freedom.

To quantize the classical gauge connection fields $\prod\limits_{\ell\in\Lambda^1}G,$ we take for a single link $\ell$ a bosonic particle on $G$. This is given in a generalized Schrödinger representation on $L^2(G)$ by the set of operators ${\{U_g,\;T_f\mid g\in G,\;f\in L^\infty(G)\}}$ where: $$\label{GSRp}
(U_g\varphi)(h):=\varphi(g^{-1}h)\quad\hbox{and}\quad\big(T_f\varphi)(h):=f(h)\varphi(h)\quad\hbox{for}\quad
\varphi\in L^2(G),$$ $g,\,h\in G$ and $f\in L^\infty(G)$, and it is irreducible in the sense that the commutant of ${U_G\cup T_{L^\infty(G)}}$ consists of the scalars. Note that there is a natural ground state unit vector $\psi_0\in L^2(G)$ given by the constant function $\psi_0(h)=1$ for all $h\in G$ (assuming that the Haar measure of $G$ is normalized). Then $U_g\psi_0=\psi_0$, and $\psi_0$ is cyclic w.r.t. the \*-algebra generated by ${U_G\cup T_{L^\infty(G)}}$ (by irreducibility).

The generalized canonical commutation relations are obtained from the intertwining relation $U_gT_fU^*_g=T\s\lambda_g(f).$ where $$\label{leftact}
\lambda:G\to\aut C(G)\, , \quad
 \lambda_g(f)(h) := f(g^{-1} h) \quad\hbox{for}\quad g,\,h\in G$$ is the usual left translation. In particular, given $X\in\mathfrak{g}$, define its associated momentum operator $$\begin{aligned}
P_X:C^\infty(G)\to C^\infty(G)\quad&\hbox{by}&\quad
P_X\varphi:=i{d\over dt}U(e^{tX})\varphi\Big|_{t=0}\,.\\[1mm]
\hbox{Then}\qquad
\big[P_X,\,T_f\big]\varphi
&=& iT\s X^R(f). \varphi\qquad\hbox{for}\quad
f,\,\varphi\in C^\infty(G),\end{aligned}$$ where $X^R\in\ot X.(G)$ is the associated right-invariant vector field. As $P_X=dU(X)$, it defines a representation of the Lie algebra $\mathfrak{g}$, and clearly $P_X\psi_0=0$.

To identify the quantum connection $\Phi(\ell)$ at link $\ell$ in this context, 