hat{H}_{E,v}$ in $\mathcal{H}_{\rm vtx}$. More precisely, $\hat{H}_{E,v}$ is
given by
$\hat{H}_{E,v}={\widehat{\sqrt{V_{v}^{-1}}}(\hat{F}_{v}+\hat{F}_{v}^{\dagger})\widehat{\sqrt{V_{v}^{-1}}}}$
with

$\hat{F}_{v}=-2\sum_{e,e^{\prime}\text{ at
}v}\epsilon(\dot{e},\dot{e}^{\prime})\epsilon^{ijk}\tr(h_{\alpha_{ee^{\prime}}}\tau_{k})\hat{J}_{i}^{v,e}\hat{J}_{k}^{v,e^{\prime}}$
(127)

where $\alpha_{ee^{\prime}}$ is a loop tangent to the two edges $e$ and
$e^{\prime}$ at the vertex $v$ up to orders $k_{e}+1$ and $k_{e^{\prime}}+1$
respectively with $k_{e}$ and $k_{e^{\prime}}$ being respectively the orders
of tangentiality of $e$ and $e^{\prime}$ at $v$ (see alesci2015hamiltonian
for more details.) By this definition, $\hat{F}_{v}$ will change the graph by
adding a loop at $v$ and, thus, $\hat{F}_{v}^{\dagger}$, by removing a loop.

A subtlety here is that the vertex Hilbert space defined in
alesci2015hamiltonian  is a little different from ours. In
alesci2015hamiltonian , the authors defines the vertex Hilbert space with
diffeomorphisms preserving $V(\gamma)$ while our work considers the
diffeomorphisms preserving $V(\ker(\gamma))$. Regardless of this difference,
the operator $\hat{H}_{E,v}$ introduced in alesci2015hamiltonian  , for $v\in
V(\ker(\gamma))$, is well-defined in our vertex Hilbert space
$\mathcal{H}_{\rm vtx}$. Indeed, due to the operator
${\widehat{\sqrt{V_{v}^{-1}}}}$ in $\widehat{\tilde{H}^{(2)}(v;\delta)}$, we
do not need to consider $\hat{H}_{E,v}$ for removable $v$, i.e. $v\notin
V(\ker(\gamma))$. According to this discussion, for $v\in V(\ker(\gamma))$, we
have

$\displaystyle\lim_{\delta\to
0}\widehat{H^{(2)}(v;\delta)}\eta(\Psi_{\gamma})=\frac{\kappa_{1}(v)}{2i\kappa\hbar\beta^{2}}N(v)\eta\left[\left(\hat{H}_{E,v}\hat{V}_{v}-\hat{V}_{v}\hat{H}_{E,v}\right)\hat{\theta}^{\dagger}(v)\hat{\theta}(v)\Psi_{\gamma}\right],$
(128)

where $\kappa_{1}(v)$ is introduced in alesci2015hamiltonian  to remove the
dependence on the partition.

Finally, for the operator $\widehat{H^{(3)}(v;\delta)}$, since
$\widehat{H^{(3)}(v;\delta)}$ for $v\in V(\ker(\gamma))$ is independent of
$\delta$ and ${\rm Diff}_{V(\ker(\gamma))}$ invariant. we have

$\lim_{\delta\to
0}\left(\widehat{H^{(3)}(v;\delta)}{}^{*}\eta(\Psi_{\gamma})\right|=\left(\eta(\hat{H}^{(3)}_{F}(v)\Psi_{\gamma})\right|$
(129)

with

$\hat{H}^{(3)}_{F}(v):=\kappa\hbar\beta N(v)\left(\sum_{e\text{ at
}v_{C}}\hat{J}_{i}^{v_{C},e}\right)\hat{\theta}^{\dagger}(v)\sigma^{i}\hat{\theta}(v).$
(130)

Let us summarize our results. According to the above discussion, we have

$\hat{A}_{F}\eta(\Psi_{\gamma}):=\lim_{\delta\to
0}\widehat{H(\delta)}{}^{*}\eta(\Psi_{\gamma})=\sum_{v\in
V(\gamma)}\eta\left(\widehat{\sqrt{V_{v}^{-1}}}\hat{H}_{v}\widehat{\sqrt{V_{v}^{-1}}}\Psi_{\gamma}\right)$
(131)

where $\hat{H}_{v}$ is given by

$\hat{H}_{v}=i\widehat{H^{(1)}(v)}-\frac{\beta}{2}\widehat{H^{(2)}(v)}-\frac{1+\beta^{2}}{2\beta}\widehat{H^{(3)}(v)}-\beta\widehat{H^{(1)}(v)},$
(132)

with

$\displaystyle\widehat{H^{(1)}(v)}$
$\displaystyle=\kappa_{2}(v)\ell_{p}^{2}\beta
N(v)\left(\hat{\mathfrak{H}}_{v}-\sum_{e\text{ at
}v}\hat{\theta}^{\dagger}(v)\sigma^{i}\hat{\theta}(v)\hat{J}_{i}^{v,e}\right),$
(133) $\displaystyle\widehat{H^{(2)}(v)}$
$\displaystyle=\frac{\kappa_{1}(v)}{2i\ell_{p}^{2}\beta^{2}}N(v)\left(\hat{H}_{E,v}\hat{V}_{v}-\hat{V}_{v}\hat{H}_{E,v}\right)\hat{\theta}^{\dagger}(v)\hat{\theta}(v),$
$\displaystyle\widehat{H^{(3)}(v)}$
$\displaystyle=\kappa_{3}(v)\ell_{p}^{2}\beta N(v)\left(\sum_{e\text{ at
}v}\hat{J}_{i}^{v,e}\right)\hat{\theta}^{\dagger}(v)\sigma^{i}\hat{\theta}(v),$

according to (126), (128) and (129) respectively. Here we again introduce the
parameters $\kappa_{2}$ and $\kappa_{3}$ as in ashtekar1997quantumII  in order
to remove the dependence on the partition. Finally, because of (76), we
finally define the fermion Hamiltonian operator $\hat{H}_{F}[N]$ on
$\mathcal{H}_{\rm vtx}$ as

$\hat{H}_{F}[N]=\frac{1}{2}(\hat{A}_{F}+\hat{A}_{F}^{\dagger}).$ (134)

Let $\hat{\mathfrak{H}}_{v}^{\prime}$ on $\mathcal{H}_{\rm vtx}$ be the
operator defined by

$\hat{\mathfrak{H}}_{v}^{\prime}\eta(\Psi_{\gamma})=\eta(\hat{\mathfrak{H}}_{v}\Psi_{\gamma}).$
(135)

Then, the only nontrivial term in $A_{F}^{\dagger}$ is the operator
$(\hat{\mathfrak{H}}_{v}^{\prime})^{\dagger}$. By the definition,
$\hat{\mathfrak{H}}_{v}$ is composed of $\hat{\mathfrak{H}}_{v}(e)$ which
annihilates the fermionic vertex contained in $e$ and closest to $v$. Thus,
the operator $(\hat{\mathfrak{H}}_{v}^{\prime})^{\dagger}$ contains the
operators, each of which is associated to an edge $e$ at $v$ and promotes a
point $v_{F}\in e$ to a fermionic vertex such that $v_{F}$ becomes the closest
to $v$ among the fermionic vertices contained in $e$. To be more precise, let
us define an operator $\hat{\mathfrak{H}}_{v}^{\\#}(e)$ (refer to (121)) as

$\widehat{\mathfrak{H}}_{v}^{\\#}(e)=\chi_{+}(|\hat{J}^{v,e}|^{2})\hat{\theta}^{\dagger}(\mathfrak{v}_{e,v})h_{e(\mathfrak{v}_{e,v})}\sigma^{i}\hat{\theta}(v)\hat{J}_{i}^{v,e}$
(136)

where $\chi_{+}$ denotes the characteristic function of
$[\frac{3}{4},\infty)$,
$|\hat{J}^{v,e}|^{2}=\delta^{kl}\hat{J}^{v,e}_{k}\hat{J}^{v,e}_{l}$,
$\mathfrak{v}_{e,v}\in e$ is arbitrary point between $v$ and the closest
fermionic vertex to $v$ in $e$, and $e(\mathfrak{v}_{e,v})\subset e$ is the
segment starting from $v$ and ending at $\mathfrak{v}_{e,v}$. Here the factor
$\chi_{+}(|\hat{J}^{v,e}|^{2})$ ensures that
$\widehat{\mathfrak{H}}_{v}^{\\#}(e)$ cannot erase the segment
$e(\mathfrak{v}_{e,v})\subset e$. In other words,
$\widehat{\mathfrak{H}}_{v}^{\\#}(e)$, acting on a state with spin
$\frac{1}{2}$ on $e(\mathfrak{v}_{e,v})$, will only change the spin
$\frac{1}{2}$ to $1$. Moreover, $\mathfrak{v}_{e,v}$ is defined such that it
becomes closest to $v$ among the fermionic vertices in $e$ after the action of
$\widehat{\mathfrak{H}}_{v}^{\\#}(e)$. With
$\widehat{\mathfrak{H}}_{v}^{\\#}(e)$,
$(\hat{\mathfrak{H}}_{v}^{\prime})^{\dagger}$ is given by

$(\hat{\mathfrak{H}}_{v}^{\prime})^{\dagger}\eta(\Psi_{\gamma})=\eta\left(\sum_{e\text{
at }v}\widehat{\mathfrak{H}}_{v}^{\\#}(e)\Psi_{\gamma}\right).$ (137)

Note that even though $\mathfrak{v}_{e,v}$ is not uniquely determined by its
definition, the operator $(\hat{\mathfrak{H}}_{v}^{\prime})^{\dagger}$ on
$\mathcal{H}_{\rm vtx}$ is independent of the choice of $\mathfrak{v}_{e,v}$,
due to the averaging operation $\eta$. Moreover,
$\chi_{+}(|\hat{J}^{v,e}|^{2})$ is introduced in (136) because (124) implies

$\widehat{\mathfrak{H}}_{v}=\sum_{e\text{ at
}v}\hat{\mathfrak{H}}_{v}(e)\chi_{+}(|\hat{J}^{v,e}|^{2}).$ (138)

Applying $\widehat{\mathfrak{H}}_{v}^{\\#}(e)$, $\hat{A}_{F}^{\dagger}$ reads

$\hat{A}_{F}^{\dagger}\eta(\Psi_{\gamma})=\sum_{v\in
V(\gamma)}\eta\left(\widehat{\sqrt{V_{v}^{-1}}}\hat{H}_{v}^{\\#}\widehat{\sqrt{V_{v}^{-1}}}\Psi_{\gamma}\right)$
(139)

where $\hat{H}_{v}^{\\#}$ is

$\hat{H}_{v}^{\\#}=-i\widehat{H^{(1)}(v)}{}^{\\#}-\frac{\beta}{2}\widehat{H^{(2)}(v)}-\frac{1+\beta^{2}}{2\beta}\widehat{H^{(3)}(v)}-\beta\widehat{H^{(1)}(v)}{}^{\\#},$
(140)

with $\widehat{H^{(2)}v)}$ and $\widehat{H^{(3)}(v)}$ given in (133), and
$\widehat{H^{(1)}(v)}{}^{\\#}$ given by

$\displaystyle\widehat{H^{(1)}(v)}{}^{\\#}$
$\displaystyle=\kappa_{2}(v)\ell_{p}^{2}\beta N(v)\left(\sum_{e\text{ at
}v}\widehat{\mathfrak{H}}_{v}^{\\#}(e)-\sum_{e\text{ at
}v}\hat{\theta}^{\dagger}(v)\sigma^{i}\hat{\theta}(v)\hat{J}_{i}^{v,e}\right).$
(141)

Let us complete this section with a discussion on the intuitive picture led by
the action of $\widehat{\mathfrak{H}}_{v}(e)$ and
$\widehat{\mathfrak{H}}_{v}^{\\#}(e)$. By definition,
$\widehat{\mathfrak{H}}_{v}^{\\#}(e)$ creates a fermionic vertex $v_{F}\in e$
so that $v_{F}$ will be closest to $v$ and carries states
$a|1,0\rangle+b|0,1\rangle$. Simultaneously,
$\widehat{\mathfrak{H}}_{v}^{\\#}(e)$ changes the fermion state
$|i_{1},i_{2}\rangle_{v}$ at $v$ in such a way that $|1,1\rangle_{v}$ becomes
$c|1,0\rangle_{v}+d|1,0\rangle_{v}$ for some constant $c$ and $d$, and
$\alpha|1,0\rangle_{v}+\beta|0,1\rangle_{v}$ for arbitrary $\alpha$ and
$\beta$ is changed to $|0,0\rangle_{v}$. Moreover, because of the holonomy
operator $h_{e(\mathfrak{v}_{e,v})}$ and flux operator $\hat{J}_{i}^{v,e}$ in
$\widehat{\mathfrak{H}}_{v}^{\\#}(e)$, the spin on the segment
$e(\mathfrak{v}_{e,v})\subset e$ and the intertwiner at $v$ are changed. These
results can be summarized as that the operator
$\widehat{\mathfrak{H}}_{v}^{\\#}(e)$ moves a fermion at $v$ to $v_{F}$, and
changes the geometry around $v$ simultaneously. For the operator
$\widehat{\mathfrak{H}}_{v}(e)$, it reverses this procedure.
$\widehat{\mathfrak{H}}_{v}(e)$ moves a fermion at the fermionic vertex
$v_{F}^{(v,e)}$, i.e., the closest fermionic vertex to $v$, to the vertex $v$,
and changes the geometry around $v$ simultaneously. Moreover, because of the
projection operator $\hat{\mathbb{P}}_{v_{F}}$ in
$\widehat{\mathfrak{H}}_{v}(e)$ (see (121)), when the fermionic vertex
$v_{F}^{(v,e)}$ carries a fermion state $|1,1\rangle_{v_{F}^{(v,e)}}$, the
fermion at $v_{F}^{(v,e)}$ cannot be moved by the operator
$\widehat{\mathfrak{H}}_{v}(e)$. To see the consequence of this fact, let us
imagine an edge $e$ with both the source $s_{e}$ and the middle point
$v_{F}^{(v,e)}$ being the fermionic vertices, where $v_{F}^{(v,e)}$ carries a
fermion state, says, $|1,0\rangle_{v_{F}^{(v,e)}}$. Acted by
$\widehat{\mathfrak{H}}_{s_{e}}^{\\#}(e)$, the fermion at $s_{e}$ will be
moved to some point $v_{F}^{\prime}$ between $s_{e}$ and $v_{F}^{(v,e)}$.
Then, acted by $\widehat{\mathfrak{H}}_{t_{e}}(e)$ twice, the fermion at
$v_{F}^{\prime}$ will be moved to $t_{e}$. Now, suppose that $v_{F}^{(v,e)}$
carries the state $|1,1\rangle_{v_{F}^{(v,e)}}$. Then, acted by
$\widehat{\mathfrak{H}}_{s_{e}}^{\\#}(e)$, the fermion at $s_{e}$ will be
again moved to $v_{F}^{\prime}$. However, the fermion at $v_{F}^{\prime}$
cannot be moved to $t_{e}$ after acted by $\widehat{\mathfrak{H}}_{t_{e}}(e)$,
due to the operator $\hat{\mathbb{P}}_{v_{F}^{(v,e)}}$ in
$\widehat{\mathfrak{H}}_{t_{e}}(e)$. Intuitively, in the second situation, the
fermion at $s_{e}$ is confined around $s_{e}$ by the fermion state
$|1,1\rangle_{v_{F}^{(v,e)}}$ at $v_{F}^{(v,e)}$. The above picture tells how
a fermion moves in loop quantum spacetime and influences the background
quantum geometry in the LQG framework.

##  V summation and outlook

This work is concerned about the model of fermion field coupled to LQG. The
Gauss and Hamiltonian constraints in this model are studied in details. In the
solution to the Gauss constraint, fermion spins and the gravitational spin
network intertwine with each other so that the fermion spins contribute to the
volume of the spin network vertices. Consequently, the closure condition
encoded in the Gauss constraint will no longer be satisfied for the gauge
invariant state with non-vanishing fermion spins. In other words, the faces
dual to the edges at a fermionic vertex with non-vanishing fermion spin could
not form a closed polyhedron, and the area defect of this unclosed polyhedron
is filled by the fermion spin. Consequently, in contract to pure-gravity case,
a 3-valence gauge invariant vertex with non-vanishing fermion spin will get
non-vanishing volume from the fermion spin. The volume of this type of
vertices is computed in details.

For the Hamiltonian constraint, the regularization and quantization procedures
are presented in details. There are several remarkable issues on the
Hamiltonian constraint operator. At first, in order to take the limit of the
regularized expression as the regulator approaches 0, we introduce the vertex
Hilbert space. By definition, the vertex Hilbert space is the space of
cylindrical functions averaged with the diffeomorphisms preserving the
unremovable vertices. Thus, the states in the vertex Hilbert space are
partially diffeomorphism invariant. The vertex Hilbert space is the dual space
to the space of cylindrical functions. Then, the regularized Hamiltonian
operator $\widehat{H(\delta)}$ can be promoted as an operator
$\widehat{H(\delta)}{}^{*}$ therein by duality. Due to the diffeomorphism
invariant feature of the vertex Hilbert space, the operators
$\widehat{H(\delta)}{}^{*}$ for different values of the regulator $\delta$ are
identical. Thus, the limit of $\widehat{H(\delta)}{}^{*}$ as $\delta$
approaches 0 can be taken. Moreover, by introducing the vertex Hilbert space,
we fix several other problematic issues. Classically, the Hamiltonian can be
divided into two parts as $H_{\mathcal{C}_{\epsilon}}$ plussing its complex
conjugate $\overline{H_{\mathcal{C}_{\epsilon}}}$. $H(\delta)$ can be promoted
to the operator $\widehat{H(\delta)}$ which is cylindrical consistent and
diffeomorphism covariant. However, its adjoint
$\widehat{H(\delta)}{}^{\dagger}$, as a candidate of the operator
corresponding to $\overline{H_{\mathcal{C}_{\epsilon}}}$, is not densely
defined, because $\widehat{H(\delta)}$ could change graphs by erasing some
segments of edges in graphs. These problems on
$\widehat{H(\delta)}{}^{\dagger}$ are finally solved by introducing the vertex
Hilbert space to define limit. In the vertex Hilbert space, the operator
$\hat{A}_{F}^{\dagger}$, as the limit of $H(\delta)$, performs in a way such
that a projection is left multiplied in $\hat{A}_{F}^{\dagger}$ as a factor.
Then, if $\hat{A}_{F}^{\dagger}$ erases segments of edges, the projection will
annihilate the resulting state, so that the adjoint of $\hat{A}_{F}^{\dagger}$
is densely defined. Finally, in the Hamiltonian constraint operator, there are
the operators $\widehat{\mathfrak{H}}_{v}^{\\#}(e)$ and
$\widehat{\mathfrak{H}}_{v}(e)$ involved. These two operators tell how a
fermion moves in LQG spacetime and influences the background loop quantum
geometry. According to our results, the operator
$\widehat{\mathfrak{H}}_{v}^{\\#}(e)$ moves a fermion at the vertex $v$ to a
point $v_{F}\in e$ so that $v_{F}$ becomes the closest fermionic vertex to
$v$, and simultaneously changes the spin on the segment connecting $v$ to
$v_{F}$ of $e$ as well as the intertwiner at $v$. This procedure will be
reversed by the operator $\widehat{\mathfrak{H}}_{v}(e)$, which moves a
fermion, located at the fermionic vertex $v_{F}^{(v,e)}$ closest to $v$ in
$e$, to the vertex $v$, and simultaneously changes the spin on the segment
connecting $v$ to $v_{F}^{(v,e)}$ in $e$ as well as the intertwiner at $v$. In
addition, $\widehat{\mathfrak{H}}_{v}(e)$ is defined to contain a projection
operator $\hat{\mathbb{P}}_{v_{F}^{(e,v)}}$. As a consequence of this
operator, the fermion located at, says, $s_{e}$ will be confined around
$s_{e}$ by the state $|1,1\rangle_{v_{F}^{(v,e)}}$ located at
$v_{F}^{(v,e)}\in e$.

Even though the current work is concerned about the graph changing feature,
the framework can be easily adapted to define a graph preserving version of
the Hamiltonian constraint operator. Then one can apply this graph preserving
operator to the lattice model of fermion coupled to LQG, so that some open
issues in lattice fermion field theory can be employed and studied. Moreover,
the properties of the fermion Hamiltonian operator are still not well
understood although we have discussed some of them. All of these will be left
as our future works.

## Acknowledgements

This work is benefited greatly from the numerous discussions with Muxin Han,
and supported by the Polish Narodowe Centrum Nauki, Grant No.
2018/30/Q/ST2/00811, and NSFC with Grants No. 11961131013 and No. 11875006.

##  Appendix A Hamiltonian analysis for fermion field

Define $P_{\pm}=\frac{1\pm\gamma^{5}}{2},$ one has

$\displaystyle P_{\pm}^{2}=P_{\pm},\ P_{+}P_{-}=P_{-}P_{+}=0,\
P_{\pm}\gamma^{\mu}=\gamma^{\mu}P_{\mp}.$ (142)

Therefore, we have

$\displaystyle\overline{\Psi}\gamma^{I}e_{I}^{\mu}P_{\pm}\nabla_{\mu}\Psi=(\Psi_{\pm})^{\dagger}\gamma^{0}\gamma^{I}e_{I}^{\mu}\Psi_{\pm}+\frac{1}{4}(\Psi_{\pm})^{\dagger}\gamma^{0}\gamma^{I}e_{I}^{\mu}\omega_{\mu
KL}\gamma^{K}\gamma^{L}\Psi_{\pm}$ (143)

with $\Psi_{\pm}:=P_{\pm}\Psi$. Let us choose the Weyl basis of the $\gamma$
matrices

$\gamma^{0}=\begin{pmatrix}0&i\mathbbm{1}_{2}\\\
i\mathbbm{1}_{2}&0\end{pmatrix},\quad\gamma^{k}=\begin{pmatrix}0&i\sigma^{k}\\\
-i\sigma^{k}&0\end{pmatrix},\quad\gamma^{5}=\begin{pmatrix}-I_{2}&0\\\
0&I_{2}\end{pmatrix}.$ (144)

Then $\Psi_{\pm}$ take the form $\Psi_{-}=(\psi,0)^{T}$ and
$\Psi_{+}=(0,\eta)^{T}$. We thus get

$\displaystyle\overline{\Psi}\gamma^{I}e_{I}^{\mu}\nabla_{\mu}\Psi=-\psi^{\dagger}\overline{\sigma}^{I}e_{I}^{\mu}\partial_{\mu}\psi+\frac{1}{4}\psi^{\dagger}\overline{\sigma}^{I}e_{I}^{\mu}\omega_{\mu
KL}\sigma^{K}\overline{\sigma}^{L}\psi-\eta^{\dagger}\sigma^{I}e_{I}^{\mu}\eta+\frac{1}{4}\eta^{\dagger}\sigma
e_{I}^{\mu}\omega_{\mu KL}\overline{\sigma}^{K}\sigma^{L}\eta$ (145)

with $\sigma^{I}=(\mathbbm{1},\text{Pauli matrax}^{i})$ and
$\overline{\sigma}^{I}=(\mathbbm{1},-\text{Pauli matrax}^{i})$.

Performing the 3+1-decomposition $\mathcal{M}=\mathbb{R}\times\Sigma$, one has

$e_{I}^{\mu}=e^{\nu}_{I}q_{\nu}^{\mu}-n^{\mu}n_{I}$ (146)

where $q_{\nu}^{\mu}$ is the projection to $\Sigma$ and
$n^{\mu}=(t^{\mu}-N^{\mu})/N$ with $N$ and $N^{\mu}$ being the lapse function
and the shift vector respectively, and $t^{\mu}$ being some time evolution
vector field given by $t^{\mu}\partial_{\mu}t=1$. Substituting (146) into
(145), we have

$\displaystyle\overline{\Psi}e_{I}^{\mu}\gamma^{I}\nabla_{\mu}\Psi=\left(\psi^{\dagger}e_{i}^{a}\sigma^{i}\mathcal{D}^{+}_{a}\psi-\eta^{\dagger}e_{i}^{a}\mathcal{D}^{-}_{a}\eta\right)-\frac{1}{N}(t^{\mu}-N^{\mu})\left(\psi^{\dagger}\mathcal{D}_{\mu}^{+}\psi+\eta^{\dagger}\mathcal{D}_{\mu}^{-}\eta\right),$
(147)

where we defined

$\mathcal{D}^{\pm}_{a}=\partial_{a}+(\Gamma_{a}^{m}\mp
K_{a}^{m})\tau_{m}=:\partial_{a}+\mathcal{A}^{\pm}_{a}.$ (148)

Defining

$\nabla_{a}=\partial_{a}+\Gamma_{a}^{m}\tau_{m}$ (149)

we can express the action of the fermion field in terms of $K_{a}^{i}$ and
$\nabla_{a}$ explicitly, which reads

$\displaystyle S_{F}=$
$\displaystyle-\frac{i}{2}\int_{\mathcal{M}}\differential^{4}x(\overline{\Psi}\gamma^{I}e_{I}^{\mu}\nabla_{\mu}\Psi-c.c)$
(150) $\displaystyle=$
$\displaystyle\frac{i}{2}\int\differential^{4}x\sqrt{q}\Bigg{(}\left(\psi^{\dagger}\partial_{t}\psi+\eta^{\dagger}\partial_{t}\eta-c.c\right)+2\Gamma_{tm}\left(\psi^{\dagger}\tau^{m}\psi+\eta^{\dagger}\tau^{m}\eta\right)-N^{a}\left(\psi^{\dagger}\nabla_{a}\psi+\eta^{\dagger}\nabla_{a}\eta-(\nabla_{a}\psi)^{\dagger}\psi-(\nabla_{a}\eta)^{\dagger}\eta\right)$
$\displaystyle-\frac{N}{\sqrt{q}}\Big{[}\psi^{\dagger}E_{i}^{a}\sigma^{i}\nabla_{a}\psi-(\nabla_{a}\psi)^{\dagger}E_{i}^{a}\sigma^{i}\psi+2\psi^{\dagger}[E^{a},K_{a}]\psi-\eta^{\dagger}E_{i}^{a}\sigma^{i}\nabla_{a}\eta+(\nabla_{a}\eta)^{\dagger}E_{i}^{a}\sigma^{i}\eta+2\eta^{\dagger}[E^{a},K_{a}]\eta\Big{]}\Bigg{)}$

Define $\xi_{A}=\sqrt[4]{q}\psi_{A}$ and $\nu_{A}=\sqrt[4]{q}\eta_{A}$ with
$A=\pm\frac{1}{2}$. Eq. (150) implies the following non-vanishing anti-Poisson
bracket,

$\displaystyle\\{\xi_{A}(x),\xi_{B}^{\dagger}(y)\\}_{+}=-i\delta_{AB}\delta(x,y)$
(151)
$\displaystyle\\{\nu_{A}(x),\nu_{B}^{\dagger}(y)\\}_{+}=-i\delta_{AB}\delta(x,y)$

For the gravitational parts, the action is

$\displaystyle
S_{H}=\frac{1}{\kappa}\int\differential^{4}x\,(E^{a}_{i}\mathcal{L}_{t}K_{a}^{i}+\frac{1}{2\sqrt{q}}NE^{a}_{i}E^{b}_{j}\Omega_{ab}^{ij}+(t\cdot\Gamma)_{m}\epsilon^{klm}K_{ak}E^{a}_{l}+2N^{b}E^{a}_{i}\nabla_{[a}K_{b]}^{i})$
(152)

Substituting the expression (150) and (152) into the total action
$S=S_{H}+S_{F}$, one can obtain the constraints governing the classical
dynamics which are expressed in terms of
$\nabla_{a}=\partial_{a}+\Gamma_{a}^{m}\tau_{m}$. Then taking advantage of
$A_{a}^{i}=\Gamma_{a}^{i}+\beta K_{a}^{i}$, one can simplify these constraints
in terms of the derivative $D_{a}=\partial_{a}+A_{a}^{i}\tau_{i}$. The results
are listed as follows. The total action reads

$S=S_{G}+S_{F}=\int\differential^{4}x(\text{simplytic structure terms
}-\lambda^{m}G_{m}-N^{a}H_{a}-NH).$ (153)

The Gaussian constraint is

$\displaystyle
G_{m}=\frac{1}{\kappa\beta}D_{a}E^{a}_{l}+\frac{1}{2}\sqrt{q}(\psi^{\dagger}\sigma_{m}\psi+\eta^{\dagger}\sigma_{m}\eta)$
(154)

The vector constraints is

$\displaystyle
H_{a}=\frac{1}{\kappa\beta}E^{b}_{i}F^{i}_{ab}+\frac{i}{2}\sqrt{q}\Big{\\{}\psi^{\dagger}D_{a}\psi-(D_{a}\psi)^{\dagger}\psi+\eta^{\dagger}D_{a}\eta-(D_{a}\eta)^{\dagger}\eta\Big{\\}}+\beta
K_{a}^{m}G_{m}.$ (155)

The scalar constraint is

$\displaystyle H=$ $\displaystyle
H_{G}+\frac{1}{2}\Big{[}i(\psi^{\dagger}E_{i}^{a}\sigma^{i}D_{a}\psi-(D_{a}\psi)^{\dagger}E_{i}^{a}\sigma^{i}\psi)-\beta
E^{a}_{i}K_{a}^{i}\psi^{\dagger}\psi-\frac{1}{\beta}(1+\beta^{2})D_{a}E^{a}_{i}\psi^{\dagger}\sigma^{i}\psi-\beta\frac{1}{\sqrt{q}}E^{a}_{i}D_{a}\Big{(}\sqrt{q}\psi^{\dagger}\sigma^{i}\psi\Big{)}$
(156)
$\displaystyle-i(\eta^{\dagger}E_{i}^{a}\sigma^{i}D_{a}\eta-(\nabla_{a}\eta)^{\dagger}E_{i}^{a}\sigma^{i}\eta)+\beta
E^{a}_{i}K_{a}^{i}\eta^{\dagger}\eta-\frac{1}{\beta}(1+\beta^{2})D_{a}E^{a}_{i}\eta^{\dagger}\sigma^{i}\eta-\beta\frac{1}{\sqrt{q}}E^{a}_{i}D_{a}\Big{(}\sqrt{q}\eta^{\dagger}\sigma^{i}\eta\Big{)}\Big{]}$

where $H_{G}$ denote the scalar constraint of pure gravity

$H_{G}=\frac{1}{2\kappa\sqrt{q}}E^{a}_{i}E^{b}_{j}\left(F_{ab}^{m}\epsilon_{m}{}^{ij}-2(1+\beta^{2})K_{[a}^{i}K_{b]}^{j}\right).$
(157)

##  Appendix B graded vector space and graded algebra

We follow the notions given in kostant1975graded . A vector space $V$ over
$\mathbb{R}$ or $\mathbb{C}$ is graded (over $\mathbb{Z}_{2}$) if there are
fixed subspaces $V_{0}$ and $V_{1}$ such that $V=V_{0}\oplus V_{1}$. An
element $v\in V$ is homogeneous if $v$ is either in $V_{0}$ or in $V_{i}$. For
all $v\in V_{i}$ with $v\neq 0$, we define their degree as

${\mathfrak{d}}(v)=i.$ (158)

Given two graded vector space $V$ and $W$, the space ${\rm Hom}(V,W)$ of
homomorphism from $V$ to $W$ is graded. An element $\alpha\in{\rm Hom}(V,W)$
is said to be homogeneous and of ${\mathfrak{d}}(\alpha)$ provided

$\alpha[V_{i}]\subset W_{i+{\mathfrak{d}}(\alpha)\,{\rm mod}\,2}.$ (159)

with $\alpha[V_{i}]$ denotes the image of $\alpha$ acting on $V_{i}$.

An algebra $(A,\cdot)$ is a graded algebra if $A$ is a graded vector space and
$A_{i}\cdot A_{j}\subset A_{i+j\,\mathrm{mod}\,2}$ where $A_{i}\cdot A_{j}$
denotes the space of elements $a_{i}\cdot a_{j}$ for all $a_{i}\in A_{i}$ and
$a_{j}\in A_{j}$. A graded algebra $A$ is a graded commutative algebra if the
product satisfies

$x\cdot y=(-1)^{{\mathfrak{d}}(x){\mathfrak{d}}(y)}y\cdot x$ (160)

where $x,y\in A$ are homogeneous. Any commutative algebra $\mathbb{A}$ is a
graded commutative algebra with the grade $\mathbb{A}_{1}=\mathbb{A}$ and
$\mathbb{A}_{0}=\\{0\\}$. An example of the graded commutative algebra is the
exterior algebra of some finite vector space $V$, i.e.

$A=\mathbb{R}\oplus V\oplus(V\wedge V)\oplus(V\wedge V\wedge
V)\oplus\cdots\oplus\bigwedge^{n}V.$ (161)

$A$ is graded as

$A_{0}=\bigoplus_{k=0}\bigwedge^{2k}V,\
A_{0}=\bigoplus_{k=0}\bigwedge^{2k+1}V.$ (162)

A graded algebra $(\mathfrak{a},[\cdot,\cdot])$ is a graded Lie algebra if the
Lie bracket satisfies

  * (1)

$[x,y]=(-1)^{1+{\mathfrak{d}}(x){\mathfrak{d}}(y)}[y,x]$;

  * (2)

$(-1)^{{\mathfrak{d}}(x){\mathfrak{d}}(z)}[[x,y],z]+(-1)^{{\mathfrak{d}}(y){\mathfrak{d}}(x)}[[y,z],x]+(-1)^{{\mathfrak{d}}(z){\mathfrak{d}}(y)}[[z,x],y]=0$.

An operation $\partial$ on a graded algebra $A$ is called a derivative if it
satisfies

$\partial(xy)=(\partial
x)y+(-1)^{{\mathfrak{d}}(\partial){\mathfrak{d}}(x)}x(\partial y)$ (163)

where ${\mathfrak{d}}(\partial)$ is defined by thinking of it as a
homomorphism on $A$. It can be checked that the operator $[x,\cdot]$ on a
graded Lie algebra $A$ for all $x\in A$ is a derivative.

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