of operators $\left( S(P)\right)_{ab}$ is a functional of the operators $P_{ab}$ and is defined implicitly through . Next we ask to find a form of ${\cal F}$ that will induce the following transformations $$\begin{aligned}
\psi^a_x &\to& \psi^{'a}_x = {\cal F} \, \psi^a_x \, {\cal F}^{\dag} = \sum_b V_{x,ab} \,\psi^b_x. \label{Psitransform}\\
\psi^{a\dag}_x &\to& \psi^{'\dag a}_x = {\cal F} \, \psi^{\dag a}_x \, {\cal F}^{\dag} = \sum_b \psi^{\dag b}\,V^\dag_{x,ba},\\
U_{x,ab} &\to& U'_{x,ab}={\cal F} \, U_{x,ab} \, {\cal F}^{\dag} = U_{x,ab}, \label{Utransform}\\
U^\dag_{x,ab} &\to& U^{'\dag}_{x,ab}={\cal F} \, U^\dag_{x,ab} \, {\cal F}^{\dag} = U^\dag_{x,ab}.\end{aligned}$$ Crucially, here the operator $\left(V_x\right)_{ab}$ is chosen to be the following functional of the gauge fields operators $\left(U_{x}\right)_{ab}$, $$\begin{aligned}
V_{x,ab}&=& \sum_{cde\cdots yz=1}^N U^\dag_{x-1,ac}\, U^\dag_{x-2,cd}\, \dots \, U^\dag_{1,yz} \, S^\dag_{zb}(P) \, e^{i\varphi_b(P) \, x}.\label{Vdef}\end{aligned}$$ This guarantees that $V$ obeys $$\sum_{bc}\left(V^\dag_x\right)_{ab} \,\left(U_{x}\right)_{bc}\,\left(V_{x+1}\right)_{cd} = e^{i\varphi_a }\delta_{ad} \label{Vdef1}$$ as an operator identity. The form of ${\cal F}$ that induces Eqs. (\[Psitransform\])–(\[Utransform\]) is then written as $${\cal F} = \exp \left( -i\sum_{xi} \rho^i_{x,F}\,  \Theta^i_x(\left\{U\right\})\right),$$ where $\Theta$ is again an operator in Hilbert space that depends on the gauge field operators $U_{x,ab}$ and that is defined through its following relation to the operator $V$ : $$\begin{aligned}
V_{x,ab}&=&\left[\exp \left( i\sum_i \lambda^i \Theta^i_x(\left\{U\right\})\right)\right]_{ab}. \label{Vdef2}\end{aligned}$$ Note that ${\cal F}$ 