# Positive mass gap of quantum Yang-Mills Fields

Adrian P. C. Lim
Email<EMAIL_ADDRESS>
###### Abstract

We construct a 4-dimensional quantum field theory on a Hilbert space,
dependent on a simple Lie Algebra of a compact Lie group, that satisfies
Wightman’s axioms. This Hilbert space can be written as a countable sum of
non-separable Hilbert spaces, each indexed by a non-trivial, inequivalent
irreducible representation of the Lie Algebra.

In each component Hilbert space, a state is given by a triple, a space-like
rectangular surface $S$ in $\mathbb{R}^{4}$, a measurable section of the Lie
Algebra bundle over this surface $S$, represented irreducibly as a matrix, and
a Minkowski frame. The inner product is associated with the area of the
surface $S$.

In our previous work, we constructed a Yang-Mills measure for a compact semi-
simple gauge group. We will use a Yang-Mills path integral to quantize the
momentum and energy in this theory. During the quantization process,
renormalization techniques and asymptotic freedom will be used. Each component
Hilbert space is the eigenspace for the momentum operator and Hamiltonian, and
the corresponding Hamiltonian eigenvalue is given by the quadratic Casimir
operator. The eigenvalue of the corresponding momentum operator will be shown
to be strictly less than the eigenvalue of the Hamiltonian, hence showing the
existence of a positive mass gap in each component Hilbert space. We will
further show that the infimum of the set containing positive mass gaps, each
indexed by an irreducible representation, is strictly positive.

In the last section, we will show how the positive mass gap will imply the
Clustering Theorem.

MSC 2020: 81T13, 81T08, 81T70
Keywords: Mass gap, Yang-Mills, Wightman’s axioms, compact simple Lie group,
renormalization, asymptotic freedom, clustering, space-like surface,
time-like surface, Casimir operator, Lorentz transformation, ${\rm
SL}(2,{{\mathbb{C}}})$,
Callan-Symanzik Equation, path integral, spinor representation

###### Contents

  1. 1 Preliminaries
    1. 1.1 Why should one read this article
  2. 2 A Description of Quantum Hilbert space
    1. 2.1 Time-like and space-like surfaces
    2. 2.2 Unitary representation of inhomogeneous ${\rm SL}(2,{{\mathbb{C}}})$
  3. 3 Quantum Field Operators
    1. 3.1 Creation operators
    2. 3.2 Domain and continuity
    3. 3.3 Cyclicity
    4. 3.4 Tempered Distribution
  4. 4 Transformation Law of the Field Operator
  5. 5 Causality
    1. 5.1 CPT Theorem
  6. 6 Yang-Mills path integrals
    1. 6.1 Hermite polynomials
    2. 6.2 Yang-Mills measure
    3. 6.3 Asymptotic freedom
    4. 6.4 Callan-Symanzik beta function
  7. 7 Hamiltonian and Momentum operator
    1. 7.1 Renormalization
    2. 7.2 Callan-Symanzik Equation
    3. 7.3 Existence of positive mass gap
  8. 8 Clustering
    1. 8.1 Vacuum Expectation
    2. 8.2 Proof of Cluster Decomposition Property
  9. A Surface Integrals
  10. B Lorentz transformation of a space-like vector

##  1 Preliminaries

Let $M$ be a 4-manifold, with $\Lambda^{q}(T^{\ast}M)$ being the $q$-th
exterior power of the cotangent bundle over the manifold $M$. Fix a Riemannian
metric $g$ on $M$ and this in turn defines an inner product
$\langle\cdot,\cdot\rangle_{q}$ on $\Lambda^{q}(T^{\ast}M)$, for which we can
define a volume form $d\omega$ on $M$. This allows us to define a Hodge star
operator $\ast$ acting on $k$-forms,
$\ast:\Lambda^{k}(T^{\ast}M)\rightarrow\Lambda^{4-k}(T^{\ast}M)$ such that for
$u,v\in\Lambda^{k}(T^{\ast}M)$, we have

$u\wedge\ast v=\langle u,v\rangle_{k}\ d\omega.$ (1.1)

An inner product on the set of smooth sections
$\Gamma(\Lambda^{k}(T^{\ast}M))$ is then defined as

$\langle u,v\rangle=\int_{M}u\wedge\ast v=\int_{M}\langle u,v\rangle_{k}\
d\omega.$ (1.2)

See [1].

Introduce a compact and simple gauge group $G$. Without loss of generality we
will assume that $G$ is a Lie subgroup of ${\rm U}(\bar{N})$,
$\bar{N}\in\mathbb{N}$. We will identify the (real) Lie Algebra $\mathfrak{g}$
of $G$ with a Lie subalgebra of the Lie Algebra $\mathfrak{u}(\bar{N})$ of
${\rm U}(\bar{N})$ throughout this article. Suppose we write the trace as
${{\rm{Tr}}}_{{\rm Mat}(\bar{N},\mathbb{C})}$, which we will abbreviate as
${{\rm{Tr}}}$ in future. Then we can define a positive, non-degenerate
bilinear form by

$\langle A,B\rangle=-{{\rm{Tr}}}_{{\rm Mat}(\bar{N},\mathbb{C})}[AB]$ (1.3)

for $A,B\in\mathfrak{g}$. Its Lie bracket will be denoted by
$[A,B]\equiv{{\rm{ad}}}(A)B$.

Let $P\rightarrow M$ be some trivial vector bundle, with structure group $G$.
The vector space of all smooth $\mathfrak{g}$-valued 1-forms on the manifold
$M$ will be denoted by $\mathcal{A}_{M,\mathfrak{g}}$. Denote the group of all
smooth $G$-valued mappings on $M$ by $\mathcal{G}$, called the gauge group.
The gauge group induces a gauge transformation on
$\mathcal{A}_{M,\mathfrak{g}}$,
$\mathcal{A}_{M,\mathfrak{g}}\times\mathcal{G}\rightarrow\mathcal{A}_{M,\mathfrak{g}}$
given by

$A\cdot\Omega:=A^{\Omega}=\Omega^{-1}d\Omega+\Omega^{-1}A\Omega$

for $A\in\mathcal{A}_{M,\mathfrak{g}}$, $\Omega\in\mathcal{G}$. The orbit of
an element $A\in\mathcal{A}_{M,\mathfrak{g}}$ under this operation will be
denoted by $[A]$ and the set of all orbits by
$\mathcal{A}_{M,\mathfrak{g}}/\mathcal{G}$.

For $A\in\mathcal{A}_{M,\mathfrak{g}}$, the curvature $dA+A\wedge A$ is a
smooth $\mathfrak{g}$-valued 2-form on $M$, whereby $dA$ is the differential
of $A$ and $A\wedge A$ is computed using the Lie Bracket of $\mathfrak{g}$ and
the wedge product on $\Lambda^{1}(T^{\ast}M)$, at each fibre of the tensor
bundle $\Lambda^{1}(T^{\ast}M)\otimes(M\times\mathfrak{g}\rightarrow M)$. The
Yang-Mills Lagrangian is given by

$S_{{\rm YM}}(A)=\int_{M}\left|dA+A\wedge A\right|^{2}\ d\omega.$

Here, the induced norm $|\cdot|$ is from the tensor product of
$\langle\cdot,\cdot\rangle_{2}$ and the inner product on $\mathfrak{g}$,
computed on each fiber of the bundle
$\Lambda^{2}(T^{\ast}M)\otimes(M\times\mathfrak{g}\rightarrow M)$. The
integral over $M$ is then defined using Equation (1.2). Note that this
Lagrangian is invariant under gauge transformations.

The 4-manifold we will consider in this article is
${{\mathbb{R}}}\times{{\mathbb{R}}}^{3}\equiv{{\mathbb{R}}}^{4}$, with tangent
bundle $T{{\mathbb{R}}}^{4}$. Note that ${{\mathbb{R}}}$ will be referred to
as the time-axis and ${{\mathbb{R}}}^{3}$ is the spatial 3-dimensional
Euclidean space. We will choose the standard Riemannian metric on
$\mathbb{R}^{4}$, denoted as $\langle\cdot,\cdot\rangle$. Let
$\\{e_{a}\\}_{a=0}^{3}$ be an orthonormal basis on
${{\mathbb{R}}}^{4}\equiv{{\mathbb{R}}}\times{{\mathbb{R}}}^{3}$, hence
defining the standard coordinates, $\vec{x}\equiv(x^{0},x^{1},x^{2},x^{3})$,
with time coordinate $x^{0}$ and spatial coordinates $(x^{1},x^{2},x^{3})$.

Let $\Lambda^{q}({{\mathbb{R}}}^{4})$ denote the fiber of the $q$-th exterior
power of the cotangent bundle over ${{\mathbb{R}}}^{4}$, and we choose the
canonical basis $\\{dx^{0},dx^{1},dx^{2},dx^{3}\\}$ for
$\Lambda^{1}({{\mathbb{R}}}^{4})$. Let $\Lambda^{1}({{\mathbb{R}}}^{3})$
denote the subspace in $\Lambda^{1}({{\mathbb{R}}}^{4})$ spanned by
$\\{dx^{1},dx^{2},dx^{3}\\}$. There is an obvious inner product defined on
$\Lambda^{1}({{\mathbb{R}}}^{4})$, i.e. $\langle dx^{a},dx^{b}\rangle=0$ if
$a\neq b$, 1 otherwise. Finally, a basis for $\Lambda^{2}({{\mathbb{R}}}^{4})$
is given by

$\\{dx^{0}\wedge dx^{1},dx^{0}\wedge dx^{2},dx^{0}\wedge dx^{3},dx^{1}\wedge
dx^{2},dx^{3}\wedge dx^{1},dx^{2}\wedge dx^{3}\\}.$

Using the volume form $d\omega=dx^{0}\wedge dx^{1}\wedge dx^{2}\wedge dx^{3}$,
the Hodge star operator $\ast$ is a linear isomorphism between
$\Lambda^{2}({{\mathbb{R}}}^{4})$ and $\Lambda^{2}({{\mathbb{R}}}^{4})$, i.e.

$\displaystyle\ast(dx^{0}\wedge dx^{1})=dx^{2}\wedge
dx^{3},\quad\ast(dx^{0}\wedge dx^{2})=dx^{3}\wedge
dx^{1},\quad\ast(dx^{0}\wedge dx^{3})=dx^{1}\wedge dx^{2}.$

We adopt Einstein’s summation convention, i.e. we sum over repeated
superscripts and subscripts. We set the speed of light $\mathbf{c}=1$. We can
define the Minkowski metric, given by

$\vec{x}\cdot\vec{y}=-x^{0}y^{0}+\sum_{i=1}^{3}x^{i}y^{i}.$ (1.4)

Note that our Minkowski metric is negative of the one used by physicists. A
vector $\vec{x}$ is time-like (space-like) if $\vec{x}\cdot\vec{x}<0$
($\vec{x}\cdot\vec{x}>0$). It is null if $\vec{x}\cdot\vec{x}=0$. When
$\vec{x}$ and $\vec{y}$ are space-like separated, it means that

$-(x^{0}-y^{0})^{2}+\sum_{i=1}^{3}(x^{i}-y^{i})^{2}\geq 0.$

A Lorentz transformation $\Lambda$ is a linear transformation mapping space-
time ${{\mathbb{R}}}^{4}$ onto itself, which preserves the Minkowski metric
given in Equation (1.4). Indeed, the Lorentz transformations form a group,
referred to as Lorentz group $L$. It has 4 components, and we will call the
component containing the identity, as the restricted Lorentz group, denoted
$L_{+}^{\uparrow}$.

###  1.1 Why should one read this article

At the time of this writing, an online search will reveal that several authors
have attempted to solve the Yang-Mills mass gap problem, which is one of the
millennium problems, as described in [2]. The problem is to construct a
Hilbert space satisfying Wightman’s axioms for a compact, simple Yang-Mills
gauge theory in 4-dimensional Euclidean space. See [3] for a complete
description of the axioms. The axioms are also stated in [4]. Furthermore,
this theory has a minimum positive mass gap. This means that besides the zero
eigenvalue, the Hamiltonian has a minimum positive eigenvalue. The momentum
operator is also a non-negative operator, and the difference between the
squares of their eigenvalues, is known as the mass gap squared. Despite all
these attempts, no solution has been widely accepted by the scientific
community. So, why should one spend time reading this article?

It is known that the set of inequivalent, non-trivial, irreducible
representation $\\{\rho_{n}:n\in\mathbb{N}\\}$ of a simple Lie Algebra
$\mathfrak{g}$ is indexed by highest weights, hence countable. See [5]. Using
all the inequivalent, non-trivial, irreducible representations of
$\mathfrak{g}$, we can construct a Hilbert space

$\mathbb{H}_{{\rm YM}}(\mathfrak{g})=\\{1\\}\oplus\bigoplus_{n\geq
1}\mathscr{H}(\rho_{n}),$

for which Wightman’s axioms are satisfied. The vacuum state will be denoted by
1 and $\\{1\\}$ will denote its linear span. The Hilbert space
$\mathscr{H}(\rho_{n})$ is defined using the non-trivial irreducible
representation $\rho_{n}$.

Experiments have shown that quantum fields are highly singular. Hence, we need
to smear the field, which we will represent it as a
$\rho(\mathfrak{g})$-valued vector field, over a rectangular surface. This
will define a state in $\mathscr{H}(\rho)$.

Each state in $\mathscr{H}(\rho)$ is described by a space-like rectangular
surface $S$ equipped with a Minkowski frame, and a measurable section of
$S\times[\rho(\mathfrak{g})\otimes{{\mathbb{C}}}]\rightarrow S$ defined on it.
See Definition 2.5. This is where the geometry comes in, as surfaces have a
well-defined physical quantity, which is the area. Incidentally, when we
compute the average flux of non-abelian Yang-Mills gauge fields through a
surface in [6], we will obtain a formula for the area of the surface. Further
justification for its connection with the Yang-Mills action will be given in
Remark 3.21.

But area is not invariant under the action of the Poincare group. To define an
unitary action by the Poincare group, we will consider time to be purely
imaginary and hence define a physical quantity on the surface $S$, associated
with the area of the surface. See Definition A.2.

We can now allow a test function to act on this ‘smeared’ field. This test
function will be represented as a field operator, to be defined later in
Section 3, and it acts on a dense subset in the Hilbert space
$\mathbb{H}_{{\rm YM}}(\mathfrak{g})$, as required in Wightman axioms.

A non-abelian Yang-Mills measure was constructed in [6]. Using a Yang-Mills
path integral, we will proceed in Section 7 to quantize momentum, which will
yield its quantum eigenvalues, associated with the quadratic Casimir operator.
A path integral approach to quantize the Yang-Mills theory was described in
[7]. During the quantization process, we will use renormalization techniques
and asympotic freedom. Incidentally, $\mathscr{H}(\rho_{n})$ is the eigenspace
for both the Hamiltonian and quantized momentum operator. We will now list
down the following reasons why we think this is a correct approach to prove
the existence of a mass gap.

The quadratic Casimir operator is dependent on the non-trivial irreducible
representation of the simple Lie Algebra $\mathfrak{g}$. It is a positive
operator and proportional to the identity. If our gauge group is ${\rm
SU}(2)$, then the quadratic Casimir operator is given by $j(j+1)$, for a given
representation $\rho:\mathfrak{su}(2)\rightarrow{\rm
End}({{\mathbb{C}}}^{2j+1})$, $j$ is a non-negative half integer or integer.
Its square root is well known to be the eigenvalue of total momentum in
quantum mechanics.

The square of the Hamiltonian will be defined later to be proportional to the
dimension of the representation times the quadratic Casimir operator. Hence,
we will correlate the Casimir operator directly with the energy levels
squared. Large values of the Casimir operator mean high energy levels. The
momentum eigenvalues will be computed via a path integral. The set containing
the eigenvalues will be discrete, and because the eigenvalues of both
operators go to infinity, both are unbounded. See subsection 7.1. By comparing
the eigenvalues of the squares of the Hamiltonian and the quantized momentum
operator, we will prove the existence of a mass gap. See subsection 7.3. Note
that it is not enough to just show that the Hamiltonian has a strictly
positive minimum eigenvalue, besides the zero eigenvalue. Our construction
will also show that the the vacuum state is an eigenstate of the Hamiltonian
and quantized momentum operator with eigenvalue 0 respectively, implying the
vacuum state is massless.

A successful quantum Yang-Mills theory should explain the following:

1.

There is a mass gap, which will explain why the strong force is short range.

2.

The theory should incorporate asymptotic freedom, i.e. at high energies and
short distances, the theory is like a free theory.

In the case of gauge group ${\rm SU}(3)$, which describes the strong force, it
must also explain the following:

3.

The theory should demonstrate quark confinement, i.e. the potential between a
quark and anti-quark grows linearly;

4.

The theory should incorporate chiral symmetry breaking, which means that the
vacuum is potentially invariant only under a certain subgroup of the full
symmetry group that acts on the quark fields.

Item 2 was proved in a prequel [6], but we only showed asymptotic freedom in
the context for short distances. In that prequel, we derived the Wilson Area
Law formula of a time-like surface $S$ using a (non-abelian) Yang-Mills path
integral. This will show quark confinement and that the potential energy
between quarks grows linearly, which is Item 3. We also like to mention that
the Area Law formula does not hold in the abelian gauge group case, as was
shown in [8].

In this article, our main goal will be to prove Item 1, and also show that
asymptotic freedom holds at high energies. Item 4 can be demonstrated in
Example 3.17. If one assumes that asymptotic freedom holds for non-abelian
simple compact gauge group, then a positive mass gap is implied, which we will
furnish the details later in this article. On pages 541-543 in [9], and also
in [10], the authors gave a qualitative explanation of asymptotic freedom,
using gauge group ${\rm SU}(2)$. See also [11].

The weak interaction is described by a ${\rm SU}(2)$ gauge theory. An
unification of the weak and electromagnetic interaction is described by the
electroweak theory, which is ${\rm SU}(2)\times{\rm U}(1)$ gauge theory, first
put forth by Sheldon Glashow in 1961, then completed later by Abdus Salam and
Steven Weinberg in 1967. Mathematically formulated as Yang-Mills fields, the
gauge bosons in this theory have to be massless. But experimentally, it was
shown that the gauge bosons responsible for weak interaction are massive,
hence short range; whereas the photons, responsible for electromagnetic
interaction, are massless, hence long range. To resolve this issue, the Higgs
mechanism was introduced. See page 330 in [12]. But to date, there is no
experimental evidence to suggest that gluons have any physical mass, even
though strong interaction is short range.

A successful quantum Yang-Mills theory that satisfies Wightman’s axioms and
exhibits a positive mass gap, will then imply and explain the short range
nature of the weak and strong interaction, without assuming the existence of a
physical non-zero mass of these gauge bosons. This will be mathematically
formulated later as the Clustering Theorem in Section 8.

This article is focused on a construction of a 4-dimensional Yang-Mills
quantum field theory that satisfies Wightman’s axioms, without assuming the
existence of a positive mass. However, the proof that a positive mass gap
exists, requires a construction of a Yang-Mills path integral, which is
detailed in [8, 6]. An abelian Yang-Mills path integral was constructed in
[8]. The former details the construction of an infinite dimensional Gaussian
probability space using Abstract Wiener space formalism, developed by Gross.
See [13]. The construction of an abelian Yang-Mills path integral will then
allow us to construct a non-abelian Yang-Mills path integral in [6]. The
latter is instrumental in proving the mass gap.

These two prequals are technical in nature. By removing all the technical
aspects of the construction, we hope that this article, will be more palatable
to both physicists and mathematicians, who only want to understand a
construction of a 4-dimensional quantum field, satisfying Wightman’s axioms.

To further convince the reader that the construction of a 4-dimensional
quantum field theory is correct, we will proceed to prove the Clustering
Theorem, that will imply that the vacuum expectation has an exponential decay
along a space-like separation, which implies that the force represented by the
Lie Algebra $\mathfrak{g}$ is short ranged in nature. Because our construction
of a 4-dimensional quantum theory for Yang-Mills fields satisfies a modified
version of Wightman’s axioms, we will give an alternative proof of the
Clustering Theorem, that fits into our context.

##  2 A Description of Quantum Hilbert space

To construct our Hilbert space, we need a compact Lie group $G$, with its real
simple Lie Algebra $\mathfrak{g}$, of which $\\{E^{\alpha}\\}_{\alpha=1}^{N}$
is an orthonormal basis using the inner product defined in Equation (1.3),
fixed throughout this article.

Extend the inner product defined in Equation (1.3) to be a sesquilinear
complex inner product, over the complexification of $\mathfrak{g}$, denoted as
$\mathfrak{g}_{{{\mathbb{C}}}}\equiv\mathfrak{g}\otimes_{{{\mathbb{R}}}}{{\mathbb{C}}}$.
Hence, it is linear in the first variable, conjugate linear in the second.

A finite dimensional representation of $\mathfrak{g}$ is a Lie Algebra
homomorphism $\rho$ of $\mathfrak{g}$ into ${\rm
End}({{\mathbb{C}}}^{\tilde{N}})$ or into ${\rm End}(V)$ for some complex
vector space $V$. All our representations will be considered to be non-trivial
and irreducible. The dimension of the representation is given by $\tilde{N}$.
Because $G$ is a compact Lie group, every finite dimensional representation of
$G$ is equivalent to a unitary representation. See Theorems 9.4 and 9.5 in
[5].

Hence, we can always assume that our Lie Algebra representation
$\rho:\mathfrak{g}\rightarrow{\rm End}({{\mathbb{C}}}^{\tilde{N}})$ is
represented as skew-Hermitian matrices. Thus, the eigenvalues of $\rho(E)$
will be purely imaginary, $E\in\mathfrak{g}$.

Suppose each non-trivial irreducible representation
$\rho_{n}:\mathfrak{g}\rightarrow{\rm End}({{\mathbb{C}}}^{\tilde{N}_{n}})$ is
indexed by $n$, whereby $n\in\mathbb{N}$ and no two representations are
equivalent. We will order them as described in subsection 7.3.

Introduce state 1, which will be referred to as the vacuum state, and
$\\{1\\}$ refers to the linear span of $1$, over the complex numbers. Let
$\mathscr{H}(\rho_{0}):=\\{1\\}$, which is an one dimensional complex inner
product space, with a complex sesquilinear inner product defined by $\langle
1,1\rangle=1$.

Define

$\mathbb{H}_{{\rm
YM}}(\mathfrak{g}):=\bigoplus_{n=0}^{\infty}\mathscr{H}(\rho_{n}),$ (2.1)

whereby $\rho_{n}:\mathfrak{g}\rightarrow{\rm
End}({{\mathbb{C}}}^{\tilde{N}_{n}})$, $n\geq 1$. The inner product defined on
this direct sum is given by

$\left\langle\sum_{n=0}^{\infty}v_{n},\sum_{n=0}^{\infty}u_{n}\right\rangle:=\sum_{n=0}^{\infty}\langle
v_{n},u_{n}\rangle,$

whereby $\langle v_{n},u_{n}\rangle$ is the inner product defined on
$\mathscr{H}(\rho_{n})$.

We will now give the full description of each Hilbert space
$\mathscr{H}(\rho)$, $\rho$ is an irreducible representation. Later, we will
see that $\mathscr{H}(\rho)$ is an eigenspace for the momentum operator and
Hamiltonian.

###  2.1 Time-like and space-like surfaces

###### Notation 2.1

We will let $I=[0,1]$ be the unit interval, and $I^{2}\equiv I\times I$.
Denote $\hat{s}=(s,\bar{s}),\ \hat{t}=(t,\bar{t})\in I^{2}$, $d\hat{s}\equiv
dsd\bar{s}$, $d\hat{t}\equiv dtd\bar{t}$. Typically, $s,\bar{s},t,\bar{t}$
will be reserved as the variable for some parametrization, i.e. $\rho:s\in
I\mapsto\rho(s)\in{{\mathbb{R}}}^{4}$.

###### Definition 2.2

(Time-like and space-like)
Let $S$ be a bounded rectangular surface in ${{\mathbb{R}}}^{4}$, contained in
some plane. By rotating the spatial axes if necessary, we may assume without
any loss of generality, a parametrization of $S$ is given by

$\left\\{(a^{0}+sb^{0},a^{1},a^{2}+sb^{2},a^{3}+tb^{3})^{T}\in{{\mathbb{R}}}^{4}:\
s,t\in I\right\\},$ (2.2)

for constants $a^{\alpha},b^{\alpha}\in{{\mathbb{R}}}$. Now, the surface $S$
is spanned by two directional vectors $(b^{0},0,b^{2},0)^{T}$ and
$(0,0,0,b^{3})^{T}$. Note that $(b^{0},0,b^{2},0)^{T}$ lie in the
$x^{0}-x^{2}$ plane and is orthogonal to $(0,0,0,b^{3})^{T}$.

We say a rectangular surface is space-like, if $|b^{0}|<|b^{2}|$, i.e. the
acute angle which the vector $(b^{0},0,b^{2},0)^{T}$ makes with the
$x^{2}$-axis in the $x^{0}-x^{2}$-plane is less than $\pi/4$.

We say a rectangular surface is time-like, if $|b^{0}|>|b^{2}|$, i.e. the
acute angle which the vector $(b^{0},0,b^{2},0)^{T}$ makes with the
$x^{0}$-axis in the $x^{0}-x^{2}$-plane is less than $\pi/4$.

Let $S$ be a rectangular surface in ${{\mathbb{R}}}^{4}$ contained in some
plane, and $TS$ denote the set of directional vectors that lie inside $S$.

Write $\vec{v}=(v^{0},v)\equiv(v^{0},v^{1},v^{2},v^{3})^{T}\in TS$, and define
$|v|^{2}=v^{1,2}+v^{2,2}+v^{3,2}$. An equivalent way to say that $S$ is time-
like is

$\inf_{\vec{0}\neq\vec{v}\in TS}\frac{|v|^{2}}{v^{0,2}}<1.$ (2.3)

And we say that $S$ is space-like if

$\inf_{\vec{0}\neq\vec{v}\in TS}\frac{|v|^{2}}{v^{0,2}}>1.$ (2.4)

###### Remark 2.3

  1. 1.

By definition, a time-like surface must contain a time-like directional vector
in it. Since under Lorentz transformation, a time-like vector remains time-
like, we see that a time-like surface remains time-like under Lorentz
transformation. Similarly, a surface is space-like means all its directional
vectors in the surface are space-like. Under Lorentz transformation, all its
directional vectors spanning $S$ remain space-like, hence a space-like surface
remains space-like under Lorentz transformation.

  2. 2.

By a boost, any space-like rectangular surface contained in a plane can be
transformed into a surface lying strictly inside
$\\{c\\}\times{{\mathbb{R}}}^{3}$, for some constant $c$.

  3. 3.

Recall $e_{0}$ spans the time-axis. By a boost, any time-like rectangular
surface contained in a plane can be transformed into a surface which is
spanned by $e_{0}$ and an orthogonal directional vector
$v\in{{\mathbb{R}}}^{3}$.

  4. 4.

Any two distinct points on a space-like surface is space-like separated; but
two distinct points on a time-like surface may not be time-like separated.

When we say surface $S$ in this article, we mean it is a countable, disjoint
union of rectangular surfaces in ${{\mathbb{R}}}^{4}$, which are space-like,
containing none, some or all of its boundary points.

###### Definition 2.4

(Surface)
Any surface $S\equiv\\{S_{u}\\}_{u\geq 1}\subset{{\mathbb{R}}}^{4}$ satisfies
the following conditions:

  * •

each component $S_{u}$ is a space-like rectangular surface contained in some
plane;

  * •

each connected component $S_{u}$ may contain none, some or all of its
boundary;

  * •

$S_{u}\cap S_{v}=\emptyset$ if $u\neq v$;

  * •

$S_{u}$ is contained in some bounded set in ${{\mathbb{R}}}^{4}$.

###### Definition 2.5

Let $S_{0}$ be a compact rectangular space-like surface inside the
$x^{2}-x^{3}$ plane. From Equation (2.2), we see that any rectangular space-
like surface $S$ contained in a plane, can be transformed to $S_{0}$ by
Lorentz transformations and translation.

Recall $\\{e_{a}\\}_{a=0}^{3}$ is an orthonormal basis on
${{\mathbb{R}}}^{4}$. We say that $\\{\hat{f}_{a}\\}_{a=0}^{3}$ is a Minkowski
frame for a compact space-like surface $S$ contained in some plane, if there
exists a sequence of Lorentz transformations $\Lambda_{1},\cdots,\Lambda_{n}$
and a translation by $\vec{a}\in{{\mathbb{R}}}^{4}$, such that

  * •

$S=\Lambda_{n}\cdots\Lambda_{1}S_{0}+\vec{a}$;

  * •

$\hat{f}_{a}=\Lambda_{n}\cdots\Lambda_{1}e_{a}\in{{\mathbb{R}}}^{4}$,
$a=0,\cdots,3$.

###### Remark 2.6

Observe that $\hat{f}_{0}$ is time-like and for $i=1,2,3$, $\hat{f}_{i}$ is
space-like, satisfying the following properties:

  * •

$\hat{f}_{a}\cdot\hat{f}_{b}=0$ if $a\neq b$; and

  * •

$\hat{f}_{0}\cdot\hat{f}_{0}=-1$ and $\hat{f}_{i}\cdot\hat{f}_{i}=1$.

Note that $\\{\hat{f}_{2},\hat{f}_{3}\\}$ spans $S$. Later we will let
$S^{\flat}$ be a time-like plane, spanned by $\\{\hat{f}_{0},\hat{f}_{1}\\}$.
Clearly, $\\{\hat{f}_{a}\\}_{a=0}^{3}$ is a basis on ${{\mathbb{R}}}^{4}$.

Each component Hilbert space $\mathscr{H}(\rho)$ will consists of vectors of
the form
$\sum_{u=1}^{\infty}\left(S_{u},f_{\alpha}^{u}\otimes\rho(E^{\alpha}),\\{\hat{f}_{a}^{u}\\}_{a=0}^{3}\right)$,
whereby $S=\bigcup_{u=1}^{\infty}S_{u}$ is some countable union of compact,
rectangular surfaces in ${{\mathbb{R}}}^{4}$, $f_{\alpha}^{u}$ will be some
(measurable) complex-valued function, which is defined on the surface $S_{u}$.
And $\\{\hat{f}_{a}^{u}\\}_{a=0}^{3}$ is a Minkowski frame for each compact
rectangular surface $S_{u}$ contained in a plane as described in Definition
2.5.

Let $\sigma:[0,1]\times[0,1]\rightarrow{{\mathbb{R}}}^{4}$ be a
parametrization of a compact $S$. The complex-valued function $f_{\alpha}$ is
measurable on $S$, if
$f_{\alpha}\circ\sigma:[0,1]^{2}\rightarrow{{\mathbb{C}}}$ is measurable.

Each of these surfaces is assumed to be space-like, as defined above. We sum
over repeated index $\alpha$, from $\alpha=1$ to $N$. One should think of
$f_{\alpha}^{u}\otimes\rho(E^{\alpha})$ as a section of the vector bundle
$S_{u}\times\rho(\mathfrak{g})_{{\mathbb{C}}}\rightarrow S_{u}$, defined over
the surface $S_{u}$. In terms of the parametrization $\sigma$, the section at
$\sigma(\hat{s})$ is given by
$f_{\alpha}^{u}(\sigma(\hat{s}))\otimes\rho(E^{\alpha})$.

###### Remark 2.7

Another way to view this vector
$\left(S_{u},f_{\alpha}^{u}\otimes\rho(E^{\alpha}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)$
is via taking its Fourier Transform into energy-momentum space. We will refer
the reader to Section 8.

Let $S$ and $\tilde{S}$ be rectangular space-like surfaces contained in a
plane. Given complex scalars $\lambda$ and $\mu$, we define the addition and
scalar multiplication as

$\displaystyle\lambda$
$\displaystyle\left(S,f_{\alpha}\otimes\rho(E^{\alpha}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)+\mu\left(\tilde{S},g_{\alpha}\otimes\rho(E^{\alpha}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)$
$\displaystyle:=\left(S\cup\tilde{S},\left(\lambda\tilde{f}_{\alpha}+\mu\tilde{g}_{\alpha}\right)\otimes\rho(E^{\alpha}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right).$
(2.5)

Here, we extend $f_{\alpha}$ to be
$\tilde{f}_{\alpha}:S\cup\tilde{S}\rightarrow{{\mathbb{C}}}$ by
$\tilde{f}_{\alpha}(p)=f_{\alpha}(p)$ if $p\in S$; $\tilde{f}(p)=0$ otherwise.
Similarly, $\tilde{g}_{\alpha}$ is an extension of $g_{\alpha}$, defined as
$\tilde{g}_{\alpha}(p)=g_{\alpha}(p)$, if $p\in\tilde{S}$,
$\tilde{g}_{\alpha}(p)=0$ otherwise.

###### Remark 2.8

For the above addition to hold, we require that the Minkowski frame
$\\{\hat{f}_{a}\\}_{a=0}^{3}$ on $S$ and $\tilde{S}$ to be identical.

Given 2 surfaces, $S$ and $\tilde{S}$, we need to take the intersection and
union of these surfaces. Now, the union of these 2 surfaces can always be
written as a disjoint union of connected sets, each such set is a space-like
surface, containing none, some or all of its boundary points. However, the
intersection may not be a surface. For example, the two surfaces may intersect
to give a curve. In such a case, we will take the intersection to be the empty
set $\emptyset$.

Given a surface $S$, let $\sigma$ be any parametrization of $S$. We can define
$\int_{S}d\rho$ using this parametrization $\sigma$ as given in Definition
A.1. Now, replace $\sigma\equiv(\sigma_{0},\sigma_{1},\sigma_{2},\sigma_{3})$
with $\acute{\sigma}=(i\sigma_{0},\sigma_{1},\sigma_{2},\sigma_{3})$ and hence
define $\int_{S}d|\acute{\rho}|$ as given in Definition A.2. With this, define
the following inner product on $\mathscr{H}(\rho)$.

###### Definition 2.9

Given a surface $S=\bigcup_{u=1}^{n}S_{u}$ equipped with a collection of
frames $\\{\hat{f}_{a}^{u}:a=0,\cdots,3\\}_{u\geq 1}$, and a set of bounded
and continuous complex-valued functions
$\\{f_{\alpha}^{u}:\alpha=1,\cdots,N\\}_{u\geq 1}$ for $u\geq 1$, defined on
$S$, form a vector
$\sum_{u=1}^{n}\left(S_{u},f_{\alpha}^{u}\otimes\rho(E^{\alpha})\\{\hat{f}_{a}^{u}\\}_{a=0}^{3}\right)$,
also referred to as Yang-Mills field. Note that for each $u$,
$f_{\alpha}^{u}\otimes\rho(E^{\alpha})$ is a section of
$S_{u}\times[\rho(\mathfrak{g})\otimes{{\mathbb{C}}}]\rightarrow S_{u}$, with
$S_{u}$ contained in some plane, equipped with a Minkowski frame
$\\{\hat{f}_{a}^{u}\\}_{a=0}^{3}$.

Let $V$ be a (complex) vector space containing such vectors, with addition and
scalar multiplication defined in Equation (2.5). The zero vector can be
written as $\left(S,0,\\{\hat{f}_{a}\\}_{a=0}^{3}\right)$ for any space-like
rectangular surface $S$ contained in a plane, equipped with any suitable
Minkowski frame $\\{\hat{f}_{a}\\}_{a=0}^{3}$.

Refer to Definition A.2. Assume that $S$ and $\tilde{S}$ be space-like
surfaces, contained in a plane. Define an inner product
$\langle\cdot,\cdot\rangle$ for
$\left(S,f_{\alpha}\otimes\rho(E^{\alpha}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)\in
V$, and
$\left(\tilde{S},g_{\beta}\otimes\rho(E^{\beta}),\\{\hat{g}_{a}\\}_{a=0}^{3}\right)\in
V$, given by

$\displaystyle\left\langle\left(S,f_{\alpha}\otimes\rho(E^{\alpha}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right),\left(\tilde{S},g_{\beta}\otimes\rho(E^{\beta}),\\{\hat{g}_{a}\\}_{a=0}^{3}\right)\right\rangle$
$\displaystyle:=\int_{S\cap\tilde{S}}[f_{\alpha}\overline{g_{\beta}}]\cdot
d|\acute{\rho}|\cdot{{\rm{Tr}}}[-\rho(E^{\alpha})\rho(E^{\beta})]$ (2.6)
$\displaystyle\equiv\sum_{\alpha=1}^{N}C(\rho)\int_{I^{2}}[f_{\alpha}\cdot\overline{g_{\alpha}}](\sigma(\hat{s}))\left|\sum_{0\leq
a<b\leq
3}\acute{\rho}_{\sigma}^{ab}(\hat{s})\left[\det\acute{J}_{ab}^{\sigma}(\hat{s})\right]\right|d\hat{s},$

provided $\hat{f}_{a}=\hat{g}_{a}$, for $a=0,\ldots,3$. Otherwise, it is
defined as zero. Here, $\sigma:I^{2}\rightarrow{{\mathbb{R}}}^{4}$ is a
parametrization of $S\cap\tilde{S}$ and $C(\rho)$ is defined later in Notation
7.1.

Denote its norm by $|\cdot|$. Let $\mathscr{H}(\rho)$ denote the Hilbert space
containing $V$.

###### Remark 2.10

The quantity $\int_{S}d\rho$ was first derived in [8], which motivates the
definition of this quantity $\int_{S}d|\acute{\rho}|$. It was obtained from
computing an abelian Yang-Mills path integral, by computing the average square
of the flux of a Yang-Mills gauge field, over a time-like or space-like
surface, using an infinite-dimensional Gaussian measure.

This quantity can also be derived from the Wilson Area Law formula in [6],
computed using a non-abelian gauge group. The reader will later observe that
this quantity $\int_{S}d|\acute{\rho}|$, plays an important role in all the 4
Wightman’s axioms.

Using this quantity is one reason why we call the states as Yang-Mills fields,
as the above inner product has its origins from a Yang-Mills path integral.
There is also a second reason for calling them Yang-Mills fields. Refer to
Remark 3.21.

###### Proposition 2.11

The Hilbert space $\mathscr{H}(\rho)$ is non-separable.

Proof. Consider a compact rectangular surface $S_{0}$ contained in the
$x^{2}-x^{3}$ plane, with $\\{e_{a}\\}_{a=0}^{3}$ as its Minkowski frame.
Then, we see that

$\left\\{\left(S_{0}+\vec{a},\rho(E^{\alpha}),\\{e_{a}\\}_{a=0}^{3}\right):\
\vec{a}\in\mathbb{R}\right\\}$

is an uncountable set of orthogonal vectors in $\mathscr{H}(\rho)$, since

$\left\langle\Big{(}S_{0}+\vec{a},\rho(E^{\alpha}),\\{e_{a}\\}_{a=0}^{3}\Big{)},\left(S_{0}+\vec{b},\rho(E^{\alpha}),\\{e_{a}\\}_{a=0}^{3}\right)\right\rangle=0,$

if $\vec{a}\neq\vec{b}$. Hence the Hilbert space is non-separable.

###  2.2 Unitary representation of inhomogeneous ${\rm SL}(2,{{\mathbb{C}}})$

Given a continuous group acting on ${{\mathbb{R}}}^{4}$, we can consider its
corresponding inhomogeneous group, whose elements are pairs consisting of a
translation and a homogeneous transformation. For example, the Poincare group
$\mathscr{P}$ containing the Lorentz group $L$, will have elements
$\\{\vec{a},\Lambda\\}$, where $\Lambda\in L$ and $\vec{a}$ will represent
translation in the direction $\vec{a}$. The multiplication law for the
Poincare group is given by

$\\{\vec{a}_{1},A_{1}\\}\\{\vec{a}_{2},A_{2}\\}=\\{\vec{a}_{1}+A_{1}\vec{a}_{2},A_{1}A_{2}\\}.$

Associated with the restricted Lorentz group $L_{+}^{\uparrow}$ is the group
of $2\times 2$ complex matrices of determinant one, denoted by ${\rm
SL}(2,{{\mathbb{C}}})$. There is an onto homomorphism $Y:{\rm
SL}(2,{{\mathbb{C}}})\rightarrow L_{+}^{\uparrow}$. Thus, given
$\Lambda\in{\rm SL}(2,{{\mathbb{C}}})$, $Y(\Lambda)\in L_{+}^{\uparrow}\subset
L$. See [3].

Instead of the Poincare group, we can consider the inhomogeneous ${\rm
SL}(2,{{\mathbb{C}}})$ in its place, which we will also denote by ${\rm
SL}(2,{{\mathbb{C}}})$, and use it to construct unitary representations. Its
elements will consist of $\\{\vec{a},\Lambda\\}$ and its multiplication law is
given by

$\\{\vec{a}_{1},\Lambda_{1}\\}\\{\vec{a}_{2},\Lambda_{2}\\}=\\{\vec{a}_{1}+Y(\Lambda_{1})\vec{a}_{2},\Lambda_{1}\Lambda_{2}\\}.$

By abuse of notation, for any $\Lambda\in{\rm SL}(2,{{\mathbb{C}}})$, we will
write $\Lambda\vec{a}$ to mean $\Lambda$ being represented as a $4\times 4$
matrix, acting on $\vec{a}\in{{\mathbb{R}}}^{4}$. This means we will write
$\Lambda_{1}\vec{a}_{2}\equiv Y(\Lambda_{1})\vec{a}_{2}$.

Given a vector $\vec{x}\in{{\mathbb{R}}}^{4}$, $\\{\vec{a},\Lambda\\}$ acts on
$\vec{x}$ by $\vec{x}\mapsto\Lambda\vec{x}+\vec{a}$. By abuse of notation, for
a surface $S$, $\\{\vec{a},\Lambda\\}$ acts on $S$ by $S\mapsto\Lambda
S+\vec{a}$, which means apply a Lorentz transformation $Y(\Lambda)$ to every
position vector on the surface $S$, followed by translation in the direction
$\vec{a}$. In terms of some parametrization
$\sigma:I^{2}\rightarrow{{\mathbb{R}}}^{4}$ for $S$, the surface $\Lambda
S+\vec{a}$ is parametrized by
$\Lambda\sigma+\vec{a}:I^{2}\rightarrow{{\mathbb{R}}}^{4}$.

In general, the only finite dimensional unitary representation of ${\rm
SL}(2,{{\mathbb{C}}})$ is the trivial representation. See Theorem 16.2 in
[14]. Thus, to construct a unitary representation, we must consider an
infinite dimensional space. See [15].

###### Definition 2.12

(Unitary Representation of the inhomogeneous ${\rm SL}(2,{{\mathbb{C}}})$)
Let $\hat{H}(\rho)$, $\hat{P}(\rho)$ be positive numbers, dependent on the
representation $\rho$, to be defined later in Definition 7.13. Let $\Lambda$
be in ${\rm SL}(2,{{\mathbb{C}}})$.

There is an unitary representation of the inhomogeneous ${\rm
SL}(2,{{\mathbb{C}}})$, $\\{\vec{a},\Lambda\\}\mapsto U(\vec{a},\Lambda)$.
Now, $U(\vec{a},\Lambda)$ acts on the Hilbert space $\mathscr{H}(\rho)$,

$\left(S,f_{\alpha}\otimes\rho(E^{\alpha}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)\longmapsto
U(\vec{a},\Lambda)\left(S,f_{\alpha}\otimes\rho(E^{\alpha}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right),$

by

$\displaystyle U$
$\displaystyle(\vec{a},\Lambda)\left(S,f_{\alpha}\otimes\rho(E^{\alpha}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)$
$\displaystyle:=\left(\Lambda
S+\vec{a},e^{-i[\vec{a}\cdot(\hat{H}(\rho_{n})\Lambda\hat{f}_{0}+\hat{P}(\rho_{n})\Lambda\hat{f}_{1})]}f_{\alpha}(\Lambda^{-1}(\cdot-\vec{a}))\otimes\rho(E^{\alpha}),\\{\Lambda\hat{f}_{a}\\}_{a=0}^{3}\right).$
(2.7)

Here, $S$ is a space-like surface contained in some plane.

###### Remark 2.13

  1. 1.

Notice that $U(\vec{a},\Lambda)$ acts trivially on $\rho(\mathfrak{g})$. In
classical Yang-Mills equation, the fields over ${{\mathbb{R}}}^{4}$, are
$\mathfrak{g}$-valued. The Lie group $G$ describes the internal symmetry, on
each fiber of the vector bundle. See [16]. When we apply a Lorentz
transformation, we expect that $G$ remains invariant under Lorentz
transformation. After all, $G$ acts fiberwise on the vector bundle over
${{\mathbb{R}}}^{4}$.

  2. 2.

Let us explain the formula on the RHS of Equation (2.7). Suppose
$\sigma:I^{2}\rightarrow S$ is a parametrization for $S$. Then
$\Lambda\sigma+\vec{a}\equiv Y(\Lambda)\sigma+\vec{a}$ will be a
parametrization for $Y(\Lambda)S+\vec{a}$. And, the field at the point
$\vec{x}:=Y(\Lambda)\sigma(\hat{s})+\vec{a}\in Y(\Lambda)S+\vec{a}$, is given
by

$\displaystyle
e^{-i[\vec{a}\cdot(\hat{H}(\rho_{n})Y(\Lambda)\hat{f}_{0}+\hat{P}(\rho_{n})Y(\Lambda)\hat{f}_{1})]}$
$\displaystyle
f_{\alpha}[Y(\Lambda^{-1})(\vec{x}-\vec{a})]\otimes\rho(E^{\alpha})$
$\displaystyle\equiv
e^{-i[\vec{a}\cdot(\hat{H}(\rho_{n})Y(\Lambda)\hat{f}_{0}+\hat{P}(\rho_{n})Y(\Lambda)\hat{f}_{1})]}f_{\alpha}[\sigma(\hat{s})]\otimes\rho(E^{\alpha}).$

  3. 3.

When there is no translation, the vector field over $Y(\Lambda)S$ is the
pushforward of the vector field $f_{\alpha}\otimes\rho(E^{\alpha})$ over $S$.

  4. 4.

When there is only translation, the unitary operator can be simplified to be

$\displaystyle U(\vec{a},1)$
$\displaystyle\left(S,f_{\alpha}\otimes\rho(E^{\alpha}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)$
$\displaystyle\equiv$ $\displaystyle
e^{-i[\vec{a}\cdot(\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1})]}\left(S+\vec{a},f_{\alpha}(\cdot-\vec{a})\otimes\rho\left(E^{\alpha}\right),\\{\hat{f}_{a}\\}_{a=0}^{3}\right),$

directly from Equation (2.7).

It is straightforward to check that this map $\\{\vec{a},\Lambda\\}\mapsto
U(\vec{a},\Lambda)$ is a representation, to be left to the reader.

###### Lemma 2.14

The map $U(\vec{a},\Lambda)$ defined on $\mathscr{H}(\rho)$ is unitary, using
the inner product $\langle\cdot,\cdot\rangle$ as defined in Definition 2.9.

Proof. It suffices to show for $S$ and $\tilde{S}$, both contained in some
space-like plane respectively. Let $\\{\hat{f}_{a}\\}_{a=0}^{3}$,
$\\{\hat{g}_{a}\\}_{a=0}^{3}$ be as defined in Definition 2.5 for $\Lambda S$
and $\Lambda\tilde{S}$ respectively.

By Definition 2.9, it suffices to prove when $\Lambda S\cap\Lambda\tilde{S}$
has non-zero area. We only consider the case when $\hat{f}_{a}=\hat{g}_{a}$,
$a=0,\cdots,3$, since the result is trivial otherwise. Thus,

$\displaystyle\left\langle
U(\vec{a},\Lambda)\left(S,f_{\alpha}\otimes\rho(E^{\alpha}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right),U(\vec{a},\Lambda)\left(\tilde{S},g_{\beta}\otimes\rho(E^{\beta}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)\right\rangle$
$\displaystyle:=\int_{[\Lambda S+\vec{a}]\cap[\Lambda\tilde{S}+\vec{a}]}\
d|\acute{\rho}|\
e^{-i[\vec{a}\cdot(\hat{H}(\rho_{n})\Lambda\hat{f}_{0}+\hat{P}(\rho_{n})\Lambda\hat{f}_{1})]}e^{i[\vec{a}\cdot(\hat{H}(\rho_{n})\Lambda\hat{f}_{0}+\hat{P}(\rho_{n})\Lambda\hat{f}_{1})]}$
$\displaystyle\hskip
156.49014pt\times[f_{\alpha}\overline{g_{\beta}}](\Lambda^{-1}(\cdot-\vec{a}))\cdot{{\rm{Tr}}}[-\rho(E^{\alpha})\rho(E^{\beta})]$
$\displaystyle=\int_{\Lambda(S\cap\tilde{S})+\vec{a}}[f_{\alpha}\overline{g_{\beta}}](\Lambda^{-1}(\cdot-\vec{a}))\cdot
d|\acute{\rho}|\cdot{{\rm{Tr}}}[-\rho(E^{\alpha})\rho(E^{\beta})]$
$\displaystyle=\int_{S\cap\tilde{S}}[f_{\alpha}\overline{g_{\beta}}](\cdot)\cdot
d|\acute{\rho}|\cdot{{\rm{Tr}}}[-\rho(E^{\alpha})\rho(E^{\beta})],$

after applying Lemma A.3.

###### Remark 2.15

In general, the directional derivative for
$\left(S,f_{\alpha}\otimes\rho(E^{\alpha}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)$
does not exist for arbitrary direction $\vec{a}$. Thus, in the computation of
the generator for translation, the derivative does not appear.

The author in [17] also talked about problems when taking the derivative of a
quantum field, at short distances. The differences in fields at different
spatial points actually diverges as the separation gets smaller. The fields
become infinitely rough at small distance scales and it means that it is
impossible experimentally to probe the field at a single point.

We have just described the Hilbert space $\mathscr{H}(\rho)$. Now let us focus
on the vacuum state, which we denoted it as 1. Recall our Yang-Mills fields
are described by a triple
$\sum_{u=1}^{\infty}\left(S_{u},f_{\alpha}^{u}\otimes\rho(E^{\alpha}),\\{\hat{f}_{a}^{u}\\}_{a=0}^{3}\right)$,
whereby $S_{u}$ is some non-empty surface.

###### Definition 2.16

The vacuum state 1 is synonymous with the empty set $\emptyset$, i.e.
$1\equiv(\emptyset)$. We define $\langle 1,1\rangle:=1$.

###### Remark 2.17

  1. 1.

On the empty set, it does not make sense to have a measurable function or
Minkowski frame defined on it.

  2. 2.

Recall that $\\{1\\}$ is a one-dimensional subspace spanned by the vacuum
state. The Lie Algebra $\mathfrak{g}$ acts trivially on $\\{1\\}$, via the
trivial representation $\rho_{0}:\mathfrak{g}\rightarrow{{\mathbb{C}}}$. Refer
to Equation (2.1).

Clearly the empty set is invariant under the Poincare action. Hence, it is
invariant under $U(\vec{a},\Lambda)$.

Given a Schwartz function $g$ on ${{\mathbb{R}}}^{4}$, we can define a quantum
field operator $\phi^{\alpha,n}(g)$, acting on the vacuum state and
$\sum_{u=1}^{\infty}\left(S,f_{\alpha}^{u}\otimes\rho_{n}(E^{\alpha}),\\{\hat{f}_{a}^{u}\\}_{a=0}^{3}\right)$
in Section 3. We will show later that $\phi^{\alpha,n}(g)$ is densely defined
on $\mathbb{H}_{{\rm YM}}(\mathfrak{g})$. Furthermore,
$\\{\phi^{\alpha,n}(g):\ 1\leq\alpha\leq\underline{N}\\}$ defines a spinor of
dimension $\underline{N}$. Using this definition and the definition of unitary
transformation, one can prove the transformation law given in Proposition 4.1,
and causality in Section 5 using this action.

###### Remark 2.18

In [4], the function $g$ is interpreted as an observable, which is elevated to
be a quantum field operator $\phi^{\alpha,n}(g)$. One can also understand it
as applying canonical quantization to a classical field $g$.

In our description of the Hilbert space containing Yang-Mills fields, nowhere
did we use the Yang-Mills action, so it is not clear if $\mathbb{H}_{{\rm
YM}}(\mathfrak{g})$ is a Hilbert space for a quantum Yang-Mills gauge theory.

The set $\\{\hat{H}(\rho_{n}),\hat{P}(\rho_{n})\\}_{n\geq 1}$, indexed by the
non-trivial irreducible, inequivalent representations, will give us a discrete
set of eigenvalues. The choices of $\hat{H}(\rho)$ and $\hat{P}(\rho)$ will be
given later, referred to as the eigenvalues of the Hamiltonian $\hat{H}$ and
momentum operator $\hat{P}$ respectively. Indeed, these discrete eigenvalues
will give us a countable spectrum for the translation operator $U(\vec{a},1)$,
provided $\vec{a}=a^{0}\hat{f}_{0}+a^{1}\hat{f}_{1}$. There are many choices
of $\\{\hat{H}(\rho),\hat{P}(\rho)\\}$ and it is not clear how we should
choose these numbers.

It is only in Wightman’s zeroth axiom, where the mass of the theory appears.
This axiom requires that $\hat{H}(\rho)^{2}-\hat{P}(\rho)^{2}=m(\rho)^{2}$,
for some mass gap $m(\rho)\geq 0$.111See Remark 8.3. Furthermore, these
eigenvalues are required to be unbounded. The mass gap problem is equivalent
to show that $m_{0}:=\inf_{n\in\mathbb{N}}m(\rho_{n})>0$, and this is only
true for a compact simple gauge group.

For each eigenstate in $\mathscr{H}(\rho_{n})$, $\hat{H}$ and $\hat{P}$ will
be multiplication by scalars $\hat{H}(\rho_{n})$ and $\hat{P}(\rho_{n})$
respectively. Then $\hat{H}^{2}-\hat{P}^{2}=m^{2}$ translates to

$\frac{\hat{P}(\rho_{n})^{2}}{\hat{H}(\rho_{n})^{2}}-1=-\frac{m(\rho_{n})^{2}}{\hat{H}(\rho_{n})^{2}}.$
(2.8)

To prove that the Hamiltonian and momentum operators are unbounded, and there
is a positive mass gap $m_{0}$, it suffices to show that for all $n\geq 1$,

$0>\frac{\hat{P}(\rho_{n})^{2}}{\hat{H}(\rho_{n})^{2}}-1\longrightarrow 0,$

as $n\rightarrow\infty$, and $\lim_{n\rightarrow\infty}m(\rho_{n})=\infty$.

To show that the operators are unbounded and the existence of a positive mass
gap, we need to make use of the Yang-Mills path integral to quantize, so that
we will obtain Equation (2.8). During the quantization process, we will use
renormalization techniques and asymptotic freedom. Note that asymptotic
freedom only holds for a non-abelian gauge group. The compact simple gauge
group will give us a quadratic Casimir operator, dependent on the
representation $\rho_{n}$. As the set containing all Casimir operators, each
corresponding to a non-equivalent irreducible representation of
$\mathfrak{g}$, is countably infinite and unbounded from above, the
Hamiltonian will be shown to be an unbounded operator.

The existence of a positive mass gap is a consequence of Equation (7.4), first
proved in [6]. To prove this equation, we need

  * •

renormalization techniques,

  * •

asymptotic freedom,

  * •

the compactness of the gauge group allows us to represent the Lie Algebra as
skew-Hermitian matrices,

  * •

the structure constants and the quadratic Casimir operator of the simple Lie
Algebra, and

  * •

the quartic term in the Yang-Mills action,

all of which are collectively responsible for the existence of a mass gap. We
also need to impose the Callan-Symanzik Equation to prove the existence of a
mass gap $m_{0}$.

###### Remark 2.19

The idea that the quartic term in the Yang-Mills action might be responsible
for the mass gap was suggested in [18].

Using a Yang-Mills path integral to define the Hamiltonian and momentum
operator eigenvalues justify our construction as a 4-dimensional Yang-Mills
quantum gauge theory. In Section 8, we will show how the positive mass gap,
will imply the Clustering Theorem.

As for the rest of the axioms, we see that it does not make use of the Yang-
Mills action. We will postpone the proof of the mass gap till Section 7. For
now, we will move on to the remaining Wightman’s axioms.

##  3 Quantum Field Operators

We will now begin our discussion on the field operators that act on the
Hilbert space $\bigoplus_{n=0}^{\infty}\mathscr{H}(\rho_{n})$, which contains
our Yang-Mills fields.

Now, every irreducible finite dimensional representation of ${\rm
SL}(2,{{\mathbb{C}}})$ is denoted by $D^{(j,k)}$, $j,k$ are non-negative
integers or half integers. This representation is known as the spinor
representation of ${\rm SL}(2,{{\mathbb{C}}})$. Under this representation, one
sees that $D^{(j,k)}(-1)=(-1)^{2(j+k)}$. Incidently, when restricted to ${\rm
SU}(2)$, $\Lambda\in{\rm SU}(2)\mapsto D^{(j,0)}(\Lambda)$ is equivalent to an
irreducible representation of ${\rm SU}(2)$. See [3].

Recall we have a Minkowski frame $\\{\hat{f}_{a}\\}_{a=0}^{3}$ defined on a
rectangular space-like surface $S$ in Definition 2.5. Now,
$\hat{f}_{a}=\Lambda e_{a}\equiv Y(\Lambda)e_{a}$ for some $\Lambda\in{\rm
SL}(2,{{\mathbb{C}}})$. Suppose we have another Minkowski frame
$\\{\hat{g}_{a}\\}_{a=0}^{3}$ on $S$ such that $\hat{f}_{a}=\hat{g}_{a}$ for
all $a$. Then, we have that $Y(\Lambda)e_{a}=Y(\tilde{\Lambda})e_{a}$ for some
$\tilde{\Lambda}\in{\rm SL}(2,{{\mathbb{C}}})$. Hence,
$Y(\Lambda)=Y(\tilde{\Lambda})$, which can be shown that
$\Lambda=\pm\tilde{\Lambda}$. When $j+k$ is an integer, we see that
$D^{(j,k)}(\pm 1)=1$, thus $D^{(j,k)}$ is a representation describing vector
bosons.

###### Definition 3.1

(Test functions)
We let $\mathscr{P}$ denote the Schwartz space consisting of infinitely
differentiable complex-valued functions on ${{\mathbb{R}}}^{4}$, which
converge to 0 at infinity faster than any powers of $|\vec{x}|$. We will refer
$f\in\mathscr{P}$ as a test function, which is bounded.

###### Notation 3.2

For $\vec{k}=(k^{0},k^{1},k^{2},k^{3})$, $k^{a}\in\\{0\\}\cup\mathbb{N}$, we
will write

$D^{\vec{k}}=\left(\frac{\partial}{\partial
x^{0}}\right)^{k^{0}}\left(\frac{\partial}{\partial
x^{1}}\right)^{k^{1}}\left(\frac{\partial}{\partial
x^{2}}\right)^{k^{2}}\left(\frac{\partial}{\partial
x^{3}}\right)^{k^{3}},\quad\vec{x}^{\vec{k}}=(x^{0})^{k^{0}}(x^{1})^{k^{1}}(x^{2})^{k^{2}}(x^{3})^{k^{3}}.$

And $|\vec{k}|=\sum_{a=0}^{3}|k^{a}|$.

###### Definition 3.3

(Norm on $\mathscr{P}$)
Let $r,s$ be whole numbers. Suppose $f\in\mathscr{P}$. With the above
notation, define a norm $\parallel\cdot\parallel_{r,s}$ on $\mathscr{P}$ as

$\parallel f\parallel_{r,s}:=\sum_{|\vec{k}|\leq r}\sum_{|\vec{l}|\leq
s}\sup_{\vec{x}\in{{\mathbb{R}}}^{4}}|\vec{x}^{\vec{k}}D^{\vec{l}}f(\vec{x})|.$

###  3.1 Creation operators

###### Definition 3.4

(Time-like plane)
Refer to Definition 2.5. Let $S$ be a connected space-like rectangular surface
contained in some plane.

Let $S^{\flat}$ be a time-like plane spanned by
$\\{\hat{f}_{0},\hat{f}_{1}\\}$, parametrized by

$\hat{s}=(s,\bar{s})\mapsto\sigma(s,\bar{s})=s\hat{f}_{0}+\bar{s}\hat{f}_{1},\
s,\bar{s}\in{{\mathbb{R}}}.$ (3.1)

And we will write $d\hat{s}=dsd\bar{s}$.

Refer to Definition A.2. Define using a Minkowski metric,
$\tilde{\eta}:\vec{v}\in{{\mathbb{R}}}^{4}\mapsto\vec{v}\cdot(\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1})\in{{\mathbb{R}}}$.
Suppose we are given a $\tilde{f}\in\mathscr{P}$. We will define a new
function
$\tilde{f}^{\\{\hat{f}_{0},\hat{f}_{1}\\}}:{{\mathbb{R}}}^{4}\rightarrow{{\mathbb{C}}}$
by

$\displaystyle\vec{x}\in{{\mathbb{R}}}^{4}$
$\displaystyle\longmapsto\tilde{f}^{\\{\hat{f}_{0},\hat{f}_{1}\\}}(\hat{H}(\rho_{n}),\hat{P}(\rho_{n}))(\vec{x}):=\int_{S^{\flat}}\frac{e^{-i\tilde{\eta}(\cdot)}}{2\pi}\tilde{f}(\vec{x}+\cdot)\
d|\acute{\rho}|$
$\displaystyle=\int_{\hat{s}\in{{\mathbb{R}}}^{2}}\frac{e^{-i[\sigma(\hat{s})\cdot(\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1})]}}{2\pi}\tilde{f}\left(\vec{x}+\sigma(\hat{s})\right)\cdot|\acute{\rho}_{\sigma}|(\hat{s})\
d\hat{s},$ (3.2)

integration over a time-like plane $S^{\flat}$, using the parametrization
given in Equation (3.1), for representation $\rho_{n}$.

###### Remark 3.5

  1. 1.

Note that $\tilde{f}^{\\{\hat{f}_{0},\hat{f}_{1}\\}}\notin\mathscr{P}$, unless
$\tilde{f}\equiv 0$.

  2. 2.

Even though we compute the integral using a given parametrization $\sigma$, it
is actually independent of the parametrization and only depends on the time-
like plane span by $\\{\hat{f}_{0},\hat{f}_{1}\\}$. Because we are doing a
Fourier Transform on time-like and one space-like variable, evaluated at
$\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1}$, we see that the
transformed function depends on $\\{\hat{f}_{0},\hat{f}_{1}\\}$, not just
$S^{\flat}$.

  3. 3.

If $\vec{x}=\sum_{a=0}^{3}x^{a}\hat{f}_{a}$, then

$\displaystyle\tilde{f}^{\\{\hat{f}_{0},\hat{f}_{1}\\}}$
$\displaystyle(\hat{H}(\rho_{n}),\hat{P}(\rho_{n}))(\vec{x})$
$\displaystyle=e^{-i[x^{0}\hat{H}(\rho_{n})-x^{1}\hat{P}(\rho_{n})]}\tilde{f}^{\\{\hat{f}_{0},\hat{f}_{1}\\}}(\hat{H}(\rho_{n}),\hat{P}(\rho_{n}))(x^{3}\hat{f}_{3}+x^{3}\hat{f}_{3}).$

Hence, $\hat{H}(\rho_{n})$ ($\hat{P}(\rho_{n})$) is the generator for
translation, in the $\hat{f}_{0}$ ($\hat{f}_{1}$) direction.

###### Definition 3.6

Let
$\mathcal{A}:=\\{F^{\alpha}:1\leq\alpha\leq\underline{N}\\}\subset\mathfrak{g}$
be a finite set, and define recursively for $n\geq 2$,
$\mathcal{A}^{n}:=[\mathcal{A}^{n-1},\mathcal{A}]$,
$\mathcal{A}^{1}:=\mathcal{A}$, such that $\mathfrak{g}$ can be spanned by
$\bigcup_{j=1}^{\underline{n}}\mathcal{A}^{j}$ for some $\underline{n}\geq 1$.

Given $\tilde{f}\in\mathscr{P}$, we now wish to describe the field operator
$\phi^{\alpha,n}(\tilde{f})$, $1\leq\alpha\leq\underline{N}$. Suppose we have
a spinor representation $A:{\rm SL}(2,{{\mathbb{C}}})\rightarrow{\rm
End}({{\mathbb{C}}}^{\underline{N}})$. For each $n\in\mathbb{N}$, the field
operator $\\{\phi^{\alpha,n}(\tilde{f}):\ 1\leq\alpha\leq\underline{N}\\}$
transforms like a spinor under the action of $A(\Lambda)$, for any
$\Lambda\in{\rm SL}(2,{{\mathbb{C}}})$. Hence, a Schwartz function $\tilde{f}$
will be ‘promoted’ to be some spinor
$\sum_{\alpha=1}^{\underline{N}}c_{\alpha}\phi^{\alpha,n}(\tilde{f})$. But
first, how does $\phi^{\alpha,n}(\tilde{f})$ act on $1$, for some Schwartz
function $\tilde{f}$?

###### Definition 3.7

(Creation operators)
Recall we indexed our non-trivial irreducible representation by a natural
number $n$. Fix a connected space-like plane $S_{0}\subset{{\mathbb{R}}}^{4}$.
We will choose $S_{0}$ to be the $x^{2}-x^{3}$ plane. Using spatial rotation,
we can rotate $S_{0}$ to be the $x^{i}-x^{j}$ plane, for $i,j=1,2,3$. Together
with translation and boost, we can transform any surface contained inside
$S_{0}$, to be any space-like surface, using the unitary representation of
$SL(2,{{\mathbb{C}}})$. By Definition 2.5, we will choose
$\\{e_{a}\\}_{a=0}^{3}$ to be a Minkowski frame for $S_{0}$.

For any $\tilde{f}\in\mathscr{P}$, we define an operator
$\phi^{\alpha,n}(\tilde{f})$, $\alpha=1,2,\cdots,\underline{N}$,
$n\in\mathbb{N}$, which acts on the vacuum state 1 by

$\displaystyle\phi^{\alpha,n}(\tilde{f})1:=$
$\displaystyle\left(S_{0},\tilde{f}^{\\{e_{0},e_{1}\\}}\otimes\rho_{n}(F^{\alpha}),\\{e_{a}\\}_{a=0}^{3}\right)$
$\displaystyle\equiv$
$\displaystyle\left(S_{0},\tilde{f}^{\\{e_{0},e_{1}\\}}(\hat{H}(\rho_{n}),\hat{P}(\rho_{n}))\otimes\rho_{n}(F^{\alpha}),\\{e_{a}\\}_{a=0}^{3}\right)\in\mathscr{H}(\rho_{n}),$

for $F^{\alpha}\in\mathcal{A}\subset\mathfrak{g}$.

In the notation above, it is understood that
$\tilde{f}^{\\{e_{0},e_{1}\\}}\equiv\tilde{f}^{\\{e_{0},e_{1}\\}}(\hat{H}(\rho_{n}),\hat{P}(\rho_{n}))$.
And we restrict the domain of
$\tilde{f}^{\\{e_{0},e_{1}\\}}(\hat{H}(\rho_{n}),\hat{P}(\rho_{n})):{{\mathbb{R}}}^{4}\rightarrow{{\mathbb{C}}}$
to be on the surface $S_{0}$.

###### Remark 3.8

The field operators can be indexed by a countably infinite set. See [19].

###  3.2 Domain and continuity

###### Definition 3.9

Let $\mathscr{S}$ denote the set consisting of countable union of space-like
rectangular surfaces in ${{\mathbb{R}}}^{4}$, each component surface is
compact or a plane. For $S\in\mathscr{S}$, let $\mathscr{P}_{S}$ denote the
set of Schwartz functions defined on $S$.

In particular, if $\sigma:I^{2}\rightarrow{{\mathbb{R}}}^{4}$ is a
parametrization of a compact space-like surface $S$, we say $f$ is a Schwartz
function on $S$ when $f\circ\sigma$ is an infinitely differentiable function
on $I^{2}$, and is compactly supported in the interior of $I^{2}$, i.e. it
decays to zero at the boundary.

If $S$ is a space-like plane, then for some parametrization
$\sigma:{{\mathbb{R}}}^{2}\rightarrow S$, we say $f$ is a Schwartz function on
$S$ when $f\circ\sigma$ is a Schwartz function on ${{\mathbb{R}}}^{2}$.

###### Definition 3.10

(Domain of field operators)
Define a domain
$\mathscr{D}\subset\bigoplus_{n=0}^{\infty}\mathscr{H}(\rho_{n})$ as

$\left\\{a_{0}1+\sum_{n,u=1}^{\infty}\left(S_{n,u},f_{n,\alpha}^{u}\otimes\rho_{n}(E^{\alpha}),\\{\hat{f}_{a}^{n,u}\\}_{a=0}^{3}\right):\
a_{0}\in{{\mathbb{C}}},\ f_{n,\alpha}^{u}\in\mathscr{P}_{S_{n,u}},\
S_{n,u}\in\mathscr{S}\right\\}.$

Given a parametrization $\sigma:I^{2}\rightarrow{{\mathbb{R}}}^{4}$ for a
surface $S$, we say that $f\in L^{2}(S)$ if $f$ is measurable on $S$ and

$\int_{I^{2}}|f\circ\sigma|^{2}(\hat{s})\left|\sum_{0\leq a<b\leq
3}\acute{\rho}_{\sigma}^{ab}(\hat{s})\left[\det\acute{J}_{ab}^{\sigma}(\hat{s})\right]\right|d\hat{s}<\infty.$

By construction of $\mathscr{H}(\rho)$, we only consider space-like surfaces,
equipped with a Minkowski frame. For any surface $S\in\mathscr{S}$,
$\mathscr{P}_{S}$ is dense inside $L^{2}(S)$. So we see that $\mathscr{D}$ is
actually a dense set inside $\mathbb{H}_{{\rm YM}}(\mathfrak{g})$, and it
contains the vacuum state.

###### Remark 3.11

In the proof of Proposition 3.20, we will show that $\mathscr{P}_{S}$ is dense
inside $L^{2}(S)$, when $S$ is compact.

From Equation (2.2), a compact space-like surface in the $x^{2}-x^{3}$ plane
can be transformed to any compact space-like surface under translation,
spatial rotation or boost. Hence, $\mathscr{S}$ remains invariant under the
action of ${\rm SL}(2,{{\mathbb{C}}})$. From Definition 2.5, a Minkowski frame
associated with a rectangular surface, is generated by Lorentz transformations
of $\\{e_{a}\\}_{a=0}^{3}$. Hence
$U(\vec{a},\Lambda)\mathscr{D}\subset\mathscr{D}$.

We can now define the field operator, whose domain is given by $\mathscr{D}$,
as follows. Recall from Definition 3.4, how we can define a new function
$\tilde{f}^{\\{\hat{f}_{0},\hat{f}_{1}\\}}$, from $\tilde{f}\in\mathscr{P}$,
using $\\{\hat{f}_{0},\hat{f}_{1}\\}$ contained in a Minkowski frame,
associated with a space-like surface $S$.

From the opening paragraph in Section 3, we saw that
$\\{\hat{f}_{a}\\}_{a=0}^{3}$ uniquely determines $\Lambda\in{\rm
SL}(2,{{\mathbb{C}}})$, up to $\pm 1$. When $j+k$ is an integer, we see that
$D^{(j,k)}(\pm\Lambda)=D^{(j,k)}(\Lambda)$.

###### Definition 3.12

Let
$\sum_{u=1}^{\infty}\left(S_{u},g_{\beta}^{u}\otimes\rho(E^{\beta}),\\{\hat{f}_{a}^{u}\\}_{a=0}^{3}\right)\in\mathscr{H}(\rho)$,
whereby each $S_{u}$ is a connected rectangular space-like surface contained
inside some plane, and $g_{\beta}^{u}\in\mathscr{P}_{S_{u}}$. Refer to
Definition 3.4. For each Minkowski frame $\\{\hat{f}_{a}^{u}\\}_{a=0}^{3}$,
let $\Lambda^{u}e_{a}=\hat{f}_{a}^{u}$ for some $\Lambda^{u}\in{\rm
SL}(2,{{\mathbb{C}}})$. The adjoint representation ${{\rm{ad}}}$ of
$\rho(\mathfrak{g})$ is defined as
${{\rm{ad}}}(\rho(E^{\alpha}))\rho(E^{\beta})=[\rho(E^{\alpha}),\rho(E^{\beta})]$.

Recall from Definition 3.6, we defined a set $\mathcal{A}\subset\mathfrak{g}$
with cardinality $\underline{N}$. Suppose we have a spinor representation
$A:{\rm SL}(2,{{\mathbb{C}}})\rightarrow{\rm
End}({{\mathbb{C}}}^{\underline{N}})$. This representation can be written as a
sum of irreducible representations of the form
$\bigoplus_{\alpha=1}^{m}D^{(j_{\alpha},k_{\alpha})}$,
$D^{(j_{\alpha},k_{\alpha})}$ is an irreducible representation of ${\rm
SL}(2,{{\mathbb{C}}})$ as described earlier. We will further assume that for
each $1\leq\alpha\leq m$, $j_{\alpha}+k_{\alpha}$ is an integer.

For $\Lambda\in{\rm SL}(2,{{\mathbb{C}}})$, let $A(\Lambda)_{\alpha}^{\beta}$
denote the entry at the $\beta$-th row, $\alpha$-th column. Given a test
function $f\in\mathscr{P}$, define a field operator $\phi^{\alpha,n}(f)$ as

$\displaystyle\phi^{\alpha,n}$
$\displaystyle(f)\sum_{u=1}^{\infty}\left(S_{u},g_{\beta}^{u}\otimes\rho_{m}(E^{\beta}),\\{\hat{f}_{a}^{u}\\}_{a=0}^{3}\right)$
$\displaystyle:=\left\\{\begin{array}[]{ll}\sum_{u=1}^{\infty}\left(S_{u},f^{\\{\hat{f}_{0}^{u},\hat{f}_{1}^{u}\\}}_{n}A(\Lambda^{u})_{\gamma}^{\alpha}\cdot
g_{\beta}^{u}\otimes\rho_{n}\left([F^{\gamma},E^{\beta}]\right),\\{\hat{f}_{a}^{u}\\}_{a=0}^{3}\right),&\hbox{$m=n$;}\\\
0,&\hbox{$m\neq n$.}\end{array}\right.$

There is an implied sum over $\gamma$ from 1 to $\underline{N}$, and over
$\beta$ from 1 to $N$. Here,
$f^{\\{\hat{f}_{0}^{u},\hat{f}_{1}^{u}\\}}_{n}\equiv
f^{\\{\hat{f}_{0}^{u},\hat{f}_{1}^{u}\\}}(\hat{H}(\rho_{n}),\hat{P}(\rho_{n}))$
and the restricted vector field
$f^{\\{\hat{f}_{0}^{u},\hat{f}_{1}^{u}\\}}_{n}\Big{|}_{S_{u}}A(\Lambda^{u})_{\gamma}^{\alpha}\otimes\rho_{n}(F^{\gamma})$
acts on $g_{\beta}^{u}\otimes\rho_{n}(E^{\beta})$, by
${{\rm{ad}}}\left(f^{\\{\hat{f}_{0}^{u},\hat{f}_{1}^{u}\\}}_{n}|_{S_{u}}A(\Lambda^{u})_{\gamma}^{\alpha}\otimes\rho_{n}(F^{\gamma})\right)$,
over the surface $S_{u}$ fiberwise, ${{\rm{ad}}}(\rho_{n}(E))$ is the adjoint
representation of $\rho_{n}(E)$.

###### Remark 3.13

  1. 1.

The alpha in $\phi^{\alpha,n}(f)$ is referred to as the spinor index. Indeed,
for each $n\in\mathbb{N}$ and $f\in\mathscr{P}$, we have that

$\\{\phi^{\alpha,n}(f):\ 1\leq\alpha\leq\underline{N}\\}$

is a spinor of dimension $\underline{N}$. If $A=D^{(s,0)}$, then we must have
that $s$ is an integer, i.e. the spinors are vector bosons in this case.

  2. 2.

By linearity, we define

$\displaystyle\phi^{\alpha,n}(f)\sum_{m=0}^{\infty}v_{m}$
$\displaystyle:=\sum_{m=0}^{\infty}\phi^{\alpha,n}(f)v_{m}$
$\displaystyle=a_{0}\Big{(}S_{0},f^{\\{e_{0},e_{1}\\}}_{n}\otimes\rho_{n}(E^{\alpha}),\\{e_{a}\\}_{a=0}^{3}\Big{)}+\phi^{\alpha,n}(f)v_{n},$

$v_{0}=a_{0}1$ is a scalar multiple of the vacuum state and
$v_{n}\in\mathscr{H}(\rho_{n})$. The domain for $\phi^{\alpha,n}(f)$ will be
$\mathscr{D}$. Note that it is a bounded operator.

  3. 3.

Suppose $g_{\beta}$ is measurable, or $L^{2}$ integrable on a space-like
rectangular surface $S$, contained in some plane. Since
$f^{\\{\hat{f}_{0},\hat{f}_{1}\\}}$ is bounded and continuous, we see that
$f^{\\{\hat{f}_{0},\hat{f}_{1}\\}}\cdot g_{\beta}$ is measurable and $L^{2}$
integrable. So,

$\displaystyle\phi^{\alpha,n}(f)$
$\displaystyle\left(S,g_{\beta}\otimes\rho_{n}(E^{\beta}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)$
$\displaystyle:=\left(S,f^{\\{\hat{f}_{0},\hat{f}_{1}\\}}_{n}\cdot
A(\Lambda)_{\gamma}^{\alpha}g_{\beta}\otimes\rho_{n}([F^{\gamma},E^{\alpha}]),\\{\hat{f}_{a}\\}_{a=0}^{3}\right),$

and $f^{\\{\hat{f}_{0},\hat{f}_{1}\\}}_{n}\cdot g_{\beta}$ is defined almost
everywhere on $S$. But we will run into problems later, when proving
Proposition 3.22. This is because multiplying a measurable function with a
tempered distribution, may not be a tempered distribution. It is thus
necessary to restrict the domain for the field operators to be on
$\mathscr{D}$, to avoid technical difficulties later on.

We showed how $\phi^{\alpha,n}(f)$ is defined on $\mathscr{D}$. We can now
define its adjoint.

###### Definition 3.14

(Annihilation operators)
Using the inner product in Definition 2.9, we define the adjoint
$\phi^{\alpha,n}(g)^{\ast}$ on a space-like surface $S$ contained in some
plane, as

$\displaystyle\phi^{\alpha,n}(g)^{\ast}$
$\displaystyle\left(S,f_{\beta}\otimes\rho_{n}(E^{\beta}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)$
$\displaystyle=$
$\displaystyle-\left(S,\overline{g^{\\{\hat{f}_{0},\hat{f}_{1}\\}}}\overline{A(\Lambda)_{\gamma}^{\alpha}}\cdot
f_{\beta}\otimes\rho_{n}([F^{\gamma},E^{\beta}]),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)$
$\displaystyle+\left\langle\left(S,f_{\beta}\otimes\rho_{n}(E^{\beta}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right),\phi^{\alpha,n}(g)1\right\rangle
1,$

whereby $\hat{f}_{a}=\Lambda e_{a}$ for $a=0,\cdots,3$, and we sum over
repeated indices $\gamma$ from 1 to $\underline{N}$, and over $\beta$ from 1
to $N$. This is because

$\left\langle{{\rm{ad}}}(\rho(E^{\alpha}))\rho(E^{\beta}),\rho(E^{\gamma})\right\rangle=-\left\langle\rho(E^{\beta}),{{\rm{ad}}}(\rho(E^{\alpha}))\rho(E^{\gamma})\right\rangle.$

And

  * •

$\phi^{\alpha,n}(g)^{\ast}$ will send
$\left(S,f_{\beta}\otimes\rho_{m}(E^{\beta}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)$
to 0 if $m\neq n$;

  * •

$\phi^{\alpha,n}(g)^{\ast}1=0$.

###### Remark 3.15

Clearly, we can choose the domain to be $\mathscr{D}$.

###### Example 3.16

Consider the Lie group ${\rm SU}(2)$. Its Lie algebra can be generated by the
three Pauli matrices. Suppose $\mathcal{A}$ as described in Definition 3.6 is
linearly independent. Hence, $2\leq|\mathcal{A}|\leq 3$. If $A=D^{(j,k)}$,
then $j=1$, $k=0$, since $j+k$ must be an integer. Thus, the Pauli matrices,
which represent $W^{\pm}$ and $Z$ bosons responsible for weak force
interactions, transform like spin 1 vector bosons.

###### Example 3.17

Consider the Lie group ${\rm SU}(3)$. Its Lie algebra can be generated by the
Gell-Mann matrices

$\displaystyle\lambda_{1}=$ $\displaystyle\left(\begin{array}[]{ccc}0&\ 1&\
0\\\ -1&\ 0&\ 0\\\ 0&\ 0&\ 0\\\
\end{array}\right),\quad\lambda_{2}=\left(\begin{array}[]{ccc}0&\ i&\ 0\\\ i&\
0&\ 0\\\ 0&\ 0&\ 0\\\
\end{array}\right),\quad\lambda_{3}=\left(\begin{array}[]{ccc}i&\ 0&\ 0\\\ 0&\
-i&\ 0\\\ 0&\ 0&\ 0\\\ \end{array}\right),$ $\displaystyle\lambda_{4}=$
$\displaystyle\left(\begin{array}[]{ccc}0&\ 0&\ i\\\ 0&\ 0&\ 0\\\ i&\ 0&\ 0\\\
\end{array}\right),\quad\lambda_{5}=\left(\begin{array}[]{ccc}0&\ 0&\ 1\\\ 0&\
0&\ 0\\\ -1&\ 0&\ 0\\\ \end{array}\right),$ $\displaystyle\lambda_{6}=$
$\displaystyle\left(\begin{array}[]{ccc}0&\ 0&\ 0\\\ 0&\ 0&\ i\\\ 0&\ i&\ 0\\\
\end{array}\right),\quad\lambda_{7}=\left(\begin{array}[]{ccc}0&\ 0&\ 0\\\ 0&\
0&\ 1\\\ 0&\ -1&\ 0\\\
\end{array}\right),\quad\lambda_{8}=\frac{1}{\sqrt{3}}\left(\begin{array}[]{ccc}i&\
0&\ 0\\\ 0&\ i&\ 0\\\ 0&\ 0&\ -2i\\\ \end{array}\right),$

each representing the gluons responsible for strong force interaction.

Observe that $\\{\lambda_{1},\lambda_{2},\lambda_{3}\\}$ is the Lie algebra of
a subgroup $H_{1}\subset{\rm SU}(3)$, isomorphic to ${\rm SU}(2)$.
Furthermore, $\lambda_{8}$ generates an abelian subgroup $H_{2}$, such that
the elements in $H_{1}$ commutes with elements in $H_{2}$, because
$\lambda_{8}$ commutes with $\\{\lambda_{1},\lambda_{2},\lambda_{3}\\}$. Let
$\tilde{H}$ be a Lie subgroup in ${\rm SU}(3)$, generated by
$\\{\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{8}\\}$.

Suppose the vacuum state $1$ is invariant under this unbroken subgroup
$\tilde{H}$ and $\mathcal{A}$ as described in Definition 3.6 is linearly
independent. This means that $3\leq|\mathcal{A}|\leq 4$ and cannot contain any
elements that are linear combination of
$\\{\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{8}\\}$, also referred to as
unbroken generators. Therefore, this ${\rm SU}(3)$ gauge theory is
spontaneously broken. Furthermore, if $A=D^{(j,k)}$, then either $j=1$, $k=0$
or $j=k=1/2$.

Choose $\mathcal{A}=\\{\lambda_{4},\lambda_{5},\lambda_{7}\\}$. A direct
computation shows that

$\displaystyle[\lambda_{7},\lambda_{4}]=$
$\displaystyle\lambda_{2},\quad[\lambda_{7},\lambda_{5}]=\lambda_{1},\quad{{\rm{ad}}}(\lambda_{4}){{\rm{ad}}}(\lambda_{5})\lambda_{7}=-\lambda_{6},$
$\displaystyle{{\rm{ad}}}(\lambda_{7}){{\rm{ad}}}(\lambda_{4}){{\rm{ad}}}(\lambda_{5})\lambda_{7}=$
$\displaystyle\lambda_{3}-\sqrt{3}\lambda_{8},\quad[\lambda_{5},\lambda_{4}]=\lambda_{3}+\sqrt{3}\lambda_{8}.$

Thus, we see that $\bigcup_{j=1}^{4}\mathcal{A}^{j}$ spans $\mathfrak{su}(3)$.
In this case, we can choose $A=D^{(1,0)}$ as an irreducible representation and
the vectors in the span of $\mathcal{A}$ are called spin 1 vectors.

If we choose
$\mathcal{A}=\\{\lambda_{4},\lambda_{5},\lambda_{6},\lambda_{7}\\}$, which are
the broken generators in $\mathfrak{su}(3)$, then $A=D^{(1/2,1/2)}$ is
equivalent to $Y$, and the vectors in the span of $\mathcal{A}$ transform like
4-vectors.

###  3.3 Cyclicity

###### Notation 3.18

Let $S=[0,1]\times[0,1]\equiv I^{2}$ and
$\mathscr{P}_{S}(n)\subset\mathscr{P}_{S}$, whereby $g\in\mathscr{P}_{S}(n)$
if $g=f_{1}f_{2}\cdots f_{n}$, each $f_{i}\in\mathscr{P}_{S}$.

Let $C_{c}(S,{{\mathbb{C}}})$ and $C(S,{{\mathbb{C}}})$ denote the set of
compactly supported continuous functions and the set of continuous functions
on $S$ respectively.

We also write $\parallel\cdot\parallel_{L^{2}}$ to denote the $L^{2}$ norm on
$S$, i.e. $\parallel f\parallel_{L^{2}}=\left[\int_{I^{2}}|f(\hat{s})|^{2}\
d\hat{s}\right]^{1/2}$.

###### Lemma 3.19

We have that $\mathscr{P}_{S}$ is dense in $L^{2}(S)$. Let $\mathcal{G}(n)$ be
the smallest algebra containing $\mathscr{P}_{S}(n)$, $n\in\mathbb{N}$ fixed.
Then, $\mathcal{G}(n)$ is dense in $\mathscr{P}_{S}$ using the $L^{2}$ norm.

Proof. Now $\mathscr{P}_{S}$ is a complex algebra and clearly it separates
interior points in $S$. Unfortunately, it does not contain the unit 1 on
$S\equiv I^{2}$. Let $\mathcal{C}$ be the smallest algebra containing $1$ and
$\mathscr{P}_{S}$.

By complex Stone Weierstrass Theorem, $\mathscr{P}_{S}$ and $\mathcal{C}$ are
respectively dense in $C_{c}(S,{{\mathbb{C}}})$ and $C(S,{{\mathbb{C}}})$.
Furthermore, since continuous functions are dense in $L^{2}(S)$, we see that
$\mathcal{C}$ will be dense in $L^{2}(S)$. To show that the space containing
polynomials of functions in $\mathscr{P}_{S}$ will generate $L^{2}(S)$, we
will show that we can approximate 1 via the $L^{2}$ norm, using a sequence of
functions in $\mathscr{P}_{S}$.

Define $\breve{\varphi}_{\delta}:[0,1]\rightarrow{{\mathbb{R}}}$ by

$\breve{\varphi}_{\delta}(t):=\left\\{\begin{array}[]{ll}\frac{1}{\delta}t,&\hbox{$0\leq
t\leq\delta$;}\\\ 1,&\hbox{$\delta<t\leq 1-\delta$;}\\\
\frac{1}{\delta}(1-t),&\hbox{$1-\delta<t\leq 1$.}\end{array}\right.$

Then, $\hat{t}=(t,\bar{t})\in
I^{2}\mapsto\phi_{\delta}(\hat{t}):=\breve{\varphi}_{\delta}(t)\breve{\varphi}_{\delta}(\bar{t})$
is continuous. Let $\epsilon>0$. We can find a $\delta>0$ such that
$\parallel\phi_{\delta}-1\parallel_{L^{2}}<\epsilon/2$.

Since $\phi_{\delta}\in C_{c}(S,{{\mathbb{C}}})$, we can find a
$g_{\epsilon}\in\mathscr{P}_{S}$ such that
$\parallel\phi_{\delta}-g_{\epsilon}\parallel_{L^{2}}<\epsilon/2$. Thus,
$\parallel 1-g_{\epsilon}\parallel_{L^{2}}<\epsilon$. This completes the claim
that $\mathscr{P}_{S}$ is dense in $L^{2}(S)$.

To prove the second statement, let $f\in\mathscr{P}_{S}$ and let $\epsilon>0$.
There exists a $M>0$ such that $|f|(\hat{s})<M$ for all $\hat{s}\in S$. Choose
a $\delta>0$ such that $\parallel
1-g_{\delta}^{n-1}\parallel_{L^{2}}<\epsilon/M$,
$g_{\delta}\in\mathscr{P}_{S}$. Let $\tilde{g}_{\epsilon}:=g_{\delta}^{n-1}f$.
Then,

$\displaystyle\parallel f-\tilde{g}_{\epsilon}\parallel_{L^{2}}$
$\displaystyle=\left[\int_{I^{2}}|1-g_{\delta}^{n-1}|^{2}(\hat{t})|f|^{2}(\hat{t})\
d\hat{t}\right]^{1/2}$ $\displaystyle\leq M\parallel
1-g_{\delta}^{n-1}\parallel_{L^{2}}<\epsilon.$

Thus, $\mathcal{G}(n)$ is dense in $\mathscr{P}_{S}$ using the $L^{2}$ norm.

Because $\mathscr{P}_{S}$ is an algebra, we see that

$\psi^{\alpha_{1},m_{1}}(g_{1})\cdots\psi^{\alpha_{k},m_{k}}(g_{k})\mathscr{D}\subset\mathscr{D},$

whereby $\psi^{\alpha_{i},k_{i}}(g_{i})=\phi^{\alpha_{i},k_{i}}(g_{i})$ or its
adjoint $\phi^{\alpha_{i},k_{i}}(g_{i})^{\ast}$.

Let $\mathscr{D}_{0}$ be a subspace inside $\mathbb{H}_{{\rm
YM}}(\mathfrak{g})$, generated by the action of $U(\vec{a},\Lambda)$ and
polynomial containing
$\phi^{\alpha_{1},n}(f_{1}),\cdots\phi^{\alpha_{r},n}(f_{r})$, on the vacuum
state 1. Clearly, $\mathscr{D}_{0}\subset\mathscr{D}$.

###### Proposition 3.20

The set $\mathscr{D}_{0}$ is dense inside $\mathbb{H}_{{\rm
YM}}(\mathfrak{g})$.

Proof. Under $U(\vec{a},\Lambda)$, we can transform any space-like rectangular
surface, into other space-like rectangular surface, via translation, spatial
rotation and boost. See Equation (2.2). Thus it suffices to show that for any
compact space-like rectangular surface $S\subset{{\mathbb{R}}}^{4}$ contained
in some plane, we can approximate any measurable section in
$S\times\rho_{n}(\mathfrak{g})_{{\mathbb{C}}}\rightarrow S$, for each fixed
$n$, using the field operators. Without loss of generality, we assume that $S$
is $I^{2}$, lying inside the $x^{2}-x^{3}$ plane.

Refer to Definition 3.6. First we assume that the span of $\mathcal{A}$ is
$\mathfrak{g}$. In this case, we see that $\underline{N}\geq N$. Recall
${{\rm{ad}}}(E)F=[E,F]\in\mathfrak{g}$. Since $\mathfrak{g}$ is simple, we see
that for a fixed $k\geq 1$,

${\rm span}\
\left\\{{{\rm{ad}}}(F^{\alpha_{1}})\cdots{{\rm{ad}}}(F^{\alpha_{k}})F^{\beta}:\
1\leq\beta\leq\underline{N},\ 1\leq\alpha_{i}\leq\underline{N},\
i=1,2,\cdots,k\right\\}=\mathfrak{g}.$

Therefore, for each $1\leq\gamma\leq N$, we can write

$E^{\gamma}=\sum_{\beta=1}^{N(\gamma)}d_{\beta}^{\gamma}{{\rm{ad}}}(F^{\alpha_{1}^{\gamma,\beta}})\cdots{{\rm{ad}}}(F^{\alpha_{m-1}^{\gamma,\beta}})F^{\alpha_{m}^{\gamma,\beta}},$

for real coefficients $d_{\beta}^{\gamma}$ and natural numbers
$1\leq\alpha_{i}^{\gamma,\beta}\leq\underline{N}$.

Let $p_{\kappa}(x)=\frac{\kappa}{\sqrt{2\pi}}e^{-\kappa^{2}x^{2}/2}$ be an
one-dimensional Gaussian function, mean 0, variance $1/\kappa^{2}$. Its
Fourier Transform is $\hat{p}_{\kappa}(q)=\kappa p_{1/\kappa}(q)$. Let
$c_{n}=\hat{p}_{1}(\hat{H}(\rho_{n}))\neq 0$,
$d_{n}=\hat{p}_{1}(\hat{P}(\rho_{n}))\neq 0$ be fixed.

For a given set of Schwartz functions $\\{f_{1},\cdots,f_{m}\\}$ defined on
$S$, extend each one to be a function

$F_{i}\in\mathscr{P}:\vec{x}\in{{\mathbb{R}}}^{4}\mapsto\frac{1}{c_{n}}p_{1}(x^{0})\frac{1}{d_{n}}p_{1}(x^{1})f_{i}(x^{2},x^{3}),\
i=1,\cdots,m.$

We also have

$F_{i}^{\\{e_{0},e_{1}\\}}(\hat{H}(\rho_{n}),\hat{P}(\rho_{n}))(0,0,x^{2},x^{3})=\frac{1}{c_{n}}\hat{p}_{1}(\hat{H}(\rho_{n}))\frac{1}{d_{n}}\hat{p}_{1}(\hat{P}(\rho_{n}))f_{i}(x^{2},x^{3})=f_{i}(x^{2},x^{3}).$

Hence,

$\displaystyle[F_{1}^{\\{e_{0},e_{1}\\}}$ $\displaystyle\cdot\cdots
F_{m}^{\\{e_{0},e_{1}\\}}](\hat{H}(\rho_{n}),\hat{P}(\rho_{n}))(0,0,x^{2},x^{3})=\prod_{i=1}^{m}f_{i}(x^{2},x^{3}).$

For any set of Schwartz functions $\\{f_{1},\cdots,f_{m}\\}$ on $S$, we can
extend them to be Schwartz functions on ${{\mathbb{R}}}^{4}$ as described
above, and thus we have

$\displaystyle\sum_{\beta=1}^{N(\gamma)}d_{\beta}^{\gamma}\phi^{\alpha_{1}^{\gamma,\beta},n}(F_{1})\cdots\phi^{\alpha_{m}^{\gamma,\beta},n}(F_{m})1$
$\displaystyle=\left(S,\prod_{i=1}^{m}f_{i}\otimes\rho_{n}(E^{\gamma}),\\{e_{a}\\}_{a=0}^{3}\right).$

If we let

$C_{m}={\rm span}\
\left\\{\phi^{\alpha_{1},n}(F_{1})\cdots\phi^{\alpha_{m},n}(F_{m})1:\
F_{i}\in\mathscr{P},\ 1\leq\alpha_{i}\leq\underline{N}\right\\},$

we see that the sum of subspaces, $\sum_{m=1}^{\infty}C_{m}$, is dense in
$L^{2}(S)\otimes\rho_{n}(\mathfrak{g})$. This follows from Lemma 3.19.

This will show that we can find a sequence of vectors in the sum
$\sum_{m=1}^{\infty}C_{m}$ and approximate any vector of the form
$\left(S,f_{\alpha}\otimes\rho_{n}(E^{\alpha}),\\{e_{a}\\}_{a=0}^{3}\right)$,
whereby $f_{\alpha}$ is measurable on a compact rectangular surface $S$ inside
$x^{2}-x^{3}$ plane.

Now suppose span of $\mathcal{A}$ is not equal to $\mathfrak{g}$. Thus,
$\underline{n}\geq 2$. By definition of $\mathcal{A}$, we see that

${\rm span}\
\left\\{{{\rm{ad}}}(F^{\alpha_{1}})\cdots{{\rm{ad}}}(F^{\alpha_{\tilde{n}-1}})F^{\alpha_{\tilde{n}}}:\
1\leq\alpha_{i}\leq\underline{N},\
1\leq\tilde{n}\leq\underline{n}\right\\}=\mathfrak{g}.$

Thus, for each $1\leq\gamma\leq N$, we can write for some
$1\leq\tilde{n}\leq\underline{n}$,

$\tilde{E}^{\gamma}=\sum_{\xi=1}^{N(\gamma)}d(\gamma,\xi){{\rm{ad}}}(F^{\alpha_{1}(\gamma,\xi)})\cdots{{\rm{ad}}}(F^{\alpha_{\tilde{n}-1}(\gamma,\xi)})F^{\alpha_{\tilde{n}}(\gamma,\xi)},$

for real coefficients $d(\gamma,\xi)$ and natural numbers
$1\leq\alpha_{i}(\gamma,\xi)\leq\underline{N}$. Here, $\\{\tilde{E}^{\alpha}:\
1\leq\alpha\leq N\\}$ is a basis for $\mathfrak{g}$.

Let $\epsilon>0$ and $f\in\mathscr{P}_{S}$. From the proof of Lemma 3.19, we
can find $g_{1},\cdots,g_{\tilde{n}}\in\mathscr{P}_{S}$ such that

$\parallel f-g_{1}\cdots
g_{\tilde{n}}\parallel_{L^{2}}<\frac{\epsilon}{C(\rho_{n})}.$

Using an earlier argument, there exists $G_{1},\cdots,G_{\tilde{n}}$ such that
$G_{i}^{\\{e_{0},e_{1}\\}}\equiv g_{i}$ for $1\leq i\leq\tilde{n}$.

Hence, we have that

$\displaystyle\sum_{\xi=1}^{N(\gamma)}$ $\displaystyle
d(\gamma,\xi)\phi^{\alpha_{1}(\gamma,\xi),n}(G_{1})\cdots\phi^{\alpha_{\tilde{n}-1}(\gamma,\xi),n}(G_{\tilde{n}-1})\phi^{\alpha_{\tilde{n}}(\gamma,\xi),n}(G_{\tilde{n}})1$
$\displaystyle=$
$\displaystyle\left(S,\prod_{i=1}^{\tilde{n}}g_{i}\otimes\rho_{n}(\tilde{E}^{\gamma}),\\{e_{a}\\}_{a=0}^{3}\right).$

A direct computation will show that

$\left\|\Big{(}S,f\otimes\rho(\tilde{E}^{\gamma}),\\{e_{a}\\}_{a=0}^{3}\Big{)}-\left(S,\prod_{i=1}^{\tilde{n}}g_{i}\otimes\rho_{n}(\tilde{E}^{\gamma}),\\{e_{a}\\}_{a=0}^{3}\right)\right\|<\epsilon.$

This completes the proof.

###### Remark 3.21

Given any
$\left(S,f_{\alpha}\otimes\rho(E^{\alpha}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)$,
we can find a sequence of
Schwartz functions
$\\{g_{\alpha}^{m}:{{\mathbb{R}}}^{4}\rightarrow{{\mathbb{C}}}\ |\
m\in\mathbb{N}\\}$, for which its partial Fourier Transform approximates
$f_{\alpha}$, for each $1\leq\alpha\leq N$. Here, for each
$g_{\alpha}^{m}:{{\mathbb{R}}}^{4}\rightarrow{{\mathbb{C}}}$, we can take the
Fourier Transform on the time and space variables, i.e. $g_{\alpha}^{m}\mapsto
g_{\alpha}^{m,\\{\hat{f}_{0},\hat{f}_{1}\\}}(\hat{H}(\rho),\hat{P}(\rho))(\vec{x})$,
$\vec{x}\in S$. As explained at the end of subsection 2.2, the eigenvalues
$\\{\hat{H}(\rho),\hat{P}(\rho)\\}$ will be defined using a Yang-Mills path
integral in subsection 7.3. Therefore,
$\left(S,f_{\alpha}\otimes\rho(E^{\alpha}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)\in\mathscr{H}(\rho)$
can be written as the limit of
$\left\\{\left(S,g_{\alpha}^{m,\\{\hat{f}_{0},\hat{f}_{1}\\}}\otimes\rho(E^{\alpha}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)\right\\}_{m=1}^{\infty}$,
using the inner product given in Definition 2.9, and hence they are referred
to as Yang-Mills fields.

###  3.4 Tempered Distribution

###### Proposition 3.22

Let $S,\tilde{S}$ be bounded space-like surfaces lying inside some plane. Let
$\left(S,g_{\beta}\otimes\rho_{n}(E^{\beta}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right),\left(\tilde{S},\tilde{g}_{\beta}\otimes\rho_{n}(E^{\beta}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)\in\mathscr{D}$.
Suppose $\hat{f}_{a}=\Lambda e_{a}$ for $\Lambda\in{\rm
SL}(2,{{\mathbb{C}}})$.

Write

$c_{\alpha}^{\gamma,\beta}=A(\Lambda)_{\delta}^{\alpha}{{\rm{Tr}}}\left[-[{{\rm{ad}}}(\rho_{n}(F^{\delta}))\rho_{n}(E^{\gamma})]\rho_{n}(E^{\beta})\right],$

with an implied sum over $\delta$. Given a test function $f\in\mathscr{P}$, we
define for a smooth parametrization $\sigma:I^{2}\rightarrow
S\cap\tilde{S}\subset{{\mathbb{R}}}^{4}$,

$\displaystyle T(f):=$
$\displaystyle\left\langle\phi^{\alpha,n}(f)\left(\tilde{S},\tilde{g}_{\gamma}\otimes\rho_{n}(E^{\gamma}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right),\left(S,g_{\beta}\otimes\rho_{n}(E^{\beta}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)\right\rangle$
$\displaystyle=$ $\displaystyle c_{\alpha}^{\gamma,\beta}\int_{I^{2}}d\hat{t}\
\left[f^{\\{\hat{f}_{0},\hat{f}_{1}\\}}\cdot\tilde{g}_{\gamma}\cdot\overline{g_{\beta}}\right](\sigma(\hat{t}))\cdot|\acute{\rho}_{\sigma}|(\hat{t}).$

Then $T$ is a linear functional on $\mathscr{P}$. Furthermore, it is a
tempered distribution.

Proof. It is clear that it is a linear functional on $\mathscr{P}$. It remains
to show that it is a tempered distribution.

Let $\sigma:I^{2}\rightarrow{{\mathbb{R}}}^{4}$ be a parametrization of
$\hat{S}:=S\cap\tilde{S}$, and write

$h=[\tilde{g}_{\gamma}\cdot\overline{g_{\beta}}]\circ\sigma\cdot\left|\acute{\rho}_{\sigma}\right|c_{\alpha}^{\gamma,\beta}.$

Then,

$\displaystyle T(f)=$
$\displaystyle\int_{I^{2}}f^{\\{\hat{f}_{0},\hat{f}_{1}\\}}(\sigma(\hat{t}))h(\hat{t})\
d\hat{t}.$

Let
$\tilde{\sigma}:\hat{s}\mapsto\tilde{\sigma}(\hat{s})=s\hat{f}_{0}+\bar{s}\hat{f}_{1}$,
$s,\bar{s}\in{{\mathbb{R}}}$, whereby $\\{\hat{f}_{a}\\}_{a=0}^{3}$ is a
Minkowski frame for $\hat{S}$. By definition,

$\displaystyle\vec{x}\longmapsto f^{\\{\hat{f}_{0},\hat{f}_{1}\\}}(\vec{x})$
$\displaystyle\equiv
f^{\\{\hat{f}_{0},\hat{f}_{1}\\}}(\hat{H}(\rho_{n}),\hat{P}(\rho_{n}))(\vec{x})$
$\displaystyle=\int_{\hat{s}\in{{\mathbb{R}}}^{2}}\frac{e^{-i[\tilde{\sigma}(\hat{s})\cdot(\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1})]}}{2\pi}f\left(\vec{x}+\tilde{\sigma}(\hat{s})\right)\left|\acute{\rho}_{\tilde{\sigma}}\right|(\hat{s})\
d\hat{s}.$

Write
$\vec{\alpha}=\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1}$. Thus

$\displaystyle T(f)=$
$\displaystyle\int_{I^{2}}f^{\\{\hat{f}_{0},\hat{f}_{1}\\}}(\sigma(\hat{t}))h(\hat{t})\
d\hat{t}$ $\displaystyle=$
$\displaystyle\int_{\hat{s},\hat{t}\in{{\mathbb{R}}}^{2}}d\hat{t}d\hat{s}\
f(\sigma(\hat{t})+\tilde{\sigma}(\hat{s}))\left|\acute{\rho}_{\tilde{\sigma}}\right|(\hat{s})\left|\acute{\rho}_{\sigma}\right|(\hat{t})\cdot\frac{e^{-i[\tilde{\sigma}(\hat{s})\cdot\vec{\alpha}]}}{2\pi}[\tilde{g}_{\gamma}\cdot\overline{g_{\beta}}]\circ\sigma(\hat{t})\cdot
c_{\alpha}^{\gamma,\beta}.$

Note that

$\displaystyle(\hat{s},\hat{t})\in{{\mathbb{R}}}^{4}$
$\displaystyle\longmapsto\left|\acute{\rho}_{\tilde{\sigma}}\right|(\hat{s})\left|\acute{\rho}_{\sigma}\right|(\hat{t})\cdot\frac{e^{-i[\tilde{\sigma}(\hat{s})\cdot\vec{\alpha}]}}{2\pi}[\tilde{g}_{\gamma}\cdot\overline{g_{\beta}}]\circ\sigma(\hat{t})\cdot
c_{\alpha}^{\gamma,\beta},$

is a tempered distribution, because
$\\{g_{\beta},\tilde{g}_{\gamma}\\}_{\beta,\gamma=1}^{N}$ are Schwartz
functions. Therefore, $T$ is a tempered distribution.

###### Remark 3.23

The map

$\displaystyle f$ $\displaystyle\in\mathscr{P}$
$\displaystyle\longmapsto\int_{\hat{s},\hat{t}\in{{\mathbb{R}}}^{4}}d\hat{t}d\hat{s}\
f(\sigma(\hat{t})+\tilde{\sigma}(\hat{s}))\left|\acute{\rho}_{\tilde{\sigma}}\right|(\hat{s})\left|\acute{\rho}_{\sigma}\right|(\hat{t})\cdot\frac{e^{-i[\tilde{\sigma}(\hat{s})\cdot\vec{\alpha}]}}{2\pi}[\tilde{g}_{\gamma}\cdot\overline{g_{\beta}}]\circ\sigma(\hat{t})$
$\displaystyle\hskip
239.00298pt\times{{\rm{Tr}}}\left[-[{{\rm{ad}}}(\cdot)\rho_{n}(E^{\gamma})]\rho_{n}(E^{\beta})\right],$

defines a $\rho_{n}(\mathfrak{g})$-valued distribution, using the inner
product on $\rho_{n}(\mathfrak{g})$ defined in Equation (7.1).

###### Corollary 3.24

Suppose $S=\tilde{S}=S_{0}$. For any $f\in\mathscr{P}$, we can write

$\displaystyle T(f)$
$\displaystyle=\int_{{{\mathbb{R}}}^{4}}f(\vec{x})\frac{e^{i[x^{0}\hat{H}(\rho_{n})-x^{1}\hat{P}(\rho_{n})]}}{2\pi}[\tilde{g}_{\gamma}\cdot\overline{g_{\beta}}](x^{2},x^{3})\
d\vec{x}\cdot{{\rm{Tr}}}\left[-[{{\rm{ad}}}(\rho_{n}(F^{\alpha}))\rho_{n}(E^{\gamma})]\rho_{n}(E^{\beta})\right].$

Proof. We note that $\\{e_{a}\\}_{a=0}^{3}$ is the default Minkowski frame. In
this case, $\Lambda=\pm 1\in{\rm SL}(2,{{\mathbb{C}}})$, and $A(\pm 1)=1$.
Furthermore, we can choose the parametrizations
$\sigma(x^{2},x^{3})=x^{2}e_{2}+x^{3}e_{3}$,
$\tilde{\sigma}(x^{0},x^{1})=x^{0}e_{0}+x^{1}e_{1}$. A direct calculation
shows $|\acute{\rho}_{\sigma}|=|\acute{\rho}_{\tilde{\sigma}}|=1$. The result
hence follows.

###### Remark 3.25

In this corollary, we see that the function that maps

$\vec{x}=(x^{0},x^{1},x^{2},x^{3})\longmapsto\frac{e^{i[x^{0}\hat{H}(\rho_{n})-x^{1}\hat{P}(\rho_{n})]}}{2\pi}[\tilde{g}_{\gamma}\cdot\overline{g_{\beta}}](x^{2},x^{3}),$

is not in $\mathscr{P}$, but rather it is a tempered distribution.

##  4 Transformation Law of the Field Operator

Recall $\mathfrak{g}$ has dimension $N$ and has an irreducible representation
$\rho:\mathfrak{g}\rightarrow{\rm End}({{\mathbb{C}}}^{\tilde{N}})$. Without
any loss of generality, consider a space-like surface $S$, contained inside
some plane. From Definition 2.5, we have a Minkowski frame
$\\{\hat{f}_{a}\\}_{a=0}^{3}$ assigned to it.

In Definition 3.6, we have a finite set $\mathcal{A}\subset\mathfrak{g}$,
which defines a spinor indexed by $1\leq\alpha\leq\underline{N}$. This spinor
transforms under the action $A(\Lambda):\phi^{\alpha,n}\mapsto
A(\Lambda)_{\beta}^{\alpha}\phi^{\beta,n}$ for $\Lambda\in{\rm
SL}(2,{{\mathbb{C}}})$.

For $\Lambda\in{\rm SL}(2,{{\mathbb{C}}})$, we consider
$\Lambda^{-1}(S-\vec{a})$, which is also a space-like surface, with
$\\{\hat{g}_{a}\\}_{a=0}^{3}=\\{\Lambda^{-1}\hat{f}_{a}\\}_{a=0}^{3}$ assigned
as a Minkowski frame to it by Definition 2.5.

###### Proposition 4.1

We have the transformation law for the field operators acting on
$\mathscr{H}(\rho)$, i.e.

$\displaystyle U(\vec{a},\Lambda)$
$\displaystyle\phi^{\alpha,n}(f)U(\vec{a},\Lambda)^{-1}\left(S,g_{\beta}\otimes\rho_{n}(E^{\beta}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)$
$\displaystyle=A(\Lambda^{-1})_{\gamma}^{\alpha}\phi^{\gamma,n}(f(\Lambda^{-1}(\cdot-\vec{a})))\left(S,g_{\beta}\otimes\rho_{n}(E^{\beta}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right),$

whereby $S$ is some rectangular space-like surface contained in some plane.

Recall $S_{0}^{\flat}$ is the $x^{0}-x^{1}$ plane. However,

$\displaystyle U$
$\displaystyle(\vec{a},\Lambda)\phi^{\alpha,n}(f)U(\vec{a},\Lambda)^{-1}1$
$\displaystyle=\left(\Lambda
S_{0}+\vec{a},e^{-i[\vec{a}\cdot(\hat{H}(\rho_{n})\hat{g}_{0}+\hat{P}(\rho_{n})\hat{g}_{1})]}f^{\\{\hat{g}_{0},\hat{g}_{1}\\}}(\Lambda^{-1}(\cdot-\vec{a}))\otimes\rho_{n}(F^{\alpha}),\\{\hat{g}_{a}\\}_{a=0}^{3}\right).$

Here, $\\{\hat{g}_{a}\\}_{a=0}^{3}=\\{\Lambda e_{a}\\}_{a=0}^{3}$ is a
Minkowski frame for $\Lambda S_{0}$.

Proof. By Definition 2.5, there is a $\tilde{\Lambda}\in{\rm
SL}(2,{{\mathbb{C}}})$ such that $\hat{f}_{a}=Y(\tilde{\Lambda})e_{a}$. Thus,
$\\{\Lambda^{-1}\tilde{\Lambda}e_{a}\\}_{a=0}^{3}=\\{\Lambda^{-1}\hat{f}_{a}\\}_{a=0}^{3}$
is a Minkowski frame for $\Lambda^{-1}(S-\vec{a})$.

Write
$d_{\beta}^{\alpha}=A(\Lambda^{-1}\tilde{\Lambda})_{\beta}^{\alpha}\equiv
A(\Lambda^{-1})_{\gamma}^{\alpha}A(\tilde{\Lambda})^{\gamma}_{\beta}$,

$\displaystyle T(\rho_{n},\vec{a})$
$\displaystyle=e^{-i[\vec{a}\cdot(\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1})]},\quad
T(\rho_{n},\vec{a})^{-1}=e^{i[\vec{a}\cdot(\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1})]},$

and
$f^{\mathcal{C}}=f^{\\{\Lambda^{-1}\hat{f}_{0},\Lambda^{-1}\hat{f}_{1}\\}}$,
with $\mathcal{C}=\\{\Lambda^{-1}\hat{f}_{0},\Lambda^{-1}\hat{f}_{1}\\}$,
$\mathcal{D}=\\{\Lambda^{-1}\hat{f}_{a}\\}_{a=0}^{3}$.

We have

$\displaystyle U$
$\displaystyle(\vec{a},\Lambda)\phi^{\alpha,n}(f)U(\vec{a},\Lambda)^{-1}\left(S,g_{\beta}\otimes\rho_{n}(E^{\beta}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)$
$\displaystyle=$ $\displaystyle
U(\vec{a},\Lambda)\phi^{\alpha,n}(f)\Big{(}\Lambda^{-1}(S-\vec{a}),T(\rho_{n},\vec{a})^{-1}g_{\beta}(\Lambda\cdot+\vec{a})\otimes\rho_{n}\left(E^{\beta}\right),\mathcal{D}\Big{)}$
$\displaystyle=$ $\displaystyle
U(\vec{a},\Lambda)\Big{(}\Lambda^{-1}(S-\vec{a}),[T(\rho_{n},\vec{a})^{-1}f^{\mathcal{C}}](\cdot)d_{\gamma}^{\alpha}\cdot
g_{\beta}(\Lambda\cdot+\vec{a})\otimes{{\rm{ad}}}(\rho_{n}(F^{\gamma}))\rho_{n}\left(E^{\beta}\right),\mathcal{D}\Big{)}$
$\displaystyle=$
$\displaystyle\Big{(}S,T(\rho_{n},\vec{a})T(\rho_{n},\vec{a})^{-1}f^{\mathcal{C}}(\Lambda^{-1}(\cdot-\vec{a}))d_{\gamma}^{\alpha}g_{\beta}(\cdot)\otimes{{\rm{ad}}}(\rho_{n}(F^{\gamma}))\rho_{n}\left(E^{\beta}\right),\\{\hat{f}_{a}\\}_{a=0}^{3}\Big{)}.$
(4.1)

Let $[\Lambda^{-1}(S-\vec{a})]^{\flat}$ be the span of
$\\{\Lambda^{-1}\hat{f}_{0},\Lambda^{-1}\hat{f}_{1}\\}$. By definition, for
any point $\vec{x}\in S$, we have that

$f^{\mathcal{C}}(\Lambda^{-1}(\vec{x}-\vec{a}))\equiv
f^{\\{\Lambda^{-1}\hat{f}_{0},\Lambda^{-1}\hat{f}_{1}\\}}(\Lambda^{-1}(\vec{x}-\vec{a})),$

means we do a partial integration using Equation (3.2) on $f$ in the time-like
plane $\vec{y}+[\Lambda^{-1}(S-\vec{a})]^{\flat}$, for
$\vec{y}=\Lambda^{-1}(\vec{x}-\vec{a})$.

Let $\sigma(s,\bar{s})=s\hat{f}_{0}+\bar{s}\hat{f}_{1}$,
$s,\bar{s}\in{{\mathbb{R}}}$. Let $\hat{g}_{0}=\Lambda^{-1}\hat{f}_{0}$ and
$\hat{g}_{1}=\Lambda^{-1}\hat{f}_{1}$. We will let
$\hat{\sigma}(\hat{s})=\Lambda^{-1}\sigma(\hat{s})=s\hat{g}_{0}+\bar{s}\hat{g}_{1}$,
$s,\bar{s}\in{{\mathbb{R}}}$.

Now, $\Lambda^{-1}\sigma\cdot\Lambda^{-1}\hat{f}_{a}=\sigma\cdot\hat{f}_{a}$
and $\acute{\rho}_{\hat{\sigma}}=\acute{\rho}_{\sigma}$. See Remark A.4.
Hence,

$\displaystyle
f^{\\{\hat{g}_{0},\hat{g}_{1}\\}}(\hat{H}(\rho_{n}),\hat{P}(\rho_{n}))(\Lambda^{-1}(\vec{x}-\vec{a}))$
$\displaystyle:=\int_{\hat{s}\in{{\mathbb{R}}}^{2}}\frac{e^{-i[\hat{\sigma}(\hat{s})\cdot(\hat{H}(\rho_{n})\hat{g}_{0}+\hat{P}(\rho_{n})\hat{g}_{1})]}}{2\pi}f(\vec{y}+\hat{\sigma}(\hat{s}))|\acute{\rho}_{\hat{\sigma}}|(\hat{s})\
d\hat{s}$
$\displaystyle=\int_{\hat{s}\in{{\mathbb{R}}}^{2}}\frac{e^{-i[\sigma(\hat{s})\cdot(\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1})]}}{2\pi}f(\vec{y}+\Lambda^{-1}\sigma(\hat{s}))|\acute{\rho}_{\sigma}|(\hat{s})\
d\hat{s}$
$\displaystyle=\int_{\hat{s}\in{{\mathbb{R}}}^{2}}\frac{e^{-i[\sigma(\hat{s})\cdot(\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1})]}}{2\pi}f(\Lambda^{-1}(\vec{x}+\sigma(\hat{s})-\vec{a}))|\acute{\rho}_{\sigma}|(\hat{s})\
d\hat{s}$
$\displaystyle=f(\Lambda^{-1}(\cdot-\vec{a}))^{\\{\hat{f}_{0},\hat{f}_{1}\\}}(\hat{H}(\rho_{n}),\hat{P}(\rho_{n}))(\vec{x}).$

Therefore, Equation (4.1) is equal to

$\displaystyle\Big{(}S,$ $\displaystyle
f(\Lambda^{-1}(\cdot-\vec{a}))^{\\{\hat{f}_{0},\hat{f}_{1}\\}}A(\Lambda^{-1})_{\gamma}^{\alpha}A(\tilde{\Lambda})_{\delta}^{\gamma}\cdot
g_{\beta}\otimes{{\rm{ad}}}(\rho_{n}(F^{\delta}))\rho_{n}(E^{\beta}),\\{\tilde{\Lambda}e_{a}\\}_{a=0}^{3}\Big{)}$
$\displaystyle=A(\Lambda^{-1})_{\gamma}^{\alpha}\phi^{\gamma,n}[f(\Lambda^{-1}(\cdot-\vec{a}))]\Big{(}S,g_{\beta}\otimes\rho_{n}(E^{\beta}),\\{\tilde{\Lambda}e_{a}\\}_{a=0}^{3}\Big{)}.$

To prove the second statement, note that $U(\vec{a},\Lambda)1=1$. Let
$\hat{g}_{a}=\Lambda e_{a}$, $a=0,\cdots,3$. Then,

$\displaystyle U(\vec{a},\Lambda)$
$\displaystyle\phi^{\alpha,n}(f)U(\vec{a},\Lambda)^{-1}1=U(\vec{a},\Lambda)\phi^{\alpha,n}(f)1$
$\displaystyle=$ $\displaystyle
U(\vec{a},\Lambda)\left(S_{0},f^{\\{e_{0},e_{1}\\}}\otimes\rho_{n}(F^{\alpha}),\\{e_{a}\\}_{a=0}^{3}\right)$
$\displaystyle=$ $\displaystyle\left(\Lambda
S_{0}+\vec{a},e^{-i[\vec{a}\cdot(\hat{H}(\rho_{n})\hat{g}_{0}+\hat{P}(\rho_{n})\hat{g}_{1})]}f^{\\{e_{0},e_{1}\\}}(\Lambda^{-1}(\cdot-\vec{a}))\otimes\rho_{n}(F^{\alpha}),\\{\hat{g}_{a}\\}_{a=0}^{3}\right),$

by definitions.

###### Remark 4.2

Relative to $\Lambda S_{0}$ and for
$\Lambda\vec{a}=a^{0}\hat{g}_{0}+a^{1}\hat{g}_{1}$, we see that multiplication
by
$e^{-i[\Lambda\vec{a}\cdot(\hat{H}(\rho_{n})\hat{g}_{0}+\hat{P}(\rho_{n})\hat{g}_{1})]}=e^{i[a^{0}\hat{H}(\rho_{n})-a^{1}\hat{P}(\rho_{n})]}$,
corresponds to a shift

$f(\cdot)^{\\{\hat{g}_{0},\hat{g}_{1}\\}}\mapsto
f(\cdot-\Lambda\vec{a})^{\\{\hat{g}_{0},\hat{g}_{1}\\}},$

when we take Fourier Transform.

##  5 Causality

We are now down to the final Wightman’s axiom. Recall that we have a countable
set $\\{\hat{H}(\rho_{n}),\hat{P}(\rho_{n})\\}_{n\in\mathbb{N}}$ to be defined
later in Definition 7.13. We will see in this section that to satisfy local
commutativity, we must have $\hat{H}(\rho_{n})^{2}-\hat{P}(\rho_{n})^{2}>0$,
for each $n$.

###### Definition 5.1

(Space-like separation)
Let $f,g\in\mathscr{P}$. The support of $f$, denoted ${\rm supp}\ f$, is the
closed set obtained by taking the complement of the largest open set in which
$f$ vanishes. We say that ${\rm supp}\ f$ and ${\rm supp}\ g$ are space-like
(time-like) separated if $f(\vec{x})g(\vec{y})=0$ for all pairs of points
$\vec{x}=(x^{0},x),\ \vec{y}=(y^{0},y)\in{{\mathbb{R}}}^{4}$ such that

$(\vec{x}-\vec{y})\cdot(\vec{x}-\vec{y})=-(x^{0}-y^{0})^{2}+\sum_{i=1}^{3}(x^{i}-y^{i})^{2}\leq\
(\geq)\ 0.$

###### Remark 5.2

  * •

Given a connected space-like rectangular surface $S$ contained in a plane, any
two distinct points $\vec{x},\vec{y}\in S$ are actually space-like separated.
By Definition 5.1, if $f$ and $g$ have supports which are space-like
separated, then we must have $f(\vec{x})g(\vec{x})=0$ for any $\vec{x}\in S$.

  * •

Note that on a time-like rectangular surface, two distinct points in it may
not be time-like separated.

###### Notation 5.3

For this section, we only consider a space-like surface $S$ contained in a
plane. It is equipped with a Minkowski frame $\\{\hat{f}_{a}\\}_{a=0}^{3}$,
which will be assumed throughout. To ease our notations, we will drop this
Minkowski frame from our notation. This means

$(S,f_{\alpha}\otimes\rho(E^{\alpha}))\equiv(S,f_{\alpha}\otimes\rho(E^{\alpha}),\\{\hat{f}_{a}\\}_{a=0}^{3}).$

###### Definition 5.4

Write the commutators of $f$ and $g$ in $\mathscr{P}$ as

$\displaystyle\lceil\phi^{\alpha,n}(f),\phi^{\beta,n}(g)\rceil$
$\displaystyle:=\phi^{\alpha,n}(f)\phi^{\beta,n}(g)-\phi^{\alpha,n}(g)\phi^{\beta,n}(f),$
$\displaystyle\lceil\phi^{\alpha,n}(f)^{\ast},\phi^{\beta,n}(g)^{\ast}\rceil$
$\displaystyle:=\phi^{\alpha,n}(f)^{\ast}\phi^{\beta,n}(g)^{\ast}-\phi^{\alpha,n}(g)^{\ast}\phi^{\beta,n}(f)^{\ast}.$

###### Remark 5.5

We are using the version of Wightman’s last axiom taken from [4], and not from
[3].

###### Lemma 5.6

Let $S$ be a space-like surface contained in a plane. Then, we have

$\displaystyle\lceil\phi^{\alpha,n}(f),\phi^{\beta,n}(g)\rceil\left(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})\right)=0,$

for any $\left(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})\right)\in\mathscr{D}$,
and $[\phi^{\alpha,n}(f),\phi^{\beta,n}(g)]1=0$.

We also have

$\displaystyle\lceil\phi^{\alpha,n}(f)^{\ast},\phi^{\beta,n}(g)^{\ast}\rceil$
$\displaystyle
1=0,\quad\lceil\phi^{\alpha,n}(f)^{\ast},\phi^{\beta,n}(g)^{\ast}\rceil\left(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})\right)=0,$

for any $\left(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})\right)\in\mathscr{D}$.

Proof. The first two commutator relations follow from Definitions 3.7, 3.12,
and that $f^{\\{\hat{f}_{0},\hat{f}_{1}\\}}\cdot
g^{\\{\hat{f}_{0},\hat{f}_{1}\\}}=g^{\\{\hat{f}_{0},\hat{f}_{1}\\}}\cdot
f^{\\{\hat{f}_{0},\hat{f}_{1}\\}}$. By taking the adjoint, the last two follow
immediately.

###### Remark 5.7

These commutation relations hold, regardless of whether the supports of $f$
and $g$ are space-like separated or not.

###  5.1 CPT Theorem

###### Definition 5.8

Define the anti-commutators and commutators of $f$ and $g$ in $\mathscr{P}$,
as

$\displaystyle\lfloor\phi^{\alpha,n}(f),\phi^{\beta,n}(g)^{\ast}\rfloor_{\pm}$
$\displaystyle:=\phi^{\alpha,n}(f)\phi^{\beta,n}(g)^{\ast}\pm\phi^{\alpha,n}(g)\phi^{\beta,n}(f)^{\ast},$
$\displaystyle\lfloor\phi^{\alpha,n}(f)^{\ast},\phi^{\beta,n}(g)\rfloor_{\pm}$
$\displaystyle:=\phi^{\alpha,n}(f)^{\ast}\phi^{\beta,n}(g)\pm\phi^{\alpha,n}(g)^{\ast}\phi^{\beta,n}(f).$

###### Lemma 5.9

Recall ${{\rm{ad}}}(\rho(E^{\alpha}))$ refers to its adjoint representation on
$\rho(\mathfrak{g})$. Suppose the Minkowski frame on $S$ is
$\hat{f}_{a}=\Lambda e_{a}$, $a=0,1,2,3$. Write
$\mathcal{C}=\\{\hat{f}_{0},\hat{f}_{1}\\}$.

For any $\left(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})\right)\in\mathscr{D}$,
we have

$\displaystyle\lfloor\phi^{\alpha,n}$
$\displaystyle(f),\phi^{\beta,n}(g)^{\ast}\rfloor_{\pm}\left(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})\right)$
$\displaystyle=-$ $\displaystyle
A(\Lambda)_{\delta}^{\alpha}\overline{A(\Lambda)_{\mu}^{\beta}}\left(S,{\rm
B}^{\pm}[f^{\mathcal{C}}\cdot\overline{g^{\mathcal{C}}}\pm
g^{\mathcal{C}}\cdot\overline{f^{\mathcal{C}}}]\cdot
h_{\gamma}\otimes{{\rm{ad}}}(\rho_{n}(F^{\delta})){{\rm{ad}}}(\rho_{n}(F^{\mu}))\rho_{n}(E^{\gamma})\right)$
$\displaystyle+\left\langle(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})),\phi^{\beta,n}(g)1\right\rangle\
\phi^{\alpha,n}(f)1$
$\displaystyle\pm\left\langle(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})),\phi^{\beta,n}(f)1\right\rangle\
\phi^{\alpha,n}(g)1,$ (5.1)

whereby ${\rm B}^{+}={\rm Re}$ and ${\rm B}^{-}={\rm Im}$ for anti-commutation
and commutation relations respectively.

And

$\displaystyle\lfloor\phi^{\alpha,n}$
$\displaystyle(f)^{\ast},\phi^{\beta,n}(g)\rfloor_{\pm}\left(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})\right)$
$\displaystyle=-$
$\displaystyle\overline{A(\Lambda)_{\delta}^{\alpha}}A(\Lambda)_{\mu}^{\beta}\left(S,{\rm
B}^{\pm}[\overline{f^{\mathcal{C}}}\cdot
g^{\mathcal{C}}\pm\overline{g^{\mathcal{C}}}\cdot f^{\mathcal{C}}]\cdot
h_{\gamma}\otimes{{\rm{ad}}}(\rho_{n}(F^{\delta})){{\rm{ad}}}(\rho_{n}(F^{\mu}))\rho_{n}(E^{\gamma})\right)$
$\displaystyle+\left\langle\left(S,g^{\mathcal{C}}A(\Lambda)_{\mu}^{\beta}\cdot
h_{\gamma}\otimes{{\rm{ad}}}(\rho_{n}(F^{\mu}))\rho_{n}(E^{\gamma})\right),\phi^{\alpha,n}(f)1\right\rangle
1$
$\displaystyle\pm\left\langle\left(S,f^{\mathcal{C}}A(\Lambda)_{\mu}^{\beta}\cdot
h_{\gamma}\otimes{{\rm{ad}}}(\rho_{n}(F^{\mu}))\rho_{n}(E^{\gamma})\right),\phi^{\alpha,n}(g)1\right\rangle
1,$ (5.2)

whereby ${\rm B}^{+}={\rm Re}$ and ${\rm B}^{-}={\rm Im}$ for anti-commutation
and commutation relations respectively.

Proof. From Definitions 3.7, 3.12 and 3.14, we see that

$\displaystyle\phi^{\alpha,n}(f)\phi^{\beta,n}(g)^{\ast}$
$\displaystyle(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma}))$ $\displaystyle=$
$\displaystyle\phi^{\alpha,n}(f)\overline{A(\Lambda)_{\mu}^{\beta}}\bigg{[}\left(S,-\overline{g^{\mathcal{C}}}\cdot
h_{\gamma}\otimes{{\rm{ad}}}(\rho_{n}(F^{\mu}))\rho_{n}(E^{\gamma})\right)$
$\displaystyle+\left\langle\left(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})\right),\phi^{\beta,n}(g)1\right\rangle
1\bigg{]}$ $\displaystyle=$
$\displaystyle-A(\Lambda)_{\delta}^{\alpha}\overline{A(\Lambda)_{\mu}^{\beta}}\left(S,f^{\mathcal{C}}\cdot\overline{g^{\mathcal{C}}}\cdot
h_{\gamma}\otimes{{\rm{ad}}}(\rho_{n}(F^{\delta})){{\rm{ad}}}(\rho_{n}(F^{\mu}))\rho_{n}(E^{\gamma})\right)$
$\displaystyle+\left\langle\left(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})\right),\phi^{\beta,n}(g)1\right\rangle\
\phi^{\alpha,n}(f)1.$

Similarly,

$\displaystyle\phi^{\alpha,n}(g)\phi^{\beta,n}(f)^{\ast}$
$\displaystyle\left(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})\right)$
$\displaystyle=$
$\displaystyle-A(\Lambda)_{\delta}^{\alpha}\overline{A(\Lambda)_{\mu}^{\beta}}\left(S,g^{\mathcal{C}}\cdot\overline{f^{\mathcal{C}}}\cdot
h_{\gamma}\otimes{{\rm{ad}}}(\rho_{n}(F^{\delta})){{\rm{ad}}}(\rho_{n}(F^{\mu}))\rho_{n}(E^{\gamma})\right)$
$\displaystyle+\left\langle\left(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})\right),\phi^{\beta,n}(f)1\right\rangle\
\phi^{\alpha,n}(g)1.$

Take the sum or difference, and we will obtain

$\displaystyle-A$
$\displaystyle(\Lambda)_{\delta}^{\alpha}\overline{A(\Lambda)_{\mu}^{\beta}}\Big{(}S,[f^{\mathcal{C}}\cdot\overline{g^{\mathcal{C}}}\pm
g^{\mathcal{C}}\cdot\overline{f^{\mathcal{C}}}]\cdot
h_{\gamma}\otimes{{\rm{ad}}}(\rho_{n}(F^{\delta})){{\rm{ad}}}(\rho_{n}(F^{\mu}))\rho_{n}(E^{\gamma})\Big{)}$
$\displaystyle+$
$\displaystyle\left\langle(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})),\phi^{\beta,n}(g)1\right\rangle\phi^{\alpha,n}(f)1\pm\left\langle(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})),\phi^{\beta,n}(f)1\right\rangle\phi^{\alpha,n}(g)1.$

Since $f^{\mathcal{C}}\cdot\overline{g^{\mathcal{C}}}\pm
g^{\mathcal{C}}\cdot\overline{f^{\mathcal{C}}}$ is real and purely imaginary
respectively, this proves Equation (5.1). The proof for Equation (5.2) is
similar, hence omitted.

###### Remark 5.10

Without any loss of generality, we assume that a time-like plane $S^{\flat}$
spanned by $\\{\hat{f}_{0},\hat{f}_{1}\\}$, is parametrized by
$\vec{y}(\hat{s})\equiv\vec{y}(s,\bar{s}):=s\hat{f}_{0}+\bar{s}\hat{f}_{1}$,
$s,\bar{s}\in{{\mathbb{R}}}$, whereby $\hat{f}_{0}\cdot\hat{f}_{0}=-1$,
$\hat{f}_{1}\cdot\hat{f}_{1}=1$, $\hat{f}_{0}\cdot\hat{f}_{1}=0$.

Suppose we now assume that ${\rm supp}\ f$ and ${\rm supp}\ g$ are disjoint
compact sets. Write $\hat{H}=\hat{H}(\rho_{n})$, $\hat{P}=\hat{P}(\rho_{n})$.
Let $\vec{x}\in S$. By definition, for any $\vec{x}\in S$,

$\displaystyle g^{\\{\hat{f}_{0},\hat{f}_{1}\\}}(\hat{H},\hat{P})(\vec{x})=$
$\displaystyle\int_{\hat{s}\in{{\mathbb{R}}}^{2}}\frac{e^{-i[\vec{y}(\hat{s})\cdot(\hat{H}\hat{f}_{0}+\hat{P}\hat{f}_{1})]}}{2\pi}g(\vec{x}+\vec{y}(\hat{s}))|\acute{\rho}_{\vec{y}}|(\hat{s})\
d\hat{s},$
$\displaystyle\overline{f^{\\{\hat{f}_{0},\hat{f}_{1}\\}}}(\hat{H},\hat{P})(\vec{x})=$
$\displaystyle\int_{\hat{t}\in{{\mathbb{R}}}^{2}}\frac{e^{i[\vec{y}(\hat{t})\cdot(\hat{H}\hat{f}_{0}+\hat{P}\hat{f}_{1})]}}{2\pi}\bar{f}(\vec{x}+\vec{y}(\hat{t}))|\acute{\rho}_{\vec{y}}|(\hat{t})\
d\hat{t}.$

Write $g_{\vec{x}}(\cdot)=g(\vec{x}+\cdot)$,
$\bar{f}_{\vec{x}}(\cdot)=\bar{f}(\vec{x}+\cdot)$. Thus,

$\displaystyle\Big{[}g^{\\{\hat{f}_{0},\hat{f}_{1}\\}}$
$\displaystyle\cdot\overline{f^{\\{\hat{f}_{0},\hat{f}_{1}\\}}}\Big{]}(\hat{H},\hat{P})(\vec{x})$
$\displaystyle=\int_{\hat{s},\hat{t}\in{{\mathbb{R}}}^{2}}\frac{e^{-i[\vec{y}(\hat{s})\cdot(\hat{H}\hat{f}_{0}+\hat{P}\hat{f}_{1})]}}{(2\pi)^{2}}g_{\vec{x}}(\vec{y}(\hat{s})+\vec{y}(\hat{t}))\bar{f}_{\vec{x}}(\vec{y}(\hat{t}))|\acute{\rho}_{\vec{y}}|(\hat{s})|\acute{\rho}_{\vec{y}}|(\hat{t})\
d\hat{s}d\hat{t}$ $\displaystyle=\int_{\hat{t}\in{{\mathbb{R}}}^{2},\hat{s}\in
D}\frac{e^{-i[\vec{y}(\hat{s})\cdot(\hat{H}\hat{f}_{0}+\hat{P}\hat{f}_{1})]}}{(2\pi)^{2}}g_{\vec{x}}(\vec{y}(\hat{s})+\vec{y}(\hat{t}))\bar{f}_{\vec{x}}(\vec{y}(\hat{t}))|\acute{\rho}_{\vec{y}}|(\hat{s})|\acute{\rho}_{\vec{y}}|(\hat{t})\
d\hat{s}d\hat{t}.$

Similarly,

$\displaystyle\Big{[}f^{\\{\hat{f}_{0},\hat{f}_{1}\\}}$
$\displaystyle\cdot\overline{g^{\\{\hat{f}_{0},\hat{f}_{1}\\}}}\Big{]}(\hat{H},\hat{P})(\vec{x})$
$\displaystyle=\int_{\hat{t}\in{{\mathbb{R}}}^{2},\hat{s}\in-D}\frac{e^{-i[\vec{y}(\hat{s})\cdot(\hat{H}\hat{f}_{0}+\hat{P}\hat{f}_{1})]}}{(2\pi)^{2}}f_{\vec{x}}(\vec{y}(\hat{s})+\vec{y}(\hat{t}))\bar{g}_{\vec{x}}(\vec{y}(\hat{t}))|\acute{\rho}_{\vec{y}}|(\hat{s})|\acute{\rho}_{\vec{y}}|(\hat{t})\
d\hat{s}d\hat{t}.$

From both expressions, we see that the integrals depend on the relative
displacement between pairs of positions in their respective supports. When the
region of integration is on $D$, it is clear that we are referring to
$g^{\\{\hat{f}_{0},\hat{f}_{1}\\}}\cdot\overline{f^{\\{\hat{f}_{0},\hat{f}_{1}\\}}}$;
when the region of integration is on $-D$, then we are referring to
$f^{\\{\hat{f}_{0},\hat{f}_{1}\\}}\cdot\overline{g^{\\{\hat{f}_{0},\hat{f}_{1}\\}}}$.
Note that the vectors in the set $\vec{y}(-D)$, are in the opposite direction
of those vectors in $\vec{y}(D)$.

Thus, by reversing time direction and taking space inversion (parity), we can
obtain its complex conjugate. The CPT theorem is being implied by these two
expressions. See Remark 5.23. Note that here, ‘C’ refers to complex
conjugation, not charge conjugation.

In general, the anti-commutators and commutators will not be equal to zero,
even when the supports are space-like separated.

###### Lemma 5.11

Let $f,g\in\mathscr{P}$ for which their supports are space-like separated, and
let $S$ be a space-like plane, equipped with a Minkowski frame
$\\{\hat{f}_{a}\\}_{a=0}^{3}$. Let $S^{\flat}$ be the span of
$\\{\hat{f}_{0},\hat{f}_{1}\\}$.

Fix a $\vec{x}\in S$. Write $g_{\vec{x}}(\cdot)=g(\vec{x}+\cdot)$,
$\bar{f}_{\vec{x}}(\cdot)=\bar{f}(\vec{x}+\cdot)$. Suppose
$\hat{H}(\rho_{n})^{2}-\hat{P}(\rho_{n})^{2}>0$.

If ${\rm supp}\ f\cap(\vec{x}+S^{\flat})=\emptyset$ or ${\rm supp}\
g\cap(\vec{x}+S^{\flat})=\emptyset$, then for any $u,v\in S^{\flat}$, we have

$\frac{e^{-i[(u-v)\cdot(\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1})]}}{(2\pi)^{2}}g_{\vec{x}}(u)\bar{f}_{\vec{x}}(v)\mp\frac{e^{-i[(v-u)\cdot(\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1})]}}{(2\pi)^{2}}f_{\vec{x}}(v)\bar{g}_{\vec{x}}(u)=0.$
(5.3)

Suppose both sets are non-empty. If $u-v$ is parallel to
$\hat{P}(\rho_{n})\hat{f}_{0}+\hat{H}(\rho_{n})\hat{f}_{1}$, then the
commutation and anti-commutation relations in Equation (5.3) hold when
$f\cdot\bar{g}$ is real and imaginary respectively.

Proof. When one of the sets is empty, then
$f_{\vec{x}}(v)\cdot\bar{g}_{\vec{x}}(u)=0$, so clearly Equation (5.3) holds.

Now consider when both are non-empty. Since

$(u-v)\cdot(\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1})=c(\hat{P}(\rho_{n})\hat{f}_{0}+\hat{H}(\rho_{n})\hat{f}_{1})\cdot(\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1})=0,$

we have

$e^{-i[(u-v)\cdot(\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1})]}=\cos\left[(u-v)\cdot(\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1})\right]=1.$

When $f_{\vec{x}}(v)\cdot\bar{g}_{\vec{x}}(u)$ is real, then the LHS of
Equation (5.3) becomes

$\left[\frac{e^{-i[(u-v)\cdot(\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1})]}}{(2\pi)^{2}}-\frac{e^{-i[(v-u)\cdot(\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1})]}}{(2\pi)^{2}}\right]f_{\vec{x}}(v)\bar{g}_{\vec{x}}(u),$

which is zero.

When $f_{\vec{x}}(v)\cdot\bar{g}_{\vec{x}}(u)$ is purely imaginary, then the
LHS of Equation (5.3) becomes

$\left[-\frac{e^{-i[(u-v)\cdot(\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1})]}}{(2\pi)^{2}}+\frac{e^{-i[(v-u)\cdot(\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1})]}}{(2\pi)^{2}}\right]f_{\vec{x}}(v)\bar{g}_{\vec{x}}(u),$

which is zero.

###### Remark 5.12

Suppose both ${\rm supp}\ f\cap(\vec{x}+S^{\flat})$ and ${\rm supp}\
g\cap(\vec{x}+S^{\flat})$ are non-empty. The lemma says that there exists a
space-like line in $S^{\flat}$, such that the LHS of Equation (5.3) is zero.

If $\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1}$ is space-like
or null, then the LHS of Equation (5.3) cannot be zero, on any space-like line
in $S^{\flat}$. This can be inferred from Lemma B.3.

Thus, it is essential that a positive mass gap exists in
$\mathscr{H}(\rho_{n})$, for the lemma to hold true.

Consider a bilinear map that sends

$\displaystyle(f,g)$ $\displaystyle\in\mathscr{P}\times\mathscr{P}$
$\displaystyle\longmapsto$
$\displaystyle\left\langle\phi^{\alpha,n}(f)\phi^{\beta,n}(g)^{\ast}\left(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})\right),(\tilde{S},\tilde{h}_{\gamma}\otimes\rho_{n}(E^{\gamma}))\right\rangle$
$\displaystyle-\left\langle\left(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})\right),\phi^{\beta,n}(g)1\right\rangle\left\langle\phi^{\alpha,n}(f)1,(\tilde{S},\tilde{h}_{\gamma}\otimes\rho_{n}(E^{\gamma}))\right\rangle.$

From Proposition 3.22, we have a tempered distribution $W(\vec{x},\vec{y})$
such that

$\displaystyle\int_{\vec{x}\in{{\mathbb{R}}}^{4}}$
$\displaystyle\int_{\vec{y}\in{{\mathbb{R}}}^{4}}W(\vec{x},\vec{y})f(\vec{x})\otimes_{{\mathbb{R}}}g(\vec{y})\
d\vec{x}d\vec{y}$ $\displaystyle=$
$\displaystyle\left\langle\phi^{\alpha,n}(f)\phi^{\beta,n}(g)^{\ast}\left(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})\right),(\tilde{S},\tilde{h}_{\gamma}\otimes\rho_{n}(E^{\gamma}))\right\rangle$
$\displaystyle-\left\langle\left(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})\right),\phi^{\beta,n}(g)1\right\rangle\left\langle\phi^{\alpha,n}(f)1,(\tilde{S},\tilde{h}_{\gamma}\otimes\rho_{n}(E^{\gamma}))\right\rangle.$

See Remark 5.14.

Indeed, writing $\vec{x}=\sum_{a=0}^{3}x^{a}\hat{f}_{a}$ and
$\vec{y}=\sum_{a=0}^{3}y^{a}\hat{f}_{a}$, we have that

$\displaystyle\left(x^{0}\hat{f}_{0}+x^{1}\hat{f}_{1},y^{0}\hat{f}_{0}+y^{1}\hat{f}_{1}\right)\longmapsto$
$\displaystyle\int_{(x^{2},x^{3})\in{{\mathbb{R}}}^{2}}\int_{(y^{2},y^{3})\in{{\mathbb{R}}}^{2}}W(\vec{x},\vec{y})f(\vec{x})\otimes_{{\mathbb{R}}}g(\vec{y})\
dx^{2}dx^{3}dy^{2}dy^{3},$ (5.4)

defines a continuous function on $S^{\flat}\times S^{\flat}$.

###### Notation 5.13

Suppose $f\cdot\bar{g}$ is real. Then we will define tempered distribution
${\rm Re}\ W$, such that

$\displaystyle\int_{\vec{x}\in{{\mathbb{R}}}^{4}}\int_{\vec{y}\in{{\mathbb{R}}}^{4}}$
$\displaystyle{\rm Re}\
W(\vec{x},\vec{y})\left[f(\vec{x})\otimes_{{\mathbb{R}}}g(\vec{y})\right]\
d\vec{x}d\vec{y}$
$\displaystyle:=\int_{\vec{x}\in{{\mathbb{R}}}^{4}}\int_{\vec{y}\in{{\mathbb{R}}}^{4}}W(\vec{x},\vec{y})\left[f(\vec{x})\otimes_{{\mathbb{R}}}g(\vec{y})\right]\
d\vec{x}d\vec{y}.$

When $f\cdot\bar{g}$ is purely imaginary, we will define tempered distribution
${\rm Im}\ W$, such that

$\displaystyle\int_{\vec{x}\in{{\mathbb{R}}}^{4}}\int_{\vec{y}\in{{\mathbb{R}}}^{4}}$
$\displaystyle{\rm Im}\
W(\vec{x},\vec{y})\left[f(\vec{x})\otimes_{{\mathbb{R}}}g(\vec{y})\right]\
d\vec{x}d\vec{y}$
$\displaystyle:=\int_{\vec{x}\in{{\mathbb{R}}}^{4}}\int_{\vec{y}\in{{\mathbb{R}}}^{4}}W(\vec{x},\vec{y})\left[f(\vec{x})\otimes_{{\mathbb{R}}}g(\vec{y})\right]\
d\vec{x}d\vec{y}.$

###### Remark 5.14

Suppose we write $f=\underline{f}+i\overline{f}$,
$g=\underline{g}+i\overline{g}$, whereby $\underline{f}={\rm Re}\ f$,
$\overline{f}={\rm Im}\ f$, $\underline{g}={\rm Re}\ g$, $\overline{g}={\rm
Im}\ g$. From Notation 5.13, we understand

$\displaystyle\int_{\vec{x}\in{{\mathbb{R}}}^{4}}$
$\displaystyle\int_{\vec{y}\in{{\mathbb{R}}}^{4}}W(\vec{x},\vec{y})\left[f(\vec{x})\otimes_{{\mathbb{R}}}g(\vec{y})\right]\
d\vec{x}d\vec{y}$ $\displaystyle=$
$\displaystyle\int_{\vec{x}\in{{\mathbb{R}}}^{4}}\int_{\vec{y}\in{{\mathbb{R}}}^{4}}{\rm
Re}\
W(\vec{x},\vec{y})\left[\underline{f}(\vec{x})\underline{g}(\vec{y})+\overline{f}(\vec{x})\overline{g}(\vec{y})\right]\
d\vec{x}d\vec{y}$
$\displaystyle+i\int_{\vec{x}\in{{\mathbb{R}}}^{4}}\int_{\vec{y}\in{{\mathbb{R}}}^{4}}{\rm
Im}\
W(\vec{x},\vec{y})\left[\overline{f}(\vec{x})\underline{g}(\vec{y})-\underline{f}(\vec{x})\overline{g}(\vec{y})\right]\
d\vec{x}d\vec{y}.$

###### Lemma 5.15

Fix a space-like plane $S$ and let the Minkowski frame on $S$ be
$\hat{f}_{a}=\Lambda e_{a}$, $a=0,1,2,3$, as defined in Definition 2.5.
Suppose $\hat{H}(\rho_{n})^{2}-\hat{P}(\rho_{n})^{2}>0$.

We have

$\displaystyle{\rm Re}\ W(\vec{x},\vec{y})-{\rm Re}\ W(\vec{y},\vec{x})$
$\displaystyle=0,$ $\displaystyle{\rm Im}\ W(\vec{x},\vec{y})+{\rm Im}\
W(\vec{y},\vec{x})$ $\displaystyle=0,$

provided $\vec{0}\neq\vec{x}-\vec{y}$ can be written as
$c_{1}(\hat{P}(\rho_{n})\hat{f}_{0}+\hat{H}(\rho_{n})\hat{f}_{1})+\sum_{i=2}^{3}c_{i}\hat{f}_{i}$
for some constants $c_{i}$’s.

Proof. Choose any $f,g\in\mathscr{P}$ such that their compact supports are
space-like separated. Write $g_{\vec{x}}(\cdot)=g(\vec{x}+\cdot)$,
$\bar{f}_{\vec{x}}(\cdot)=\bar{f}(\vec{x}+\cdot)$, and
$\hat{H}=\hat{H}(\rho_{n})$, $\hat{P}=\hat{P}(\rho_{n})$.

From the proof in Lemma 5.9,

$\displaystyle\phi^{\alpha,n}(f)$
$\displaystyle\phi^{\beta,n}(g)^{\ast}\left(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})\right)-\left\langle\left(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})\right),\phi^{\beta,n}(g)1\right\rangle\
\phi^{\alpha,n}(f)1$ $\displaystyle=$
$\displaystyle-A(\Lambda)_{\delta}^{\alpha}\overline{A(\Lambda)_{\mu}^{\beta}}\Big{(}S,[f^{\\{\hat{f}_{0},\hat{f}_{1}\\}}\cdot\overline{g^{\\{\hat{f}_{0},\hat{f}_{1}\\}}}]\cdot
h_{\gamma}\otimes{{\rm{ad}}}(\rho_{n}(F^{\delta})){{\rm{ad}}}(\rho_{n}(F^{\mu}))\rho_{n}(E^{\gamma})\Big{)}.$

Refer to the calculations in Remark 5.10. If we swap $f$ with $g$ in the above
expression and take the sum or difference, we will have

$-A(\Lambda)_{\delta}^{\alpha}\overline{A(\Lambda)_{\mu}^{\beta}}\Big{(}S,A^{\pm}\cdot
h_{\gamma}\otimes{{\rm{ad}}}(\rho_{n}(F^{\delta})){{\rm{ad}}}(\rho_{n}(F^{\mu}))\rho_{n}(E^{\gamma})\Big{)},$

whereby

$\displaystyle A^{\pm}(\vec{z})=$
$\displaystyle\int_{\hat{s},\hat{t}\in{{\mathbb{R}}}^{2}}\frac{e^{-i[(\vec{y}(\hat{s})-\vec{y}(\hat{t}))\cdot(\hat{H}\hat{f}_{0}+\hat{P}\hat{f}_{1})]}}{(2\pi)^{2}}f_{\vec{z}}(\vec{y}(\hat{s}))\bar{g}_{\vec{z}}(\vec{y}(\hat{t}))|\acute{\rho}_{\vec{y}}|(\hat{s})|\acute{\rho}_{\vec{y}}|(\hat{t})\
d\hat{s}d\hat{t}$
$\displaystyle\pm\int_{\hat{s},\hat{t}\in{{\mathbb{R}}}^{2}}\frac{e^{-i[(\vec{y}(\hat{t})-\vec{y}(\hat{s}))\cdot(\hat{H}\hat{f}_{0}+\hat{P}\hat{f}_{1})]}}{(2\pi)^{2}}g_{\vec{z}}(\vec{y}(\hat{t}))\bar{f}_{\vec{z}}(\vec{y}(\hat{s}))|\acute{\rho}_{\vec{y}}|(\hat{s})|\acute{\rho}_{\vec{y}}|(\hat{t})\
d\hat{s}d\hat{t}.$

Since their supports are disjoint, by definition of $W(\vec{y},\vec{x})$, we
note that swapping the arguments $\vec{x}$ and $\vec{y}$, is equivalent to
swapping $f$ and $g$, i.e. $f(\vec{x})\bar{g}(\vec{y})\mapsto
g(\vec{y})\bar{f}(\vec{x})$ in their respective integrals.

Consider when
$\vec{0}\neq\vec{x}-\vec{y}=c_{1}(\hat{P}\hat{f}_{0}+\hat{H}\hat{f}_{1})+\sum_{i=2}^{3}c_{i}\hat{f}_{i}$,
whereby not all the $c_{i}$’s are zero. By Lemma 5.11, we see that if $c_{2}$
or $c_{3}$ is non-zero, then we must have
$W(\vec{x},\vec{y})=W(\vec{y},\vec{x})=0$.

Consider when $c_{1}\neq 0$ and $f\bar{g}$ is real. By Equation (5.3), we have
that

$\displaystyle{\rm Re}\ W(\vec{x},\vec{y})-{\rm Re}\ W(\vec{y},\vec{x})=0,$

for the real part.

Now consider when $c_{1}\neq 0$ and $f\bar{g}$ is purely imaginary. By
Equation (5.3), we have that

$\displaystyle{\rm Im}\ W(\vec{x},\vec{y})+{\rm Im}\ W(\vec{y},\vec{x})=0,$

for the imaginary part.

###### Remark 5.16

In the Wightman’s axiom for local commutativity, it is required that
$\lfloor\phi^{\alpha,n}(f),\phi^{\beta,n}(g)^{\ast}\rfloor_{\pm}$ is zero when
their supports are space-like separated. This is not true in general, even if
$f\cdot\bar{g}$ is real or purely imaginary. The same remark applies to
$\lfloor\phi^{\alpha,n}(f)^{\ast},\phi^{\beta,n}(g)\rfloor_{\pm}$, as we will
see in Lemmas 5.18 and 5.20.

Indeed, when $(\vec{x}+S^{\flat})\cap\ {\rm supp}\ f$ or
$(\vec{x}+S^{\flat})\cap\ {\rm supp}\ g$ is empty for every $\vec{x}\in S$,
then we have
$\lfloor\phi^{\alpha,n}(f),\phi^{\beta,n}(g)^{\ast}\rfloor_{\pm}=\lfloor\phi^{\alpha,n}(f)^{\ast},\phi^{\beta,n}(g)\rfloor_{\pm}=0$,
acting on $(S,f_{\alpha}\otimes E^{\alpha})$.

The above lemma says that if we consider the orthogonal projection $P_{0}$, on
the orthogonal complement of $\\{1\\}$, then the operators commute or anti-
commute on the hyperplane spanned by space-like vectors
$\\{\hat{P}(\rho_{n})\hat{f}_{0}+\hat{H}(\rho_{n})\hat{f}_{1},\hat{f}_{2},\hat{f}_{3}\\}$.
This fact will be used in the proof of Lemma 8.23.

Consider another bilinear map that sends

$\displaystyle(f,g)$
$\displaystyle\in\mathscr{P}\times\mathscr{P}\longmapsto\left\langle\phi^{\alpha,n}(f)^{\ast}\phi^{\beta,n}(g)\left(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})\right),(\tilde{S},\tilde{h}_{\gamma}\otimes\rho_{n}(E^{\gamma}))\right\rangle.$

From Proposition 3.22, we have a tempered distribution
$\tilde{W}(\vec{x},\vec{y})$ such that

$\displaystyle\int_{\vec{x}\in{{\mathbb{R}}}^{4}}$
$\displaystyle\int_{\vec{y}\in{{\mathbb{R}}}^{4}}\tilde{W}(\vec{x},\vec{y})f(\vec{x})\otimes_{{\mathbb{R}}}g(\vec{y})\
d\vec{x}d\vec{y}$ $\displaystyle=$
$\displaystyle\left\langle\phi^{\alpha,n}(f)^{\ast}\phi^{\beta,n}(g)\left(S,h_{\gamma}\otimes\rho_{n}(E^{\gamma})\right),(\tilde{S},\tilde{h}_{\gamma}\otimes\rho_{n}(E^{\gamma}))\right\rangle.$

Similar to Equation (5.4), we can define a continuous complex-valued function
on $S^{\flat}\times S^{\flat}$ from it.

###### Remark 5.17

Suppose we write $f=\underline{f}+i\overline{f}$,
$g=\underline{g}+i\overline{g}$, whereby $\underline{f}={\rm Re}\ f$,
$\overline{f}={\rm Im}\ f$, $\underline{g}={\rm Re}\ g$, $\overline{g}={\rm
Im}\ g$. Define tempered distributions ${\rm Re}\ \tilde{W}$ and ${\rm Im}\
\tilde{W}$, similar to how we defined ${\rm Re}\ W$ and ${\rm Im}\ W$ in
Notation 5.13. We understand

$\displaystyle\int_{\vec{x}\in{{\mathbb{R}}}^{4}}$
$\displaystyle\int_{\vec{y}\in{{\mathbb{R}}}^{4}}\tilde{W}(\vec{x},\vec{y})\left[f(\vec{x})\otimes_{{\mathbb{R}}}g(\vec{y})\right]\
d\vec{x}d\vec{y}$ $\displaystyle=$
$\displaystyle\int_{\vec{x}\in{{\mathbb{R}}}^{4}}\int_{\vec{y}\in{{\mathbb{R}}}^{4}}{\rm
Re}\
\tilde{W}(\vec{x},\vec{y})\left[\underline{f}(\vec{x})\underline{g}(\vec{y})+\overline{f}(\vec{x})\overline{g}(\vec{y})\right]\
d\vec{x}d\vec{y}$
$\displaystyle+i\int_{\vec{x}\in{{\mathbb{R}}}^{4}}\int_{\vec{y}\in{{\mathbb{R}}}^{4}}{\rm
Im}\
\tilde{W}(\vec{x},\vec{y})\left[\underline{f}(\vec{x})\overline{g}(\vec{y})-\overline{f}(\vec{x})\underline{g}(\vec{y})\right]\
d\vec{x}d\vec{y}.$

###### Lemma 5.18

Fix a space-like plane $S$ and let $\\{\hat{f}_{a}\\}_{a=0}^{3}$ be a basis as
defined in Definition 2.5. Suppose
$\hat{H}(\rho_{n})^{2}-\hat{P}(\rho_{n})^{2}>0$.

We have

$\displaystyle{\rm Re}\ \tilde{W}(\vec{x},\vec{y})-{\rm Re}\
\tilde{W}(\vec{y},\vec{x})$ $\displaystyle=0,$ $\displaystyle{\rm Im}\
\tilde{W}(\vec{x},\vec{y})+{\rm Im}\ \tilde{W}(\vec{y},\vec{x})$
$\displaystyle=0,$

provided $\vec{0}\neq\vec{x}-\vec{y}$ can be written as
$c_{1}(\hat{P}(\rho_{n})\hat{f}_{0}+\hat{H}(\rho_{n})\hat{f}_{1})+\sum_{i=2}^{3}c_{i}\hat{f}_{i}$
for some constants $c_{i}$’s.

Proof. Proof is similar to Lemma 5.15, hence omitted.

Finally consider the following bilinear map that sends

$\displaystyle(f,g)$
$\displaystyle\in\mathscr{P}\times\mathscr{P}\longmapsto\left\langle\phi^{\alpha,n}(f)^{\ast}\phi^{\beta,n}(g)1,(\tilde{S},\tilde{h}_{\gamma}\otimes\rho_{n}(E^{\gamma}))\right\rangle.$

From Proposition 3.22, we have a tempered distribution
$\check{W}(\vec{x},\vec{y})$ such that

$\displaystyle\int_{\vec{x}\in{{\mathbb{R}}}^{4}}$
$\displaystyle\int_{\vec{y}\in{{\mathbb{R}}}^{4}}\check{W}(\vec{x},\vec{y})f(\vec{x})\otimes_{{\mathbb{R}}}g(\vec{y})\
d\vec{x}d\vec{y}$ $\displaystyle=$
$\displaystyle\left\langle\phi^{\alpha,n}(f)^{\ast}\phi^{\beta,n}(g)1,(\tilde{S},\tilde{h}_{\gamma}\otimes\rho_{n}(E^{\gamma}))\right\rangle.$

Similar to Equation (5.4), we can obtain a continuous complex-valued function
defined on $S^{\flat}\times S^{\flat}$, from the integral.

###### Remark 5.19

Suppose we write $f=\underline{f}+i\overline{f}$,
$g=\underline{g}+i\overline{g}$, whereby $\underline{f}={\rm Re}\ f$,
$\overline{f}={\rm Im}\ f$, $\underline{g}={\rm Re}\ g$, $\overline{g}={\rm
Im}\ g$. Define tempered distributions ${\rm Re}\ \check{W}$ and ${\rm Im}\
\check{W}$, similar to how we defined ${\rm Re}\ W$ and ${\rm Im}\ W$ in
Notation 5.13. We understand

$\displaystyle\int_{\vec{x}\in{{\mathbb{R}}}^{4}}$
$\displaystyle\int_{\vec{y}\in{{\mathbb{R}}}^{4}}\check{W}(\vec{x},\vec{y})\left[f(\vec{x})\otimes_{{\mathbb{R}}}g(\vec{y})\right]\
d\vec{x}d\vec{y}$ $\displaystyle=$
$\displaystyle\int_{\vec{x}\in{{\mathbb{R}}}^{4}}\int_{\vec{y}\in{{\mathbb{R}}}^{4}}{\rm
Re}\
\check{W}(\vec{x},\vec{y})\left[\underline{f}(\vec{x})\underline{g}(\vec{y})+\overline{f}(\vec{x})\overline{g}(\vec{y})\right]\
d\vec{x}d\vec{y}$
$\displaystyle+i\int_{\vec{x}\in{{\mathbb{R}}}^{4}}\int_{\vec{y}\in{{\mathbb{R}}}^{4}}{\rm
Im}\
\check{W}(\vec{x},\vec{y})\left[\underline{f}(\vec{x})\overline{g}(\vec{y})-\overline{f}(\vec{x})\underline{g}(\vec{y})\right]\
d\vec{x}d\vec{y}.$

###### Lemma 5.20

Fix a space-like plane $S$ and let $\\{\hat{f}_{a}\\}_{a=0}^{3}$ be a basis as
defined in Definition 2.5. Suppose
$\hat{H}(\rho_{n})^{2}-\hat{P}(\rho_{n})^{2}>0$.

We have

$\displaystyle{\rm Re}\ \check{W}(\vec{x},\vec{y})-{\rm Re}\
\check{W}(\vec{y},\vec{x})$ $\displaystyle=0,$ $\displaystyle{\rm Im}\
\check{W}(\vec{x},\vec{y})+{\rm Im}\ \check{W}(\vec{y},\vec{x})$
$\displaystyle=0,$

provided $\vec{0}\neq\vec{x}-\vec{y}$ can be written as
$c_{1}(\hat{P}(\rho_{n})\hat{f}_{0}+\hat{H}(\rho_{n})\hat{f}_{1})+\sum_{i=2}^{3}c_{i}\hat{f}_{i}$
for some constants $c_{i}$’s.

Proof. Proof is similar to Lemma 5.15, hence omitted.

###### Remark 5.21

Note that $\phi^{\alpha,n}(f)^{\ast}1=0$ by definition, thus we always have

$\phi^{\alpha,n}(f)\phi^{\beta,n}(g)^{\ast}1\pm\phi^{\alpha,n}(g)\phi^{\beta,n}(f)^{\ast}1=0.$

The Lemmas 5.6, 5.15, 5.18 and 5.20 are collectively known as the local
commutation and anti-commutation relations.

Indeed, we see that the relations hold, because
$\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1}$ is time-like. If
it is space-like or null vector, then we see local commutativity only holds
for space-like directions in the span of $\\{\hat{f}_{2},\hat{f}_{3}\\}$.

Now,
$\\{\hat{P}(\rho_{n})\hat{f}_{0}+\hat{H}(\rho_{n})\hat{f}_{1},\hat{f}_{2},\hat{f}_{3}\\}$
is a set of space-like vectors. In fact, for any vector given by a linear
combination of these three vectors, there exists a sequence of translations
and Lorentz transformations that rotates it by $\pi$ radians in
${{\mathbb{R}}}^{4}$, which is time and space inversion. Refer to Lemma B.1
and Remark B.2.

###### Corollary 5.22

(CPT Theorem)
Fix a space-like plane $S$, with a time-like plane $S^{\flat}$ spanned by
$\\{\hat{f}_{0},\hat{f}_{1}\\}$, whereby $\\{\hat{f}_{a}\\}_{a=0}^{3}$ is a
basis defined in Definition 2.5. By abuse of notation, write

$W={\rm Re}\ W+\sqrt{-1}\ {\rm Im}\ W,\ \ \tilde{W}={\rm Re}\
\tilde{W}+\sqrt{-1}\ {\rm Im}\ \tilde{W},\ \ \check{W}={\rm Re}\
\check{W}+\sqrt{-1}\ {\rm Im}\ \check{W}.$

Suppose $\hat{H}(\rho_{n})^{2}-\hat{P}(\rho_{n})^{2}>0$. We have that

$\displaystyle W(\vec{y},\vec{x})$
$\displaystyle=\overline{W}(\vec{x},\vec{y}),\quad\tilde{W}(\vec{y},\vec{x})=\overline{\tilde{W}}(\vec{x},\vec{y}),\quad\check{W}(\vec{y},\vec{x})=\overline{\check{W}}(\vec{x},\vec{y}),$

provided
$\vec{0}\neq\vec{x}-\vec{y}=c_{1}(\hat{P}(\rho_{n})\hat{f}_{0}+\hat{H}(\rho_{n})\hat{f}_{1})+\sum_{i=2}^{3}c_{i}\hat{f}_{i}$,
for constants $c_{i}$’s.

Proof. Immediate from the statements in Lemmas 5.15, 5.18 and 5.20.

###### Remark 5.23

In Lemma 8.13, we will see that we can write
$W(\vec{y},\vec{x})=\mathscr{W}(\vec{\xi})$, for some distribution
$\mathscr{W}$, and $\vec{\xi}=\vec{y}-\vec{x}$. The same remark applies to
$\tilde{W}(\vec{y},\vec{x})$ and $\check{W}(\vec{y},\vec{x})$.

Thus, the above corollary says that by taking time inversion and space
inversion (parity) of a space-like vector, i.e. $\vec{\xi}\mapsto-\vec{\xi}$,
is equivalent to taking the complex conjugation. This is the content of the
CPT Theorem. Refer also to Remark 5.10. But, this only applies if $\vec{\xi}$
lies in the hyperplane spanned by
$\\{\hat{P}(\rho_{n})\hat{f}_{0}+\hat{H}(\rho_{n})\hat{f}_{1},\hat{f}_{2},\hat{f}_{3}\\}$.

In [3], it is actually stated that

$\displaystyle\phi^{\alpha,n}(f)\phi^{\beta,n}(g)\pm\phi^{\beta,n}(g)\phi^{\alpha,n}(f)$
$\displaystyle=0,$
$\displaystyle\phi^{\alpha,n}(f)^{\ast}\phi^{\beta,n}(g)\pm\phi^{\beta,n}(g)\phi^{\alpha,n}(f)^{\ast}$
$\displaystyle=0,$

if their respective supports are space-like separated. But this is only true
iff
${\rm supp}\ f^{\\{\hat{f}_{0},\hat{f}_{1}\\}}\cap{\rm supp}\
g^{\\{\hat{f}_{0},\hat{f}_{1}\\}}=\emptyset$.

It would be ideal, that Lemmas 5.15, 5.18 and 5.20 hold provided
$0\neq\vec{x}-\vec{y}$ is any space-like vector. But, this cannot be true in
general. What we have shown instead, is that the commutation and anti-
commutation relations hold in a three dimensional subspace, containing space-
like vectors.

We would suggest that the final Wightman’s axiom be replaced by the following.

###### Axiom 5.24

Let $H$ be a Hilbert space, as described in Wightman’s zeroth axiom. Let
$f,g\in\mathscr{P}$ and define tempered distributions
$\Omega,\widetilde{\Omega},\widehat{\Omega}:\mathscr{P}\times\mathscr{P}\rightarrow{{\mathbb{C}}}$,
such that

$\displaystyle\int_{{{\mathbb{R}}}^{4}\times{{\mathbb{R}}}^{4}}\Omega(\vec{x},\vec{y})f(\vec{x})g(\vec{y})\
d\vec{x}d\vec{y}$
$\displaystyle:=\left\langle\phi^{\alpha}(f)\phi^{\beta}(g)\Phi,\Psi\right\rangle,$
$\displaystyle\int_{{{\mathbb{R}}}^{4}\times{{\mathbb{R}}}^{4}}\widetilde{\Omega}(\vec{x},\vec{y})f(\vec{x})g(\vec{y})\
d\vec{x}d\vec{y}$
$\displaystyle:=\left\langle\phi^{\alpha}(f)\phi^{\beta}(g)^{\ast}\Phi,\Psi\right\rangle-\left\langle\Phi,\phi^{\beta}(g)1\right\rangle\left\langle\phi^{\alpha}(f)1,\Psi\right\rangle,$
$\displaystyle\int_{{{\mathbb{R}}}^{4}\times{{\mathbb{R}}}^{4}}\widehat{\Omega}(\vec{x},\vec{y})f(\vec{x})g(\vec{y})\
d\vec{x}d\vec{y}$
$\displaystyle:=\left\langle\phi^{\alpha}(f)^{\ast}\phi^{\beta}(g)\Phi,\Psi\right\rangle,$

whereby the field operators $\phi^{\alpha}(f)$, $\phi^{\beta}(g)$ act on
$\Phi,\Psi\in\mathscr{D}\subset H$.

Then, there exists a three-dimensional subspace $V$ in ${{\mathbb{R}}}^{4}$,
such that

$\displaystyle\Omega(\vec{x},\vec{y})=\pm\Omega(\vec{y},\vec{x}),\quad\widetilde{\Omega}(\vec{x},\vec{y})=\pm\widetilde{\Omega}(\vec{y},\vec{x}),\quad\widehat{\Omega}(\vec{x},\vec{y})=\pm\widehat{\Omega}(\vec{y},\vec{x}),$

if $\vec{0}\neq\vec{x}-\vec{y}\in V$. Here, any non-zero vector $v\in V$ is
space-like.

##  6 Yang-Mills path integrals

We have completed the description of Wightman’s axioms. We have seen that to
satisfy local commutativity, the vector
$\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1}$ must be time-like,
thus $\hat{H}(\rho_{n})^{2}-\hat{P}(\rho_{n})^{2}=m_{n}^{2}>0$, for each
$n\geq 1$. Local commutativity already implies the existence of a positive
mass gap in each component Hilbert space $\mathscr{H}(\rho_{n})$. Indeed, we
will see that $\hat{H}(\rho_{n})\hat{f}_{0}+\hat{P}(\rho_{n})\hat{f}_{1}$
defines a time-like222In subsection 8.2, we will see that the mass gap $m_{n}$
is the generator for translation in the $\tilde{f}_{0}^{n}$ direction. vector
$m_{n}\tilde{f}_{0}^{n}$, $\tilde{f}_{0}^{n}\cdot\tilde{f}_{0}^{n}=-1$, in
Definition 8.28, which is crucial to prove the Clustering Theorem 8.34.

But how do we choose
$\\{(\hat{H}(\rho_{n}),\hat{P}(\rho_{n})):n\in\mathbb{N}\\}$? In the next
section, we will explain how we are going to compute the eigenvalues for the
Hamiltonian and momentum operator. To prove the existence of a positive mass
gap, we need to further show that $\inf_{n\in\mathbb{N}}m_{n}>0$.

To do the quantization, we need to turn to Yang-Mills path integrals, which we
will now summarize the construction done in [8] and [6].

###  6.1 Hermite polynomials

Consider the inner product space $\mathcal{S}_{\kappa}(\mathbb{R}^{4})$,
consisting of functions of the form $f\sqrt{\phi_{\kappa}}$, whereby
$\phi_{\kappa}(\vec{x})=\kappa^{4}e^{-\kappa^{2}|\vec{x}|^{2}/2}/(2\pi)^{2}$
is a Gaussian function and $f$ is a polynomial in
$\vec{x}=(x^{0},x^{1},x^{2},x^{3})\in{{\mathbb{R}}}^{4}$. Its inner product is
given by

$\left\langle
f\sqrt{\phi_{\kappa}},g\sqrt{\phi_{\kappa}}\right\rangle=\int_{{{\mathbb{R}}}^{4}}fg\cdot\phi_{\kappa}\
d\lambda,$

$\lambda$ is Lebesgue measure on $\mathbb{R}^{4}$.

Suppose $h_{i}/\sqrt{i}$ is a normalized Hermite polynomial of degree $i$ on
${{\mathbb{R}}}$. Let $\overline{\mathcal{S}}_{\kappa}({{\mathbb{R}}}^{4})$ be
the smallest Hilbert space containing
$\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})$ and hence

$\left\\{\frac{h_{i}(\kappa x^{0})h_{j}(\kappa x^{1})h_{k}(\kappa
x^{2})h_{l}(\kappa x^{3})}{\sqrt{i!j!k!l!}}\sqrt{\phi_{\kappa}(\vec{x})}\
\Big{|}\ \vec{x}=(x^{0},x^{1},x^{2},x^{3})\in{{\mathbb{R}}}^{4},\ i,j,k,l\geq
0\right\\}$

forms an orthonormal basis. Note its dependence on $\kappa>0$.

Recall we chose the standard metric on $T{{\mathbb{R}}}^{4}$, thus the volume
form on ${{\mathbb{R}}}^{4}$ is given by $d\omega=dx^{0}\wedge dx^{1}\wedge
dx^{2}\wedge dx^{3}$. Using the Hodge star operator and the above volume form,
we will define an inner product on
$\mathcal{\mathcal{S}}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{4})$
from Equation (1.2). Explicitly, it is given by

$\left\langle\sum_{0\leq a<b\leq 3}f_{ab}\otimes dx^{a}\wedge
dx^{b},\sum_{0\leq a<b\leq 3}\hat{f}_{ab}\otimes dx^{a}\wedge
dx^{b}\right\rangle=\sum_{0\leq a<b\leq 3}\left\langle
f_{ab},\hat{f}_{ab}\right\rangle.$ (6.1)

Write $\partial_{a}=\partial/\partial x^{a}$. Given
$f=\sum_{i=1}^{3}f_{i}\otimes
dx^{i}\in\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})$,
the differential $df$ is given by

$df=\sum_{i=1}^{3}\partial_{0}f_{i}\otimes dx^{0}\wedge dx^{i}+\sum_{1\leq
i<j\leq 3}(\partial_{i}f_{j}-\partial_{j}f_{i})dx^{i}\wedge dx^{j}.$ (6.2)

###### Definition 6.1

Recall $\\{E^{\alpha}\in\mathfrak{g}:\ 1\leq\alpha\leq N\\}$ is an orthonormal
basis in $\mathfrak{g}$. Define

$c_{\gamma}^{\alpha\beta}=-{{\rm{Tr}}}\left[E^{\gamma}[E^{\alpha},E^{\beta}]\right],\
E^{\alpha},E^{\beta},E^{\gamma}\in\mathfrak{g}.$

The term $c_{\gamma}^{\alpha\beta}$ is referred to as the structure constant.

###### Proposition 6.2

Suppose $A=\sum_{\alpha=1}^{N}\sum_{i=1}^{3}a_{i,\alpha}\otimes dx^{i}\otimes