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In this section we resolve the Gauss law constraint of and write the electric field operators $E^{''}$ in terms of the fermion color charge densities $\rho_F$. For that purpose we first specify the basis of the color group generators $\lambda^i$. We choose the $N-1$ traceless generators $\lambda^{i=1,2,\dots,N-1}$ to span the traceless Cartan sub-algebra, and the remaining $N(N-1)$ generators to have no diagonal entries. In this basis we see that $$\left(e^{-i\varphi}\right)^{\rm Adj.}_{ij} \equiv 2\, e^{i(\varphi_a-\varphi_b)}\, \lambda^i_{ab}\, \lambda^j_{ba}  = \left[ \begin{array}{cc}
\delta_{ij} & {\rm if}\,\,i,j \in [1,N-1],\\
0 & \hspace{0.5cm} {\rm if}\,\, i\in  [1,N-1] {\rm \,\, but\, \, not}\, j, \, \, {\rm and \,vice\,versa}  
\end{array}
\right. .$$ As a result, if we focus on the Cartan sub-algebra (let us denote its generators by $i=I$), and Fourier transform the operators $E''$ and $\rho$ according to $$\begin{aligned}
E^{''I}_x &=& \frac1{\sqrt{L_s}} \, \sum_p \,e^{ipn} E^{''I}_p,\label{FTE''}\\
\rho^{I}_x &=& \frac1{\sqrt{L_s}} \, \sum_p \,e^{ipn} \rho^{I}_p,\label{FTrho},\end{aligned}$$ with $p=\frac{2\pi n}{L_s}\, ; \, n=0,1,\dots,L_s-1$, then we can resolve the nonzero momentum components of $E^{''I}_{p\neq 0}$ : $$\begin{aligned}
E^{''I}_{p} &=& \frac{\rho^I_{F,p}}{1-e^{-ip}}, \quad {\rm for} \quad p\neq 0. \label{E''p}\end{aligned}$$ For $p=0$ Gauss law cannot be resolved, and instead becomes a constraint on the fermionic global color charges : $$\begin{aligned}
\forall I=1,2,\dots,N-1 