Yang-Mills Measure and Axial Gauge Fixing on ℝ^4
Adrian P. C. Lim

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Let $\mathfrak{g}$ be a semi-simple Lie algebra. For a $\mathfrak{g}$-valued 1-form $A$, consider the Yang-Mills action S_YM(A) = ∫_ℝ^4  |dA + A ∧A |^2 using the standard metric on $T^\ast\mathbb{R}^4$. Using axial gauge fixing, we want to make sense of the following path integral,

\begin{equation}
{\rm Tr}\ \int_{A \in \mathcal{A}_{\mathbb{R}^4, \mathfrak{g}} /\mathcal{G}} \mathcal{T} \exp \left[  \int_{C} A\right] e^{-\frac{1}{2}S_{YM}(A)}\ DA, \nonumber
\end{equation}

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 $\mathcal{T}$ is the time ordering operator.

We will construct an Abstract Wiener space for which we can define the Yang-Mills path integral rigorously, both in the Abelian and non-Abelian cases. Subsequently, we will then derive the Wilson Area Law formula, in both the Abelian and non-Abelian cases from these definitions. 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 a linear potential.

MSC 2010: 81T13, 81T08

Keywords: Yang-Mills measure, axial gauge fixing, area law, quark confinement, vortex

§ PRELIMINARIES

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 $P \rightarrow M$ 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 $\Tr \equiv \Tr_{{\rm Mat}(\bar{N}, \mathbb{C})}$.
Then we can define a positive, non-degenerate bilinear form by ⟨A, B ⟩= -_Mat(N̅, ℂ)[AB]  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.

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 ·Ω:= A^Ω = Ω^-1dΩ+ Ω^-1AΩfor $A \in \mathcal{A}_{M, \mathfrak{g}}$, $\Omega \in \mathcal{G}$. 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 $\mathcal{A}/\mathcal{G}$.

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 $\ast$ acting on $k$-forms, $\ast: \Lambda^k(T^\ast M) \rightarrow \Lambda^{4-k}(T^\ast M)$ such that for $u, v \in \Lambda^k(T^\ast M)$, we have u ∧∗v = ⟨u, v ⟩_q dω.  An inner product on the set of smooth sections $\Gamma(\Lambda^k(T^\ast M))$ is then defined as ⟨u, v ⟩= ∫_M ⟨u, v ⟩_q dω.

Given $u \otimes E \in \Lambda^q(T^\ast M) \otimes \mathfrak{g}$, we write |u ⊗E|^2 = -[ E ·E ] u ∧∗u = -[ E ·E ]⟨u, u ⟩_q dω. Hence for $A \in \mathcal{A}_{M, \mathfrak{g}}$, the Yang-Mills action is given by S_YM(A) = ∫_M  |dA + A ∧A |^2.  Note that this action is invariant under gauge transformations.

Let $C$ be a simple closed curve in the manifold $M$. The holonomy operator of $A$, computed along the curve $C$, is given by 𝒯 exp[  ∫_C A], whereby $\mathcal{T}$ is the time ordering operator. See Definition <ref> for the definition of $\mathcal{T}$.

It is of interest to make sense of the following path integral,
1/Z∫_A ∈𝒜_M, 𝔤 /𝒢 𝒯 exp[  ∫_C A] e^-1/2S_YM(A) DA,  whereby $DA$ is some Lebesgue type of measure on the space of $\mathfrak{g}$-valued 1-forms, modulo gauge transformations and Z = ∫_A ∈𝒜_M, 𝔤 /𝒢e^-1/2S_YM(A) DA.

Let $\Lambda^p(\bR^n)$ be the $p$-th exterior power of the vector space $\bR^n$, $n=3,4$. We also denote the smooth sections of a bundle $P$ by $\Gamma(P)$. In this article, when $n=3$, $p=1$; and when $n=4$, $p=2$.

From now on, we only consider $M = \bR^4$ and take the principal bundle $P$ over $\bR^4$ to be the trivial bundle. On $\bR^4$, fix the coordinate axis and choose global coordinates $\{x^0, x^1, x^2, x^3\}$ and let $\{e_a\}_{a=0}^3$ be the standard orthonormal basis in $\bR^4$. We will also choose the standard Riemannian metric on $\mathbb{R}^4$. Now let $T^\ast\bR^4 \rightarrow \bR^4$ denote the trivial cotangent bundle over $\bR^4$, i.e. $T^\ast \bR^4 \cong \bR^4 \times \Lambda^1(\bR^4)$ and $\Lambda^1(\bR^3)$ denote the subspace in $\Lambda^1(\bR^4)$ spanned by $\{dx^1, dx^2, dx^3\}$. There is an obvious inner product defined on $\Lambda^1(\bR^3)$, i.e. $\langle dx^i, dx^j \rangle = 0$ if $i \neq j$, 1 otherwise, which it inherits from the standard metric on $T^\ast \bR^4$.

Using axial gauge fixing, every $A \in \mathcal{A}_{\bR^4, \mathfrak{g}} /\mathcal{G}$ can be gauge transformed into a $\mathfrak{g}$-valued 1-form, of the form $A = \sum_\alpha \sum_{j=1}^3a_{j, \alpha} \otimes  dx^j \otimes E^\alpha $,  subject to the conditions a_1,α(0, x^1, 0, 0) = 0, a_2, α(0, x^1, x^2, 0)=0, a_3,α(0, x^1, x^2, x^3) = 0. Hence it suffices to consider the Yang-Mills integral in Expression (<ref>) to be over the space of $\mathfrak{g}$-valued 1-forms of the form $A = \sum_\alpha \sum_{j=1}^3a_{j, \alpha} \otimes  dx^j \otimes E^\alpha $, whereby $a_{j,\alpha}: \bR^4 \rightarrow \bR$ is smooth.

Its curvature is then given by

\begin{align}
dA + A \wedge A
=&  \sum_\alpha\sum_{1\leq i<j \leq 3}a_{i;j, \alpha} \otimes  dx^i \wedge dx^j \otimes E^\alpha + \sum_{\alpha,\beta}\sum_{1 \leq i<j\leq 3}a_{i,\alpha}a_{j,\beta} \otimes dx^i \wedge  dx^j \otimes [E^\alpha, E^\beta] \nonumber \\
&+ \sum_\alpha\sum_{j=1}^3  a_{0:j, \alpha} \otimes dx^0 \wedge dx^j \otimes E^\alpha, \nonumber
\end{align}

for $a_{i;j,\alpha} := (-1)^{ij}[\partial_i a_{j, \alpha} - \partial_j a_{i, \alpha}]$, $a_{0:j, \alpha} := \partial_0 a_{i, \alpha}$.

Note that $\Lambda^2(T^\ast \bR^4) \cong \bR^4 \times \Lambda^2(\bR^4)$. Using the global 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 $\Lambda^2 (\bR^4)$. Using the standard metric on $\bR^4$, 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 $\Gamma(\Lambda^2( T^\ast \bR^4))$ as in Equation (<ref>).

For $i,j,k = 1,2,3$, we will define $\epsilon_{ijk} = (-1)^{|\sigma(ijk)|}$ if all are distinct, $\sigma(ijk)$ is a permutation in $S_3$, and $|\sigma(ijk)|$ is the number of transpositions; 0 otherwise.

Define c_γ^αβ = -[E^γ[E^α, E^β] ].

\begin{align}
dA &+ A \wedge A \nonumber \\
=& \sum_\gamma\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 \nonumber \\
&\hspace{8cm}+ \sum_{j=1}^3  a_{0:j, \gamma}\otimes  dx^0 \wedge dx^j \bigg]\otimes E^\gamma. \nonumber
\end{align}

\begin{align}
\int_{\bR^4}|dA + A \wedge A|^2 =& \sum_{i<j}\int_{\bR^4}\bigg[ \sum_{\alpha}a_{i;j, \alpha}^2 + \sum_\gamma\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}} \nonumber \\
&+2\sum_{\alpha< \beta, \gamma}a_{i;j,\gamma}a_{i,\alpha}a_{j,\beta}c_\gamma^{\alpha\beta} \bigg]\ d\omega+ \sum_j\int_{\bR^4}\sum_{\alpha} a_{0:j,\alpha}^2\ d\omega.
\label{e.ym.2}
\end{align}

It is conventional wisdom to interpret $\exp\left[ -\frac{1}{2}\sum_{i<j}\int_{\bR^4}d\omega\ \sum_{\alpha}\left( a_{i;j, \alpha}^2 +  a_{0:j,\alpha}^2 \right) \right]DA$ as a Gaussian measure.

§.§ Abelian Case

Consider first a 2-dimensional Euclidean space $\bR^2$, with Abelian group $G = U(1)$, and  $\mathfrak{u}(1) \cong \bR \otimes \sqrt{-1}$. Let $S$ be a surface with boundary $C = \partial S$. Write $A = \sum_{i=1}^2 A_i \otimes dx^i \in \Gamma(\Lambda^1(T^\ast\bR^2))$, $A_i: \bR^2 \rightarrow \bR$.
Using Stokes' Theorem,

\begin{align}
\int_C \sum_{i=1}^2 A_i \otimes dx^i =& \int_{\partial S}  \sum_{i=1}^2 A_i dx^i = \int_S dA = \int_{\bR^2} dA \cdot1_S = \left\langle dA, 1_S \right\rangle. \nonumber
\end{align}

Here, $\langle \cdot, \cdot \rangle$ is the $L^2$ inner product on the space of Lebesgue integrable functions on $\bR^2$. Thus, the Yang-Mills path integral becomes 1/∫_A e^-|dA|^2/2DA∫_A e^√(-1)⟨dA, 1_S ⟩e^-1/2|dA|^2DA. Now, make a heuristic change of variables, $A \mapsto dA$, hence we have

\begin{align*}
\frac{1}{\det{d^{-1}}\int_A e^{-|dA|^2/2}D[dA]}&\det{d^{-1}}\int_{A} e^{\sqrt{-1}\langle dA, 1_S \rangle}e^{-\frac{1}{2}|dA|^2}D[dA] \\
&= \frac{1}{\int_A e^{-|dA|^2/2}D[dA]}\int_{A} e^{\sqrt{-1}\langle dA, 1_S \rangle}e^{-\frac{1}{2}|dA|^2}D[dA].
\end{align*}

This is a Gaussian integral of the form given in Lemma <ref>, hence we can define the path integral as $\exp\left[ -|1_S|^2/2\right] = \exp\left[ -|S|/2\right]$, whereby $|S|$ is the area of the surface $S$.

Now, we move up to $\bR^4$, still using $G = U(1)$. The above argument still applies, except that now $1_S$ has norm 0. This will yield 1 for any surface $S$, which means we have to redefine our path integral to obtain non-trivial results.

Let $\chi_x$ be the evaluation map, i.e. given a (smooth) function $f: \bR^4 \rightarrow \bR$, we will write $f(x) = \langle f, \chi_x \rangle$. To the physicists, $\chi_x$ is just the Dirac Delta function. Furthermore, given a 2-form written in the form $F \equiv \sum_{0\leq i<j \leq 3} f_{ij} dx^i \wedge dx^j$, with $f_{ij}$ being smooth functions on $\bR^4$; when we write $\langle F, \chi_x \otimes dx^a \wedge dx^b \rangle$, we mean

\begin{align}
\left\langle F, \chi_x \otimes dx^a \wedge dx^b \right\rangle =& \left\langle \sum_{0\leq i<j \leq 3} f_{ij} dx^i \wedge dx^j, \chi_x \otimes dx^a \wedge dx^b \right\rangle \nonumber\\
=& \left\langle  f_{ab} dx^a \wedge dx^b, \chi_x \otimes dx^a \wedge dx^b \right\rangle = f_{ab}(x) \in \bR. \label{e.dr.2}
\end{align}

Choose a surface $S$ such that $\partial S = C$. Suppose $\sigma: [0,1]^2 \rightarrow \bR^4$ is any parametrization for $S$. Using axial gauge fixing, we only consider 1-forms of the form $A = \sum_{i=1}^3 A_i \otimes dx^i$. Write $A_{i;j} := (-1)^{ij}[\partial_i A_{j} - \partial_j A_{i}]$, $A_{0:j} = \partial_j A_0$. We also write $A_{0;j} = A_{0:j}$. By Stokes' Theorem,

\begin{align}
\int_C \sum_{i=1}^3A_i \otimes dx^i =& \int_{\partial S}\sum_{i=1}^3 A_i \otimes dx^i = \int_S dA \nonumber \\
=& \int_{[0,1]^2}ds dt\ \left[\sum_{1\leq i < j \leq 3}[A_{i;j}(\sigma)|J_{ij}^\sigma|](s,t) + \sum_{j=1}^3 [A_{0:j}(\sigma)|J_{0j}^\sigma|](s,t) \right] \nonumber \\
=& \int_{[0,1]^2}ds dt\ \left\langle dA, \sum_{0\leq i < j \leq 3}\chi_{\sigma(s,t)}|J_{ij}^\sigma|(s,t)\otimes dx^i \wedge dx^j \right\rangle
:= \left\langle dA, \tilde{\nu}_S \right\rangle, \label{e.dr.3}
\end{align}

where ν̃_S = ∫_[0,1]^2 ∑_0≤i < j ≤3dsdt χ_σ(s,t)|J_ij^σ|(s,t)⊗dx^i ∧dx^j, is a linear functional on the space of smooth sections of 2-forms in $\Lambda^2( T^\ast \bR^4)$. And $J_{ij}^\sigma$ is defined in Definition <ref>.

Here, we wish to point out that the inner product in question is given by Equation (<ref>) and therefore,

\begin{align}
\left\langle dA, \sum_{0\leq i < j \leq 3}\chi_{\sigma(s,t)}|J_{ij}^\sigma|(s,t)\otimes dx^i \wedge dx^j \right\rangle =& \left\langle dA, \sum_{0\leq i < j \leq 3}\chi_{\sigma(s,t)}\otimes dx^i \wedge dx^j\right\rangle |J_{ij}^\sigma|(s,t) \label{e.dr.1}\\
=& \sum_{0\leq i < j \leq 3}\left\langle A_{i;j}, \chi_{\sigma(s,t)}\right\rangle |J_{ij}^\sigma|(s,t) \nonumber\\
=& \sum_{0\leq i < j \leq 3} A_{i;j}(\sigma(s,t))\ |J_{ij}^\sigma|(s,t). \nonumber
\end{align}

Thus, we want to make sense of
1/Z∫_A e^√(-1)⟨dA, ν̃_S ⟩e^-1/2|dA|^2 DA for some normalization constant. We have to do a change of variables, i.e. $A \mapsto dA$. Then we have 1/Z∫_A e^√(-1)⟨dA, ν̃_S⟩e^-1/2|dA|^2 (d^-1) D[dA], which is some undefined constant $\det(d^{-1})$. After dividing away this constant, we are left with 1/Z̅∫_A e^√(-1)⟨dA, ν̃_S⟩e^-1/2|dA|^2 D[dA], which we can define as a Gaussian integral, for some normalization constant $\bar{Z}$. Now, apply a similar argument as in the $\bR^2$ case, the Yang-Mills path integral will yield $\exp\left[ -|\tilde{\nu}_S|^2/2 \right]$.

There are a few problems with this argument. First of all, $\chi_x$ is not an honest function, in fact it is a generalized function. Secondly, when we construct a Gaussian measure on $\Gamma(\Lambda^2( T^\ast \bR^4))$, we need to enlarge the space and consider generalized sections in $\Lambda^2( T^\ast \bR^4)$. As such, the term $|\nu_S|^2$ will face serious problems as it is not clear how to define a product of generalized functions.

Because of the above problems, if we wish to carry out the above heuristic argument, we cannot define the path integral over real-valued 2-forms. We will instead construct an Abstract Wiener space, consisting of $\Lambda^1(\bR^3)$-valued holomorphic functions in $\bC^4$. The path integral given in Equation (<ref>) depends on the curve $C \subset M \equiv \bR^4$, or more correctly a choice of a surface $S$ whose boundary is $\partial S = C$. Hence, we need to embed the surface $S \subset \bR^4$ inside $\bC^4$, by embedding $\bR^4$ inside $\bC^4$ in the standard way, so that we can define the path integral using the Abstract Wiener space setting. Such an approach was used in [9] and [10] to construct the Chern-Simons Path integral and obtain link invariants in $\bR^3$ respectively.

This work is motivated by trying to apply the Abstract Wiener Space construction used in [9]. However, there are significant differences in the Chern-Simons Path integral and the Yang-Mills Path integral. Hence, we need to modify the construction used in [9] accordingly, so that a rigorous definition can be given to the Yang-Mills Path integral. See Definition <ref>.

The non-Abelian case will be dealt with in Section <ref>. See Definition <ref> for a rigorous definition of the Yang-Mills Path integral for the non-Abelian case. From both the Abelian and non-Abelian cases, we will then derive the Wilson Area Law Formulas from these definitions. The results are given by Theorems <ref> and <ref>. Finally, Section <ref> will focus on applications of the Area Law formula.

We wish to make the following final remark. Dimension 3 is key for the Chern-Simons Path integral formalism to work effectively. In the Yang-Mills Path integral formalism, dimension 4 is the key. So where did we use the fact that the dimension is 4? It is actually used in Appendix <ref>.

Suppose we have two Hilbert spaces, $H_1$ and $H_2$. We consider the tensor product $H_1 \otimes H_2$. The inner product on the tensor product $H_1 \otimes H_2$ is given by ⟨u_1⊗u_2, v_1⊗v_2⟩_H_1 ⊗H_2 = ⟨u_1, v_1⟩_H_1⟨u_2, v_2 ⟩_H_2. This definition of the inner product on the tensor product of Hilbert spaces will be assumed throughout this article.
See also subsection <ref>.

Finally, we always use $\langle \cdot, \cdot \rangle$ to denote an inner product.

§ CONSTRUCTION OF WIENER MEASURE

Throughout the rest of this article, we adopt the following notation.

For $x \in \bR^4$, 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 $z \equiv (z_0, z_1, z_2, z_3) \in \bC^4$.

Throughout this article, this Gaussian function will play a key role in our calculations.
As mentioned earlier on, the Dirac Delta function is a generalized function, but we can approximate it using a Gaussian function $\phi_\kappa$. 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 $\bR^4$. In our construction of the Yang-Mills path integral, we will make use of this Gaussian function in this construction, thus all our Yang-Mills path integral define later, will depend on the parameter $\kappa$.

§.§ Schwartz Space

We let $p_r$ denote the $n$-tuple $(m_1,m_2,\ldots, m_n)$, $m_1, \ldots m_n \geq 0$ are integers with $\sum_{j=1}^n m_j = r$. And we write $p_r! := m_1!m_2!\cdots m_n!$. For $z = (z_1,z_2, \ldots z_n) \in \mathbb{C}^n$, $z^{p_r} := z_1^{m_1}z_2^{m_2}\cdots z_n^{m_n}$. Let $\mathcal{P}_r$ denote the set of all such $n$-tuples, i.e. 𝒫_r = {(m_1,m_2,…, m_n)| ∑_j=1^n m_j=r }. Let $\mathcal{P} = \bigcup_{r=0}^\infty \mathcal{P}_r$.

There is an ordering which we will adopt in the rest of the article. We will write $p_r \leq p_{\bar{r}}$, if in the order of priority, $r \leq \bar{r}$, followed by $m_1 \leq \bar{m}_1$, $m_2 \leq \bar{m}_2, \ldots, m_n \leq \bar{m}_n$.

Consider the Schwartz space $\mathcal{S}_\kappa(\mathbb{R}^4)$, with the Gaussian function $\phi_\kappa$, $ \sqrt{\phi_{\kappa}}(x) = \kappa^2 e^{-\kappa^2|x|^2/4}/(2\pi)$. Any $f \in \mathcal{S}_\kappa(\mathbb{R}^4)$ can be written in the form f(x) = p(x)√(ϕ_κ)(x), whereby $p$ is a polynomial in $(x^0, x^1, x^2, x^3)$. The inner product $\langle \cdot, \cdot \rangle$ defined on the Schwartz space is given by ⟨f, g ⟩= ∫_^4 f·g dλ,  $d\lambda$ is Lebesgue measure on $\mathbb{R}^4$. Let $\overline{\mathcal{S}}_\kappa(\bR^4)$ be the smallest Hilbert space containing $\mathcal{S}_\kappa(\mathbb{R}^4)$, using this inner product.

The Hermite polynomials $\{h_i\}_{i \geq 0}$ form an orthogonal set in $L^2(\mathbb{R}, d\mu)$ with the Gaussian measure $d\mu(x_0) \equiv e^{-x_0^2/2}dx_0/\sqrt{2\pi}$. Let H_p_r(x) := h_i(x^0) h_j(x^1)h_k(x^2)h_l(x^3) , p_r = (i,j,k,l) ∈𝒫_r,be a product of Hermite polynomials and $H_{p_r}^\kappa := H_{p_r}(\kappa\cdot)$.

We have the normalized Hermite polynomials $H_{p_r}/\sqrt{p_r!}$ with respect to the Gaussian measure $e^{-\sum_{i=0}^3|x^i|^2 /2}d\lambda/(2\pi)^2 $. Then  ⋃_r=0^∞{  H_p_r(κx^0, κx^1, κx^2, κx^3)√(ϕ_κ)/√(p_r!): p_r ∈𝒫_r }
is an orthonormal basis for $\overline{\mathcal{S}}_\kappa(\bR^4)$.

We will write
𝒮_κ(^4)⊗Λ^1(^3) = {∑_a=1^3 f_a ⊗dx^a: f_a ∈𝒮_κ(^4)} and
𝒮_κ(^4)⊗Λ^2(^4) = {∑_0≤a< b ≤3 f_ab ⊗dx^a∧dx^b: f_ab ∈𝒮_κ(^4)}.

Here, $\mathcal{S}_\kappa(\bR^4)\otimes \Lambda^1(\bR^3)$ is the space of smooth 1-forms over $\bR^4$ and this space consists of $\Lambda^1(\bR^3)$-valued functions in $\bR^4$, integrable with respect to Lebesgue measure.

In a similar fashion, $\overline{\mathcal{S}}_\kappa(\bR^4)\otimes \Lambda^2(\bR^4)$ is the space of smooth 2-forms over $\bR^4$ and this space consists of $\Lambda^2(\bR^4)$-valued functions in $\bR^4$, integrable with respect to Lebesgue measure.

Note that these Schwartz spaces are dependent on $\kappa$. Recall we use the standard metric on $T\bR^4$ and the volume form on $\bR^4$ is given by $d\omega = dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3$. Using the Hodge star operator and the above volume form, we will define an inner product on $\overline{\mathcal{S}}_\kappa(\bR^4)\otimes \Lambda^2(\bR^4)$ by ⟨∑_0≤a< b ≤3 f_ab ⊗dx^a∧dx^b, ∑_0≤a< b ≤3 f̂_ab ⊗dx^a∧dx^b ⟩= ∑_0≤a< b ≤3 ⟨f_ab, f̂_ab ⟩.

Define a bilinear form on $\mathcal{S}_\kappa(\bR^4)\otimes \Lambda^1(\bR^3)$, by
$\langle df, dg \rangle$, $f, g \in \mathcal{S}_\kappa(\bR^4)\otimes \Lambda^1(\bR^3)$. Note that $df, dg \in \overline{\mathcal{S}}_\kappa(\bR^4)\otimes \Lambda^2(\bR^4)$, and we compute the bilinear form of $df$ and $dg$ using the inner product in $\overline{\mathcal{S}}_\kappa(\bR^4)\otimes \Lambda^2(\bR^4)$.

The bilinear form $\langle d\cdot, d \cdot \rangle$ on $\mathcal{S}_\kappa(\bR^4)\otimes \Lambda^1(\bR^3)$ is an inner product.

The only thing we have to show is that $\langle d\omega, d\omega \rangle = 0$ imply $\omega = 0$.
Now, recall using axial gauge fixing, $\omega = \sum_{i=1}^3 \omega_{i} \otimes dx^i $. Therefore, $0= |d\omega|^2 \geq |\partial_0 \omega|^2$ and thus $|\partial_0 \omega_{i}|^2 = 0$ for each $i= 1, 2, 3$. This means that $\omega_{i}$ is independent of $x_0$. But since $\omega_{i}$ is $L^2$ integrable over $x_0$, hence $\omega_{i} \equiv 0$ for each $i$. Therefore $\omega \equiv 0$.

Complete $\mathcal{S}_\kappa(\bR^4)\otimes \Lambda^1(\bR^3)$ into a Hilbert space, denoted by $\overline{\mathcal{S}_\kappa(\bR^4)\otimes \Lambda^1(\bR^3)}$. In future, we will denote the inner product on this Hilbert space by $\langle d\cdot, d\cdot \rangle$.

§.§ Segal Bargmann Transform

In the construction of the Abstract Wiener space, we need to enlarge the space $\overline{\mathcal{S}_\kappa(\bR^4)\otimes \Lambda^1(\bR^3)}$, by choosing a suitable measurable norm.

But by doing so, we will end up with distribution valued forms. We do not want to consider this, as this will lead to difficulties in defining $\tilde{\nu}_S$ given in Equation (<ref>).
To avoid this problem, we will use the Segal Bargmann Transform and map the Schwartz space to the space of $\Lambda^1(\bR^3)$-valued holomorphic functions, over in $\bC^4$.

Thus, path integral is not defined on the space of $\Lambda^1(\bR^3)$-valued Schwartz functions in $\bR^4$, but rather, over the space of $\Lambda^1(\bR^3)$-valued holomorphic functions in $\bC^4$.

Let $z = (z_0, z_1, z_2, z_3) \in \bC^4$, each $z_i \in \bC$. Consider the real vector space spanned by $\{z^n:\ z \in \bC\}_{n=0}^\infty$, integrable with respect to the Gaussian measure, equipped with a sesquilinear complex inner product, given by

\begin{align}
\langle z^{r},& z^{r'}\rangle
= \frac{1}{\pi}\int_{\mathbb{C}}   z^{r} \cdot \overline{z^{r'}}e^{-|z^2|}
\ dx\ dp,\ z = x + \sqrt{-1}p.  \label{e.l.2}
\end{align}

Note that $\overline{z}$ means complex conjugate. Denote this (real) inner product space by $H^2(\bC)$, which consists of polynomials in $z$. An orthonormal basis is given by { z^n/√(n!): n ≥0 }.

Let $H^2(\bC^4) \equiv H^2(\bC)^{\otimes^4}$, which consists of polynomials in $z=(z_0, z_1, z_2, z_3)$. Write H^2(^4)⊗Λ^1(^3) = {∑_a=1^3 f_a ⊗dx^a: f_a ∈H^2(^4)}. Note that this is a space of complex-valued holomorphic sections in the trivial bundle $\Lambda^1(\bR^3) \rightarrow \bC^4$, integrable using a Gaussian measure.

Let $\mathcal{H}^2(\bC^4)$ be the smallest Hilbert space containing $H^2(\bC^4)$. Also denote ℋ^2(^4)⊗Λ^2(^4) = {∑_0 ≤a< b ≤3 f_ab ⊗dx^a ∧dx^b: f_ab ∈ℋ^2(^4)}. Note that this is a space of complex valued holomorphic sections in the trivial bundle $\Lambda^2(\bR^4) \rightarrow \bC^4$, integrable using a Gaussian measure.

We will further define an inner product on $\mathcal{H}^2(\bC^4)\otimes \Lambda^2(\bR^4)$ by ⟨∑_0≤a< b ≤3 f_ab ⊗dx^a∧dx^b, ∑_0≤a< b ≤3 f̂_ab ⊗dx^a∧dx^b ⟩= ∑_0≤a< b ≤3 ⟨f_ab, f̂_ab ⟩. Here, $f_{ab}$ and $\hat{f}_{ab}$ are in $\mathcal{H}^2(\bC^4)$.

We will continue to use $\langle\cdot, \cdot \rangle $ to denote the inner product on $\mathcal{H}^2(\bC^4)\otimes \Lambda^2(\bR^4)$ and
write $d\lambda_4 = \frac{1}{\pi^4}e^{-\sum_{i=0}^3|z_i|^2} \prod_{i=0}^3 dx^i dp^i$. Note that ⟨f_ab, f̂_ab ⟩= ∫_^4 f_abf̂_ab dλ_4.

Recall the inner product we are using in $\overline{\mathcal{S}_\kappa(\bR^4)\otimes \Lambda^1(\bR^3)}$ is $\langle d \cdot, d\cdot \rangle$. We are now going to construct an isometry between $\mathcal{S}_\kappa(\bR^4)\otimes \Lambda^1(\bR^3)$ and $H^2(\bC^4)\otimes \Lambda^1(\bR^3)$, using $\langle d \cdot, d\cdot \rangle$.

The Hermite polynomials satisfy the following property h_n+1(x) = xh_n(x) - h_n'(x). Therefore, d/dx( h_n(x) e^-x^2/4 ) = [ xh_n(x) - h_n+1(x) - x/2h_n(x) ] e^-x^2/4. But we also have $h_n' = nh_{n-1}$, which means that $xh_n(x) = h_{n+1}(x) + nh_{n-1}(x)$. Thus,

\begin{align}
\frac{d}{dx}\left( h_n(x) e^{-x^2/4} \right) =& \left[ \frac{1}{2}h_{n+1}(x) + \frac{n}{2}h_{n-1}(x) - h_{n+1}(x) \right] e^{-x^2/4} \nonumber \\
=& \left[\frac{n}{2}h_{n-1}(x)-\frac{1}{2}h_{n+1}(x) \right]e^{-x^2/4}. \nonumber
\end{align}

Define an operator $\mathfrak{d}_a$ by 𝔡_a z_a^n= n/2z_a^n-1 - 1/2z_a^n+1,  for $a = 0, 1, 2, 3$.

For any $f_a \in H^2(\bC^4)$, define ∑_a=1^3 f_a ⊗dx^a := ∑_a=1^3 _0 f_0 dx^0 ∧dx^a + ∑_1 ≤i < j ≤3(-1)^ij[_i f_j - _j f_i ]dx^i ∧dx^j. Furthermore, define a bilinear form $\langle \cdot, \cdot \rangle_{\fd,\kappa}$ in $H^2(\bC^4)\otimes \Lambda^1(\bR^3)$ by ⟨∑_a f_a ⊗dx^a, ∑_b g_b ⊗dx^b ⟩_,κ = κ^2⟨∑_a f_a ⊗dx^a, ∑_b g_b ⊗dx^b ⟩.

Just as we have the differential operator $d$ acting on 1-forms in $\mathcal{S}_\kappa(\bR^4) \otimes \Lambda^1(\bR^3)$, the analogous operator will be $\fd$ acting on $\Lambda^1(\bR^3)$-valued holomorphic functions, which are in $H^2(\bC^4) \otimes \Lambda^1(\bR^3)$.

The Segal Bargmann transform $\Psi_\kappa$ actually maps Hermite polynomials in $\bR^4$ to complex holomorphic functions in $\bC^4$, equipped with the Gaussian measure as defined in Equation (<ref>).

(Segal Bargmann Transform)

Refer to Notation <ref>. Using the Segal Bargmann Transform $\Psi_\kappa: \overline{\mathcal{S}}_\kappa(\bR^4) \rightarrow \mathcal{H}^2(\bC^4)$, we map Ψ_κ:  H_p_r(κx^0, κx^1, κx^2, κx^3)√(ϕ_κ)/√(p_r!) ⟼z^p_r/√(p_r)! ≡z_0^i_0z_1^i_1z_2^i_2z_3^i_3/√(i_0!i_1!i_2!i_3!), if $p_r = (i_0, i_2, i_2, i_3)$.
Clearly, it is an isometry.

We can extend this isometry to tensor and direct products and by abuse of notation, use the same symbol.
That is, $\Psi_\kappa:  \overline{\mathcal{S}}_\kappa(\bR^4) \otimes \Lambda^2(\bR^4) \rightarrow \mathcal{H}^2(\bC^4) \otimes \Lambda^2(\bR^4)$ by Ψ_κ[ ∑_0≤a< b≤3 f_a,b ⊗dx^a ∧dx^b ] = ∑_0≤a<b ≤3 Ψ_κ[f_a,b] ⊗dx^a ∧dx^b.

By abuse of notation, we use the same symbol to define an isometry that sends this Schwartz space $\mathcal{S}_\kappa(\bR^4) \otimes \Lambda^1(\bR^3)$ into $H^2(\bC^4) \otimes \Lambda^1(\bR^3)$, over in $\bC^4$. Note that $H^2(\bC^4) \otimes \Lambda^1(\bR^3)$ is complex-valued holomorphic sections of the trivial bundle $\Lambda^1(\bR^3) \rightarrow \bC^4$, integrable using a Gaussian measure. Alternatively, one can view this space as $\Lambda^1(\bR^3)$-valued holomorphic functions in $\bC^4$, which are integrable with respect to the Gaussian measure.

By definition of $\fd_a$, we note that Ψ_κ(d f) = κ(Ψ_κf), f ∈𝒮_κ(^4) ⊗Λ^1(^3). Thus we have the following result.

We have $\langle \cdot, \cdot \rangle_{\fd,\kappa}$ is an inner product. Thus, we let $|\cdot|_{\fd,\kappa}$ denote the norm using this inner product and $(H^2(\bC^4)\otimes \Lambda^1(\bR^3), \langle \cdot, \cdot \rangle_{\fd,\kappa})$ denote this inner product space.

It follows from $\langle \Psi_\kappa \omega, \Psi_\kappa \omega\rangle_{\fd,\kappa} = \langle d\omega, d\omega \rangle$.

Henceforth, complete $(H^2(\bC^4)\otimes \Lambda^1(\bR^3), \langle \cdot, \cdot \rangle_{\fd,\kappa})$ into a Hilbert space, denoted by $\overline{H^2(\bC^4)\otimes \Lambda^1(\bR^3)}$. Note that the Segal Bargmann Transform $\Psi_\kappa$ extends into an isometry between
$\overline{\mathcal{S}_\kappa(\bR^4)\otimes \Lambda^1(\bR^3)}$ and $\overline{H^2(\bC^4)\otimes \Lambda^1(\bR^3)}$.

§.§ Abstract Wiener space

Let $(H, \langle \cdot, \cdot \rangle)$ be a (real) infinite dimensional Hilbert space.

* Let $\mathcal{F}$ be a partial ordered set of finite dimensional orthogonal projections onto $H$, i.e. $P > Q$ if $QH \subseteq PH$.

* Let $P \in \mathcal{F}$. Given any Borel subset $F  \subseteq PH$, define for $\kappa > 0$,  μ_κ( x ∈P^-1(F) )= ( κ/2π)^l/2∫_y ∈F e^-κ|y|^2 /2 dy ,  where $l$ is the dimension of $P H$.

* A semi-norm $\parallel \cdot \parallel$ in $H$ is called measurable if for every $\epsilon > 0$, there exists a $P_0 \in \mathcal{F}$ such that μ_κ(∥P x ∥> ϵ) < ϵfor all $P \perp P_0$ and $P \in \mathcal{F}$.

The Hilbert space we are going to complete will be $\overline{H^2(\bC^4)\otimes \Lambda^1(\bR^3)}$. Recall the inner product in this Hilbert space is given by Equation (<ref>). As explained in [6], an infinite dimensional Gaussian measure does not exist in $\overline{H^2(\bC^4) \otimes \Lambda^1(\bR^3)}$. We need to complete the space $\overline{H^2(\bC^4) \otimes \Lambda^1(\bR^3)}$ into a Banach space, denoted $B(\bR^4; \fd)$, using a measurable norm. This space $B(\bR^4; \fd)$ is an Abstract Wiener space in the sense of Gross and it supports a Gaussian measure required for our definition of the path integral.

Refer to Notation <ref>. Let V_a, r = span { z^p_s ⊗dx^a: s ≤r-1, p_s ∈𝒫_s}. The following observations are useful.

* For $a = 1, 2, 3$, $\fd [f \otimes dx^a] = \sum_{i\neq a} \fd_i f \otimes dx^i \wedge dx^a$.

* For any $p_r = (i_0, i_1, i_2, i_3)$, it is clear that the only vectors in $V_{a,r}$ which are not orthogonal to $z^{p_r}$, are of the form $z^{q_{r-2}} \otimes dx^a$, where $q_{r-2}$ are of the form (i_0-2, i_1, i_2, i_3), (i_0, i_1-2, i_2, i_3), (i_0, i_1, i_2-2, i_3) and (i_0, i_1, i_2, i_3-2). If $i_k - 2 < 0$, we take it to be 0.

* Last but not least, $z^{p_r} \otimes dx^a$ is orthogonal to $z^{q_r} \otimes dx^a$ for any $p_r \neq q_r$.

Let $\hat{z}^{p_r} \otimes dx^a$ denote the projection of the vector $z^{p_r} \otimes dx^a$ which is orthogonal to $V_{a,r}$ using the inner product $\langle \cdot, \cdot \rangle_{\fd,\kappa}$. And for $p_r = (i, j, k, l) \in \mathcal{P}_r$, we write $p_r^{\alpha,\pm} \in \mathcal{P}_{r\pm 2}$ by adding $\pm2$ to the $\alpha$-entry in $p_r$. If the $\alpha$-entry is 0 or 1, then $p_r^{\alpha,-}$ is the empty set.

Let $a=1,2,3$. For any orthogonal projection $P$, we let $P_a$ be the orthogonal projection onto the range of $P$ which intersects $\overline{H^2(\bC^4) \otimes dx^a}$, closed using $|\cdot|_{\fd,\kappa}$.

Note that { ẑ^p_r ⊗dx^a/|ẑ^p_r ⊗dx^a|_,κ: p_r ∈𝒫_r, r ≥0} is an orthonormal set and we can write

\begin{align}
\hat{z}^{p_r} \otimes dx^a = z^{p_r} \otimes dx^a - \sum_{\alpha=0}^3 \left\langle z^{p_r} \otimes dx^a, \frac{\hat{z}^{p_r^{\alpha,-}} \otimes dx^a}{|\hat{z}^{p_r^{\alpha,-}}\otimes dx^a|_{\fd,\kappa}}\right\rangle_{\fd,\kappa}\frac{\hat{z}^{p_r^{\alpha,-}}\otimes dx^a}{|\hat{z}^{p_r^{\alpha,-}}\otimes dx^a|_{\fd,\kappa}}. \label{e.x.1}
\end{align}

* We have |z^p_r ⊗dx^a|_,κ ≥κ/2√(p_r! ). Similarly, |ẑ^p_r ⊗dx^a|_,κ ≥κ/2√(p_r! ).
* We have |z^p_r ⊗dx^a|_, κ ≤3(r+1)κ√(p_r!).  Thus,
|ẑ^p_r| ≤|z^p_r| + 6(r+1)^2∑_α=0^3 |z^p_r^α,-|.  Here, $|\hat{z}^{p_r}|$ refers to the absolute value of of $\hat{z}^{p_r}$.

By definition of $\fd_0$, we see that |_0 z_0^n|^2 = n^2/4(n-1)! + 1/4(n+1)! .  Hence, $|z^{p_r} \otimes dx^a|_{\fd,\kappa} \geq \frac{\kappa}{2}\sqrt{p_{r}! }$. From Equation (<ref>), we see that $\fd (z^{p_r} \otimes dx^a)$ will contribute a term $f \equiv z_0^{i_0+1}z_1^{i_1}z_2^{i_2}z_3^{i_3} \otimes dx^0 \wedge dx^a$ when we apply the operator $\fd$, if $p_r = (i_0, i_1, i_2, i_3)$. Note that $\fd (z^{p_r^{\alpha,-}}\otimes dx^a)$ will not contribute any term that will cancel the term $f$. Using Equation (<ref>), |z^p_r^α,-⊗dx^a|^2_, κ ≥κ^2/4p_r!. This gives us the lower bound.
When we apply $\fd$ to $z^{p_r}\otimes dx^a$, we will obtain 3 terms, and from Equation (<ref>), we obtain |z^p_r⊗dx^a|_, κ ≤3κ×[ r/2√(p_r!) + √(p_r! ·(r+1))/2 ]≤3κ(r+1)√(p_r!). Together with Equation (<ref>), we see that |z^p_r ⊗dx^a/|ẑ^p_r^α,-⊗dx^a|_,κ |_, κ ≤3κ(r+1)√(p_r!) /κ√(p_r!/r^2)/2 ≤6(r+1)^2. Therefore,
|ẑ^p_r| ≤|z^p_r| + 6(r+1)^2∑_α=0^3 |z^p_r^α,-|.

Now, we are ready to choose a measurable norm, so as to define an infinite dimensional Gaussian measure on $\overline{H^2(\bC^4) \otimes \Lambda^1(\bR^3)}$.

(Measurable norm)

Let x_a ⊗dx^a = ∑_r∑_p_r∈𝒫_rc_a, p_r ẑ^p_r/|ẑ^p_r⊗dx^a|_,κ ⊗dx^a and $c_{a, p_r} \in \bR$.
Introduce a norm $\parallel \cdot \parallel$ by setting ‖∑_a=1^3 x_a ⊗dx^a ‖= sup_z∈B(0,1/2) ∑_a=1^3 ∑_r ∑_p_r ∈𝒫_r |c_p_r,a|[ |z^p_r| + 6(r+1)^2∑_α=0^3 |z^p_r^α,-|].  Here, $B(0,1/2)$ is the ball with radius $1/2$, center 0 in $\mathbb{C}^4$.

The norm $\parallel \cdot \parallel$ is measurable in $(\overline{H^2(\bC^4)\otimes \Lambda^1(\bR^3)}, \langle \cdot, \cdot \rangle_{\fd,\kappa})$.

Now $H^2(\bC^4) \otimes dx^b \equiv \{ f \otimes dx^b:\ f \in H^2(\bC^4)\}$ is a subspace in $\overline{H^2(\bC^4)\otimes \Lambda^1(\bR^3)}$ and we denote its closure, using $\langle \cdot, \cdot \rangle_{\fd,\kappa}$, by $\overline{H^2(\bC^4) \otimes dx^b}$.

It suffices to show that for any projection $P = \sum_{a=1}^3 P_a$ orthogonal to $P_o$, μ_κ( ∥P_a x ∥≥ϵ) ≤ϵ/3. We will determine $P_o$ later.
Let $\epsilon > 0$ and choose a $\tilde{N}$ large such that ∑_k ≥Ñ 102k^3(k+1)^2  2^-k/2< κ^1/4ϵ/√(3) and let $P_{o,b}$ be a finite dimensional orthogonal projection onto the span of { z^p_r ⊗dx^b: p_r ∈𝒫_r, r ≤Ñ }. Let $V \subseteq \overline{H^2(\bC^4) \otimes dx^b}$ be a finite dimensional subspace in the orthogonal complement of the range of $P_{o,b}$ and let $\{\beta_{1,b}, \ldots \beta_{l,b}\}$ be an orthonormal basis for $V$. Let $P_{b}$ denote the projection onto $V\subseteq \overline{H^2(\bC^4) \otimes dx^b}$. Now, there are at most $k^{3}$ $4$-tuples $(m_1,m_2, m_3, m_4)$ with $\sum_{j=1}^4 m_j = k$. It is possible to write each basis vector as a linear combination β_s,b =  ∑_p_r ≥q^s a^s_p_r,b ẑ^p_r ⊗dx^b/|ẑ^p_r⊗dx^b|_,κ, with $q^{1} < q^{2} < \ldots < q^{l}$ and each $q^t$ is a $4$-tuple $(m^t_1,m^t_2,m^t_3, m^t_4)$ with $\sum_{j=1}^4 m^t_j \geq \tilde{N}$. Observe that $|a^{s}_{p_r,b}| \leq 1$.

Any projected vector $P_bx$ can be written as $P_bx =  \sum_s c_{s}(x) \beta_{s,b}$. By definition (See Equation (<ref>).), $c_s(\cdot): \overline{H^2(\bC^4) \otimes dx^b} \rightarrow \mathbb{R}$ is a Gaussian random variable with variance $1/\kappa$. Let $\mathbb{E}$ denote the expectation of  a Gaussian random variable.

\begin{align}
\mu_{\kappa}\left( \parallel P_bx \parallel > \epsilon \right)
\leq& \mu_{\kappa}\left( \sup_{\vec{z}\in B(0,1/2)}\sum_s \sum_{p_r \geq q^s} |c_{s} a^{s}_{p_r,b}|\left[ |z^{p_r}|+ 6(r+1)^2\sum_{\alpha=0}^3|z^{p_r^{\alpha,-}}| > \epsilon \right]\right) \nonumber \\
\leq& \frac{1}{\epsilon}\sup_{z\in B(0,1/2)}\sum_s \sum_{p_r \geq q^s} \mathbb{E}|c_{s} a^{s}_{p_r,b}| \left[|z^{p_r}|+ 6(r+1)^2\sum_{\alpha=0}^3|z^{p_r^{\alpha,-}}| \right] \nonumber \\
\leq& \frac{1}{\epsilon}\sup_{z\in B(0,1/2)}\sum_s (\mathbb{E}|c_{s}|^2 )^{1/2} \sum_{p_r \geq q^s}\left[|z^{p_r}|+ 6(r+1)^2\sum_{\alpha=0}^3|z^{p_r^{\alpha,-}}| \right] \nonumber \\
\leq& \frac{1}{\sqrt\kappa\epsilon}\sum_{s\geq \tilde{N}}\frac{1}{\sqrt{2^s}} \sum_{k\geq \tilde{N}} \frac{17k^{3}6(k+1)^2}{\sqrt{2^k}} \nonumber \\
\leq& \frac{1}{\sqrt\kappa\epsilon}  \left(\sum_{k \geq \tilde{N}}  102(k+1)^2k^{3} 2^{-k/2}\right)^2 < \epsilon/3. \nonumber
\end{align}

The proof is thus complete by choosing $P_o = \sum_{a=1}^3 P_{o,a}$.

Complete $(\overline{H^2(\bC^4)\otimes \Lambda^1(\bR^3)}, \langle \cdot, \cdot \rangle_{\fd,\kappa})$ into a Banach space $B(\bR^4; \fd)$, and form a triple,

$(i, \overline{H^2(\bC^4)\otimes \Lambda^1(\bR^3)},B(\bR^4; \fd))$, an Abstract Wiener Space. Identify y ∈B(^4; )^∗⊆H^2(ℂ^4)⊗Λ^1(^3)⊆B(^4; ) with an element in $\overline{H^2(\mathbb{C}^4)\otimes \Lambda^1(\bR^3)}$ and denote the pairing $(x, y)_\sharp = y(x)$. Here, $y \in B(\bR^4; \fd)^\ast$.

We will always write $(\cdot, \cdot)_\sharp$ to denote a pairing in this article.

(Gaussian measure on Abstract Wiener Space)

* Define $\tilde{\mu}_{\theta}$, a measure on $B(\bR^4; \fd)$ with variance $1/\theta$, by
\begin{align*} \tilde{\mu}_{\theta}\Big\{x \in B(\bR^4; \fd): \Big( (x,y_1)_\sharp, & \ldots, (x,y_k)_\sharp \Big) \in F \Big\} \\
&= \mu_{\theta} \Big\{x \in \overline{H^2(\mathbb{C}^4)\otimes \Lambda^1(\bR^3)}: \Big( \langle x,y_1 \rangle, \ldots, \langle x,y_n \rangle \Big) \in F \Big\}. \nonumber
\end{align*}

The $y_j$'s are in $B(\bR^4; \fd)^\ast$.

* $\left\{x \in B(\bR^4; \fd): \Big( (x,y_1)_\sharp, \ldots, (x,y_n)_\sharp \Big) \in F \right\}$ is called a cylinder set in $B(\bR^4; \fd)$. Let $\mathcal{R}_{B(\bR^4; \fd)}$ be the collection of cylinder sets in $B(\bR^4; \fd)$.

* It was shown by Gross that $\tilde{\mu}_{\theta}$ is $\sigma$-additive in the $\sigma$-field generated by $\mathcal{R}_{B(\bR^4; \fd)}$.

* Extend $\tilde{\mu}_{\theta}$ over the Borel field of $B(\bR^4; \fd)$.

* It can be shown that the $\sigma$-field generated by $\mathcal{R}_{B(\bR^4; \fd)}$ is equal to the Borel field of $B(\bR^4; \fd)$.

Now, any $x \in B(\bR^4; \fd)$ can be written as x = ∑_a∑_r∑_p_r∈𝒬_r c_p_r,a ẑ^p_r⊗dx^a/|ẑ^p_r⊗dx^a|_, κ ,  convergence in the sense of $\parallel \cdot \parallel$.  This space $B(\bR^4; \fd)$ can be described explicitly. Let ℋ^2(ℂ^4)_ℂ = ℋ^2(ℂ^4)⊗_ℝ ℂ   and   B( ^4; )^∗_ℂ = B(^4; )^∗ ⊗_ℝ ℂ.

Let $x$ be given by Equation (<ref>).

* For $w \in \mathbb{C}^n$, define ζ_a(w) : x = ∑_a x_a ⊗dx^a ∈B(^4; ) ↦⟨x(w), dx^a ⟩= x_a(w). Then $\zeta_a(w)$ is in $ B(\bR^4; \fd)^{\ast}_{\mathbb{C}}$.

* For $w \in \mathbb{C}^n$, define ξ̃_ab^κ(w) : x = ∑_a x_a ⊗dx^a ∈B(^4; ) ↦⟨κx(w), dx^a ∧dx^b ⟩= κ[_a x_b - _b x_a]. Then $\tilde{\xi}_{ab}^\kappa(w)$ is in $ B(\bR^4; \fd)^{\ast}_{\mathbb{C}}$.

* Let $R \geq 2|w|+1$. Choose $M > 0$ such that for all $r > M$, $R^r \leq \sqrt{ \lfloor(r/4)\rfloor!}$. Note that $p_r! \geq \lfloor(r/4)\rfloor!$. Then,

\begin{align}
|\langle x(w), dx^a \rangle| \leq& \left|\sum_r\sum_{p_r\in \mathcal{Q}_r} \frac{c_{p_r,a} \hat{w}^{p_r}}{ |\hat{w}^{p_r}\otimes dx^a|_{\fd,\kappa}} \right| \nonumber \\
\leq & R^{M}\sum_{r \leq M}\sum_{p_r\in \mathcal{Q}_r} |c_{p_r,a}|\left[ \left| \left( \frac{w}{R} \right)^{p_r} \right| + 6(r+1)^2 \sum_{\alpha=0}^3\left| \left( \frac{w}{R} \right)^{p_r^{\alpha,-}} \right|\right]\frac{2}{\kappa\sqrt{p_r !}} \nonumber \\
& +  \sum_{r > M} \sum_{p_r\in \mathcal{Q}_r}|c_{p_r,a}|\left[ \left| \left( \frac{w}{R} \right)^{p_r} \right| + 6(r+1)^2\sum_{\alpha=0}^3\left| \left( \frac{w}{R} \right)^{p_r^{\alpha,-}} \right|\right] \frac{2R^r}{\kappa\sqrt{p_r !}} \nonumber \\
<& \frac{2(R^{M}+1)}{\kappa} \parallel x \parallel. \label{e.r.1}
\end{align}

This shows that $\zeta(w): x \rightarrow \langle x(w), dx_a \rangle$ is a bounded complex functional on $B(\bR^4; \fd)_{\mathbb{C}}$.

* Let $R \geq 2|w|+1$. Choose $M > 0$ such that for all $r > M$, $4(r+1)R^r \leq \sqrt{ \lfloor(r/4)\rfloor!}$. Note that $p_r! \geq \lfloor(r/4)\rfloor!$. From Equation (<ref>), note that |_a z_a^n| ≤(n+1)|z_a^n-1| ≤2(n+1)|z_a|^n, |z_a| < 1/2. Then,

\begin{align}
|\langle \kappa&\fd x(w), dx^a \wedge dx^b \rangle| \nonumber \\
\leq& \kappa\left|\sum_r\sum_{p_r\in \mathcal{Q}_r} c_{p_r,b} \frac{\fd_a\hat{w}^{p_r}}{|\hat{w}^{p_r}\otimes dx^b|_{\fd,\kappa}} \right| + \kappa\left|\sum_r\sum_{p_r\in \mathcal{Q}_r} c_{p_r,a} \frac{\fd_b\hat{w}^{p_r}}{|\hat{w}^{p_r}\otimes dx^a|_{\fd,\kappa}} \right|\nonumber \\
\leq & R^{M}\sum_{r \leq M}\sum_{p_r\in \mathcal{Q}_r} \left[|c_{p_r,b}|+|c_{p_r,a}|\right]\left[ \left| \left( \frac{w}{R} \right)^{p_r} \right| + 6(r+1)^2\sum_{\alpha=0}^3\left| \left( \frac{w}{R} \right)^{p_r^{\alpha,-}} \right|\right]\frac{4(r+1)}{\sqrt{p_r !}} \nonumber \\
& +  \sum_{r > M} \sum_{p_r\in \mathcal{Q}_r}\left[|c_{p_r,a}| + |c_{p_r,b}|\right]\left[ \left| \left( \frac{w}{R} \right)^{p_r} \right| + 6(r+1)^2\sum_{\alpha=0}^3\left| \left( \frac{w}{R} \right)^{p_r^{\alpha,-}} \right|\right] \frac{4(r+1)R^r}{\sqrt{p_r !}} \nonumber \\
<& (R^{M}+1) \parallel x \parallel.
\end{align}

This shows that $\tilde{\xi}_{ab}^\kappa(w): x \rightarrow \langle \fd x(w), dx^a \wedge dx^b \rangle$ is a bounded complex functional on $B(\bR^4; \fd)_{\mathbb{C}}$.

Both $\zeta_a(w)$ and $\tilde{\xi}_{ab}(x)$ are like the Dirac Delta functions, except that over in $(\overline{H^2(\bC^4)\otimes \Lambda^1(\bR^3)}, \langle \cdot, \cdot \rangle_{\fd,\kappa})$, they are honest linear functionals in $B(\bR^4; \fd)_{\mathbb{C}}^\ast$. We refer $\zeta_a(w)$ as an evaluation form, for the rest of this article. This functional will not be used for the Abelian case; it will only be used in the non-Abelian case.

The other linear functional  $\tilde{\xi}_{ab}(x)$ plays an important role in the definition of the Yang-Mills path integral, which we will in future refer to it as an evaluation differential form. We just showed that it lies in the Hilbert space $\overline{H^2(\bC^4) \otimes \Lambda^1(\bR^3)}$.

This evaluation differential form does the following. For each $x \in \bC^4$, it sends ξ̃_ab(x): Ψ_κ[∑_1≤i ≤3g_i ⊗dx^i ] ∈H^2(^4) ⊗Λ^1(^3) ⟼Ψ_κ[∂_a g_b - ∂_b g_a](x) ∈. Here, $\sum_{1\leq i \leq 3}g_i \otimes dx^i \in \mathcal{S}_\kappa(\bR^4) \otimes \Lambda^1(\bR^3)$. We are going to use it to construct $\tilde{\nu}_S$, which we talked about earlier in Equation (<ref>).

The support of $\tilde{\mu}_{\theta}$ is on continuous $\Lambda^1(\bR^3)$-valued holomorphic functions on $\mathbb{C}^4$.

It suffices to prove for the case when $x = f \otimes dx^a$.
Fix a $w_0 = (t_1,t_2,t_3, t_4)$.
Now for any $w\in \mathbb{C}$, ⟨x(w_0 + w) - x(w_0), dx^a ⟩= ( x, ζ_a(w_0+w) - ζ_a(w_0) )_♯for $x \in B(\bR^4; \fd)_{\mathbb{C}}$. From Equation (<ref>), it is clear that $\zeta_a(\cdot)$ is continuous at 0 using the operator norm.

Fix a $z_0 \neq0 \in \mathbb{C}$ and let $|z_0| \leq R$. Let $D = \{z \in \mathbb{C}: |z - z_0|< R/4\} \subseteq \mathbb{C}$ and $\gamma$ be the boundary of $D$. By Cauchy integral formula, z_0^k = ∫_γz^k/z - z_0 dz. Thus, for $z_1 $ close to $z_0$, i.e. $|z_0 - z_1| < R/8$, |z_1^k - z_0^k| = | ∫_γz^k(z_1-z_0)/(z-z_1)(z-z_0) dz| ≤2π(2R+1)^k|z_1 - z_0|/(R/8)^2. Thus, if write $p_r = (k_1,k_2,k_3, k_4)$ and $w+w_0 = (y_1,y_2,y_3, y_4)$, then

\begin{align}
|(w+ w_0)^{p_r} - w_0^{p_r}| &= \left|(t_1^{k_1}-y_1^{k_1})t_2^{k_2}t_3^{k_3} t_4^{k_4} + y_1^{k_1}(t_2^{k_2}-y_2^{k_2})t_3^{k_3} t_4^{k_4} + \cdots +y_1^{k_1}y_2^{k_2}y_3^{k_3}(t_4^{k_4}-y_4^{k_4}) \right| \nonumber \\
&\leq \frac{2\pi \cdot 4(2|w_0| + 1)^r|w|}{(|w_0|/8)^2} . \label{e.r.6}
\end{align}

Let $x = \sum_r\sum_{p_r\in \mathcal{Q}_r} c_{p_r} \hat{w}^{p_r}\otimes dx^a/|\hat{w}^{p_r} \otimes dx^a|_{\fd,\kappa} $. From Equation (<ref>), choose $R > 0$ such that |(w + w_0)^p_r - w_0^p_r| + 6(r+1)^2∑_α=0^3 |(w + w_0)^p_r^α,- - w_0^p_r^α,-| ≤R^r|w| for all $w \in B(w_0,2|w_0|)$. Then, if $ |w|<\epsilon $,

\begin{align}
\sup_{w \in B(0,\epsilon)}&|\langle x(w+ w_0) - x(w_0), dx^a \rangle| \nonumber \\
\leq& \sup_{w \in B(0,\epsilon)}   \sum_r\sum_{p_r\in \mathcal{P}_r} |c_{p_r}| \frac{2|\widehat{(w+w_0)}^{p_r} - \widehat{w_0}^{p_r}|}{\kappa\sqrt{p_r!}} \nonumber \\
&+\sup_{w \in B(0,\epsilon)}   \sum_{\alpha=0}^3\sum_r\sum_{p_r\in \mathcal{P}_r} |c_{p_r}| \frac{6(r+1)^2 \times 2|\widehat{(w+w_0)}^{p_r^{\alpha,-}} - \widehat{w_0}^{p_r^{\alpha,-}}|}{\kappa\sqrt{p_r!}} \nonumber \\
\leq& \frac{2}{\kappa}\sup_{w \in B(0,\epsilon)}   \sum_r\sum_{p_r\in \mathcal{P}_r} \frac{1}{2^r}|c_{p_r}| \frac{\left(2^rR^r|w|\right)}{\sqrt{p_r!}} \nonumber \\
&+\frac{2}{\kappa}\sup_{w \in B(0,\epsilon)}   \sum_{\alpha=0}^3\sum_r\sum_{p_r\in \mathcal{P}_r} \frac{6(r+1)^2}{2^{r-2}}|c_{p_r}| \frac{\left(2^{r-2}R^{r-2}|w|\right)}{\sqrt{p_r!}} \nonumber \\
\leq& \frac{2 c(w_0)}{\kappa}\cdot \epsilon \parallel x \parallel \nonumber
\end{align}

for some constant $c(w_0)$.

Thus for any $\epsilon > 0$

\begin{align}
\tilde{\mu}_{\kappa}\left( \sup_{w \in B(w_0,1/k)}|\langle x(w_0 + w) - x(w_0), dx^a \rangle | > \epsilon \right) =&\ \tilde{\mu}_{\kappa}\left( \sup_{w \in B(w_0,1/k)}|(x, \zeta_a(w_0+w) - \zeta_a(w_0))_\sharp | > \epsilon \right) \nonumber \\
\leq&\ \tilde{\mu}_{\kappa}\left( \frac{2c(w_0)}{\kappa k}\parallel x \parallel > \epsilon \right)
\longrightarrow \ 0 \nonumber
\end{align}

as $k$ goes to infinity. Let E_k = {x: sup_w ∈B(w_0,1/k)|(x, ζ_a(w_0+w) - ζ_a(w_0))_♯| > ϵ}. Choose an increasing subsequence $\{ r_k \}_{k=1}^\infty$ in $\mathbb{N}$ such that $\sum_k \tilde{\mu}(E_{r_k}) < \infty$.  Then, by the choice of $r_k$, $\sum_k \tilde{\mu}(E_{r_k}) < \infty$ and hence by the Borel Cantelli Lemma, ⋂_q=1^∞⋃_p = q^∞E_r_p is a null set with respect to the measure $\tilde{\mu}_\theta$. Hence $\tilde{\mu}_\theta$-almost surely, for each $x$, there exists a $r_k(x)$ such that sup_w ∈B(w_0,1/r_k)|⟨x(w_0 + w) - x(w_0), dx^a ⟩| < ϵ.

Let $\chi_w(z) = e^{\bar{w}z}$, where $w, z \in \bC^4$. We have |ζ_a(w)|_,κ ≤2/κe^|w|^2/2.

Now by definition of $\fd_a$, we note that for any $n \geq 0$, ⟨_a z_a^n/√(n!), _a z_a^n/√(n!) ⟩≥n+1/4⟨z_a^n+1/√((n+1)!),  z_a^n+1/√((n+1)!) ⟩. Thus, it follows that $|\langle A, A \rangle|^{1/2} \leq \frac{2}{\kappa}|A|_{\fd,\kappa}$, if $A = f \otimes dx^a$ for $f \in H^2(\bC^4)$.

A simple computation will show that $\langle A, \zeta_a(w) \rangle_{\fd,\kappa} = \langle A(w), dx^a \rangle = \langle f, \chi_w \rangle = f(w)$. Then we have

\begin{align}
|A(w)| =& \left|\left\langle f, \chi_w \right\rangle \right| \leq |f| |\chi_w| \leq \frac{2}{\kappa}|A|_{\fd,\kappa}| \chi_w|. \nonumber
\end{align}

A direct computation will show that $|\chi_w|= \langle \chi_w, \chi_w \rangle^{1/2} = e^{|w|^2/2}$ and this completes the proof.

§.§ A note on Tensor Products

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 the tensor product of Hilbert spaces, with the inner product ⟨f_1 ⊗g_1, f_2 ⊗g_2 ⟩= ⟨f_1, f_2 ⟩_1 ⟨g_1 , g_2 ⟩_2. Now suppose $H_2$ is finite dimensional with dimension $n$. Then, $H_1 \otimes H_2 \cong H_1^{\times^n}$. 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 measure $\tilde{\mu}^{\times^n}$.

Suppose we have an Abstract Wiener space, $B$, with a Hilbert space $H \subset B$ which is dense and let $B^\ast \subset H$ be the dual space to $B$. Let $f \in B^\ast$, $(x, f)_\sharp \in \bR$. Complexify $B$ into $B \otimes_\bR \mathbb{C}$, and extend $f$ to be a complex valued linear functional on $B \otimes_\bR \mathbb{C}$.

When we write $B^{\otimes^k}$, we mean the $k$ 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 ∑_α∑_β(x_1^α⊗x_2^α⊗⋯⊗x_k^α, z_1^β⊗z_2^β⊗⋯⊗z_k^β)_♯= ∑_α∑_β(x_1^α, z_1^β)_♯(x_2^α, z_2^β)_♯⋯(x_k^α, z_k^β)_♯. By abuse of notation, we will use the same symbol $(\cdot, \cdot)_\sharp$ to denote the pairing used in the $k$ tensor product of $B$.

In the rest of the article, we equip $B(\bR^4; \fd)$ with a Wiener measure $\tilde{\mu}$ with variance 1. Note that $B(\bR^4; \fd)$ is actually a probability space with Wiener measure as the probability measure. We will often write for a continuous function $f: B(\bR^4; \fd) \rightarrow \bR$, the expectation of $f$ as [f] ≡∫_B(^4; )f dμ̃.

§ DEFINITION OF THE YANG-MILLS PATH INTEGRAL IN ABELIAN CASE

Consider $G$ Abelian, i.e. $G = U(1)$ The Lie algebra $\mathfrak{u}(1)$ will be isomorphic to $\bR \otimes \sqrt{-1}$. Let $ A = \sum_{j=1}^3 A_j \otimes dx^j,\ A_j \in C^\infty(\bR^4) $. In this section, we want to define a functional integral of the form 1/Z∫_A e^∫_C ∑_j=1^3 A_j ⊗dx^j ⊗ie^-|dA|^2/2 DA, i = √(-1), in the Abelian case, since $A \wedge A = 0$.  Let $\sigma: [0,1]^2\equiv I^2 \rightarrow \bR^4$ be any parametrization of the surface $S$ such that $\partial S = C$, $C$ is a simple closed curve.

Using Stokes Theorem, $\int_C \sum_{j=1}^3 A_j \otimes dx^j = \int_S dA$, where $A = \sum_{j=1}^3 A_j \otimes dx^j$ and $C = \partial S$. Motivated by the case in $\bR^2$, we will make sense of 1/Z∫_A e^∫_S dA ⊗ie^-|dA|^2/2 DA for a surface $S$. However, there are many surfaces $S$ with $\partial S = C$. And depending on the surface $S$, we will obtain a different result. Thus, unless we specify a surface $S$, the Yang-Mills path integral is not well-defined for a closed loop $C$.

Now recall our construction of $\overline{H^2(\bC^4) \otimes \Lambda^1(\bR^3)}$ which is contained inside $B(\bR^4; \fd)$ from Subsection <ref>. Our path integral will be made sense of over $B(\bR^4; \fd)$. To do that, we embed $\bR^4 \hookrightarrow \bC^4$ in the standard way. Given the curve $C \in \bR^4$, we scale it by a factor of $\kappa/2$ inside $\bC^4$, hence the surface $S$ is scaled by a factor of $(\kappa/2)^2$. And we replace $d$ by $\kappa\fd$. Therefore, the Yang-Mills path integral we wish to make sense of is of the form 1/Z∫_A ∈H^2(^4) ⊗Λ^1(^3) e^i∫_S κAe^-|κA|^2/2 DA  for a surface $S$.

Recall the linear functional $\tilde{\nu}_S$ defined in Equation (<ref>), which is essential in defining the path integral. Using Proposition <ref>, there exists a $\tilde{\nu}_S^\kappa$ such that ∫_S  κA = ( A, ν̃_S^κ)_♯. Explicitly, if we let $\sigma$ be a parametrisation of $S$, then ν̃_S^κ= ∫_I^2dsdt ∑_0≤a < b ≤3 κ^2/4|J_ab^σ|(s,t)ξ̃_ab^κ(κσ(s,t)/2) , whereby $\left(\sum_{1\leq i \leq 3}f_{i}\otimes dx^i, \tilde{\xi}_{ab}^\kappa(w)  \right)_\sharp = \kappa(\fd_{a}f_b - \fd_b f_a)(w) $ and $|J_{ab}^\sigma|$ is defined in Definition <ref>. The factor $\kappa^2/4$ comes from the scaling of the curve $C$ in $\bC^4$ by $\kappa/2$.

However, instead of using $\tilde{\xi}_{ab}^\kappa(w)$, we replace it with $\xi_{ab}^\kappa(w)$, where $\xi_{ab}^\kappa(w) = \psi(w) \tilde{\xi}_{ab}^\kappa(w)$. So define ν_S^κ= ∫_I^2ds dt ∑_0≤a < b ≤3 κ^2/4|J_ab^σ|(s,t)ξ_ab^κ(κσ(s,t)/2) .  The factor $\psi_w$ can be regarded as a form of normalization constant, necessary to obtain the Area Law Formula when we take the limit as $\kappa$ goes to infinity.

In the definition of the Chern Simons path integral in [9], this factor $\psi_w$ was also used.

(Definition of Yang-Mills path integral in the Abelian case)

Let $\nu_S^\kappa$ be given by Equation (<ref>). Hence,
we now define the Yang-Mills path integral in the Abelian case in the Quantum space $B(\bR^4; \fd)  \otimes i$, as

\begin{align}
Y(\bR^4, \kappa; S, i) :=& \mathbb{E}_{YM} \Bigg[ \exp\Bigg[ \frac{i}{\kappa}\left( \cdot, \nu_S^\kappa \right)_\sharp \Bigg] \Bigg] \nonumber \\
:=& \int_{A \in B(\bR^4; \fd)} \exp\left[  \frac{1}{\kappa}\left(A, \nu_S^\kappa \otimes i  \right)_\sharp \right] d\tilde{\mu}(A).
\label{e.a.1}
\end{align}

The factor $\dfrac{1}{\kappa}$ is a constant required to obtain non-trivial limits as $\kappa$ goes to infinity.

We will now take the limit as $\kappa$ goes to infinity, which gives us the Wiilson Area Law formula.

Consider the Abelian Lie group $U(1)$. Define the Yang-Mills path integral using Equation (<ref>). Therefore,

\begin{align}
\mathbb{E}_{YM} &\left[ \exp\left[ \frac{1}{\kappa} \sum_{0\leq a<b  \leq 3}\int_{I^2}\frac{\kappa^2}{4}|J_{ab}^{\sigma}|(s,t) \left(\cdot,\xi_{ab}^\kappa(\kappa\sigma(s,t)/2)\otimes   i \right)_\sharp ds dt\right]\right]  \nonumber \\
=& \exp\left[ -\frac{1}{2}\left|\sum_{0\leq a<b  \leq 3}\int_{I^2} dsdt\ \frac{\kappa}{4}|J_{ab}^{\sigma}|(s,t)\xi_{ab}^\kappa(\kappa\sigma(s,t)/2)  \right|_{\fd,\kappa}^2\right] \nonumber \\
\longrightarrow& \exp\left[ -\frac{1}{8}\int_S \rho_S \right], \label{e.qc.2}
\end{align}

as $\kappa$ goes to infinity.

Recall $\psi_w = \psi(w) = e^{-|w|^2/2}/\sqrt{2\pi}$ and $\chi_w(z) = e^{\bar{w}z}$. For $v,w \in \bR^4 \subset \bC^4$, ⟨ψ_wχ_w , ψ_v χ_v ⟩= ψ_wψ_v e^wv = 1/2πe^-|w-v|^2/2. And we can write
\begin{align}
\left\langle \sum_{i=1}^3\sum_r \sum_{p_r} c_{p_r,i} z^{p_r}\otimes dx^i, \xi_{ab}^\kappa(w) \right\rangle_{\fd,\kappa} =& \kappa^2\left\langle \fd \sum_{i=1}^3\sum_r \sum_{p_r} c_{p_r,i} z^{p_r}\otimes dx^i, \fd \xi_{ab}^\kappa(w) \right\rangle \nonumber \\
=& \kappa\sum_r \sum_{p_r} \psi(w)[c_{p_r,b} \fd_a w^{p_r} - c_{p_r,a} \fd_b w^{p_r}]. \nonumber
\end{align}

We leave to the reader to check that for any $f \in \mathcal{H}^2(\bC^4)$, $\chi_w(z) \equiv e^{\bar{w}z}$ is the unique vector such that $\langle f, \chi_w \rangle = f(w)$ for any $w \in \bC^4$. And the subspace spanned by { _a z^p_r: p_r ∈𝒫_r, r ≥0 }, is dense inside $\mathcal{H}^2(\bC^4)$.
Therefore, ξ_ab^κ(w) = 1/κψ_w χ_w ⊗dx^a ∧dx^b. Thus, ⟨ξ_ab^κ(w), ξ_ab^κ(ŵ) ⟩= 1/2πκ^2e^-|w - ŵ|^2/2.

From Lemma <ref>, RHS of Equation (<ref>) is equal

\begin{align}
\exp&\left[ -\left|\frac{1}{\kappa}\sum_{0\leq a<b \leq 3}\int_{I^2}dsdt\ |J_{ab}^\sigma|(s,t)\frac{\kappa^2}{4}\xi_{ab}^\kappa(\kappa\sigma(s,t)/2) \right|_{\fd,\kappa}^2/2\right] \nonumber \\
&=\exp\left[ -\frac{1}{2}\sum_{0\leq a<b \leq 3}\int_{I^2 \times I^2} ds dt d\bar{s}d\bar{t}\ \frac{\kappa^2}{16}|J_{ab}^{\sigma}|(s,t)|J_{ab}^{\sigma}|(\bar{s},\bar{t})\kappa^2\left\langle \fd \xi_{ab}^\kappa(\kappa\sigma(s,t)/2), \fd \xi_{ab}^\kappa(\kappa\sigma(\bar{s}, \bar{t})/2) \right\rangle \right]\nonumber \\
&=\exp\left[ -\frac{1}{8(2\pi)}\sum_{0\leq a<b \leq 3}\int_{I^2 \times I^2} ds dt d\bar{s}d\bar{t}\ \frac{\kappa^2}{4}|J_{ab}^{\sigma}|(s,t)|J_{ab}^{\sigma}|(\bar{s},\bar{t})e^{-\kappa^2|\sigma(s,t) - \sigma(\bar{s}, \bar{t})|^2/8} \right]\nonumber \\
&\longrightarrow \exp\left[ -\frac{1}{8} \sum_{0\leq  a<b \leq 3}\int_{I^2} ds dt\ |J_{ab}^\sigma|(s,t) \rho_S^{ab}(\sigma(s,t))\right] = e^{-\frac{1}{8}\int_S \rho_S }, \nonumber
\end{align}

the limit now follows from Corollary <ref>.

From the discussion in Appendix <ref>, we note that $\int_S \rho_S$ is the area of the surface $S$.
Thus, in the Abelian case, our definition of the Yang-Mills path integral is actually dependent on the surface $S$ chosen, since different surfaces with the same boundary will have different areas.

Now that we have settled the Abelian case, we will now move on to the non-Abelian case.

§ NON-ABELIAN CASE

Let $G$ be a Lie group and $\mathfrak{g}$ be its Lie algebra. We will define any inner product on $\mathfrak{g}$ using Equation (<ref>). Write $F = dA + A \wedge A$, whereby $A \in \Gamma(\Lambda^1 (T^\ast \bR^3) \otimes \mathfrak{g})$. In the non-Abelian case, we want to make sense of 𝒯∫_A e^∫_C ∑_i=1^3A_i ⊗dx^i  e^-1/2|F|^2 DA, where $\mathcal{T}$ is the time ordering operator. From the Abelian case, we know that we need to consider a surface $S$ such that the boundary of $S$ is $C$. Therefore, our approach would be to rewrite $\int_C \sum_{i=1}^3A_i \otimes dx^i $ in terms of a surface $S$. Unfortunately, we cannot apply Stokes Theorem here, unless $G$ is Abelian group. This is one major obstacle that we need to overcome.

Now recall that $\mathcal{T}\exp[\int_C \sum_{i=1}^3A_i\otimes dx^i]$ gives us the holonomy operator by parallel translating a frame along the closed loop $C$. Given a surface $S$, we break up $S$ into small little surfaces $\{S_i\}$, such that each surface $S_i$ has area approximately $\Delta_is^2$. Furthermore, let $C_i$ be the boundary of $S_i$. Let $s_i$ be some point in $C_i$. Since $\mathcal{T}$ commutes all the holonomy operators, using such an approximation gives

\begin{align}
\mathcal{T}& \exp\left[ \int_{C} \sum_{i=1}^3A_i \otimes dx^i \right] =  \mathcal{T} \exp\left[ \sum_i\int_{C_i} \sum_{j=1}^3A_j\otimes dx^j \right]
= \mathcal{T} \bigotimes_{i}\exp\left[ \int_{C_i} \sum_{j=1}^3A_j\otimes dx^j \right] \nonumber \\
=& \mathcal{T} \bigotimes_i \left[ I + \sum_\alpha[F_\alpha E^\alpha](s_i) \Delta_i s^2 + O(\Delta_i s^3) \right] \nonumber \\
=& \mathcal{T}\bigotimes_i\exp\left[ \log[I + \sum_\alpha[F_\alpha E^\alpha](s_i) \Delta_i s^2 + O(\Delta_i s^3)] \right] \nonumber \\
=& \mathcal{T}\exp\left[ \sum_i \sum_\alpha[F_\alpha E^\alpha](s_i) \Delta_i s^2 + O(\Delta_i s^3 )\right]
\longrightarrow \mathcal{T}\exp\left[ \int_S \sum_\alpha F_\alpha E^\alpha \right]. \nonumber
\end{align}

Here, $[F_\alpha E^\alpha](s_i) = \langle F(s_i), E^\alpha\rangle$.

Notice that we are now able to write the holonomy operator as a surface integral. Unfortunately, there is a problem with the above argument. The time ordering operator $\mathcal{T}$ only orders the operators indexed by points along the curve $C$. However, on the surface $S$, we approximate the surface by a finite number of surfaces $\{S_i\}$, each with a boundary curve $C_i$. Thus, the time ordering operator does not know how to order the operators indexed by internal curves $C_i$. As such, the above heuristic argument is seriously flawed.

Fortunately, the above argument can be made rigorous if we can extend the definition of $\mathcal{T}$ on $C$ to the surface $S$. If we can order the operators indexed by points in $S$, then we can apply the above argument and rewrite the holonomy operator in terms of the surface $S$. We will devote one full section, Section <ref>, to explain how we can extend the definition of the time ordering operator $\mathcal{T}$. The correct expression for the holonomy operator, written as a surface integral, is given by Corollary <ref> in Section <ref>.

§ TIME ORDERING OPERATOR

In this and the next section, we allow $M$ to be any 4-manifold.
We kick off this section with the definition of the time ordering operator.

(Time ordering operator)

For any permutation $\sigma \in S_r$, 𝒯(A(s_σ(1))⋯A(s_σ(r))) = A(s_1)⋯A(s_r), s_1 > s_2 > …> s_r.

(Grid $Z_n$)

Consider the grid $Z_n := \{(i,j): i,j= 0, 1/n, 2/n, \ldots 1\}$, which we identify it to be a graph. The vertex set $V(Z_n)$ consists of points labeled in the grid, $(i/n, j/n)$, $0\leq i,j \leq n$. The horizontal edges are the lines joining $(x,y)$ to $(x+1/n, y)$ and the vertical edges are lines joining $(x,y)$ to $(x, y+1/n)$. Let $\partial Z_n$ denote the boundary of $Z_n$, a subgraph of $Z_n$ that contains the vertices $(x,y)$ such that either $x = 0,1$ or $y = 0,1$. Note that $\partial Z_n$ is a cycle.

Define a directed cycle $\overrightarrow{\partial Z_n}$ by assigning arrows to the edges in $\partial Z_n$ in a counterclockwise direction. Further define a directed graph on $Z_n$, denoted $\overrightarrow{Z_n}$ as follows. Let $\mathbf{p}$, $\mathbf{q}$ denote vertices in $Z_n$. The set of vertices in $\overrightarrow{Z_n}$ is the same as the set of vertices in $Z_n$. The set of directed edges consist of all the edges going from $\mathbf{p}$ to $\mathbf{q}$ and back, if either $\mathbf{p}$ or $\mathbf{q}$ is not a vertex inside $\partial Z_n$, union all the directed edges in $\overrightarrow{\partial Z_n}$.

To simplify notations, we will write $e_{i+}^j$ ($e_{i-}^j$) to denote a directed edge joining $(i/n, j/n)$ to $((i+1)/n, j/n)$ ($((i-1)/n, j/n)$). Similarly, we will write $e_{i}^{j+}$ ($e_{i}^{j-}$) to denote a directed edge joining $(i/n, j/n)$ to $(i/n, (j+1)/n)$ ($(i/n, (j-1)/n)$).

Let $C$ be a simple closed curve and $S$ be a surface such that $\partial S = C$. Let $\sigma: [0,1] \times [0,1] \rightarrow M$ be any parametrization for the surface $S$. An edge $x$ to $y$ in $Z_n$ is mapped under $\sigma$ to the corresponding edge joining $\sigma(x)$ to $\sigma(y)$ in $\sigma(Z_n) \subset S$. Thus using this parametrization, we can form a grid $\sigma(Z_n)$ on the surface $S$, with $\sigma(\partial Z_n)$ as the boundary. We will also write $\sigma(\overrightarrow{Z_n})$ to denote the directed graph in the surface $S$ using $\sigma$.  Finally when we say we move from a point $x$ to $y$ in $Z_n$, we mean moving from $\sigma(x)$ to $\sigma(y)$ in $\sigma(Z_n)$.

We want to move on this grid $\sigma(Z_n)$, starting from the vertex $\sigma(0,0)$, subject to the condition that we traverse the edges in $\sigma(\partial Z_n)$ once and only once, in a counterclockwise direction, and all other edges twice, in opposite directions. To describe this movement, we first define 4 basic moves.

* Let $R_n(x,y)$ denote moving from $\sigma(x,y)$ to $\sigma(x+1/n, y)$.

* Let $U_n(x,y)$ denote moving from $\sigma(x,y)$ to $\sigma(x, y+1/n)$.

* Let $\Lambda_n(x,y)$ denote moving from $\sigma(x,y)$ upwards to $\sigma(x, y+1/n)$, then left to $\sigma(x-1/n, y+1/n)$, then downwards to $\sigma(x-1/n, y)$.

* Let $C_n(x,y)$ denote moving from $\sigma(x,y)$ to the left to $\sigma(x-1/n, y)$, then downwards to $\sigma(x-1/n, y-1/n)$, then to the right to $\sigma(x, y-1/n)$.

Finally, we write $U_n\Lambda_n^{n-1}R_n^{n-1}(1/n,y)$ to mean starting from $\sigma(1/n,y)$, we first apply $R_n$ a total of $n-1$ times, followed by $\Lambda_n$ a total of $n-1$ times, then Move $U_n$ once. The final point is at $\sigma(1/n, y+1/n)$.

On $\overrightarrow{\partial Z_n}$, there is an obvious ordering on the edges, by letting the directed edge joining $\sigma(0,0)$ to $\sigma(1/n, 0)$ to be the first edge. Now, we will show how to extend this ordering to all the directed edges in $\overrightarrow{Z_n}$. We will order the edges in the order we move along the edges in the grid, as described below:

* Starting from $\sigma(0,0)$, we apply Move $R_n$ once, so we end up at $\sigma(1/n,0)$.

* At $\sigma(1/n,0)$, we apply $U_n\Lambda_n^{n-1}R_n^{n-1}(1/n,0)$, so we end up at $\sigma(1/n, 1/n)$.

* Repeat $U_n\Lambda_n^{n-1}R_n^{n-1}$ a total of $n-2$ times, so we end up at $\sigma(1/n, (n-1)/n)$.

* Now we apply $U_n$, so we move one unit up to $\sigma(1/n, 1)$.

* Apply Move $C_n$ a total of $n-1$ times, until we end up at $\sigma(1/n, 1/n)$.

* Finally, move left to $\sigma(0, 1/n)$ and then down to $\sigma(0,0)$.

(Time ordering operator $\hat{\mathcal{T}}$)
We now extend our time ordering operator $\mathcal{T}$ to $\hat{\mathcal{T}}$ on the directed graph $\sigma(\overrightarrow{Z_n})$, and order the edges according to how we move along the grid $\sigma(Z_n)$ described above.

Let $C$ be a simple closed curve and $S$ be a surface such that $\partial S = C$. On the surface $S$, let $\sigma: I^2 \rightarrow M$ be any parametrization of $S$.
Let $A_{i\pm}^j$ denote parallel translation along the path $\sigma(i,j)$ to $\sigma(i\pm 1,j)$. Likewise, let $A_i^{j\pm}$ denote parallel translation along the path $\sigma(i,j)$ to $\sigma(i,j\pm 1)$. The parallel translation in question here is prescribed by a choice of $A \in \mathcal{A}_{M, \mathfrak{g}}$.

Suppose we have a set of matrices $B_i$ indexed by $i$. We will write

\begin{align}
\overrightarrow{\prod}_{i=l}^m B_i  \equiv B_l B_{l+1} \cdots B_m, \nonumber \\
\overleftarrow{\prod}_{i=l}^m B_i  \equiv B_m B_{m-1} \cdots B_l .\nonumber
\end{align}

Recall the holonomy operator, defined by $\mathcal{T}e^{\int_C A}$, whereby $A \in \mathcal{A}_{M, \mathfrak{g}}$. Using the directed graph $\sigma(\overrightarrow{\partial Z_n})$, we will write it as 𝒯e^∫_C A  = ∏_i=1^n A_0^i-·∏_i=1^n A_i-^n ·∏_i=0^n-1 A_n^i+·∏_i=0^n-1 A_i+^0. We remark that $\overleftarrow{\prod}_{i=0}^{n-1} A_{i+}^0$ means parallel translation from $\sigma(0,0)$ to $\sigma(1,0)$, $\overleftarrow{\prod}_{i=0}^{n-1} A_n^{i+}$ means parallel translation from $\sigma(1,0)$ to $\sigma(1,1)$, $\overrightarrow{\prod}_{i=1}^{n} A_{i-}^n$ means parallel translation from $\sigma(1,1)$ to $\sigma(0,1)$ and finally $\overrightarrow{\prod}_{i=1}^n A_0^{i-}$ means parallel translation from $\sigma(0,1)$ down to $\sigma(0,0)$.

We have that

\begin{align}
\mathcal{T}e^{\int_C A} =& \overrightarrow{\prod}_{i=1}^n A_0^{i-}\cdot \overrightarrow{\prod}_{i=1}^{n} A_{i-}^n \cdot \overleftarrow{\prod}_{i=0}^{n-1} A_n^{i+}\cdot \overleftarrow{\prod}_{i=0}^{n-1} A_{i+}^0 \nonumber \\
=& \hat{\mathcal{T}}\prod_{i=0, \ldots, n-1 \atop j=0, \ldots n-1}A_{i+}^j \cdot \prod_{i=1, \ldots, n \atop j=1, \ldots n}A_{i-}^j
\cdot \prod_{i=1, \ldots, n \atop j=0, \ldots n-1}A_{i}^{j+}
\cdot \prod_{i=0, \ldots n-1 \atop j=1, \ldots n}A_{i}^{j-}
. \nonumber
\end{align}

We need to show that parallel translation along the directed edges in $\sigma(\overrightarrow{Z_n})$ will give the same holonomy operator as parallel translation along $\sigma(\overrightarrow{\partial Z_n})$.

First, observe that we begin by parallel translation from $\sigma(0,0)$ using move $R_n$, applied $n$ times until we arrive at $\sigma(1,0)$. The parallel translation operator can be written as ∏_i=0^n-1 A_i+^0.

From $\sigma(1,0)$, we apply moves $\Lambda_n$ a total of $n-1$ times. Now, from any point $\sigma(i/n, 0)$, when we parallel translate using Move $\Lambda_n$, the parallel translation operator is given by $A_{i-1}^{1-}A_{i-}^{1}A_i^{0+}$. Therefore, from $\sigma(1,0)$, when we parallel translate along the edges using Moves $\Lambda_n$ applied $n-1$ times, the translation operator is given by ∏_i=2^n A_i-1^1-A_i-^1A_i^0+ = A_1^1-·∏_i=2^n A_i-^1 ·A_n^0+,  since $A_{i}^{0+}A_{i}^{1-}$ is the identity.

From $\sigma(1/n, 0)$, we move up to $\sigma(1/n, 1/n)$, then move right along the edges until we arrive at $\sigma(1, 1/n)$. The parallel translation operator is given by ∏_i=1^n-1 A_i+^1 ·A_1^0+.  Combine Equations (<ref>) and (<ref>) together, from $\sigma(1,0)$, we move along the internal edges in $Z_n$ and arrive at $\sigma(1, 1/n)$, the parallel translation operator is given by ∏_i=1^n-1 A_i+^1 ·A_1^0+ ·A_1^1-·∏_i=2^n A_i-^1 ·A_n^0+ = A_n^0+.

Now, we repeat the same argument, from $\sigma(1, j/n)$, as we move along the internal edges in $Z_n$ using moves $R_n^{n-1}U_n\Lambda_n^{n-1}$ and arrive at $\sigma(1, (j+1)/n)$, the parallel translation operator is given by  ∏_i=1^n-1 A_i+^j+1 ·A_1^j+ ·A_1^(j+1)-·∏_i=2^n A_i-^j+1 ·A_n^j+ = A_n^j+.

Hence, from the point $\sigma(1,0)$, we move along the internal edges in $Z_n$ and end up at $\sigma(1,(n-1)/n)$, apply Equation (<ref>) a total of $n-1$ times, we obtain $\overleftarrow{\prod}_{i=0}^{n-2} A_n^{i+}$. Now, from $\sigma(1, (n-1)/n)$, we apply Move $\Lambda_n$ a total of $n-1$ times to arrive at $\sigma(1/n, (n-1)/n)$, we obtain the translation operator ∏_i=2^n A_i-1^n-A_i-^nA_i^(n-1)+ = A_1^n-·∏_i=2^n A_i-^n ·A_n^(n-1)+.  Next move up one unit to $\sigma(1/n, n)$. Thus, the translation operator from $\sigma(1,0)$ to $\sigma(1/n, 1)$, moving along the internal edges, is given by A_1^(n-1)+·∏_i=2^n A_i-1^n-A_i-^nA_i^(n-1)+ ·∏_i=0^n-2 A_n^i+ = ∏_i=2^n A_i-^n ·∏_i=0^n-1 A_n^i+,  using Equation (<ref>).

From $\sigma(1/n, 1)$, we apply Move $C_n$ a total of $n-1$ times. From $\sigma(1/n, j/n)$, the translation operator along Move $C_n$ is given by $A_{0+}^{j-1}A_0^{j-}A_{1-}^j$, ending at the point $\sigma(1/n, (j-1)/n)$. Thus, from $\sigma(1/n, 1)$, parallel transport to $\sigma(1/n, 1/n)$ by applying Move $C_n$ successively, we obtain the parallel transport operator ∏_i=2^n A_0+^i-1A_0^i-A_1-^i = A_0+^1 ·∏_i=2^n A_0^i- ·A_1-^n.

Finally, we move left from $\sigma(1/n, 1/n)$ to $\sigma(0, 1/n)$ and down to $\sigma(0,0)$. The parallel translation operator along these 2 paths is given by $A_0^{1-}A_{1-}^1$. Now put Equations (<ref>), (<ref>), (<ref>) together, we obtain the translation operator

\begin{align}
A_0^{1-}A_{1-}^1 & \cdot A_{0+}^1 \cdot \overrightarrow{\prod}_{i=2}^n A_0^{i-} \cdot A_{1-}^n\cdot \overrightarrow{\prod}_{i=2}^n A_{i-}^{n}  \cdot \overleftarrow{\prod}_{i=0}^{n-1} A_n^{i+} \cdot  \overleftarrow{\prod}_{i=0}^{n-1} A_{i+}^0 \nonumber \\
=&\overrightarrow{\prod}_{i=1}^n A_0^{i-} \cdot \overrightarrow{\prod}_{i=1}^n A_{i-}^{n}
\cdot \overleftarrow{\prod}_{i=0}^{n-1} A_n^{i+} \cdot  \overleftarrow{\prod}_{i=0}^{n-1} A_{i+}^0. \nonumber
\end{align}

This completes the proof.

§ HOLONOMY OPERATOR ON A GRID

In this section, we need not restrict ourselves to $\mathbb{R}^4$ and the following discussion holds in general on any Riemannian 4-manifold $M$. Let $\pi: P \rightarrow M$ be a trivial principal bundle over $M$ with a structure group $G$. For any $\mathfrak{g}$-valued 1-form $A$ on $M$, let $R = dA + A \wedge A$ denote its curvature, which is a $\mathfrak{g}$-valued 2-form. Recall we define $\sigma: [0,1]^2 \rightarrow M$ to be a parametrization of a surface $S$ in $M$. Suppose we choose a group element $u \in G$ at the point $m \in \sigma([0,1]^2) \subset M$. We will write $\Omega_u := u^{-1}R(\sigma', \dot{\sigma})u$, where $\sigma' = d\sigma/ds$ and $\dot{\sigma} = d\sigma/dt$.

The directed graph $\sigma(\overrightarrow{Z_n})$ was defined in Section <ref>. For some $A \in \mathcal{A}_{M, \mathfrak{g}}$, $\mathcal{T}\exp\left[ \oint_C A \right]$ represents the parallel translation along a closed curve $C$. We will now try to rewrite this expression in terms of its curvature, by first doing a Riemannian sum approximation over $\sigma(\overrightarrow{Z_n})$, and then taking the limit as $n$ goes to infinity. To simplify a bit of writing, we will let $\epsilon = 1/n$ in this section.

(Frame and Curvature)
Fix a group element $u_0 \equiv u_0^0 \in G$ at $\sigma(0,0)$. Now we will write B_1^j ≡∏_l=2^n A_l-1^j+1-A_l-^j+1A_l^j+ ·∏_l=1^n-1 A_l+^j.

Define for $i\geq 1$, $0 \leq j \leq n$, a group element $u_i^j \in G$ at $\sigma(i/n, j/n)$ by u_i^j := ∏_k=1^i-1 A_k+^j ·∏_l=0^j-1 A_1^l+ B_1^l ·A_0+^0 u_0. Let $\Omega_i^j \in \mathfrak{g}$ denote the curvature at $\sigma(i/n, j/n)$, $\Omega_i^j := u_i^{j,-1}R(\sigma'(i/n, j/n), \dot{\sigma}(i/n, j/n))u_i^j$, $i \geq 1$. We will also let $\Box_i^j$ denote the directed square spanned by the edges $\{ e_{i+}^j, e_{i+1}^{j+}, e_{i+1}^{(j+1)-}, e_{(i+1)-}^j \}$.

The curvature $R$ can be written as $R = \sum_{0\leq a < b \leq 3} R_{ab} dx^a \wedge dx^b$ in local coordinates $\{x^a\}_{a=0}^3$. Given a group element $u_i^j$ at the point $\sigma(i/n, j/n)$, we will write Ω_i,ab^j = u_i^j,-1R_ab(σ'(i/n,j/n), σ̇(i/n,j/n))u_i^j.

To ease the notation in the next few paragraphs, we will write JΩ_i^j ≡∑_0≤a< b ≤3 [|J_ab^σ|Ω_i,ab^j](σ'(i/n,j/n), σ̇(i/n,j/n)).

By definition of curvature,

\begin{align}
A_{i}^{(j+1)-}A_{(i+1)-}^{j+1}A_{i+1}^{j+}A_{i+}^ju_i^j =& u_i^j + \epsilon^2 \sum_{0\leq a<b \leq 3}[|J_{ab}^\sigma| R_{ab}](\sigma'(i/n,j/n), \dot{\sigma}(i/n,j/n))u_i^j + O(\epsilon^3) \nonumber \\
:=& u_i^j + \epsilon^2 u_i^{j}J\Omega_{i}^j + O(\epsilon^3). \nonumber
\end{align}

That is, when we parallel translate a group element $u_i^j$ along $\Box_i^j$, we obtain $u_i^j + \epsilon^2u_i^jJ\Omega_i^j + O(\epsilon^3)$.

Fix a $0 \leq j \leq n-1$. From the fixed group element $u_0$ at $\sigma(0,0)$, we parallel translate $u_0$ all the way to the right to $\sigma(1, 0)$, then move up $j$ times, to $\sigma(1, j/n)$, the group element obtained is equivalent to $u_n^j$. Recall from $\sigma(1, j/n)$, we apply Move $\Lambda_n$ a total of $n-2$ times, to arrive at $\sigma(1/n, j/n)$. Our goal is to obtain an expression for the group element at $\sigma(1/n, j/n)$, in terms of $u_i^j$.

Let $\hat{u}_i^j$ be the group element at $\sigma(i/n, j/n)$, obtained by parallel translating the group element $u_n^j$ by Move $\Lambda_n$ a total of $n-i$ times, defined by û_i^j = ∏_k=i+1^n A_k-1^(j+1)-A_k-^j+1 A_k^j+·u_n^j.  Then, we can write û_i^j = u_i^j∏_k=i^n-1[I + JΩ_k^j ϵ^2+ O(ϵ^3) ],  for $1\leq i \leq n-1$ and $0\leq j \leq n-1$.

The proof is by induction. We begin at $\sigma(n/n, j/n)$. Next, we apply  $A_{n-1}^{(j+1)-}A_{n-}^{j+1}A_{n}^{j+}$ to it. Because $u_n^j = A_{(n-1)+}^j u_{n-1}^j$ by definition,

\begin{align}
A_{n-1}^{(j+1)-}A_{n-}^{j+1}A_{n}^{j+} u_{n}^j =& A_{n-1}^{(j+1)-}A_{n-}^{j+1}A_{n}^{j+} A_{(n-1)+}^j u_{n-1}^j \nonumber \\
=& u_{n-1}^j + \epsilon^2 u_{n-1}^j J\Omega_{n-1}^j + O(\epsilon^3) \nonumber \\
=& u_{n-1}^j\left[1 + \epsilon^2  J\Omega_{n-1}^j + O(\epsilon^3) \right]. \nonumber
\end{align}

Suppose we have a group element $\hat{u}_{i+1}^{j}$ at $\sigma((i+1)/n, j/n)$. Apply $A_{i}^{(j+1)-}A_{(i+1)-}^{j+1}A_{i+1}^{j+}$ to $\hat{u}_{i+1}^j$ defined by Equation (<ref>), we get

\begin{align}
A_{i}^{(j+1)-}&A_{(i+1)-}^{j+1}A_{i+1}^{j+} A_{i+}^j \cdot A_{(i+1)-}^j u_{i+1}^j\overrightarrow{\prod}_{k=i+1}^{n-1}\left[ 1 + \epsilon^2 J\Omega_{k}^j + O(\epsilon^3) \right] \nonumber \\
=& A_{i}^{(j+1)-}A_{(i+1)-}^{j+1}A_{i+1}^{j+} A_{i+}^j u_i^j\overrightarrow{\prod}_{k=i+1}^{n-1}\left[ 1 + \epsilon^2 J\Omega_{k}^j + O(\epsilon^3) \right] \nonumber \\
=& u_i^j\left[ 1 +  \epsilon^2    J\Omega_{i}^j + O(\epsilon^3)\right]\overrightarrow{\prod}_{k=i+1}^{n-1}\left[ 1 + \epsilon^2 J\Omega_{k}^j + O(\epsilon^3) \right] . \nonumber
\end{align}

By Definition <ref>, $u_1^j = \overleftarrow{\prod}_{l=0}^{j-1} A_1^{l+} B_1^l \cdot A_{0+}^0 u_0$. From the definition of $\hat{u}_i^j$ defined in Equation (<ref>), we have û_1^j = ∏_l=0^j-1A_1^l+·A_0+^0 u_0∏_l=0^j∏_k=1^n-1[ 1+ ϵ^2JΩ_k^l + O(ϵ^3)], for $0 \leq j \leq n-1$.

The proof is by induction. When $j=0$, the result is trivial. Now using Equation (<ref>) and the fact that $u_i^{j+1} = \overleftarrow{\prod}_{l=1}^{i-1} A_{l+}^{j+1} \cdot A_1^{j+} \hat{u}_1^j$,

\begin{align}
\hat{u}_1^{j+1} =&  A_1^{j+} \hat{u}_1^j \cdot \overrightarrow{\prod}_{k=1}^{n-1}\left[1 + J\Omega_k^{j+1} \epsilon^2+ O(\epsilon^3)) \right]\nonumber \\
=& A_1^{j+} \cdot \overleftarrow{\prod}_{l=0}^{j-1}A_1^{l+}\cdot A_{0+}^0 u_0\overleftarrow{\prod}_{l=0}^j\overrightarrow{\prod}_{k=1}^{n-1}\left[ 1+ \epsilon^2J\Omega_k^l + O(\epsilon^3)\right]\cdot \overrightarrow{\prod}_{k=1}^{n-1}\left[1 + J\Omega_k^{j+1} \epsilon^2+ O(\epsilon^3)) \right]\nonumber \\
=&\overleftarrow{\prod}_{l=0}^{j}A_1^{l+}\cdot A_{0+}^0 u_0\overleftarrow{\prod}_{l=0}^{j+1}\overrightarrow{\prod}_{k=1}^{n-1}\left[ 1+ \epsilon^2J\Omega_k^l + O(\epsilon^3)\right]. \nonumber
\end{align}

In particular, at $\sigma(1/n, (n-1)/n)$, the group element $\hat{u}_1^{n-1}$ is given by û_1^n-1 = ∏_l=0^n-2A_1^l+·A_0+^0 u_0∏_l=0^n-1∏_k=1^n-1[ 1+ ϵ^2JΩ_k^l + O(ϵ^3)].  From $\sigma(1/n, (n-1)/n)$, we want to parallel translate $\hat{u}_1^{n-1}$ back to $\sigma(0,0)$, by first translate it up to $\sigma(1/n, 1)$, then apply Move $C_n$ a total of $n-1$ times to be at $\sigma(1/n, 1/n)$, then translate to the left and then down to $\sigma(0,0)$. Let $J\hat{\Omega}_1^j$ denote the curvature at $\sigma(1/n, j/n)$, JΩ̂_i^j := ũ_i^j,-1∑_0≤a< b ≤3[|J_ab^σ|R_ab](σ̇(1/n, j/n), σ'(1/n, j/n))ũ_i^j, 0≤j ≤n-1.

Let $\bar{u}_1^j$ be the group element at $\sigma(1/n, j/n)$, obtained by parallel translating the group element $A_{1}^{(n-1)+}\hat{u}_1^{n-1}$ by Move $C_n$ a total of $n-j$ times, defined by u̅_1^j = ∏_k=j+1^n A_0+^k-1A_0^k- A_1-^k·A_1^(n-1)+û_1^n-1. Then, we can write u̅_1^j = ∏_l=0^j-1A_1^l+·A_0+^0 u_0∏_l=j^n-1[ 1 + ϵ^2JΩ̂_1^l + O(ϵ^3) ] ·∏_l=0^n-1∏_k=1^n-1[ 1+ ϵ^2JΩ_k^l + O(ϵ^3)],  for $1\leq j \leq n-1$.

Let $\tilde{u}_1^j = \overleftarrow{\prod}_{l=0}^{j-1}A_1^{l+}\cdot A_{0+}^0 u_0$.
The proof is by induction. We begin at $\sigma(1/n, 1)$ with group element $\bar{u}_1^n$. Next, we apply  $A_{0+}^{n-1}A_{0}^{n-} A_{1-}^{n}$ to $\bar{u}_1^n$. Then we have from Equation (<ref>),

\begin{align}
A_{0+}^{n-1}A_{0}^{n-} A_{1-}^{n} A_{1}^{(n-1)+}\hat{u}_1^{n-1} =& A_{0+}^{n-1}A_{0}^{n-} A_{1-}^{n} A_{1}^{(n-1)+}\tilde{u}_1^{n-1}\overleftarrow{\prod}_{l=0}^{n-1}\overrightarrow{\prod}_{k=1}^{n-1}\left[ 1+ \epsilon^2J\Omega_k^l + O(\epsilon^3)\right] \nonumber \\
=& \tilde{u}_1^{n-1}\left[1 + \epsilon^2  J\hat{\Omega}_{1}^{n-1} + O(\epsilon^3) \right]\overleftarrow{\prod}_{l=0}^{n-1}\overrightarrow{\prod}_{k=1}^{n-1}\left[ 1+ \epsilon^2J\Omega_k^l + O(\epsilon^3)\right]. \nonumber
\end{align}

Suppose we have a group element $\bar{u}_{1}^{j+1}$ at $\sigma(1/n, (j+1)/n)$. Apply $A_{0+}^{j}A_{0}^{(j+1)-} A_{1-}^{j+1}$ to $\bar{u}^{j+1}_1$ defined by Equation (<ref>), we get

\begin{align}
A_{0+}^{j}&A_{0}^{(j+1)-} A_{1-}^{j+1} \cdot \tilde{u}_1^{j+1}\overrightarrow{\prod}_{l=j+1}^{n-1}\left[ 1 + \epsilon^2J\hat{\Omega}_1^l + O(\epsilon^3) \right] \cdot \overleftarrow{\prod}_{l=0}^{n-1}\overrightarrow{\prod}_{k=1}^{n-1}\left[ 1+ \epsilon^2J\Omega_k^l + O(\epsilon^3)\right]\nonumber \\
=& A_{0+}^{j}A_{0}^{j+1-} A_{1-}^{j+1} A_{1}^{j+} \tilde{u}_1^j \cdot \overrightarrow{\prod}_{l=j+1}^{n-1}\left[ 1 + \epsilon^2J\hat{\Omega}_1^l + O(\epsilon^3) \right] \cdot \overleftarrow{\prod}_{l=0}^{n-1}\overrightarrow{\prod}_{k=1}^{n-1}\left[ 1+ \epsilon^2J\Omega_k^l + O(\epsilon^3)\right]\nonumber \\
=& \tilde{u}_1^j\overrightarrow{\prod}_{l=j}^{n-1}\left[ 1 + \epsilon^2J\hat{\Omega}_1^l + O(\epsilon^3) \right] \cdot \overleftarrow{\prod}_{l=0}^{n-1}\overrightarrow{\prod}_{k=1}^{n-1}\left[ 1+ \epsilon^2J\Omega_k^l + O(\epsilon^3)\right].\nonumber
\end{align}

In particular, we have at $\sigma(1/n, 1/n)$,
u̅_1^1 =  A_1^0+A_0+^0 u_0∏_l=1^n-1[ 1 + ϵ^2JΩ̂_1^l + O(ϵ^3) ] ·∏_l=0^n-1∏_k=1^n-1[ 1+ ϵ^2JΩ_k^l + O(ϵ^3)]. Let the final group element at $\sigma(0,0)$ be $\hat{u}_0$.

We have û_0 =  u_0∏_l=0^n-1[ 1 + ϵ^2JΩ̂_1^l + O(ϵ^3) ] ·∏_l=0^n-1∏_k=1^n-1[ 1+ ϵ^2JΩ_k^l + O(ϵ^3)].

From $\sigma(1/n, 1/n)$, we parallel translate $\bar{u}_1^1$ to the left to $\sigma(0, 1/n)$, then down to $\sigma(0,0)$. Thus we need to apply $A_{0}^{1-}A_{1-}^1$ to $\bar{u}_1^1$. But note that

\begin{align}
A_{0}^{1-}A_{1-}^1 & A_1^{0+}A_{0+}^0 u_0\overrightarrow{\prod}_{l=1}^{n-1}\left[ 1 + \epsilon^2J\hat{\Omega}_1^l + O(\epsilon^3) \right] \cdot \overleftarrow{\prod}_{l=0}^{n-1}\overrightarrow{\prod}_{k=1}^{n-1}\left[ 1+ \epsilon^2J\Omega_k^l + O(\epsilon^3)\right] \nonumber \\
=& u_0 \overrightarrow{\prod}_{l=0}^{n-1}\left[ 1 + \epsilon^2J\hat{\Omega}_1^l + O(\epsilon^3) \right] \cdot \overleftarrow{\prod}_{l=0}^{n-1}\overrightarrow{\prod}_{k=1}^{n-1}\left[ 1+ \epsilon^2J\Omega_k^l + O(\epsilon^3)\right] \nonumber
\end{align}

and this completes the proof.

(Time ordering operator on $I^2$.)
Let $\tilde{\mathcal{T}}$ be a time ordering operator on $I^2$, 𝒯̃A(s_1, t_1) A(s_2, t_2)⋯A(s_m, t_m) = A(s_τ(1),t_τ(1))A(s_τ(2), t_τ(2)) ⋯A(s_τ(m), t_τ(m)), where we order first according to $t_{\tau(1)}>  t_{\tau(2)} > \cdots > t_{\tau(m)}$, followed by $s_{\tau(1)}\leq  s_{\tau(2)} \leq \cdots \leq s_{\tau(m)}$.

(Parallel translation of $u_{s,t}$.)
Let $\sigma: I^2 \rightarrow M$ be a parametrization of a surface $S$. Given $(s,t) \in I^2$, $0< s, t < 1$, we define a path $P_{s,t}$ from $\sigma(0,0)$ to $\sigma(s,t)$ as follows:
Starting from $\sigma(0,0)$, we move along the path $\sigma(\cdot, 0)$ until $\sigma(1,0)$. Then from $\sigma(1,0)$, we move along the path $\sigma(1, \cdot)$ to $\sigma(1,t)$. Then from $\sigma(1,t)$, we move along $\sigma(1-\cdot, t)$ until we arrive at $\sigma(s,t)$. Note that $P_{s,t}$ is an open path.

Define $u_{s,t}: \bR^d \rightarrow \pi^{-1}(\sigma(s,t))$ as a parallel translation of a group element $u_0\in G$ from $\sigma(0,0)$ to $\sigma(s,t)$ along the path $P_{s,t}$. Note that we can also write it as $
u_{s,t} = \mathcal{T} e^{\int_{P_{s,t}} A } u_0$, whereby $A \in \mathcal{A}_{M, \mathfrak{g}}$.

Define $\Omega_{ab}^\sigma$ as Ω_ab^σ(s,t) = u_s,t^-1 R_ab(σ'(s,t), σ̇(s,t) )u_s,t.

Let $C$ be a closed curve and $S$ be a surface such that $\partial S = C$. Let $\sigma: I^2 \rightarrow M$ be a parametrization of $S$. Refer to Definitions <ref>, <ref> and Notation <ref>.  Then we have
\begin{align}
\mathcal{T} e^{\int_C A }u_0 =& u_0 \overrightarrow{\prod}_{l=0}^{n-1}\left[ 1 + \epsilon^2J\hat{\Omega}_1^l + O(\epsilon^3) \right] \cdot \overleftarrow{\prod}_{l=0}^{n-1}\overrightarrow{\prod}_{k=1}^{n-1}\left[ 1+ \epsilon^2J\Omega_k^l + O(\epsilon^3)\right] \nonumber \\
=& u_0 \overrightarrow{\prod}_{l=0}^{n-1}e^{ \epsilon^2J\hat{\Omega}_1^l + O(\epsilon^3) } \cdot \overleftarrow{\prod}_{l=0}^{n-1}\overrightarrow{\prod}_{k=1}^{n-1}e^{ \epsilon^2J\Omega_k^l + O(\epsilon^3)} \nonumber \\
\longrightarrow&\ u_0 \cdot \tilde{\mathcal{T}}\exp\left[ \int_{I^2} \sum_{0\leq a < b \leq 3}|J_{ab}^\sigma|(s,t)\Omega_{ab}^\sigma(s,t) ds dt \right]. \nonumber
\end{align}

Hence the holonomy operator is given as  𝒯 e^∫_C A =  𝒯̃exp[ ∫_I^2 ∑_0≤a < b ≤3|J_ab^σ|(s,t)Ω_ab^σ(s,t) ds dt ].

§ YANG-MILLS PATH INTEGRAL

In the Abelian case, we defined a measure $\tilde{\mu}$, given in Equation (<ref>), over a Banach space $B(\bR^4; \fd)$. Recall that $B(\bR^4; \fd)_{\bC} = B(\bR^4; \fd) \otimes_{\bR} \bC$. In the non-Abelian case, we will construct a measure over $B(\bR^4; \fd)_{\bC} \otimes \mathfrak{g}$, which is absolutely continuous with respect to the Wiener measure $\tilde{\mu}^{\times^{2N}}$. See Definition <ref>. The other main result in this section is that the holonomy operator defined in Corollary <ref> is integrable with respect to this measure. See Theorem <ref>.

Let $\sigma: I^2 \rightarrow M$ be a parametrization of a surface $S$. Based on our discussion in Section <ref>, our integrand should be of the form $e^{\int_S u^{-1}[ dA + A \wedge A] u}$, whereby $u$ is parallel translation, even though the correct form is given by Equation (<ref>).

Let $H^2(\bC^4) \otimes_{\bR} \bC = H^2(\bC^4)_\bC$. On $\overline{H^2(\bC^4)_\bC \otimes \Lambda^1(\bR^3 )} \otimes \mathfrak{g}$, the Gaussian term in the Yang-Mills integral in Equation (<ref>) is replaced by

\begin{align*}
\exp&\left( -\frac{1}{2}  \int_{\bC^4}d\lambda_4 |\kappa\fd A + A \wedge A|^2 \right)DA \nonumber \\
&= \exp \left( -\frac{1}{2}\int_{\bC^4}d\lambda_4 \langle \kappa\fd A, A \wedge A \rangle + \langle A \wedge A, \kappa\fd A \rangle + |A \wedge A|^2 \right) \exp\left(-\frac{1}{2} \int_{\bC^4}d\lambda_4 |\kappa\fd A|^2 \right)DA. \nonumber
\end{align*}

Using the construction of the Abstract Wiener space in Section <ref>, we will interpret $e^{-\frac{1}{2}\int_{\bC^4} d\lambda_4 |\kappa\fd A|^2}DA$ as the Wiener measure over $B(\bR^4; \fd)_{\bC} \otimes \mathfrak{g}$. We will now explain how to interpret the remaining term exp( -1/2 ∫_^4dλ_4 ⟨κA, A ∧A ⟩+ ⟨A ∧A, κA ⟩+ |A ∧A|^2 ),  which will be shown later to be integrable with respect to Wiener measure.

The main obstruction to define the Yang-Mills measure for the non-Abelian group is to define the cubic and quartic terms. Before we explain how we are going to make sense of the cubic and quartic terms ∫_^4dλ_4 ⟨κA , A ∧A ⟩,   ∫_^4dλ_4 ⟨A ∧A, κA ⟩,  ∫_^4dλ_4 |A ∧A|^2 respectively, we need to recall Proposition <ref>.

Recall there is an evaluation differential form $\tilde{\xi}_{ab}(w)$ that we used to define $\fd A$, $A = \sum_{i=1}^3 A_i \otimes dx^i$, i.e. $\kappa [\fd_a A_b - \fd_b A_a](w) = (A, \tilde{\xi}_{ab}(w))_\sharp$, or ⟨κA(w), dx^a ∧dx^b ⟩= (A, ξ̃_ab^κ(w))_♯.

So how are we going to make sense of $A_i(w)$? From Proposition <ref>, there is an evaluation form $\zeta_i(w)$, that sends $\sum_{i=1}^3 A_i \otimes dx^i \in B(\bR^4; \fd)_{\bC}$ to $A_i(w)$. That is, for $w \in \bC^4$, we write $A_{i}(w) = (A, \zeta_i(w))_\sharp$. Observe that we can now write the product [A_iA_j](w) = A_i(w) A_j(w) = (A ⊗A, ζ_i(w) ⊗ζ_j(w))_♯.

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 $\bC^4$ and $\overline{A_{i,\alpha}}$ denotes its complex conjugate. The trick is to write [A_i,αA_j,βA_i,α̂A_j,β̂](w) = ⟨A ⊗A ⊗A ⊗A, χ_i,α,w ⊗χ_j,β,w ⊗χ_i,α̂,w ⊗χ_i,β̂,w ⟩, where $\chi_{i,\alpha, w} = \zeta_i(w) \otimes E^\alpha$, $\{E^\alpha\}_{\alpha=1}^N$ is an orthonormal basis in $\mathfrak{g}$ and A = ∑_i=1^3∑_α=1^N A_i,α⊗dx^i ⊗E^α, A = ∑_i=1^3∑_α=1^N A_i,α⊗dx^i ⊗E^α.

Thus, ∫_w ∈^4 [A_i,αA_j,βA_i,α̂A_j,β̂](w) = ⟨A^⊗^2⊗A^⊗^2, ∫_w ∈^4dω χ_i,α,w ⊗χ_j,β,w ⊗χ_i,α̂,w ⊗χ_j,β̂,w ⟩. This means that we have to consider the tensor products of $A$ in order to define the Yang-Mills integral. A similar approach will be used to define the cubic term.

Write $A_{i,\alpha} \equiv A_{i,\alpha}\otimes dx^i \otimes E^\alpha$ and $A = \sum_{i=1}^3 \sum_{\alpha=1}^N A_{i,\alpha}$.

To ease our notations, we will write for $j=1,2,3$, $\zeta_{j,w} \equiv \zeta_j(w)$ and ( A_i_1,α_1 ⊗⋯A_i_4, α_4, π̃_w^⊗^4 ) = ∏_j=1^4 ( A_i_j, α_j, π̃_i_j,α_j, w )_♯, whereby $\tilde{\pi}_{i_j, \alpha_j, w} = \zeta_{i_j,w} \otimes E^{\alpha_j}$. We will drop the subscripts $i_j$ and $\alpha_j$ and write as $\tilde{\pi}_w$.

Similarly, we will write ( A_i_1, α_1 ⊗⋯A_i_3, α_3, ( ξ̃_ab^κ(w) ⊗E^α_1 ) ⊗π̃_w^⊗^2 )_♯= ( A_i_1, α_1, ξ̃_ab^κ(w)  ⊗E^α_1 )_♯∏_j=2^3 ( A_i_j, α_j, π̃_i_j, α_j, w)_♯and ( A_i_1, α_1 ⊗⋯A_i_3 α_3,   π̃_w^⊗^2 ⊗( ξ̃_ab^κ(w) ⊗E^α_3 ))_♯= ( A_i_3, α_3, ξ̃_ab^κ(w) ⊗E^α_3 )_♯∏_j=1^2 ( A_i_j, α_j, π̃_i_j, α_j, w)_♯.

But just like in the Abelian case, we do not consider $\tilde{\xi}_{ab}^\kappa(w)$. Instead, we introduce a factor $\psi_w$, defined in Notation <ref>. We will also write $\pi_w = \psi_w \tilde{\pi}_w$, i.e. that is we replace $\tilde{\xi}_{ab}^\kappa(w)$ and $\tilde{\pi}_w$ with ξ_ab^κ(w) = ψ_wξ̃_ab^κ(w)  and  π_w = ψ_wπ̃_w  respectively. This factor $\psi_w$ is required to obtain the necessary convergence as we will see later on.

Let dλ_4(w) =  1/π^4e^-|w|^2∏_i=1^4dp_i dq_i, $w_i = p_i + \sqrt{-1}q_i$.

Refer to Equation (<ref>). For A = ∑_i,αA_i,α ∈B(^4; )_ ⊗𝔤, we will write $\int_{\bC^4}d\lambda_4|A\wedge A|^2$ as

\begin{align}
\int_{\bC^4}&d\lambda_4|A\wedge A|^2 \nonumber \\
:=&\sum_\gamma\sum_{1\leq i<j \leq 3}  \sum_{\alpha < \beta \atop \hat{\alpha} < \hat{\beta}}\frac{1}{2} c_\gamma^{\alpha \beta}c_\gamma^{\hat{\alpha} \hat{\beta}} \bigg[\left( A_{i,\alpha} \otimes A_{j,\beta}\otimes A_{i,\hat{\alpha}}  \otimes A_{j,\hat{\beta}}, \int_{w \in \bC^4}d\lambda_4(w)\tilde{\pi}_w^{\otimes^4} \right)_{\sharp,34} \nonumber \\
&\hspace{4cm}+ \left( A_{i,\alpha} \otimes A_{j,\beta}\otimes A_{i,\hat{\alpha}}  \otimes A_{j,\hat{\beta}}, \int_{w \in \bC^4}d\lambda_4(w)\tilde{\pi}_w^{\otimes^4} \right)_{\sharp,12} \bigg], \label{e.x.6}
\end{align}

where the complex-valued linear functionals $(\cdot, \cdot)_{\sharp, 12}$, $(\cdot, \cdot)_{\sharp, 34}$ on $(B(\bR^4; \fd)_{\bC}\otimes \mathfrak{g})^{\otimes^4}$ are defined as

\begin{align}
(a_1\otimes a_2\otimes a_3\otimes a_4, x_1\otimes x_2 \otimes x_3 \otimes x_4)_{\sharp, 34} :=& \prod_{i=1}^2(a_i, x_i)_\sharp \times \prod_{i=3}^4\overline{(a_i, x_i)_\sharp}, \nonumber \\
(a_1\otimes a_2\otimes a_3\otimes a_4, x_1\otimes x_2 \otimes x_3 \otimes x_4)_{\sharp, 12} :=& \prod_{i=1}^2\overline{(a_i, x_i)_\sharp} \times \prod_{i=3}^4 (a_i, x_i)_{\sharp}, \nonumber
\end{align}

and $\overline{(a_i, x_i)_\sharp}$ means complex conjugate of $(a_i, x_i)_\sharp$.

Note that in writing Equation (<ref>), we use the fact that for a set of complex numbers $z_i$, ∑_i ≠j z_i z_j = ∑_i ≠j 1/2( z_i z_j + z_iz_j ).

Similarly, we interpret

\begin{align}
\int_{\bC^4}&d\lambda_4 \left\langle \kappa\fd A, A \wedge A \right\rangle \nonumber \\
&:= \sum_{\gamma}\sum_{k=1}^3\sum_{1\leq i < j \leq 3}\sum_{\alpha < \beta} c_\gamma^{\alpha\beta}\left( A_{k,\gamma} \otimes A_{i,\alpha} \otimes A_{j,\beta}, \int_{w \in \bC^4}d\lambda_4(w)\left( \tilde{\xi}_{ij}^\kappa(w)  \otimes E^\gamma\right) \otimes \tilde{\pi}_w^{\otimes^2} \right)_{\sharp,23}, \nonumber \\
\int_{\bC^4}&d\lambda_4 \left\langle A \wedge A, \kappa\fd A \right\rangle \nonumber \\
&:= \sum_{\gamma}\sum_{k=1}^3\sum_{1\leq i < j \leq 3}\sum_{\alpha < \beta}c_\gamma^{\alpha\beta}\left( A_{i,\alpha} \otimes A_{j,\beta} \otimes A_{k,\gamma} , \int_{w \in \bC^4}d\lambda_4(w)\ \tilde{\pi}_w^{\otimes^2} \otimes \left( \tilde{\xi}_{ij}^\kappa(w)  \otimes E^\gamma\right) \right)_{\sharp,3},  \nonumber
\end{align}

where the complex-valued linear functionals $(\cdot, \cdot)_{\sharp, 23}$, $(\cdot, \cdot)_{\sharp, 3}$ on $B(\bR^4; \fd)_{\bC} \otimes \mathfrak{g}))^{\otimes^3}$ are defined as

\begin{align}
(a_1\otimes a_2\otimes a_3, x_1\otimes x_2 \otimes x_3 )_{\sharp, 23} :=& (a_i, x_i)_\sharp \times \prod_{i=2}^3\overline{(a_i, x_i)_\sharp}, \nonumber \\
(a_1\otimes a_2\otimes a_3, x_1\otimes x_2 \otimes x_3 )_{\sharp, 3} :=& \prod_{i=1}^2(a_i, x_i)_\sharp \times  \overline{(a_3, x_3)_{\sharp}}. \nonumber
\end{align}

Refer to Definition <ref>. Recall Ω_ab^σ(s,t) = u_s,t^-1R_ab(σ'(s,t), σ̇(s,t))u_s,t = u_s,t^-1[ dA + A ∧A ]_ab(σ'(s,t), σ̇(s,t))u_s,t. Suppose $P_{s,t} \equiv (P_{s,t}^0, P_{s,t}^1, P_{s,t}^2, P_{s,t}^3)$ and $P_{s,t}^{i,\prime}(\tau) = dP_{s,t}^i(\tau)/d\tau$, $i=1, 2, 3$. Define

\begin{align}
u_{s,t} &= \mathcal{T} e^{\int_{P_{s,t}}\sum_{i, \alpha} A_{i,\alpha} }u_0 := \mathcal{T} e^{ \sum_{1\leq i\leq 3}\sum_\alpha (A_{i,\alpha}  , \int_{I} d\tau P_{s,t}^{i,\prime}(\tau) \pi_{P_{s,t}(\tau)} )_\sharp \otimes E_{s,t}^\alpha}u_0 . \nonumber
\end{align}

See Definition <ref>. Note that $E_{s,t}^\alpha \equiv E^\alpha$ and is indexed by $(s,t)$ and we order these matrices according to the time ordering defined in Definition <ref>. And it is understood that (A_i,α, π_P_s,t(τ) )_♯≡(A_i,α ⊗E^α, ζ_i,P_s,t(τ) ⊗E^α)_♯= A_i,α(P_s,t(τ)).

Let $S$ be a surface such that $\partial S = C$ and embed $C \subset \bR^4$ inside $\bC^4$ and scale it with a factor $\kappa/2$. Let $\sigma: I^2 \rightarrow \bR^4$ be any parametrization of $S$. Let $\rho: \mathfrak{g} \rightarrow {\rm End}(\bC^{\tilde{N}})$ be a representation of $\mathfrak{g}$. Write $A_\alpha = \sum_{k=1}^3 A_{k,\alpha} $.
We interpret $\Tr\ \mathcal{T}\exp\left[\int_{C} \sum_{i,\alpha}A_{i, \alpha} \right]$ using Equation (<ref>) as

\begin{align}
\mathcal{J}_S^\kappa& \left( \left\{ A_{i,\alpha}  \right\}_{i, \alpha}\right) \nonumber \\
&:= \Tr\ \tilde{\mathcal{T}}\exp \Bigg[\frac{1}{\kappa}\frac{\kappa^2}{4}\int_{I^2}ds dt\ u_{s,t}^{-1} \Bigg(\sum_{0\leq i<j \leq 3} |J_{ij}^{\sigma}|(s,t) \sum_\alpha\left(   A_{\alpha}, \xi_{ab}^\kappa(\kappa\sigma(s,t)/2)\otimes E^\alpha \right)_\sharp \otimes \rho(E^\alpha) \nonumber \\
&\hspace{2cm}+ \sum_{1\leq i < j \leq 3} |J_{ij}^{\sigma}|(s,t)\sum_{\alpha < \beta}\sum_\gamma c_\gamma^{\alpha\beta}\left(A_{i,\alpha} \otimes A_{j,\beta} , \pi_{\kappa\sigma(s,t)/2}^{\otimes^2}  \right)_\sharp \otimes \rho(E^\gamma)\Bigg) u_{s,t}\Bigg] \label{e.j.1}
\end{align}

and Expression (<ref>) is interpreted as

\begin{align}
\mathcal{Y}^\kappa& \left( \left\{ A_{i,\alpha}  \right\}_{i, \alpha}\right) \nonumber \\
:=& \exp\Bigg\{ -\frac{1}{2}\sum_{\gamma}\sum_{k=1}^3\sum_{1\leq i < j \leq 3}\sum_{\alpha < \beta}c_\gamma^{\alpha\beta}\left( A_{k,\gamma} \otimes A_{i,\alpha} \otimes A_{j,\beta}, \int_{w \in \bC^4}d\lambda_4(w)\left( \xi_{ij}^\kappa(w) \otimes E^\gamma \right) \otimes \pi_w^{\otimes^2} \right)_{\sharp,23} \nonumber \\
&\hspace{1cm}- \frac{1}{2}\sum_{\gamma}\sum_{k=1}^3\sum_{1\leq i < j \leq 3}\sum_{\alpha < \beta}c_\gamma^{\alpha\beta}\left( A_{i,\alpha} \otimes A_{j,\beta}\otimes A_{k,\gamma}, \int_{w \in \bC^4}d\lambda_4(w)\pi_w^{\otimes^2} \otimes \left(\xi_{ij}^\kappa(w) \otimes E^\gamma \right) \right)_{\sharp,3} \nonumber \\
&\hspace{1cm}- \frac{1}{2}\sum_\gamma\sum_{1\leq i<j \leq 3}  \sum_{\alpha < \beta \atop \hat{\alpha} < \hat{\beta}}\frac{c_\gamma^{\alpha \beta}c_\gamma^{\hat{\alpha} \hat{\beta}}}{2}  \Bigg[\left( A_{i,\alpha} \otimes A_{j,\beta}\otimes A_{i,\hat{\alpha}}  \otimes A_{j,\hat{\beta}}, \int_{w \in \bC^4}d\lambda_4(w)\pi_w^{\otimes^4} \right)_{\sharp,34} \nonumber \\
&\hspace{5cm}+ \left( A_{i,\alpha} \otimes A_{j,\beta}\otimes A_{i,\hat{\alpha}}  \otimes A_{j,\hat{\beta}}, \int_{w \in \bC^4}d\lambda_4(w)\pi_w^{\otimes^4} \right)_{\sharp,12} \Bigg]
\Bigg\}  . \label{e.j.2}
\end{align}

Note that we replace $\tilde{\pi}$ with $\pi$ and $\tilde{\xi}_{ab}^\kappa$ with $\xi_{ab}^\kappa$.

Consider the probability space $B(\bR^4; \fd)_{\bC} \otimes \mathfrak{g}$ equipped with Wiener measure $\tilde{\mu}^{\times^{2N}}$. For any $\kappa > 0$, $\mathbb{E}\left[\mathcal{Y}^\kappa \right] := \int_{B(\bR^4; \fd)_{\bC} \otimes \mathfrak{g}} \mathcal{Y}^\kappa d\tilde{\mu}^{\times^{2N}}$ is finite.

Write $A_\alpha = \sum_{k=1}^3 A_{k,\alpha} $. Observe that

\begin{align}
\sum_\gamma&\sum_{1\leq i<j \leq 3}\bigg\{\sum_{k=1}^3\sum_{\alpha < \beta}c_{\gamma}^{\alpha \beta}\left( A_{k,\gamma} \otimes A_{i,\alpha} \otimes A_{j,\beta}, \left( \xi_{ij}^\kappa(w)\otimes  E^\gamma \right) \otimes \pi_w^{\otimes^2} \right)_{\sharp,23} \nonumber \\
&\hspace{2cm}+ \sum_{k=1}^3\sum_{\alpha < \beta}c_{\gamma}^{\alpha \beta}\left(  A_{i,\alpha} \otimes A_{j,\beta}\otimes A_{k,\gamma}, \pi_w^{\otimes^2} \otimes \left( \xi_{ij}^\kappa(w) \otimes  E^\gamma \right) \right)_{\sharp,3} \bigg\} \nonumber \\
&+ \frac{1}{2}\sum_\gamma\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( A_{i,\alpha} \otimes A_{j,\beta}\otimes A_{i,\hat{\alpha}}  \otimes A_{j,\hat{\beta}}, \int_{w \in \bC^4}d\lambda_4(w)\pi_w^{\otimes^4} \right)_{\sharp,34} \nonumber \\
&\hspace{2cm}+\left( A_{i,\alpha} \otimes A_{j,\beta}\otimes A_{i,\hat{\alpha}}  \otimes A_{j,\hat{\beta}}, \int_{w \in \bC^4}d\lambda_4(w)\pi_w^{\otimes^4} \right)_{\sharp,12} \bigg] \nonumber \\
&= \sum_\gamma \sum_{1\leq i<j \leq 3}\Big|\left(A_{\gamma}, \xi_{ij}^\kappa(w) \otimes E^\gamma\right)_\sharp+ \sum_{\alpha < \beta}c_\gamma^{\alpha\beta}\left(A_{i,\alpha} \otimes A_{j,\beta}, \pi_w \otimes \pi_w \right)_\sharp \Big|^2 \nonumber \\
&\hspace{2cm}- \sum_{1\leq i<j \leq 3}\sum_{\alpha}\left|\left( A_{\alpha}, \xi_{ij}^\kappa(w)\otimes E^\alpha\right)_\sharp \right|^2. \nonumber
\end{align}

\begin{align}
\mathcal{Y}^\kappa & \left( \left\{ A_{i,\alpha} \right\}_{i, \alpha}\right) \nonumber \\
=&\exp\Bigg[ -\frac{1}{2}\sum_\gamma\sum_{1\leq i< j \leq 3}\int_{w \in \bC^4}d\lambda_4(w)\Big| \left(A_{\gamma}, \xi_{ij}^\kappa(w)\otimes E^\gamma \right)_\sharp+ \sum_{\alpha < \beta}c_\gamma^{\alpha\beta}\left(A_{i,\alpha} \otimes A_{j,\beta}, \pi_w \otimes \pi_w \right)_\sharp \Big|^2 \nonumber \\
&\hspace{3cm}+ \frac{1}{2}\int_{w \in \bC^4}d\lambda_4(w)\sum_{1\leq i<j\leq 3}\sum_{\alpha}|\left( A_{\alpha}, \xi_{ij}^\kappa\otimes E^\alpha\right)_\sharp|^2 \Bigg] \nonumber \\
\leq& \exp\left[ \frac{1}{2}\int_{w \in \bC^4}d\lambda_4(w)\sum_{1\leq i<j \leq 3}\sum_{\alpha}|\left( A_{\alpha}, \xi_{ij}^\kappa(w)\otimes E^\alpha\right)_\sharp|^2 \right]. \nonumber
\end{align}

Therefore it suffices to show that $\exp\left[ \frac{1}{2}\int_{w \in \bC^4}d\lambda_4(w)|\left( A_{\alpha}, \xi_{ij}^\kappa(w)\otimes E^\alpha\right)_\sharp|^2 \right]$ is integrable.

First, note that $\langle \chi_w, \chi_w \rangle = e^{|w|^2}$. From Equation (<ref>), |∑_1≤i < j ≤3ξ_ij^κ(w) |_,κ^2 = 1/2πe^-|w|^2∑_1≤i ≤3|χ_w|^2 = 3/2π.  Hence,

\begin{align}
\int_{w \in \bC^4}d\lambda_4(w) \left|\left(\cdot, \sum_{1\leq i < j \leq 3}\xi_{ij}^\kappa(w)\right)_\sharp\right|^2  =& \frac{3}{(2\pi)\pi^4}\int_{w \in \bC^4}  e^{-|w|^2}\left|\left(\cdot, \frac{\sum_{1\leq i < j \leq 3}\xi_{ij}^\kappa(w)}{|\sum_{1\leq i < j \leq 3}\xi_{ij}^\kappa|_{\fd,\kappa}} \right)_\sharp \right|^2 \prod_{i=1}^4 dp_i dq_i . \nonumber
\end{align}

Therefore we want to compute 𝔼[ exp[ 1/23/(2π)π^4∫_w ∈^4e^-|w|^2|N_w|^2 ∏_i=1^4 dp_i dq_i] ], whereby $\{N_w = (\cdot, \hat{\varphi_w})_\sharp \}$ is a family of complex-valued random variables, $N_w$ equal in distribution to $X_w + iY_w$, $X_w$ and $Y_w$ are independent Normal distributions, with $\bE[X^2 + Y^2] = 1$. And φ̂_̂ŵ = ∑_1≤i < j ≤3ξ_ij^κ(w)/|∑_1≤i < j ≤3ξ_ij^κ|_,κ.

On 3/π^4∫_w ∈^4dλ_4(w)e^-|w|^2 = 3/π^4∫_w ∈^4 e^-|w|^2 ∏_i=1^4 dp_i dq_i = 3, we approximate it by a lower Riemann sum using some partition function $\beta: \{1,2,\cdots M(n)\} \rightarrow \bC^4$, 3 ≥3/π^4 ∑_j=1^M(n) e^-|β(j)|^21/n^8 ⟶3/π^4∫_w ∈^4 e^-|w|^2 ∏_i=1^4 dp_i dq_i  as $n \rightarrow \infty$. Note that $M(n) = O(n^{16})$ and $\beta(j) \in \bC^4$.

Now, let $\{N_{\beta(j)}\}_{j=1}^{M(n)}$ be equal in distribution to $\{ \hat{N}_k \}_{k=1}^{M(n)}$, whereby $\{\hat{N}_k\}_{k=1}^{M(n)}$ is a finite set of independent random variables, with $\hat{N}_k$ each equal to some $X + iY$, $X$ and $Y$ are independent Normal distributions with the condition that $\bE[X^2 + Y^2] = 1$.

Suppose $N_{\beta(j)} = \sum_{k=1}^{M(n)} \alpha_{j, k} \hat{N}_{k}$, equality in terms of distribution and $\sum_{k=1}^{M(n)} |\alpha_{j,k}|^2 = 1$. Define a matrix $\mathcal{A}$ by 𝒜_j^k = √(3(2π)π^4)e^-|β(j)|^2/21n^4α_j,k. Then,

\begin{align}
\frac{3}{(2\pi)\pi^4}\sum_{j=1}^{M(n)} e^{-|\beta(j)|^2}\frac{1}{n^8}N_{\beta(j)}^2 =& \frac{3}{(2\pi)\pi^4}\sum_{j=1}^{M(n)} e^{-|\beta(j)|^2}\frac{1}{n^8} \left| \sum_{k=1}^{M(n)} \alpha_{j, k} \hat{N}_{k} \right|^2 \nonumber \\
=& \left\langle \mathcal{A} (\hat{N}_1, \ldots, \hat{N}_{M(n)})^T, \mathcal{A} (\hat{N}_1, \ldots, \hat{N}_{M(n)})^T \right\rangle \nonumber \\
=& \left\langle \mathcal{A}^\ast \mathcal{A} (\hat{N}_1, \ldots, \hat{N}_{M(n)})^T,  (\hat{N}_1, \ldots, \hat{N}_{M(n)})^T \right\rangle. \label{e.c.2}
\end{align}

All the equality is in terms of distributions.

\begin{align}
\left[\mathcal{A}^\ast \mathcal{A} \right]_i^i =& \sum_k [\mathcal{A}^\ast]_i^k \mathcal{A}_{k}^i
= \sum_k |\mathcal{A}_k^i|^2 = \frac{3}{(2\pi)\pi^{4}}\sum_{k} e^{-|\beta(k)|^2}\dfrac{1}{n^8}|\alpha_{k,i}|^2 \nonumber
\end{align}

and thus [ 𝒜^∗𝒜 ] = 3/(2π)π^4∑_i ∑_k e^-|β(k)|^21n^8|α_k,i|^2 = 3/(2π)π^4∑_k e^-|β(k)|^21n^8.

From Equation (<ref>), we see that $\parallel \mathcal{A}^\ast \mathcal{A} \parallel \leq 3/2\pi < 1$. Hence,

\begin{align}
\mathbb{E} &\left[ \exp\left[ \frac{1}{2}\frac{3}{(2\pi)\pi^4} \sum_{j=1}^{M(n)} e^{-|\beta(j)|^2}\frac{1}{n^8} N_{\beta(k)}^2 \right] \right]\nonumber \\
&= \mathbb{E}\left[ \exp\left[ \frac{1}{2}\left\langle \mathcal{A}^\ast \mathcal{A} (\hat{N}_1, \ldots, \hat{N}_{M(n)})^T,  (\hat{N}_1, \ldots, \hat{N}_{M(n)})^T \right\rangle\right] \right]\nonumber \\
&=\left( \frac{1}{\det\left(1 - \mathcal{A}^\ast\mathcal{A} \right)} \right)^{1/2}\leq \exp\left[ \frac{c}{2}\Tr\left[ \mathcal{A}^\ast \mathcal{A} \right] \right] \nonumber \\
&=\exp\left[ \frac{1}{2}\frac{3c}{(2\pi)\pi^4}\sum_{j=1}^{M(n)} e^{-|\beta(j)|^2}\frac{1}{n^8} \right]\longrightarrow \exp\left[ \frac{1}{2}\frac{3c}{(2\pi)\pi^4}\int_{w \in \bC^4} e^{-|w|^2} \prod_{i=1}^4 dp_i dq_i\right] < \infty. \nonumber
\end{align}

The first, second and third equality follows from Equations (<ref>), (<ref>) and (<ref>) respectively. The inequality follows from Lemma <ref>. Since this holds for any $n$, we thus have 𝔼[ exp[ 1/23/(2π)π^4∫_w ∈^4e^-|w|^2N_w^2 ∏_i=1^4 dp_i dq_i] ]< ∞.

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

For any $p < 2\pi/3$, there exists a $c \equiv c(p) = (1-3p/2\pi)^{-1}$ such that  𝔼 [ [𝒴^κ]^p ] ≤exp[ 1/23pc/(2π)π^4∫_w ∈^4 e^-|w|^2 ∏_i=1^4 dp_i dq_i].

Consider the probability space $B(\bR^4; \fd)_{\bC} \otimes \mathfrak{g}$ equipped with Wiener measure $\tilde{\mu}^{\times^{2N}}$. Define $\mathcal{J}_S^\kappa$ as in Equation (<ref>).
For large $\kappa$, we have

\begin{align}
\mathbb{E}_{YM}^\kappa \left[ \mathcal{J}_S^\kappa \right]
:=& \frac{1}{\mathbb{E}\left[ \mathcal{Y}^\kappa\right]}\mathbb{E}\left[ \mathcal{J}_S^\kappa \mathcal{Y}^\kappa\right] < \infty. \nonumber
\end{align}

By Holder's Inequality and previous Lemma <ref>, it suffices to show that $\mathbb{E}\left[ \mathcal{J}_S^\kappa\right] < \infty$ for $\kappa$ sufficiently large. Now since $S$ and $G$ are compact, $|u|$ is bounded above and below. Thus we will show that

\begin{align}
\exp\Bigg[ p\frac{\kappa}{4}\int_{I^2}&ds dt\ \sum_{0\leq i< j \leq 3}|J_{ij}^{\sigma}|(s,t)\left|\sum_\alpha\left(  A_{\alpha} ,  \xi_{ij}^\kappa(\kappa \sigma(s,t)/2) \otimes E^\alpha \right)_\sharp \right|\nonumber \\
&+ \sum_{1\leq i<j \leq 3}|J_{ij}^{\sigma}|(s,t)\sum_\gamma \sum_{\alpha < \beta}|c_\gamma^{\alpha\beta}|\left|\left(A_{i,\alpha} \otimes A_{j,\beta} , \pi_{\kappa\sigma(s,t)/2}^{\otimes^2}  \right)_\sharp \right|\Bigg] \label{e.y.4}
\end{align}

is integrable for large values of $p$. Because

\begin{align}
\exp\left[\frac{p\kappa}{4}\int_{I^2}ds dt\ \sum_{0\leq i< j\leq 3}|J_{ij}^{\sigma}|(s,t) \left|\sum_\alpha \left(  A_{\alpha}, \xi_{ij}^\kappa(\kappa\sigma(s,t)/2) \otimes E^\alpha \right)_\sharp \right| \right] \nonumber
\end{align}

is definitely integrable for any $\kappa$ and for any large value of $p$, using Holder's Inequality again, it suffices to show that

\begin{align}
\exp\left[ \frac{p\kappa}{4}\int_{I^2}ds dt\ \sum_{1\leq i<j \leq 3}|J_{ij}^{\sigma}|(s,t) \sum_\gamma\sum_{\alpha < \beta}|c_\gamma^{\alpha\beta}|\left| \left(A_{i,\alpha} \otimes A_{j,\beta} , \pi_{\kappa\sigma(s,t)/2}^{\otimes^2} \right)_\sharp \right|\right] \label{e.y.1}
\end{align}

is integrable if $\kappa$ is made large.

Now from Corollary <ref>,

\begin{align}
|\pi_i(w)|_{\fd,\kappa} \leq \psi(w)|\zeta_i(w)|_{\fd,\kappa} \leq \frac{2}{\kappa\sqrt{2\pi}}. \label{e.p.7}
\end{align}

Thus M_i,κσ(s,t)/2^α:= |1/|π_i,α,κσ(s,t)/2|_,κ(·, π_i,α,κσ(s,t)/2)_♯| is equal to $|Z|$ in distribution, where $Z$ is the standard normal.

Using Equation (<ref>), we have

\begin{align}
&\left|\left(\frac{\kappa}{4} \right)\int_{I^2}ds dt\ \sum_{1\leq i<j \leq 3}|J_{ij}^{\sigma}|(s,t)\sum_\gamma \sum_{\alpha < \beta}|c_\gamma^{\alpha\beta}||\pi_i(w)|_{\fd,\kappa} |\pi_j(w)|_{\fd,\kappa} M^\alpha_{i,\kappa\sigma(s,t)/2} M^\beta_{j,\kappa\sigma(s,t)/2} \right|\nonumber \\
&\leq\left(\frac{\kappa}{4} \right)\frac{4}{(2\pi) \kappa^2}\int_{I^2}ds dt\ \sum_{1\leq i<j \leq 3}|J_{ij}^{\sigma}|(s,t) \sum_\gamma\left|\sum_{\alpha < \beta}|c_\gamma^{\alpha\beta}|M^\alpha_{i,\kappa\sigma(s,t)/2} M^\beta_{\kappa\sigma(s,t)/2} \right|.  \nonumber
\end{align}

Let F_n := 4(κ/4)/ (2π)κ^2∑_1≤i<j ≤3∑_p, q=1^n |J_ij^σ|(p/n,q/n)/n^2 ∑_γ|∑_α< β|c_γ^αβ| M^α_i,κσ(p/n,q/n)/2 ⊗M^β_j,κσ(p/n,q/n)/2| be a Riemannian sum approximation to (κ/4 )4/(2π) κ^2∫_I^2ds dt ∑_1≤i<j ≤3|J_ij^σ|(s,t) ∑_γ|∑_α< β|c_γ^αβ|M^α_i,κσ(s,t)/2 M^β_κσ(s,t)/2|.  It suffices to show that $\sup_n\mathbb{E}[F_n]$ is bounded above.

For each $\gamma$, define an upper triangular matrix $\mathcal{B}(\gamma)$ by $\mathcal{B}(\gamma)_{\alpha}^\beta = |c_\gamma^{\alpha \beta}|$. Note that

\begin{align*}
\sum_{\alpha < \beta}&|c_\gamma^{\alpha\beta}|M^\alpha_{i,\kappa\sigma(s,t)/2} M^\beta_{j,\kappa\sigma(s,t)/2} \\
&= \left\langle \mathcal{B}(\gamma) (M^1_{i,\kappa\sigma(s,t)/2}, \ldots, M^N_{i,\kappa\sigma(s,t)/2})^T, (M^1_{j,\kappa\sigma(s,t)/2}, \ldots, M^N_{j,\kappa\sigma(s,t)/2})^T \right\rangle. \nonumber
\end{align*}

\begin{align}
\left| \sum_{\alpha < \beta}|c_\gamma^{\alpha\beta}|M^\alpha_{i,\kappa\sigma(s,t)/2} M^\beta_{j,\kappa\sigma(s,t)/2} \right| \leq& N\parallel \mathcal{B}(\gamma) \parallel \sqrt{\sum_\alpha M_{i,\kappa \sigma(s,t)/2}^{\alpha,2}}\sqrt{\sum_\alpha M_{j,\kappa \sigma(s,t)/2}^{\alpha,2}} \nonumber \\
\leq& N\parallel \mathcal{B}(\gamma) \parallel \left[ \sum_\alpha M_{i,\kappa \sigma(s,t)/2}^{\alpha,2} + \sum_\alpha M_{j,\kappa \sigma(s,t)/2}^{\alpha,2} \right]. \nonumber
\end{align}

\begin{align}
\mathbb{E}&\left[ \exp\left\{ \frac{4N(\kappa/4)}{(2\pi) \kappa^2}\sum_{1\leq a<b \leq 3}\sum_{p, q=1}^n \frac{|J_{ab}^{\sigma}|(p/n,q/n)}{n^2}\sum_\gamma \left|\sum_{\alpha < \beta}|c_\gamma^{\alpha\beta}| M_{a,\kappa \sigma(p/n,q/n)/2}^{\alpha} \otimes M_{b,\kappa \sigma(p/n,q/n)/2}^{\beta}\right| \right\}\right] \nonumber \\
\leq& \mathbb{E}\Bigg[\exp\Bigg\{ \left(\frac{\kappa}{4} \right)\frac{4N}{(2\pi) \kappa^2} \sum_{1\leq a<b \leq 3}\sum_{p=1}^n \sum_{q=1}^n \frac{1}{n^2}\left| |J_{ab}^{\sigma}|(p/n,q/n)\right|\nonumber \\
&\hspace{2.5cm}\times \sum_\gamma \parallel \mathcal{B}(\gamma) \parallel \left[ \sum_\alpha M_{a,\kappa \sigma(p/n,q/n)/2}^{\alpha,2} + \sum_\alpha M_{b,\kappa \sigma(p/n,q/n)/2}^{\alpha,2} \right] \Bigg\}\Bigg]. \label{e.y.2}
\end{align}

Choose $\kappa$ large enough so that we can apply Lemma <ref>. Then using a similar argument used in Lemma <ref>, the RHS of Equation (<ref>) is less than or equal to

\begin{align}
\exp&\left[ \left(\frac{\kappa}{4} \right)\frac{4Nc}{(2\pi) \kappa^2} \sum_{1\leq a<b \leq 3}\sum_{p=1}^n \sum_{q=1}^n \frac{1}{n^2} |J_{ab}^{\sigma}|(p/n,q/n) 2\sum_\gamma \parallel \mathcal{B}(\gamma) \parallel  \right] \nonumber \\
\longrightarrow&\exp\left[\left(\frac{\kappa}{4} \right)\frac{8Nc}{4(2\pi) \kappa^2}\sum_\gamma\parallel \mathcal{B}(\gamma) \parallel  \sum_{1\leq a<b \leq 3}\int_{I^2}\left| \sigma_a'\dot{\sigma}_b - \sigma_b'\dot{\sigma}_a\right|(s,t) ds dt \right] < \infty, \nonumber
\end{align}

as $n \rightarrow \infty$. This completes the proof.

(Definition of the Yang-Mills Path integral.)

Let $C$ be a closed curve in $\bR^4$ and let $S$ be a surface with $\partial S = C$. Let $\rho$ be a representation of $\mathfrak{g}$. Consider the probability space $B(\bR^4; \fd)_{\bC} \otimes \mathfrak{g}$ equipped with Wiener measure $\tilde{\mu}^{\times^{2N}}$.

We define the Yang-Mills Path integral for the Wilson loop, as

\begin{align}
\frac{1}{Z}\Tr\ \int_{A} \mathcal{T} e^{\int_C \sum_{i}A_i \otimes dx^i }e^{-\frac{1}{2}\int_{\bR^4}|dA + A \wedge A|^2/2}  DA
:=& \mathbb{E}_{YM}^\kappa\left[ \mathcal{J}_s \right] = \frac{1}{\mathbb{E} \left[ \mathcal{Y}^\kappa\right]}\mathbb{E}\left[ \mathcal{J}_S^\kappa \cdot \mathcal{Y}^\kappa\right], \nonumber
\end{align}

where $\mathcal{J}_S^\kappa$ and $\mathcal{Y}^\kappa$ were defined in Definition <ref>. Because the path integral will be dependent on the chosen surface $S$, we will also write Y(^4, κ; S, ρ) = 𝔼_YM^κ[ 𝒥_s ].

§ WILSON AREA LAW FORMULA

Recall we derive the Area Law Formula for the Abelian case, in Section <ref>. Now we will show how we can obtain the Area Law Formula for the non-Abelian case, using our definition of the Yang-Mills Path integral, given in Definition <ref>.

From the proof of Theorem <ref>, we see that Expression (<ref>) converges pointwise to 1 as $\kappa \rightarrow 0$. We will now show that as $\kappa \rightarrow \infty$, the Yang-Mills Path integral actually converges to a Gaussian type of integral, similar in form to Expression (<ref>) in Section <ref>. To prove this, we will show that $(\cdot, \pi_w)_\sharp \rightarrow 0$ as a linear functional on $B(\bR^4; \fd)_{\bC}$, hence implying $\mathcal{Y}^\kappa \rightarrow 1$.

Refer to Equation (<ref>). We will show that as $\kappa \rightarrow \infty$,

\begin{align}
Y_1^\kappa  \left( \left\{ A_{i,\alpha}  \right\}_{i, \alpha}\right)
:=& \frac{1}{2}\sum_{\gamma}\sum_{k=1}^3\sum_{1\leq i < j \leq 3}\sum_{\alpha < \beta}c_\gamma^{\alpha\beta}\left( A_{k,\gamma} \otimes A_{i,\alpha} \otimes A_{j,\beta}, \int_{w \in \bC^4}d\lambda_4(w)\ \left(\xi_{ij}^\kappa(w)\otimes E^\gamma  \right)  \otimes \pi_w^{\otimes^2} \right)_{\sharp,23}, \nonumber \\
Y_2^\kappa \left( \left\{ A_{i,\alpha} \right\}_{i, \alpha}\right)
:=& \frac{1}{2}\sum_{\gamma}\sum_{k=1}^3\sum_{1\leq i < j \leq 3}\sum_{\alpha < \beta}c_\gamma^{\alpha\beta}\left( A_{i,\alpha} \otimes A_{j,\beta}\otimes A_{k,\gamma}, \int_{w \in \bC^4}d\lambda_4(w)\ \pi_w^{\otimes^2} \otimes \left(\xi_{ij}^\kappa(w)\otimes E^\gamma \right) \right)_{\sharp,3}, \nonumber
\end{align}

\begin{align}
Y_3^\kappa  \left( \left\{ A_{i,\alpha} \right\}_{i, \alpha}\right) :=& \frac{1}{2}\sum_\gamma\sum_{1\leq i<j \leq 3}  \sum_{\alpha < \beta \atop \hat{\alpha} < \hat{\beta}} \frac{1}{2}c_\gamma^{\alpha \beta}c_\gamma^{\hat{\alpha} \hat{\beta}} \Bigg[\left( A_{i,\alpha} \otimes A_{j,\beta}\otimes A_{i,\hat{\alpha}}  \otimes A_{j,\hat{\beta}}, \int_{w \in \bC^4}d\lambda_4(w)\ \pi_w^{\otimes^4} \right)_{\sharp,34} \nonumber \\
&+ \left( A_{i,\alpha} \otimes A_{j,\beta}\otimes A_{i,\hat{\alpha}}  \otimes A_{j,\hat{\beta}}, \int_{w \in \bC^4}d\lambda_4(w)\ \pi_w^{\otimes^4} \right)_{\sharp,12} \Bigg]
\nonumber
\end{align}

all tend to 0.

For $i=1,2,3$, $Y_i^\kappa \rightarrow 0$ as $\kappa \rightarrow \infty$.

To show that $Y_1^\kappa$ and $Y_2^\kappa$ goes to 0, it suffices to show that

\begin{align}
\int_{w \in \bC^4}d\lambda_4(w) \left(\cdot, \sum_{1\leq i<j \leq 3}\left( \xi_{ij}^\kappa(w)\otimes E^\gamma \right) \otimes \pi_w^{\otimes^2} \right)_\sharp,\ \int_{w \in \bC^4}d\lambda_4(w) \left(\cdot, \sum_{1\leq i<j \leq 3} \pi_w^{\otimes^2} \otimes \left( \xi_{ij}^\kappa(w)\otimes E^\gamma \right)   \right)_\sharp \nonumber
\end{align}

all go to 0.
From Equations (<ref>) and (<ref>), we have

\begin{align}
&\left|\int_{w \in \bC^4}d\lambda_4(w) \left(\cdot, \sum_{1\leq i<j \leq 3}\left( \xi_{ij}^\kappa(w)\otimes E^\gamma \right)  \otimes \pi_w^{\otimes^2} \right)_\sharp \right|_{\fd,\kappa} \nonumber \\
&\leq \int_{w \in \bC^4}d\lambda_4(w) \left|\left(\cdot, \sum_{1\leq i<j \leq 3}\left(\xi_{ij}^\kappa(w)\otimes E^\gamma \right)  \right)_\sharp\right|_{\fd,\kappa} \left| (\cdot, \pi_w)_\sharp\right|_{\fd,\kappa}^2 \leq \int_{w \in \bC^4}d\lambda_4 \sqrt{\frac{3}{2\pi}} \left( \frac{2}{\kappa\sqrt{2\pi}} \right)^2 \longrightarrow 0\nonumber
\end{align}

as $\kappa \rightarrow 0$. The other term is similar.

To show $Y_3^\kappa$ converges to 0, it suffices to show that $\int_{w \in \bC^4}d\lambda_4(w)  \pi_w^{\otimes^4}$ goes to 0 in $|\cdot|_{\fd,\kappa}$ as $\kappa \rightarrow \infty$. Using Equation (<ref>), we have
| ∫_w ∈^4dλ_4(w) π_w^⊗^4|_,κ ≤∫_w ∈^4dλ_4(w) |π_w^⊗^4|_,κ = ∫_w ∈^4dλ_4(w) ( 2/κ√(2π) )^4  = 16/(2π)^2κ^4 ⟶0as $\kappa$ goes to $\infty$.

Recall we fix an orthonormal basis $\{E^\alpha\}_{\alpha=1}^N$ for $\mathfrak{g}$. Define a set $\mathcal{S} = \{ E_{s,t}^\alpha\}_{(s,t) \in S,\ \alpha=1, \ldots, N}$, $S$ is possibly uncountable set. Note that $\mathcal{S}$ is only a set, and we have $E_{s,t}^\alpha \equiv E^\alpha$. We will write $f \otimes E_{s,t}^\alpha$ to mean $(f \otimes E^\alpha) \otimes E_{s,t}^\alpha$, where $f \otimes E^\alpha \in (B(\bR^4; \fd)_{\bC} \otimes \mathfrak{g})^\ast$ and $E_{s,t}^\alpha \in \mathcal{S}$.

In the following, we will extend our real-valued inner product to be a $\mathcal{S}$-valued inner product. Thus, ⟨f ⊗E_s,t^α, g ⊗E_u,v^β⟩_,κ := ⟨f ⊗E^α, g ⊗E^β⟩_,κ ⊗E_s,t^α⊗E_u,v^β. Likewise, we will also write (b ⊗E_s,t^α, f ⊗E_u,v^β)_♯= (b ⊗E^α, f ⊗E^β)_♯⊗E_s,t^α⊗E_u,v^βfor $b \in B(\bR^4; \fd)_{\bC}$ and $f \in B(\bR^4; \fd)_{\bC}^\ast$. Similar definition for $\langle \cdot, \cdot \rangle$.

Let $C$ be a closed curve in $\bR^4$ and fix a surface $S$ such that $\partial S = C$. Define $\mu(A \otimes B) = A \cdot B$, for $A, B \in M_n(\bR)$.
As $\kappa \rightarrow \infty$, we have
𝔼_YM^κ[ 𝒥_S^κ] ⟶exp[1/8∫_Sρ_S ⊗μ( ∑_αρ(E^α) ⊗ρ(E^α)) ] . Here, $\rho$ is some representation of $\mathfrak{g}$.

Observe that $\mathcal{Y}^\kappa = \exp\left[ -\frac{1}{2}(Y_1^\kappa + Y_2^\kappa + Y_3^\kappa) \right]$ and by Equation (<ref>), $\{ \mathcal{Y}^\kappa\}_{\kappa>0}$ is actually $L^p$ integrable for some $p > 1$. Hence it is uniformly integrable and because of Lemma <ref>, $\mathbb{E}\left[ \mathcal{Y}^\kappa\right]$ converges to 1. Hence it suffices to show that $\mathbb{E} \left[ \mathcal{J}_S^\kappa \mathcal{Y}^\kappa \right]$ converges to the desired limit.

Now for $p > 1$, $(1 + x)^p \leq 2^{p-1}(1 + |x|^p)$ by Jensen's Inequality. Because $\mathcal{Y}^\kappa \rightarrow 1$ pointwise and by Equation (<ref>), we have that $\mathbb{E}\left[ (1 - \mathcal{Y}^\kappa)^p \right] \rightarrow 0$ by uniform integrability. For some $p > 1$, using Holder's Inequality,

\begin{align}
\mathbb{E} \left[ \mathcal{J}_S^\kappa \left(1-\mathcal{Y}^\kappa \right)\right] \leq \mathbb{E} \left[ \mathcal{J}_S^{\kappa,q} \right]^{1/q} \mathbb{E}\left[\left(1-\mathcal{Y}^\kappa \right)^p\right]^{1/p} \longrightarrow 0 \nonumber
\end{align}

as $\kappa \rightarrow \infty$ if $\mathbb{E}\left[ \mathcal{J}_S^{\kappa,q}\right]$ converges. The rest of the proof is now focus on proving this convergence, for $q=1$ without any loss of generality.

Write $E_{s,t}^\alpha \equiv \rho(E_{s,t}^\alpha)$ and

\begin{align}
V_1^\kappa  \left( \left\{ A_{i,\alpha} \right\}_{i, \alpha}\right) =& \frac{\kappa}{4}\int_{I^2}ds dt\  \Bigg[\sum_{0\leq i<  j \leq 3} |J^\sigma_{ij}|(s,t)  \sum_\alpha\left(   A_{\alpha} , \xi_{ij}^\kappa(\kappa\sigma(s,t)/2) \otimes E^\alpha \right)_\sharp \otimes E_{s,t}^\alpha \Bigg] , \nonumber \\
V_2^\kappa \left( \left\{ A_{i,\alpha} \right\}_{i, \alpha}\right)=& \frac{\kappa}{4}\int_{I^2}ds dt\  \Bigg[\sum_{0\leq i< j \leq 3} |J^\sigma_{ij}|(s,t)  \sum_\alpha \left[u_{s,t}^{-1}, E^\alpha \right]\left(  A_{\alpha} , \xi_{ij}^\kappa(\kappa\sigma(s,t)/2) \otimes E^\alpha \right)_\sharp \otimes E_{s,t}^\alpha  \Bigg] u_{s,t}, \nonumber \\
V_3^\kappa \left( \left\{ A_{i,\alpha} \right\}_{i, \alpha}\right)=& \frac{\kappa}{4}\int_{I^2}ds dt\ u_{s,t}^{-1} \Bigg[\sum_{1\leq i < j \leq 3} |J^\sigma_{ij}|(s,t)\sum_\gamma \sum_{\alpha < \beta}c_\gamma^{\alpha\beta}\left(A_{i,\alpha} \otimes A_{j,\beta} , \pi_{\kappa\sigma(s,t)/2}^{\otimes^2}  \right)_\sharp \otimes E_{s,t}^\gamma \Bigg] u_{s,t}. \nonumber
\end{align}

\begin{align}
u_{s,t} &= \mathcal{T} e^{\int_{P_{s,t}} \sum_{i,\alpha}A_{i,\alpha} }u_0 := \mathcal{T} \exp\left[ \sum_{1\leq i\leq 3}\sum_\alpha (A_{i,\alpha} , \int_{I} d\tau P_{s,t}^{i,\prime}(\tau) \pi_{p_{s,t}(\tau)} \otimes dx^i \otimes E^\alpha)_\sharp \otimes E_{s,t}^\alpha \right]u_0 . \nonumber
\end{align}

Note that $\mathcal{J}_S^\kappa = \exp\left[ V_1^\kappa + V_2^\kappa + V_3^\kappa \right]$.
Apply Fubini's Theorem,

\begin{align}
\mathbb{E}\ &\Tr\ \tilde{\mathcal{T}}\ \left[ \exp\left[ V_1^\kappa + V_2^\kappa + V_3^\kappa\right] \right] = \Tr\ \tilde{\mathcal{T}}\ \mathbb{E} \left[ \exp\left[ V_1^\kappa + V_2^\kappa + V_3^\kappa \right] \right]. \nonumber
\end{align}

Since $G$ is assumed to be compact, we have $u_{s,t} = 1 + O(1/\kappa)$ from Equation (<ref>). Thus, we have $V_2^\kappa, V_3^\kappa$ all converge to 0 pointwise.

Apply Lemma <ref>,

\begin{align}
\bE\left[ \left| e^{V_1^\kappa+V_2^\kappa + V_3^\kappa} - e^{V_1^\kappa } \right|\right]
\leq \bE \left[ |V_2^\kappa + V_3^\kappa | \exp\left[ 2|V_1^\kappa| + |V_2^\kappa| + |V_3^\kappa|\right] \right]. \nonumber
\end{align}

Now, $\exp\left[ 2|V_1^\kappa| + |V_2^\kappa| + |V_3^\kappa| \right]$ is less than or equal to Expression (<ref>) for $p = 4$ and in the proof of Theorem <ref>, we showed that if $\kappa$ is large enough, then Expression (<ref>) is in fact integrable for any values of $p$. Thus, { |V_2^κ+ V_3^κ| exp[ 2|V_1^κ| + |V_2^κ| + |V_3^κ|] }_κ≥1 is uniformly integrable and since $V_2^\kappa$, $V_3^\kappa$ all converge to 0, we have $\bE\left[ \left| e^{V_1^\kappa+V_2^\kappa + V_3^\kappa} - e^{V_1^\kappa } \right|\right]$ converge to 0 as $\kappa$ goes to infinity.

Thus it remains to compute the limit of $\Tr\ \tilde{\mathcal{T}}\ \bE\left[ e^{V_1^\kappa} \right]$. Write F^κ= ∑_α∑_0≤a< b ≤3 ∫_I^2ds dt |J^σ_ab|(s,t) ξ_ab^κ(κσ(s,t)/2) ⊗ρ(E_s,t^α). Now apply Lemma <ref>,

\begin{align}
\bE\left[ e^{V_1^\kappa} \right] =
&\exp\left[ \frac{\kappa^2}{32}\left\langle F^\kappa, F^\kappa\right\rangle_{\fd,\kappa}\right] \nonumber \\
\longrightarrow& \exp\left[ \frac{1}{8}\sum_{0\leq a < b \leq 3}\int_{I^2} ds dt\ |J_{ab}^\sigma|(s,t) \rho_S^{ab}(\sigma(s,t)) \otimes \left[\sum_\alpha \rho(E^\alpha) \otimes \rho(E^\alpha) \right](s,t)\right], \nonumber
\end{align}

using calculations from Section <ref>. Hence, 𝔼[ exp[V_1^κ+ V_2^κ+ V_3^κ] ] ⟶exp[ 1/8∑_0≤a <b ≤3∫_I^2 ds dt J_ab^σ(s,t) ρ_S^ab(σ(s,t)) ⊗[∑_αρ(E^α) ⊗ρ(E^α) ](s,t)].

Write $E = \sum_\alpha \rho(E^\alpha) \otimes \rho(E^\alpha)$, $g(s,t) = \frac{1}{8}\sum_{0\leq a<b  \leq 3}\rho_S^{ab}(\sigma(s,t))|J_{ab}^\sigma|(s,t)$.

By definition of $\tilde{\mathcal{T}}$ (See Definition <ref>),

\begin{align}
\tilde{\mathcal{T}}&\exp\left[ \frac{1}{8}\sum_{0\leq a < b \leq 3}\int_{I^2} ds dt\ |J_{ab}^\sigma|(s,t) \rho_S^{ab}(\sigma(s,t)) \otimes \sum_\alpha \rho(E^\alpha) \otimes \rho(E^\alpha) (s,t)\right] \nonumber \\
=& \tilde{\mathcal{T}}\sum_{n=0}^\infty \frac{1}{n!}\int_{I^{2n}} \prod_{i=1}^n g(s_i,t_i)ds_i dt_i \otimes \left(E(s,t) \right)^{\otimes^n} \nonumber \\
=& \sum_{n=0}^\infty n!\int_{s_1\leq \cdots \leq s_n}\prod_{i=1}^n ds_i \int_{t_1 > \cdots > t_n} \prod_{i=1}^n g(s_i, t_i)dt_i \otimes  \mu(E)(s_1, t_1) \cdots   \mu(E)(s_n, t_n) \nonumber \\
=& \sum_{n=0}^\infty \frac{1}{n!}\int_{I^{2n}} \prod_{i=1}^n g(s_i,t_i)ds_i dt_i  \otimes \left[ \mu(E) \right]^n\nonumber \\
=& \sum_{n=0}^\infty \frac{1}{n!} \left( \frac{1}{8}\int_S\rho_S \right)^n \otimes \left(\mu\sum_\alpha  \rho(E^\alpha) \otimes \rho(E^\alpha)  \right)^n
= \exp\left[\frac{1}{8}\int_S\rho_S \otimes \mu\left( \sum_\alpha \rho(E^\alpha) \otimes \rho(E^\alpha)\right) \right]. \nonumber
\end{align}

Hence we have

\begin{align}
\mathbb{E}_{YM}^\kappa\left[ \mathcal{J}_S^\kappa \right] \longrightarrow \Tr \exp\left[\frac{1}{8}\int_S\rho_S \otimes \mu\left( \sum_\alpha \rho(E^\alpha) \otimes \rho(E^\alpha)\right) \right] \nonumber
\end{align}

as $\kappa$ goes to infinity.

§.§ Some Important Examples

The Yang-Mills path integral is dependent on the parameter $\kappa$. We will now take the limit as $\kappa$ go to infinity, in the cases of $G = SU(N)$ and $G = SO(N)$.

(Special elements in $\bigotimes^2 M_N(\mathbb{C})$.)
Define $I$, $J$, $K$ in $\bigotimes^2M_N(\mathbb{C})$, I^ab_cd = δ^a_cδ^b_d J^ab_cd := δ^a_dδ^b_c K^ab_cd = δ^a_bδ^c_d. $K$ commutes with $J$. Note that if $\mathbb{I}$ is the identity matrix in $M_N(\mathbb{C})$, then $I = \mathbb{I} \otimes \mathbb{I}$. Furthermore, $J \cdot J = I$, $K\cdot K = NK$ and $K\cdot J = K = J \cdot K$.

Example 1 ($SU(N)$)

Suppose our Lie group is $G = SU(N)$. Considering its standard representation, one shows that $\sum_{\alpha \in \sigma} E^\alpha \otimes E^{\alpha} = \frac{1}{N}I - J$. Hence, $\mu(\sum_\alpha E^\alpha \otimes E^\alpha) = \frac{1}{N}\mathbb{I} - N \mathbb{I}$, and

\begin{align}
\exp\left[\frac{1}{8}\int_S \rho_S\cdot \mu\left( \sum_\alpha E^\alpha \otimes E^\alpha \right) \right] =& \Tr \exp\left(  \frac{1}{8}\left(\frac{1}{N}-N \right)\int_S \rho_S \right)\mathbb{I} = N\exp\left(  \frac{1}{8}\left( \frac{1}{N}-N \right)\int_S \rho_S \right). \nonumber
\end{align}

Therefore, lim_κ→∞𝔼_YM^κ[ 𝒥_S^κ] = Nexp(  1/8( 1/N-N )∫_S ρ_S ).

Example 2 ($SO(N)$)

Now consider $G = SO(N)$. Considering its standard representation, then $\sum_{\alpha \in \sigma} E^\alpha \otimes E^{\alpha} = (K - J)/2$. Hence, $\mu(\sum_\alpha E^\alpha \otimes E^\alpha) = [1 - N ]\mathbb{I}/2$, and

\begin{align}
\exp\left[\frac{1}{8}\int_S \rho_S\cdot \mu\left( \sum_\alpha E^\alpha \otimes E^\alpha \right)\right] =& \Tr \exp\left(  \frac{1}{16}\left( 1-N \right)\int_S \rho_S \right)\mathbb{I} = N\exp\left(  \frac{1}{16}\left( 1-N \right)\int_S \rho_S \right). \nonumber
\end{align}

Therefore, lim_κ→∞𝔼_YM^κ[ 𝒥_S^κ] = Nexp(  1/16( 1-N )∫_S ρ_S ).

§ APPLICATIONS

The main application of computing the Wilson Loop observable using Yang-Mills measure 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 Wilson Loop observable using the Yang-Mills action can be used to explain why this potential is linear. However, to compute the Wilson Loop observable is highly nontrivial and difficult. Note that perturbative methods using Feynman diagrams cannot be used to compute it, lest derive the area formula.

Another important application for this is to explain superconductivity, by replacing quarks with magnetic monopoles. Due to our limited understanding on this subject, we prefer to refer the interested reader to [12] for a more concise description on this subject matter.

§.§ Quark Confinement

Let $C$ be a rectangular contour with spatial length $R$ and timelike length $T$. Let $S$ be a minimal surface bounded by $C$, which in this case will be a rectangular region of area $RT$. Then, we will write $W(R, T; \rho) := \lim_{\kappa \rightarrow \infty}Y(\bR^4, \kappa; S, \rho)$. From Theorem <ref>, we have W(R, T; ρ) =  exp[ RT/8μ( ∑_αρ(E^α) ⊗ρ(E^α) ) ].

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 Wilson Loop observable computed along the rectangular $C$, the potential is given by V(R) = -lim_T →∞logW(R, T+1; ρ)/W(R, T; ρ). This formula can be derived from Equation (19.11) in [12].

When $G$ is Abelian, from Equation (<ref>), $V(R) = R/8$. In the case of the standard representation of $SU(N)$, we can see from Equation (<ref>) that V(R) = R/8( N - 1/N ), and in the case of the standard representation of $SO(N)$, we see that V(R) = R/16( N - 1 ) from Equation (<ref>). 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.

§.§ Superconductivity

In electromagnetism, we have the electric field $E \equiv \sum_{i=1}^3 E_i dx^0 \wedge dx^i$ and the magnetic field $B \equiv \sum_{i=1}^3 \epsilon_{ijk}B_{k} dx^i \wedge dx^j $. Now, we are going to interchange $E$ with the $B$ field, i.e. we define $\bar{E}$ and $\bar{B}$ as E̅ = ∑_i=1^3 ϵ_ijk E_k dx^i ∧dx^j  and  B̅ = ∑_i=1^3 B_i dx^0 ∧dx^i. This electric-magnetic duality transformation motivates the next definition.

Let $S$ be a bounded and connected surface with boundary $\partial S$, a simple connected closed curve. Recall $S$ defines F^κ= ∑_α∑_0≤a< b ≤3 ∫_I^2ds dt |J^σ_ab|(s,t) ξ_ab^κ(κσ(s,t)/2) ⊗E_s,t^α. We define a vortex of $S$ in $\bR^4$,which is a surface $\bar{S}$ defined by F̅^κ= ∑_α∑_0≤a,b,c,d ≤3 ∫_I^2ds dt ϵ_abcd|J^σ_cd|(s,t) ξ_ab^κ(κσ(s,t)/2) ⊗E_s,t^α. Here, $\epsilon_{abcd} = (-1)^{|\sigma(abcd)|}$ if $a, b, c, d$ all distinct for a permutation $\sigma(abcd)$ in $S_4$, $|\sigma(abcd)|$ counts the number of transpositions in the permutation; 0 otherwise.

Note that $F^\kappa$ and $\bar{F}^\kappa$ are $\mathfrak{g}$-valued evaluation differential forms and $\bar{F}^\kappa$ should be interpreted as the dual of $F^\kappa$. Furthermore, $|F^\kappa|_{\fd,\kappa} = |\bar{F}^\kappa|_{\fd,\kappa}$, so both $S$ and $\bar{S}$ have the same area and we can write F^κ= (cosθ)F̅^κ+ (sinθ)γ,  whereby $|\gamma|_{\fd,\kappa} = 1$ and cosθ= ⟨F^κ, F̅^κ⟩_,κ/|F^κ|_,κ^2,  sinθ= 1/|F^κ|_,κ√(|F^κ|_,κ^4 - ⟨F^κ, F̅^κ⟩_,κ^2).

Let ${\rm lk}(S, \bar{S})$ denote the linking number between a surface $S$ and its vortex, which counts the total number of points the vortex intersect the surface $S$.

It is straightforward to show that (κ/2)^4⟨F^κ, F̅^κ⟩_,κ →(2π)^2 lk(S, S̅)∑_αE^α⊗E^α.  So, we have a dual statement to Theorem <ref>.

Let $C$ be a closed curve in $\bR^4$ and fix a surface $S$ such that $\partial S = C$. Define its vortex $\bar{S}$, with $\partial \bar{S} = \bar{C}$.
As $\kappa \rightarrow \infty$, we have
𝔼_YM^κ[ 𝒥_S̅^κ] ⟶exp[1/8∫_Sρ_S ⊗μ( ∑_αE^α⊗E^α) ] .

We just note that from Equation (<ref>), $\kappa^2 \langle F^\kappa, \bar{F}^\kappa \rangle_{\fd,\kappa} \rightarrow 0$. Then from Equation (<ref>), it follows that the limit coincides with the limit of $\mathbb{E}_{YM}^\kappa\left[ \mathcal{J}_{S}^\kappa \right]$. This completes the proof.

In the case of quark confinement, the force between the pair of quarks is electric in nature. In the case of a type II superconductor, a similar phenomenon is present, except that we focus on the vortex, which is of magnetic in nature. Here, we have a magnetic monopole and an antimonopole pair which are connected via a vortex. From Theorem <ref>, we have an area law for the vortex, which implies that the interaction energy between the magnetic monopole and antimonopole pair increases linearly with their separation distance.

Thus, for a closed curve $C$, the Wilson Loop observable along $C$ is interpreted as electric confinement; for the boundary of a vortex $\bar{C}$, the Wilson Loop observable along $\bar{C}$ is interpreted as magnetic confinement. The former is due to magnetic superconductivity, the later is due to electric superconductivity.

§ USEFUL CALCULATIONS

§.§ Gaussian Integrals

The following two lemmas are trivial, but are central to our calculations.

For $\theta > 0$,
θ/√(2π)∫_ e^αxe^-θ^2 x^2/2dx = e^α^2/2θ^2.

Let $A$ be a $n \times n$ matrix and $x = (x_1, \ldots, x_n)^T$. If $\parallel A^\ast A \parallel < 1/2$, then
(1/√(2π))^n∫_^n exp[ ⟨Ax, Ax ⟩] e^-|x|^2/2 ∏_i=1^n dx_i = ( 1/[1-2A^∗A] )^1/2.

§.§ Surface Integrals

Fix a closed curve $C$ and let $S$ be a surface embedded in $\bR^4$ such that $\partial S = C$. Let $\sigma\equiv ( \sigma_0, \sigma_1, \sigma_2, \sigma_3): [0,1]^2 \equiv I^2 \rightarrow M$ be a parametrization of a surface $S$. Here, $\sigma' = \partial \sigma/\partial s$ and $\dot{\sigma} = \partial \sigma/\partial t$. Let $x \equiv (x_0, x_1, x_2, x_3)$.

Let $\sigma: [0,1]^2 \equiv I^2 \rightarrow \bR^4$ be a parametrization of a surface $S \subset \bR^4$.
Define Jacobian matrices,

\begin{align}
J_{ij}^\sigma(s,t) = \left(
\begin{array}{cc}
\sigma_i'(s,t) & \dot{\sigma}_i(s,t) \\
\sigma_j'(s,t) & \dot{\sigma}_j(s,t) \\
\end{array}
\right), \nonumber
\end{align}

and write $|J^\sigma_{ij}| = |\det{J^\sigma_{ij}}|$ and $W_{ab}^{ cd} := J_{cd}J_{ab}^{-1}$, $a, b, c, d$ all distinct. Note that $W_{cd}^{ab} = (W_{ab}^{cd})^{-1}$.

For $a,b, c, d$ all distinct, define $\rho_S^{ab}: S \rightarrow \bR$ by

\begin{align}
\rho_S^{ab} =& \frac{1}{\sqrt{\det\left[ 1+ W_{ab}^{cd,T}W_{ab}^{cd}\right]}}. \nonumber
\end{align}

Assume that in a small neighborhood $U \subset [0,1]^2$, $\sigma_a', \dot{\sigma}_a, \sigma_b', \dot{\sigma}_b \neq 0$ and let $\sigma(\bp) = x$, $\bp \in U$. We have for distinct $a, b, c, d$,
lim_κ→∞∫_U  κ^2/4e^-κ^2|σ(s,t)-x|^2/8 (σ'_aσ̇_b - σ_b'σ̇_a)(s,t) ds dt
= 2π/√([1 + W_ab^cd,TW_ab^cd()]).

Write $\underline{s}=s-s_0$, $\underline{t}=t-t_0$ and $J_{ab} \equiv J_{ab}^\sigma(\bp)$. Let $\sigma_i(s_0, t_0) = x_i$. Using Taylor's theorem,
σ_i(s,t)=x_i + s σ_i'() + tσ̇_i() + O((|s|+ |t|)^2). Let (s,t)^T = (

(u, v)^T = (

Use a transformation, i.e. $u = \underline{s}\sigma_a'(\bp) + \underline{t}\dot{\sigma}_a(\bp)$, $v = \underline{s}\sigma_b'(\bp) + \underline{t}\dot{\sigma}_b(\bp)$. And let $0 \in V$ be the range of $U$ under this transformation. Then, we have

\begin{align}
& \int_{U} \frac{\kappa^2}{4}e^{-\kappa^2|\sigma(s,t)-x|^2/8} (\sigma'_a\dot{\sigma}_b - \sigma_b'\dot{\sigma}_a)(s,t) ds dt \nonumber \\
=& \int_U \frac{\kappa^2}{4}e^{-\kappa^2 \left[|J_{ab}(\underline{s},\underline{t})^T|^2 + |J_{cd}(\underline{s},\underline{t})^T|^2+ O\left((|\underline{s}|+ |\underline{t}|)^3 \right)\right]/8} \left[ (\sigma'_a\dot{\sigma}_b - \sigma_b'\dot{\sigma}_a) (\bp)+ O\left(|\underline{s}|+ |\underline{t}|\right) \right] ds dt \nonumber \\
=& \int_V \frac{\kappa^2}{4}e^{-\kappa^2\left[(u^2+v^2) + O\left((|u|+ |v|)^3\right) \right]/8} e^{-\kappa^2 \left[ |J_{cd}J_{ab}^{-1}(u,v)^T|^2 + O\left((|u|+ |v|)^3\right) \right]/8} (1 + O(|u|+ |v|)) du dv \nonumber \\
=& \int_V \frac{\kappa^2}{4}e^{-\kappa^2  \left[\langle (1 + J_{ab}^{T,-1} J_{cd}^T J_{cd}J_{ab}^{-1})(u,v)^T, (u,v)^T \rangle + O((|u|+ |v|)^3) \right] /8} (1 + O(|u|+ |v|)) du dv \nonumber \\
&\longrightarrow_{\kappa \rightarrow \infty} \frac{2\pi}{\sqrt{\det\left[1 + J_{ab}^{T,-1} J_{cd}^T J_{cd}J_{ab}^{-1}\right]}}=  \frac{2\pi|J_{ab}|}{\sqrt{\det\left[J_{ab}^TJ_{ab} +  J_{cd}^T J_{cd}\right]}}. \nonumber
\end{align}

Let $\sigma$ be any parametrization for a surface $S \subset \bR^4$.
As $\kappa \rightarrow \infty$, we have

\begin{align}
\sum_{0 \leq a<b \leq 3}&\frac{\kappa^2}{4} \int_{I^2 \times I^2} ds dt d\bar{s}d\bar{t}\ e^{-\kappa^2|\sigma(s,t)-\sigma(\bar{s},\bar{t})|^2/8} |J_{ab}^\sigma|(s,t)|J_{ab}^\sigma|(\bar{s},\bar{t})  \nonumber \\
\longrightarrow& 2\pi \sum_{0 \leq a<b \leq 3}\int_{I^2}\rho_S^{ab}(\sigma(s,t))|J_{ab}^\sigma|(s,t)
\ ds dt := 2\pi \int_S \rho_S,\nonumber
\end{align}

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

This follows from Lemma <ref>, details omitted.