E^{\alpha}\in\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}$.
Write $a_{0:j,\gamma}=\partial a_{j,\gamma}/\partial x^{0}$ and
$a_{i;j,\alpha}=\partial a_{i,\alpha}/\partial x^{j}-\partial
a_{j,\alpha}/\partial x^{i}$. Then,

$\displaystyle dA+A\wedge A=$
$\displaystyle\sum_{\gamma=1}^{N}\Bigg{[}\sum_{j=1}^{3}a_{0:j,\gamma}\otimes
dx^{0}\wedge dx^{j}+\sum_{1\leq i<j\leq 3}a_{i;j,\gamma}\otimes dx^{i}\wedge
dx^{j}$ $\displaystyle\hskip 28.45274pt+\sum_{1\leq i<j\leq
3}\sum_{1\leq\alpha,\beta\leq
N}a_{i,\alpha}a_{j,\beta}c_{\gamma}^{\alpha\beta}\otimes dx^{i}\wedge
dx^{j}\Bigg{]}\otimes E^{\gamma}.$ (6.3)

Proof. By direct computation, using Equation (6.2).

###  6.2 Yang-Mills measure

We will now summarize the following results proved in [6]. After applying
axial gauge fixing to the time-axis, the only $\mathfrak{g}$-valued gauge
fields $A$ over ${{\mathbb{R}}}^{4}$, which we need to consider are in
$L^{2}({{\mathbb{R}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}$.
Instead of making sense of a path integral over in the Hilbert space
$L^{2}({{\mathbb{R}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}$,
we will make sense of a Yang-Mills measure over
$\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}$,
of the form

$\frac{1}{Z}e^{-\frac{1}{2}\int_{{\mathbb{R}^{4}}}|dA+A\wedge A|^{2}\
d\omega}D[dA],$ (6.4)

whereby

$Z=\int_{\\{dA:\
A\in\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}\\}}e^{-\frac{1}{2}\int_{{\mathbb{R}^{4}}}|dA+A\wedge
A|^{2}\ d\omega}D[dA],$

and $D[dA]$ is ‘Lebesgue measure’ on
$\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{4})\otimes\mathfrak{g}$,
which does not exist.

###### Remark 6.3

  1. 1.

This is analogous to how Balaban in his series of papers from 1984 to 1989,
defined a finite dimensional integral over a finite lattice gauge of spacing
$\epsilon$, synonymous with $1/\kappa$. A complete citation of Balaban’s work
can be found in [8].

  2. 2.

The norm $|\cdot|$ on
$\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{4})\otimes\mathfrak{g}$
is from the tensor inner product, taken from Equations (6.1) and (1.3).

Using Equation (6.1) and Proposition 6.2,

$\displaystyle\int_{{{\mathbb{R}}}^{4}}|dA+A\wedge A|^{2}\ d\omega=$
$\displaystyle\sum_{1\leq i<j\leq
3}\int_{{{\mathbb{R}}}^{4}}\Bigg{[}\sum_{\alpha=1}^{N}a_{i;j,\alpha}^{2}+\sum_{\gamma=1}^{N}\sum_{\genfrac{}{}{0.0pt}{}{\alpha,\beta}{\hat{\alpha},\hat{\beta}}}a_{i,\alpha}a_{j,\beta}a_{i,\hat{\alpha}}a_{j,\hat{\beta}}c_{\gamma}^{\alpha\beta}c_{\gamma}^{\hat{\alpha}\hat{\beta}}$
$\displaystyle+2\sum_{\gamma=1}^{N}\sum_{\alpha,\beta}a_{i;j,\gamma}a_{i,\alpha}a_{j,\beta}c_{\gamma}^{\alpha\beta}\Bigg{]}\
d\omega+\sum_{j=1}^{3}\int_{{{\mathbb{R}}}^{4}}\sum_{\alpha=1}^{N}a_{0:j,\alpha}^{2}\
d\omega.$ (6.5)

###### Remark 6.4

Those terms that contain the structure constants in Equation (6.5) will
henceforth be referred to as the interaction terms in the Yang-Mills
Lagrangian.

It is conventional wisdom to interpret

$\exp\left[-\frac{1}{2}\sum_{\alpha=1}^{N}\int_{{{\mathbb{R}}}^{4}}d\omega\
\sum_{1\leq i<j\leq
3}a_{i;j,\alpha}^{2}+\sum_{j=1}^{3}a_{0:j,\alpha}^{2}\right]D[dA]$

as a Gaussian measure. To define this measure, we will work with holomorphic
sections of 2-forms over ${{\mathbb{C}}}^{4}$, instead of working on
$\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{4})$.
We will go over the construction, done in [8].

Consider the real vector space spanned by $\\{z^{n}:\
z\in{{\mathbb{C}}}\\}_{n=0}^{\infty}$, integrable with respect to the Gaussian
measure, equipped with a real inner product, given by

$\displaystyle\langle z^{r},$ $\displaystyle
z^{r^{\prime}}\rangle=\frac{1}{\pi}\int_{\mathbb{C}}z^{r}\cdot\overline{z^{r^{\prime}}}e^{-|z^{2}|}\
dx\ dp,\ z=x+\sqrt{-1}p.$ (6.6)

Note that $\overline{z}$ means complex conjugate. Denote this (real) inner
product space by $H^{2}({{\mathbb{C}}})$, which consists of polynomials in
$z$. An orthonormal basis is hence given by

$\left\\{\frac{z^{n}}{\sqrt{n!}}:\ n\geq 0\right\\}.$

Consider the tensor product $H^{2}({{\mathbb{C}}})^{\otimes^{4}}$, which we
will also denote it as $H^{2}({{\mathbb{C}}}^{4})$, to consist of all
polynomials in $z=(z_{0},z_{1},z_{2},z_{3})\in{{\mathbb{C}}}^{4}$, equipped
with the tensor inner product from Equation (6.6). Denote the closure of
$H^{2}({{\mathbb{C}}}^{4})$ using this tensor inner product by
$\mathcal{H}^{2}({{\mathbb{C}}}^{4})$.

The Segal Bargmann Transform maps the inner product space
$\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})$ to $H^{2}({{\mathbb{C}}}^{4})$,
i.e.

$\Psi_{\kappa}:\frac{h_{i}(\kappa\cdot)}{\sqrt{i!}}\frac{h_{j}(\kappa\cdot)}{\sqrt{j!}}\frac{h_{k}(\kappa\cdot)}{\sqrt{k!}}\frac{h_{l}(\kappa\cdot)}{\sqrt{l!}}\sqrt{\phi_{\kappa}}\longmapsto\frac{z_{0}^{i}}{\sqrt{i!}}\frac{z_{1}^{j}}{\sqrt{j!}}\frac{z_{2}^{k}}{\sqrt{k!}}\frac{z_{3}^{l}}{\sqrt{l!}}.$

Given any $f_{i,\alpha}\otimes dx^{i}\otimes
E^{\alpha}\in\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}$,
$f_{i,\alpha}\in\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})$, we map it inside
$H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}$
by

$\Psi_{\kappa}:f_{i,\alpha}\otimes dx^{i}\otimes
E^{\alpha}\longmapsto\Psi_{\kappa}(f_{i,\alpha})\otimes dx^{i}\otimes
E^{\alpha}.$

We will term $\Psi_{\kappa}$ as a renormalization flow, mapping a continuous
sequence of vector spaces, into a fixed vector space
$H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}$.

We will further define an inner product on
$\mathcal{H}^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{4})$ by

$\left\langle\sum_{0\leq a<b\leq 3}f_{ab}\otimes dx^{a}\wedge
dx^{b},\sum_{0\leq a<b\leq 3}\hat{f}_{ab}\otimes dx^{a}\wedge
dx^{b}\right\rangle=\sum_{0\leq a<b\leq 3}\left\langle
f_{ab},\hat{f}_{ab}\right\rangle.$ (6.7)

We will continue to use $\langle\cdot,\cdot\rangle$ to denote the tensor inner
product on
$\mathcal{H}^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{4})\otimes\mathfrak{g}$,
together with Equation (1.3).

###### Definition 6.5

For each $a=0,1,2,3$, define a linear operator ${\mathfrak{d}}_{a}$ acting on
$H^{2}({{\mathbb{C}}}^{4})$ as

$\mathfrak{d}_{a}\left[z_{a}^{p}\prod_{\genfrac{}{}{0.0pt}{}{b\neq
a}{b=0,\cdots,3}}z_{b}^{q_{b}}\right]=\left[\frac{p}{2}z_{a}^{p-1}-\frac{1}{2}z_{a}^{p+1}\right]\cdot\prod_{\genfrac{}{}{0.0pt}{}{b\neq
a}{b=0,\cdots,3}}z_{b}^{q_{b}}.$

The Segal Bargmann Transform maps $\partial_{a}$ to the linear operator
$\kappa{\mathfrak{d}}_{a}$. As a consequence, we will have the following
renormalization rule on $H^{2}({{\mathbb{C}}}^{4})$. For each $\kappa>0$, we
will add an extra factor $\kappa$ to ${\mathfrak{d}}_{a}$, i.e.
${\mathfrak{d}}_{a}\mapsto\kappa{\mathfrak{d}}_{a}$.

In [8], we defined an operator
${\mathfrak{d}}:H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\rightarrow
H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{4})$,

$\ {\mathfrak{d}}\sum_{i=1}^{3}f_{i}\otimes
dx^{i}=\sum_{i=1}^{3}[{\mathfrak{d}}_{0}f_{i}]\otimes dx^{0}\wedge
dx^{i}+\sum_{1\leq i<j\leq
3}[{\mathfrak{d}}_{i}f_{j}-{\mathfrak{d}}_{j}f_{i}]\otimes dx^{i}\wedge
dx^{j}.$

Extend the Segal Bargmann Transform
$\Psi_{\kappa}:\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{4})\otimes\mathfrak{g}\rightarrow
H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{4})\otimes\mathfrak{g}$,
i.e.

$\Psi_{\kappa}:\ \sum_{1\leq i<j\leq 3}f_{i,j,\alpha}\otimes dx^{i}\wedge
dx^{j}\otimes E^{\alpha}\longmapsto\sum_{1\leq i<j\leq
3}\Psi_{\kappa}[f_{i,j,\alpha}]\otimes dx^{i}\wedge dx^{j}\otimes E^{\alpha}.$

Thus, $\Psi_{\kappa}[dA]=\kappa{\mathfrak{d}}\Psi_{\kappa}[A]$, due to the
renormalization rule. As such, the Segal Bargmann Transform $\Psi_{\kappa}$ is
an isometry (up to a constant $\kappa^{2}$) between their respective spaces,
i.e.

$\langle
dA,dA\rangle=\kappa^{2}\langle{\mathfrak{d}}\Psi_{\kappa}[A],{\mathfrak{d}}\Psi_{\kappa}[A]\rangle,$

$A\in\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}$,
using their respective tensor inner products.

Balaban used a renormalization flow to map a finite gauge lattice of spacing
$\epsilon$, to a finite gauge lattice of unit spacing. The analogous procedure
here would be to use the Segal Bargmann Transform, to define Expression 6.4
over in
$H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{4})\otimes\mathfrak{g}$,
as

$\frac{1}{Z}e^{-\frac{1}{2}\int_{{{\mathbb{C}}}^{4}}|\kappa{\mathfrak{d}}A+A\wedge
A|^{2}d\lambda_{4}}D[{\mathfrak{d}}A],$ (6.8)

whereby

$Z=\int_{\\{{\mathfrak{d}}A:\ A\in
H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}\\}}e^{-\frac{1}{2}\int_{{{\mathbb{C}}}^{4}}|\kappa{\mathfrak{d}}A+A\wedge
A|^{2}d\lambda_{4}}D[{\mathfrak{d}}A],$

and $D[{\mathfrak{d}}A]$ is ‘Lebesgue measure’ on
$H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}$,
which does not exist. Note that $d\lambda_{4}$ is a 4-dimensional Gaussian
measure on ${{\mathbb{C}}}^{4}$, by using the tensor inner product from
Equation (6.6).

The factor $\kappa$ is due to the above renormalization rule. As a result of
this factor, we complete the inner product space into a Banach space equipped
with a Wiener measure of variance $1/\kappa^{2}$, as was constructed in [8]
and [6].

###### Theorem 6.6

Let the span of $\\{dx^{0}\wedge dx^{i},1\leq i\leq 3\\}$ and the span of
$\\{dx^{i}\wedge dx^{j},1\leq i<j\leq 3\\}$ be denoted by
$\ast\Lambda^{2}({{\mathbb{R}}}^{3})$ and $\Lambda^{2}({{\mathbb{R}}}^{3})$
respectively. Define

$\mathbb{H}:=\left\\{[{\mathfrak{d}}_{0}H^{2}({{\mathbb{C}}}^{4})]\otimes\left[\ast\Lambda^{2}({{\mathbb{R}}}^{3})\right]\right\\}\oplus\left\\{H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{3})\right\\}\subset
H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{4}),$

${\mathfrak{d}}_{0}H^{2}({{\mathbb{C}}}^{4})$ denotes the range of
${\mathfrak{d}}_{0}$.

In [8], we completed it into a Banach space ${\mathbb{B}}$ using a supremum
norm, equipped with a $\sigma$-algebra defined on it. On this Banach space, we
constructed a product Wiener measure denoted as $\tilde{\mu}_{\kappa^{2}}$,
which is an infinite dimensional Gaussian measure of variance $1/\kappa^{2}$,
therefore making $({\mathbb{B}},\tilde{\mu}_{\kappa^{2}})$ a probability
space.

When we consider the tensor inner product space
$\mathbb{H}\otimes\mathfrak{g}$, then the completion of it into a Banach space
will be ${\mathbb{B}}\otimes\mathfrak{g}$. The product Wiener measure on it,
hereby denoted as $\tilde{\mu}_{\kappa^{2}}^{\times^{N}}$, will make
${\mathbb{B}}\otimes\mathfrak{g}$ into a probability space. We will denote the
expectation on this probability space using ${{\mathbb{E}}}$.

Because of the renormalization flow $\\{\Psi_{\kappa}:\kappa>0\\}$, we showed
in [6] that there exists a sequence of positive functions defined on the
Wiener space ${\mathbb{B}}\otimes\mathfrak{g}$, denoted by
$\\{\mathcal{Y}^{\kappa}:\kappa>0\\}$, such that we can define a measure as
$\mathcal{Y}^{\kappa}d\tilde{\mu}_{\kappa^{2}}^{\times^{N}}$. Hence, we will
interpret the Yang-Mills measure in Expression 6.8 as

$\frac{1}{Z}e^{-\frac{1}{2}\int_{{{\mathbb{C}}}^{4}}|\kappa{\mathfrak{d}}A+A\wedge
A|^{2}d\lambda_{4}}D[{\mathfrak{d}}A]:=\frac{\mathcal{Y}^{\kappa}d\tilde{\mu}_{\kappa^{2}}^{\times^{N}}}{\int_{{\mathbb{B}}\otimes\mathfrak{g}}\mathcal{Y}^{\kappa}d\tilde{\mu}_{\kappa^{2}}^{\times^{N}}}=\frac{\mathcal{Y}^{\kappa}d\tilde{\mu}_{\kappa^{2}}^{\times^{N}}}{{{\mathbb{E}}}[\mathcal{Y}^{\kappa}]},$

which is also a probability measure on the Banach space
${\mathbb{B}}\otimes\mathfrak{g}$.

###### Notation 6.7

For each $\kappa>0$, the Yang-Mills measure is a probability measure, so it
would be more convenient to use expectation to denote this integral. For any
measurable and bounded function $F$ on ${\mathbb{B}}\otimes\mathfrak{g}$, we
will write

$\mathbb{E}_{{\rm
YM}}^{\kappa}[F]:=\frac{1}{\int_{{\mathbb{B}}\otimes\mathfrak{g}}\mathcal{Y}^{\kappa}\
d\tilde{\mu}_{\kappa^{2}}^{\times^{N}}}\int_{{\mathbb{B}}\otimes\mathfrak{g}}F\mathcal{Y}^{\kappa}\
d\tilde{\mu}_{\kappa^{2}}^{\times^{N}}.$

###### Remark 6.8

This sequence of Yang-Mills measure is analogous to how Balaban defined a
sequence of finite-dimensional integrals on a finite lattice gauge, with
spacing $\epsilon$, which were renormalized into finite lattice gauge with
unit spacing. One should think of $\epsilon$ as synonymous to $1/\kappa$.
Refer to [8] and [6] for details.

We will term the vectors in ${\mathbb{B}}\otimes\mathfrak{g}$ as Yang-Mills
gauge fields. When we complete the space into a Banach space using a supremum
norm, we showed in [8] that this Banach space consists of holomorphic
$\Lambda^{2}({{\mathbb{R}}}^{4})\otimes\mathfrak{g}_{{\mathbb{C}}}$-valued
functions, over the complex space ${{\mathbb{C}}}^{4}$. Or one can refer this
Banach space as containing holomorphic sections of the complexified trivial
bundle
${{\mathbb{C}}}^{4}\times[\Lambda^{2}({{\mathbb{R}}}^{4})\otimes\mathfrak{g}_{{\mathbb{C}}}]\rightarrow{{\mathbb{C}}}^{4}$.

###  6.3 Asymptotic freedom

In physics, asymptotic freedom refers to the phenomenon whereby the coupling
constant $c$ goes to zero as the momentum scale (or energy scale) increases.
See page 425 in [9]. This coupling constant, which depends on momentum,
satisfies a certain differential equation. See page 459 in [4]. Thus, it must
vary continuously with momentum.

In physics literature, the interaction terms in Equation (6.5) is actually due
to the non-linear term in Equation (6.3). Physicists would introduce a
coupling constant $c$ to the non-linear term, so that they can make the
interaction terms small. See [4]. When the coupling constant $c$ is zero, we
obtain the free theory.

In a compact semi-simple Yang-Mills gauge theory, the structure constant
$c^{\alpha\beta}_{\gamma}$ of a semi-simple Lie Algebra is non-zero, only if
$\alpha$, $\beta$ and $\gamma$ are all distinct, and this introduces
interaction terms into the Yang-Mills Lagrangian, making the Yang-Mills path
integrals impossible to compute analytically. The coupling constant $c$ is
introduced so that one can apply perturbation theory to compute the path
integrals. When $c$ is small, that means one can use Feynmann diagrams to
compute the path integrals. This happens when the energy scale is large.

Unfortunately, perturbation methods are no longer valid if the energy scale is
small. It is believed that non-abelian gauge theories exhibit asymptotic
freedom. It was showed in [11] that for renormalizable quantum field theories
in 4-dimensional space-time, only non-abelian gauge theories are
asymptotically free. On page 541 in [9], one sees that for a non-abelian gauge
group, the coupling constant $c(k)^{2}$ decreases at a rate of $\dfrac{1}{\ln
k^{2}}$, whereby $k$ is the momentum scale.

In [6], we set the coupling constant $c=1/\kappa$, and we approximate the
Dirac delta function with a Gaussian function
$\kappa^{4}e^{-\kappa^{2}|\vec{x}|^{2}/2}/(2\pi)^{2}$, so the variance is
given by $1/\kappa^{2}$. To resolve points separated by short distances, we
need a small variance, which means that $\kappa$ is large. Therefore, at short
distances, the coupling constant is small.

###  6.4 Callan-Symanzik beta function

To determine how $\kappa$ varies with the energy scale, we need to impose a
Callan-Symanzik Equation, by first introducing a beta function $\beta(c)$,
which depends on the coupling constant. The purpose is to formulate
renormalization conditions in the energy scale, instead of the renormalization
scale $\kappa$.

In the next section, we will define the quadratic Casimir operator, for each
irreducible representation $\rho_{n}:\mathfrak{g}\rightarrow{\rm
End}({{\mathbb{C}}}^{\tilde{N}_{n}})$. Because the Lie Algebra is simple, the
Casimir operator is a constant $C_{2}(\rho_{n})$ times the identity. We will
interpret the increasing sequence $\\{\frac{1}{4}C_{2}(\rho_{n}):\
n\in\mathbb{N}\\}$ as the set of quantized energy levels squared in the
theory.

Instead of using $C_{2}(\rho_{n})$, it is actually more appropriate to use
$\tilde{N}_{n}$, the dimension of the representation $\rho_{n}$. Hence, we
will define the beta function as,

$\beta(c)=\frac{\partial c}{\partial[\ln\tilde{N}_{n}]},$ (6.9)

to determine how the coupling constant varies. Compare this with Equation
12.90 in [9]. We will compute the $\beta$ function in the next section, using
a path integral. Because asymptotic freedom holds, beta must be chosen to be
negative.

###### Remark 6.9

The dimension $\tilde{N}_{n}$ is defined using the highest weights of an
irreducible representation, from the Weyl dimension formula. As such, the set
$\\{\tilde{N}_{n}:\ n\in\mathbb{N}\\}$ is unbounded. In the definition of the
beta function, we treat $\tilde{N}_{n}$ as a continuous variable.

##  7 Hamiltonian and Momentum operator

###  7.1 Renormalization

###### Notation 7.1

(Casimir operator)
Let $\mathfrak{g}$ be a semi-simple Lie Algebra. For an irreducible
representation $\rho:\mathfrak{g}\rightarrow{\rm
End}({{\mathbb{C}}}^{\tilde{N}})$ such that $\rho(\mathfrak{g})$ consists of
skew-Hermitian matrices, we define $C(\rho)\in{{\mathbb{R}}}$ such that

${{\rm{Tr}}}[\rho(E^{\alpha})\rho(E^{\beta})]=C(\rho){{\rm{Tr}}}[E^{\alpha}E^{\beta}].$
(7.1)

Also define

$\mathscr{E}(\rho):=-\sum_{\alpha=1}^{N}\rho(E^{\alpha})\rho(E^{\alpha})$

to be its (quadratic) Casimir operator. When $\mathfrak{g}$ is simple, the
Casimir operator is a constant multiple of the identity. We write
$C_{2}(\rho)$ to denote this constant.

###### Remark 7.2

When $\mathfrak{g}$ is simple, note that $C(\rho)$ satisfies
$C_{2}(\rho)\tilde{N}=NC(\rho)$.

###### Definition 7.3

Suppose $e^{0}$ is a directional vector in the time direction, with length
$|e^{0}|=T$, and $a\in{{\mathbb{R}}}^{3}$ be any directional spatial vector,
with length $|a|$. Let $\sigma:I^{2}\rightarrow{{\mathbb{R}}}^{4}$ be some
parametrization of a compact time-like rectangular surface $R[a,T]$ contained
in a plane, spanned by $e^{0}$ and $a$, of dimensions height T and length
$|a|$. When $T=1$, we will write $R[a,1]=R[a]$.

Define a $\rho(\mathfrak{g})$\- valued random variable
$\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}$, which sends
$\sum_{\alpha=1}^{N}\sum_{j=1}^{3}{\mathfrak{d}}_{0}A_{j,\alpha}\otimes
dx^{0}\wedge dx^{j}\otimes E^{\alpha}\in{\mathbb{B}}\otimes\mathfrak{g}$ to

$\frac{1}{\kappa}\sum_{\alpha=1}^{N}\frac{\kappa^{2}}{4}\int_{\hat{s}\in[0,1]^{2}}d\hat{s}\
\sum_{j=1}^{3}|J_{0j}^{\sigma}|(\hat{s})\kappa[\psi\cdot{\mathfrak{d}}_{0}A_{j,\alpha}](\kappa\sigma(\hat{s})/2)\otimes\rho(E^{\alpha}),$
(7.2)

whereby $\psi(w):=\frac{1}{\sqrt{2\pi}}e^{-|w|^{2}/2}$, and
$|J^{\sigma}_{0i}|$ is defined in Definition A.1.

###### Remark 7.4

This $\rho(\mathfrak{g})$-valued random variable on
${\mathbb{B}}\otimes\mathfrak{g}$ first appeared in [6]. The factor $\psi(w)$
is known as a renormalization factor. Its importance was explained in [8]. The
factor $1/\sqrt{2\pi}$ can be replaced with any number
$0<\tilde{c}<1/\sqrt{2}$.

The factors of $\kappa$ which appeared in Expression 7.2 are all due to
renormalization:

  1. 1.

The factor $\kappa$ in front of $\psi\cdot{\mathfrak{d}}_{0}A_{j,\alpha}$ is
due to the renormalization rule.

  2. 2.

We embed ${{\mathbb{R}}}^{4}$ inside ${{\mathbb{C}}}^{4}$, by
$\vec{x}\in{{\mathbb{R}}}^{4}\hookrightarrow\kappa\vec{x}/2$. Hence, the
factor $\kappa/2$ in parenthesis is due to this renormalization
transformation, which will also give us an extra factor $\kappa^{2}/4$ for the
surface integral.

The factor $1/\kappa$ outside of the integral, is due to asymptotic freedom,
whereby we set the coupling constant $c=1/\kappa$.

For a gauge field $A\in
H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}$,
we can interpret

$c\sum_{\alpha=1}^{N}\int_{\hat{s}\in[0,1]^{2}}d\hat{s}\
\sum_{j=1}^{3}|J_{0j}^{\sigma}|(\hat{s})\left[{\mathfrak{d}}_{0}A_{j,\alpha}\right](\sigma(\hat{s}))\otimes\rho(E^{\alpha}),$
(7.3)

as measuring the field strength of ${\mathfrak{d}}A+A\wedge A$, over the time-
like rectangular surface
$R[a,T]\subset{{\mathbb{R}}}^{4}\hookrightarrow{{\mathbb{C}}}^{4}$. It has
dimension of energy and it is similar to Expression 7.2, but without the
factors of $\kappa$ and renormalization factor $\psi(w)$.

###### Remark 7.5

Note that we introduced the coupling constant $c$ to the quantity in
Expression 7.3, instead to the non-linear term in Equation (6.3). So, by
applying the renormalization techniques and asymptotic freedom explained in
Remark 7.4, to $c\int_{R[a]}{\mathfrak{d}}A+A\wedge A$, we will obtain
Expression 7.2.

Write

$\left\langle\nu_{R[a,T]}^{\kappa,\rho}\right\rangle^{2}=-{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}\mathcal{Y}^{\kappa}\right],$

whereby $\mathcal{Y}^{\kappa}$ was defined in Theorem 6.6. We interpret
$\sqrt{\left\langle\nu_{R[a,T]}^{\kappa,\rho}\right\rangle^{2}}$ as measuring
the average flux passing through the time-like rectangular surface over a time
interval $T$, using the non-abelian Yang-Mills measure, and hence it has
dimension of energy. See also Equation 19.11 in [20].

###### Remark 7.6

Our term $\left({\mathfrak{d}}A,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}$ is
actually a skew-Hermitian matrix. When we square it, it will be a non-positive
definite matrix. We need to put a negative sign in front, to make it non-
negative definite.

For an irreducible representation $\rho$ for $\mathfrak{g}$ and
$A=\sum_{\alpha=1}^{N}\sum_{i=1}^{3}a_{i,\alpha}\otimes dx^{i}\otimes
E^{\alpha}$, $a_{i,\alpha}\in\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})$, we
will write

$A^{\rho}:=\sum_{\alpha=1}^{N}\sum_{i=1}^{3}a_{i,\alpha}\otimes
dx^{i}\otimes\rho(E^{\alpha})\in\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{4})\otimes\rho(\mathfrak{g}).$

In [6], we made sense of a non-abelian Yang-Mills path integral using the
renormalization flow $\\{\Psi_{\kappa}:\kappa>0\\}$,

$\displaystyle\frac{1}{Z}\int_{\\{dA\in\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{4})\otimes\mathfrak{g}\\}}\exp\left[c\int_{R[a]}d[A^{\rho}]\right]e^{-\frac{1}{2}S_{{\rm
YM}}(A)}$ $\displaystyle\ D[dA]$ $\displaystyle:={{\mathbb{E}}}_{{\rm
YM}}^{\kappa}\Bigg{[}\exp\left[\left(\cdot,\nu_{R[a]}^{\kappa,\rho}\right)_{\sharp}\right]\Bigg{]},$

whereby $D[dA]$ is some Lebesgue type of measure and

$Z=\int_{\\{dA\in\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{4})\otimes\mathfrak{g}\\}}e^{-\frac{1}{2}S_{{\rm
YM}}(A)}\ DA.$

Taking the limit as $\kappa\rightarrow\infty$, it will give us the Wilson Area
Law formula.

And in the same article, we also showed that the average flux squared

$\displaystyle-{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a]}^{\kappa,\rho_{n}}\right)_{\sharp}^{2}\mathcal{Y}^{\kappa}\right]=\frac{|a|}{4}\otimes\mathscr{E}(\rho_{n})+\epsilon(n,\kappa),$
(7.4)

whereby the error term $\epsilon(n,\kappa)$, is a negative definite matrix, if
$\kappa\geq\kappa_{0}$ for some $\kappa_{0}\in\mathbb{N}$. Its trace norm is
given by

$\parallel\epsilon(n,\kappa)\parallel=O\left(C(\rho_{n})/\kappa^{4}\right).$
(7.5)

###### Remark 7.7

In the proof of Equation (7.4) in [6], we did a series expansion for
$\mathcal{Y}^{\kappa}$, in terms of $1/\kappa$. This accounts for the
$1/\kappa^{4}$ term in the trace norm estimate. Because of this series
expansion, we see that indeed $\epsilon(n,1/c)$ is differentiable in $c$.

Due to the compactness of the gauge group, we can assume that
$\rho_{n}(\mathfrak{g})$ consists of skew-Hermitian matrices, so the LHS of
Equation (7.4) is indeed non-negative definite. But because its representation
$\rho_{n}$ is irreducible, we can prove that it is in fact positive definite.
Besides using renormalization techniques and applying asymptotic freedom, the
structure constants of a simple Lie Algebra and the quartic term in the Yang-
Mills action, are all instrumental in proving this equality. Refer to the
proof in [6].

###### Remark 7.8

In Equation (7.4), we see that $|a|$ is the area of the rectangular time-like
surface $R[a]$. In general, we will obtain the area $|a|T$ for a time-like
surface $R[a,T]$.

As such, the Yang-Mills path integral will give us the area density $d\rho$
given in Definition A.1, which in turn allows us to construct a unitary
representation of ${\rm SL}(2,{{\mathbb{C}}})$, which acts on space-like
surfaces in ${{\mathbb{R}}}^{4}$.

Recall we used the quantized values $\hat{H}(\rho_{n})$, $\hat{P}(\rho_{n})$
in Equation (3.2). They will be defined via a Yang-Mills path integral, given
by Equation (7.4). For a given space-like surface $S$, associated with it is a
set $\\{\hat{f}_{0},\hat{f}_{1}\\}$ contained in a Minkowski frame, spanning a
time-like plane $S^{\flat}$. From Definition 2.5, this set can be transformed
from $\\{e_{0},e_{1}\\}$, a set spanning the $x^{0}-x^{1}$ plane. Without any
loss of generality, we will quantize energy and momentum using a time-like
rectangular surface $R[a]$, contained in a plane parallel to the time-axis.

###  7.2 Callan-Symanzik Equation

To define our sequence of masses $\\{m_{n}:\ n\in\mathbb{N}\\}$, we need to
determine the beta function, so that we can see how $\kappa$ correlates with
$\tilde{N}_{n}$. The beta function is found typically by solving a Callan-
Symanzik Equation.

The set $\\{\rho_{n}(E^{\alpha})/\sqrt{C(\rho_{n})}\\}_{\alpha=1}^{N}$ forms
an orthonormal basis. If we replace each $\rho_{n}(E^{\alpha})$ with
$\rho_{n}(E^{\alpha})/\sqrt{C(\rho_{n})}$ in Equation (7.2), we will obtain
from Equation (7.4),

$\displaystyle-\frac{1}{C(\rho_{n})}{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a]}^{\kappa,\rho_{n}}\right)_{\sharp}^{2}\mathcal{Y}^{\kappa}\right]=\frac{|a|}{4}\otimes\frac{C_{2}(\rho_{n})}{C(\rho_{n})}\mathbb{I}_{n}+\frac{1}{C(\rho_{n})}\epsilon(n,\kappa),$
(7.6)

whereby $\mathbb{I}_{n}$ is a $\tilde{N}_{n}\times\tilde{N}_{n}$ identity
matrix for an irreducible representation $\rho_{n}:\mathfrak{g}\rightarrow{\rm
End}({{\mathbb{C}}}^{\tilde{N}_{n}})$.

Note that $\dfrac{C_{2}(\rho)}{C(\rho)}=\dfrac{N}{\tilde{N}}$, which varies
according to the representation $\rho$. Set $|a|=4$. Since
$\\{\rho_{n}(E^{\alpha})/\sqrt{C(\rho_{n})}\\}_{\alpha=1}^{N}$ is an
orthonormal basis for every $n\in\mathbb{N}$, from Equation (7.2) and that
$\mathcal{Y}^{\kappa}$ is independent of any representation used, it is not
difficult to see that ${{\rm{Tr}}}\ \epsilon(n,\kappa)=C(\rho_{n}){{\rm{Tr}}}\
\epsilon(1,\kappa)/C(\rho_{1})$ for all $n\in\mathbb{N}$.

When we take the trace on the RHS of Equation (7.6), we will obtain

$\tilde{N}_{n}\frac{C_{2}(\rho_{n})}{C(\rho_{n})}+\frac{1}{C(\rho_{n})}{{\rm{Tr}}}\
\epsilon(n,\kappa)\equiv N+\frac{1}{C(\rho_{1})}{{\rm{Tr}}}\
\epsilon(1,\kappa).$

Write $\epsilon(\kappa):=\epsilon(1,\kappa)/C(\rho_{1})$. Recall,
$c=1/\kappa$. Using the trace estimate given in Equation (7.5), for each
$n\in\mathbb{N}$, we will define

$\displaystyle G_{n}^{(2)}(c,e):=$
$\displaystyle\frac{\tilde{N}_{n}}{e}+{{\rm{Tr}}}\ \epsilon(1/c)$
$\displaystyle=$
$\displaystyle\frac{\tilde{N}_{n}}{e}-c^{4}\bar{\lambda}+f(c^{5}),$

whereby $\bar{\lambda}$ is some positive number, and the remainder term is
denoted by $f(c^{5})$, which has bounded derivatives in $c$, all independent
of $n$. See Remark 7.7.

Furthermore, there exists a constant $\tilde{C}_{1}$, independent of $n$, such
that $|\bar{\lambda}|\leq\tilde{C}_{1}$ and
$|f(c^{5})|\leq\tilde{C}_{1}c^{5}$.

###### Remark 7.9

The integral $G_{n}^{(2)}(c,\tilde{N}_{n}/N)$ is actually a $2$-point Green’s
function, which follows from taking the trace of Equation (7.6).

Recall the definition of the beta function, $e\dfrac{\partial c}{\partial
e}=\beta(c)$, with $e\equiv\tilde{N}$. We will impose the following Callan-
Symanzik Equation

$\left[e\frac{\partial}{\partial e}+\beta(c)\frac{\partial}{\partial
c}+2\gamma(c)\right]G_{n}^{(2)}(c,e)=0.$ (7.7)

See Equation (12.41) in [9]. This equation asserts that there exist two scalar
functions $\beta(c)$ and $\gamma(c)$, related to the shifts in the coupling
constant and the field strength, that compensates for the shift in the ‘new’
renormalization scale $e$.

###### Proposition 7.10

For $c$ small, there exists a scalar-valued function
$\beta(c)=-c/4+\lambda(c)$ and $\gamma=1/2$ that solves Equation (7.7). Note
that $|\lambda(c)|\leq\tilde{C}_{4}c^{2}$, for some constant $\tilde{C}_{4}$
independent of $n$.

Proof. Now

$G_{n}^{(2)}(c,e)=\frac{\tilde{N}_{n}}{e}-c^{4}\bar{\lambda}+f(c^{5}).$

A direct computation shows that

$\frac{\partial}{\partial
e}G_{n}^{(2)}(c,e)=-\frac{\tilde{N}_{n}}{e^{2}},\quad\frac{\partial}{\partial
c}G_{n}^{(2)}(c,e)=-4c^{3}\bar{\lambda}+\tilde{f}(c^{4}),$

$\tilde{f}(c^{4})\equiv\partial f(c^{5})/\partial c$ is scalar valued and
$|\tilde{f}(c^{4})|\leq\tilde{C}_{2}c^{4}$, $\tilde{C}_{2}$ is a constant
independent of $n$.

Plug into Equation (7.7), we have that

$-\frac{\tilde{N}_{n}}{e}-4\beta(c)c^{3}\bar{\lambda}+2\gamma(c)G_{n}^{(2)}(c,e)+\beta(c)\tilde{f}(c^{4})=0.$

To satisfy the above Callan-Symanzik Equation, we must have that
$\beta(c)=-c/4+\lambda(c)$, $\gamma(c)=1/2$, $\lambda(c)$ to be determined.

Hence, we have

$-\frac{c}{4}\tilde{f}(c^{4})+f(c^{5})-4c^{3}\lambda(c)\bar{\lambda}+\lambda(c)\tilde{f}(c^{4})=0.$

Because the error term $\epsilon(\kappa)$ in Equation (7.4) is a negative
definite matrix, we see that $\bar{\lambda}$, which is independent of $n$, is
positive. Furthermore, we have a constant $\tilde{C}_{3}$ such that

$\frac{1}{c^{4}}|\tilde{f}(c^{4})|+\frac{1}{c^{5}}|f(c^{5})|\leq\tilde{C}_{3},$

whereby this constant $\tilde{C}_{3}$ is independent of $n$.

Therefore, if $c$ is small enough, then
$-4c^{3}\bar{\lambda}+\tilde{f}(c^{4})$ is indeed non-zero. Hence,

$\lambda(c)=\frac{1}{-4c^{3}\bar{\lambda}+\tilde{f}(c^{4})}\left[\frac{c}{4}\tilde{f}(c^{4})-f(c^{5})\right],$

is scalar valued, and is such that $|\lambda(c)|\leq\tilde{C}_{4}c^{2}$,
$0<\tilde{C}_{4}$ is independent of $n$.

###### Definition 7.11

The Callan-Symanzik beta function is given as
$\beta(c)=-\dfrac{c}{4}+\lambda(c)$, whereby
$|\lambda(c)|\leq\tilde{C}_{4}c^{2}$.

This will allow us to correlate the coupling constant $c$ with the dimension
$\tilde{N}_{n}$. Its solution will be found in the next subsection.

###  7.3 Existence of positive mass gap

In Wightman’s zeroth axiom, we note that $\hat{H}^{2}-\hat{P}^{2}=m^{2}\geq
0$. Rewriting this equation, the Hamiltonian and momentum eigenvalue equation
is equivalent to

$\frac{\hat{P}^{2}}{\hat{H}^{2}}-1=-\frac{m^{2}}{\hat{H}^{2}}.$ (7.8)

In this equation, it should be understood that $\hat{P}$ and $\hat{H}$ are
eigenvalues of their respective operators on
$\bigoplus_{n=1}^{\infty}\mathscr{H}(\rho_{n})$.

We now need to define momentum and Hamiltonian eigenvalues, that satisfy the
above equation, in such a way that the operators are unbounded. We will see
later that $m^{2}\rightarrow\infty$ and
$0>\hat{P}^{2}/\hat{H}^{2}-1\rightarrow 0$, all implying their respective
eigenvalues tend towards infinity.

The error term in Equation (7.4) comes from the interaction terms in the Yang-
mills action in Equation (6.5). Equation (7.5) gives us its trace bound, and
because of the quartic term, it also gives us a positive mass gap. Indeed,
Equation (7.4) is actually a ‘continuous’ version of the above eigenvalue
equation.

Let us review our setup. We have a compact gauge group $G$ with a (real)
simple Lie Algebra $\mathfrak{g}$, henceforth considered as a sub-Lie Algebra
in $\mathfrak{u}(\bar{N})$, for which we can define an inner product on
$\mathfrak{g}$. Hence, we will assume that $\rho(E)$ is skew-Hermitian, i.e.
$-\rho(E)=\rho(E)^{\ast}$. We also let $\\{\alpha_{1},\cdots,\alpha_{l}\\}$ be
a simple system of roots for $\mathfrak{g}_{{\mathbb{C}}}$. Furthermore, let
$\\{H_{1},H_{2},\cdots,H_{l}\\}$ be a basis for a Cartan subalgebra
$\mathfrak{h}\subset\mathfrak{g}$. It may not be orthonormal, so we can define
an orthonormal set $\\{E^{\alpha}\\}_{\alpha=1}^{l}$ as

$E^{\alpha}=\sum_{\beta=1}^{l}a_{\alpha,\beta}H_{\beta},\ \ 1\leq\alpha\leq
l.$

Define an invertible $l\times l$ matrix
$B=\\{a_{\alpha,\beta}\\}_{\alpha,\beta=1}^{l}$. For any vector
$u\in{{\mathbb{R}}}^{l}$, there exists a constant $c>0$ such that
$|Bu|_{2}^{2}\geq c|u|_{2}^{2}$, $|\cdot|_{2}$ is the standard Euclidean norm.
Extend $\\{E^{\alpha}:\ 1\leq\alpha\leq l\\}$ to be an orthonormal basis
$\\{E^{\alpha}\\}_{\alpha=1}^{N}$ in $\mathfrak{g}$.

Each inequivalent irreducible representation is indexed uniquely by the
highest weight

$\lambda_{\rho}\equiv\left(\lambda_{\rho}(H_{1}),\cdots,\lambda_{\rho}(H_{l})\right),$

which is a $l$-tuple of non-negative half-integers or integers, i.e.
$\lambda_{\rho}(H_{i})\geq 0$. We will write
$|\lambda_{\rho}|_{2}:=\left(\sum_{i=1}^{l}\lambda_{\rho}(H_{i})^{2}\right)^{1/2}$.

For each representation $\rho:\mathfrak{g}\rightarrow{\rm
End}({{\mathbb{C}}}^{\tilde{N}})$, its corresponding Casimir operator is given
by

$\mathscr{E}(\rho)=-\sum_{\alpha=1}^{N}\rho(E^{\alpha})\rho(E^{\alpha})=C_{2}(\rho)I,\
\ C_{2}(\rho)\geq 0.$

Since $\rho(E^{\alpha})$ is skew-Hermitian, we have

$\left\langle-\rho(E^{\alpha})\rho(E^{\alpha})v,v\right\rangle=\left\langle\rho(E^{\alpha})v,\rho(E^{\alpha})v\right\rangle\geq
0$

for any vector $v\in{{\mathbb{C}}}^{\tilde{N}}$.

Because $\rho(H)$ is skew-Hermitian, its eigenvalues are purely imaginary. Let
$v$ be a unit weight vector corresponding to the highest weight, i.e.
$\langle\rho(H)v,\rho(H)v\rangle=\lambda_{\rho}(H)^{2}$, $H\in\mathfrak{h}$.

In terms of the highest weight, we have

$\displaystyle\left\langle C_{2}(\rho)v,v\right\rangle$
$\displaystyle=\sum_{\alpha=1}^{N}\left\langle\rho(E^{\alpha})v,\rho(E^{\alpha})v\right\rangle\geq\sum_{\alpha=1}^{l}\left|\sum_{\beta=1}^{l}a_{\alpha,\beta}\lambda_{\rho}(H_{\beta})v\right|_{2}^{2}$
$\displaystyle=\sum_{\beta=1}^{l}\sum_{\gamma=1}^{l}\sum_{\alpha=1}^{l}\lambda_{\rho}(H_{\beta})a_{\alpha,\beta}a_{\alpha,\gamma}\lambda_{\rho}(H_{\gamma})\geq
c|\lambda_{\rho}|_{2}^{2}.$

Therefore, we see that $\\{C_{2}(\rho_{n}):n\in\mathbb{N}\\}$ is unbounded,
because the highest weight is a $l$-tuple consisting of non-negative half-
integers or integers.

We can thus list the inequivalent irreducible representations as a sequence

$\left\\{\rho_{n}:\mathfrak{g}\rightarrow{\rm
End}({{\mathbb{C}}}^{\tilde{N}_{n}})\right\\}_{n=1}^{\infty},\quad{\rm such\
that}\quad 0<C_{2}(\rho_{n})\leq C_{2}(\rho_{n+1}),$

for all $n\geq 1$.

Recall for each inequivalent irreducible non-trivial representation
$\rho:\mathfrak{g}\rightarrow{\rm End}({{\mathbb{C}}}^{\tilde{N}})$, we
defined a Hilbert space $\mathscr{H}(\rho)$. Using the above sequence,
construct a Hilbert space
$\\{1\\}\oplus\bigoplus_{n=1}^{\infty}\mathscr{H}(\rho_{n})$, for which the
Wightman’s axioms are satisfied.

Define the Hamiltonian eigenvalue $\hat{H}(\rho_{n})$ to be

$\hat{H}(\rho_{n})^{2}:=\frac{\tilde{N}_{n}}{4}C_{2}(\rho_{n})=\frac{N}{4}C(\rho_{n})>0,$

for each irreducible non-trivial representation
$\rho_{n}:\mathfrak{g}\rightarrow{\rm End}({{\mathbb{C}}}^{\tilde{N}_{n}})$.
See Remark 7.2.

When $n\in\mathbb{N}$ is large, we see that the energy level is
correspondingly large. By asymptotic freedom, the coupling constant $c$ should
weaken with large values of $n$. Since $c=1/\kappa$, this means that $\kappa$
must be large when $n$ is large. To define the momentum eigenvalues, we need
to choose an unbounded sequence $\\{\kappa_{n}:n\in\mathbb{N}\\}$ for Equation
(7.4).

Recall we solved for the beta function in subsection 7.2. By Definition 7.11,
we have that

$\frac{\partial c}{\partial[\ln\tilde{N}]}=-\frac{c}{4}+\lambda(c),\ \
|\lambda(c)|\leq\tilde{C}_{4}c^{2}.$

Now,

$\displaystyle\frac{dc}{d[\ln\tilde{N}]}=-\frac{c}{4}+\lambda(c)\quad\Longrightarrow\frac{dc}{c-4\lambda(c)}=-\frac{d[\ln\tilde{N}]}{4}.$

Write $\mu(c)=-4\lambda(c)/c$. Thus,

$\displaystyle\frac{1}{c}\frac{dc}{1+\mu(c)}=-\frac{d[\ln\tilde{N}]}{4}\quad\Longrightarrow\left[\frac{1}{c}\sum_{k=0}^{\infty}(-1)^{k}\mu(c)^{k}\right]dc=-\frac{d[\ln\tilde{N}]}{4},$

whereby $\mu(c)$ is such that $|\mu(c)|\leq\tilde{C}_{5}c$. Integrate, we
obtain

$\ln c+\tilde{\mu}(c)=-\frac{1}{4}\ln\tilde{N}+C,$

$C$ is some constant. Note that $\tilde{\mu}(c)$ is a scalar function such
that $|\tilde{\mu}(c)|\leq\tilde{C}_{6}c$. Exponentiate, we have

$ce^{\tilde{\mu}(c)}=\frac{1}{\hat{C}}\tilde{N}^{-1/4}\ \Longrightarrow\
c(1+\bar{\mu}(c))=\frac{1}{\hat{C}}\tilde{N}^{-1/4},$

whereby $0<\hat{C}$ is some positive constant and
$\bar{\mu}(c)=e^{\tilde{\mu}(c)}-1$ is such that
$|\bar{\mu}(c)|\leq\tilde{C}_{7}c$. Thus, we have that
$c\tilde{N}^{1/4}=\dfrac{1}{\hat{C}}\dfrac{1}{1+\bar{\mu}(c)}$ and

$\displaystyle\left|\frac{1}{\hat{C}}\frac{1}{1+\bar{\mu}(c)}\right|\leq\frac{1}{\hat{C}}\left[1+\tilde{C}_{8}|\bar{\mu}(c)|\right]\leq\frac{1}{\hat{C}}\left[1+\tilde{C}_{8}\tilde{C}_{7}c\right]$
$\displaystyle\leq\frac{1}{\hat{C}}\left[1+\tilde{C}_{7}\tilde{C}_{8}\frac{1}{\hat{C}}\tilde{N}^{-1/4}\left(1+\tilde{C}_{8}\tilde{C}_{7}\right)\right]=\frac{1}{\hat{C}}+\frac{\tilde{N}^{-1/4}}{\hat{C}^{2}}\tilde{C}_{7}\tilde{C}_{8}\left(1+\tilde{C}_{7}\tilde{C}_{8}\right).$

Thus

$\tilde{N}^{1/4}|\bar{\mu}(c)|\leq
c\tilde{N}^{1/4}\tilde{C}_{7}\leq\frac{\tilde{C}_{7}}{\hat{C}}+\frac{\tilde{C}_{7}}{\hat{C}^{2}}\tilde{C}_{7}\tilde{C}_{8}\left(1+\tilde{C}_{7}\tilde{C}_{8}\right).$

Recall $c=1/\kappa$. Hence,

$\displaystyle\kappa=$
$\displaystyle\frac{1}{c}=\hat{C}\tilde{N}^{1/4}(1+\bar{\mu}(c))$
$\displaystyle=$ $\displaystyle\hat{C}\tilde{N}^{1/4}+\tilde{R}(c),$

whereby the remainder term $\tilde{R}(c)$ is such that

$|\tilde{R}(c)|\leq\tilde{C}_{7}+\frac{\tilde{C}_{7}}{\hat{C}}\tilde{C}_{7}\tilde{C}_{8}\left(1+\tilde{C}_{7}\tilde{C}_{8}\right)=\tilde{C}_{7}+\frac{1}{\hat{C}}\tilde{C}_{9}.$

In the above calculations, $\tilde{C}_{k}$ are positive constants, $5\leq
k\leq 9$, all independent of $\tilde{N}$. And we see that if $\hat{C}>1$, then
we have that $|\tilde{R}(c)|\leq\tilde{C}_{7}+\tilde{C}_{9}$, independent of
$\hat{C}$ and $\tilde{N}$.

We will henceforth define for $n\in\mathbb{N}$,
$\kappa_{n}:=\hat{C}\tilde{N}_{n}^{1/4}+\hat{C}_{n}$, for some fixed positive
constant $\hat{C}>1$ and $|\hat{C}_{n}|\leq\tilde{C}_{7}+\tilde{C}_{9}$ for
all $n\in\mathbb{N}$.

Set $|a|=1$. Start with the representation
$\rho_{1}:\mathfrak{g}\rightarrow{\rm End}({{\mathbb{C}}}^{\tilde{N}_{1}})$.
We choose a constant $\hat{C}$ large enough, such that when we plug in
$\hat{C}\tilde{N}_{n}^{1/4}+\hat{C}_{1}=\kappa_{1}\geq\kappa_{0}$ into
Equation (7.4), we have

$0<m_{1}^{2}:=\parallel\epsilon(1,\kappa_{1})\parallel=\frac{\tilde{N}_{1}}{4}C_{2}(\rho_{1})-\left\|{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a]}^{\kappa_{1},\rho_{1}}\right)_{\sharp}^{2}\mathcal{Y}^{\kappa}\right]\right\|,$

$\parallel\cdot\parallel$ is a matrix trace norm defined as $\parallel
A\parallel={{\rm{Tr}}}\ |A|$. Since the mass gap $m_{1}>0$, we define the
quantized momentum eigenvalue
$\hat{P}(\rho_{1})^{2}:=\hat{H}(\rho_{1})^{2}-m_{1}^{2}$.

Take the trace on Equation (7.4),

$\left\|{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a]}^{\kappa,\rho_{n}}\right)_{\sharp}^{2}\mathcal{Y}^{\kappa}\right]\right\|=\frac{\tilde{N}_{n}}{4}C_{2}(\rho_{n})-\parallel\epsilon(n,\kappa)\parallel.$
(7.9)

Rewrite this, we will obtain

$\frac{4}{NC(\rho_{n})}\left\|{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a]}^{\kappa,\rho_{n}}\right)_{\sharp}^{2}\mathcal{Y}^{\kappa}\right]\right\|-1=-\frac{4\parallel\epsilon(n,\kappa)\parallel}{NC(\rho_{n})}.$

Compare this with Equation (7.8). Note that we made use of Remark 7.2 and the
trace norm for matrices. It remains to plug in $\kappa=\kappa_{n}$ in the RHS
of the equation, to determine the momentum eigenvalues.

Consider $n\geq 2$. For any irreducible representation
$\rho_{n}:\mathfrak{g}\rightarrow{\rm End}({{\mathbb{C}}}^{\tilde{N}_{n}})$,
we define $m_{n}$, such that

$m_{n}^{2}:=-{{\rm{Tr}}}[\epsilon(n,\kappa_{n})]\equiv\parallel\epsilon(n,\kappa_{n})\parallel.$
(7.10)

If one examines the proof of Equation (7.4) in [6], we have that for all
$n\in\mathbb{N}$,

$0<\frac{\underline{c}}{\kappa_{n}^{4}}<-\frac{1}{C(\rho_{n})}{{\rm{Tr}}}[\epsilon(n,\kappa_{n})]\leq\frac{\bar{c}}{\kappa_{n}^{4}},$
(7.11)

for constants $\underline{c},\bar{c}$ independent of $\rho_{n}$.

Hence, the trace $-{{\rm{Tr}}}[\epsilon(n,\kappa_{n})]$ is of the order
$C(\rho_{n})/\kappa_{n}^{4}$. In terms of the dimension scale, we see that it
is of the order $C(\rho_{n})/\hat{C}^{4}\tilde{N}_{n}\equiv
C_{2}(\rho_{n})/N\hat{C}^{4}$. From Equation (7.11), we must have that
$m_{n}^{2}$ is of the order $\dfrac{C_{2}(\rho_{n})}{\hat{C}^{4}}>0$, which
tends to infinity, as $n\rightarrow\infty$, because
$\\{C_{2}(\rho_{n}):n\in\mathbb{N}\\}$ is unbounded,

Define the quantized momentum eigenvalue, in the direction of a unit vector
$a$ as

$\displaystyle\hat{P}(\rho_{n})^{2}:=$ $\displaystyle-{{\rm{Tr}}}\
{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a]}^{\kappa_{n},\rho_{n}}\right)_{\sharp}^{2}\mathcal{Y}^{\kappa}\right]$
$\displaystyle=$
$\displaystyle\frac{\tilde{N}_{n}}{4}C_{2}(\rho_{n})-\parallel\epsilon(n,\kappa_{n})\parallel=\frac{\tilde{N}_{n}}{4}C_{2}(\rho_{n})-m_{n}^{2}>0,$

from Equations (7.9) and (7.10). Thus, the average of the flux through the
time-like rectangular surface $R[a]$ using the Yang-Mills measure, quantize
the momentum in the direction $a$.

Note that $m_{n}^{2}/\hat{H}(\rho_{n})^{2}$, is of the order
$\dfrac{1}{\tilde{N}_{n}\hat{C}^{4}}>0$. Hence, we see that the momentum
operator eigenvalues will go to infinity, from

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

as $n\rightarrow\infty$.

###### Remark 7.12

Note that we interpret $\tilde{N}C_{2}(\rho)$ as energy squared. Because of
the beta function, we see that $\kappa$ increases with the dimension of the
irreducible representation. In the case of ${\rm SU}(2)$ or ${\rm SU}(3)$, we
see that the Casimir operator is large, when the dimension of the
representation is large. More generally, the Weyl dimension formula says that
when the dimension of the representation $\rho$ is large, then
$|\lambda_{\rho}|$ will be large, which also implies that the Casimir constant
$C_{2}(\rho)$ will be large. Since the coupling constant $c=1/\kappa$, the
coupling constant weakens at high energies, which the physicists will term as
asymptotic freedom.

###### Definition 7.13

(Hamiltonian and momentum operator)
For each $n\in\mathbb{N}$, let $\rho_{n}$ be an irreducible non-trivial
representation for a simple Lie Algebra $\mathfrak{g}$. Recall we defined
$\hat{H}(\rho_{n})^{2}=\frac{\tilde{N}_{n}}{4}C_{2}(\rho_{n})>0$ and
$\hat{P}(\rho_{n})$ such that

$\hat{H}(\rho_{n})^{2}-\hat{P}(\rho_{n})^{2}=m_{n}^{2}>0,$

for some positive mass gap $m_{n}>0$ defined by Equation (7.10).

From Definition 2.5, we have a basis for ${{\mathbb{R}}}^{4}$. Define a
Hamiltonian $\hat{H}(\vec{a},\rho)$, in the direction
$\vec{a}\in{{\mathbb{R}}}^{4}$, as
$\hat{H}(\vec{a},\rho):=(\vec{a}\cdot\hat{f}_{0})\hat{H}(\rho)\hat{f}_{0}$.
Explicitly, if $\vec{a}=\sum_{b=0}^{3}a^{b}\hat{f}_{b}$, we will write
$\hat{H}(\vec{a},\rho):=-a^{0}\hat{H}(\rho)\hat{f}_{0}$.

We define a momentum operator, $\hat{P}(\vec{a},\rho)$, in the direction
$\vec{a}\in{{\mathbb{R}}}^{4}$, as
$\hat{P}(\vec{a},\rho):=(\vec{a}\cdot\hat{f}_{1})\hat{P}(\rho)\hat{f}_{1}$.
Explicitly, if $a=\sum_{b=0}^{3}a^{b}\hat{f}_{b}$, we will write
$\hat{P}(\vec{a},\rho):=a^{1}\hat{P}(\rho)\hat{f}_{1}$. Thus

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

We also define $\hat{H}(\rho_{0})=\hat{P}(\rho_{0})=0$, thus for the vacuum
state, $\hat{P}1=\hat{H}1=0$. The momentum operator $\hat{P}$ and Hamiltonian
$\hat{H}$ act on the Hilbert space $\mathbb{H}_{{\rm YM}}(\mathfrak{g})$, as

$\displaystyle\hat{P}\sum_{n=0}^{\infty}v_{n}:=\sum_{n=1}^{\infty}\hat{P}(\rho_{n})v_{n},\quad\hat{H}\sum_{n=0}^{\infty}v_{n}:=\sum_{n=1}^{\infty}\hat{H}(\rho_{n})v_{n},$

provided the sum converges. Because $\\{C_{2}(\rho_{n})\\}_{n\geq 1}$ is
unbounded and the eigenvalues are real, we see that $\hat{P}$ and $\hat{H}$
are unbounded self-adjoint operators.

###### Theorem 7.14

(Positive mass Gap Theorem)
Consider the Hilbert space defined in Equation (2.1). The quantized momentum
operator $\hat{P}$ and Hamiltonian $\hat{H}$ are non-negative, unbounded self-
adjoint operators on this Hilbert space. They annihilate the vacuum state 1
and their other eigenspaces are $\mathscr{H}(\rho_{n})$, $n\geq 1$, with
corresponding eigenvalues $\hat{P}(\rho_{n})$ and $\hat{H}(\rho_{n})$
respectively, such that,
$\hat{H}(\rho_{n})^{2}-\hat{P}(\rho_{n})^{2}=m_{n}^{2}$ for some positive mass
gap $m_{n}>0$.

These eigenvalues were chosen because of the Callan-Symanzik Equation, and we
have that $\lim_{n\rightarrow\infty}m_{n}^{2}=\infty$. Thus
$m_{0}:=\inf_{n\in\mathbb{N}}m_{n}>0$, showing the existence of a positive
mass gap in a 4-dimensional quantum compact and simple Yang-Mills gauge
theory.

###### Remark 7.15

In [21], they explicitly state that the mass gap refers to the gap in the
spectrum of the Hamiltonian operator. But this cannot be correct, as the mass
gap refers to the difference between the Hamiltonian and the momentum
operator. A strictly positive mass gap is required to show the short range
nature of the weak force.

Our discussion on the construction of a 4-dimensional quantum Yang-Mills
simple and compact gauge group is now complete. Does our construction apply to
an abelian gauge group? Answer is no.

When we have an abelian group ${\rm U}(1)$, we can define a non-trivial
representation $F_{a}:u\in\mathfrak{u}(1)\mapsto 0\neq a\in{{\mathbb{R}}}$,
for any $a$. It is clear that $F_{a}$ is a representation of
$\mathfrak{u}(1)$. Essentially, there is only one irreducible representation
of $\mathfrak{u}(1)$, up to a constant. As a consequence, using the
construction as outlined, the momentum operator and Hamiltonian, will not be
unbounded operators.

An abelian gauge group describes the quantum electromagnetic force. As photons
are massless, we see that $(\hat{H},\hat{P},0,0)$ must be a null-vector. As a
result, the commutation and anti-commutation relations discussed in Section 5
will only hold on a 2-dimensional subspace.

A positive mass gap does not exist for an abelian gauge group ${\rm U}(1)$.
The existence of a mass gap is due to the quartic term of the interaction term
in the Yang-Mills action. And this gives us Equation (7.4), proved in [6]
using asymptotic freedom and the structure constants of the Lie Algebra
$\mathfrak{g}$, which do not apply to an abelian gauge group. Incidently, it
will also give us the area law formula. Because of the absence of asymptotic
freedom, the area law formula will not hold in the case of an abelian gauge
group, as was shown in [8].

##  8 Clustering

One important feature of the weak and strong force is that both are short
range. This is formulated mathematically as the Clustering Theorem. To further
validate that our construction of a 4-dimensional quantum field over in
${{\mathbb{R}}}^{4}$, satisfying Wightman’s axioms with a mass gap is correct,
we will devote this section to the proof of the Clustering Theorem.

The first version was proved by Ruelle in [19]. The proof that it decays
exponentially is taken from [22]. Our main reference for the Cluster
Decomposition Property will be from [3], and we will follow closely the
notations used in there.

###### Remark 8.1

We will use the standard orthonormal basis $\\{e_{a}\\}_{a=0}^{3}$, whereby
$\\{e_{2},e_{3}\\}$ span $S_{0}$, and $\\{e_{0},e_{1}\\}$ will span a time-
like plane $S_{0}^{\flat}$. The coordinates used in this section will be
pertaining to the standard orthonormal basis. Thus,
$\vec{a}=(a^{0},a^{1},a^{2},a^{3})$ means the vector
$\sum_{b=0}^{3}a^{b}e_{b}\in{{\mathbb{R}}}^{4}$.

###### Definition 8.2

(Fourier Transform)
We define the Fourier Transform of a $L^{2}$ function
$f:{{\mathbb{R}}}\rightarrow{{\mathbb{R}}}$ by

$\hat{f}(p)=\mathcal{F}[f](p):=\frac{1}{\sqrt{2\pi}}\int_{{{\mathbb{R}}}}e^{-ipx}f(x)\
dx.$

On ${{\mathbb{R}}}^{4}$, our Fourier Transform is given by

$\hat{f}(\vec{p})=\mathcal{F}[f](\vec{p}):=\frac{1}{(2\pi)^{2}}\int_{{{\mathbb{R}}}^{4}}e^{-i\vec{p}\cdot\vec{x}}f(\vec{x})\
d\vec{x}.$

See Equation (8.1).

Recall we fixed a surface $S_{0}$, which is the $x^{2}-x^{3}$ plane, and we
have the quantum state
$\Big{(}S_{0},f_{\alpha}\otimes\rho(E^{\alpha}),\\{e_{a}\\}_{a=0}^{3}\Big{)}$.
But how does one understand it? From the definition of the field operators in
Section 3, we see that $f_{\alpha}$ is synonymous with
$F_{\alpha}\left(\hat{H}(\rho),\hat{P}(\rho),x^{2},x^{3}\right)$, whereby
$F_{\alpha}\left(\hat{H}(\rho),\hat{P}(\rho),\cdot,\cdot\right):S_{0}\rightarrow{{\mathbb{C}}}$,
is a Schwartz function on $S_{0}$.

Let $\hat{F}_{\alpha}$ be the Fourier Transform of $F_{\alpha}$. By taking the
Fourier Transform of $F_{\alpha}$ over in $S_{0}$,

$\displaystyle\left(\frac{1}{\sqrt{2\pi}}\right)^{2}$
$\displaystyle\int_{S_{0}}e^{-i[q^{2}x^{2}+q^{3}x^{3}]}F_{\alpha}\left(\hat{H}(\rho),\hat{P}(\rho),x^{2},x^{3}\right)\
dx^{2}dx^{3}\otimes\rho(E^{\alpha})$ $\displaystyle=$
$\displaystyle\hat{F}_{\alpha}\left(\hat{H}(\rho),\hat{P}(\rho),q^{2},q^{3}\right)\otimes\rho(E^{\alpha}).$

Hence one can view the quantum state as an operator-valued tempered
distribution, over in momentum-energy space.

###### Remark 8.3

The momentum coordinates
$\left(\hat{H}(\rho),\hat{P}(\rho),q^{2},q^{3}\right)$ are with respect to the
basis $\\{e_{a}\\}_{a=0}^{3}$. In general, if $S$ is a space-like plane
equipped with a Minkowski frame $\\{\hat{f}_{a}\\}_{a=0}^{3}$, then the
Fourier Transform will yield
$\hat{F}_{\alpha}\left(\hat{H}(\rho),\hat{P}(\rho),q^{2},q^{3}\right)\otimes\rho(E^{\alpha})$,
but with the coordinates pertaining to the basis
$\\{\hat{f}_{a}\\}_{a=0}^{3}$.

Thus, it is clear that $\hat{F}_{\alpha}$ depends on the basis
$\\{\hat{f}_{a}\\}_{a=0}^{3}$. We wish to point out that it is also dependent
on $\vec{a}=a^{0}\hat{f}_{0}+a^{1}\hat{f}_{1}$ in space-time, which is a
position vector in the space-like plane $S$. This is because its Fourier
Transform is

$\displaystyle\hat{F}$
$\displaystyle\left(\hat{H}(\rho)\hat{f}_{0}+\hat{P}(\rho)\hat{f}_{1}+q^{2}\hat{f}_{2}+q^{3}\hat{f}_{3}\right)$
$\displaystyle:=\frac{e^{-i[a^{0}\hat{H}(\rho)-a^{1}\hat{P}(\rho)]}}{2\pi}\int_{\hat{s}\in{{\mathbb{R}}}^{2}}e^{-i(sq^{2}+\bar{s}q^{3})}f^{\\{\hat{f}_{0},\hat{f}_{1}\\}}(\hat{H}(\rho),\hat{P}(\rho))(s\hat{f}_{2}+\bar{s}\hat{f}_{3})\
d\hat{s}$ (8.1) $\displaystyle\equiv
e^{-i[a^{0}\hat{H}(\rho)-a^{1}\hat{P}(\rho)]}\hat{f}\left(\hat{H}(\rho)\hat{f}_{0}+\hat{P}(\rho)\hat{f}_{1}+q^{2}\hat{f}_{2}+q^{3}\hat{f}_{3}\right),$

from Equation (3.2). Also refer to Item 3 in Remark 3.5.

Hence,
$\hat{F}_{\alpha}\left(\hat{H}(\rho)\hat{f}_{0}+\hat{P}(\rho)\hat{f}_{1}+q^{2}\hat{f}_{2}+q^{3}\hat{f}_{3}\right)\otimes\rho(E^{\alpha})$
allows us to reconstruct
$\left(S,f_{\alpha}\otimes\rho(E^{\alpha}),\\{\hat{f}_{a}\\}_{a=0}^{3}\right)$.

For a Schwartz function $f\in\mathscr{P}$, ${\rm supp}\ \hat{f}$ may or may
not lie in the positive light cone in energy-momentum space. Thus,
$\hat{H}(\rho)^{2}-\hat{P}(\rho)^{2}-q^{2,2}-q^{2,3}$ may not be greater than
or equal to 0. But this does not matter, as we are still able to prove the
Clustering Decomposition Property in Theorem 8.32. Even though it is not
required or necessary to impose the condition that
$\hat{H}(\rho)^{2}-\hat{P}(\rho)^{2}-q^{2,2}-q^{2,3}\geq 0$, this approach is
not quite correct.

In the proof of Clustering Theorem 8.34, we will see that by a suitable choice
of a Minkowski frame, we will instead use $\left(m,q^{1},q^{2},q^{3}\right)$
as the coordinates in a 4-dimensional mass-momentum space after taking Fourier
Transform. The translation operator in a time-like direction will be generated
by mass, not energy. The momentum coordinates $q^{i}$’s will be free, and the
existence of a positive mass gap $m_{0}$ says that the total energy is given
by $m^{2}+\sum_{i=1}^{3}|q^{i}|^{2}\geq m_{0}^{2}$.

Refer to Definition 7.13. When we take the Fourier Transform of
$U(\vec{a},1)\left(S_{0},f_{\alpha}\otimes\rho(E^{\alpha}),\\{e_{a}\\}_{a=0}^{3}\right)$,
a similar set of calculations will give us

$e^{-i\vec{a}\cdot\vec{\alpha}}e^{-i[q^{2}a^{2}+q^{3}a^{3}]}\hat{F}_{\alpha}\left(\hat{H}(\rho),\hat{P}(\rho),q^{2},q^{3}\right)\otimes\rho(E^{\alpha}),$

$\vec{\alpha}=(\hat{H}(\rho),\hat{P}(\rho),0,0)$.

Recall, $\\{0\\}\cup\\{\hat{H}(\rho_{n}),\hat{P}(\rho_{n})\\}_{n\geq 1}$ are
the discrete eigenvalues of the Hamiltonian and momentum operator. By the SNAG
theorem, we see that when $\vec{a}\in S_{0}^{\flat}$, i.e.
$\vec{a}=a^{0}e_{0}+a^{1}e_{1}$, the spectrum of the translation operator
$U(\vec{a},1)$ on $\mathbb{H}_{{\rm YM}}(\mathfrak{g})$ is
$\\{1\\}\cup\\{e^{i[a^{0}\hat{H}(\rho_{n})-a^{1}\hat{P}(\rho_{n})]}\\}_{n\geq
1}$, which is discrete.

Now, recall $S_{0}^{\flat}$ is the $x^{0}-x^{1}$ plane and we have the field
operator $\phi^{\alpha,n}(f)$, $f\in\mathscr{P}$, defined in Definitions 3.7
and 3.12. We can define a distribution, denoted as
$\left\langle\phi^{\alpha,n}(\vec{x})1,\phi^{\beta,n}(\vec{y})1\right\rangle$,
via sending $(f,g)\in\mathscr{P}\times\mathscr{P}$ to

$\left\langle\phi^{\alpha,n}(f)1,\phi^{\beta,n}(g)1\right\rangle:=C(\rho_{n}){{\rm{Tr}}}[-F^{\alpha}F^{\beta}]\int_{S_{0}}\left[f^{\\{e_{0},e_{1}\\}}\overline{g^{\\{e_{0},e_{1}\\}}}\right](\hat{H}(\rho_{n}),\hat{P}(\rho_{n}))(\hat{s})\
d\hat{s}.$

From this, we can define an $\rho_{n}(\mathfrak{g})$-valued distribution
$\phi^{\alpha,n}(\vec{x})$, and write

$\phi^{\alpha,n}(f)1=\int_{\vec{x}\in{{\mathbb{R}}}^{4}}d\vec{x}\
f(\vec{x})\phi^{\alpha,n}(\vec{x})1,$

when it acts on vacuum state. See [3, 19].

Using the transformation law for the field operator, we see that

$U(\vec{a},1)\phi^{\alpha,n}(\vec{x})U(\vec{a},1)^{-1}=\phi^{\alpha,n}(\vec{x}+\vec{a}).$

###### Remark 8.4

We present an alternative way to understand quantum fields. One will be
tempted to view $\phi^{\alpha,n}(\vec{x})1$ heuristically as

$\frac{1}{2\pi}e^{i[x^{0}\hat{H}(\rho_{n})-x^{1}\hat{P}(\rho_{n})]}\delta(\cdot-(x^{2},x^{3}))\otimes\rho_{n}(F^{\alpha}),$

whereby $\delta$ is the 2-dimensional Dirac delta function. But this is not
entirely correct.

Suppose we replace the Dirac delta function with a Gaussian function
$p_{\kappa}^{\vec{x}}(\cdot)=(\kappa^{3}/2\sqrt{2\pi})\exp[-\kappa^{2}|\vec{x}-\cdot|^{2}/4]/(2\pi)$.
When we take the inner product,

$\int_{\vec{z}\in{{\mathbb{R}}}^{4}}p_{\kappa}^{\vec{x}}(\vec{z})p_{\kappa}^{\vec{y}}(\vec{z})\
d\vec{z}=\frac{\kappa^{2}}{4(2\pi)}\exp[-\kappa^{2}|\vec{x}-\vec{y}|^{2}/8].$

Write $x^{+}=(0,0,x^{2},x^{3})$, $\hat{H}\equiv\hat{H}(\rho_{n})$ and
$\hat{P}\equiv\hat{P}(\rho_{n})$. Thus, we can approximate
$\langle\phi^{\alpha,n}(\vec{x})1,\phi^{\beta,n}(\vec{y})1\rangle$ with

$\left\langle\frac{1}{2\pi}e^{i[x^{0}\hat{H}-x^{1}\hat{P}]}p_{\kappa}^{x^{+}}\otimes\rho_{n}(F^{\alpha}),\frac{1}{2\pi}e^{i[y^{0}\hat{H}-y^{1}\hat{P}]}p_{\kappa}^{y^{+}}\otimes\rho_{n}(F^{\beta})\right\rangle,$

as

$\displaystyle\frac{1}{(2\pi)^{2}}$ $\displaystyle
e^{i[\hat{H}(x^{0}-y^{0})-\hat{P}(x^{1}-y^{1})]}\int_{\vec{z}\in{{\mathbb{R}}}^{4}}p_{\kappa}^{x^{+}}(\vec{z})p_{\kappa}^{y^{+}}(\vec{z})\
d\vec{z}\cdot\left\langle\rho_{n}(F^{\alpha}),\rho_{n}(F^{\beta})\right\rangle$
$\displaystyle=$
$\displaystyle\frac{\kappa^{2}}{4(2\pi)^{3}}e^{i[\hat{H}(x^{0}-y^{0})-\hat{P}(x^{1}-y^{1})]}\exp[-\kappa^{2}|x^{+}-y^{+}|^{2}/8]\cdot\left\langle\rho_{n}(F^{\alpha}),\rho_{n}(F^{\beta})\right\rangle.$

Taking the limit, we see that we can understand the distribution
$\left\langle\phi^{\alpha,n}(\vec{x})1,\phi^{\beta,n}(\vec{y})1\right\rangle$
as

$\frac{1}{(2\pi)^{2}}e^{i[\hat{H}(x^{0}-y^{0})-\hat{P}(x^{1}-y^{1})]}\cdot\delta(x^{+}-y^{+})\cdot
C(\rho_{n}){{\rm{Tr}}}[-F^{\alpha}F^{\beta}],\ x^{+},y^{+}\in S_{0}.$

Therefore, we can approximate
$\langle\phi^{\alpha,n}(f),\phi^{\beta,n}(g)\rangle$ as

$\displaystyle\frac{1}{(2\pi)^{2}}$
$\displaystyle\int_{\vec{x},\vec{y}\in{{\mathbb{R}}}^{4}}f(\vec{x})\overline{g(\vec{y})}\left\langle
e^{i(x^{0}\hat{H}-x^{1}\hat{P})}p_{\kappa}^{x^{+}}\otimes\rho_{n}(F^{\alpha}),e^{i(y^{0}\hat{H}-y^{1}\hat{P})}p_{\kappa}^{y^{+}}\otimes\rho_{n}(F^{\beta})\right\rangle
d\vec{x}d\vec{y}$ $\displaystyle=$
$\displaystyle\frac{\kappa^{2}}{4}\int_{\hat{s},\hat{t}\in
S_{0}}f^{\\{e_{0},e_{1}\\}}(\hat{s})\overline{g^{\\{e_{0},e_{1}\\}}}(\hat{t})\frac{1}{2\pi}\exp[-\kappa^{2}|\hat{s}-\hat{t}|^{2}/8]d\hat{s}d\hat{t}\cdot\left\langle\rho_{n}(F^{\alpha}),\rho_{n}(F^{\beta})\right\rangle,$

which approaches to

$\int_{\hat{s}\in
S_{0}}[f^{\\{e_{0},e_{1}\\}}\overline{g^{\\{e_{0},e_{1}\\}}}](\hat{s})d\hat{s}\cdot\left\langle\rho_{n}(F^{\alpha}),\rho_{n}(F^{\beta})\right\rangle,$

when $\kappa\rightarrow\infty$. This justifies the inner product given in
Definition 2.9. These calculations can also be found in [8].

Observe that

$\lim_{\kappa\rightarrow\infty}\int_{\vec{z}\in{{\mathbb{R}}}^{4}}p_{\kappa}^{\vec{x}}(\vec{z})f(\vec{z})\
d\vec{z}=0,$

for any $f\in\mathscr{P}$. Thus, $\phi^{\alpha,n}(\vec{x})1$ itself has no
meaning. But, we can give meaning to
$\int_{\vec{x}\in{{\mathbb{R}}}^{4}}d\vec{x}\
f(\vec{x})\phi^{\alpha,n}(\vec{x})1$.

###  8.1 Vacuum Expectation

###### Notation 8.5

Let $\\{f_{1},\cdots,f_{r}\\}$ and $\\{g_{1},\cdots,g_{s}\\}$ be two sets of
compactly supported functions in $\mathscr{P}$. Let
$\psi^{\alpha_{\tau},n}(f_{\tau})$ be either the creation operator
$\phi^{\alpha_{\tau},n}(f_{\tau})$ or annihilation operator
$\phi^{\alpha_{\tau},n}(f_{\tau})^{\ast}$. Similar notation for
$\psi^{\beta_{\theta},n}(g_{\theta})$. Without loss of generality, we assume
that $r\geq s$.

Fix a $n\in\mathbb{N}$. Throughout this subsection 8.1, we will write
$\hat{H}(\rho_{n})=\hat{H}$, $\hat{P}(\rho_{n})=\hat{P}$. Define

$\displaystyle A_{r}^{n}=$
$\displaystyle\psi^{\alpha_{1},n}(f_{1})\cdots\psi^{\alpha_{r},n}(f_{r}),\quad
B_{s}^{n}=\psi^{\beta_{1},n}(g_{1})\cdots\psi^{\beta_{s},n}(g_{s}).$

Consider the vacuum expectation

$\left\langle A_{r}^{n}P_{0}B_{s}^{n}1,1\right\rangle\equiv\left\langle
A_{r}^{n}B_{s}^{n}1,1\right\rangle-\left\langle
A_{r}^{n}1,1\right\rangle\left\langle B_{s}^{n}1,1\right\rangle.$

Here, $P_{0}$ is the orthogonal projection onto $\bigoplus_{n\geq
1}\mathscr{H}(\rho_{n})$.

To write down an explicit formula, we need the following notation.

###### Notation 8.6

Define $\left\\{h_{\theta}\in\mathscr{P}\right\\}_{\theta=1}^{r+s}$ as
follows. When $1\leq\theta\leq r$, then

$h_{\theta}=\left\\{\begin{array}[]{ll}f_{\theta},&\hbox{{\rm if}
$\psi^{\alpha_{\theta},n}(f_{\theta})=\phi^{\alpha_{\theta},n}(f_{\theta})$;}\\\
-\overline{f}_{\theta},&\hbox{{\rm if}
$\psi^{\alpha_{\theta},n}(f_{\theta})=\phi^{\alpha_{\theta},n}(f_{\theta})^{\ast}$.}\end{array}\right.$

When $r+1\leq\theta\leq r+s$, then

$h_{\theta}=\left\\{\begin{array}[]{ll}g_{\theta-r},&\hbox{{\rm if}
$\psi^{\beta_{\theta-r},n}(g_{\theta-r})=\phi^{\beta_{\theta-r},n}(g_{\theta-r})$;}\\\
-\overline{g}_{\theta-r},&\hbox{{\rm if}
$\psi^{\beta_{\theta-r},n}(g_{\theta-r})=\phi^{\beta_{\theta-r},n}(g_{\theta-r})^{\ast}$.}\end{array}\right.$

Similarly, define
$\left\\{\tilde{h}_{\theta}\in\mathscr{P}\right\\}_{\theta=1}^{r+s}$ as
follows. When $1\leq\theta\leq s$, then

$\tilde{h}_{\theta}=\left\\{\begin{array}[]{ll}g_{\theta},&\hbox{{\rm if}
$\psi^{\alpha_{\theta},n}(g_{\theta})=\phi^{\alpha_{\theta},n}(g_{\theta})$;}\\\
-\overline{g}_{\theta},&\hbox{{\rm if}
$\psi^{\alpha_{\theta},n}(g_{\theta})=\phi^{\alpha_{\theta},n}(g_{\theta})^{\ast}$.}\end{array}\right.$

When $s+1\leq\theta\leq r$, then

$\tilde{h}_{\theta}=\left\\{\begin{array}[]{ll}f_{\theta-s},&\hbox{{\rm if}
$\psi^{\alpha_{\theta},n}(f_{\theta-s})=\phi^{\alpha_{\theta},n}(f_{\theta-s})$;}\\\
-\overline{f}_{\theta-s},&\hbox{{\rm if}
$\psi^{\alpha_{\theta},n}(f_{\theta-s})=\phi^{\alpha_{\theta},n}(f_{\theta-s})^{\ast}$.}\end{array}\right.$

When $r+1\leq\theta\leq r+s$, then

$\tilde{h}_{\theta}=\left\\{\begin{array}[]{ll}f_{\theta-s},&\hbox{{\rm if}
$\psi^{\beta_{\theta-r},n}(f_{\theta-s})=\phi^{\beta_{\theta-r},n}(f_{\theta-s})$;}\\\
-\overline{f}_{\theta-s},&\hbox{{\rm if}
$\psi^{\beta_{\theta-r},n}(f_{\theta-s})=\phi^{\beta_{\theta-r},n}(f_{\theta-s})^{\ast}$.}\end{array}\right.$

For those $\theta\in\\{1,2,\cdots,r+s\\}$, such that

$\displaystyle\psi^{\alpha_{\theta},n}(h)$
$\displaystyle=\phi^{\alpha_{\theta},n}(h)^{\ast},\ 1\leq\theta\leq r,$
$\displaystyle\psi^{\beta_{\theta-r},n}(h)$
$\displaystyle=\phi^{\beta_{\theta-r},n}(h)^{\ast},\ r+1\leq\theta\leq r+s,$

we will say that $\theta$ is adjoint. Then, we will write

$\chi(\theta)=\left\\{\begin{array}[]{ll}-1,&\hbox{$\theta$ is adjoint;}\\\
1,&\hbox{otherwise.}\end{array}\right.$

For $\vec{x}=(x^{0},x^{1},x^{2},x^{3})$, we will write
$d\vec{x}=dx^{0}dx^{1}dx^{2}dx^{3}$. And we will write
$\vec{x}=(x^{-},x^{+})$, $x^{-}=(x^{0},x^{1})$, $x^{+}=(x^{2},x^{3})$,
$dx^{-}=dx^{0}dx^{1}$, $dx^{+}=dx^{2}dx^{3}$.

A partition of $R=\\{1,2,\cdots,r+s\\}$ is given by
$Q=\\{A_{1},\ldots,A_{n(Q)}\\}$, whereby

  * •

$R=\bigcup_{l=1}^{n(Q)}A_{l}$;

  * •

$A_{l}\cap A_{\hat{l}}=\emptyset$, if $l\neq\hat{l}$;

  * •

$A_{l}=\\{z,z+1,\cdots,z+k\\}$, for some $z\in R$ and some $0\leq k\leq r+s$.

Let $\Gamma$ be the set of all such possible partitions of $R$. For a
partition $Q=\\{A_{1},\ldots,A_{n(Q)}\\}\in\Gamma$, we will write

$\displaystyle\int_{Q}\\{h_{\theta}\\}_{\theta=1}^{r+s}$
$\displaystyle:=\prod_{l=1}^{n(Q)}\left\\{\int_{S_{0}}\left[\prod_{\theta\in
A_{l}}\int_{y_{\theta}^{-}\in{{\mathbb{R}}}^{2}}\frac{e^{i\chi(\theta)[y_{\theta}^{0}\hat{H}-y_{\theta}^{1}\hat{P}]}}{2\pi}h_{\theta}(y_{\theta}^{-},y^{+})dy_{\theta}^{-}\right]dy^{+}\right\\}$
$\displaystyle=\prod_{l=1}^{n(Q)}\left\\{\int_{S_{0}}\left[\prod_{\theta\in
A_{l}}\int_{y_{\theta}^{0},y_{\theta}^{1}\in{{\mathbb{R}}}}\frac{e^{i\chi(\theta)[y_{\theta}^{0}\hat{H}-y_{\theta}^{1}\hat{P}]}}{2\pi}h_{\theta}(y_{\theta}^{0},y_{\theta}^{1},y^{2},y^{3})dy_{\theta}^{0}dy_{\theta}^{1}\right]dy^{2}dy^{3}\right\\}.$

Note that it is a product of $n(Q)$ integrals and
$A_{l}=\\{z,z+1,\cdots,z+k\\}$ for some $k$.

By abuse of notation, we will write $R=\\{1,2,\cdots,r+s\\}\in\Gamma$, which
is itself a partition of $R$.

From Definitions 3.7, 3.12, 3.14, we see that we can write

$\displaystyle\left\langle A_{r}^{n}P_{0}B_{s}^{n}1,1\right\rangle$
$\displaystyle=\sum_{Q\in\Gamma}c_{Q}\int_{Q}\\{h_{\theta}\\}_{\theta=1}^{r+s},$
(8.2)

for some set of complex coefficients
$\\{c_{Q}\in{{\mathbb{C}}}\\}_{Q\in\Gamma}$.

###### Remark 8.7

It is not difficult to see from Definitions 3.12 and 3.14, that if $Q$
contains a subset $\\{z\\}$, $z\in R$, then we must have $c_{Q}=0$.

###### Definition 8.8

Refer to Notation 8.5. Recall we assume $r\geq s$. Define

$\displaystyle C_{r}^{n}$
$\displaystyle=\psi^{\alpha_{1},n}(g_{1})\cdots\psi^{\alpha_{s},n}(g_{s})\psi^{\alpha_{s+1},n}(f_{1})\cdots\psi^{\alpha_{r},n}(f_{r-s}),$
$\displaystyle D_{s}^{n}$
$\displaystyle=\psi^{\beta_{1},n}(f_{r-s+1})\cdots\psi^{\beta_{s},n}(f_{r}).$

Refer to Notation 8.6. Define the following tempered distribution

$\displaystyle\mathscr{W}^{n}$
$\displaystyle:f_{1}\otimes_{{\mathbb{R}}}\cdots\otimes_{{\mathbb{R}}}f_{r}\otimes_{{\mathbb{R}}}g_{1}\otimes_{{\mathbb{R}}}\cdots\otimes_{{\mathbb{R}}}g_{s}\longmapsto\left\langle
A_{r}^{n}P_{0}B_{s}^{n}1,1\right\rangle.$

Here,

$\displaystyle\mathscr{W}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r},\\{\vec{x}_{\theta}\\}_{\theta=r+1}^{r+s}\right)$
$\displaystyle\equiv\mathscr{W}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{s},\\{\vec{x}_{\theta}\\}_{\theta=s+1}^{s+r}\right)$
$\displaystyle:=\mathscr{W}^{n}\left(\vec{x}_{1},\cdots,\vec{x}_{r},\vec{x}_{r+1},\cdots,\vec{x}_{r+s}\right),$

is a tempered distribution such that
($\vec{x}=(x^{0},x)=(x^{0},x^{1},x^{2},x^{3})\in{{\mathbb{R}}}^{4}$)

$\displaystyle\left\langle A_{r}^{n}B_{s}^{n}1,1\right\rangle-\left\langle
A_{r}^{n}1,1\right\rangle\left\langle
B_{s}^{n}1,1\right\rangle\equiv\left\langle
A_{r}^{n}P_{0}B_{s}^{n}1,1\right\rangle$
$\displaystyle=\int_{{{\mathbb{R}}}^{4}\times\cdots\times{{\mathbb{R}}}^{4}}\mathscr{W}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r},\\{\vec{x}_{\theta}\\}_{\theta=r+1}^{r+s}\right)\bigotimes_{\tau=1}^{r+s}p_{\tau}(\vec{x}_{\tau})\cdot\prod_{\tau=1}^{r+s}d\vec{x}_{\tau}$
$\displaystyle=c_{R}\int_{S_{0}}\left[\prod_{\tau=1}^{r}\int_{x_{\tau}^{-}\in{{\mathbb{R}}}^{2}}E(x_{\tau}^{-})h_{\tau}(x_{\tau}^{-},x^{+})dx_{\tau}^{-}\cdot\prod_{\theta=r+1}^{r+s}\int_{x_{\theta}^{-}\in{{\mathbb{R}}}^{2}}E(x_{\theta}^{-})h_{\theta}(x_{\theta}^{-},x^{+})\
dx_{\theta}^{-}\right]dx^{+}$ $\displaystyle+\sum_{\genfrac{}{}{0.0pt}{}{Q\neq
R}{Q\in\Gamma}}c_{Q}\int_{Q}\\{h_{\theta}\\}_{\theta=1}^{r+s},$ (8.3)

whereby
$E(x_{\tau}^{-})=E(x_{\tau}^{0},x_{\tau}^{1})=\dfrac{e^{i\chi(\tau)[x_{\tau}^{0}\hat{H}-x_{\tau}^{1}\hat{P}]}}{2\pi}$.
Note that $\chi(\tau)$ was defined in Notation 8.6 and $p_{\tau}=f_{\tau}$ if
$1\leq\tau\leq r$; $p_{\tau}=g_{\tau-r}$ if $r+1\leq\tau\leq r+s$.

If we switch the sets such that

$\mathscr{W}^{n}:g_{1}\otimes_{{\mathbb{R}}}\cdots\otimes_{{\mathbb{R}}}g_{s}\otimes_{{\mathbb{R}}}f_{1}\otimes_{{\mathbb{R}}}\cdots\otimes_{{\mathbb{R}}}f_{r}\longmapsto\left\langle
C_{r}^{n}P_{0}D_{s}^{n}1,1\right\rangle,$

then

$\displaystyle\left\langle C_{r}^{n}D_{s}^{n}1,1\right\rangle-\left\langle
C_{r}^{n}1,1\right\rangle\left\langle
D_{s}^{n}1,1\right\rangle\equiv\left\langle
C_{r}^{n}P_{0}D_{s}^{n}1,1\right\rangle$
$\displaystyle=\int_{{{\mathbb{R}}}^{4}\times\cdots\times{{\mathbb{R}}}^{4}}\mathscr{W}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{s},\\{\vec{x}_{\theta}\\}_{\theta=s+1}^{s+r}\right)\bigotimes_{\tau=1}^{s+r}\tilde{p}_{\tau}(\vec{x}_{\tau})\cdot\prod_{\tau=1}^{s+r}d\vec{x}_{\tau}$
$\displaystyle=c_{R}\int_{S_{0}}\left[\prod_{\tau=1}^{s}\int_{x_{\tau}^{-}\in{{\mathbb{R}}}^{2}}E(x_{\tau}^{-})\tilde{h}_{\tau}(x_{\tau}^{-},x^{+})dx_{\tau}^{-}\cdot\prod_{\theta=s+1}^{s+r}\int_{x_{\theta}^{-}\in{{\mathbb{R}}}^{2}}E(x_{\theta}^{-})\tilde{h}_{\theta}(x_{\theta}^{-},x^{+})\
dx_{\theta}^{-}\right]dx^{+}$ $\displaystyle+\sum_{\genfrac{}{}{0.0pt}{}{Q\neq
R}{Q\in\Gamma}}c_{Q}\int_{Q}\\{\tilde{h}_{\theta}\\}_{\theta=1}^{s+r}.$ (8.4)

Note that $\tilde{p}_{\tau}=g_{\tau}$ if $1\leq\tau\leq s$;
$\tilde{p}_{\tau}=f_{\tau-s}$ if $s+1\leq\tau\leq s+r$.

###### Remark 8.9

Refer to Remark 8.22.

Write $B_{s}^{n,\vec{a}}1$ as

$\displaystyle\psi^{\beta_{1},n}(g_{1}(\cdot-\vec{a}))\cdots\psi^{\beta_{s-1},n}(g_{s-1}(\cdot-\vec{a}))\Big{(}S_{0}+\vec{a},g_{s}^{\\{e_{0},e_{1}\\}}(\cdot-\vec{a})\otimes\rho_{n}(F^{\beta_{s}}),\\{e_{a}\\}_{a=0}^{3}\Big{)},$

and
$\psi^{\beta_{\tau},n}(g_{\tau})_{U(\vec{a})}:=U(\vec{a},1)\psi^{\beta_{\tau},n}(g_{\tau})U(\vec{a},1)^{-1}$,
for $1\leq\tau\leq s$.

By definition of the field operator and its adjoint, we have from Proposition
4.1,

$\displaystyle P_{0}U(\vec{a},1)B_{s}^{n}1=$ $\displaystyle
P_{0}\psi^{\beta_{1},n}(g_{1})_{U(\vec{a})}\psi^{\beta_{2},n}(g_{2})_{U(\vec{a})}\cdots\psi^{\beta_{s-1},n}(g_{s-1})_{U(\vec{a})}U(\vec{a},1)\psi^{\beta_{s},n}(g_{s})1$
$\displaystyle=$ $\displaystyle
e^{i[a^{0}\hat{H}-a^{1}\hat{P}]}P_{0}B_{s}^{n,\vec{a}}1=e^{i[a^{0}\hat{H}-a^{1}\hat{P}]}\left[B_{s}^{n,\vec{a}}1-\left\langle
B_{s}^{n}1,1\right\rangle 1\right],$

for $\vec{a}=\sum_{b=0}^{3}a^{b}e_{b}$.

Hence, we have

$\left\langle
A_{r}^{n}P_{0}U(\vec{a},1)B_{s}^{n}1,1\right\rangle=e^{i[a^{0}\hat{H}-a^{1}\hat{P}]}\left[\left\langle
A_{r}^{n}B_{s}^{n,\vec{a}}1,1\right\rangle-\left\langle
A_{r}^{n}1,1\right\rangle\left\langle B_{s}^{n}1,1\right\rangle\right].$ (8.5)

When $\vec{a}\notin S_{0}$, by our construction, we see that $\left\langle
A_{r}^{n}P_{0}U(\vec{a},1)B_{s}^{n}1,1\right\rangle$ is zero, because
$S_{0}\cap(S_{0}+\vec{a})=\emptyset$.

By definition, we have

$\displaystyle e^{i[a^{0}\hat{H}-a^{1}\hat{P}]}$
$\displaystyle\int_{{{\mathbb{R}}}^{2}}\frac{e^{i[s\hat{H}-t\hat{P}]}}{2\pi}f\left(s,t,x^{2},x^{3}\right)\
dsdt$ $\displaystyle=$
$\displaystyle\int_{{{\mathbb{R}}}^{2}}\frac{e^{i[(s+a^{0})\hat{H}-(t+a^{1})\hat{P}]}}{2\pi}f\left(s,t,x^{2},x^{3}\right)\
dsdt$ $\displaystyle=$
$\displaystyle\int_{{{\mathbb{R}}}^{2}}\frac{e^{i[s\hat{H}-t\hat{P}]}}{2\pi}f\left(s-a^{0},t-a^{1},x^{2},x^{3}\right)\
dsdt$ $\displaystyle=$ $\displaystyle
f\left(\cdot-(a^{0},a^{1},0,0)\right)^{\\{e_{0},e_{1}\\}}(\hat{H},\hat{P})(0,0,x^{2},x^{3}).$
(8.6)

Hence, after a translation by $\vec{a}=(a^{0},a^{1},0,0)$, we can consider
$e^{i[a^{0}\hat{H}-a^{1}\hat{P}]}\left(S_{0}+\vec{a},f^{\\{e_{0},e_{1}\\}}(\cdot-\vec{a})\otimes\rho(F^{\alpha}),\\{e_{0},e_{1}\\}\right)$,
as
$\left(S_{0},f(\cdot-\vec{a})^{\\{e_{0},e_{1}\\}}\otimes\rho(F^{\alpha}),\\{e_{a}\\}_{a=0}^{3}\right)$.
Thus, we will define the following.

###### Notation 8.10

Let $\vec{a}=(a^{0},a)\in{{\mathbb{R}}}^{4}$. Write

$\mathscr{W}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r},\\{\vec{x}_{\theta}+\vec{a}\\}_{\theta=r+1}^{r+s}\right):=\mathscr{W}^{n}(\vec{x}_{1},\cdots,\vec{x}_{r},\vec{x}_{r+1}+\vec{a},\cdots,\vec{x}_{r+s}+\vec{a}).$

And write $h_{\theta}^{\vec{a}}(\vec{y})=h_{\theta}(y^{0},y)$ if
$1\leq\theta\leq r$;
$h_{\theta}^{\vec{a}}(\vec{y})=h_{\theta}(y^{0}-a^{0},y-a)$ if
$r+1\leq\theta\leq r+s$.

###### Definition 8.11

Refer to Definition 7.13, Notations 8.6 and 8.10. Write
$\vec{a}=(a^{0},a)=(a^{0},a^{1},a^{2},a^{3})\in{{\mathbb{R}}}^{4}$.

Using Equations (8.2), (8.5) and (8.6), we will define

$\displaystyle H^{n}(\vec{a})$
$\displaystyle:=\int_{{{\mathbb{R}}}^{4}\times\cdots\times{{\mathbb{R}}}^{4}}\mathscr{W}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r},\\{\vec{x}_{\theta}\\}_{\theta=r+1}^{r+s}\right)\bigotimes_{\tau=1}^{r}p_{\tau}(\vec{x}_{\tau})\cdot\bigotimes_{\theta=r+1}^{r+s}p_{\theta}(\vec{x}_{r+\theta}-\vec{a})\cdot\prod_{\tau=1}^{r+s}d\vec{x}_{\tau}$
$\displaystyle\equiv\int_{{{\mathbb{R}}}^{4}\times\cdots\times{{\mathbb{R}}}^{4}}\mathscr{W}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r},\\{\vec{x}_{\theta}+\vec{a}\\}_{\theta=r+1}^{r+s}\right)\bigotimes_{\tau=1}^{r+s}p_{\tau}(\vec{x}_{\tau})\cdot\prod_{\tau=1}^{r+s}d\vec{x}_{\tau}$
$\displaystyle=c_{R}\int_{S_{0}}\left[\prod_{\tau=1}^{r}\int_{x_{\tau}^{-}\in{{\mathbb{R}}}^{2}}E(x_{\tau}^{-})h_{\tau}(x_{\tau}^{-},x^{+})dx_{\tau}^{-}\cdot\prod_{\theta=r^{+}}^{r+s}\int_{x_{\theta}^{-}\in{{\mathbb{R}}}^{2}}E(x_{\theta}^{-})h_{\theta}^{\vec{a}}(x_{\theta}^{-},x^{+})dx_{\theta}^{-}\right]\
dx^{+}$ $\displaystyle+\sum_{\genfrac{}{}{0.0pt}{}{Q\neq
R}{Q\in\Gamma}}c_{Q}\int_{Q}\\{h_{\theta}^{\vec{a}}\\}_{\theta=1}^{r+s},$
(8.7)

where
$E(x_{\tau}^{-})=E(x_{\tau}^{0},x_{\tau}^{1})=\dfrac{e^{i\chi(\tau)[x_{\tau}^{0}\hat{H}-x_{\tau}^{1}\hat{P}]}}{2\pi}$
and $p_{\tau}=f_{\tau}$ if $1\leq\tau\leq r$; $p_{\tau}=g_{\tau-r}$ if
$r+1\leq\tau\leq r+s$.

###### Remark 8.12

We will use $H^{n}(\vec{a})$, instead of $\left\langle
A_{r}^{n}P_{0}U(\vec{a},1)B_{s}^{n}1,1\right\rangle$, to prove the both the
Cluster Decomposition Property and Clustering Theorem. When $\vec{a}\in
S_{0}$, then they are equal to each other.

###### Lemma 8.13

Refer to Notation 8.5, whereby we have the two sets
$\\{\psi^{\alpha_{\tau},n}\\}_{\tau=1}^{r}$ and
$\\{\psi^{\beta_{\theta},n}\\}_{\theta=1}^{s}$. Suppose the sum
$\sum_{\tau=1}^{r}\chi(\tau)+\sum_{\theta=r+1}^{r+s}\chi(\theta)=0$, $\chi$
was defined in Notation 8.6.

Define for $1\leq i\leq r+s-1$, $\vec{\xi}_{i}=\vec{x}_{i}-\vec{x}_{i+1}$,
which are relative coordinates. Let $\vec{a}=(a^{0},a^{1},a^{2},a^{3})$,
pertaining to the standard orthonormal basis.

  1. 1.

For any $1\leq\alpha\leq r+s$, $\mathscr{W}^{n}$ is independent of the
variables $\\{x_{\alpha}^{2},x_{\alpha}^{3}\\}$.

  2. 2.

There exists a tempered distribution $W^{n}$, such that

$\mathscr{W}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r},\\{\vec{x}_{\theta}+\vec{a}\\}_{\theta=r+1}^{r+s}\right)=W^{n}\left(\vec{\xi}_{1},\cdots,\vec{\xi}_{r-1},\vec{\xi}_{r}-\vec{a},\vec{\xi}_{r+1},\cdots,\vec{\xi}_{r+s-1}\right).$

  3. 3.

Their Fourier Transforms are related by

$\displaystyle\widehat{\mathscr{W}}^{n}$
$\displaystyle\left(\vec{p}_{1},\cdots,\vec{p}_{r+s}\right)$
$\displaystyle=(2\pi)^{4}\delta\left(\sum_{\tau=1}^{r+s}\vec{p}_{\tau}\right)\widehat{W}^{n}\left(\vec{p}_{1},\vec{p}_{1}+\vec{p}_{2},\cdots,\vec{p}_{1}+\vec{p}_{2}+\cdots+\vec{p}_{r+s-1}\right).$
(8.8)

Proof. Recall we have a set of Schwartz functions
$\\{f_{1},\cdots,f_{r},g_{1},\cdots g_{s}\\}\subset\mathscr{P}$. Since
$\sum_{\tau=1}^{r}\chi(\tau)+\sum_{\theta=r+1}^{r+s}\chi(\theta)=0$, the set
of adjoint integers $\\{\delta_{1},\cdots,\delta_{c}\\}$ is in one to one
correspondence with the set of non-adjoint integers
$\\{\epsilon_{1},\cdots,\epsilon_{c}\\}$. We will pair each $\delta_{i}$ with
$\epsilon_{i}$. Note that for each $i=1,\cdots,c$, $c=\frac{r+s}{2}$, we can
write

$\vec{x}_{\delta_{i}}-\vec{x}_{\epsilon_{i}}=\vec{\xi}_{\delta_{i}}+\vec{\xi}_{\delta_{i}+1}+\cdots+\vec{\xi}_{\epsilon_{i}-1},$
(8.9)

when $\delta_{i}<\epsilon_{i}$. When $\delta_{i}>\epsilon_{i}$, we have

$\vec{x}_{\delta_{i}}-\vec{x}_{\epsilon_{i}}=-\left(\vec{\xi}_{\epsilon_{i}}+\vec{\xi}_{\epsilon_{i}+1}+\cdots+\vec{\xi}_{\delta_{i}-1}\right).$
(8.10)

Hence, we can write

$\sum_{\tau=1}^{r}\chi(\tau)[x_{\tau}^{0}\hat{H}-x_{\tau}^{1}\hat{P}]+\sum_{\theta=r+1}^{r+s}\chi(\theta)[x_{\theta}^{0}\hat{H}-x_{\theta}^{1}\hat{P}]=-\sum_{i=1}^{r+s-1}c_{i}[\xi_{i}^{0}\hat{H}-\xi_{i}^{1}\hat{P}],$
(8.11)

for some integers $c_{i}$’s. Therefore,

$\prod_{\tau=1}^{r}E(x_{\tau}^{-})\prod_{\theta=r+1}^{r+s}E(x_{\theta}^{-})=\frac{1}{(2\pi)^{r+s}}\exp\left[-i\sum_{j=1}^{r+s-1}c_{j}[\xi_{j}^{0}\hat{H}-\xi_{j}^{1}\hat{P}]\right].$
(8.12)

Thus, Remark 8.7 and Equation (8.7) imply that

  1. 1.

$\mathscr{W}^{n}$ is independent of any chosen variables
$\\{x_{\alpha}^{2},x_{\alpha}^{3}\\}$, by letting
$x^{+}=(x_{\alpha}^{2},x_{\alpha}^{3})$ in the equation;

  2. 2.

there is a tempered distribution $W^{n}$ such that

$W^{n}\left(\vec{\xi}_{1},\cdots,\vec{\xi}_{r+s-1}\right):=\mathscr{W}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r},\\{\vec{x}_{\theta}\\}_{\theta=r+1}^{r+s}\right).$

We also have

$\mathscr{W}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r},\\{\vec{x}_{\theta}+\vec{a}\\}_{\theta=r+1}^{r+s}\right)=W^{n}\left(\vec{\xi}_{1},\cdots,\vec{\xi}_{r-1},\vec{\xi}_{r}-\vec{a},\vec{\xi}_{r+1},\cdots,\vec{\xi}_{r+s-1}\right),$

by definition of $\vec{\xi}_{j}$’s.

The proof for Equation (8.8) can be found in [3], hence omitted.

###### Lemma 8.14

Refer to Definition 7.13. Assume the sum
$\sum_{\tau=1}^{r}\chi(\tau)+\sum_{\theta=r+1}^{r+s}\chi(\theta)=0$. Recall in
the proof of Lemma 8.13, we defined the integers $c_{i}$’s in Equation (8.11).
For $\vec{a}\in{{\mathbb{R}}}^{4}$, write $a^{-}=(a^{0},a^{1},0,0)$,
$a^{+}=(0,0,a^{2},a^{3})$, $da^{-}=da^{0}da^{1}$, and $da^{+}=da^{2}da^{3}$.
For any $1\leq t\leq r+s-1$, we have

$\int_{{{\mathbb{R}}}^{2}}e^{-ia^{-}\cdot
q^{-}}W^{n}\left(\vec{\xi}_{1},\cdots,\vec{\xi}_{t}-a^{-},\vec{\xi}_{t+1},\cdots,\vec{\xi}_{r+s-1}\right)\
da^{-}$

vanishes if $q^{-}\neq-c_{t}(\hat{H}(\rho_{n}),\hat{P}(\rho_{n}),0,0)$.

Proof. Let $\vec{\alpha}=(\hat{H}(\rho_{n}),\hat{P}(\rho_{n}),0,0)$, and
$t^{+}\equiv t+1$. Using Equations (8.7) and (8.12), we see that

$W^{n}\left(\vec{\xi}_{1},\cdots,\vec{\xi}_{t}-a^{-},\vec{\xi}_{t+1},\cdots,\vec{\xi}_{r+s-1}\right)=e^{-ic_{t}a^{-}\cdot\vec{\alpha}}W^{n}\left(\vec{\xi}_{1},\cdots,\vec{\xi}_{t},\vec{\xi}_{t+1},\cdots,\vec{\xi}_{r+s-1}\right).$
(8.13)

If we take the Fourier Transform,

$\displaystyle\int_{{{\mathbb{R}}}^{2}}$ $\displaystyle e^{-iq^{-}\cdot
a^{-}}W^{n}\left(\vec{\xi}_{1},\cdots,\vec{\xi}_{t}-a^{-},\vec{\xi}_{t^{+}},\cdots,\vec{\xi}_{r+s-1}\right)\
da^{-}$ $\displaystyle=$
$\displaystyle\int_{{{\mathbb{R}}}^{2}}e^{-ia^{-}\cdot
q^{-}}e^{-ic_{t}a^{-}\cdot\vec{\alpha}}W^{n}\left(\vec{\xi}_{1},\cdots,\vec{\xi}_{t},\vec{\xi}_{t^{+}},\cdots,\vec{\xi}_{r+s-1}\right)\
da^{-}$ $\displaystyle=$ $\displaystyle 2\pi
W^{n}\left(\vec{\xi}_{1},\cdots,\vec{\xi}_{t},\cdots,\vec{\xi}_{r+s-1}\right)\cdot\delta\left(q^{-}+c_{t}(\hat{H}(\rho_{n}),\hat{P}(\rho_{n}),0,0)\right).$

This completes the proof.

###### Remark 8.15

If one examines the proof, we note that the Fourier Transform of

$a^{-}\longmapsto F^{-}(a^{-})\equiv
W^{n}\left(\vec{\xi}_{1},\cdots,\vec{\xi}_{r}-a^{-},\vec{\xi}_{r+1},\cdots,\vec{\xi}_{r+s-1}\right),$

is simply a multiple of the Dirac delta function
$\delta\left(\lambda_{r}^{-}+c_{r}(\hat{H}(\rho_{n}),\hat{P}(\rho_{n}))\right)$.
From Equation (8.8), $\lambda_{r}^{-}=\sum_{i=1}^{r}p_{i}^{-}$,
$p_{i}^{-}\equiv(p_{i}^{0},p_{i}^{1})$ is the momenta variable conjugate to
position variable $x_{i}^{-}\equiv(x_{i}^{0},x_{i}^{1})$.

A similar calculation will show that the Fourier Transform of

$\displaystyle a^{-}\longmapsto F^{+}(a^{-})$
$\displaystyle=\mathscr{W}^{n}\left(\\{\vec{x}_{\theta}+a^{-}\\}_{\theta=1}^{s},\\{\vec{x}_{\tau}\\}_{\tau=s+1}^{s+r}\right)$
$\displaystyle\equiv
W^{n}\left(\vec{\xi}_{1},\cdots,\vec{\xi}_{s}+a^{-},\vec{\xi}_{s+1},\cdots,\vec{\xi}_{r+s-1}\right),$

is simply a multiple of the Dirac delta function
$\delta\left(\lambda_{s}^{-}-c_{s}(\hat{H}(\rho_{n}),\hat{P}(\rho_{n}))\right)$.
In this case, the support is on the set
$\left\\{\lambda_{s}^{-}=c_{s}[\hat{H}(\rho_{n})e_{0}+\hat{P}(\rho_{n})e_{1}]\right\\}$,
$c_{s}\in\mathbb{Z}$.

###### Definition 8.16

Refer to Notation 8.5. We say that the set
$\\{\psi^{\alpha_{\tau},n}\\}_{\tau=1}^{r}$ is regular if there exists a
unique $0\leq k\leq r$ such that for any $h\in\mathscr{P}$,

  * •

$\psi^{\alpha_{\tau},n}(h)=\phi^{\alpha_{\tau},n}(h)^{\ast}$, for every
$1\leq\tau\leq k$, and

  * •

$\psi^{\alpha_{\tau},n}(h)=\phi^{\alpha_{\tau},n}(h)$, for every
$k+1\leq\tau\leq r$.

###### Proposition 8.17

Refer to Notation 8.5, whereby we have the two sets
$\\{\psi^{\alpha_{\tau},n}\\}_{\tau=1}^{r}$ and
$\\{\psi^{\beta_{\theta},n}\\}_{\theta=1}^{s}$, $r\geq s$. Suppose

  * •

both sets are regular as defined in Definition 8.16;

  * •

the former has $0\neq\bar{k}$ annihilation operators, and $\bar{l}:=r-\bar{k}$
creation operators;

  * •

the latter has $\underline{k}$ annihilation operators, and
$0\neq\underline{l}:=s-\underline{k}$ creation operators;

  * •

the sum $\sum_{\tau=1}^{r}\chi(\tau)+\sum_{\theta=r+1}^{r+s}\chi(\theta)=0$,
$\chi$ was defined in Notation 8.6. That is, we have equal number of
annihilation and creation operators, i.e.

$\bar{k}+\underline{k}=\bar{l}+\underline{l}.$ (8.14)

Suppose ${\rm min}\ \\{\underline{k},\bar{l}\\}\neq 0$. If $\bar{k}>\bar{l}$,
then the coefficients $c_{i}$’s in Equation (8.11) can all be chosen to be
positive. In particular, we have that $c_{r}>0$. If $\bar{k}<\bar{l}$, then
the coefficients $c_{i}$’s, $2\bar{k}+1\leq i\leq r$, are all negative. In
particular, we have that $c_{r}<0$.

When ${\rm min}\ \\{\underline{k},\bar{l}\\}=0$, the coefficients $c_{i}$’s in
Equation (8.11) can all be chosen to be positive. In particular, we have that
$c_{r}>0$.

Proof. First consider ${\rm min}\ \\{\underline{k},\bar{l}\\}\neq 0$. We have
the sets $A_{1}:=\\{1,\cdots,\bar{k}\\}$ and
$A_{2}:=\\{r+1,\cdots,r+\underline{k}\\}$, both containing integers which are
adjoint. And we have the sets $B_{1}:=\\{\bar{k}+1,\cdots,\bar{k}+\bar{l}\\}$
and $B_{2}:=\\{r+\underline{k}+1,\cdots,r+\underline{k}+\underline{l}\\}$,
both containing integers which are not adjoint.

We will pair the numbers in $A_{1}\cup A_{2}$, with numbers in $B_{1}\cup
B_{2}$, since the cardinality of both sets are the same. Assume
$\bar{k}\geq\bar{l}$. Define a bijective map $G:A_{1}\cup A_{2}\longrightarrow
B_{1}\cup B_{2}$ by

$\displaystyle G(a)=\left\\{\begin{array}[]{ll}a+\bar{k},&\hbox{$1\leq
a\leq\bar{l}$;}\\\ a+\bar{k}+\underline{k},&\hbox{$\bar{l}+1\leq
a\leq\bar{k}$;}\\\ a+\underline{k}+\bar{k}-\bar{l}\equiv
a+\underline{l},&\hbox{$r+1\leq a\leq r+\underline{k}$.}\end{array}\right.$

Using this map, we see that we can pair in such a way that each $\delta_{i}\in
A_{1}\cup A_{2}$ is paired with a unique $\epsilon_{i}\in B_{1}\cup B_{2}$,
with each $\delta_{i}<\epsilon_{i}$. Equation (8.9) holds for all the
$1\leq\delta_{i}\leq\bar{k}$ and $r+1\leq\delta_{i}\leq r+\underline{k}$. The
map $G$ shows that all the $c_{i}$’s in Equation (8.11), except at $i=r$, are
all positive integers. But if we assume $\bar{k}>\bar{l}$, we will have
$G:\bar{l}+1\mapsto r+\underline{k}+1$, thus we see that $c_{r}>0$.

Assume $\bar{k}<\bar{l}$. Define a bijective map $\tilde{G}:A_{1}\cup
A_{2}\longrightarrow B_{1}\cup B_{2}$ by

$\displaystyle\tilde{G}(a)=\left\\{\begin{array}[]{ll}a+\bar{k},&\hbox{$1\leq
a\leq\bar{k}$;}\\\ a+\bar{k}-\bar{l},&\hbox{$r+1\leq a\leq
r+\bar{l}-\bar{k}$;}\\\ a+\bar{k}-\bar{l}+\underline{k}\equiv
a+\underline{l},&\hbox{$r+\bar{l}-\bar{k}+1\leq a\leq
r+\underline{k}$.}\end{array}\right.$

In this case, we see that for $r+1\leq\delta_{i}\leq r+\bar{l}-\underline{k}$,
we have that $\delta_{i}>\tilde{G}(\delta_{i})=\epsilon_{i}$. Equation (8.10)
holds instead, for such $\delta_{i}$ and corresponding $\epsilon_{i}$. Thus,
the coefficients $c_{i}$’s, $2\bar{k}+1\leq i\leq r$ in Equation (8.11) are
negative. In particular, we have that $c_{r}<0$.

Now we consider ${\rm min}\ \\{\underline{k},\bar{l}\\}=0$. When $\bar{l}=0$,
this implies that $r=s$ and thus $r=\bar{k}=\underline{l}=s$. Define a
bijective map $\hat{G}:\\{1,\cdots,r\\}\longrightarrow\\{r+1,\cdots,2r\\}$ as
$G(a)=a+r$. Then clearly, $a<\hat{G}(a)$ and since $\hat{G}:r\mapsto r+s=2r$,
all the coefficients in Equation (8.11) are strictly positive. In particular,
$c_{r}>0$.

When $\underline{k}=0$, then we must have $\bar{k}=\bar{l}+\underline{l}$.
Define a bijective map $\dot{G}$ that sends $a$ to $a+\bar{k}$, for $1\leq
a\leq\bar{k}$. Again, $a<\dot{G}(a)$ and because $\dot{G}:\bar{k}\mapsto r+s$,
all the coefficients in Equation (8.11) are strictly positive. In particular,
$c_{r}>0$.

###### Remark 8.18

When $\bar{k}=\bar{l}$, we see that $c_{r}=0$ in the proof, which implies that
the support in Lemma 8.14 is at $\\{\vec{q}=\vec{0}\\}$. This will pose a
problem later, when we try to prove the Clustering Theorem 8.34. Hence, we do
not consider this particular case.

When $\bar{k}=0$ or $\underline{l}=0$, then $\mathscr{W}^{n}\equiv 0$, which
is trivial.

###### Remark 8.19

Assume the conditions in Proposition 8.17 hold. We will further assume one of
the following cases hold:

  1. 1.

${\rm min}\ \\{\underline{k},\bar{l}\\}=0$; or

  2. 2.

${\rm min}\ \\{\underline{k},\bar{l}\\}\neq 0$ and $\bar{k}>\bar{l}$; or

  3. 3.

${\rm min}\ \\{\underline{k},\bar{l}\\}\neq 0$, $\bar{k}<\bar{l}$ and
$2\bar{k}+1\leq s$.

When we assume one of the above 3 cases holds, Proposition 8.17 says that
${\rm sgn}(c_{s})={\rm sgn}(c_{r})$. Then the supports described in Remark
8.15 are on respective positive and negative cones in energy-momentum space.
The fact that the supports are separated is key to prove the Clustering
Decomposition Property, as demonstrated in [3] and [22].

In general, it is not true that

$\mathscr{W}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r},\\{\vec{x}_{\theta}\\}_{\theta=r+1}^{r+s}\right)=\pm\mathscr{W}^{n}\left(\\{\vec{x}_{\theta}\\}_{\theta=r+1}^{r+s},\\{\vec{x}_{\tau}\\}_{\tau=1}^{r}\right).$

The commutation and anti-commutation relations in Lemmas 5.15, 5.18 and 5.20
hold only for real or purely imaginary functions respectively. Thus, we need
to consider in a similar fashion.

###### Remark 8.20

Proposition 8.29 will imply that the above equality holds, provided
$x_{\theta}^{+}-x_{\tau}^{+}\neq 0$ for any $1\leq\tau\leq r$ and
$r+1\leq\theta\leq r+s$.

###### Notation 8.21

For a natural number $r$, let the set $X_{r}:=\\{1,2,\cdots,r\\}$. Decompose
$X_{r}$ into 2 sets $\kappa_{r}$, $\bar{\kappa}_{r}$, i.e.
$\kappa_{r}\cup\bar{\kappa}_{r}=X_{r}$ and
$\kappa_{r}\cap\bar{\kappa}_{r}=\emptyset$, one of it possibly empty. Let
$\Omega_{r}$ denote the set containing all such decompositions of $X_{r}$.

Recall we have two sets $\\{f_{1},\cdots,f_{r}\\}$ and
$\\{g_{1},\cdots,g_{s}\\}$. Assume that each $f_{i}$ ($g_{j}$) is either real
or purely imaginary. This defines 2 disjoint sets $\kappa_{r}$ ($\pi_{s}$),
and $\bar{\kappa}_{r}$ ($\bar{\pi}_{s}$), whereby $f_{i}$ ($g_{j}$) is real if
$i\in\kappa_{r}$ ($j\in\pi_{s}$), otherwise $f_{i}$ ($g_{j}$) is purely
imaginary if $i\in\bar{\kappa}_{r}$ ($j\in\bar{\pi}_{s}$).

Then, we will write for $(\kappa_{r},\bar{\kappa}_{r})\in\Omega_{r}$,
$(\pi_{s},\bar{\pi}_{s})\in\Omega_{s}$, tempered distributions such that

$\displaystyle\mathscr{W}_{\kappa_{r},\pi_{s}}^{n}$
$\displaystyle:f_{1}\otimes_{{\mathbb{R}}}\cdots\otimes_{{\mathbb{R}}}f_{r}\otimes_{{\mathbb{R}}}g_{1}\otimes_{{\mathbb{R}}}\cdots\otimes_{{\mathbb{R}}}g_{s}\longmapsto\left\langle
A_{r}^{n}P_{0}B_{s}^{n}1,1\right\rangle,$
$\displaystyle\mathscr{W}_{\pi_{s},\kappa_{r}}^{n}$
$\displaystyle:g_{1}\otimes_{{\mathbb{R}}}\cdots\otimes_{{\mathbb{R}}}g_{s}\otimes_{{\mathbb{R}}}f_{1}\otimes_{{\mathbb{R}}}\cdots\otimes_{{\mathbb{R}}}f_{r}\longmapsto\left\langle
C_{r}^{n}P_{0}D_{s}^{n}1,1\right\rangle,$

for these particular sets $\\{f_{1},\cdots,f_{r}\\}$ and
$\\{g_{1},\cdots,g_{s}\\}$. Refer back to Equations (8.3) and (8.4).

###### Remark 8.22

Recall from Notation 8.5, we have the sets $\\{f_{1},\cdots,f_{r}\\}$ and
$\\{g_{1},\cdots,g_{s}\\}$. Write for $\tau=1,\cdots,r$,
$f_{\tau}=\underline{f}_{\tau}+i\overline{f}_{\tau}$, whereby
$\underline{f}_{\tau}={\rm Re}\ f_{\tau}$ and $\overline{f}_{\tau}={\rm Im}\
f_{\tau}$. Similarly, write for $\theta=1,\cdots,s$,
$g_{\theta}=\underline{g}_{\theta}+i\overline{g}_{\theta}$, whereby
$\underline{g}_{\theta}={\rm Re}\ g_{\theta}$ and $\overline{g}_{\theta}={\rm
Im}\ g_{\theta}$.

Refer to Notation 8.21. We can now understand

$\displaystyle\int_{{{\mathbb{R}}}^{4}\times\cdots\times{{\mathbb{R}}}^{4}}\mathscr{W}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r},\\{\vec{x}_{\theta}\\}_{\theta=r+1}^{r+s}\right)\bigotimes_{\tau=1}^{r+s}p_{\tau}(\vec{x}_{\tau})\cdot\prod_{\tau=1}^{r+s}d\vec{x}_{\tau}$
$\displaystyle=\sum_{\genfrac{}{}{0.0pt}{}{(\kappa_{r},\bar{\kappa}_{r})\in\Omega_{r}}{(\pi_{s},\bar{\pi}_{s})\in\Omega_{s}}}c_{\kappa_{r},\pi_{s}}\int_{{{\mathbb{R}}}^{4}\times\cdots\times{{\mathbb{R}}}^{4}}\mathscr{W}_{\kappa_{r},\pi_{s}}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r},\\{\vec{x}_{\theta}\\}_{\theta=r+1}^{r+s}\right)\prod_{\tau=1}^{r+s}q^{\\{\kappa_{r},\pi_{s}\\}}_{\tau}(\vec{x}_{\tau})\cdot\prod_{\tau=1}^{r+s}d\vec{x}_{\tau},$
(8.15)

whereby for each integral term in the above sum, indexed by
$\\{\kappa_{r},\pi_{s}\\}$,

  * •

$c_{\kappa_{r},\pi_{s}}$ takes values in $\\{\pm 1,\pm\sqrt{-1}\\}$;

  * •

for $1\leq\tau\leq r$, $q^{\\{\kappa_{r},\pi_{s}\\}}_{\tau}$ is
$\underline{f}_{\tau}$ if $\tau\in\kappa_{r}$, otherwise
$q^{\\{\kappa_{r},\pi_{s}\\}}_{\tau}$ is $\overline{f}_{\tau}$;

  * •

for $r+1\leq\tau\leq r+s$, $q^{\\{\kappa_{r},\pi_{s}\\}}_{\tau}$ is
$\underline{g}_{\tau-r}$ if $\tau-r\in\pi_{s}$, otherwise
$q^{\\{\kappa_{r},\pi_{s}\\}}_{\tau}$ is $\overline{g}_{\tau-r}$.

To apply the commutation or anti-commutation relations in Lemmas 5.6, 5.15,
5.18 and 5.20 to switch the variables, we need to assume regularity of the
sets in Definition 8.16. Making this assumption and using the above notations,
we will have the following lemma. The proof involves applying successively,
the commutation and anti-commutation relations in the lemmas, and the details
will be left to the reader.

###### Lemma 8.23

Assume the two sets $\\{\psi^{\alpha_{\tau},n}\\}_{\tau=1}^{r}$ and
$\\{\psi^{\beta_{\theta},n}\\}_{\theta=1}^{s}$, $r\geq s$, both are regular.
We have that

$\mathscr{W}_{\kappa_{r},\pi_{s}}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r},\\{\vec{x}_{\theta}\\}_{\theta=r+1}^{r+s}\right)=\pm\mathscr{W}_{\pi_{s},\kappa_{r}}^{n}\left(\\{\vec{x}_{\theta}\\}_{\theta=r+1}^{r+s},\\{\vec{x}_{\tau}\\}_{\tau=1}^{r}\right),$
(8.16)

provided for any $1\leq\tau\leq r$, $r+1\leq\theta\leq r+s$, we have
$0\neq\vec{x}_{\tau}-\vec{x}_{\theta}$ lies in the hyperplane spanned by
$\\{\hat{P}(\rho_{n})e_{0}+\hat{H}(\rho_{n})e_{1},e_{2},e_{3}\\}$. The $\pm$
sign depends on how many anti-commutation relations in Lemmas 5.15, 5.18 and
5.20 are applied.

###  8.2 Proof of Cluster Decomposition Property

###### Definition 8.24

Let $\varphi_{1}$, $\varphi_{2}$ be Schwartz functions on
${{\mathbb{R}}}^{4r}$ and ${{\mathbb{R}}}^{4s}$ respectively, $r\geq s$.
Define

$\displaystyle T_{1}^{n}$
$\displaystyle:=\int_{{{\mathbb{R}}}^{4}\times\cdots\times{{\mathbb{R}}}^{4}}\prod_{\tau=1}^{r}d\vec{x}_{\tau}\cdot\varphi_{1}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r}\right)\cdot\prod_{\tau=1}^{r}\psi^{\alpha_{\tau},n}(\vec{x}_{\tau}),$
$\displaystyle T_{2}^{n}$
$\displaystyle:=\int_{{{\mathbb{R}}}^{4}\times\cdots\times{{\mathbb{R}}}^{4}}\prod_{\theta=r+1}^{r+s}d\vec{x}_{\theta}\cdot\varphi_{2}\left(\\{\vec{x}_{\theta}\\}_{\theta=r+1}^{r+s}\right)\cdot\prod_{\theta=r+1}^{r+s}\psi^{\beta_{\theta-r},n}(\vec{x}_{\theta}).$

And define

$\displaystyle\mathcal{T}_{1}^{n}$
$\displaystyle\left(\\{\vec{x}_{\tau}\\}_{\tau=s+1}^{r}\right)$
$\displaystyle:=\int_{{{\mathbb{R}}}^{4}\times\cdots\times{{\mathbb{R}}}^{4}}\prod_{\theta=r+1}^{r+s}d\vec{x}_{\theta}\cdot\varphi_{1}\left(\\{\vec{x}_{\tau}\\}_{\tau=s+1}^{r},\\{\vec{x}_{\theta}\\}_{\theta=r+1}^{r+s}\right)\cdot\prod_{\theta=r+1}^{r+s}\psi^{\beta_{\theta-r},n}(\vec{x}_{\theta}),$
$\displaystyle\mathcal{T}_{2}^{n}$
$\displaystyle\left(\\{\vec{x}_{\tau}\\}_{\tau=s+1}^{r}\right)$
$\displaystyle:=\int_{{{\mathbb{R}}}^{4}\times\cdots\times{{\mathbb{R}}}^{4}}\prod_{\tau=1}^{s}d\vec{x}_{\tau}\cdot\varphi_{2}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{s}\right)\cdot\prod_{\tau=1}^{r}\psi^{\alpha_{\tau},n}(\vec{x}_{\tau}),$

whereby $\psi^{\alpha_{\tau},n}(\vec{x}_{\tau})$ is either
$\phi^{\alpha_{\tau},n}(\vec{x}_{\tau})$ or its adjoint operator. Likewise for
$\psi^{\beta_{\theta-r},n}(\vec{x}_{\theta})$.

We will need to assume the following:

  * •

both the sets $\\{\psi^{\alpha_{\tau},n}\\}_{\tau=1}^{r}$ and
$\\{\psi^{\beta_{\theta},n}\\}_{\theta=1}^{s}$ are regular, as defined in
Definition 8.16;

  * •

$\sum_{\tau=1}^{r}\chi(\tau)+\sum_{\theta=r+1}^{r+s}\chi(\theta)=0$, i.e.
there are an equal number of creation and annihilation operators.

Suppose $\vec{a}\in{{\mathbb{R}}}^{4}$. Define

$\displaystyle T_{1}^{n}\left(\vec{a}\right)$
$\displaystyle:=U(\vec{a},1)T_{1}^{n}U(\vec{a},1)^{-1},$ $\displaystyle
T_{2}^{n}\left(\vec{a}\right)$
$\displaystyle:=U(\vec{a},1)T_{2}^{n}U(\vec{a},1)^{-1}.$

By default, we will write $T_{i}^{n}\equiv T_{i}^{n}(\vec{0})$. Define
similarly

$\mathcal{T}_{1}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=s+1}^{r},\vec{a}\right),\quad\mathcal{T}_{2}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=s+1}^{r},\vec{a}\right),$

replacing $T_{i}^{n}$ with $\mathcal{T}_{i}^{n}$ respectively in the above
formulas.

Refer to Definition 8.24. Define

$\displaystyle T_{1}:=\sum_{n=1}^{\infty}c_{1,n}T_{1}^{n},\quad
T_{2}:=\sum_{n=1}^{\infty}c_{2,n}T_{2}^{n},$
$\displaystyle\mathcal{T}_{1}\left(\\{\vec{x}_{\tau}\\}_{\tau=s+1}^{r}\right):=\sum_{n=1}^{\infty}c_{1,n}\mathcal{T}_{1}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=s+1}^{r}\right),$
$\displaystyle\mathcal{T}_{2}\left(\\{\vec{x}_{\tau}\\}_{\tau=s+1}^{r}\right):=\sum_{n=1}^{\infty}c_{2,n}\mathcal{T}_{2}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=s+1}^{r}\right),$

each $\\{c_{i,n}\\}_{n=1}^{\infty}$ is a sequence of numbers in $l^{2}$ space,
such that $\sum_{n=1}^{\infty}|c_{i,n}|^{2}C(\rho_{n})^{2k}<\infty$, for any
$k\in\mathbb{N}$.

###### Remark 8.25

Note that for $j=1,2$, $|T_{j}^{n}1|^{2}=O(C(\rho_{n}))$,
$|\mathcal{T}_{j}^{n}1|^{2}=O(C(\rho_{n}))$. Because for any $k\in\mathbb{N}$,
$\sum_{n=1}^{\infty}|c_{j,n}|^{2}C(\rho_{n})^{2k}<\infty$, we see that

$|\hat{H}^{k}T_{j}(\cdot,\vec{a})1|^{2}\leq\sum_{n=1}^{\infty}C(\rho_{n})^{2k}|c_{j,n}|^{2}|T_{j}^{n}(\cdot)1|^{2}<\infty.$
(8.17)

Similar bound for $\mathcal{T}_{j}$. This is required to prove the Clustering
Decomposition Property 8.32 and Clustering Theorem 8.34.

Indeed,

$\displaystyle\Big{\langle}$ $\displaystyle
T_{1}\left(\vec{y}\right)T_{2}\left(\vec{x}\right)1,1\Big{\rangle}$
$\displaystyle=\sum_{n=1}^{\infty}c_{1,n}c_{2,n}\left\langle
U(\vec{y},1)T_{1}^{n}U(\vec{y},1)^{-1}U(\vec{x},1)T_{2}^{n}U(\vec{x},1)^{-1}1,1\right\rangle$
$\displaystyle=\sum_{n=1}^{\infty}c_{1,n}c_{2,n}\left\langle
T_{1}^{n}U(\vec{x}-\vec{y},1)T_{2}^{n}1,1\right\rangle,$

which converges by Equation (8.17).

Similarly,

$\displaystyle\Big{\langle}$
$\displaystyle\mathcal{T}_{2}\left(\\{\vec{x}_{\tau}\\}_{\tau=s+1}^{r},\vec{x}\right)\mathcal{T}_{1}\left(\\{\vec{x}_{\tau}\\}_{\tau=s+1}^{r},\vec{y}\right)1,1\Big{\rangle}$
$\displaystyle=\sum_{n=1}^{\infty}c_{1,n}c_{2,n}\left\langle\mathcal{T}_{2}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=s+1}^{r},\vec{x}\right)\mathcal{T}_{1}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=s+1}^{r},\vec{y}\right)1,1\right\rangle,$

which also converges.

###### Definition 8.26

For $\vec{a}=\vec{x}-\vec{y}\in S_{0}$, we will define

$\displaystyle h_{12}^{n}\left(\vec{a}\right):=$ $\displaystyle\left\langle
T_{1}^{n}\left(\vec{y}\right)P_{0}T_{2}^{n}\left(\vec{x}\right)1,1\right\rangle$
$\displaystyle=$ $\displaystyle\left\langle
T_{1}^{n}U(\vec{y},1)^{-1}U(\vec{x},1)T_{2}^{n}1,1\right\rangle-\langle
T_{1}^{n}1,1\rangle\langle T_{2}^{n}1,1\rangle$ $\displaystyle=$
$\displaystyle\left\langle
T_{1}^{n}U(\vec{a},1)T_{2}^{n}1,1\right\rangle-\langle
T_{1}^{n}1,1\rangle\langle T_{2}^{n}1,1\rangle.$

Similarly, define ($s^{+}=s+1$)

$\displaystyle h_{21}^{n}$ $\displaystyle\left(\vec{a}\right)$
$\displaystyle:=$
$\displaystyle\int_{{{\mathbb{R}}}^{4}\times\cdots\times{{\mathbb{R}}}^{4}}\prod_{\tau=s^{+}}^{r}d\vec{x}_{\tau}\
\left\langle\mathcal{T}_{2}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=s^{+}}^{r},\vec{x}\right)P_{0}\mathcal{T}_{1}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=s^{+}}^{r},\vec{y}\right)1,1\right\rangle$
$\displaystyle=$
$\displaystyle\left\langle\int_{{{\mathbb{R}}}^{4(r-s)}}\prod_{\tau=s^{+}}^{r}d\vec{x}_{\tau}\
\mathcal{T}_{2}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=s^{+}}^{r}\right)U(-\vec{a},1)\mathcal{T}_{1}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=s^{+}}^{r}\right)1,1\right\rangle$
$\displaystyle-\int_{{{\mathbb{R}}}^{4(r-s)}}\prod_{\tau=s^{+}}^{r}d\vec{x}_{\tau}\left\langle\mathcal{T}_{1}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=s^{+}}^{r}\right)1,1\right\rangle\left\langle\mathcal{T}_{2}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=s^{+}}^{r}\right)1,1\right\rangle.$

Using Definition 8.11, we will extend the definitions of
$h_{12}^{n}\left(\vec{a}\right)$ and $h_{21}^{n}\left(\vec{a}\right)$, for
$\vec{a}\in S_{0}$, to all of $\vec{a}\in{{\mathbb{R}}}^{4}$.

From Equation (8.7), we will have ($r^{+}=r+1$)

$\displaystyle h_{12}^{n}$ $\displaystyle\left(\vec{a}\right)$
$\displaystyle:=$
$\displaystyle\int_{{{\mathbb{R}}}^{4}\times\cdots\times{{\mathbb{R}}}^{4}}\prod_{\tau=1}^{r+s}d\vec{x}_{\tau}\
\mathscr{W}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r},\\{\vec{x}_{\theta}+\vec{a}\\}_{\theta=r^{+}}^{r+s}\right)\varphi_{1}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r}\right)\varphi_{2}\left(\\{\vec{x}_{\theta}\\}_{\theta=r^{+}}^{r+s}\right).$
(8.18)

Similarly, we will define ($s^{+}=s+1$)

$\displaystyle h_{21}^{n}$ $\displaystyle\left(\vec{a}\right)$
$\displaystyle:=$
$\displaystyle\int_{{{\mathbb{R}}}^{4}\times\cdots\times{{\mathbb{R}}}^{4}}\prod_{\tau=1}^{r+s}d\vec{x}_{\tau}\
\mathscr{W}^{n}\left(\\{\vec{x}_{\theta}+\vec{a}\\}_{\theta=r^{+}}^{r+s},\\{\vec{x}_{\tau}\\}_{\tau=1}^{r}\right)\varphi_{1}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r}\right)\varphi_{2}\left(\\{\vec{x}_{\theta}\\}_{\theta=r^{+}}^{r+s}\right).$
(8.19)

###### Definition 8.27

Define for any $\vec{a}\in{{\mathbb{R}}}^{4}$,

$h_{12}\left(\vec{a}\right):=\sum_{n=1}^{\infty}c_{1,n}c_{2,n}h_{12}^{n}\left(\vec{a}\right),\quad
h_{21}\left(\vec{a}\right):=\sum_{n=1}^{\infty}c_{1,n}c_{2,n}h_{21}^{n}\left(\vec{a}\right).$

###### Definition 8.28

Recall from Definition 7.13, we defined the set
$\\{\hat{H}(\rho_{n}),\hat{P}(\rho_{n})\\}_{n\in\mathbb{N}}$, such that
$\hat{H}(\rho_{n})^{2}-\hat{P}(\rho_{n})^{2}=m_{n}^{2}$, with $m_{n}>0$ as the
mass gap in $\mathscr{H}(\rho_{n})$.

Let $\Lambda_{n}$ be a Lorentz transformation that maps the standard
orthonormal basis $e_{a}\mapsto\tilde{f}_{a}^{n}:=\Lambda_{n}e_{a}$, whereby
$\tilde{f}_{0}^{n}$ and $\tilde{f}_{1}^{n}$ are

$\frac{\hat{H}(\rho_{n})}{m_{n}}e_{0}+\frac{\hat{P}(\rho_{n})}{m_{n}}e_{1},\quad\frac{\hat{P}(\rho_{n})}{m_{n}}e_{0}+\frac{\hat{H}(\rho_{n})}{m_{n}}e_{1}$

respectively, and $\tilde{f}_{2}^{n}=e_{2}$, $\tilde{f}_{3}^{n}=e_{3}$, which
together spans the plane $S_{0}$.

###### Proposition 8.29

Write $\vec{a}=(a^{-},a^{+})$, whereby $a^{-}=(a^{0},a^{1})$,
$a^{+}=(a^{2},a^{3})$. If $x_{\tau}^{+}-x_{\theta}^{+}\neq a^{+}$ for any
$1\leq\tau\leq r$ and $r+1\leq\theta\leq r+s$, then we have that

$\mathscr{W}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r},\\{\vec{x}_{\theta}+\vec{a}\\}_{\theta=r+1}^{r+s}\right)=0.$

.

Proof. It is not difficult to see from Equation (8.13) that

$\displaystyle\mathscr{W}^{n}$
$\displaystyle\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r},\\{\vec{x}_{\theta}+\vec{a}\\}_{\theta=r+1}^{r+s}\right)$
$\displaystyle=$ $\displaystyle e^{-i\vec{a}\cdot
c_{r}m_{n}\tilde{f}_{0}^{n}}\prod_{\theta=1}^{r+s}E(x_{\theta}^{-})\cdot\mathscr{W}_{0}^{n}\left(\\{x^{+}_{\tau}\\}_{\tau=1}^{r},\\{x_{\theta}^{+}+a^{+}\\}_{\theta=r+1}^{r+s}\right),$
(8.20)

whereby
$E(x_{\theta}^{-})=E(x_{\theta}^{0},x_{\theta}^{1})=\dfrac{e^{i\chi(\theta)[x_{\theta}^{0}\hat{H}-x_{\theta}^{1}\hat{P}]}}{2\pi}$
and ($0^{-}\equiv(0,0)$)

$\mathscr{W}_{0}^{n}\left(\\{x^{+}_{\tau}\\}_{\tau=1}^{r},\\{x_{\theta}^{+}+a^{+}\\}_{\theta=r+1}^{r+s}\right)=\mathscr{W}^{n}\left(\\{0^{-},x^{+}_{\tau}\\}_{\tau=1}^{r},\\{0^{-},x_{\theta}^{+}+a^{+}\\}_{\theta=r+1}^{r+s}\right).$
(8.21)

Because we subtract off $\left\langle A_{r}^{n}1,1\right\rangle\left\langle
B_{s}^{n}1,1\right\rangle$ in Equation (8.3), we see that $c_{Q}$ defined in
Equation (8.2) is non-zero, iff there exists a set $A_{l}\in Q$ which contains
both $r$ and $r+1$. From Equation (8.7), we have that
$\mathscr{W}_{0}^{n}\equiv 0$, if $x_{\tau}^{+}-x_{\theta}^{+}\neq a^{+}$ for
any $1\leq\tau\leq r$ and $r+1\leq\theta\leq r+s$, and this completes the
proof.

###### Remark 8.30

From Item 2 in Lemma 8.13, we see that $\mathscr{W}^{n}=0$ iff for any
partition $Q=\\{A_{1},\ldots,A_{n(Q)}\\}\in\Gamma$, there exists some
$\theta\neq r$ such that $\\{\theta,\theta+1\\}\subset A_{l}$ for some $1\leq
l\leq n(Q)$ and $\vec{\xi}_{\theta}\neq 0$, or $\vec{\xi}_{r}\neq\vec{a}$.

###### Corollary 8.31

Refer to Definition 8.28. Recall the sets
$\kappa_{r}\subseteq\\{1,\cdots,r\\}$, $\pi_{s}\subseteq\\{1,\cdots,s\\}$, as
explained in Notation 8.21.

Suppose

$\mathscr{W}_{\kappa_{r},\pi_{s}}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r},\\{\vec{x}_{\theta}+\vec{a}\\}_{\theta=r+1}^{r+s}\right)=\pm\mathscr{W}_{\pi_{s},\kappa_{r}}^{n}\left(\\{\vec{x}_{\theta}+\vec{a}\\}_{\theta=r+1}^{r+s},\\{\vec{x}_{\tau}\\}_{\tau=1}^{r}\right)$

holds for any space-like vector
$\vec{a}=a^{0}\tilde{f}_{0}^{n}+a^{2}\tilde{f}_{2}^{n}+a^{3}\tilde{f}_{3}^{n}$.
Then it is necessary and sufficient that
$\mathscr{W}_{\kappa_{r},\pi_{s}}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r},\\{\vec{x}_{\theta}+\vec{a}\\}_{\theta=r+1}^{r+s}\right)$
and
$\mathscr{W}_{\pi_{s},\kappa_{r}}^{n}\left(\\{\vec{x}_{\theta}+\vec{a}\\}_{\theta=r+1}^{r+s},\\{\vec{x}_{\tau}\\}_{\tau=1}^{r}\right)$
both must vanish.

Recall from Definition 3.3, we defined the norm
$\parallel\cdot\parallel_{p,q}$ for Schwartz functions on
${{\mathbb{R}}}^{4}$. We can generalize this definition, for a Schwartz
function defined on the Euclidean space ${{\mathbb{R}}}^{r}$, for any
$r\in\mathbb{N}$.

By considering $h_{12}-h_{21}$, we can mimic the proof of Theorem 3-4 (Cluster
Decomposition Property) in [3], or the lemma in [19], to prove that
$\left|h_{12}\left(\vec{a}\right)\right|$ decays at the rate of
$\frac{1}{|a^{+}|^{p}}$ for $|a^{+}|$ large and for any $p\in\mathbb{N}$.
However, using Lemma 8.23 is inadequate to prove this result. We instead need
local commutativity as given by Corollary 8.31 and assuming one of the 3 cases
hold in Remark 8.19, to prove the result. We will however, prove the following
slightly stronger result, using Proposition 8.29 and without assuming any of
the cases in Remark 8.19 hold.

###### Theorem 8.32

(Cluster Decomposition Property)
Recall $m_{0}$ is the mass gap of the 4-dimensional quantum Yang-Mills theory.
Suppose the assumptions in Definition 8.24 hold, i.e. regularity of the two
sets and equal number of creation and annihilation operators. Let
$\vec{a}=\sum_{b=0}^{3}a^{b}e_{b}$ be a space-like vector, with
$a^{-}\equiv(a^{0},a^{1})$ and $a^{+}\equiv(a^{2},a^{3})$. Then, we have for
some positive constant $C$,

$\displaystyle\left|h_{12}\left(\vec{a}\right)\right|$
$\displaystyle=\left|\sum_{n=1}^{\infty}c_{1,n}c_{2,n}h_{12}^{n}\left(\vec{a}\right)\
\right|$
$\displaystyle\leq\frac{C}{m_{0}^{l}+|a^{+}|^{k}}\left(\parallel\varphi_{1}\parallel_{p,q}\parallel\varphi_{2}\parallel_{p,q}\right),$
(8.22)

for any $k,l\in\mathbb{N}$, provided $p,q\in\mathbb{N}$ are large enough.

Proof. Recall the sets $\kappa_{r}\subseteq\\{1,\cdots,r\\}$,
$\pi_{s}\subseteq\\{1,\cdots,s\\}$, as explained in Notation 8.21. Consider
the real vector spaces $\mathcal{Y}_{\kappa_{r}}$ and
$\widetilde{\mathcal{Y}}_{\pi_{s}}$, each spanned by tensor products of
functions of the form

$f_{1}\otimes_{{\mathbb{R}}}f_{2}\otimes_{{\mathbb{R}}}\cdots\otimes_{{\mathbb{R}}}f_{r}\quad{\rm
and}\quad g_{1}\otimes_{{\mathbb{R}}}\cdots\otimes_{{\mathbb{R}}}g_{s}$

respectively, whereby each $f_{i}$ ($g_{j}$) is real if $i\in\kappa_{r}$
($j\in\pi_{s}$), purely imaginary if $i\notin\kappa_{r}$ ($j\notin\pi_{s}$).

Instead of proving the result for general $\varphi_{i}$’s, from Equation
(8.15), it suffices to prove it for $\varphi_{1}\in\mathcal{Y}_{\kappa_{r}}$
and $\varphi_{2}\in\widetilde{\mathcal{Y}}_{\pi_{s}}$, whereby we need to
change the definition of $h_{12}^{n}$, by replacing $\mathscr{W}^{n}$ with
$\mathscr{W}_{\kappa_{r},\pi_{s}}^{n}$ in Equation (8.18).

Without any loss of generality, we assume that $C(\rho_{n})>1$, otherwise we
can replace $C(\rho_{n})$ in the following inequalities with $(1+C(\rho_{n}))$
instead. Write

$\varphi\left(\\{\vec{x}_{\theta}\\}_{\theta=1}^{r+s}\right)=\varphi_{1}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r}\right)\otimes_{{\mathbb{R}}}\varphi_{2}\left(\\{\vec{x}_{\theta}\\}_{\theta=r^{+}}^{r+s}\right).$

Let $q^{-}=(q^{0},q^{1})$ be coordinates pertaining to the $\\{e_{0},e_{1}\\}$
basis, with $x_{\tau}^{-}\cdot
q_{\tau}^{-}=-x_{\tau}^{0}q_{\tau}^{0}+x_{\tau}^{1}q_{\tau}^{1}$, and define
the following partial Fourier Transforms,

$\displaystyle\begin{split}\tilde{\varphi}_{1}\left(\\{q_{\tau}^{-},x_{\tau}^{+}\\}_{\tau=1}^{r}\right)&:=\int_{{{\mathbb{R}}}^{2r}}\prod_{\tau=1}^{r}\frac{e^{-i\chi(\tau)x_{\tau}^{-}\cdot
q_{\tau}^{-}}}{2\pi}\cdot\varphi_{1}\left(\\{x_{\tau}^{-},x_{\tau}^{+}\\}_{\tau=1}^{r}\right)\
\prod_{\tau=1}^{r+s}dx_{\tau}^{-},\\\
\tilde{\varphi}_{2}\left(\\{q_{\theta}^{-},x_{\theta}^{+}\\}_{\theta=r+1}^{r+s}\right)&:=\int_{{{\mathbb{R}}}^{2s}}\prod_{\theta=r+1}^{r+s}\frac{e^{-i\chi(\theta)x_{\theta}^{-}\cdot
q_{\theta}^{-}}}{2\pi}\cdot\varphi_{2}\left(\\{x_{\theta}^{-},x_{\theta}^{+}\\}_{\theta=r+1}^{r+s}\right)\
\prod_{\theta=r+1}^{r+s}dx_{\theta}^{-},\end{split}$ (8.23)

and

$\tilde{\varphi}\left(\\{q_{\tau}^{-},x_{\tau}^{+}\\}_{\tau=1}^{r+s}\right):=\tilde{\varphi}_{1}\left(\\{q_{\tau}^{-},x_{\tau}^{+}\\}_{\tau=1}^{r}\right)\tilde{\varphi}_{2}\left(\\{q_{\theta}^{-},x_{\theta}^{+}\\}_{\theta=r+1}^{r+s}\right).$

Suppose we let $H_{n}^{-}=(\hat{H}(\rho_{n}),\hat{P}(\rho_{n}))$ and recall we
defined $\mathscr{W}_{0}^{n}$ in Equation (8.21). Then, we see that

$\displaystyle h_{12}^{n}(\vec{a})$
$\displaystyle=e^{-ic_{r}m_{n}\vec{a}\cdot\tilde{f}_{0}^{n}}\int_{{{\mathbb{R}}}^{2(r+s)}}\prod_{\tau=1}^{r+s}dx_{\tau}^{+}\
\mathscr{W}_{0}^{n}\left(\\{x_{\tau}^{+}\\}_{\tau=1}^{r},\\{x_{\theta}^{+}+a^{+}\\}_{\theta=r^{+}}^{r+s}\right)\tilde{\varphi}\left(\\{H_{n}^{-},x_{\tau}^{+}\\}_{\tau=1}^{r+s}\right),$

which shows that it is a constant in the $\tilde{f}_{1}^{n}$ direction.

Since the Fourier transform of a Schwartz function remains a Schwartz
function, we see that for some $\bar{n}\in\mathbb{N}$, independent of $n$,

$\left|\tilde{\varphi}\left(\\{q_{\theta}^{-},x_{\theta}^{+}\\}_{\theta=1}^{r+s}\right)\right|\leq
C(\rho_{n})^{\bar{n}}\frac{\parallel\varphi\parallel_{\hat{k},\hat{l}}}{\sum_{\theta=1}^{r+s}(|q_{\theta}^{0}|^{2}+|q_{\theta}^{1}|^{2})^{l/2}+\left(|x_{\theta}^{2}|^{2}+|x_{\theta}^{3}|^{2}\right)^{k/2}},$

for some $\hat{k}\geq k$, $\hat{l}\geq l$ large enough. In particular,
$(\hat{H}(\rho_{n})^{2}+\hat{P}(\rho_{n})^{2})^{l/2}\geq m_{n}^{l}$. Thus, we
have that

$\left|\tilde{\varphi}\left(\\{H_{n}^{-},x_{\theta}^{+}\\}_{\theta=1}^{r+s}\right)\right|\leq
C(\rho_{n})^{\bar{n}}\frac{\parallel\varphi\parallel_{\hat{k},\hat{l}}}{\sum_{\theta=1}^{r+s}m_{n}^{l}+\left(x_{\theta}^{2,2}+x_{\theta}^{3,2}\right)^{k/2}}.$

There exists a set containing polynomially bounded continuous functions
$\\{G_{\vec{m}}^{n}:\ 0\leq|\vec{m}|\leq\mathcal{N}\\}$, such that

$\displaystyle
e^{-ic_{r}m_{n}\vec{a}\cdot\tilde{f}_{0}^{n}}\sum_{|\vec{m}|\leq\mathcal{N}}\int_{{{\mathbb{R}}}^{2(r+s)}}$
$\displaystyle\prod_{\tau=1}^{r+s}dx_{\tau}^{+}D^{\vec{m}}G_{\vec{m}}^{n}\left(\\{x_{\tau}^{+}\\}_{\tau=1}^{r+s};a^{+}\right)\tilde{\varphi}\left(\\{H_{n}^{-},x_{\theta}^{+}\\}_{\theta=1}^{r+s}\right)$
$\displaystyle=$ $\displaystyle h_{12}^{n}\left(\vec{a}\right),$

and

$\displaystyle\sum_{|\vec{m}|\leq\mathcal{N}}D^{\vec{m}}G_{\vec{m}}^{n}\left(\\{\\{x_{\tau}^{+}\\}_{\tau=1}^{r+s};a^{+}\right)\equiv$
$\displaystyle\mathscr{W}_{0}^{n}\left(\\{x_{\tau}^{+}\\}_{\tau=1}^{r},\\{x_{\theta}^{+}+a^{+}\\}_{\theta=r+1}^{r+s}\right),$

with

$\sum_{|\vec{m}|\leq\mathcal{N}}|G_{\vec{m}}^{n}|\left(\\{x_{\tau}^{+}\\}_{\tau=1}^{r+s};a^{+}\right)\leq
C(\rho_{n})^{\tilde{k}}\left[|a^{+}|^{\alpha}+\left(\sum_{\tau=1}^{r+s}|x_{\tau}^{+}|^{2}\right)^{\gamma/2}\right],$

for constants $\tilde{k},\alpha,\gamma$, all independent of $n$.

Choose $R_{0}=|a^{+}|/2>0$, because $\vec{a}$ is space-like. For
$0<\epsilon<|a^{+}|/2$ small, the set

$\\{R>R_{0}-\epsilon\\}:=\left\\{\\{x_{\tau}^{+}\\}_{\tau=1}^{r+s}:\
\sum_{\tau=1}^{r+s}|x_{\tau}^{+}|^{2}>(R_{0}-\epsilon)^{2}\right\\}\subset{{\mathbb{R}}}^{2(r+s)}.$

Hence, we can apply Proposition 8.29 and
$D^{\vec{m}}G_{\vec{m}}^{n}\left(\cdot;a^{+}\right)$ vanishes on its
complement. Thus we have

$\displaystyle\sum_{|\vec{m}|\leq\mathcal{N}}$
$\displaystyle\int_{{{\mathbb{R}}}^{2(r+s)}}\prod_{\tau=1}^{r+s}dx_{\tau}^{+}\
D^{\vec{m}}G_{\vec{m}}^{n}\left(\\{x_{\tau}^{+}\\}_{\tau=1}^{r+s};a^{+}\right)\tilde{\varphi}\left(\\{q_{\theta}^{-},x_{\theta}^{+}\\}_{\theta=1}^{r+s}\right)$
$\displaystyle=\sum_{|\vec{m}|\leq\mathcal{N}}\int_{\\{R>R_{0}-\epsilon\\}}\prod_{\tau=1}^{r+s}dx_{\tau}^{+}\
D^{\vec{m}}G_{\vec{m}}^{n}\left(\\{x_{\tau}^{+}\\}_{\tau=1}^{r+s};a^{+}\right)\tilde{\varphi}\left(\\{q_{\theta}^{-},x_{\theta}^{+}\\}_{\theta=1}^{r+s}\right).$

Using spherical coordinates,

$\displaystyle\bigg{|}h_{12}^{n}\left(\vec{a}\right)\bigg{|}$
$\displaystyle\leq\sum_{|\vec{m}|\leq\mathcal{N}}\int_{\\{R>R_{0}-\epsilon\\}}\prod_{\tau=1}^{r+s}dx_{\tau}^{+}\
\left|G_{\vec{m}}^{n}\left(\\{x_{\tau}^{+}\\}_{\tau=1}^{r+s};a^{+}\right)D^{\vec{m}}\tilde{\varphi}\left(\\{H_{n}^{-},x_{\theta}^{+}\\}_{\theta=1}^{r+s}\right)\right|$
$\displaystyle\leq
C(\rho_{n})^{\tilde{K}}\int_{\\{R>R_{0}-\epsilon\\}}\left[|a^{+}|^{\alpha}+R^{\gamma}\right]\frac{\parallel\varphi\parallel_{p,q}}{(r+s)|m_{n}|^{l}+R^{k+\bar{l}}}\frac{R^{2(r+s)}}{R}dRd\Omega,$

for some norm $\parallel\cdot\parallel_{p,q}$, and $k,l,\bar{l}$ can be any
natural numbers less than $p$, $q$. And $\tilde{K}\geq\tilde{k}+\bar{n}$ is
some natural number, independent of $n$.

By choosing $p\geq k+\bar{l},q\geq l$ and $\tilde{K}$ large enough, we see
that the above inequality is less than some positive constant times

$\displaystyle
C(\rho_{n})^{\tilde{K}}\frac{\parallel\varphi_{1}\parallel_{p,q}\parallel\varphi_{2}\parallel_{p,q}}{m_{n}^{l}+|a^{+}|^{k}},$

for any $|a^{+}|$.

Since the above equation holds for any non-compact $\varphi_{i}$, we have for
each $n$,

$\displaystyle\mathscr{W}_{\kappa_{r},\pi_{s}}^{n}$
$\displaystyle\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r},\\{\vec{x}_{\theta}+\vec{a}\\}_{\theta=r^{+}}^{r+s}\right)\longrightarrow
0,$

as $|a^{+}|\equiv|(a^{2},a^{3})|\rightarrow\infty$, decaying at the rate of
$\dfrac{1}{m_{n}^{l}+|a^{+}|^{k}}$, for any power $k,l\in\mathbb{N}$.

For some fixed integer $\tilde{K}>0$,
$C=O\left(\sum_{n=1}^{\infty}|c_{1,n}c_{2,n}|C(\rho_{n})^{\tilde{K}}\right)$
from Equation (8.17). Since $m_{0}\leq m_{n}$ for all $n\in\mathbb{N}$, we
have

$\displaystyle\bigg{|}h_{12}\left(\vec{a}\right)\bigg{|}$
$\displaystyle\leq\sum_{n=1}^{\infty}|c_{1,n}c_{2,n}|\bigg{|}h_{12}^{n}\left(\vec{a}\right)\bigg{|}$
$\displaystyle\leq\frac{C}{m_{0}^{l}+|a^{+}|^{k}}\left(\parallel\varphi_{1}\parallel_{p,q}\parallel\varphi_{2}\parallel_{p,q}\right).$

###### Notation 8.33

Refer to Notation 8.6. Let $\Omega_{r}$ denote the set consisting of
partitions in $\Gamma$, such that
$Q=\\{A_{1},\ldots,A_{n(Q)}\\}\in\Omega_{r}$, iff there exists a (unique)
$1\leq z\leq n(Q)$, such that both $r,r+1\in A_{z}$.

Recall $H_{n}^{-}=(\hat{H}(\rho_{n}),\hat{P}(\rho_{n}))$. Write for $1\leq
i\leq n(Q)$, $n_{0}:=0$, $n_{i}=\sum_{k=1}^{i}|A_{k}|$ and

$\\{y_{j}^{+}\\}^{|A_{i}|}=\Big{(}\underbrace{H_{n}^{-},y_{i}^{+},H_{n}^{-},y_{i}^{+},\cdots,H_{n}^{-},y_{i}^{+}}_{|A_{i}|\
{\rm copies}\ {\rm of}\ (H_{n}^{-},y_{i}^{+})}\Big{)}.$

Refer to Equation (8.23) for the notations

$\tilde{\varphi}_{1}\left(\\{H_{n}^{-},x_{\tau}^{+}\\}_{\tau=1}^{r}\right)\quad{\rm
and}\quad\tilde{\varphi}_{2}\left(\\{H_{n}^{-},x_{\theta}^{+}\\}_{\theta=r+1}^{r+s}\right).$

For a given partition $Q=\\{A_{1},\ldots,A_{n(Q)}\\}\in\Omega_{r}$, we will
write

$\displaystyle\tilde{\varphi}_{Q,1}^{n}\left(\\{x_{\theta}^{+}\\}_{\theta=n_{z-1}+1}^{r}\right)$
$\displaystyle:=\int_{{{\mathbb{R}}}^{2(z-1)}}\prod_{i=1}^{z-1}dy_{i}^{+}\tilde{\varphi}_{1}\left(\\{y_{j}^{+}\\}^{|A_{1}|},\\{y_{j}^{+}\\}^{|A_{2}|},\cdots,\\{y_{j}^{+}\\}^{|A_{z-1}|},\\{H_{n}^{-},x_{\theta}^{+}\\}_{\theta=n_{z-1}+1}^{r}\right).$

Similarly, for $m_{z}:=r+s-\sum_{k=1}^{z}|A_{k}|$, we will write ($z^{+}=z+1$)

$\displaystyle\tilde{\varphi}_{Q,2}^{n}\left(\\{x_{\theta}^{+}\\}_{\theta=r^{+}}^{m_{z}}\right)$
$\displaystyle:=\int_{{{\mathbb{R}}}^{2(n(Q)-z)}}\prod_{i=z^{+}}^{n(Q)}dy_{i}^{+}\tilde{\varphi}_{2}\left(\\{H_{n}^{-},x_{\theta}^{+}\\}_{\theta=r^{+}}^{m_{z}},\\{y_{j}^{+}\\}^{|A_{z^{+}}|},\\{y_{j}^{+}\\}^{|A_{z+2}|},\cdots,\\{y_{j}^{+}\\}^{|A_{n(Q)}|}\right).$

Note that $\\{n_{z-1}+1,\cdots,r,r+1,\cdots,m_{z}\\}=A_{z}\in Q$. Finally, we
will write

$\tilde{\varphi}_{Q,1}^{n}(x_{r}^{+}):=\tilde{\varphi}_{Q,1}^{n}\left(\\{x_{r}^{+}\\}_{\theta=n_{z-1}+1}^{r}\right),\quad\tilde{\varphi}_{Q,2}^{n}(x_{r+1}^{+}):=\tilde{\varphi}_{Q,2}^{n}\left(\\{x_{r+1}^{+}\\}_{\theta=r+1}^{m_{z}}\right).$

From Equations (8.7) and (8.18), we will leave to the reader to verify that
for any $\vec{a}\equiv(a^{-},a^{+})\in{{\mathbb{R}}}^{4}$,

$\displaystyle
h_{12}(\vec{a})=e^{-ic_{r}m_{n}\vec{a}\cdot\tilde{f}_{0}^{n}}\sum_{Q\in\Omega_{r}}c_{Q}\int_{{{\mathbb{R}}}^{2}}dx^{+}\
\tilde{\varphi}_{Q,1}^{n}(x^{+})\tilde{\varphi}_{Q,2}^{n}(x^{+}-a^{+})$
$\displaystyle=e^{-ic_{r}m_{n}\vec{a}\cdot\tilde{f}_{0}^{n}}\sum_{Q\in\Omega_{r}}c_{Q}\int_{{{\mathbb{R}}}^{2}}dx_{r}^{+}\int_{{{\mathbb{R}}}^{2}}dx_{r^{+}}^{+}\
\tilde{\varphi}_{Q,1}^{n}(x_{r}^{+})\delta(a^{+}+x_{r^{+}}^{+}-x_{r}^{+})\tilde{\varphi}_{Q,2}^{n}(x_{r^{+}}^{+}),$
(8.24)

whereby $\delta$ is Dirac delta function. Note that the coefficients $c_{Q}$
were defined in Equation (8.2).

###### Theorem 8.34

(Clustering Theorem)
Suppose the assumptions in Definition 8.24 hold with $\varphi_{i}$ is
compactly supported for $i=1,2$. Assume that $0\neq c_{r}$ in Equation (8.11).

Recall $m_{0}$ is the mass gap of the 4-dimensional quantum Yang-Mills theory.
There exist positive constants $\epsilon$, $C$, and $p_{i},q_{i}\in\mathbb{N}$
for $i=1,2$, such that

$\displaystyle\left|h_{12}\left(\vec{a}\right)\right|$
$\displaystyle=\left|\sum_{n=1}^{\infty}c_{1,n}c_{2,n}h_{12}^{n}\left(\vec{a}\right)\right|$
$\displaystyle\leq
C\frac{e^{-m_{0}|a^{+}|}}{|a^{+}|-\epsilon}\parallel\varphi_{1}\parallel_{p_{1},q_{1}}\parallel\varphi_{2}\parallel_{p_{2},q_{2}},$
(8.25)

provided $|a^{+}|>\epsilon$, with $\epsilon$ depending only on the support of
$\varphi_{1}$ and $\varphi_{2}$. And,
$C=O\left(\sum_{n=1}^{\infty}|c_{1,n}c_{2,n}|C(\rho_{n})^{\tilde{k}}e^{m_{0}\epsilon}\right)$
for some fixed integer $\tilde{k}>0$. Note that $a^{+}=(a^{2},a^{3})$ are
coordinates with respect to $\\{e_{2},e_{3}\\}$, spanning the $x^{2}-x^{3}$
plane $S_{0}$.

Proof. Just like in the proof of Theorem 8.32, it suffices to prove it for
$\varphi_{1}\in\mathcal{Y}_{\kappa_{r}}$ and
$\varphi_{2}\in\widetilde{\mathcal{Y}}_{\pi_{s}}$. Recall in Definition 8.28,
we defined a new basis $\\{\tilde{f}_{a}^{n}\\}_{a=0}^{3}$, which depends on
the representation $\rho_{n}$ on $\mathfrak{g}$. Let
$(a^{0},a^{1},a^{2},a^{3})$ be coordinates with respect to
$\\{\tilde{f}_{a}^{n}\\}_{a=0}^{3}$. Further recall that $h_{12}^{n}$ is
independent of the $a^{1}$ coordinate, so we will henceforth write
$h_{12}^{n}(\vec{a})=h_{12}^{n}(a^{0},a^{+})$, $a^{+}=(a^{2},a^{3})$.

Define

$D(\varphi_{1},\varphi_{2})=\left\\{\vec{x}_{\tau}-\vec{y}_{\theta}\in{{\mathbb{R}}}^{4}|\
\\{\vec{x}_{\tau}\\}_{\tau=1}^{r}\in{\rm supp}\ \varphi_{1},\
\\{\vec{y}_{\theta}\\}_{\theta=r+1}^{r+s}\in{\rm supp}\ \varphi_{2}\right\\},$

and consider orthogonally projecting it onto the $x^{2}-x^{3}$ plane spanned
by $\\{e_{2},e_{3}\\}$, this projected set denoted by $D^{0}$. It is bounded.

Denote a compact set $D^{1}$, which is the convex closure of $D^{0}$, which
lies in the plane spanned by $\\{e_{2},e_{3}\\}$. Then, we see that there
exists an $\epsilon$ such that $D_{1}$ lies in an open disc of radius
$\epsilon$, center at the origin.

Let $g$ be any compact Schwartz function on ${{\mathbb{R}}}$. Consider the
following elliptic equation333We used an elliptic equation, because the total
energy squared is then given by $(c_{r}m_{n})^{2}+\sum_{i=1}^{3}q^{i,2}$,
which is mass squared plus total momentum squared. The generator for
translation in the time-like direction $\tilde{f}_{0}^{n}$, is given by the
mass eigenvalue $-c_{r}m_{n}$, and not energy eigenvalue.

$\left(\frac{\partial^{2}}{\partial
a^{0,2}}-\hat{\mathcal{P}}\right)\Psi_{n}(a^{0},a):=\left(\sum_{i=0}^{3}\frac{\partial^{2}}{\partial
a^{i,2}}\right)\Psi_{n}(a^{0},a)=h_{12}^{n}(a^{0},a^{+})g(a^{1}).$

Refer to Notation 8.33. We leave to the reader to verify from Equation (8.24),
($r^{+}\equiv r+1$)

$\displaystyle\Psi_{n}$
$\displaystyle(a^{0},a)=e^{ic_{r}m_{n}a^{0}}\tilde{G}(n;a),\ {\rm and}$
$\displaystyle\tilde{G}$ $\displaystyle(n;a)$ $\displaystyle=$
$\displaystyle\sum_{Q\in\Omega_{r}}c_{Q}\int_{{\mathbb{R}}}d\xi^{1}\int_{{{\mathbb{R}}}^{2\times
2}}dx_{r}^{+}dx_{r^{+}}^{+}\tilde{\varphi}_{Q,1}^{n}(x_{r}^{+})G_{n}(a^{1}-\xi^{1};x_{r}^{+},x_{r^{+}}^{+}+a^{+})\tilde{\varphi}_{Q,2}^{n}(x_{r^{+}}^{+})g(\xi^{1}),$

whereby $G_{n}(a^{1}-\xi^{1};x_{r}^{+},x_{r+1}^{+}+a^{+})$ is a Green’s
function that solves

$\displaystyle\left(-(c_{r}m_{n})^{2}+\sum_{i=1}^{3}\frac{\partial^{2}}{\partial
a^{i,2}}\right)G_{n}(a^{1}-\xi^{1};x_{r}^{+},x_{r+1}^{+}+a^{+})$
$\displaystyle=\delta(a^{1}-\xi^{1},x_{r}^{+}-x_{r+1}^{+}-a^{+})$
$\displaystyle\equiv\delta(a-\xi_{r}),$

whereby $\delta$ is the Dirac delta function, $a=(a^{1},a^{+})$ and
$\xi_{r}=(\xi^{1},\xi_{r}^{+})\equiv(\xi^{1},x_{r}^{+}-x_{r+1}^{+})$.

Write $\omega=|c_{r}m_{n}|\neq 0$ and $q=(q^{1},q^{2},q^{3})$ be the momentum
coordinates dual to $a\equiv(a^{1},a^{2},a^{3})$. Using Fourier Transform
techniques, the solution is given by

$G_{n}(a^{1}-\xi^{1};x_{r}^{+},x_{r+1}^{+}+a^{+})=-\frac{1}{(2\pi)^{3/2}}\int_{{{\mathbb{R}}}^{3}}dq\
\frac{e^{iq\cdot(a-\xi_{r})}}{\omega^{2}+|q|^{2}}.$

Note that the integrand is invariant under spatial rotation. Without any loss
of generality, we can assume that
$(a-\xi_{r})\equiv(a^{1}-\xi^{1},a^{+}-\xi_{r}^{+})$ lies in the $x^{3}$-axis,
or in the $e_{3}\equiv\tilde{f}_{3}^{n}$ positive direction. Hence,
$q\cdot(a-\xi_{r})=|q|R\cos\theta$, whereby $R=|(a-\xi_{r})|$ and $\theta$ is
the azimuthal angle a vector in 3-dimensional space makes with the positive
$x^{3}$\- axis.

Integrate out the polar and azimuthal angles in spherical coordinates, we have

$\displaystyle G_{n}(a^{1}-\xi^{1};x_{r}^{+},x_{r+1}^{+}+a^{+})$
$\displaystyle=-\frac{1}{(2\pi)^{1/2}}\frac{1}{iR}\int_{0}^{\infty}\lambda\frac{e^{iR\lambda}-e^{-iR\lambda}}{\omega^{2}+\lambda^{2}}\
d\lambda$
$\displaystyle=-\frac{1}{(2\pi)^{1/2}}\frac{1}{iR}\int_{-\infty}^{\infty}\frac{\lambda
e^{iR\lambda}}{(\lambda-i\omega)(\lambda+i\omega)}\ d\lambda.$

We will evaluate this integral using Cauchy’s Residue Theorem, by considering
an upper semi-circle, which contains a simple pole at $z=i\omega$ in its
interior. We choose the upper semi-circle because on the upper half plane,
$|e^{iRz}|\leq 1$ has an exponential decay. Finally taking its radius to be
infinity, the result is that the integral is equal to

$G_{n}(a^{1}-\xi^{1};x_{r}^{+},x_{r+1}^{+}+a^{+})=-\frac{2\pi
i}{(2\pi)^{1/2}}\frac{1}{iR}\frac{i\omega
e^{-R\omega}}{2i\omega}=-\sqrt{2\pi}\frac{e^{-R|c_{r}m_{n}|}}{2R},$

with
$R^{2}=|a^{1}-\xi^{1}|^{2}+|a^{2}+\xi_{r}^{2}|^{2}+|a^{3}+\xi_{r}^{3}|^{2}$.

Therefore, ($G_{n}(a-\xi):=G_{n}(a^{1}-\xi^{1};x^{-},x^{+}+a^{+})$)

$\displaystyle\left|h_{12}^{n}(\vec{a})g\left(a^{1}\right)\right|=\left|\left[\frac{\partial^{2}}{\partial
a^{0,2}}-\hat{\mathcal{P}}\right]\Psi_{n}(a^{0},a)\right|$
$\displaystyle=\left|\sum_{Q\in\Omega_{r}}c_{Q}\int_{{\mathbb{R}}}d\xi^{1}\int_{{{\mathbb{R}}}^{4}}d\vec{x}\
\tilde{\varphi}_{Q,1}^{n}(x^{-})G_{n}(a-\xi)\left[(c_{r}m_{n})^{2}+\hat{\mathcal{P}}\right]\tilde{\varphi}_{Q,2}^{n}(x^{+})g(\xi^{1})\right|$
$\displaystyle\leq\sum_{Q\in\Omega_{r}}|c_{Q}|\Bigg{\\{}\int_{{{\mathbb{R}}}}\
d\xi^{1}\left[|g(\xi^{1})|+\frac{1}{m_{0}^{2}}|g^{\prime\prime}(\xi^{1})|\right]\cdot\sqrt{\frac{\pi}{2}}\frac{e^{-m_{0}|a^{+}+x^{+}-y^{+}|}}{|a^{+}+x^{+}-y^{+}|}$
$\displaystyle\hskip
68.28644pt\times\int_{{{\mathbb{R}}}^{2}}dy^{+}\left|\tilde{\varphi}_{Q,1}^{n}(y^{+})\right|\cdot\int_{{{\mathbb{R}}}^{2}}dx^{+}\left|\left[(c_{r}m_{n})^{2}+\hat{\mathcal{P}}\right]\tilde{\varphi}_{Q,2}^{n}(x^{+})\right|\Bigg{\\}}$
$\displaystyle\leq C(\rho_{n})^{\tilde{k}}e^{m_{0}\epsilon}\parallel
g\parallel_{p_{3},q_{3}}\parallel\varphi_{1}\parallel_{p_{1},q_{1}}\parallel\varphi_{2}\parallel_{p_{2},q_{2}}\cdot\frac{e^{-m_{0}|a^{+}|}}{|a^{+}|-\epsilon},$

for some natural numbers $\tilde{k}$, $p_{i}$’s, $q_{i}$’s, provided
$|x^{+}-y^{+}|\leq\epsilon<|a^{+}|$ is large enough. Note that all these
natural numbers are independent of $n$, $a^{+}$ and the Schwartz functions. We
can assume that $C(\rho_{n})\geq 1$, otherwise we will replace it with
$1+C(\rho_{n})$ in the above inequality.

Because the above inequality holds for any compact Schwartz function
$g:{{\mathbb{R}}}\rightarrow{{\mathbb{R}}}$ and any $a^{1}\in{{\mathbb{R}}}$,
we have the inequality

$|h_{12}^{n}(\vec{a})|\leq
C(\rho_{n})^{\tilde{k}}e^{m_{0}\epsilon}\parallel\varphi_{1}\parallel_{p_{1},q_{1}}\parallel\varphi_{2}\parallel_{p_{2},q_{2}}\cdot\frac{e^{-m_{0}|a^{+}|}}{|a^{+}|-\epsilon},$

for $|a^{+}|>\epsilon$.

Since the above inequality holds for any $n\in\mathbb{N}$, we see that
Equation (8.25) follows immediately from triangle inequality and Equation
(8.17).

###### Remark 8.35

When the space like separation $\vec{a}$ lies in the $x^{2}-x^{3}$ plane or
$S_{0}$, then $h_{12}(\vec{a})$ is equivalent to the vacuum expectation
$\left\langle
T_{1}^{n}\left(\vec{y}\right)P_{0}T_{2}^{n}\left(\vec{x}\right)1,1\right\rangle$
from Definition 8.26. Suppose we are given two sets of cluster points
$\\{\vec{x}_{\theta}\in{{\mathbb{R}}}^{4}\\}_{\theta=1}^{r}$,
$\\{\vec{y}_{\tau}\in{{\mathbb{R}}}^{4}\\}_{\tau=1}^{s}$, and project them
orthogonally onto $S_{0}$, as $A=\\{x_{\theta}^{+}\in
S_{0}\\}_{\theta=1}^{r}$, $B=\\{y_{\tau}^{+}\in S_{0}\\}_{\tau=1}^{s}$
respectively. Clustering Theorem hence says that the vacuum expectation decays
exponentially, at a rate dependent on the mass gap $m_{0}$, provided the
space-like separation $a^{2}e_{2}+a^{2}e_{3}\in S_{0}$ between $A,B\subset
S_{0}$, is large enough. In other words,

$\displaystyle\left|\mathscr{W}^{n}\left(\\{\vec{x}_{\tau}\\}_{\tau=1}^{r},\\{\vec{x}_{\theta}+\vec{a}\\}_{\theta=r^{+}}^{r+s}\right)\right|\leq
C(n)e^{-m_{0}|a^{+}|},$

for some constant $C(n)$ dependent on $n$, and $a^{+}$ larger than the space-
like separation between the two sets in $S_{0}$.

Note that Equation (8.22) holds, even for space-like distance $|a^{+}|$ small
and close to zero. When there is no minimum mass gap $m_{0}$, the given upper
bound is not meaningful for small $|a^{+}|$. For $|a^{+}|$ small and for each
component mass gap $m_{n}>1$, we see that the heavier the mass gap $m_{n}$,
the faster is the rate of decay. Of course, when $|a^{+}|$ is large, then the
impact of $m_{0}$ to the inverse power decay law is minimal.

An inverse power decay is definitely not as rapid a decline, as an exponential
decay, which was proved in [22]. They showed that the decay is at most
$\frac{1}{|a^{+}|^{3/2}}e^{-m_{0}|a^{+}|}$, which is stronger than our result.
But do note that the authors proved exponential decay, by using a homogeneous
wave equation, for the cases when $\varphi_{1}$, $\varphi_{2}$ are both
compactly supported, and the space-like separation needs to be greater than
some $\epsilon$. In contrast, we used an elliptic equation, reason given in
the footnote. Any of the three cases in Remark 8.19 will imply that $c_{r}\neq
0$. The rate of decay $\frac{1}{|a^{+}|}e^{-m_{0}|a^{+}|}$ which we have
shown, is asymptotically the same as the Yukawa potential, who used it as a
basis to describe the nuclear force. Refer to [9].

For large space-like separation, an exponential decay will be a lot faster
than the inverse power decay given in Equation (8.22). But for small space-
like separation, an inverse power decay might be better.

Local commutativity in Wightman’s last axiom allows us to define a Lorentz
transformation $\Lambda_{n}$ in Definition 8.28, which implies an exponential
decay $e^{-m_{n}|a^{+}|}$, but depending on each $n\in\mathbb{N}$. We will
need a Yang-Mills path integral, to prove the existence of a minimum positive
mass gap $m_{0}\leq m_{n}$, which gives us a best possible exponential decay
$e^{-m_{0}|a^{+}|}$, independent of the representation $\rho_{n}$ on the
simple Lie algebra $\mathfrak{g}$, for the vacuum expectation.

##  Appendix A Surface Integrals

Let $\sigma\equiv(\sigma_{0},\sigma_{1},\sigma_{2},\sigma_{3}):[0,1]^{2}\equiv
I^{2}\rightarrow{{\mathbb{R}}}^{4}$ be a parametrization of a surface
$S\subset{{\mathbb{R}}}^{4}$. Here, $\sigma^{\prime}=\partial\sigma/\partial
s$ and $\dot{\sigma}=\partial\sigma/\partial t$.

###### Definition A.1

For $a,b=0,1,2,3$, define Jacobian matrices,

$\displaystyle
J_{ab}^{\sigma}(s,t)=\left(\begin{array}[]{cc}\sigma_{a}^{\prime}(s,t)&\dot{\sigma}_{a}(s,t)\\\
\sigma_{b}^{\prime}(s,t)&\dot{\sigma}_{b}(s,t)\\\ \end{array}\right),\ a\neq
b,$ (A.3)

and write $|J^{\sigma}_{ab}|=\sqrt{[\det{J^{\sigma}_{ab}}]^{2}}$ and
$W_{ab}^{cd}:=J_{cd}^{\sigma}J_{ab}^{\sigma,-1}$, $a,b,c,d$ all distinct. Note
that $W_{cd}^{ab}=(W_{ab}^{cd})^{-1}$.

For $a,b,c,d$ all distinct, define
$\rho_{\sigma}^{ab}:I^{2}\rightarrow{{\mathbb{R}}}$ by

$\displaystyle\rho_{\sigma}^{ab}=$
$\displaystyle\frac{1}{\sqrt{\det\left[1+W_{ab}^{cd,T}W_{ab}^{cd}\right]}}\equiv\frac{|J_{ab}^{\sigma}|}{\sqrt{\det\left[J_{ab}^{\sigma,T}J_{ab}^{\sigma}+J_{cd}^{\sigma,T}J_{cd}^{\sigma}\right]}}.$
(A.4)

And

$\int_{S}d\rho:=\sum_{0\leq a<b\leq
3}\int_{I^{2}}\rho_{\sigma}^{ab}(s,t)|J_{ab}^{\sigma}|(s,t)\ dsdt,$

which is independent of the parametrization $\sigma$ used.

We leave to the reader to check that $\int_{S}d\rho$ gives us the area of the
surface $S$. Area of a surface is invariant under spatial rotation, but it is
not invariant under boost.

To construct an unitary representation of the Lorentz group, we need to
consider imaginary time axis. Instead of using $\rho_{\sigma}^{ab}$ as defined
in Equation (A.4) by some parametrization $\sigma$ for $S$, we will replace
the time component $\sigma_{0}$ with imaginary time $i\sigma_{0}$.

###### Definition A.2

Let $\sigma:[0,1]^{2}\equiv I^{2}\rightarrow{{\mathbb{R}}}^{4}$ be a
parametrization of a surface $S\subset{{\mathbb{R}}}^{4}$. For $a,b=0,1,2,3$
and $a<b$, define Jacobian matrices,

$\displaystyle\acute{J}_{ab}^{\sigma}(s,t)=\left\\{\begin{array}[]{ll}\ \ \ \
J_{ab}^{\sigma}(s,t),&\hbox{$a\neq 0$;}\\\
\left(\begin{array}[]{cc}i\sigma_{a}^{\prime}(s,t)&i\dot{\sigma}_{a}(s,t)\\\
\sigma_{b}^{\prime}(s,t)&\dot{\sigma}_{b}(s,t)\\\
\end{array}\right),&\hbox{$a=0$.}\end{array}\right.$ (A.9)

For $a,b,c,d$ all distinct, define
$\acute{\rho}_{\sigma}^{ab}:I^{2}\rightarrow{{\mathbb{C}}}$ by

$\displaystyle\acute{\rho}_{\sigma}^{ab}:=$
$\displaystyle\frac{\det\acute{J}_{ab}^{\sigma}}{\sqrt{\det\left[\acute{J}_{ab}^{\sigma,T}\acute{J}_{ab}^{\sigma}+\acute{J}_{cd}^{\sigma,T}\acute{J}_{cd}^{\sigma}\right]}},$

and

$\displaystyle\int_{S}\ d\acute{\rho}:=$ $\displaystyle\sum_{0\leq a<b\leq
3}\int_{I^{2}}\acute{\rho}_{\sigma}^{ab}(s,t)[\det\acute{J}_{ab}^{\sigma}(\hat{s})]\
d\hat{s},$ $\displaystyle\int_{S}d|\acute{\rho}|:=$
$\displaystyle\int_{I^{2}}\left|\sum_{0\leq a<b\leq
3}\acute{\rho}_{\sigma}^{ab}(\hat{s})[\det\acute{J}_{ab}^{\sigma}](\hat{s})\right|\
d\hat{s}.$

We will also write

$\displaystyle\acute{\rho}_{\sigma}(\hat{s})=$ $\displaystyle\sum_{0\leq
a<b\leq
3}\acute{\rho}_{\sigma}^{ab}(\hat{s})[\det\acute{J}_{ab}^{\sigma}](\hat{s}),$
$\displaystyle|\acute{\rho}_{\sigma}|(\hat{s})=$
$\displaystyle\left|\sum_{0\leq a<b\leq
3}\acute{\rho}_{\sigma}^{ab}(\hat{s})[\det\acute{J}_{ab}^{\sigma}](\hat{s})\right|.$

###### Lemma A.3

Let $S$ be a compact time-like or space-like surface, contained in a plane.
Then $\int_{S}d\acute{\rho}$ and $\int_{S}d|\acute{\rho}|$ remain invariant
under any Lorentz transformation $\Lambda:S\mapsto\hat{S}=\Lambda S$,
$\Lambda$ is a $4\times 4$ Lorentz matrix.

Proof. Let $\sigma:I^{2}\rightarrow S$ be a parametrization of $S$. Then
$\hat{\sigma}=\Lambda\sigma$ is a parametrization of $\hat{S}=\Lambda S$. We
leave to the reader to verify that

$\sum_{0\leq a<b\leq
3}\acute{\rho}_{\hat{\sigma}}^{ab}[\det\acute{J}_{ab}^{\hat{\sigma}}]=\sqrt{[\sigma^{\prime}\cdot\sigma^{\prime}][\dot{\sigma}\cdot\dot{\sigma}]-[\sigma^{\prime}\cdot\dot{\sigma}]^{2}}.$

Since $\vec{x}\cdot\vec{y}$ is invariant under Lorentz transformation, we thus
have

$\displaystyle\sum_{0\leq a<b\leq
3}\acute{\rho}_{\hat{\sigma}}^{ab}[\det\acute{J}_{ab}^{\hat{\sigma}}]$