# Wilson Area Law formula on $\mathbb{R}^{4}$

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

Let $\mathfrak{g}$ be the Lie Algebra of a compact semi-simple gauge group.
For a $\mathfrak{g}$-valued 1-form $A$, consider the Yang-Mills action

$S_{{\rm YM}}(A)=\int_{\mathbb{R}^{4}}\left|dA+A\wedge A\right|^{2}\ d\omega$

using the standard metric on $T\mathbb{R}^{4}$. We want to make sense of the
following path integral,

${\rm Tr}\
\int_{A\in\mathcal{A}_{\mathbb{R}^{4},\mathfrak{g}}/\mathcal{G}}\exp\left[c\int_{S}dA\right]e^{-\frac{1}{2}S_{{\rm
YM}}(A)}\ DA,$

whereby $DA$ is some Lebesgue type of measure on the space of
$\mathfrak{g}$-valued 1-forms, modulo gauge transformations
$\mathcal{A}_{\mathbb{R}^{4},\mathfrak{g}}/\mathcal{G}$. Here, $S$ is some
compact flat rectangular surface.

Using an Abstract Wiener space, we can define a Yang-Mills path integral
rigorously, for a compact semi-simple gauge group. Subsequently, we will then
derive the Wilson area law formula from the definition, using renormalization
techniques and asymptotic freedom.

One of the most important applications of the Area Law formula will be to
explain why the potential measured between a quark and antiquark is linear.

MSC 2020: 81T13, 81T08
Keywords: Yang-Mills measure, axial gauge fixing, area law, renormalization,
asymptotic freedom

##  1 Preliminaries

This is a sequel to the article [1], which constructs a Yang-Mills measure
using an abelian gauge group, in ${{\mathbb{R}}}^{4}$. We will now consider a
compact semi-simple gauge group. Balaban from 1984 to 1989, published a series
of papers, defining a Yang-Mills path integral using finite dimensional
lattice gauge. The references for these papers can be found in [1].

Instead of using lattice gauge approximation, we will use the Segal Bargmann
Transform, to construct a path integral on the Hilbert space containing
$\mathfrak{g}$-valued holomorphic functions. Renormalization techniques used
in the abelian gauge group case will also be applied in the non-abelian case.
But the most notable difference is that the Wilson Area Law formula does not
hold in the abelian case; however by assuming asymptotic freedom, we will show
that it holds in the non-abelian case.

In Balaban’s series of papers, asymptotic freedom was not used. Rivasseau and
his co-authors in [2], applied asymptotic freedom to construct a Yang- Mills
integral in 4 dimensional space, but the analysis is presented specifically
for $SU(N)$ gauge group. Both Rivasseau and Balaban considered infra-red
cutoff in their analysis, and proved ultraviolet stability. But they did not
show that the Area Law formula holds. We will also refer the reader to [3] for
a comprehensive review on work done in non-abelian Yang-Mills path integrals.

Consider a 4-manifold $M$ and a principal bundle $P$ over $M$, with structure
group $G$. We assume that $G$ is compact and semi-simple. Without loss of
generality we will assume that $G$ is a Lie subgroup of $U(\bar{N})$,
$\bar{N}\in\mathbb{N}$ and $\pi:P\rightarrow M$ is a trivial bundle. We will
identify the Lie Algebra $\mathfrak{g}$ of $G$ with a Lie subalgebra of the
Lie Algebra $\mathfrak{u}(\bar{N})$ of $U(\bar{N})$ throughout this article.
Suppose we write ${{\rm{Tr}}}\equiv{{\rm{Tr}}}_{{\rm
Mat}(\bar{N},\mathbb{C})}$. Then we can define a positive, non-degenerate
bilinear form by

$\langle A,B\rangle=-{{\rm{Tr}}}_{{\rm Mat}(\bar{N},\mathbb{C})}[AB]$ (1.1)

for $A,B\in\mathfrak{g}$. Let $\\{E^{\alpha}\\}_{\alpha=1}^{N}$ be an
orthonormal basis in $\mathfrak{g}$, which will be fixed throughout this
article. Refer to [4], [5] and [6].

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

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

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

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

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

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

$\langle u,v\rangle=\int_{M}u\wedge\ast v=\int_{M}\langle u,v\rangle_{k}\
d\omega.$ (1.3)

See [7].

Given $u\otimes
E\in\Lambda^{q}(T^{\ast}M)\otimes(M\times\mathfrak{g}\rightarrow M)$, we write

$|u\otimes E|^{2}\ d\omega=-{{\rm{Tr}}}\left[E\cdot E\right]\ u\wedge\ast
u=-{{\rm{Tr}}}\left[E\cdot E\right]\langle u,u\rangle_{q}\ d\omega,$

computed at each fiber of the tensor bundle. Hence for
$A\in\mathcal{A}_{M,\mathfrak{g}}$, the Yang-Mills action is given by

$S_{{\rm YM}}(A)=\int_{M}\left|dA+A\wedge A\right|^{2}\ d\omega.$ (1.4)

Note that this action is invariant under gauge transformations.

###### Notation 1.1

From now on, we only consider $M={{\mathbb{R}}}^{4}$ and take the principal
bundle $P$ over ${{\mathbb{R}}}^{4}$ to be the trivial bundle. Let
$\Lambda^{p}({{\mathbb{R}}}^{n})$ be the $p$-th exterior power of the dual to
the vector space ${{\mathbb{R}}}^{n}$, $n=3,4$.

On ${{\mathbb{R}}}^{4}$, fix the coordinate axis and choose standard
coordinates $\\{x^{0},x^{1},x^{2},x^{3}\\}$. We will also choose the standard
Riemannian metric on $T\mathbb{R}^{4}$. Now let
$T^{\ast}{{\mathbb{R}}}^{4}\rightarrow{{\mathbb{R}}}^{4}$ denote the trivial
cotangent bundle over ${{\mathbb{R}}}^{4}$, i.e.
$T^{\ast}{{\mathbb{R}}}^{4}\cong{{\mathbb{R}}}^{4}\times\Lambda^{1}({{\mathbb{R}}}^{4})$.
Note that $\Lambda^{1}({{\mathbb{R}}}^{4})$ is spanned by
$\\{dx^{0},dx^{1},dx^{2},dx^{3}\\}$ and $\Lambda^{1}({{\mathbb{R}}}^{3})$
denote the subspace in $\Lambda^{1}({{\mathbb{R}}}^{4})$, spanned by
$\\{dx^{1},dx^{2},dx^{3}\\}$.

Using axial gauge fixing, every
$A\in\mathcal{A}_{{{\mathbb{R}}}^{4},\mathfrak{g}}/\mathcal{G}$ can be gauge
transformed into a $\mathfrak{g}$-valued 1-form, of the form
$A=\sum_{\alpha=1}^{N}\sum_{j=1}^{3}a_{j,\alpha}\otimes dx^{j}\otimes
E^{\alpha}$, subject to the conditions

$a_{1,\alpha}(0,x^{1},0,0)=0,\ a_{2,\alpha}(0,x^{1},x^{2},0)=0,\
a_{3,\alpha}(0,x^{1},x^{2},x^{3})=0.$

Hence it suffices to consider the Yang-Mills integral in Expression (1.8) to
be over the space of $\mathfrak{g}$-valued 1-forms of the form
$A=\sum_{\alpha=1}^{N}\sum_{j=1}^{3}a_{j,\alpha}\otimes dx^{j}\otimes
E^{\alpha}$, whereby
$a_{j,\alpha}:{{\mathbb{R}}}^{4}\rightarrow{{\mathbb{R}}}$ is smooth.

Now,

$dA=\sum_{\alpha=1}^{N}\sum_{j=1}^{3}a_{0:j,\alpha}\otimes dx^{0}\wedge
dx^{j}\otimes E^{\alpha}+\sum_{\alpha=1}^{N}\sum_{1\leq i<j\leq
3}a_{i;j,\alpha}\otimes dx^{i}\wedge dx^{j}\otimes E^{\alpha},$ (1.5)

for

$a_{i;j,\alpha}:=\partial_{i}a_{j,\alpha}-\partial_{j}a_{i,\alpha},\quad
a_{0:j,\alpha}:=\partial_{0}a_{j,\alpha}.$ (1.6)

Its curvature will hence be given by

$\displaystyle dA+A\wedge A=$
$\displaystyle\sum_{\alpha=1}^{N}\sum_{j=1}^{3}a_{0:j,\alpha}\otimes
dx^{0}\wedge dx^{j}\otimes E^{\alpha}+\sum_{\alpha=1}^{N}\sum_{1\leq i<j\leq
3}a_{i;j,\alpha}\otimes dx^{i}\wedge dx^{j}\otimes E^{\alpha}$
$\displaystyle+\sum_{1\leq\alpha,\beta\leq N}\sum_{1\leq i<j\leq
3}a_{i,\alpha}a_{j,\beta}\otimes dx^{i}\wedge
dx^{j}\otimes[E^{\alpha},E^{\beta}].$

See [8]. Note that $[\cdot,\cdot]$ is the Lie Bracket.

###### Notation 1.2

Note that
$\Lambda^{2}(T^{\ast}{{\mathbb{R}}}^{4})\cong{{\mathbb{R}}}^{4}\times\Lambda^{2}({{\mathbb{R}}}^{4})$.
Using the standard coordinates $\\{x^{0},x^{1},x^{2},x^{3}\\}$, we fix an
orthonormal basis $\\{dx^{1}\wedge dx^{2},dx^{3}\wedge dx^{1},dx^{1}\wedge
dx^{2},dx^{0}\wedge dx^{1},dx^{0}\wedge dx^{2},dx^{0}\wedge dx^{3}\\}$ in
$\Lambda^{2}({{\mathbb{R}}}^{4})$, using the standard metric on
$T{{\mathbb{R}}}^{4}$. The corresponding volume form is given by
$d\omega=dx^{0}\wedge dx^{1}\wedge dx^{2}\wedge dx^{3}$.

From the Hodge star operator and the above volume form, we can define an inner
product on the set of smooth sections
$\Gamma(\Lambda^{2}(T^{\ast}{{\mathbb{R}}}^{4}))$ as in Equation (1.3).

###### Definition 1.3

Define

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

They are also referred to as the structure constants in physics.

###### Remark 1.4

Note that $c_{\gamma}^{\alpha\beta}\neq 0$, provided $\gamma$, $\alpha$ and
$\beta$ are all distinct. See [9].

Then,

$\displaystyle dA$ $\displaystyle+A\wedge A$ $\displaystyle=$
$\displaystyle\sum_{\gamma=1}^{N}\bigg{[}\sum_{1\leq i<j\leq
3}a_{i;j,\gamma}\otimes dx^{i}\wedge dx^{j}+\sum_{\alpha,\beta}\sum_{1\leq
i<j\leq 3}a_{i,\alpha}a_{j,\beta}c_{\gamma}^{\alpha\beta}\otimes dx^{i}\wedge
dx^{j}$ $\displaystyle\hskip 227.62204pt+\sum_{j=1}^{3}a_{0:j,\gamma}\otimes
dx^{0}\wedge dx^{j}\bigg{]}\otimes E^{\gamma}.$

Thus,

$\displaystyle\int_{{{\mathbb{R}}}^{4}}|dA+A\wedge A|^{2}\ d\omega=$
$\displaystyle\sum_{\gamma=1}^{N}\sum_{1\leq i<j\leq
3}\int_{{{\mathbb{R}}}^{4}}\bigg{[}a_{i;j,\gamma}^{2}+\sum_{\alpha,\beta,\hat{\alpha},\hat{\beta}}a_{i,\alpha}a_{j,\beta}a_{i,\hat{\alpha}}a_{j,\hat{\beta}}c_{\gamma}^{\alpha\beta}c_{\gamma}^{\hat{\alpha}\hat{\beta}}$
$\displaystyle+2\sum_{\alpha,\beta=1}^{N}a_{i;j,\gamma}a_{i,\alpha}a_{j,\beta}c_{\gamma}^{\alpha\beta}\bigg{]}\
d\omega+\sum_{\gamma=1}^{N}\sum_{1\leq j\leq
3}\int_{{{\mathbb{R}}}^{4}}a_{0:j,\gamma}^{2}\ d\omega.$ (1.7)

###### Definition 1.5

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 vector, with
length $|a|$. Let $\sigma:I^{2}\rightarrow{{\mathbb{R}}}^{4}$ be some
parametrization of a compact rectangular surface $R[a,T]$, spanned by $e^{0}$
and $a$, of dimensions height T and length $|a|$.

On the surface $R[a,T]$, we see that

$\int_{R[a,T]}dA+A\wedge A=\int_{R[a,T]}dA.$

For a coupling constant $c$, we made sense of

$\frac{1}{Z}\int_{A\in\mathcal{A}_{{{\mathbb{R}}}^{4},\mathfrak{g}}/\mathcal{G}}\exp\left[c\int_{R[a,T]}dA\right]e^{-\frac{1}{2}\int_{{{\mathbb{R}}}^{4}}|dA|^{2}\
d\omega}D[dA],$

in [1], for an abelian gauge group.

Let $\rho:\mathfrak{g}\rightarrow{\rm End}({{\mathbb{C}}}^{\tilde{N}})$ be an
irreducible representation of $\mathfrak{g}$. Throughout this article, we
assume that $\rho(\mathfrak{g})$ consists of skew-Hermitian matrices. Write
$A^{\rho}:=\sum_{\alpha=1}^{N}\sum_{i=1}^{3}A_{i,\alpha}\otimes
dx^{i}\otimes\rho(E^{\alpha})$, which is $\rho(\mathfrak{g})$-valued.

Motivated by the abelian case, we want to make sense of the following path
integral,

$\frac{1}{Z}{{\rm{Tr}}}\int_{A\in\mathcal{A}_{{{\mathbb{R}}}^{4},\mathfrak{g}}/\mathcal{G}}\exp\left[c\int_{R[a,T]}d[A^{\rho}]\right]e^{-\frac{1}{2}S_{{\rm
YM}}(A)}\ D[dA],$ (1.8)

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

$Z=\int_{A\in\mathcal{A}_{{{\mathbb{R}}}^{4},\mathfrak{g}}/\mathcal{G}}e^{-\frac{1}{2}S_{{\rm
YM}}(A)}\ DA,$

and $c$ is a coupling constant. Note that $d[A^{\rho}]$ is given by Equation
(1.5), except that we replace $E^{\alpha}$ in the said equation with
$\rho(E^{\alpha})$.

##  2 Non-abelian Yang-Mills measure

###### Notation 2.1

For $x\in{{\mathbb{R}}}^{4}$, we let
$\phi_{\kappa}(x)=\kappa^{4}e^{-\kappa^{2}|x|^{2}/2}/(2\pi)^{2}$, which is a
Gaussian function with variance $1/\kappa^{2}$. Define a function
$\psi_{z}\equiv\psi(z)$, where
$\psi(z)=\frac{1}{\sqrt{2\pi}}e^{-\sum_{i=1}^{4}|z_{i}|^{2}/2}$ and
$z\equiv(z_{0},z_{1},z_{2},z_{3})\in{{\mathbb{C}}}^{4}$.

###### Remark 2.2

In [1], we explained that the Gaussian function $\phi_{\kappa}$ will
approximate the Dirac Delta function, which is a generalized function. The
larger the value of $\kappa$, the better is this approximation. The variance
$1/\kappa^{2}$, denotes how well we can resolve a point in
${{\mathbb{R}}}^{4}$. In all our Yang-Mills path integrals to be define later,
it will depend on the parameter $\kappa$.

A finite dimensional gauge lattice with lattice spacing $\epsilon$ is commonly
used to define a finite dimensional path integral, of which one has to take
the limit as $\epsilon$ goes down to zero, to define an infinite dimensional
path integral. One should think of $1/\kappa$ as being synonymous with the
lattice spacing $\epsilon$. We refer the reader to [1] for a brief account on
the use of lattice gauge theory, for the purpose of defining a path integral.

###  2.1 Wiener measure

In [1], we considered the (real) vector space denoted by
$\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})$, $\kappa>0$. A vector in the said
space is written in the form $p_{i}\cdot\sqrt{\phi_{\kappa}}$, whereby $p_{i}$
is a polynomial on ${{\mathbb{R}}}^{4}$. We used the $L^{2}$ inner product on
this space.

The Segal Bargmann Transform $\Psi_{\kappa}$ maps it into the (real) vector
space of polynomials, over in ${{\mathbb{C}}}^{4}$, denoted as
$H^{2}({{\mathbb{C}}}^{4})$. On ${{\mathbb{C}}}^{4}$, we will write

$z=(z_{0},z_{1},z_{2},z_{3})\in{{\mathbb{C}}}^{4}\quad{\rm and}\quad
d\lambda_{4}=\frac{1}{\pi^{4}}e^{-\sum_{i=0}^{3}|z_{i}|^{2}}\prod_{i=0}^{3}dx^{i}dp^{i},$

be a Gaussian measure, whereby $z_{i}\equiv x^{i}+\sqrt{-1}p^{i}$.

The real inner product on $H^{2}({{\mathbb{C}}}^{4})$ is given by

$\langle f,g\rangle:=\int_{{{\mathbb{C}}}^{4}}f(z)\bar{g}(z)\ d\lambda_{4},$

$\bar{g}(z)$ means the complex conjugate of $g(z)$. Let the Hilbert space
$\mathcal{H}^{2}({{\mathbb{C}}}^{4})$ be the completion of
$H^{2}({{\mathbb{C}}}^{4})$ and
$\mathcal{H}^{2}({{\mathbb{C}}}^{4})\otimes_{{{\mathbb{R}}}}{{\mathbb{C}}}=\mathcal{H}^{2}({{\mathbb{C}}}^{4})_{{\mathbb{C}}}$,
the complexification of the Hilbert space consisting of polynomials over in
${{\mathbb{C}}}^{4}$. The inner product will now be extended to be
sesquilinear.

We also call $\\{\Psi_{\kappa}:\kappa>0\\}$ a renormalization flow between

$\left\\{(\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4}),\langle\cdot,\cdot\rangle):\kappa>0\right\\}\quad{\rm
and}\quad\left\\{(H^{2}({{\mathbb{C}}}^{4}),\langle\cdot,\cdot\rangle)\right\\}.$

The Segal Bargmann Transform is an isometry between the two inner product
spaces. We extend this isometry to be between
$\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{4})$
and $H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{4})$, using
the tensor inner product defined on the respective spaces.

We explained in [1], by defining a measurable norm $|\cdot|$, we complete
$\mathcal{H}^{2}({{\mathbb{C}}}^{4})$ into a Banach space
$B({{\mathbb{C}}}^{4})$, and form an Abstract Wiener space
$(i,\mathcal{H}^{2}({{\mathbb{C}}}^{4}),B({{\mathbb{C}}}^{4}))$, whereby $i$
is an inclusion map and $B({{\mathbb{C}}}^{4})$ is equipped with a Wiener
measure. This Banach space consists of holomorphic functions over
${{\mathbb{C}}}^{4}$.

###  2.2 Tensor products

Suppose we have an Abstract Wiener space, $(i,H,B)$, with a Hilbert space
$H\subset B$ which is dense and let the dual space $B^{\ast}$ be identified
with a dense subspace in $H$. By abuse of notation, we will write
$B^{\ast}\subset H$ to denote this identification. Denote the inner product on
$H$ with $\langle\cdot,\cdot\rangle$.

We will identify $f\in B^{\ast}$ with $(\cdot,f)_{\sharp}$ as a random
variable on $(B,\mu)$, $\mu$ is Wiener measure on $B$ with variance $\lambda$.
Thus for $x\in B$, $(x,f)_{\sharp}\in{{\mathbb{R}}}$. Note that the random
variable $(\cdot,f)_{\sharp}$ is equal to a Gaussian random variable in
distribution, with mean 0 and variance $\lambda\parallel
f\parallel^{2}\equiv\lambda\langle f,f\rangle$.

For $f\in B^{\ast}_{{\mathbb{C}}}\equiv
B^{\ast}\otimes_{{\mathbb{R}}}{{\mathbb{C}}}\subset
H\otimes_{{\mathbb{R}}}{{\mathbb{C}}}\equiv H_{{\mathbb{C}}}$,
$(\cdot,f)_{\sharp}$ can be written as an infinite sum of random variables
$\sum_{n=1}^{\infty}c_{n}(\cdot,f_{n})_{\sharp}$, using complex coefficients
$c_{n}$, each such real random variable $(\cdot,f_{n})$ on $(B,\mu)$ is equal
to a Gaussian random variable in distribution.

Write

$\displaystyle\left(\cdot,f\right)_{\sharp}=$
$\displaystyle\sqrt{-1}\left(\cdot,g_{1}\right)_{\sharp}+\left(\cdot,g_{2}\right)_{\sharp}=\left[\frac{\langle
g_{1},g_{2}\rangle}{\langle
g_{1},g_{1}\rangle}+\sqrt{-1}\right]\left(\cdot,g_{1}\right)_{\sharp}+\left(\cdot,g_{2}-\frac{\langle
g_{1},g_{2}\rangle}{\langle g_{1},g_{1}\rangle}g_{1}\right)_{\sharp}.$

Thus, $|(\cdot,f)_{\sharp}|^{2}$ is equal in distribution to
$|Z_{1}+\sqrt{-1}Z_{2}|^{2}$, whereby each $Z_{i}$ is a (real) Gaussian random
variable with mean 0. Extend $\langle\cdot,\cdot\rangle$ to be a sesquilinear
inner product on the complexified Hilbert space $H_{{\mathbb{C}}}$. Then the
complex normal random variable $Z_{1}+\sqrt{-1}Z_{2}$ has variance
$\lambda\langle f,f\rangle$.

Let $H_{i}\equiv(H_{i},\langle\cdot,\cdot\rangle_{i})$ be Hilbert spaces,
$i=1,2$. When we write $H_{1}\otimes H_{2}$, we define it as the tensor
product of Hilbert spaces, with the inner product

$\langle f_{1}\otimes g_{1},f_{2}\otimes g_{2}\rangle=\langle
f_{1},f_{2}\rangle_{1}\langle g_{1},g_{2}\rangle_{2}.$

Now suppose $H_{2}$ is finite dimensional with dimension $n$. Then,
$H_{1}\otimes H_{2}\cong H_{1}^{\times^{n}}$, direct product of $n$ copies of
$H_{1}$. Suppose $B_{1}^{\ast}\subset H_{1}\subset B_{1}$ is an Abstract
Wiener space containing $H_{1}$ as a dense subspace, with Wiener measure
$\tilde{\mu}$. We will construct an Abstract Wiener space $B_{1}\otimes
H_{2}$, containing $H_{1}\otimes H_{2}$, with the product Wiener measure
$\tilde{\mu}^{\times^{n}}\equiv\tilde{\mu}\times\cdots\times\tilde{\mu}$, $n$
copies of $\tilde{\mu}$.

When we write $B^{\otimes^{k}}$, we mean the $k$-th tensor product of $B$. An
element $x$ in $B^{\otimes^{k}}$ can be written in the form
$\sum_{\alpha}x_{1}^{\alpha}\otimes x_{2}^{\alpha}\otimes\cdots\otimes
x_{k}^{\alpha}$, $x_{i}^{\alpha}\in B$. Likewise, an element $z\in
B^{\otimes^{k},\ast}$ can be written in the form
$\sum_{\beta}z_{1}^{\beta}\otimes z_{2}^{\beta}\otimes\cdots\otimes
z_{k}^{\beta}$, where $z_{i}^{\beta}\in B^{\ast}$.

When we write $(x,z)_{\sharp}$, we mean

$\sum_{\alpha}\sum_{\beta}(x_{1}^{\alpha}\otimes
x_{2}^{\alpha}\otimes\cdots\otimes x_{k}^{\alpha},z_{1}^{\beta}\otimes
z_{2}^{\beta}\otimes\cdots\otimes
z_{k}^{\beta})_{\sharp}=\sum_{\alpha}\sum_{\beta}(x_{1}^{\alpha},z_{1}^{\beta})_{\sharp}(x_{2}^{\alpha},z_{2}^{\beta})_{\sharp}\cdots(x_{k}^{\alpha},z_{k}^{\beta})_{\sharp}.$

###  2.3 Triple and quartic terms

The reference for the following remarks can be found in [1].

###### Remark 2.3

  1. 1.

We define a linear operator
${\mathfrak{d}}_{a}:H^{2}({{\mathbb{C}}}^{4})\rightarrow
H^{2}({{\mathbb{C}}}^{4})$, for $a=0,1,2,3$, whereby

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

We denote the range of ${\mathfrak{d}}_{a}$ by
${\mathfrak{d}}_{a}H^{2}({{\mathbb{C}}}^{4})$. With this operator, we can
define an operator
${\mathfrak{d}}:H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}\rightarrow
H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{4})\otimes\mathfrak{g}$,
analogous to the differential operator $d$, as

$\displaystyle{\mathfrak{d}}\sum_{\alpha=1}^{N}\sum_{i=1}^{3}A_{i,\alpha}\otimes
dx^{i}\otimes E^{\alpha}=$
$\displaystyle\sum_{\alpha=1}^{N}\sum_{i=1}^{3}{\mathfrak{d}}_{0}A_{i,\alpha}\otimes
dx^{0}\wedge dx^{i}\otimes E^{\alpha}$
$\displaystyle+\sum_{\alpha=1}^{N}\sum_{1\leq i<j\leq
3}[{\mathfrak{d}}_{i}A_{j,\alpha}-{\mathfrak{d}}_{j}A_{i,\alpha}]\otimes
dx^{i}\wedge dx^{j}\otimes E^{\alpha},$ (2.1)

$A_{i,\alpha}\in H^{2}({{\mathbb{C}}}^{4})$ for $i=1,2,3$.

  2. 2.

Recall we had a renormalization flow

$\left\\{\Psi_{\kappa}:\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\rightarrow
H^{2}({{\mathbb{C}}}^{4})\ |\ \kappa>0\right\\}.$

Because
$\Psi_{\kappa}[\partial_{a}f]=\kappa{\mathfrak{d}}_{a}\Psi_{\kappa}[f]$, hence
we need to apply a renormalization rule, which assigns a $\kappa$ to both
${\mathfrak{d}}$ and ${\mathfrak{d}}_{a}$, when defining our path integral.

  3. 3.

In [1], we defined an $L^{2}$ inner product $\langle f,g\rangle$, for
$f,g\in\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})$. We can extend this inner
product into a tensor inner product, for $\mathfrak{g}$-valued 2-forms, over
in ${{\mathbb{R}}}^{4}$. For
$A=\sum_{\alpha=1}^{N}\sum_{i=1}^{3}a_{i,\alpha}\otimes dx^{i}\otimes
E^{\alpha}$, the tensor inner product is

$\langle dA,dA\rangle=\sum_{\gamma=1}^{N}\sum_{1\leq i<j\leq
3}\int_{{{\mathbb{R}}}^{4}}a_{i;j,\gamma}^{2}\
d\omega+\sum_{\gamma=1}^{N}\sum_{1\leq j\leq
3}\int_{{{\mathbb{R}}}^{4}}a_{0:j,\gamma}^{2}\ d\omega,$

on the inner product space
$\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{4})\otimes\mathfrak{g}$.
See Equation (1.6). We extend the Segal Bargmann Transform, so

$\Psi_{\kappa}[A]=\sum_{\alpha=1}^{N}\sum_{i=1}^{3}\Psi_{\kappa}[a_{i,\alpha}]\otimes
dx^{i}\otimes E^{\alpha}.$

  4. 4.

For

$F=\sum_{\alpha=1}^{N}\sum_{i=1}^{3}f_{i,\alpha}\otimes dx^{i}\otimes
E^{\alpha},\quad G=\sum_{\alpha=1}^{N}\sum_{i=1}^{3}g_{i,\alpha}\otimes
dx^{i}\otimes E^{\alpha}$

in
$H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}$,
we define a tensor inner product

$\displaystyle\langle{\mathfrak{d}}F,{\mathfrak{d}}G\rangle=$
$\displaystyle\sum_{\alpha=1}^{N}\sum_{i=1}^{3}\int_{{{\mathbb{C}}}^{4}}d\lambda_{4}\
[{\mathfrak{d}}_{0}f_{i,\alpha}]\overline{[{\mathfrak{d}}_{0}g_{i,\alpha}]}$
$\displaystyle+\sum_{\alpha=1}^{N}\sum_{1\leq i<j\leq
3}\int_{{{\mathbb{C}}}^{4}}d\lambda_{4}\
[{\mathfrak{d}}_{i}f_{j,\alpha}-{\mathfrak{d}}_{j}f_{i,\alpha}]\overline{[{\mathfrak{d}}_{i}g_{j,\alpha}-{\mathfrak{d}}_{j}g_{i,\alpha}]}.$
(2.2)

Note that $|g|^{2}\equiv g\cdot\bar{g}$, $\bar{g}$ is the complex conjugate of
a holomorphic function $g$.

  5. 5.

The Segal Bargmann Transform $\Psi_{\kappa}$ is an isometry (up to a constant
$\kappa^{2}$) between their respective spaces, i.e.

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

$A\in\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}$.

  6. 6.

We defined a closed subspace $B_{0}({{\mathbb{C}}}^{4})\subset
B({{\mathbb{C}}}^{4})$, using the norm $\parallel
x\parallel:=|x|+|{\mathfrak{d}}_{0}x|$, which can be shown to be a measurable
norm. Hence, we can equip $B_{0}({{\mathbb{C}}}^{4})$ with a Wiener measure.
The map ${\mathfrak{d}}_{0}$ can be shown to be an injective map on
$B_{0}({{\mathbb{C}}}^{4})$ and we can define a norm on its range
${\mathfrak{d}}_{0}H^{2}({{\mathbb{C}}}^{4})\subset{\mathfrak{d}}_{0}B_{0}({{\mathbb{C}}}^{4})\subset
B({{\mathbb{C}}}^{4})$ by
$\parallel{\mathfrak{d}}_{0}x\parallel:=|x|+|{\mathfrak{d}}_{0}x|$. Hence
${\mathfrak{d}}_{0}$ is an isometry map and thus
$({\mathfrak{d}}_{0}B_{0}({{\mathbb{C}}}^{4}),\parallel\cdot\parallel)$ is
also a Banach space, equipped with a Wiener measure.

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 the inner product space

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

Observe that for $A\in
H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})$, we have that

$\displaystyle\sum_{i=1}^{3}{\mathfrak{d}}_{0}A_{i}\otimes dx^{0}\wedge
dx^{1}\in$
$\displaystyle[{\mathfrak{d}}_{0}H^{2}({{\mathbb{C}}}^{4})]\otimes\left[\ast\Lambda^{2}({{\mathbb{R}}}^{3})\right],$
$\displaystyle\sum_{1\leq i<j\leq
3}[{\mathfrak{d}}_{i}A_{j}-{\mathfrak{d}}_{j}A_{i}]\otimes dx^{i}\wedge
dx^{j}\in$ $\displaystyle
H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{3}),$

and the said inner product given by Equation (2.2).

We can complete it into a Banach space

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

using a measurable norm. It can be shown that this Banach space contains the
dense inner product space
$H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{4})$, and it has
a product Wiener measure defined on it.

Just like how Balaban in his series of papers, defined a finite dimensional
integral on a finite gauge lattice of spacing $\epsilon$, we will define a
path integral over in
$\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}$
from Expression 1.8 as

$\frac{1}{Z}{{\rm{Tr}}}\int_{\\{dA:\
A\in\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}\\}}\exp\left[c\int_{R[a,T]}d[A^{\rho}]\right]e^{-\frac{1}{2}S_{{\rm
YM}}(A)}\ D[dA],$ (2.3)

$Z$ is some normalization constant. Note that $\epsilon$ is synonymous with
$1/\kappa$ and ${{\rm{Tr}}}$ means taking the matrix trace.

After Balaban defined a finite dimensional integral, he applied a
renormalization flow and mapped the finite gauge lattice of spacing $\epsilon$
to a finite gauge lattice of unit spacing. We will do the same thing here, but
using the Segal Bargmann Transform $\Psi_{\kappa}$ instead.

###### Remark 2.4

We can also view Segal Bargmann Transform as analogous to Fourier Transform,
which is how physicists will attempt to define a path integral over in
momentum space.

Because of the isometry given by Item 5 in Remark 2.3, we will make sense of

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

as

$\frac{1}{Z}\exp\left(-\frac{1}{2}\int_{{{\mathbb{C}}}^{4}}d\lambda_{4}|\kappa{\mathfrak{d}}A|^{2}\right)D[{\mathfrak{d}}A]$
(2.4)

over in $\mathbb{H}$, instead of over in
$\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{4})$.
For each $\kappa$, we will interpret Expression 2.4 as a Wiener measure
$\tilde{\mu}_{\kappa^{2}}$ with variance $1/\kappa^{2}$, over the Banach space
${\mathbb{B}}$, containing $\mathbb{H}$. Note that
$({\mathbb{B}},\tilde{\mu}_{\kappa^{2}})$ is actually a probability space, the
$\sigma$ algebra is the Borel field of ${\mathbb{B}}$.

Refer to the isometry given by Item 5 in Remark 2.3. We will extend this
isometry to a product of triple or quartic terms. Therefore, we will make
sense of $\frac{1}{Z}e^{-\frac{1}{2}S_{{\rm YM}}(A)}\ D[dA]$ in Expression
2.3, by replacing it with

$\displaystyle\frac{1}{Z}\exp$
$\displaystyle\left(-\frac{1}{2}\int_{{{\mathbb{C}}}^{4}}d\lambda_{4}|\kappa{\mathfrak{d}}A+A\wedge
A|^{2}\right)D[{\mathfrak{d}}A]=D_{1}D_{2},$

whereby

$\displaystyle\begin{split}D_{1}&:=\exp\left(-\frac{1}{2}\int_{{{\mathbb{C}}}^{4}}d\lambda_{4}\langle\kappa{\mathfrak{d}}A,A\wedge
A\rangle+\langle A\wedge A,\kappa{\mathfrak{d}}A\rangle+|A\wedge
A|^{2}\right),\\\
D_{2}&:=\frac{1}{Z}\exp\left(-\frac{1}{2}\int_{{{\mathbb{C}}}^{4}}d\lambda_{4}|\kappa{\mathfrak{d}}A|^{2}\right)D[{\mathfrak{d}}A],\end{split}$
(2.5)

all on $\mathbb{H}\otimes\mathfrak{g}$, $Z$ is some normalization constant.

For $A=\sum_{\alpha=1}^{N}\sum_{i=1}^{3}A_{i,\alpha}\otimes dx^{i}\otimes
E^{\alpha}\in
H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}$,
we have ${\mathfrak{d}}A$ as given in Equation (2.7),

$A\wedge A=\sum_{\alpha,\beta=1}^{N}\sum_{1\leq i<j\leq
3}A_{i,\alpha}A_{j,\beta}\otimes dx^{i}\wedge
dx^{j}\otimes[E^{\alpha},E^{\beta}],$

which can be written as

$A\wedge A=\sum_{\gamma=1}^{N}\sum_{\alpha,\beta=1}^{N}\sum_{1\leq i<j\leq
3}c_{\gamma}^{\alpha\beta}A_{i,\alpha}A_{j,\beta}\otimes dx^{i}\wedge
dx^{j}\otimes E^{\gamma}.$

Therefore,

$|A\wedge A|^{2}=\sum_{\gamma=1}^{N}\sum_{1\leq i<j\leq
3}\sum_{\alpha,\beta\atop\hat{\alpha},\hat{\beta}}c_{\gamma}^{\alpha\beta}c_{\gamma}^{\hat{\alpha}\hat{\beta}}A_{i,\alpha}A_{j,\beta}\overline{A_{i,\hat{\alpha}}}\overline{A_{j,\hat{\beta}}},$

and

$\displaystyle\langle{\mathfrak{d}}A,A\wedge A\rangle=$
$\displaystyle\sum_{\gamma=1}^{N}\sum_{1\leq i<j\leq
3}\sum_{\alpha,\beta=1}^{N}c_{\gamma}^{\alpha\beta}[{\mathfrak{d}}_{i}A_{j,\gamma}-{\mathfrak{d}}_{j}A_{i,\gamma}]\overline{A_{i,\alpha}A_{j,\beta}},$
$\displaystyle\langle A\wedge A,{\mathfrak{d}}A\rangle=$
$\displaystyle\sum_{\gamma=1}^{N}\sum_{1\leq i<j\leq
3}\sum_{\alpha,\beta=1}^{N}c_{\gamma}^{\alpha\beta}A_{i,\alpha}A_{j,\beta}\overline{[{\mathfrak{d}}_{i}A_{j,\gamma}-{\mathfrak{d}}_{j}A_{i,\gamma}]}.$

But we would like to interpret them as tensor products of complex random
variables on the Banach space ${\mathbb{B}}\otimes\mathfrak{g}$, which
contains $\mathbb{H}\otimes\mathfrak{g}$. To do this, we first need to review
one linear functional, defined in [1].

###### Definition 2.5

For any $f\in\mathcal{H}^{2}({{\mathbb{C}}}^{4})$, $\chi_{w}(z)\equiv
e^{\bar{w}\cdot z}$ is the unique vector such that $\langle
f,\chi_{w}\rangle=f(w)$ for any $w\in{{\mathbb{C}}}^{4}$. Note that
$\bar{w}\cdot z=\sum_{a=0}^{3}\overline{w_{a}}z_{a}$, $\overline{w_{a}}$ is
the complex conjugate of $w_{a}\in{{\mathbb{C}}}$. Because $\chi_{w}\in
B({{\mathbb{C}}}^{4})_{{\mathbb{C}}}^{\ast}\subset\mathcal{H}({{\mathbb{C}}}^{4})_{{\mathbb{C}}}$,
we see that it is also in $B_{0}({{\mathbb{C}}}^{4})_{{\mathbb{C}}}^{\ast}$
and $[{\mathfrak{d}}_{0}B_{0}({{\mathbb{C}}}^{4})]_{{\mathbb{C}}}^{\ast}$.
Thus, $(\cdot,\chi_{w})_{\sharp}$ is a complex random variable.

Define

$\tilde{\xi}_{ab}(w):=(-1)^{ab}\chi_{w}\otimes dx^{a}\wedge dx^{b},$

and $\tilde{\xi}_{ab,\gamma}(w):=(-1)^{ab}\tilde{\xi}_{ab}(w)\otimes
E^{\gamma}$.

For $A=\sum_{i=1}^{3}A_{i}\otimes dx^{i}$, this linear functional
$(\cdot,\tilde{\xi}_{ab}(w))_{\sharp}$ sends ${\mathfrak{d}}A$ to

$(-1)^{ab}[{\mathfrak{d}}_{a}A_{b}-{\mathfrak{d}}_{b}A_{a}](w)=\langle{\mathfrak{d}}A(w),(-1)^{ab}dx^{a}\wedge
dx^{b}\rangle=({\mathfrak{d}}A,\tilde{\xi}_{ab}(w))_{\sharp},$

$A_{0}\equiv 0$. Refer to [1].

And how are we going to make sense of $A_{i}(w)$? In [1], we showed that the
map ${\mathfrak{d}}_{0}:B_{0}({{\mathbb{C}}}^{4})\rightarrow
B({{\mathbb{C}}}^{4})$ is a bounded and injective map.

###### Proposition 2.6

For each $w\in{{\mathbb{C}}}^{4}$, we can define a bounded linear functional
$f_{w}:{\mathfrak{d}}_{0}B_{0}({{\mathbb{C}}}^{4})\rightarrow{{\mathbb{C}}}$,

$f_{w}({\mathfrak{d}}_{0}x):=(x,\chi_{w})_{\sharp},$

and thus there is a
$\zeta(w)\in[{\mathfrak{d}}_{0}B_{0}({{\mathbb{C}}}^{4})]_{{\mathbb{C}}}^{\ast}\subset\mathcal{H}({{\mathbb{C}}}^{4})_{{\mathbb{C}}}$
such that

$f_{w}({\mathfrak{d}}_{0}x)=({\mathfrak{d}}_{0}x,\zeta(w))_{\sharp}=(x,\chi_{w})_{\sharp}=x(w).$

Proof. Note that
$(\cdot,\chi_{w})_{\sharp}:(B({{\mathbb{C}}}^{4}),|\cdot|)\rightarrow{{\mathbb{C}}}$
is a bounded linear functional. For any
$y\in{\mathfrak{d}}_{0}B_{0}({{\mathbb{C}}}^{4})\subset
B({{\mathbb{C}}}^{4})$, we can find an unique $x\in B_{0}({{\mathbb{C}}}^{4})$
such that $y={\mathfrak{d}}_{0}x$. Hence,

$\displaystyle|f_{w}(y)|=$
$\displaystyle|f_{w}({\mathfrak{d}}_{0}x)|=|x(w)|=|(x,\chi_{w})_{\sharp}|$
$\displaystyle\leq$ $\displaystyle C|x|\leq C\parallel x\parallel.$

This shows that $f_{w}$ is a bounded linear functional on
${\mathfrak{d}}_{0}B_{0}({{\mathbb{C}}}^{4})$.

###### Definition 2.7

Write $\zeta_{i}(w):=\zeta(w)\otimes(dx^{0}\wedge dx^{i})$, $i=1,2,3$. From
the previous proposition, $\zeta_{i}(w)$ lies inside
${\mathbb{B}}_{{\mathbb{C}}}^{\ast}$. Note that
$(\cdot,\zeta_{i}(w))_{\sharp}$ is a complex random variable on the Banach
space and for $A=\sum_{i=1}^{3}A_{i}\otimes dx^{i}\in
H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})$,

$\left(\cdot,\zeta_{i}(w)\right)_{\sharp}:\ {\mathfrak{d}}A\longmapsto
A_{i}(w)\equiv\left({\mathfrak{d}}_{0}A_{i},\zeta(w)\right)_{\sharp}.$

For $w\in{{\mathbb{C}}}^{4}$, we will write
$A_{i}(w)=({\mathfrak{d}}A,\zeta_{i}(w))_{\sharp}$. Observe that we can now
write the product

$[A_{i}A_{j}](w)=A_{i}(w)A_{j}(w)=\left({\mathfrak{d}}A\otimes{\mathfrak{d}}A,\zeta_{i}(w)\otimes\zeta_{j}(w)\right)_{\sharp}.$

And, the complex conjugate

$\overline{({\mathfrak{d}}A,\zeta_{i}(w))_{\sharp}}=\overline{A_{i}(w)}=A_{i}(\bar{w})=({\mathfrak{d}}A,\zeta_{i}(\bar{w}))_{\sharp}.$
(2.6)

###### Remark 2.8

Note that $A_{i}\in B_{0}({{\mathbb{C}}}^{4})$, which consists of holomorphic
functions over on ${{\mathbb{C}}}^{4}$, with real coefficients. Hence, the
product $A_{i}A_{j}$ is actually well defined and Equation (2.6) holds.

###### Notation 2.9

To ease our notations, we will write for $j=1,2,3$,
$\zeta_{j,w}\equiv\zeta_{j}(w)$ and
$\tilde{\pi}_{i,\alpha,w}=\tilde{\pi}_{i,\alpha}(w)=\zeta_{i}(w)\otimes
E^{\alpha}$, $\\{E^{\alpha}\\}_{\alpha=1}^{N}$ is an orthonormal basis in
$\mathfrak{g}$. So

$\left({\mathfrak{d}}A,\tilde{\pi}_{i,\alpha,w}\right)_{\sharp}=\left({\mathfrak{d}}_{0}A_{i,\alpha},\zeta(w)\right)_{\sharp}.$

Note that $A=\sum_{\alpha=1}^{N}\sum_{i=1}^{3}A_{i,\alpha}\otimes
dx^{i}\otimes E^{\alpha}\in
B_{0}({{\mathbb{C}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}$.

We can now explain how to define the quartic term, given by
$A_{i,\alpha}A_{j,\beta}\overline{A_{i,\hat{\alpha}}}\overline{A_{j,\hat{\beta}}}$,
whereby $A_{i,\alpha}$ is a holomorphic function over ${{\mathbb{C}}}^{4}$ and
$\overline{A_{i,\alpha}}$ denotes its complex conjugate. The trick is to write

$\displaystyle[A_{i,\alpha}$ $\displaystyle
A_{j,\beta}\overline{A_{i,\hat{\alpha}}}\overline{A_{j,\hat{\beta}}}](w)$
$\displaystyle=\langle{\mathfrak{d}}A,\zeta_{i}(w)\otimes E^{\alpha}\rangle\
\langle{\mathfrak{d}}A,\zeta_{j}(w)\otimes E^{\beta}\rangle\
\overline{\langle{\mathfrak{d}}A,\zeta_{i}(w)\otimes E^{\hat{\alpha}}\rangle}\
\overline{\langle{\mathfrak{d}}A,\zeta_{j}(w)\otimes E^{\hat{\beta}}\rangle}$
$\displaystyle=:\left\langle{\mathfrak{d}}A\otimes{\mathfrak{d}}A\otimes{\mathfrak{d}}A\otimes{\mathfrak{d}}A,\tilde{\pi}_{i,\alpha,w}\otimes\tilde{\pi}_{j,\beta,w}\otimes\tilde{\pi}_{i,\hat{\alpha},\bar{w}}\otimes\tilde{\pi}_{j,\hat{\beta},\bar{w}}\right\rangle,$

for

$\displaystyle{\mathfrak{d}}A=$
$\displaystyle{\mathfrak{d}}\sum_{i=1}^{3}\sum_{\alpha=1}^{N}A_{i,\alpha}\otimes
dx^{i}\otimes E^{\alpha},$ $\displaystyle=$
$\displaystyle\sum_{\alpha=1}^{N}\left[\sum_{i=1}^{3}{\mathfrak{d}}_{0}A_{i,\alpha}\otimes
dx^{0}\wedge dx^{i}+\sum_{1\leq i<j\leq
3}[{\mathfrak{d}}_{i}A_{j,\alpha}-{\mathfrak{d}}_{j}A_{i,\alpha}]\otimes
dx^{i}\wedge dx^{j}\right]\otimes E^{\alpha}.$ (2.7)

Therefore,

$\displaystyle\int_{w\in{{\mathbb{C}}}^{4}}$
$\displaystyle[A_{i,\alpha}A_{j,\beta}\overline{A_{i,\hat{\alpha}}A_{j,\hat{\beta}}}](w)d\lambda_{4}(w)$
$\displaystyle=\int_{w\in{{\mathbb{C}}}^{4}}\
\left\langle{\mathfrak{d}}A^{\otimes^{4}},\tilde{\pi}_{i,\alpha,w}\otimes\tilde{\pi}_{j,\beta,w}\otimes\tilde{\pi}_{i,\hat{\alpha},\bar{w}}\otimes\tilde{\pi}_{j,\hat{\beta},\bar{w}}\right\rangle\
d\lambda_{4}(w)$
$\displaystyle=\left\langle{\mathfrak{d}}A^{\otimes^{4}},\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\
\tilde{\pi}_{i,\alpha,w}\otimes\tilde{\pi}_{j,\beta,w}\otimes\tilde{\pi}_{i,\hat{\alpha},\bar{w}}\otimes\tilde{\pi}_{j,\hat{\beta},\bar{w}}\right\rangle.$

This means that we interpret the quartic term in the Yang-Mills integral, as a
tensor product of functionals. A similar approach will be used to define the
cubic term.

For $A=\sum_{\alpha=1}^{N}\sum_{i=1}^{3}A_{i,\alpha}\otimes dx^{i}\otimes
E^{\alpha}\in
B_{0}({{\mathbb{C}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}$,
we will interpret $\int_{{{\mathbb{C}}}^{4}}d\lambda_{4}|A\wedge A|^{2}$ as

$\displaystyle\sum_{\gamma=1}^{N}$ $\displaystyle\sum_{1\leq i<j\leq
3}\sum_{\alpha,\beta\atop\hat{\alpha},\hat{\beta}}c_{\gamma}^{\alpha\beta}c_{\gamma}^{\hat{\alpha}\hat{\beta}}\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\
\bigg{[}\left({\mathfrak{d}}A^{\otimes^{4}},\tilde{\pi}_{i,\alpha}\otimes\tilde{\pi}_{j,\beta}\otimes\tilde{\pi}_{i,\hat{\alpha}}\otimes\tilde{\pi}_{j,\hat{\beta}}(w)\right)_{\sharp,34}\bigg{]}$
$\displaystyle\equiv$ $\displaystyle\sum_{\gamma=1}^{N}\sum_{1\leq i<j\leq
3}\sum_{\alpha,\beta\atop\hat{\alpha},\hat{\beta}}c_{\gamma}^{\alpha\beta}c_{\gamma}^{\hat{\alpha}\hat{\beta}}\bigg{[}\left({\mathfrak{d}}A^{\otimes^{4}},\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\tilde{\pi}_{i,\alpha}\otimes\tilde{\pi}_{j,\beta}\otimes\tilde{\pi}_{i,\hat{\alpha}}\otimes\tilde{\pi}_{j,\hat{\beta}}(w)\right)_{\sharp,34}\bigg{]},$
(2.8)

where the complex-valued random variable on
$\left({\mathbb{B}}\otimes\mathfrak{g},\tilde{\mu}_{\kappa^{2}}^{\times^{N}}\right)$
is defined as

$\displaystyle\left(B^{\otimes^{4}},\tilde{\pi}_{i_{1},\alpha_{1}}\otimes\tilde{\pi}_{i_{2},\alpha_{2}}\otimes\tilde{\pi}_{i_{3},\alpha_{3}}\otimes\tilde{\pi}_{i_{4},\alpha_{4}}(w)\right)_{\sharp,34}:=$
$\displaystyle\prod_{j=1}^{2}\left(B,\tilde{\pi}_{i_{j},\alpha_{j},w}\right)_{\sharp}\times\prod_{j=3}^{4}\left(B,\tilde{\pi}_{i_{j},\alpha_{j},\bar{w}}\right)_{\sharp}.$

Similarly, we interpret
$\int_{{{\mathbb{C}}}^{4}}d\lambda_{4}\left\langle\kappa{\mathfrak{d}}A,A\wedge
A\right\rangle$ and $\int_{{{\mathbb{C}}}^{4}}d\lambda_{4}\left\langle A\wedge
A,\kappa{\mathfrak{d}}A\right\rangle$ as

$\displaystyle\begin{split}\sum_{\gamma=1}^{N}&\sum_{1\leq i<j\leq
3}\sum_{\alpha,\beta}c_{\gamma}^{\alpha\beta}\left({\mathfrak{d}}A^{\otimes^{3}},\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\kappa\tilde{\xi}_{ij,\gamma}\otimes\tilde{\pi}_{i,\alpha}\otimes\tilde{\pi}_{j,\beta}(w)\right)_{\sharp,23},\\\
\sum_{\gamma=1}^{N}&\sum_{1\leq i<j\leq
3}\sum_{\alpha,\beta}c_{\gamma}^{\alpha\beta}\left({\mathfrak{d}}A^{\otimes^{3}},\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\
\tilde{\pi}_{i,\alpha}\otimes\tilde{\pi}_{j,\beta}\otimes\kappa\tilde{\xi}_{ij,\gamma}(w)\right)_{\sharp,3},\end{split}$
(2.9)

respectively, where the respective complex-valued random variables on
$\left({\mathbb{B}}\otimes\mathfrak{g},\tilde{\mu}_{\kappa^{2}}^{\times^{N}}\right)$
are defined as

$\displaystyle\left(B^{\otimes^{3}},\kappa\tilde{\xi}_{ij,\alpha_{1}}\otimes\tilde{\pi}_{i_{2},\alpha_{2}}\otimes\tilde{\pi}_{i_{3},\alpha_{3}}(w)\right)_{\sharp,23}:=$
$\displaystyle\kappa\left(B,\tilde{\xi}_{ij}(w)\otimes
E^{\alpha_{1}}\right)_{\sharp}\times\prod_{j=2}^{3}\left(B,\tilde{\pi}_{i_{j},\alpha_{j},\bar{w}}\right)_{\sharp},$
$\displaystyle\left(B^{\otimes^{3}},\tilde{\pi}_{i_{1},\alpha_{1}}\otimes\tilde{\pi}_{i_{2},\alpha_{2}}\otimes\kappa\tilde{\xi}_{ij,\alpha_{3}}(w)\right)_{\sharp,3}:=$
$\displaystyle\prod_{j=1}^{2}\left(B,\tilde{\pi}_{i_{j},\alpha_{j},w}\right)_{\sharp}\times\kappa\left(B,\tilde{\xi}_{ij}(\bar{w})\otimes
E^{\alpha_{3}}\right)_{\sharp}.$

###### Remark 2.10

For $B=\sum_{0\leq a<b\leq 3}B_{ab}\otimes dx^{a}\wedge
dx^{b}\in{\mathbb{B}}$, $B_{ab}$ is holomorphic. In fact,
$\left(B,\tilde{\xi}_{ij}(w)\right)_{\sharp}=(-1)^{ij}B_{ij}(w)$ and its
complex conjugate
$\overline{\left(B,\tilde{\xi}_{ij}(w)\right)_{\sharp}}=(-1)^{ij}B_{ij}(\bar{w})$.

But just like in the abelian case considered in [1], we do not consider
$\tilde{\xi}_{ab}(w)$ and $\tilde{\pi}_{i,\alpha}(w)$. Instead, we will
introduce a renormalization factor $\psi_{w}$, defined in Notation 2.1. That
is we replace $\tilde{\xi}_{ab}(w)$ and $\tilde{\pi}_{i,\alpha}(w)$ with

$\xi_{ab}(w)=\psi_{w}\tilde{\xi}_{ab}(w)\ \ {\rm and}\ \
\pi_{i,\alpha}(w)\equiv\pi_{i,\alpha,w}=\psi_{w}\tilde{\pi}_{i,\alpha,w}$

respectively.

###### Remark 2.11

  1. 1.

We explained in [1], the necessity of this renormalization factor $\psi_{w}$.

  2. 2.

A direct computation will show that
$\parallel\xi_{ab}(w)\parallel^{2}=\langle\xi_{ab}(w),\xi_{ab}(w)\rangle=1/2\pi$.

  3. 3.

In [1], we showed that $\langle f,f\rangle\leq
40\langle{\mathfrak{d}}_{0}f,{\mathfrak{d}}_{0}f\rangle$, $f\in
H^{2}({{\mathbb{C}}}^{4})$. By definition of $\zeta(w)$ in Proposition 2.6, we
see that

$\parallel\pi_{i,\alpha}(w)\parallel^{2}=\langle\pi_{i,\alpha}(w),\pi_{i,\alpha}(w)\rangle^{2}\leq\frac{40}{2\pi}.$

###### Definition 2.12

Apply the renormalization rule and add in the renormalization factor. For
$B\in{\mathbb{B}}\otimes\mathfrak{g}$, we will define

$\displaystyle Y_{1}^{\kappa}$ $\displaystyle\left(B\right)$
$\displaystyle:=\sum_{\gamma=1}^{N}\sum_{1\leq i<j\leq
3}\sum_{\alpha,\beta}c_{\gamma}^{\alpha\beta}\left(B^{\otimes^{3}},\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\
\kappa\xi_{ij,\gamma}\otimes\pi_{i,\alpha}\otimes\pi_{j,\beta}(w)\right)_{\sharp,23},$
$\displaystyle Y_{2}^{\kappa}$ $\displaystyle\left(B\right)$
$\displaystyle:=\sum_{\gamma=1}^{N}\sum_{1\leq i<j\leq
3}\sum_{\alpha,\beta}c_{\gamma}^{\alpha\beta}\left(B^{\otimes^{3}},\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\
\pi_{i,\alpha}\otimes\pi_{j,\beta}\otimes\kappa\xi_{ij,\gamma}(w)\right)_{\sharp,3},$

and

$\displaystyle Y_{3}^{\kappa}$ $\displaystyle\left(B\right)$
$\displaystyle:=\sum_{\gamma=1}^{N}\sum_{1\leq i<j\leq
3}\sum_{\alpha,\beta\atop\hat{\alpha},\hat{\beta}}c_{\gamma}^{\alpha\beta}c_{\gamma}^{\hat{\alpha}\hat{\beta}}\left(B^{\otimes^{4}},\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\
\pi_{i,\alpha}\otimes\pi_{j,\beta}\otimes\pi_{i,\hat{\alpha}}\otimes\pi_{j,\hat{\beta}}(w)\right)_{\sharp,34},$

all being random variables on
$\left({\mathbb{B}}\otimes\mathfrak{g},\tilde{\mu}_{\kappa^{2}}^{\times^{N}}\right)$,
using Expressions 2.8 and 2.9.

From Expression 2.5, we replace ${\mathfrak{d}}A$ with $B$, so we interpret
the expression over in ${\mathbb{B}}\otimes\mathfrak{g}$. Using the
construction of the Abstract Wiener space in [1], we will interpret

$D_{2}=\frac{1}{Z}e^{-\frac{1}{2}\int_{{{\mathbb{C}}}^{4}}d\lambda_{4}|\kappa{\mathfrak{d}}A|^{2}}D[{\mathfrak{d}}A]$

as product Wiener measure $\tilde{\mu}_{\kappa^{2}}^{\times^{N}}$ over
${\mathbb{B}}\otimes\mathfrak{g}$, for some normalization constant $Z$.

And we interpret

$D_{1}=\exp\left(-\frac{1}{2}\int_{{{\mathbb{C}}}^{4}}d\lambda_{4}\langle\kappa{\mathfrak{d}}A,A\wedge
A\rangle+\langle A\wedge A,\kappa{\mathfrak{d}}A\rangle+|A\wedge
A|^{2}\right)$

as

$\mathcal{Y}^{\kappa}\left(B\right):=\exp\left\\{-\frac{1}{2}\left[Y_{1}^{\kappa}\left(B\right)+Y_{2}^{\kappa}\left(B\right)+Y_{3}^{\kappa}\left(B\right)\right]\right\\},$
(2.10)

over in ${\mathbb{B}}\otimes\mathfrak{g}$.

In Lemma 3.1, we will show that it is integrable with respect to product
Wiener measure $\tilde{\mu}_{\kappa^{2}}^{\times^{N}}$.

##  3 Yang-Mills Path Integral

In the abelian case considered in [1], we defined a Wiener (probability)
measure $\tilde{\mu}_{\kappa^{2}}$, on a Banach space ${\mathbb{B}}$. We will
write ${{\mathbb{E}}}$ to denote expectation on this probability space
$({\mathbb{B}},\tilde{\mu}_{\kappa^{2}})$.

In the non-abelian case, we will now construct a measure on
${\mathbb{B}}\otimes\mathfrak{g}$, which is absolutely continuous with respect
to the product Wiener measure
$\tilde{\mu}_{\kappa^{2}}^{\times^{N}}\equiv\tilde{\mu}_{\kappa^{2}}\times\cdots\times\tilde{\mu}_{\kappa^{2}}$.
That is, we will define a sequence of Yang-Mills measures on

${\mathbb{B}}\otimes\mathfrak{g},\quad{\rm
as}\quad\left\\{\mathcal{Y}^{\kappa}d\tilde{\mu}_{\kappa^{2}}^{\times^{N}}:\
\kappa>0\right\\}.$

The sequence of Yang-Mills measures is indexed by $\kappa$, which is due to
the renormalization flow $\\{\Psi_{\kappa}:\ \kappa>0\\}$ defined in Item 2 in
Remark 2.3. We will continue to write ${{\mathbb{E}}}$ to denote expectation
on this probability space
$({\mathbb{B}}\otimes\mathfrak{g},\tilde{\mu}_{\kappa^{2}}^{\times^{N}})$.

###### Lemma 3.1

Refer to Definition 2.12 for $\mathcal{Y}^{\kappa}$. Consider the probability
space ${\mathbb{B}}\otimes\mathfrak{g}$ equipped with a product Wiener measure
$\tilde{\mu}_{\kappa^{2}}^{\times^{N}}$. For any $\kappa>0$,

$\mathbb{E}\left[\mathcal{Y}^{\kappa}\right]:=\int_{{\mathbb{B}}\otimes\mathfrak{g}}\mathcal{Y}^{\kappa}d\tilde{\mu}_{\kappa^{2}}^{\times^{N}}$

is finite.

Proof. Observe that
$Y_{1}^{\kappa}\left(B\right)+Y_{2}^{\kappa}\left(B\right)+Y_{3}^{\kappa}\left(B\right)$
is equal to

$\displaystyle\sum_{\gamma=1}^{N}\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\
\Bigg{[}\sum_{1\leq i<j\leq
3}\Big{|}\left(B,\kappa\xi_{ij,\gamma}(w)\right)_{\sharp}+\sum_{\alpha,\beta}c_{\gamma}^{\alpha\beta}$
$\displaystyle\left(B^{\otimes^{2}},\pi_{i,\alpha,w}\otimes\pi_{j,\beta,w}\right)_{\sharp}\Big{|}^{2}$
$\displaystyle-$ $\displaystyle\sum_{1\leq i<j\leq
3}\left|\left(B,\kappa\xi_{ij,\gamma}(w)\right)_{\sharp}\right|^{2}\Bigg{]}.$

Thus,

$\displaystyle\mathcal{Y}^{\kappa}\left(B\right)=$
$\displaystyle\exp\Bigg{[}-\frac{1}{2}\left[Y_{1}^{\kappa}\left(B\right)+Y_{2}^{\kappa}\left(B\right)+Y_{3}^{\kappa}\left(B\right)\right]\Bigg{]}$
$\displaystyle\leq$
$\displaystyle\exp\left[\frac{1}{2}\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\sum_{\alpha=1}^{N}\sum_{1\leq
i<j\leq
3}\left|\left(B,\kappa\xi_{ij,\alpha}(w)\right)_{\sharp}\right|^{2}\right].$

Thus it suffices to show

$\exp\left[\frac{1}{2}\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\left|\left(\cdot,\kappa\xi_{ij}(w)\right)_{\sharp}\right|^{2}\right]$

is integrable on ${\mathbb{B}}$.

Using Jensen’s inequality, we have

$\displaystyle\exp$
$\displaystyle\left[\frac{1}{2}\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\left|\left(\cdot,\kappa\xi_{ij}(w)\right)_{\sharp}\right|^{2}\right]$
$\displaystyle\leq\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\exp\left[\frac{1}{2}\left|\left(\cdot,\kappa\xi_{ij}(w)\right)_{\sharp}\right|^{2}\right].$

Now, we can write the complex-valued random variable

$\displaystyle\left(\cdot,\xi_{ij,\alpha}(w)\right)_{\sharp}=$
$\displaystyle\sqrt{-1}\left(\cdot,f\right)_{\sharp}+\left(\cdot,g\right)_{\sharp}$
$\displaystyle=$ $\displaystyle\left[\frac{\langle f,g\rangle}{\langle
f,f\rangle}+\sqrt{-1}\right]\left(\cdot,f\right)_{\sharp}+\left(\cdot,g-\frac{\langle
f,g\rangle}{\langle f,f\rangle}f\right)_{\sharp},$ (3.1)

whereby $f,g\in[{\mathbb{B}}\otimes\mathfrak{g}]^{\ast}$. Thus,
$\\{\left(\cdot,\xi_{ij,\alpha}(w)\right)_{\sharp}:\ 1\leq\alpha\leq N,1\leq
i<j\leq 3\\}$ is a set of i.i.d. complex random variables with mean 0, and
$|\left(\cdot,\xi_{ij,\alpha}(w)\right)_{\sharp}|^{2}$ is equal to
$|Z_{1}+\sqrt{-1}Z_{2}|^{2}$ in distribution, each $Z_{i}$ is a real normal
random variable with mean 0.

From Equation (3.1), write

$c=\frac{\langle f,g\rangle}{\langle f,f\rangle}+\sqrt{-1},\quad
W_{1}=\left(\cdot,f\right)_{\sharp},\quad W_{2}=\left(\cdot,g-\frac{\langle
f,g\rangle}{\langle f,f\rangle}f\right)_{\sharp}.$

Then $|(\cdot,\kappa\xi_{ij}(w))_{\sharp}|\leq|c||W_{1}|+|W_{2}|$, so by
Jenson’s inequality,

$|(\cdot,\kappa\xi_{ij}(w))_{\sharp}|^{2}\leq
4\left[\frac{1}{2}|c||W_{1}|+|W_{2}|\right]^{2}\leq
2[|c|^{2}|W_{1}|^{2}+|W_{2}|^{2}].$

From Remark 2.11,
${{\mathbb{E}}}[|(\cdot,\kappa\xi_{ij}(w))_{\sharp}|^{2}]=1/2\pi$. Because
$W_{1}$ is independent of $W_{2}$, note that
${{\mathbb{E}}}[|(\cdot,\kappa\xi_{ij}(w))_{\sharp}|^{2}]=|c|^{2}{{\mathbb{E}}}[|W_{1}|^{2}]+{{\mathbb{E}}}[|W_{2}|^{2}]$.
Hence, for some constants $a^{2}+b^{2}=1$, which depend on $w$, we have

$2\pi|(\cdot,\kappa\xi_{ij}(w))_{\sharp}|^{2}\leq
2[a^{2}\tilde{W}_{1}^{2}+b^{2}\tilde{W}_{2}^{2}],$

whereby for each $i=1,2$,

  * •

$\tilde{W}_{i}$ is a (real) random variable on
$\left({\mathbb{B}},\tilde{\mu}_{\kappa^{2}}\right)$,

  * •

$\tilde{W}_{i}$ is equal in distribution to a standard normal $Z_{i}$ and

  * •

$Z_{1}$ is independent of $Z_{2}$.

Using Fubini’s Theorem,

$\displaystyle\mathbb{E}$
$\displaystyle\left[\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\exp\left[\frac{1}{2}\left|\left(\cdot,\kappa\xi_{ij}(w)\right)_{\sharp}\right|^{2}\right]\right]$
$\displaystyle=\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\mathbb{E}\left[\exp\left[\frac{1}{2}\left|\left(\cdot,\kappa\xi_{ij}(w)\right)_{\sharp}\right|^{2}\right]\right]$
$\displaystyle\leq\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\mathbb{E}\left[\exp\left[\frac{1}{2\pi}[a^{2}\tilde{W}_{1}^{2}+b^{2}\tilde{W}_{2}^{2}]\right]\right]\leq\mathbb{E}\left[\exp\left[\frac{1}{2\pi}[Z_{1}^{2}+Z_{2}^{2}]\right]\right]$
$\displaystyle\leq\frac{1}{2\pi}\int_{x\in{{\mathbb{R}}}}e^{x^{2}/2\pi}e^{-x^{2}/2}dx\cdot\int_{y\in{{\mathbb{R}}}}e^{y^{2}/2\pi}e^{-y^{2}/2}dy$
$\displaystyle=\left[\frac{1}{\sqrt{1-(1/\pi)}}\right]^{2}<\infty.$

By modifying the above proof, we can actually prove the following result.

###### Corollary 3.2

For any $1\leq p<\pi$, we have

$\mathbb{E}\left([\mathcal{Y}^{\kappa}]^{p}\right)\leq\left(\frac{1}{\sqrt{1-p/\pi}}\right)^{2N}.$
(3.2)

###### Remark 3.3

If we replace the factor $1/\sqrt{2\pi}$ in $\psi_{w}$ with any
$0<\tilde{c}<1/\sqrt{2}$, then the corollary holds true for any $1\leq
p<1/2\tilde{c}^{2}$.

###### Notation 3.4

(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 $\rho(\mathfrak{g})$ is a set
containing skew-Hermitian matrices, we define $C(\rho)\in{{\mathbb{R}}}$ such
that

${{\rm{Tr}}}[\rho(E^{\alpha})\rho(E^{\beta})]=C(\rho){{\rm{Tr}}}[E^{\alpha}E^{\beta}].$
(3.3)

Also define

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

to be its (quadratic) Casimir operator.

###### Definition 3.5

Refer to Definitions 1.5 and A.1. Let
$\sigma:I^{2}\rightarrow{{\mathbb{R}}}^{4}$ be any parametrization of
$R[a,T]$. In terms of this parametrization, we can write

$\displaystyle c\int_{R[a,T]}{\mathfrak{d}}A=$ $\displaystyle
c\sum_{\alpha=1}^{N}\int_{I^{2}}d\hat{s}\
\sum_{j=1}^{3}|J_{0j}^{\sigma}|(\hat{s})[{\mathfrak{d}}_{0}A_{j,\alpha}](\sigma(\hat{s}))\otimes
E^{\alpha}$ $\displaystyle=$ $\displaystyle
c\sum_{\alpha=1}^{N}\int_{I^{2}}d\hat{s}\
\sum_{j=1}^{3}|J_{0j}^{\sigma}|(\hat{s})\left({\mathfrak{d}}A,\tilde{\xi}_{0j}(\sigma(\hat{s}))\otimes
E^{\alpha}\right)_{\sharp}\otimes E^{\alpha}.$ (3.4)

Besides the renormalization rule mentioned in Remark 2.3 and adding a
renormalization factor $\psi_{w}$ as discussed in Remark 2.11, we also need a
renormalization transformation as discussed in [1]. We embed
${{\mathbb{R}}}^{4}$ inside ${{\mathbb{C}}}^{4}$, scaled by a factor
$\kappa/2$. Thus, a surface $R[a,T]\subset{{\mathbb{R}}}^{4}$ will be embedded
inside ${{\mathbb{C}}}^{4}$, but scaled with a factor $\kappa/2$.

Finally, we also need to introduce asymptotic freedom, whereby the coupling
constant $c$ goes down to zero, as $\kappa$ goes to infinity. Specifically, we
will define $c=1/\kappa$.

###### Remark 3.6

Because of renormalization, note that we have to make the following changes to
the RHS of Equation 3.4.

  * •

By introducing the renormalization factor $\psi_{w}$, we replace
$\tilde{\xi}_{ab}$ in RHS of Equation (3.4) with
$\xi_{ab}=\psi\cdot\tilde{\xi}_{ab}$ .

  * •

By the renormalization rule
${\mathfrak{d}}_{a}\mapsto\kappa{\mathfrak{d}}_{a}$, we have to add an extra
$\kappa$ in front of $\xi_{0i}$ in RHS of Equation (3.4).

  * •

Because of the renormalization transformation, we have to scale
$\sigma(s,t)$by a factor $\kappa/2$. Hence we have to replace

$\sigma(s,t)\mapsto\kappa\sigma(s,t)/2.$

As a consequence, we need to add an extra $\kappa^{2}/4$ to the surface
integral in RHS of Equation (3.4).

Let $c=1/\kappa$. The idea that the coupling constant goes to 0 as $\kappa$
goes to infinity, is known as asymptotic freedom. See [1] and [10].

###### Definition 3.7

Suppose $\rho:\mathfrak{g}\rightarrow{\rm End}({{\mathbb{C}}}^{\tilde{N}})$ is
an irreducible representation of a semi-simple Lie Algebra $\mathfrak{g}$, and
$\rho(\mathfrak{g})$ consists of skew-Hermitian matrices.

Replace the coupling constant $c$ with $1/\kappa$. After applying the
renormalization rule, transformation and adding a factor to Equation (3.4),
define

$\displaystyle\nu_{R[a,T]}^{\kappa,\alpha}$
$\displaystyle:=\frac{1}{\kappa}\int_{I^{2}}d\hat{s}\
\frac{\kappa^{2}}{4}\sum_{j=1}^{3}|J_{0j}^{\sigma}|(\hat{s})\kappa\xi_{0j}(\kappa\sigma(\hat{s})/2)\otimes
E^{\alpha},$
$\displaystyle\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}$
$\displaystyle:=\sum_{\alpha=1}^{N}\left(\cdot,\nu_{R[a,T]}^{\kappa,\alpha}\right)_{\sharp}\otimes\rho(E^{\alpha}),\quad\left(\cdot,\nu_{R[a,T]}^{\kappa}\right)_{\sharp}:=\sum_{\alpha=1}^{N}\left(\cdot,\nu_{R[a,T]}^{\kappa,\alpha}\right)_{\sharp}\otimes
E^{\alpha},$

and
$\mathcal{J}_{R[a,T]}^{\kappa,\rho}:=\exp\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}\right]$.

###### Remark 3.8

By abuse of notation, for $B\in{\mathbb{B}}\otimes\mathfrak{g}\subset
B({{\mathbb{C}}}^{4})\otimes\Lambda^{2}({{\mathbb{R}}}^{4})\otimes\mathfrak{g}$,

$\left(B,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}:=\frac{1}{\kappa}\sum_{\alpha=1}^{N}\int_{I^{2}}d\hat{s}\
\frac{\kappa^{2}}{4}\sum_{j=1}^{3}|J_{0j}^{\sigma}|(\hat{s})\left(B,\kappa\xi_{0j}(\kappa\sigma(\hat{s})/2)\otimes
E^{\alpha}\right)_{\sharp}\otimes\rho(E^{\alpha}),$

so $\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}$ and
$\left(\cdot,\nu_{R[a,T]}^{\kappa}\right)_{\sharp}$ are actually
$\rho(\mathfrak{g})$-valued and $\mathfrak{g}$-valued random variables
respectively. Compare with Equation (3.4).

###### Proposition 3.9

Let $|\cdot|$ be the spectral norm of a matrix. We have

${{\mathbb{E}}}\left[\left|\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}\right|^{4}\right]\leq
C(\rho)^{2}N^{4}C,$ (3.5)

for some constant $C$ independent of $\kappa$.

Proof. Note that on the probability space
$({\mathbb{B}}\otimes\mathfrak{g},\tilde{\mu}_{\kappa^{2}}^{\times^{N}})$,

$\frac{1}{\parallel\nu_{R[a,T]}^{\kappa,\alpha}\parallel}\left(\cdot,\kappa\nu_{R[a,T]}^{\kappa,\alpha}\right)_{\sharp}$

has mean 0, and its norm squared is equal to $Z^{2}$ in distribution, $Z$ is a
standard normal random variable.

Using Jensen’s inequality,

$\left|\sum_{\alpha=1}^{N}f_{a}\right|^{4}=N^{4}\left|\frac{1}{N}\sum_{\alpha=1}^{N}f_{a}\right|^{4}\leq
N^{3}\sum_{\alpha=1}^{N}|f_{a}|^{4}.$

And from Equation (3.3),

$|\rho(E^{\alpha})\rho(E^{\alpha})|\leq C(\rho).$

Thus,

$\displaystyle{{\mathbb{E}}}\left[\left|\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}\right|^{4}\right]\leq$
$\displaystyle
C(\rho)^{2}N^{3}\sum_{\alpha=1}^{N}{{\mathbb{E}}}\left[\left|\left(\cdot,\nu_{R[a,T]}^{\kappa,\alpha}\right)_{\sharp}\right|^{4}\right]$
$\displaystyle=$ $\displaystyle
3C(\rho)^{2}N^{3}\sum_{\alpha=1}^{N}\left[\frac{1}{\kappa^{2}}\left\langle\nu_{R[a,T]}^{\kappa,\alpha},\nu_{R[a,T]}^{\kappa,\alpha}\right\rangle\right]^{2}.$

Using the calculations from the main theorem in [1], we have

$\displaystyle\frac{1}{\kappa^{2}}\left\langle\nu_{R[a,T]}^{\kappa,\alpha},\nu_{R[a,T]}^{\kappa,\alpha}\right\rangle\longrightarrow\frac{1}{4}\sum_{1\leq
i\leq 3}\int_{I^{2}}dsdt\
\rho_{\sigma}^{0i}(s,t)|J_{0i}^{\sigma}|(s,t)=\frac{|a|T}{4}.$ (3.6)

Hence, we see that
${{\mathbb{E}}}\left[\left|\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}\right|^{4}\right]$
can be bounded by a constant independent of $\kappa$, multiplied by
$N^{4}C(\rho)^{2}$.

###### Definition 3.10

(Definition of the Yang-Mills Path integral.)
Let $R[a,T]$ be a compact rectangular surface described in Definition 1.5. Let
$\rho$ be an irreducible representation of $\mathfrak{g}$. Consider the
probability space ${\mathbb{B}}\otimes\mathfrak{g}$ equipped with product
Wiener measure $\tilde{\mu}_{\kappa^{2}}^{\times^{N}}$.

From Expression 1.8, we define a sequence of Yang-Mills Path integrals using
the renormalization flow

$\left\\{\Psi_{\kappa}:\mathcal{S}_{\kappa}({{\mathbb{R}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}\longrightarrow
H^{2}({{\mathbb{C}}}^{4})\otimes\Lambda^{1}({{\mathbb{R}}}^{3})\otimes\mathfrak{g}\
|\ \kappa>0\right\\},$

as

$\displaystyle\left\\{{{\mathbb{E}}}_{{\rm
YM}}^{\kappa}\left[\exp\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}\right]\right]:=\frac{1}{\mathbb{E}\left[\mathcal{Y}^{\kappa}\right]}\mathbb{E}\left[\mathcal{J}_{R[a,T]}^{\kappa,\rho}\cdot\mathcal{Y}^{\kappa}\right]:\
\kappa>0\right\\},$

where $\mathcal{Y}^{\kappa}$ was defined in Definition 2.12.

We will also write

${{\mathbb{E}}}_{{\rm YM}}^{\kappa}\left[{{\rm{Tr}}}\
\exp\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}\right]\right]\equiv\frac{1}{\mathbb{E}\left[\mathcal{Y}^{\kappa}\right]}\mathbb{E}\left[{{\rm{Tr}}}\
\exp\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}\right]\cdot\mathcal{Y}^{\kappa}\right].$

Here, it is understood that the expectation is taken on the probability space
$({\mathbb{B}}\otimes\mathfrak{g},\tilde{\mu}_{\kappa^{2}}^{\times^{N}})$.

##  4 Wilson Area Law Formula

We will now show how one can obtain the Area Law Formula for a compact semi-
simple gauge group, using our definition of the Yang-Mills Path integral,
given in Definition 3.10.

On a finite lattice gauge with spacing $\epsilon$, one will define a finite
dimensional integral, dependent on $\epsilon$, synonymous with $1/\kappa$. The
Yang-Mills path integral is then defined by taking the limit as $\epsilon$
goes down to 0. The reader may refer to [11], or other references on lattice
gauge approximation in [1]. On a lattice gauge, Gross in [10] explained that
the coupling constant $c(\epsilon)$ should depend on $\epsilon$, and as
$\epsilon$ goes down to 0, $c(\epsilon)$ should go down to 0. This idea is
also known as asymptotic freedom.

We constructed a sequence of functionals
$\\{\mathcal{Y}^{\kappa}\\}_{\kappa>0}$ and
$\\{\mathcal{J}_{R[a,T]}^{\kappa,\rho}\\}_{\kappa>0}$, each defined on a
Banach space ${\mathbb{B}}\otimes\mathfrak{g}$, equipped with product Wiener
measure $\tilde{\mu}_{\kappa^{2}}^{\times^{N}}$ with variance $1/\kappa^{2}$.
For each $\kappa>0$, the path integral defined in Definition 3.10 is analogous
to a finite dimensional integral on a finite lattice gauge with spacing
$\epsilon$. It remains to take $\kappa$ going to infinity.

When the gauge group is non-abelian, asymptotic freedom applies. See [9]. In
Definition 3.7, we set the coupling constant $c=1/\kappa$. This is crucial as
it allows us to obtain the Area Law formula. (See Remark 4.5.) Without
asymptotic freedom, the Area Law formula does not hold. Refer to [1] for the
case of abelian gauge group.

When $\xi_{ab}(w)$ and $\zeta_{i}(w)$ are complex-valued random variables with
mean zero on ${\mathbb{B}}$, with probability measure
$\tilde{\mu}_{\kappa^{2}}$, we have

$\left|\left(\cdot,\frac{\kappa\xi_{ab}(w)}{\parallel\xi_{ab}(w)\parallel}\right)_{\sharp}\right|^{2},\quad\left|\left(\cdot,\frac{\kappa\zeta_{i}(w)}{\parallel\zeta_{i}(w)\parallel}\right)_{\sharp}\right|^{2},$

both are equal to $|Z_{1}+\sqrt{-1}Z_{2}|^{2}$ in distribution, each $Z_{i}$
is a (real) normal random variable. See Equation (3.1).

Both $\parallel\xi_{ab}(w)\parallel$ and $\parallel\zeta_{i}(w)\parallel$ are
less than or equal to $\sqrt{40/2\pi}$, from Remark 2.11. Hence, we see that

${{\mathbb{E}}}[|Y_{1}^{\kappa}|^{2l}]=O(1/\kappa^{4l}),\quad{{\mathbb{E}}}[|Y_{2}^{\kappa}|^{2l}]=O(1/\kappa^{4l}),\quad{{\mathbb{E}}}[|Y_{3}^{\kappa}|^{n}]=O(1/\kappa^{4n}),$
(4.1)

and when $i=1,2$, ${{\mathbb{E}}}[Y_{i}^{\kappa,2l+1}]=0$, for
$l,n\in\mathbb{N}$. This is because the expectation, of an odd product of
Gaussian random variables with mean 0, is zero.

###### Lemma 4.1

There exists a $\kappa_{0}>0$, independent of $\rho$, such that for all
$\kappa>\kappa_{0}$, we have

${{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}(\mathcal{Y}^{\kappa}-1)\right]>0,$

i.e. it is positive definite.

Here, the expectation is taken with respect to the Banach space
${\mathbb{B}}\otimes\mathfrak{g}$ equipped with product Wiener measure
$\tilde{\mu}_{\kappa^{2}}^{\times^{N}}$. In fact,

$\left|{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}(\mathcal{Y}^{\kappa}-1)\right]\right|=O(C(\rho)/\kappa^{4}),$

whereby $|\cdot|$ is the trace norm of a matrix.

Proof. Note that because $\rho(\mathfrak{g})$ consists of skew-Hermitian
matrices, the square $\left(B,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}$
is non-positive definite.

Refer to Definition 2.12. From Equation (4.1), the important terms we need to
consider are $Y_{3}^{\kappa}$ and the squares of $Y_{1}^{\kappa}$ and
$Y_{2}^{\kappa}$. First, we claim that
${{\mathbb{E}}}\left[-\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}Y_{3}^{\kappa}\right]$
is positive definite.

From Equation (4.7), we see that

${{\mathbb{E}}}\left[-\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}\right]=\mathscr{E}(\rho)+O\left(\tilde{C}e^{-C\kappa}\right),$
(4.2)

for positive constants $\tilde{C}$ and $C$.

Consider the random variable

$A\in\left({\mathbb{B}}\otimes\mathfrak{g},\tilde{\mu}_{\kappa^{2}}^{\times^{N}}\right)\longmapsto\inf_{\parallel
v\parallel=1}\left\langle-\left(A,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}v,v\right\rangle.$

The RHS of the mapping computes the smallest eigenvalue of
$-\left(A,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}$. The smallest
eigenvalue is zero iff the determinant is zero. For each $\alpha=1,\cdots,N$,
$\left(\cdot,\nu_{R[a,T]}^{\kappa,\alpha}\right)_{\sharp}$ follows a Gaussian
distribution. Hence, $\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}$
is equal in distribution to
$\sum_{\alpha=1}^{N}\mathcal{N}_{\alpha}\rho(E^{\alpha})$, whereby the set
$\\{\mathcal{N}_{\alpha}:\ 1\leq\alpha\leq N\\}$ is a set of i.i.d. real-
valued Gaussian random variables.

Note that
$\det\left[\sum_{\alpha=1}^{N}\mathcal{N}_{\alpha}\rho(E^{\alpha})\right]=0$
defines a polynomial equation in the Gaussian variables $N_{\alpha}$’s.
Because $\mathfrak{g}$ is semi-simple and $\rho$ is its irreducible
representation, we see that the Casimir operator $\mathscr{E}(\rho)$ in
Equation (4.2) has a non-zero determinant. Hence this polynomial is not the
zero function, unless all the $\mathcal{N}_{\alpha}$’s are all zero.

Thus, $\left\\{(x_{1},\cdots,x_{N})\in{{\mathbb{R}}}^{N}:\
\det\left[\sum_{\alpha=1}^{N}x_{\alpha}\rho(E^{\alpha})\right]=0\right\\}$ has
Lebesgue measure zero. Therefore,
$\det\left[\sum_{\alpha=1}^{N}\mathcal{N}_{\alpha}\rho(E^{\alpha})\right]\neq
0$ with probability 1, which implies that $\inf_{\parallel
v\parallel=1}\left\langle-\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}v,v\right\rangle$
is non-zero with probability 1.

This will also imply that

$V:={{\mathbb{E}}}\left[\inf_{\parallel
v\parallel=1}\left\langle-\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}v,v\right\rangle\cdot
Y_{3}^{\kappa}\right],$

is also greater than 0, since the structure constants are non-zero and
$Y_{3}^{\kappa}>0$. Furthermore, for any unit vector $w$, we have

$\left\langle{{\mathbb{E}}}\left[-\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}Y_{3}^{\kappa}\right]w,w\right\rangle={{\mathbb{E}}}\left[\left\langle-\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}Y_{3}^{\kappa}w,w\right\rangle\right]\geq
V.$

Thus,

$\inf_{\parallel
w\parallel=1}\left\langle{{\mathbb{E}}}\left[-\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}Y_{3}^{\kappa}\right]w,w\right\rangle\geq
V>0,$

which proves the claim.

The complex random variable $(\cdot,\pi_{i,\alpha}(w))_{\sharp}$ is
independent of $(\cdot,\pi_{i,\beta}(w))_{\sharp}$, if $\alpha\neq\beta$. And

${{\mathbb{E}}}\left[(\cdot,\pi_{i,\alpha}(w))_{\sharp}\overline{(\cdot,\pi_{j,\alpha}(w))_{\sharp}}\right]={{\mathbb{E}}}\left[(\cdot,\pi_{i,\alpha}(w))_{\sharp}(\cdot,\pi_{j,\alpha}(\bar{w}))_{\sharp}\right]=\frac{1}{\kappa^{2}}\langle\pi_{i,\alpha}(w),\pi_{j,\alpha}(w)\rangle,$

so $(\cdot,\pi_{i,\alpha}(w))_{\sharp}$ has variance $O(1/\kappa^{2})$. Note
that $\langle\cdot,\cdot\rangle$ is a sesquilinear complex inner product.

On the probability space
$({\mathbb{B}}\otimes\mathfrak{g},\tilde{\mu}_{\kappa^{2}}^{\times^{N}})$, we
have

$\displaystyle\int_{{\mathbb{B}}\otimes\mathfrak{g}}\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}Y_{3}^{\kappa}\
d\tilde{\mu}_{\kappa^{2}}^{\times^{N}}\equiv$
$\displaystyle{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}Y_{3}^{\kappa}\right]$
$\displaystyle=$ $\displaystyle\sum_{\gamma=1}^{N}\sum_{1\leq i<j\leq
3}\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}\ C_{ij}^{\gamma}(w),$

whereby

$\displaystyle C_{ij}^{\gamma}(w):=$
$\displaystyle{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}\left|\sum_{\alpha,\beta}c_{\gamma}^{\alpha\beta}(\cdot,\pi_{i,\alpha}(w))_{\sharp}(\cdot,\pi_{j,\beta}(w))_{\sharp}\right|^{2}\right].$

Clearly, we see that $C_{ij}^{\gamma}<0$, since the structure constants are
non-zero and the square of
$\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}$ is negative definite,
with probability 1.

Now, if the structure constant $c_{\gamma}^{\alpha\beta}$ is non-zero, then
all the $\alpha,\beta$ and $\gamma$ must be distinct. And
$(\cdot,\xi_{ij}(w))_{\sharp}$ is independent of
$(\cdot,\xi_{\hat{i}\hat{j}}(w))_{\sharp}$, provided
$(i,j)\neq(\hat{i},\hat{j})$, with

${{\mathbb{E}}}\left[\left(\cdot,\xi_{ij}(w)\right)_{\sharp}\overline{\left(\cdot,\xi_{ij}(w)\right)_{\sharp}}\right]=\frac{1}{\kappa^{2}}\parallel\xi_{ij}(w)\parallel^{2}=\frac{1}{2\pi\kappa^{2}},$

on the probability space
$({\mathbb{B}}\otimes\mathfrak{g},\tilde{\mu}_{\kappa^{2}}^{\times^{N}})$.

Write

$\widetilde{\sum}=\sum_{\gamma=1}^{N}\sum_{1\leq i<j\leq
3},\quad\nu^{\kappa}=\nu_{R[a,T]}^{\kappa,\rho}.$

Now, $\\{(\cdot,\xi_{ij,\gamma}(w))_{\sharp}:1\leq i<j\leq 3,\ 1\leq\gamma\leq
N\\}$ is a set of i.i.d. complex random variables, independent of

  * •

$\left(\cdot,\nu_{R[a,T]}^{\kappa}\right)_{\sharp}$, because they are
independent of $(\cdot,\xi_{0j,\gamma}(w))_{\sharp}$, for $j=1,2,3$;

  * •

$(\cdot,\pi_{i,\alpha}(w))_{\sharp}$, because
$\zeta_{i}(w)=\zeta(w)\otimes(dx^{0}\wedge dx^{i})$.

Therefore,

$\displaystyle{{\mathbb{E}}}$
$\displaystyle\left[\left|\left(\cdot,\nu^{\kappa}\right)_{\sharp}Y_{2}^{\kappa}\right|^{2}\right]$
$\displaystyle=$
$\displaystyle{{\mathbb{E}}}\left[\left|\left(\cdot,\nu^{\kappa}\right)_{\sharp}\widetilde{\sum}\sum_{\alpha,\beta}c_{\gamma}^{\alpha\beta}\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\
(\cdot,\pi_{i,\alpha}(w))_{\sharp}(\cdot,\pi_{j,\beta}(w))_{\sharp}(\cdot,\kappa\xi_{ij,\gamma}(\bar{w}))_{\sharp}\right|^{2}\right]$
$\displaystyle=$
$\displaystyle\widetilde{\sum}{{\mathbb{E}}}\left[\left|\left(\cdot,\nu^{\kappa}\right)_{\sharp}\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\
\sum_{\alpha,\beta}c_{\gamma}^{\alpha\beta}(\cdot,\pi_{i,\alpha}(w))_{\sharp}(\cdot,\pi_{j,\beta}(w))_{\sharp}(\cdot,\kappa\xi_{ij,\gamma}(\bar{w}))_{\sharp}\right|^{2}\right].$

By Jensen’s inequality and independence of the random variables, we have

$\displaystyle{{\mathbb{E}}}$
$\displaystyle\left[\left|\left(\cdot,\nu^{\kappa}\right)_{\sharp}Y_{2}^{\kappa}\right|^{2}\right]$
$\displaystyle\leq$
$\displaystyle\widetilde{\sum}\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\
{{\mathbb{E}}}\left[\left|\left(\cdot,\nu^{\kappa}\right)_{\sharp}\sum_{\alpha,\beta}c_{\gamma}^{\alpha\beta}(\cdot,\pi_{i,\alpha}(w))_{\sharp}(\cdot,\pi_{j,\beta}(w))_{\sharp}(\cdot,\kappa\xi_{ij,\gamma}(\bar{w}))_{\sharp}\right|^{2}\right]$
$\displaystyle=$
$\displaystyle\widetilde{\sum}\int_{{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\
{{\mathbb{E}}}\left[\left|(\cdot,\kappa\xi_{ij,\gamma}(w))_{\sharp}\right|^{2}\right]\cdot{{\mathbb{E}}}\left[\left|\left(\cdot,\nu^{\kappa}\right)_{\sharp}\sum_{\alpha,\beta}c_{\gamma}^{\alpha\beta}(\cdot,\pi_{i,\alpha}(w))_{\sharp}(\cdot,\pi_{j,\beta}(w))_{\sharp}\right|^{2}\right]$
$\displaystyle=$ $\displaystyle\sum_{\gamma=1}^{N}\sum_{1\leq i<j\leq
3}\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\
\parallel\xi_{ij}(w)\parallel^{2}|C_{ij}^{\gamma}(w)|.$

Since

$\parallel\xi_{ij}(w)\parallel^{2}=\frac{1}{2\pi},$

we see that

${{\mathbb{E}}}\left[\left|\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}Y_{2}^{\kappa}\right|^{2}\right]\leq\frac{1}{2\pi}\sum_{\gamma=1}^{N}\sum_{1\leq
i<j\leq 3}\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\ |C_{ij}^{\gamma}(w)|.$

Similarly,

${{\mathbb{E}}}\left[\left|\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}Y_{1}^{\kappa}\right|^{2}\right]\leq\frac{1}{2\pi}\sum_{\gamma=1}^{N}\sum_{1\leq
i<j\leq 3}\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\ |C_{ij}^{\gamma}(w)|.$

By Jensen’s Inequality,

$\displaystyle{{\mathbb{E}}}$
$\displaystyle\left[\left|\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}Y_{1}^{\kappa}+\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}Y_{2}^{\kappa}\right|^{2}\right]$
$\displaystyle=$ $\displaystyle
4{{\mathbb{E}}}\left[\left|\frac{\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}}{2}(Y_{1}^{\kappa}+Y_{2}^{\kappa})\right|^{2}\right]\leq
2{{\mathbb{E}}}\left[\left|\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}\right|^{2}|Y_{1}^{\kappa}|^{2}+\left|\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}\right|^{2}|Y_{2}^{\kappa}|^{2}\right]$
$\displaystyle\leq$ $\displaystyle\frac{4}{2\pi}\sum_{\gamma=1}^{N}\sum_{1\leq
i<j\leq 3}\int_{w\in{{\mathbb{C}}}^{4}}d\lambda_{4}(w)\ |C_{ij}^{\gamma}(w)|.$

Therefore, we see that

$\frac{1}{4}{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}(Y_{1}^{\kappa}+Y_{2}^{\kappa})^{2}\right]-{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}Y_{3}^{\kappa}\right]>0,$
(4.3)

and is of order $1/\kappa^{4}$, from the bounds in Equation (4.1).

Define

$\displaystyle F^{\kappa}:=$
$\displaystyle\mathcal{Y}^{\kappa}-1+\frac{1}{2}\sum_{i=1}^{3}Y_{i}^{\kappa}-\frac{1}{8}\left(\sum_{i=1}^{3}Y_{i}^{\kappa}\right)^{2}$
$\displaystyle=$
$\displaystyle\sum_{k=3}^{\infty}\left(-\frac{1}{2}\right)^{k}\frac{1}{k!}\left(\sum_{i=1}^{3}Y_{i}^{\kappa}\right)^{k}.$

Corollary 3.2 says that $\mathcal{Y}^{\kappa}$ is actually $L^{p}$ integrable,
for any $1<p<\pi$. Hence
$\left|\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}\right|F^{\kappa}$
is also $L^{p}$ integrable from Proposition 3.9, for some $p>1$. From
Equations (3.2) and (3.5), we have

${{\mathbb{E}}}\left[\left|\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}\right|^{p}|F^{\kappa}|^{p}\right]\leq
C_{p},$ (4.4)

for some constant $C_{p}$ independent of $\kappa$.

Write

$\mathbb{I}_{M}:=1_{\\{\sum_{i=1}^{3}|Y_{i}^{\kappa}|\leq M\\}}\quad{\rm
and}\quad|\nu^{\kappa,2}|=\left|\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}\right|.$

For such value of $p$, let $q>0$ such that $1/q+1/p=1$. For any positive
number $M$, we have

$\displaystyle{{\mathbb{E}}}$
$\displaystyle\left|\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}F^{\kappa}\right|$
$\displaystyle=$
$\displaystyle{{\mathbb{E}}}\left|1_{\\{\sum_{i=1}^{3}|Y_{i}^{\kappa}|>M\\}}\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}F^{\kappa}\right|+{{\mathbb{E}}}\left|1_{\\{\sum_{i=1}^{3}|Y_{i}^{\kappa}|\leq
M\\}}\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}F^{\kappa}\right|$
$\displaystyle\leq$
$\displaystyle\left[{{\mathbb{E}}}\left|1_{\\{\sum_{i=1}^{3}|Y_{i}^{\kappa}|>M\\}}\right|\right]^{1/q}\left({{\mathbb{E}}}\left[|\nu^{\kappa,2}|^{p}|F^{\kappa}|^{p}\right]\right)^{1/p}+{{\mathbb{E}}}\left|\mathbb{I}_{M}\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}F^{\kappa}\right|,$
(4.5)

after applying Holder’s Inequality.

Apply Chebyshev’s Inequality, the RHS of Equation (4.5) is less than or equal
to

$\displaystyle\frac{1}{M^{2n/q}}$
$\displaystyle\left[{{\mathbb{E}}}\left|\left(\sum_{i=1}^{3}|Y_{i}^{\kappa}|\right)^{2n}\right|\right]^{1/q}\left({{\mathbb{E}}}\left[|\nu^{\kappa,2}|^{p}|F^{\kappa}|^{p}\right]\right)^{1/p}+{{\mathbb{E}}}\left|\mathbb{I}_{M}\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}F^{\kappa}\right|$
$\displaystyle\leq\frac{3^{2n-1}}{M^{2n/q}}\left[{{\mathbb{E}}}\sum_{i=1}^{3}|Y_{i}^{\kappa}|^{2n}\right]^{1/q}\left({{\mathbb{E}}}\left[|\nu^{\kappa,2}|^{p}|F^{\kappa}|^{p}\right]\right)^{1/p}+{{\mathbb{E}}}\left|\mathbb{I}_{M}\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}F^{\kappa}\right|$
$\displaystyle\leq
C(n)\frac{1}{M^{2n/q}\kappa^{4n/q}}\left({{\mathbb{E}}}\left[|\nu^{\kappa,2}|^{p}|F^{\kappa}|^{p}\right]\right)^{1/p}+{{\mathbb{E}}}\left|\mathbb{I}_{M}\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}F^{\kappa}\right|,$

for some constant $C(n)$, dependent only on $n\in\mathbb{N}$.

Choose $n$ such that $n/q\geq 2$. From Equation (4.4), we see that

${{\mathbb{E}}}\left|\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}F^{\kappa}\right|=O(1/M^{4}\kappa^{8})+{{\mathbb{E}}}\left|1_{\\{\sum_{i=1}^{3}|Y_{i}^{\kappa}|\leq
M\\}}\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}F^{\kappa}\right|.$
(4.6)

And,

$\displaystyle{{\mathbb{E}}}$
$\displaystyle\left|1_{\\{\sum_{i=1}^{3}|Y_{i}^{\kappa}|\leq
M\\}}\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}F^{\kappa}\right|$
$\displaystyle\leq\sum_{k=3}^{\infty}\frac{1}{2^{k}k!}{{\mathbb{E}}}\left[1_{\\{\sum_{i=1}^{3}|Y_{i}^{\kappa}|\leq
M\\}}\left|\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}\right|\left(\sum_{i=1}^{3}|Y_{i}^{\kappa}|\right)^{k}\right]$
$\displaystyle\leq\sum_{k=3}^{\infty}\frac{M^{k-2}}{2^{k}k!}{{\mathbb{E}}}\left|\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}\left(\sum_{i=1}^{3}|Y_{i}^{\kappa}|\right)^{2}\right|$
$\displaystyle=M\left(\sum_{k=0}^{\infty}\frac{M^{k}}{2^{k+3}(k+3)!}\right)\times{{\mathbb{E}}}\left|\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}\left(\sum_{i=1}^{3}|Y_{i}^{\kappa}|\right)^{2}\right|=O(Me^{M}/\kappa^{4}),$

the last equality follows from Equation (4.1).

Using this bound in Equation (4.6), we see that

$\left|{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}F^{\kappa}\right]\right|=O(1/M^{4}\kappa^{8})+O(Me^{M}/\kappa^{4}).$

Together with the bounds in Equation (4.1), we see that

$\displaystyle{{\mathbb{E}}}$
$\displaystyle\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}(\mathcal{Y}^{\kappa}-1)\right]$
$\displaystyle=$
$\displaystyle-\frac{1}{2}{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}Y_{3}^{\kappa}\right]+\frac{1}{8}{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}\left(\sum_{i=1}^{2}Y_{i}^{\kappa}\right)^{2}\right]$
$\displaystyle+\frac{1}{8}{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}Y_{3}^{\kappa,2}+2\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}Y_{3}^{\kappa}\sum_{i=1}^{2}Y_{i}^{\kappa}\right]+{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}F^{\kappa}\right]$
$\displaystyle=$
$\displaystyle-\frac{1}{2}{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}Y_{3}^{\kappa}\right]+\frac{1}{8}{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}\left(\sum_{i=1}^{2}Y_{i}^{\kappa}\right)^{2}\right]$
$\displaystyle+O(Me^{M}/\kappa^{4})+O(1/\kappa^{6})+O(1/M^{4}\kappa^{8}).$

Choose a small $M>0$. Then we see from Equation (4.3), that there exists a
large $\kappa_{0}>1$ such that for all $\kappa>\kappa_{0}$, we have
$0<{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}(\mathcal{Y}^{\kappa}-1)\right]$.

Refer to Definition 3.7. We have

${{\rm{Tr}}}\
{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}(\mathcal{Y}^{\kappa}-1)\right]\leq
C(\rho){{\rm{Tr}}}\
{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a,T]}^{\kappa}\right)_{\sharp}^{2}(\mathcal{Y}^{\kappa}-1)\right].$

Hence

$\left|{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}(\mathcal{Y}^{\kappa}-1)\right]\right|=O\left(C(\rho)/\kappa^{4}\right),$

where we use the bounds in Equation (4.1). This completes the proof.

###### Remark 4.2

Note that Equation (4.3) still holds true, if we replace the factor
$1/\sqrt{2\pi}$ in $\psi_{w}$ with any $0<\tilde{c}<1/\sqrt{2}$. Also see
Remark 3.3.

###### Corollary 4.3

For $\kappa>\kappa_{0}$, we have

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

whereby $\epsilon(\rho,\kappa)$ is negative definite. Furthermore,
$|\epsilon(\rho,\kappa)|=O\left(C(\rho)/\kappa^{4}\right)$, $|\cdot|$ is the
trace norm.

Proof. In [1], we showed that

$\left\|\frac{\kappa}{4}\int_{I^{2}}d\hat{s}\
\sum_{j=1}^{3}|J_{0j}^{\sigma}|(\hat{s})\xi_{0j}(\kappa\sigma(\hat{s})/2)\right\|^{2}=\frac{|a|T}{4}+O\left(\tilde{C}e^{-\kappa\bar{C}}\right),$
(4.7)

for constants $\tilde{C},\bar{C}$ both dependent only on $R[a,T]$, but
independent of $\kappa$.

Hence,

$\displaystyle{{\mathbb{E}}}\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}\right]=$
$\displaystyle\frac{1}{\kappa^{2}}\sum_{\alpha=1}^{N}\left\|\nu_{R[a,T]}^{\kappa,\alpha}\right\|^{2}\rho(E^{\alpha})\rho(E^{\alpha})$
$\displaystyle=$
$\displaystyle\frac{|a|T}{4}\sum_{\alpha=1}^{N}\rho(E^{\alpha})\rho(E^{\alpha})+O\left(C(\rho)e^{-\kappa\bar{C}}\right).$

Together with the statement in Lemma 4.1, we will have our result.

###### Theorem 4.4

Refer to Definition 3.10. As $\kappa\rightarrow\infty$, we have

$\mathbb{E}_{{\rm
YM}}^{\kappa}\left[\mathcal{J}_{R[a,T]}^{\kappa,\rho}\right]\longrightarrow\exp\left[-\frac{1}{8}\sum_{0\leq
a<b\leq 3}\int_{I^{2}}dsdt\
\rho_{\sigma}^{ab}(s,t)J_{ab}^{\sigma}(s,t)\mathscr{E}(\rho)\right].$

Here, $\mathscr{E}(\rho)$ is the Casimir operator and

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

which is the area of the rectangular surface $R[a,T]$.

Proof.

The proof in Lemma 4.1 can be adapted to show that

$\left|\mathbb{E}\left[\mathcal{J}_{R[a,T]}^{\kappa,\rho}(1-\mathcal{Y}^{\kappa})\right]\right|=O(1/\kappa^{2}),$

by replacing $\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}^{2}$ with
$\mathcal{J}_{R[a,T]}^{\kappa,\rho}$.

Hence we have

$\displaystyle\lim_{\kappa\rightarrow\infty}\int_{{\mathbb{B}}\otimes\mathfrak{g}}\mathcal{J}_{R[a,T]}^{\kappa,\rho}\mathcal{Y}^{\kappa}\
d\tilde{\mu}_{\kappa^{2}}^{\times^{N}}$
$\displaystyle=\lim_{\kappa\rightarrow\infty}\mathbb{E}\left[\mathcal{J}_{R[a,T]}^{\kappa,\rho}\right].$

Refer to Definition 3.7. Now,
$\left\\{\left(\cdot,\nu_{R[a,T]}^{\kappa,\alpha}\right)_{\sharp}:1\leq\alpha\leq
N\right\\}$ is a set of i.i.d. real Gaussian random variables. From Equation
(3.6), we have

$\displaystyle{{\mathbb{E}}}\left[e^{\left(\cdot,\nu_{R[a,T]}^{\kappa,\rho}\right)_{\sharp}}\right]=$
$\displaystyle\exp\left[\frac{1}{2\kappa^{2}}\sum_{\alpha=1}^{N}\left\langle\nu_{R[a,T]}^{\kappa,\alpha},\nu_{R[a,T]}^{\kappa,\alpha}\right\rangle\rho(E^{\alpha})\rho(E^{\alpha})\right]$
$\displaystyle\longrightarrow$ $\displaystyle\exp\left[-\frac{1}{8}\sum_{1\leq
i\leq 3}\int_{I^{2}}dsdt\
\rho_{\sigma}^{0i}(s,t)|J_{0i}^{\sigma}|(s,t)\mathscr{E}(\rho)\right],$

as $\kappa\rightarrow\infty$.

From the proof of Lemma 4.1, we see that

$\displaystyle{{\mathbb{E}}}[\mathcal{Y}^{\kappa}]=1+O(1/\kappa^{4}),$

which converges to 1 as $\kappa$ goes to infinity.

Hence,

$\displaystyle\mathbb{E}_{{\rm
YM}}^{\kappa}\left[\mathcal{J}_{R[a,T]}^{\kappa,\rho}\right]$
$\displaystyle=\frac{1}{{{\mathbb{E}}}[\mathcal{Y}^{\kappa}]}{{\mathbb{E}}}\left[\mathcal{J}_{R[a,T]}^{\kappa,\rho}\mathcal{Y}^{\kappa}\right]$
$\displaystyle\longrightarrow\exp\left[-\frac{1}{8}\sum_{1\leq i\leq
3}\int_{I^{2}}dsdt\
\rho_{\sigma}^{0i}(s,t)|J_{0i}^{\sigma}|(s,t)\mathscr{E}(\rho)\right],$

as $\kappa\rightarrow\infty$.

###### Remark 4.5

If one examines the above proof carefully, asymptotic freedom is required to
show that
$\lim_{\kappa\rightarrow\infty}\mathbb{E}\left(\exp\left[\left(\cdot,\nu_{R[a,T]}^{\kappa,\alpha}\right)_{\sharp}\right]\right)$
exists and is equal to the exponential of the area of the surface. Asymptotic
freedom and renormalization allow one to compute the path integral in
Definition 3.10 via perturbation series expansion, when $\kappa$ is large.

In Remark 2.2, we explained that $1/\kappa$ denote the spread of a Gaussian
function, used to approximate a Dirac Delta function, where $1/\kappa$ is also
synonymous with the lattice spacing $\epsilon$ in lattice gauge theories. When
the distance between distinct points is small, one needs a higher $\kappa$, to
resolve them. Thus, we can say that non-abelian Yang-Mills theory is
asymptotically free at short distances.

We need to normalize $\chi_{w}$, and choose a $0<\tilde{c}<1$ such that the
norm $\parallel\tilde{c}\psi_{w}\chi_{w}\parallel<1$. In this article, we
chose $\tilde{c}=1/\sqrt{2\pi}$. In [1], we explained why we need the
renormalization factor $\frac{1}{\sqrt{2\pi}}\psi_{w}$. In Quantum Field
Theory, when using Fourier transform to go into momentum space, one will
introduce a cutoff function, so as to resolve the ultra-violet divergence
issue. This was necessary to prove Corollary 3.2 and Lemma 4.1. Also refer to
Remarks 3.3 and 4.2. Indeed, Corollary 3.2 and Lemma 4.1 still hold by
choosing any $\tilde{c}<1/\sqrt{2}$.

##  5 Application

The main purpose of computing the Yang-Mills path integral is to derive the
Wilson Area Law formula. A quark is never observed in isolation due to quark
confinement. A quark and an antiquark are bounded together in a meson via the
strong force. Unlike the Coulomb potential, which decays inversely
proportional to the distance, the strong force actually gets stronger as the
separation between the quarks increases. It is confirmed experimentally that
the potential varies proportionally to the distance between quarks.

The path integral given by Expression 1.8 can be used to explain why this
potential is linear. However, to compute it is highly nontrivial and
difficult. Note that perturbative methods using Feynman diagrams cannot be
used to compute it, lest derive the area formula.

Let $S$ be a flat rectangular surface, whose boundary is a rectangular contour
$C$ with spatial length $R$ and timelike length $T$, thus $S$ has area $RT$.
Then, we will write

$W(R,T;\rho):=\lim_{\kappa\rightarrow\infty}{{\mathbb{E}}}_{{\rm
YM}}^{\kappa}\left[{{\rm{Tr}}}\
\exp\left[(\cdot,\nu_{S}^{\kappa,\rho})_{\sharp}\right]\right].$

From Theorem 4.4, we have

$W(R,T;\rho)={{\rm{Tr}}}\
\exp\left[\frac{RT}{8}\sum_{\alpha=1}^{N}\rho(E^{\alpha})\rho(E^{\alpha})\right].$

A meson is made up of a quark and antiquark. Now, these 2 quarks go in
opposite direction along the spatial length $R$ and after time $T$, are
attracted to each other along the spatial length $R$, hence tracing a
rectangular contour $C$ described in the previous paragraph. We want to
calculate the potential energy between the quark and antiquark, as a function
of distance $R$. Using the path integral computed on the surface $S$ with
boundary $C$, the potential is given by

$V(R)=-\lim_{T\rightarrow\infty}\log\frac{W(R,T+1;\rho)}{W(R,T;\rho)}.$

This formula can be derived from Equation (19.11) in [12].

In the case of the standard representation of ${\rm SU}(N)$, we can see that

$\sum_{\alpha=1}^{N^{2}-1}E^{\alpha}E^{\alpha}=\left(\frac{1}{N}-N\right)\mathbb{I},$

$\mathbb{I}$ is the $N\times N$ identity matrix. See [13]. Therefore,

$V(R)=\frac{R}{8}\left(N-\frac{1}{N}\right).$

In the case of the standard representation of ${\rm SO}(N)$, we see that

$\sum_{\alpha=1}^{N(N-1)/2}E^{\alpha}E^{\alpha}=\frac{1-N}{2}\mathbb{I},$

hence

$V(R)=\frac{R}{16}\left(N-1\right).$

Compare with Equation (19.6) in [12]. Thus, we see that quark confinement
implies that the potential energy between the quarks is linear in nature.

##  Appendix A Definition

Let $S$ be a surface embedded in ${{\mathbb{R}}}^{4}$ and
$\sigma\equiv(\sigma_{0},\sigma_{1},\sigma_{2},\sigma_{3}):[0,1]^{2}\equiv
I^{2}\rightarrow{{\mathbb{R}}}^{4}$ be its parametrization. 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)

###### Definition A.2

Define

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

which gives us the area of the surface $S$.

###### Remark A.3

  1. 1.

Note that $\int_{S}d\rho$ is independent of the choice of orthonormal basis
and parametrization used, by a straightforward calculation. See [7].

  2. 2.

Dimension 4 is required to write the formula for $\int_{S}d\rho$.

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