the subspace $C^\infty(G)\subset L^2(G)$ of the $\ell^{\rm th}$ factor. Note that if we set $g=\exp(t\Psi(\ell))$ in ${\widehat}{U}_g^{(\ell)}:=\un\otimes\big(\un\otimes\cdots\otimes\un\otimes{U}_g^{(\ell)}\otimes\un\otimes\cdots\otimes\un\big)$ where $U^{(\ell)}_g$ is the $U_g$ operator on the $\ell^{\rm th}$ factor, then this is a unitary one parameter group w.r.t. $t$, with generator the gauge momentum operator ${\widehat}{P}^{(\ell)}_{\Psi(\ell)}$. Thus the quantum G–electrical field ${\widehat}{E}_r$ is a map from $\Lambda^1$ to operators on the dense domain $\cl H._F\otimes\mathop{\bigotimes}\limits_{\ell\in\Lambda^1}C^\infty(G)$, given by ${\widehat}{E}_r(\ell):={\widehat}{P}^{(\ell)}_{Y_r}$.

Next, to define gauge transformations, recall from Equation (\[ClassGTr\]) the action of the local gauge group $\gauc \Lambda=\prod\limits_{x\in \Lambda^0}G=\{\zeta:\Lambda^0\to G\}$ on the classical configuration space. For the Fermion algebra we define an action $\alpha^1:\gauc \Lambda \to\aut{\mathfrak F}_{\Lambda}$ by $$\alpha_\zeta^1(a(f)):=a(\zeta\cdot f)\qquad \hbox{where}\qquad (\zeta\cdot f)(x):={\zeta(x)}f(x)\quad \hbox{for all}\quad x\in\Lambda^0,$$ and $f\in\prod\limits_{x\in\Lambda^0}\Cn$ since $f\mapsto \zeta\cdot f$ defines a unitary on $\prod\limits_{x\in\Lambda^0}\Cn$ where $\zeta\in\gauc \Lambda$. As ${\mathfrak F}_{\Lambda}$ has up to unitary equivalence only one irreducible representation, it follows that ${\pi_F:{\mathfrak F}_{\Lambda}}\to\cl B.(\cl H._F)$ is equivalent to the Fock representation, hence it is covariant w.r.t. $\alpha^1$, i.e. there is a (continuous) unitary representation $U^F:\gauc \Lambda\to \cl U.(\cl H._F)$ such that $$\pi_F(\alpha^1_\zeta(A))=U^F_\zeta\pi_F(A)U^F_{\zeta^{-1}}\quad\hbox{for}\quad A\in {\mathfrak F}_{\Lambda}.$$

On the other hand, if the classical configuration space $G$ corresponds to a link $\ell=(x_\ell,y_\ell)$, then the gauge transformation is $\zeta\cdot g=
\zeta(x_\ell)\,g\,\zeta(y_\ell)^{-1}$ for 